 Hello, I'm Keith Devlin. Welcome to this online course on mathematical thinking. The goal of the course is to help you develop a valuable mental ability, a powerful way of thinking that people have developed over 3,000 years. What I want to do today is get you ready for the course and tell you a little bit about the way the course will work. I'm doing this because for most of you this will be a very different perspective than what mathematics is. Apart from the final two lectures, there's very little mathematical content in the course and you won't learn any new mathematical procedures. But mathematical thinking is essential if you want to make the transition from high school math to university level mathematics. The quickest way to learn what mathematical thinking is is to take a course like this. So, by the time we're finished, you should know what it is. Now, let me give you an analogy. If we compare mathematics with the automotive world, school math corresponds to learning to drive. In the automotive equivalent to college mathematics, in contrast, you will learn how a car works, how to maintain and repair it, and if you pursue the subject far enough, how to design and build your own car. The only prerequisite for the course is completion of high school mathematics. That means many people could take the course and find it valuable. In particular, a key feature of mathematical thinking is thinking outside the box. In contrast, the key to success in high school math was to learn to think inside the box. It's because thinking outside the box is such a valuable ability in today's world that this course could be valuable to many people. But my primary student is someone in their final year of high school or their first year at college or university who's thinking of majoring in mathematics or a math dependent subject. If that's you, then you'll probably find a transition from high school mathematics to college level, pure abstract mathematics, difficult. I certainly did, and so did most mathematicians I now. Not because the mathematics gets harder, but once you've successfully made the transition, I think you'll agree that college math is in many ways easier. What causes the problem is the change in emphasis. At high school, the focus is primarily on mastering procedures to solve various kinds of problems. That gives the subject very much the flavour of a cookbook full of mathematical recipes, thinking inside boxes. At university, the focus is on learning to think a different way, to think like a mathematician, thinking outside the box. That's not true of all college math courses. Those designed for science and engineering students are often very much in the same vein as the courses you had in high school. It's the courses that form the bulk of the mathematics major that are different. But some of those courses are usually required for all advanced work in science and engineering. So if you are a student in those disciplines, you may also find yourself faced with this different kind of mathematics. If you did well at math and school, you probably got good at recognising different kinds of problems. So you could apply different techniques you learned. At university, you have to learn how to approach a new problem, one that doesn't quite fit any template you're familiar with. It comes down to learning how to think about a problem in a certain way. The first key step is learn to stop looking for a formula to apply of a procedure to follow. Approaching a new problem by looking for a template, a work to example in a textbook or presented on YouTube and then just changing the numbers often won't work. Sometimes it will. So all that work you did at high school won't go to waste. But it isn't enough for many of your college math courses. If you can't solve a problem by looking for a template to follow or a formula to plug some numbers into or a procedure to apply, what do you do? The answer is you think about the problem a certain way. Not the form of the problem. That's probably what you were taught to do at school and it served you well there. Rather, you have to look at what the problem actually says. That sounds as though it ought to be easy, but most of us initially find it extremely hard and very frustrating. It doesn't come quickly or easily. You have to work at it. You're going to have to accept going a lot slower than you're used to. Most of the time you won't feel as though you're making any progress. Your goal has to be understanding, not doing. You should definitely try all the exercises. They're there to aid your understanding. You should also work with others. Few of us can master this crucial shift in thinking on our own. That part's crucial. A lot of what we'll be doing is not so much focused on right and wrong, but on learning how to think about a problem. Yeah, sure, at the end of the day, solutions are right and wrong, but usually there are many different right answers or different ways to the right answer and many wrong ones. When you're learning how to think mathematically, it's how and why you got something right and wrong that's important. The only way to find that out, to find out how well you're doing, is for somebody else to look at your attempt and critique your way. It's not possible to automate the grading process. Maybe one day artificial intelligence will have advanced far enough for a course like this to be automated, though frankly I doubt it. But right now you need feedback from other people. For a regular class here at Stanford, the professor and the graduate student TAs grade students' work and provide feedback. With an open online course like this, where there are many thousands of students, that's not possible. So we have to go about things a different way. I've designed the course so that the benefit comes primarily from doing the work and discussing it with other students. Getting it right is important in mathematics, but in a massively open online course, a MOOC like this, there's no way to guarantee that. Incidentally, not being sure if we are right until others have seen our work is very familiar to we mathematicians. Even very famous mathematicians have had the experience of thinking they've solved the problem and writing it up, writing up their solution and sending it off for publication, only for an anonymous referee to find an error. In mathematics, there is such a thing as right and wrong, but deciding between them can be very difficult. So even the professionals have to live with never being sure whether they're right or not. Part of this introductory session is a reading assignment. It gives you a bit of the history that should explain why today's mathematics students need a course such as this. In lecture one, there'll be a short quiz on the reading. Let me say a few words about the quizzes. If you were in one of my regular physical classes, I'd talk with you to find out if you had understood the material sufficiently to progress. But with a massively open online course, a MOOC, that's not possible. You have to monitor yourself. The quizzes are one way to help you do that. I'll say a little bit more about the quizzes later, but before you start lecture one, please read a file called background reading. It's just over six pages long. It's a PDF file, so you can download it and read it offline. Well, now you know what an intellectual quiz looks like. The next thing I suggest you do is view the video finding your way around the course website. It will point you to some particularly important things to check out before you start the course. I'm going to start off with a question. What is mathematics? That might seem strange, given you've probably spent several years being taught math. But for all the time schools devote to the teaching of mathematics, very little, if any, is spent trying to convey just what the subject is about. Instead, the focus is on learning and applying various procedures to solve math problems. Well, that's a bit like explaining soccer by saying it's a series of maneuvers you execute to get the ball into the goal. Both accurately describe various key features, but they miss the what's and the why of the big picture. If all you want to do is learn new mathematical techniques to apply in different circumstances, then you can probably get by without knowing what math is really about. But if that's the case, then this isn't the course for you. One thing you should realise is that a lot of school mathematics dates back to medieval times, with pretty well all the rest coming from the 17th century at the very latest. Actually nothing from the last 300 years has found its way into the classroom. Yet the world we live in has changed dramatically in the last 10 years, let alone the last 300. Most of the changes in mathematics over the centuries were just expansion. But in the 19th century, there was a major change in the nature of mathematics. First, it became much more abstract. Second, the primary focus shifted from calculation and following procedures to one of analysing relationships. The change in emphasis wasn't arbitrary. It came about through the increasing complexity of what became the world we are familiar with. Procedures and computation did not go away. They're still important, but in today's world they're not enough. You need understanding. In our education system, the change in emphasis in mathematics usually comes when you transition from high school to university. In the 1980s, I was one of a number of mathematicians who advocated a new meme to capture what mathematics is today. The science of patterns. According to that description, the mathematician identifies and analyses abstract patterns. There can be numerical patterns, patterns of shape, patterns of motion, patterns of behaviour, voting patterns in a population, patterns of repeating chance events and so on. There can be either real or imagined patterns, visual or mental, static or dynamic, qualitative or quantitative, utilitarian or recreational. There can arise from the world around us, from the pursuit of science, or from the inner workings of the human mind. Different kinds of pattern give rise to different branches of mathematics. For example, arithmetic and number theory study the patterns of counting and number. Geometry studies the patterns of shape. Calculus allows us to handle patterns of motion. Logic studies patterns of reasoning. Probability theory deals with patterns of chance. Topology studies patterns of closeness and position. Fracol geometry studies the self-similarity found in the natural world, and so on and so on and so on. One major consequence of the increasing abstraction and complexity of the mathematics in the 19th century was that methods developed to solve important real-world problems had consequences that were counterintuitive. Let me give you one example. It's called the Banach-Tarski paradox. It says you can, in theory, take a sphere and cut it up in such a way that you can reassemble it to form two identical spheres, each the same size as the original one. And that wasn't the only surprise. Mathematicians had to learn to trust the mathematics above our intuitions, just as physicists did with the discovery of relativity theory and quantum mechanics. Of course, if you are going to trust mathematics above intuition and common sense, you'd better be sure the math is right. This is why the mathematicians in the 19th and early 20th centuries developed the precise way of thinking and calling mathematical thinking. What this course is about. Now for that quiz I promised you during my course introduction. Did you read that short document called background reading? If not, I suggest you pause the video right now and then come back after you've looked at it. The correct answer is money, for the first question. People certainly measured land and they used various kinds of yardstick, but they didn't use numbers. And they certainly counted seasons. But you can count without numbers. You can count with notches in sticks and you can count with pebbles and so forth. Our ancestors only invented abstract numbers in order to get money. At least that's the best available evidence that we have. We think happened about 10,000 years ago. For the second question, topology studies patterns of closeness. If you thought it was geographical terrain, you were confusing topology with topography. How did you do? Well for this one, the main focus in the 19th century became concepts and relationships. That was a revolution in mathematics that took place in Germany. And for this question, at least according to me and many of my colleagues I should point out, the main mathematical ability today is being able to adapt old methods or develop new ones. Yes, use of technology is important. Yes, you need mastery of basic skills. But the crucial ability in today's world is adapting old methods or developing new ones. Okay, how did you do on that quiz? Yup, I know it wasn't a math quiz. It was really there just to get you used to the in lecture quiz format. I gave the rationale for those quizzes on that very first quiz. As I wrote there, the intention is that you should find the answers to the quizzes immediately obvious. If you do, that's a sign that you are sufficiently engaged and not trying to move too fast. If you find you have to spend time on a quiz question or go back and look at the lecture again, then you will know that you are not engaging sufficiently closely and you need to slow down. The secret to this entire course is reflection, not completion. Doing all the in lecture quizzes is required to get a certificate of completion for the course, but does not contribute to your final grade. As those of us giving and taking these first generation MOOCs have discovered, in lecture quizzes are valuable. Without the regular feedback from an instructor or a TA, neither of which is possible with a course having many thousands of students, the individual student has to monitor his or her own progress. The quizzes, though simple, have proved to be very useful in that regard. By the way, I know that some of you don't have constant broadband access and you have to download the videos and watch them locally, so you can't do all the quizzes. Well, they're all reproduced on the course website with minor changes, so you can do them there when you go online. They're grouped by lecture. You should definitely try to do the quizzes as close in time to watching the lecture, as that will help you decide if you are moving at a good pace, which for this course means not too fast. We professional mathematicians despair at school systems that impose tight time limits on completing mathematics tests and encourage fast working. Real math takes time. Before I dive into the first topic, let me say a couple of things about what to expect as we get into the course. The main thing to realise is that a lot of what we do probably won't seem like doing math, since the focus is on how to think mathematically, not how to apply standard techniques to solve problems. For most of you, if you've had a fairly standard high school mathematics education, this is a big shift. Second, even when we do math that looks familiar to you, you'll spend most of the time thinking rather than writing things down. If at all possible, you should work with somebody else or in a small group. Learning to think a different way is a lot harder than learning a new technique and few of us can do it alone. If you find you need help or if you think there's a mistake in something I've said or written, the first thing to do is work with your study group or in the course discussion forum. If, after discussion, your group thinks there's a mistake, post it on the forum to see what others think. That way, issues that arouse a lot of interest will rapidly rise to the attention of me and my teaching assistants. Okay, now we've gotten you orientated and disposed of the preliminaries, let's get down to work in earnest. The first topic is being precise about how we use language. The American Melanoma Foundation in its 2009 fact sheet states that one American dies of melanoma almost every hour. Like many mathematicians, I can't help being amused by such claims. Not because we mathematicians lack sympathy for a tragic loss of life. What we find amusing is that if you take the sentence literally, it does not at all mean what the AMF intended. What that sentence actually claims is that there is one American, let's call this person X, who has the misfortune to say nothing of the remarkable ability of almost instant resurrection to die of melanoma every hour. The sentence the AMF writer should have written is this, almost every hour an American dies of melanoma. You see the difference? Mishuses of language like this are fairly common. So much so that they really aren't misuses. Everybody reads the first sentence as having the meaning captured accurately by the second. Such sentences have become figures of speech. Apart from mathematicians and others whose profession requires precision of language, hardly anyone ever notices that the first sentence, when read literally, actually makes an absurd claim. When people use language in everyday context to talk about everyday circumstances, they share a common knowledge of the world. And that common knowledge can be relied upon to determine the intended meaning. But when mathematicians use language in their work, there often is no shared common understanding. Moreover, in mathematics, the need for precision is paramount. That means that when mathematicians use language in doing mathematics, they rely upon the literal meaning. They have to. As a result, mathematicians have to be aware of the literal meaning of the language they use. This is why beginning students of mathematics in college are generally given a crash course in the precise use of language. That might sound like a huge undertaking given the enormous breadth of language, but the use of language in mathematics is so constrained that the task actually turns out to be relatively small. Modern pure mathematics is primarily concerned with precise statements about mathematical objects. Mathematical objects are things like integers, real numbers, sets, functions, et cetera. Here are some mathematical statements. There are infinitely many prime numbers. For every real number a, the equation x squared plus a equals 0 has a real root. The square root of 2 is irrational. If pn denotes the number of primes less than or equal to the natural number n, then as n becomes very large, p of n approaches n divided log e of n. Not only are mathematicians interested in statements like these, they are above all interested in knowing which statements are true and which are false. The truth or falsity in each case is demonstrated not by observation or measurement or experiment as in the natural sciences, but by a proof. In this course, we look at some different ways of proving statements. In the case of my four examples, one, three and four are true, but two is false. Let me show you a proof of the first statement. It's due to the ancient Greek mathematician Euclid who lived around 350 BCE. We show that if we list the primes p1, p2, p3, etc., the list continues forever. Well, suppose we've reached some stage n, so we've listed p1, p2, p3 up to pn. Can we find another prime to continue the list? If we can always find another prime, then the list goes on forever and we've shown that there are infinitely many primes. Well, can we? Well, here's a clever trick that Euclid described in his famous book Elements in 350 BCE. Look at the number n defined as follows. Set n equal to p1 times p2 times p3 all the way up to pn, multiply them all together and add one. Clearly, n is bigger than pn. So if n happens to be prime, we found a prime number bigger than pn and we can continue the list. If n is not prime, then it's going to be divisible by a prime, say p. Now, p cannot be any of the primes p1, p2, p3 up to pn because if you divide any of those primes into n, that prime will divide into this part and then there's a remainder of one. So p cannot be any of those. Why? Because dividing them leaves a remainder of one. So p is bigger than pn. That means we've found a prime number bigger than pn. Either way, if n is prime or if it's not prime, we've shown that there's a bigger prime than pn. Which means the list can always be continued and that proves that there are infinitely many primes. Let's just take another look at what we've done. We start with a list of all of the primes and we try to list all of the primes. We have to show that we can do that and keep going. Okay, so we start with a list of the primes. We assume we've reached some stage n, n could be a 10, a million, a billion, a trillion, whatever. We've reached some stage and we show that we can always find another prime bigger than the last one. How do we do that? Well, there's a clever idea. We look at this number, big n, which means you multiply the first little n primes. That's a little n there. You multiply them together and you add one. n is certainly bigger than the last one in that sequence. So if big n is prime, then it's a prime bigger than pn. Now, we're not seeing that big n would be the next prime after pn. In fact, it almost certainly wouldn't be because big n is a lot bigger than these numbers. So this number is a lot bigger than pn. So this isn't going to be the next prime. Almost certainly. But that doesn't matter. We've shown that there is another prime and whatever the next prime is, we'll put it onto the list. The alternative was that n wasn't prime, in which case it's divisible by a prime and we call it p. Now, that prime p can't be any of these. Why? Well, this is why we defined n the way we did. If you divide n by any of these primes, you're left with a remainder of one. So the prime that divides n can't be any of these. It must be a different one. If it's a different one, it's bigger than pn. So again, we've found a prime that's bigger than pn. Is this prime p the next prime after pn? Well, it might be, but there's no reason to assume it is and it doesn't matter. The point is we've found a prime bigger than pn, so once again, the list can be continued. Either way, the list can be continued. This is the clever trick that makes it work. Defining n that way and we define n that way to make sure that if there's a prime dividing big n, it won't be equal to any of those. Till was proved. There are infinitely many primes. How about that? So we've proved the first of our four examples. There are infinitely many prime numbers. That one's true. What about the second one? Well, it says that for every real number a, the equation x squared plus a equals 0 has a real root. That turns out to be false. To show that it's false, all you need to do is find a single real number a for which the equation does not have a real root. Well, why don't we just take a equals 1? Then we know that the equation x squared plus 1 equals 0 does not have a root because there's no number that you can square, no real number that you can square, such that when you add 1 to it, you get 0. The square of any real number is positive. When you take a positive number and add 1, you end up with a positive number. Because there is a real number a for which the equation does not have a root, that shows that the statement that for every real number there's a root is false. What about number 3? Well, that one turns out to be true, and we're going to prove that's true later in the course. The fourth one is a rather complicated looking statement about the distribution of the prime numbers. That's a very famous result that was proved at the, about just over 100 years ago, at the end of the 19th century. It's known as the prime number theorem. So this one is true, the prime number theorem. And there we are. Well, the correct answer is false. The proof certainly involved looking at this number, but it didn't require that this number be prime. It was rather different. So if you thought the answer was true here, I would strongly advise you to go back and look at that proof again. Because we most definitely did not assume that this number was prime. Clearly, before we can prove whether a certain statement is true or false, we must be able to understand precisely what the statement says. Above all, mathematics is a very precise subject where exactness of expression is required. This already creates a difficulty since words tend to be ambiguous and in real life our use of language is really precise. In particular, when we use language in an everyday setting, we often rely on context to determine what our words convey. For example, an American can truthfully say July is a summer month, but that would be false if spoken by an Australian. The word summer means the same in both statements, namely the hottest three months of the year. But it refers to one part of the year in America and another in Australia. In everyday life we use context and our general knowledge of the world and of our lives to fill in missing information in what is written or said and to eliminate the false interpretations that can result from ambiguities. For example, we would need to know something about the context in order to correctly understand the statement the man saw the woman with a telescope. Who had the telescope, the man or the woman? Ambiguities in newspaper headlines, which are generally written in great haste, can sometimes result in unintended but amusing second readings. Among my favourites that have appeared over the years are sisters reunited after ten years in checkout line at Safeway. Large hall appears in High Street. City authorities are looking into it. Mayor says bus passengers should be belted. OK, to systematically make the English language precise so people can communicate effectively by defining exactly what each word is to mean would be an impossible task. It would also be unnecessary since people generally do just fine by relying on context and background knowledge. But in mathematics things are different. Precision is crucial and it cannot be assumed that all parties have the same contextual and background knowledge in order to remove ambiguities. Moreover, since mathematical results are regularly used in science and engineering, the cost of miscommunication through an ambiguity can be high, possibly fatal. At first it might seem like a Herculium task to make the use of language in mathematics sufficiently precise. But fortunately it turns out to be very doable though a bit tricky in places. And what makes it possible is the special highly restricted nature of mathematical statements. Almost every key statement of mathematics, the axioms, conjectures, hypotheses and theorems is a positive or negative version of one of four linguistic forms. Object A has property P. Every object of type T has property P. There is an object of type T having property P. If statement A, then statement B. Or else the statement is a simple combination of substatements of these forms using the connecting words, which we call combinators, and, or, and not. For example, three is a prime number, or ten is not a prime number. Every polynomial equation has a complex root. Or it's not the case that every polynomial equation has a real root. There is a prime number between 20 and 25. There's no even number beyond two that is prime. If P is a prime of the form 4n plus 1, then P is a sum of two squares. In their everyday work, mathematicians often use more fluent variants of such statements, such as, not every polynomial equation has a real root, or no even number is prime except for two. But those are just that variants. Incidentally, the final statement about the primes of the form 4n plus 1 is a celebrated theorem of the early 19th century mathematician Carl Friedrich Gauss. The ancient Greek mathematicians seem to have been the first to notice that all mathematical statements can be expressed using one of these simple forms. And they made a systematic study of the key language terms involved, namely, and, or, not, implies, for all, and there exists. The Greek mathematicians provided universally accepted meanings of these key terms and analysed their behaviour. When that study is carried out in a mathematically formal way, it's known as formal logic, or mathematical logic. The study of mathematical logic is a well-established branch of mathematics studded and used to this day in university departments of mathematics, computer science, philosophy and linguistics. It gets a lot more complicated than the original work carried out in ancient Greece by Aristotle and his followers and by the stoic logicians. But that's well outside our present interest. It's time for another quiz. Let me stress what I wrote to accompany the first in-lecture quiz. The quizzes are designed so that if you're progressing in a manner that will get you through the entire course, you'll find the quiz is easy. In most cases, the answer will be obvious. Their role is so that if you find you're getting some quiz questions wrong, or if you have to check back through the lecture before you answer, then you'll know you need to put in more focus on mastering the material. You have to do all the quizzes to get the course completion certificate, but your course grade comes from the weekly problem sets and the final exam. So here's the quiz. The ancient Greeks were the ones who began the formal study of language and reasoning that became the branch of mathematics known as formal logic. And that brings us to the end of the first lecture. Your next task is to complete course assignment one. You probably find it easier to print off the PDF file and work on it when you have time. It's really hard to say how much time you should allow for the course assignments as people work at very different rates. The first one is meant to be pretty light, so I'm guessing that an hour or so should be enough for most of you, but I really can't be sure. If you're ever uncertain about anything in a lecture, a reading or an assignment, discuss it with your study team or go on to the course discussion forum. In fact, even if you don't feel uncertain, you should discuss the course material regularly with other students. Many students do fine at high school working on their own. In fact, the best students usually work that way all the time. I did when I was at high school, but when I got to university, I soon discovered that working alone was the worst possible strategy. University mathematics is not focused on learning procedures to solve problems. It's about thinking a certain way and mastering a new way of thinking is best learned by working in groups. In a MOOC where you don't have an instructor or a TA to regularly check on your progress, working with a study group is pretty well essential. The folks at Coursera are working on developing their platform to facilitate group work, but that's not available to us now. So you'll have to form and manage your study groups yourself. Of course, the course discussion forum provides a starting point, but you should use whatever medium you prefer to keep in touch with other students. Forming a group on Facebook, Google groups, Yahoo etc. are all obvious ways to do it, and you can share files and work together on Google Docs. Check out the course website for information, recommendations, suggestions and updates. And keep in touch with the rest of the class by following activities on the forums. Good luck, and I'll see you next time. Welcome to the second lecture. I hope you're making progress with the first assignment. I'll post my answers to some of the assignment questions later. Consult the course schedule on the website and check regularly for announcements of any changes. You should not expect to solve all the problems in an assignment in a single session, or even before the next lecture. What you should do before the next lecture is attempt each question. That's what I mean when I say complete the assignment. Remember, the goal of this course is to acquire a certain way of thinking, not to solve problems by a given deadline. The only way to develop a new way of thinking is to keep trying to think in different ways. Without guidance, that would be unlikely to get you anywhere, of course. But the point of a course like this is to provide that guidance, and the assignments are designed to guide your thinking attempts in productive directions. Okay, let's proceed. As a first step in becoming more precise about our use of language in mathematical contexts, we'll develop precise, unambiguous definitions of the key connecting words and, all, and not. The other terms we need to make precise implies equivalence for all and their exist, and more tricky, and we'll handle them later. Let's start with and. We often want to combine two statements into a single statement using the word and. So we need to analyse the way the word and works. The standard abbreviation that mathematicians use for this is an inverted V known as the wedge. Sometimes you'll see the old familiar ampersand used, but I'm going to stick to the common mathematical practice of using the wedge. For example, we might want to say, pi is bigger than three and less than 3.2. We could do this as follows. We could write pi is bigger than three and pi is less than 3.2. In fact, for this example where we're just talking about the position of numbers on the real line, there's an even simpler notation. We would typically write three less than pi, less than 3.2. But as an example illustrating the use of the word and this one is fine. What does it mean? Well, if we have two statements, phi and psi, phi and psi means that they're both true. The official term for an expression like this is it's the conjunction of phi and psi. Relative to the conjunction, the two constituents, phi and psi, are called the conjuncts of phi and psi. What are the circumstances under which a conjunction phi and psi is true? Well, if phi and psi are individually true, then the conjunction phi and psi will be true. Under what circumstances will phi and psi be false? Well, if either phi is false or psi is false, or they're both false. This might seem very self-evident and trivial, but already this definition leads to a rather surprising conclusion. And here it is. According to our definition, phi and psi means the same as psi and phi. Both mean that phi and psi are both true. In mathematical pilots, conjunction is commutative. But that's not the case for the use of the word and in everyday English. For example, John took the free kick and the ball went into the net. That doesn't mean the same as the sentence the ball went into the net and John took the free kick. They're both conjunctions and the two conjunctions are the same. One of them is John took the free kick. The other one is the ball went into the net. But anyone who's familiar with soccer realises that these two sentences have very different meanings. The fact is in everyday English, the word and is not commutative. Sometimes it is, but not always. Let's see what you make of this one. Let A be the sentence it rained on Saturday and let B be the sentence it snored on Saturday. Question. Does the conjunction A and B accurately reflect the meaning of the sentence it rained and snored on Saturday? Well, what do you think? Yes or no? Although I can think of situations in which the answer would be no, in general I would be inclined to say the answer is yes. A useful way to represent a definition like this is with a propositional truth table. What we do is we list component statements, in this case, will be phi and psi. And they're going to go together to make the conjunction phi and psi. And now we're going to do a table that lists all the possible truth-false combinations for phi, psi and phi and psi. So let me see. Phi could be true and psi could be true. Or phi could be true and psi could be false. Or phi could be false and psi could be true. Or phi could be false and psi could be false. The next step is to list in the final column t or f according to our definition of what the conjunction means. Why don't you see if you can fill in t or f in each of those four boxes to represent the definition of phi and psi that we've given? According to the definition, phi and psi is going to be true whenever phi is true and psi is true. That's the first rule. So there's a t going to go here. But that's the only condition under which phi and psi is true. In all other circumstances it's false. So the entries for these are all f. So in one simple table we've captured the entire definition of phi and psi. This emphasises the fact that the truth of a conjunction depends only on the truth or falsity of the two conjuncts. The definition was entirely in terms of truth and falsity. What phi and psi meant was irrelevant. It was only about truth and falsity. That's going to be the case for all the definitions that we're going to give in order to make language precise. They're going to depend upon truth or falsity, not upon meanings or logical connections. Now let's look at the combinator or. We want to be able to assert that statement a is true or statement b is true. For instance, we might want to say a is greater than zero or the equation x squared plus a equals zero has a real root. Or maybe we want to say a b equals zero if a equals zero or b equals zero. Those are both statements that we get when we combine two sub-statements with the word or. Both statements are in fact true, but there's a difference between them. The meaning of or is not the same in the first sentence as it is in the second sentence. In the first sentence there's no possibility of both parts being true at the same time. Either a is going to be positive or this equation will have a real root. They can't both occur. If a is positive then this equation does not actually have a real root. In the case of the second sentence they could both occur together. To get a b equals zero it's enough if a is zero it's enough if b is zero or they can both be zero. So these two are different. In the first case we have an exclusive or in the second case we have an inclusive or. Incidentally it doesn't matter if you try to enforce the exclusivity by putting an either in front of it. If you look at the way the word either operates if you say either this or that then what happens is that the either simply reinforces an exclusive or if one happens to be there. In the case of the second one you could say a b equals zero if either a equals zero or b equals zero and in fact that doesn't enforce the exclusivity at all. We just accept the fact that they could both be true. In other words the word or in everyday English is ambiguous and we rely upon the context to disambiguate. In mathematics it's different we simply can't afford to have ambiguity floating around. We have to make a choice between either the exclusive or or the inclusive or and for various reasons it turns out to be more convenient in mathematics to adopt the inclusive use. A mathematical symbol we use to denote the inclusive or is a v symbol. It's known as the disjunctive symbol so given two sentences phi and psi phi v psi means phi or psi or both. This sentence phi or psi is called a disjunction of phi and psi and relative to the disjunction the constituents phi and psi are called the disjuncts. Remember phi or psi in mathematics means at least one of those two is true. They could both be true. For example the following rather silly statement is true three is less than five or one equals zero. I can't imagine a mathematician actually writing that down except as an example as I'm doing now. Silly examples like this are actually quite useful in mathematics because they help us understand exactly what a definition means. This thing is true even though one of the disjuncts is apparently false. So this emphasises the fact that for a disjunction to be true all you need to do is find one of the disjuncts which is true. It doesn't matter if one or more of the other disjuncts is seemingly false. Okay. Let's see how well we do understand that. Here's a quick quiz. Let A be the sentence it will rain tomorrow. And let B be the sentence it will be dry tomorrow. Here's the question. Does the disjunction A or B accurately reflect the meaning of the sentence tomorrow it will rain or it will be dry all day? Well, what do you think? The answer is clearly no. If that comes as a surprise to you you need to think about the definition of A a little bit longer and see what's going on here. I'll leave you to that one. To wrap up this discussion of disjunction let's see if we can complete the truth table for phi or psi. Okay. If you've got this one right your truth table should look like this. True, true, true, false. The disjunction is true if both are true if one is true or if the other is true. The only time when a disjunction is false is when both disjuncts are false. Okay. So now we've sorted out the meaning of the word or. The next word I want to look at is not. If psi is a sentence then we want to be able to say that psi is false. So given psi we want to create the sentence not psi. The standard abbreviation mathematicians use today is this symbol which is like a negation symbol with a little vertical hook. Older textbooks you'll find will use a tilde and that's not the one I'm going to use I'm going to stick with the modern notation this sort of negative sign with the hook. And we call this the negation of psi. If psi is true then a negation of psi is false. And if psi is false then a negation of psi is true. We often use special notations in particular circumstances. For example we would typically write x not equal to y instead of not x equals y. But you have to be a little bit careful. For example I would write not the case a less than x less than or equal to b. You might be tempted to write something like a not less than x not less than or equal to b. I would advise against that. That one is better than this one. This one is completely unambiguous. It means that it's not the case that x is between a and b in that fashion. This one well it could mean I mean you could agree that it means that but it's really ambiguous as to exactly what's going on here. So I would say avoid things like that use something like this. We should always go for clarity in the case of mathematics. Remember the whole point of this precision that we're trying to introduce is to avoid ambiguities to avoid confusions because in more advanced situations all we're going to have to rely upon is the language. And then we need to make sure that we're using language in a non ambiguous and reliable way. Negation might seem pretty straightforward and in many ways it is to be real. If we took something like not the case that pi is less than three then that's pretty straightforward. That means pi is greater than or equal to three. Okay, that's easy. No problems there. Let me give you one that's not quite so obvious. Look at this sentence. All phone cars are badly made. What's the negation of this sentence? Let me give you four possibilities. Possibility one or possibility A. All phone cars are well made. Possibility B. All phone cars are not badly made. Possibility C. At least one phone car is well made. Possibility D. At least one phone car is not badly made. Well I'm not giving you this as a quiz but I would like you to think for a minute as to which one of these you think is a negation of that original sentence. Or maybe you think it's something else. Maybe you think it's something to do with domestic cars. Domestic is after all the negation of phone. So what do you think it is? Well let's look at them. A is actually a very common one for beginners to pick. I've been teaching this material for many years now. It's one of the examples I've always gave. It's in the textbook I've written for this course and previous textbook I've written and this one is a common answer that I often get. But if you think about what the sentence really means it's obviously not this one. This is not the negation. Why? Is the original sentence true? No of course it's not. There are many good cars that are well made. So it's not the case that all phone cars are badly made. So this sentence is in fact a false sentence. We know that just by our knowledge of the world. If that sentence is false then its negation is going to be true. But this isn't true. It's not the case that all phone cars are well made. That's false. So that can't be the negation. What about B? Same reasoning. That can't be the negation. Because it's simply not the case that all phone cars are not badly made. That's a false statement. That's false so they can't be the negation of a false sentence. The negation of a false sentence is going to have to be true. So whatever the negation of this original sentence is that negation will have to be something that's true. And we know what's true and false in terms of cars being well made. Well is this one true? Yeah that's true. Is this one true? Well these are both true. So these are both possibilities for the negation of that. And this is still not a quiz but I'm going to leave you for a little while to think about this one. Which one of these do you think actually is the negation of this? We'll come back to this. I'm going to introduce some formal notation from sort of algebraic notation and eventually we'll be able to reason precisely to say which one of these two here or maybe a different thing is the actual negation of this. But let me stress a point I made a minute ago and didn't write anything down. Look at the following sentence. All domestic cars are well made. I've actually had students over the years who have thought that that was in fact the negation of this. I know why they're saying that because they're saying this says something about all foreign cars and this says something about all cars that are not foreign. So there is a sort of negation going on between these two but it's not the negation of the original sentence. How do I know it's not the negation of the original sentence? Because the original sentence is false therefore whatever the negation is is going to have to be true but this isn't true. This is also false and because this is false it can't possibly have been a negation of the original sentence. And in fact this one really falls a long way from being the negation of that. For the following reason the original sentence is about foreign cars. That's what it's talking about. It has nothing to do with domestic cars. It's purely talking about foreign cars. So the negation can only possibly talk about foreign cars. These were good candidates for the negation because they talked about foreign cars. This one isn't even in the ballpark for being a negation because it's not talking about foreign cars it's talking about domestic cars. Negating a word in a sentence is not at all the same as negating the sentence. So this one is a really bad choice. Let me finish with a very simple quiz. Let me ask you to fill in the truth table for negation. This one's a much simpler table because there's only one statement involved. Phi and then we're going to negate it. So very simple truth table. What do you think the values are? This one was an easy one. If Phi is true, the negation is false. If Phi is false, the negation is true. And with that you should be in a position to complete assignment too. That last example about the negation of the sentence all foreign cars are badly made should, I think, illustrate why we're devoting time to making simple bits of language precise. To figure out what the correct negation is we relied on our knowledge of the everyday world. That's fine for statements about the everyday world we're familiar with. But in a lot of mathematics, we're dealing with an unfamiliar world and we can't fall back on what we already know. We have to rely purely on the language we use to describe that world. When we've taken our study of language far enough we'll be able to look at that foreign cars statement again and use rigorous mathematical reasoning to determine exactly what its negation is. Well, that brings us to the end of the first week. How are you getting on? For most of you, this will seem like a very strange course and certainly won't look much like mathematics. That's because you've only been exposed to school math. This course is about the transition to university level mathematics, which in some ways is very different. There isn't much material and as a result the lectures are short. I'm not providing you with new methods of procedures I'm trying to help you learn to think a different way. Doing that is mostly up to you. It has to be. If you're at all like me and pretty well every other mathematician I know you're going to find it hard and frustrating and it's going to take some time. You definitely need to connect to other students and start working together. If you're able to scan pages of work into PDF or use your smartphone to take good clear photos of your work I advise you to start showing your work to other students to get their feedback. Send images as email attachments. Put them on Google Docs or upload them to whatever networking site you choose. You should definitely attempt all the assignments that I give out after each lecture. Doing those assignments, both on your own and in collaboration with others is really the heart of this course. Yeah sure, you can watch the lecture several times but you'll find that it almost never tells you the answer or even how to get the answer in the way that you're familiar with from high school. It's like learning to ride a bike. Someone can ride up and down in front of you for hours telling you how they do it but you won't learn to ride from watching them or having them explain it to you. You have to keep trying it for yourself and failing until it eventually clicks. Now this is a very different way of learning than you're used to, at least in mathematics. As well as the assignments, there's also a weekly problem set. The problem sets comprise assignment questions that count directly towards your grade. Each set has a submission deadline. Because this course is designed for many thousands of students it's impossible for me or my TAs to look at everyone's work and provide feedback. So we have to rely on automated grading. This means that the questions are posed in multiple choice format but these are not at all like the in lecture quizzes. Those are supposed to be answerable on the spot. The problem set questions will require considerable time. This is not ideal. For the material in this course whether you get particular questions right or wrong is pretty insignificant. It's your thinking process that's important but we can't check that automatically. Asking you to answer multiple choice questions is like checking your health by taking your temperature. It tells us something and can alert you and us that something is wrong but it's pretty limited. Still checking temperatures is better than nothing and the same is true for the problem set grading. What I'd like you to do is to try to grade your own work and that of others in whatever study group you form and you should definitely try to get into one. There's a mechanism for doing this. It's called calibrated peer review. It's been tried a number of times and does appear to offer learning benefits. It's the method we're going to use to grade the final exam. So anyone who wants to earn a distinction in this course is going to have to learn and use it eventually. The advantage of using it now is that it does yield positive learning outcomes. Check out the description on the course website and then give it a shot. When assignment won, most of the questions are just there for you to think about and to discuss with your fellow students and I'm going to leave you to do that and sort them out on your own, discuss them on the forums or with whatever group you put together and please do put together a group. What I will do is say a little bit about Question 8. Question 8 is related to Euclid's proof that there are infinitely many primes and at a crucial point in that proof we looked at this number where p1 through pn enumerates the first 10 primes. We multiply them together and we add one and I mentioned during the proof that this number is not always prime and Question 8 asks how would you prove that it's not always prime? And the answer is you have to find a number n such that when you do multiply the first 10 primes and add one, it's not prime. Well, how would you go about doing that? Well, the most obvious way is to just start searching. So you would start by looking at the first two primes say, multiply them together, add one, you get seven. Well, that is prime. Try another one, two times three times five, add one. That's 31. That's also prime. Try another one, two times three times five times seven plus one. That's 211. That's also prime. At this point, you're probably starting to lose heart. But if you keep on just two more primes, two, three, five, seven, 11 and 13, multiply them together and add one, you end up with 30,031, which is not prime because it's a multiple of 59 and 509. Okay? So here we've got an example of a number of this form which is not prime. And in order to show that not every number of that form is prime, all you have to do is find a single example of a number of that form which isn't prime, and we've done it. So that's the answer to question eight. That's how you prove that those numbers are not always prime. Okay? Well, that's really all I want to say about assignment one. Assignment two has a little bit more meat to it in the sense that there are more things that I'll need to show you. Okay, let's see what we've got. Well, number one has been done for you. Number two, the simple word of like that is to say seven, less than or equal to p less than 12. The next one, we would write as five less than x less than seven. The next one, well, if x is, let me see, if x is less than four, then it's automatically less than six. So the second conjunct here is superfluous. We could just write that as x less than four. What about the next one? Well, y squared less than nine means, let me see, negative three is less than y is less than three. That's what the first, that's what the second conjunct means. Okay? But if y satisfies this condition, then certainly y is less than four. So the first conjunct is superfluous. And the only one that counts is the second one. So we could simplify that as y squared less than nine. Okay, that was just our thinking on the way. Well, I hope this one. Well, x is greater than or equal to zero, and it's less than or equal to zero. There's only one possibility, and that's x equals zero. Well, for number three, in order to show that the conjunction is true, what you would do is show that all of phi one, phi two, et cetera, up to phi n are true. And for number four, to show that the conjunction is false, is you show that one of phi one, phi two, phi n is false. So you have to find one of these, at least one of these, which is false, and then it follows us, follows that the conjunction is false. Okay, well, that's that one. Okay, well, with part A, if pi is bigger than ten, then it's automatically bigger than three. So in terms of the disjunct, pi bigger than three dominates. That says more. If we draw a picture, we've got zero, we've got three, we've got ten. And the first one says that x is to the right of this point. And the second disjunct says that x is to the right of that point. And the disjunction will be true if at least one of them is true. And at least one of them will be true if we start at three. First of all, only the first disjunct is correct. And then when we get beyond this point, both disjuncts are correct. So that's the first one that works. This is that either x is less than zero, or x is greater than zero. So the simplest way to write that is to say that x is not equal to zero. It's either less than zero, it's negative, or it's bigger than zero, it's positive. This says that x equals zero or x is greater than zero. The standard of abbreviation for that is x greater than or equal to zero. And once you've got x greater than or equal to zero, that's going to dominate over the first disjunct in terms of a disjunction. So you're going to have x greater than or equal to zero. That makes a more general claim of the two. And for this one, let me make note that x squared greater than nine, this part, means that either x is greater than three, or x is less than negative three. Well, one of these two disjuncts is the one that's here. So this one is superfluous. So that's the one that counts. So x squared greater than nine is a simple way of saying it. And again, as I was, as with the previous case, as with numbers one and two, as I was going through them, I actually articulated what these things are. And these were really just to sort of prompt you to thinking about how you would express them in English. OK, well, that's that one. Well, in number seven, to show that a disjunction is true, you have to show that one of them is true, at least one of them. And for number eight, to show that it's false, you have to show that all of phi one, phi two, et cetera, to phi n are false. OK? Well, that's number seven and eight taken care of. Turning to number nine now, to say that it's not the case that pi is greater than 3.2, is to say that pi is less than or equal to 3.2. When you negate a strict greater than, you get less than or equal to. To say that it's not the case that x is negative, is to say that x is greater than or equal to zero. To say that it's not the, well, these are real numbers from the context of the expression. We can assume that these are talking about real numbers. And for a real number, every real number has a square which is strictly positive with one exception. And that exception is zero. So the only real number for which it's not the case that x squared is strictly positive is x equals zero. The standard of deviation for not x equals one is x not equal to one. And when you negate a negation, it takes you back to the original statement. And again, as I went through them, I essentially answered number 10. Turning to question 11 now, here are the answers that I get. Dollar and yarn both strong. OK, dollar strong and yarn strong. I think that one's fairly straightforward. What about this one? Well, as these words, despite and but, by my mind, those are just nuanced forms of conjunction. They both mean and. So we've got the yarn weak and there's a trade agreement and the dollar strong. So the despite and the but, they sort of get at the whether these things are contrary to our expectations or consistent with our expectations. But they still say that all of these three things hold together. OK, this one I think is fairly straightforward. They can't both be strong at the same time. So it's not the case that the dollar strong and the yarn strong. For part D, well, if that doesn't prevent that, it means that happens and both of those fail. So the trade agreement signed, but the dollar falls and the yarn falls. The trade agreement holding doesn't prevent the falling of the dollar and the yarn. That means the dollar and the yarn do fall. And then for the last one, US-China trade agreement fails, but both currencies remain strong. Again, the but, I think, is hand, it's conjunction. So we've got the trade agreement failing, the currencies both remain strong, the dollar strong, and the yarn strong. OK, those are my answers. I would hope you get more or less the same. You might end up writing things slightly differently. But even though we had to think a little bit about what these words mean, to my mind, there's no real dispute as to whether these are correct or not. OK, well, how did you do on that one? We're judging from the discussions on the course forum. In problem set one, the questions that cause people the most difficulty were numbers six and seven. Those are those questions about Alice. And let's just see what's going on here. Notice that in each of these, we've got the same phrase, works in a bank, works in a bank, works in a bank, works in a bank, works in a bank. But for four of these, there's an additional requirement. Not only does she work in a bank, but something else, works in a bank, something else, works in a bank, something else, works in a bank, something else. y casu. Mae'r rhai dyn nhw, mae'n rhai dyn ni yw i fod ymyledd, mae'r peth yn bwysig beth ydy'r mynd i'r hanes i'n bach yn gwybodaeth. Mae'n ddydd fod yn bach yn ei wneud i gydaeryddiad y gallu hanes a ddweud i ddechrau oes, mwy o'r Bwysig half aleis Oestrwf gydwch yn bach yn gwybodaeth yma. Mae rhai ddyn nhw.... maen nhw yn ddidd yn lluniedig yma, yma yn ddidd yn lluniedig. Mae'r hanes i'n ddidd yn lluniedig. most likely to be true. Now it's possible that in a particular instance, another one has the same likelihood. We're not saying that there's a unique most likely one, but we are saying there's a most likely one. Incidentally, it doesn't matter whether you express in terms of the word likely or the word probable or probability, you can do it numerically, you can do it however you want. The issue is not what the word is, the issue is the information you have. And in each of these cases, you have an extra restriction. And whenever you've got an extra restriction with a conjunction, and it's critical that there's a conjunction here, when you've got an extra restriction with a conjunction, it makes it less likely. Because instead of just having to satisfy one thing, you have to satisfy two things. And when you satisfy two things, it makes it less likely to happen. So regardless of whether you use the word likely, probable, whatever you want to do, the most likely one is this one because this is the easiest one to satisfy. And this actually I think is a great example of mathematical thinking. I mean this is a superb example of the kind of thing I'm talking about in this course. Because we're arguing purely in terms of mathematics. In fact, it's the mathematics of the word and the reason this is in this problem set is because this is all about the way and works. So if you really understand how and works, this one just drops right out. That's why it's there. It's to make you really reflect on conjunction and the power of conjunction. So it's, there you go. It's all to do with the amount of information you get. Incidentally, all the probability theory, although I say all, I mean probability theory is a powerful theory. But what probability theory does is adds numbers to the amount of information we have. Okay? And in this case, the extra information is more restrictive. And when you assign probabilities to that, that gets reflected. Okay? Well, that was number six. Let's take a look at number seven. Well, number seven is slightly different because we don't have a common expression in all of the five in the case of works in a bank. In this one, we have works in a bank and something else. This one we've got is a rock star. This one we've got works in a bank and something else. This one we've got works in a bank. There we've got works in a bank. Okay. The other thing here is being quiet, being honest and so forth. But if you look through these, this one has works in a bank and something else. So this is less likely than working in a bank. This has been a rock star. This is honest and so this one is also less likely than working in a bank. So if we're looking for the most likely thing, these make them less likely. So let's just sort of ignore that part. That makes it less likely. It's already more likely that she works in a bank. This one has nothing to do with more. This one she works in a bank. So we've got either working in a bank. You know, this is more likely than it was originally. So I'm making things more likely. Here I've got Alice been a rock star. In this one, either she's a rock star or she works in a bank. So this is the most general one. This is more likely. Because there are two chances here, either working in a bank or being a rock star. Here there's just one of them. Here's just one of them. Here's just one of them. And here's just one of them. Okay? So the most likely one is this one because of disjunction. There are two ways that she could satisfy this, by being a rock star and working in a bank. In each of these, there's one of them. And I've ignored the thing that makes it even less likely. So having modified that, I've made B more likely. Having modified D, I've made it more likely. So now I've just got works in a bank, works in a bank, rock star, works in a bank. But there's only one of them in each of that. In this one, there are two of them. So again, purely by using mathematical thinking, I've arrived at the one that's most likely. And as in number six, it doesn't matter whether you take expressions in terms of probabilities or relative probabilities. However you want, incidentally, probabilities. There are many different types of probability. There's frequentist probability. There's Bayesian probability. There's subjective probability. There's epistemic probability. Assigning numbers to the information you have and calling them probabilities is a pretty tricky thing. There are many different ways of doing it. And it's just a very complicated issue. So those of you who actually tried to use probability theory, that's fine. You can do it. But it actually gets more complicated. Not least because you end up having to talk about whether things are independent events and so forth. And the point is you don't need any of that. Remember at the beginning I said this course is not about rushing in and applying techniques or procedures. It's about thinking basically about the issues. And you don't need the machinery here. If you think about the issues, it comes out. Remember my exhortations. Look at a problem, ask yourself what it means, what is it saying, and try to reason in terms of the problem itself. Don't look for techniques to apply. You often need to do that. But in this case that's not what the course is about. And many of the questions I'm giving in this course should be solved not by applying a technique, but by just thinking about the problem. And let me finish this tutorial on the problem session. And these are the only two problems I want to look at. Let me finish this tutorial session with a different problem of the same kind where you can solve it just by thinking about the problem, not applying a technique. And here is that problem. You've arrived late at an airport. You're rushing to catch your plane. Unfortunately, your gate is at the far end of the terminal. It always seems to be at the far end of the terminal, doesn't it? The fastest you can walk is a constant speed of four miles per hour. For part of the way, there's a moving walkway moving at two miles per hour. You decide to take the walkway and continue to walk. So you're going to walk all the way, but part of the when there's a walkway, you're going to walk on the walkway. So you can go faster, right? Just as you're about to step onto the walkway, you notice your shoelace is loose. And the last thing you want to do is get your shoelace caught in the mechanism of the walkway. So you've got a choice. The stop and fashion your shoelace just before you step on the walkway or you step on the walkway and then stop to fashion your shoelace whilst the walkway is moving you along. And there's no other options. Those are the two options you have. Which option will get you to the gate fastest? Now there are two ways you could try to go about this. You could try to argue in terms of speeds and adding the speeds together and so forth. And if you do that, you're going to run into a problem. There's going to be some information that you need that you don't have. For example, I haven't told you how long the walkway is and how long you had to walk in general. So there's a problem if you try to use standard methods. However, you do have enough information to solve the problem. But if you try to apply relative speeds, the kind of methods you were taught to use in the high school for solving problems about speeds, relative speeds and so forth, you're going to run into problems. And you don't need that. If you think about the problem, you should be able to solve it. Well, I'm just going to leave you with that one. It's an interesting problem. I think it's a rather neat problem. There are many problems like this that you really can just solve by mathematical thinking. And the point is this again is all about this important notion mathematical thinking. It's powerful if you can master mathematical thinking. Many problems can be resolved without going into the complexity of applying mathematical techniques. Mathematical techniques are what you need when mathematical thinking alone doesn't work. But often you can get by with just mathematical thinking. So good luck on that one and enjoy it. It's a lot of fun. Welcome to the third lecture. I hope you made good progress with the assignment from the last session. Our next step in becoming more precise about our use of language for use in mathematics is to take a close look at the meaning of the word implies. This turns out to be pretty tricky. Brace yourself for several days of confusion until the ideas sort themselves out in your mind. Just like learning to ride a bike where all the actual learning occurs before you finally get the hang of it, so too most of the benefit from understanding the way language is used in mathematics comes from trying to figure it out. The benefit in this case is helping to develop your mathematical thinking ability. And it's the process of trying to understand the issues that does that for you. Yeah, sure. Once you've sorted this language stuff out, you'll be able to use language more accurately. But by that stage, you're just using it. The part that helps you learn how to think mathematically is largely over. We need other tasks to develop your mathematical thinking ability further. So as you get into this, bear in mind that the payoff for struggling with the issues is significant in terms of being able to think like a mathematician. Here we go. In mathematics, we frequently encounter expressions of the form phi implies psi. Indeed, implication is the means by which we prove results in mathematics starting with observations or axioms. So we'd better understand how the word implies behaves. In particular, how does the truth or falsity of a statement phi implies psi depend upon the truth or falsity of phi and of psi? Well, the obvious answer is to say that phi implies psi if the truth of psi follows from the truth of phi. But is that what we want? Well, let me give you an example. Let phi by the statement root two is a rational. And let psi by the statement zero is less than one. And let's ask ourselves, is the statement phi implies psi true? Well, phi is true. As I've mentioned once already, we're going to prove that later in the course. And we all know that psi is true. Zero is certainly less than one. So we have the truth of this and we have the truth of that. Does that mean that phi implies psi? Obviously not. There's no relationship between phi and psi. This takes some effort to prove, as you'll see. This, we all know this one. So yes, this is true and that's true, but the truth of this doesn't follow from the truth of that. And now we realize there's a complexity with implication that we didn't meet before when we were dealing with and, with or and with not. And the complexity is that implication involves causality. And causality is an issue of great complexity that philosophers have been discussing for generations. Now we're facing a problem. It didn't arise before because when we were dealing with conjunction and disjunction, it didn't matter whether there was any kind of relationship between the two conjuncts or the two disjuncts. For example, let's look at the sentence, Julius Caesar is dead. And let's conjoin it with the sentence one plus one equals three, the mathematical sentence. And let's do the same thing with disjunction. Forming a conjunction and disjunction didn't require any kind of relationship between these two. These were actually clearly independent. One's a statement about a long dead individual and the other one is a mathematical statement. Incidentally, while we've got these in front of us, let me just give you a quick quiz. Is the first one true or is it false? Is the second one true or is it false? What do you think? Well, remember what the definition was. A conjunction is true if both conjuncts are true. This one's true but this one's false so the conjunction's false. This one's true, this one's false. All you need for a disjunction to be true is at least one of the disjunction to be true so that one's true. And the fact that there's no meaningful relationship between the two conjuncts in this case or the two disjuncts in that case played no role in determining what the truth value was. It was purely in terms of truth and falsity. But it's not the same with implication involves causality. So let me just express that explicitly. Implication has a truth part and a causation part. What we're going to do is ignore that part. We're going to leave that to the philosophers if you like and we're just going to focus on the truth part. Now that might sound to be a very rash thing to do. Throwing away causation, are we going to be left with anything useful? It turns out we are. It might seem a dangerous thing to do to serve where this important constituent of implication but it turns out that when we focus on the truth part we're left with enough to serve our needs in mathematics. So much so that we're going to give this a name. We're going to call it the conditional or sometimes the material conditional. That's the part we're going to focus on. So we're going to split implication into two parts conditional and causation. The first part, the conditional, we're going to define entirely in terms of truth values and the second part we're going to leave to the philosophers. The symbol that we use normally for conditional or at least a symbol I'm going to use is a double arrow. Lots of textbooks use a single arrow but that's got, that has many uses in mathematics. We're certainly going to use that for something else later on in the course. So to avoid confusion I'm going to stick to this notation for the conditional. So I'm going to write conditional expressions like this. That's the truth part of phi implies psi. When we have a conditional we call phi the antecedent and we call psi the consequent and we're going to formally define the truth of phi conditional psi in terms of the truth of phi and the truth of psi. Now you might worry that by throwing away a causation we're going to be left with a notion that's really of no use whatsoever. That actually is not the case. Even though we're throwing away something of great significance hanging on to the truth part leaves us something very useful. And the reason is whenever we have a genuine implication which are actually the only circumstances in which we're ultimately going to be interested whenever we have a genuine implication the truth behaviour of the conditional is the correct one. It really does capture what happens with truth and falsity when we have a genuine implication. That probably seems a bit mysterious at this stage but when we start to look at some examples I think it should become clear what I mean. The advantage is that a conditional is always defined. For real implication you've got that issue of causation. The square root of 2 and the 0 less than one example. The truth of falsity wasn't the issue. It was whether there was a relationship between those two statements. That's a complicated issue. But because we're going to define the conditional purely in terms of the truth value of the two constituents, the antecedent and the consequent it turns out that the conditional will always be defined. When we do have a genuine implication the definition of the conditional will agree with the way implication behaves. And when we don't have genuine implication the conditional will still be defined and so we can proceed. Again this probably seems very mysterious when I describe it in this way but as we develop some examples I hope you'll be able to understand what I'm trying to get at. I think we need to catch our breath. Let me do that by giving you a quiz. The truth of the conditional from phi to psi is defined in terms of 1, the truth of phi and psi or 2, whether phi causes psi or both. Which is it? It's number one. We define the truth of a conditional in terms of the truth and falsity of the antecedent and the consequent. And because we define the truth of the conditional in terms of truth and falsity in that way it has a truth table. So let's see if we can figure out what it is. Well I'm not giving you this as a quiz by the way. I'm going to work this one through. This is tricky. We're not out of the woods yet by any means. So I'm going to lead you through this one. This part we've already looked at. We define the conditional as the truth part of implication. And implication has a property that a true implication leads to a true conclusion from a true assumption. So because we take the conditional from real implication we have to have truth all the way to the top level. Now the fun begins. We have to fill in these three values. And when I say fun, and I mean fun, let's look at that first row of the truth table. Suppose phi is the statement n is bigger than seven. And suppose psi is the statement n squared is bigger than 40. If phi is true, in other words if n is bigger than seven, then n squared is certainly bigger than 40. In fact it's bigger than 49. So certainly in this case phi does imply psi. So that thing is true. If n is bigger than seven, then n squared is certainly bigger than 40. So that's true. We have truth everywhere. So this is consistent with the truth table. Now let's look at a different example. Phi with a statement, our old friend Julius Caesar is dead. And that's psi with a statement, pi is bigger than three. Phi is true, psi is true. According to the truth table it follows that phi conditional psi is true. In other words Julius Caesar is dead, conditional pi is bigger than three. Now if you read this as Julius Caesar is dead implies pi bigger than three, then you're in a nonsensical situation. But remember this isn't implication. This is just the truth part of implication. And in terms of the truth part there's no problem. That's true, that's true, the conditional is true. In the first example there is a meaningful relationship between phi and psi. When we know that n is bigger than seven then we can conclude that n squared is certainly bigger than 40. There's a connection between the two. And in that case the behavior of the conditional is certainly consistent with what's really going on. In the second case there's no connection between the two. The conditional is true but it's got nothing to do with one thing following from the other. The value of doing this is even though this has no meaning in terms of implication its truth value is defined. In both cases we have a well defined truth value. In the first case it's a meaningful truth value. That does follow from that. In the second case it's purely a defined truth value. But that's not going to cause us any problems. Because we're never going to encounter this kind of thing in mathematics. We encounter this kind of thing all the time. But we're not going to encounter this kind of thing. So all we've done is we've extended the notion to be defined under all circumstances. And we've done it in a way that's consistent with the behavior we want when something meaningful is going on. This is actually quite common in mathematics to extend the domain of definition of something so that it's always defined. So long as it has the correct behavior, the correct definition for the meaningful cases and providing we do the definition correctly it really doesn't cause any problems. In fact it solves a lot of problems and eliminates a lot of difficulties if we extend the definition so that it covers all cases. This is just something we do in mathematics all the time. It may seem strange when you first meet it, but it is a part of modern advanced mathematics. Incidentally, if you think this is just playing games, let me mention that the computer system that controls that aircraft that you'll be flying in next time depends upon the fact that operations like this are always well defined. That software control system doesn't depend upon knowing whether Julius Caesar is dead or things like that. It doesn't depend on those kind of facts of the world. Computer systems by and large don't depend upon understanding causation, which is just as well because they don't. What computer systems depend upon is that things are always accurately and precisely defined. And this expression, phi conditional psi, occurs all over the place in software systems. So quite literally your life depends upon the fact that this is always well defined. It doesn't depend upon the fact that the computer doesn't know whether Julius Caesar is dead or not. Okay, time to look at the second row of the truth table. What goes in here? T or F? We've only got two to choose from. I guess we could just make a guess, but we're not going to do that. We're going to figure it out and get the right answer. But what would happen if we put a T here? And we put it in very lightly because I'm not sure it's going to be a T. If this were true, and we think about it in terms of genuine implication because we are trying to capture the truth behaviour of genuine implication, remember. So if it was the case that phi really did imply psi, if that statement was true when we interpreted it as real implication, then the truth of psi would follow from the truth of phi. That's how we began, remember. That's what real implication means. Real implication means the truth of this would follow from the truth of that. So if that were a T and it was real implication, then when we have a T here, we would have to have a T there. But we don't. We've got an F. So we can't have a T here because if we put a T here, the conditional is contrary. It contradicts real implication. And we're trying to extend implication to be defined in all cases when there's no causation. So this has to be an F. If we put a T there, we're in trouble. Our notion doesn't agree with real implication. In order that the conditional agrees with real implication, that has to be an F. If it's a true, if it's a truth, then we would have a true antecedent and a false consequence from a true implication. So we've argued backwards to conclude that this has to be F. Let me write that down just to make sure everyone's following what I'm trying to say. If there were a genuine implication, phi implies psi, and if that implication were true, then psi would have to be true if phi were true. So we cannot have phi true and psi false if phi implies psi is true. That means that in the case where phi is true and psi is false, we have to have a false here. Let me write that down. Are you confused? This is tricky. There's no getting around the fact that sorting out implication and extracting the conditional from implication is tricky. You're probably going to have to replay the video several times and look at what's been written and think about it in order to sort this out. Now remember when I first met this, it took me quite a while to get on top of it. It really isn't an easy thing to do. So you certainly have my sympathy that you're going to have to struggle with this thing. It's probably one of the hardest things in the course to really come to terms with the way we define the meaning of the conditional. Okay, final straight. Let's see if we can fill in the last two inches in the truth table for the conditional. When we've done that, we'll all go off in search of some aspirin. Now if you're like me, you have no intuitions as to what to put here. And the reason you have no intuition is that even though you're used to dealing with implication, you've never dealt with an implication where the antecedent was false. You're not only ever interested in drawing conclusions from true assumptions. So you never dealt with a true one of these guys when this guy was false. You don't have any intuitions. You do, however, have intuitions about this guy. And the reason that's going to help us out is that negation swaps around F and T. So corresponding to the F's here, when we look at this guy, we'll have truths. We'll have T's for this. So you are used to having to deal with this when this is true. And that will be equivalent to dealing with this when that's false. Because negation swaps truth and falsity. So the trick, or at least the idea by which we're going to figure out what goes here, is to stop looking at implication and look at not implication. Or phi does not imply psi if even though phi is true, psi is nevertheless false. Just think about that for a minute. Phi does not imply psi if even though phi is true, psi is nevertheless false. That's how you know that this guy holds. You know that phi doesn't imply psi if you can check that phi is true, but psi is nevertheless false. That's the only circumstance under which you can conclude this thing is true. In all other circumstances, this guy will be false. So let me write that down. This guy is true if you have phi true and psi false. In all other circumstances, this guy will be false. You still with me? Because we're only one line away from the conclusion now. Let me just rewrite that as follows. Because negation swaps f and t, this guy will be false when this guy is true. So if we have t and f, which we have here, then negation is false, which means this guy is true. In all other circumstances, this guy is true. So those are truth. And we're done. Phew, that was difficult. In fact, to my mind, this is the most difficult part of the entire course. This is not easy stuff. It's going to take you some time before you sort it out in your mind. You're going to have to keep working at this. Look through the lectures several times and think about it. It will take effort, but you'll get there. We all get there in the end, but it's just not easy. In the meantime, you've got this far. We should start thinking about opening a bottle of champagne. But before you crack open the bottle, let me leave you with one more little quiz. Which of the following are true? Phi conditional psi is true whenever 1, phi and psi are both true. 2, phi is false and psi is true. 3, phi and psi are both false. And 4, phi is true and psi is false. And I want you to check all that are true. Well, which of these four are the case? Which of these four conditions tell you when phi and psi is true? The answer is 1, 2 and 3. Phi conditional psi is true either when 1 is true or when 2 is true or when 3 is true. The only time when phi conditional psi is false is when 4 holds. So the correct answer is 1, 2 and 3 and phi is not. OK, now you can open that champagne. Let me sum up what we've done. We've defined a notion, the conditional, that captures only part of what implies means. To avoid difficulties, we base our definition solely on the notion of truth and falsity. Our definition agrees with our intuition concerning implication in all meaningful cases. The definition for a true antecedent is based on an analysis of the truth values of genuine implication. The definition for a false antecedent is based on a truth value analysis of the notion does not imply. In defining the conditional the way we do, we do not end up with a notion that contradicts the notion of genuine implication. Rather we obtain a notion that extends genuine implication to cover those cases where a claim of implication is irrelevant because the antecedent is false or meaningless when there's no real connection between the antecedent and the consequence. In the meaningful case where there is a relationship between phi and psi and in addition where phi is true, namely the cases covered by the first two rows of the truth table, the truth value of the conditional will be the same as the truth value of the actual implication. Remember it's the fact that the conditional always has a well-defined truth value that makes this notion important in mathematics. Since in mathematics we can't afford to have statements with undefined truth values floating around. Let's finish this lecture with a short quiz. When you've done that, you should complete assignment three. I've kept assignment three fairly short since I expect you'll need most of your time simply understanding our analysis of implication and the definition of the conditional. Remember to discuss these assignments with other students. Don't struggle for too long on your own. On this kind of material it's much more effective to work with others. Here's the quiz. The answer to the first one is that it's true. The antecedent is true and the consequence is true. So the conditional is true. In fact there's a deeper result going on here. Providing you take a positive number instead of pi, any positive number, then if the square of that positive number is bigger than two, that number must be bigger than 1.2 because the square root of two is 1.4 etc etc etc. So for positive numbers, it doesn't have to be pi, it can be anything. Any positive number whose square is bigger than two must be bigger than 1.2. And that would be a case of genuine causation, genuine implication. But in terms of the conditional, it's enough that the antecedent is true and the consequence is true. Now this one, it's also true. Now the consequence is false. It certainly does not equal three. But the antecedent is false. And if you have a false antecedent, a conditional is always true. Pi squared is most certainly not less than zero. So you've got false, false, that makes a conditional true. Number three, that one's false. The antecedent is true and the consequence is false. And you cannot obtain a false conclusion from a true assumption. But about the next one, well that one's true. The antecedent is true and the consequence is true. But about this one, do triangles have four sides? No. Do squares have five sides? No. But anything with a false antecedent is true. So that's true. You've got false conditional false and that's always true. What about this one? Well we don't know when Euclid's birthday was. At least I don't know when Euclid's birthday was. I suspect you don't either. Rectangles certainly have four sides however. Either this is true or it's false. Either way, since the consequence is true, the thing is true. We've got an unknown yielding of true consequence. So either we have true-true, in which case it's true, or we have false-true, in which case it's true. Okay, how did you get on with that? How did you get on with assignment three? I kept it short since I wanted to give you a lot of time to come to terms with the tricky notion of implication. This lecture is going to be fairly short as well. Though the assignment that goes with it will be the longest of the entire course. I'm not expecting you to complete it all by the next lecture though. The next topic I want to look at is logical equivalence. Equivalence is closely related to implication. Two statements are said to be equivalent or more fully logically equivalent if each implies the other. Equivalence is a central notion in mathematics. Many mathematical results are proofs that two statements are equivalent. In fact, equivalence is to logic as equations are to arithmetic and algebra. And you already know that equations play a central role in mathematics. Just as we had to introduce a formal version of implication that avoids the complex issue of causation, namely the conditional, we have to introduce an analogous version of equivalence. It's called the biconditional. Fortunately, we did all the difficult work with implication. Now we can reap the benefits of those efforts. Two statements, phi and psi, are said to be logically equivalent or just equivalent if each implies the other. Since we have a formal version of implication, namely the conditional, it follows that there's a formal version of equivalence. We call it the biconditional. The biconditional of phi and psi is denoted by means of a double headed arrow like this. And formally, the biconditional is an abbreviation of phi conditional psi and psi conditional phi. Since the conditional is defined in terms of truth values, it follows that the biconditional is defined in terms of truth values. If you work out the truth table for phi conditional psi and for psi conditional phi and then you work out the conjunction, you'll get the truth table for phi biconditional psi. And if you do that, what you will find is that phi biconditional psi is true, if phi and psi are both true or if they're both false. One way to show that two statements phi and psi are equivalent is to show they have the same truth tables. Actually, to avoid confusion, let me change those to capital phi and capital psi. Because I want to use the low case phi and psi for something else. Because here's the example I'm going to look at. I want to show that phi co-injoined with psi or not phi is equivalent to phi conditional psi. So this is actually going to be my capital phi and this is going to be my capital psi. If I was teaching this material at a high school level, I'd be very careful to choose different letters to denote everything. But we're looking at college university level mathematics now and university mathematicians, professional mathematicians, frequently use upper and lower case symbols in the same context and part of being able to master university level mathematics is actually getting used to disambiguate the notations. Okay, what we have to do is work out the truth table for this, the truth table for that and show that they're the same. So we're going to have to start with phi, with psi. I'm going to need to work out phi and psi. I'm going to need to work out not phi. Then I'm going to need to work out phi and psi disjoined with not phi. And then I'm going to have to compare that with phi conditional psi. Okay, as usual we start with the four combinations, T, T, Tf, Ft, Ff. By the way, it doesn't matter which order you write these in so long as you get all four possible combinations. The way I've written it is the way that most mathematicians write it. Conjunction. Conjunction is true if the two conjuncts are true. T, T gives T. Here there's an F, so you're going to get an F. Here there's an F, so you're going to get an F. We have two F, so you're going to get an F. Negation simply flips T and F. So we have T, T, Ff, becomes F, F, T, T. Now I'm going to disjoin this with that. And disjunction picks out at least one T. Well, there's a T there, so I'm going to get a T here. There's no T, so I'm going to get an F. There's a T there which gives me a T. And there's a T there which gives me a T. I know the truth values for phi conditional psi. There were T, F, T, T. If you look back at the earlier work you'll see that that's what we worked at before. And now I just compare that column with that column. T, F, T, T is the same, T, F, T, T. Because these two columns are the same, I can conclude that that is equivalent to that. I should mention that proving equivalence by means of truth tables is very unusual. It's only a very special case of equivalence. In general, proving equivalence is really quite hard. You have to look at what the two statements mean and develop a proof based on their meaning. Equivalence itself is not too difficult a notion to deal with. What is problematic is mastering the various nomenclatures that are associated with implication. We start out with the notion phi implies psi. And I'm really going to be thinking of genuine implication here, but you can interpret everything I say in terms of the conditional. We'll look at that, I'll talk about that a little bit later. If mathematicians always describe this situation in this way, then life would be very simple. But we don't. There are many different expressions we use to describe phi implies psi. Some of them are intuitively obvious and some of them are actually counterintuitive when you first meet them. The following all mean phi implies psi. One, if phi then psi. Okay, that's very obvious. Two, phi is sufficient for psi. In order to conclude that psi, it suffices to know phi. On the face of it, that's fairly straightforward, but in complicated situations people can sometimes get misled by the use of the word sufficient. The use of the word sufficient is by no means the most problematic in the context of implication. This one causes people a lot of problems at first. Phi only if psi. That's not the same as if psi and phi. The reason I'm emphasizing that is that the if here goes with the psi just as it does here. Let's put quotes from that one. The if goes with the psi, the if goes with the psi. But when it's expressed this way with an only if, it flips round the order of the implication. For example, some of you may know that I'm a very keen cyclist. On the other hand, I've never qualified nor have I ever ridden in the Tour de France. However, I could honestly say a person can ride in the Tour de France only if they have a bicycle. Well that's true because if you don't have a bicycle, you can't ride in the Tour de France. So a person can ride in the Tour de France only if they have a bicycle. Well that's true for me. But it doesn't mean the same as if you have a bicycle, then you can ride in the Tour de France. I actually do have a bicycle. I have several. But I can't ride in the Tour de France. So it's true to say Keith could ride in the Tour de France only if you had a bicycle because that's true for everybody. You can only ride in the Tour de France if you have a bicycle. Otherwise you can't ride in it. But it's not the same as saying it this way around that if you have a bicycle then you can ride in the Tour de France. So I've got to be a little bit careful. These are often confused, which is why I'm stressing the distinction. Okay, let's look at this one because this one that I'm going to write next is one that I've already encountered in my discussions about the Tour de France. Psy if Psy. Notice that we've flipped the order now. Psy is the antecedent. Psy is a consequent. Let me write those down just to remind us. Psy is the antecedent. Psy is a consequent. And those are the roles going to be played by Psy and Psy in all of these. So Psy is the antecedent. Psy is a consequent. Psy is the antecedent. Psy is a consequent. Psy is the antecedent. Psy is a consequent. Fy. Cyfan o gded freshly anticedon. Cyfan o gded�eth rydyn ni yn youredd iddyntod ti'n ei bod Rhyw oed yn rhaid o gae calwyddi. gan oeddiwch o'r anodd neu'r anodd o'r anodd neu'r anodd neu'r anodd. Mae gennym holl gael eich dŵr i'r ffordd neu'r anodd. Mae'r anodd neu'r anodd neu'r anodd. Rwy'n dweud o'r hyn arall, sy'n dweud i'r hyn, mae'n iawn, oeddiwch ymddangos o'r hyn o'r hyn oeddiwn. Mae'n dweud i'r hyn oeddiwch, mae'n dweud i'r hyn oeddiwch oeddiwch. PSY when ever five. And finally PSY is necessary for five. Again, the order is flipped round. That's still a consequent. That's still an antecedent. PSY is the composition of five. In order to ride in the Tour de France it's necessary that you have a bicycle. It's not sufficient that you have a bicycle. To ride in the Tour de France it's necessary that you have a bicycle but it's not sufficient that you have a bicycle. ac y gallwn nhw'n yn fawr yw'r cyffredinol. Ond maethwch i'r eich cyngorau am gyliffr ddigonol i'r ddim. Mae'n ddysgu'r consyau gyda'r acell cyrych hefyd. Roedd sy'n gwybod bryddoedd dros amser, eich meddwl sy'n meddwl, a'r mwyaf, yn anolwg, yn allwedig, a'r ddysgu sy'n meddwl yma. Yn neud hynny, mae'n meddwl y malwchbeithio. Yn haf meddwl y maen nhw'n meddwl o dymlu o ddweud o ran ymddangas unig cref. ond ond o'r blwyddyn a chael diolch yn ddim yn y fwyllt. Ysgol gweithio'r sgwbl wneud, mae'n dechrau yn gweithio, mae'n drefnio'r bwysig yn y dyfodol. A o'r ddweud y mynd yn ystyried o'r cyfnodol sydd yma ar y cyfeiriannol, mae'n ddweud y mynd yn ystyried o'r cyfeiriannol. Mae'r cyfeiriannol i'w ffau, mae'r cyfeiriannol i'w'r cyfeiriannol, yn amlwg, mae'r cyfeiriannol i'w ddiddorol, ekrifolansau, fel y bwysig y gallai gyda Dysgrifetau. Roeddwn i ni registernwyd y cwerthol o'i gwrth dechreu ymdeithas. Y bwysig y gallai gyda Dysgrifetau efo'r cyflwyneth o'r ysgolau, yn y gallai ddod o'u cyflwyneth. Felly yr efo'r cyflwynhau efo'r gwrthau efo'r math 들. Felly efo'r egoi gwneud, y llythydd, y llythdoedd yw'r cyflwynethau. to. Fi is necessary and sufficient for Psi. This combination necessary and sufficient is very common in mathematics. B, Fi if and only if Psi. That's also very common in mathematics if and only if. Notice that we're combining necessary and sufficient. With necessary we have Psi before Fi. With sufficient we have Psi before Psi. And that gets us the implication in both directions. And equivalence means implication in both directions. So the fact that it's in both directions is captured by the fact that here we have the Psi before the Psi and here we have the Psi before the Psi. Similarly with B. If and only if combines only if where Psi comes before Psi. With if where Psi comes before Psi. So in both of these cases we have an implication from Psi to Psi and from Psi to Psi. Final remark, this expression is often abbreviated IFF. IFF is a standard mathematician's abbreviation for if and only if. So if and only if or if if means the two things are equivalent. Okay once you've mastered this terminology you should be able to read and make sense of pretty well any mathematics that you come across. That doesn't mean to say you understand the mathematics itself but at least you should be able to understand what it's talking about. And that's the first step towards understanding the mathematics itself. And that's all there is to say about this. The rest is really up to you to spend some time mastering the concepts and the associated terminology. Time for a quiz. This quiz comes in four parts. So how did you do? Well let's see what's going on here. Which of the following conditions is necessary for the natural number n to be a multiple of ten? So the question we have to ask ourselves is, does n being a multiple of ten imply a statement? To be necessary n being a multiple of ten has to imply the statement. Well let's see. Does n being a multiple of ten imply that it's a multiple of five? Yes it does. So that one's necessary. Does n being a multiple of ten imply that it's a multiple of twenty? No. I mean ten is itself a multiple of ten. But ten is certainly not a multiple of twenty. Does n being a multiple of ten imply that n is even and that it's a multiple of five? Yes. Does n being a multiple of ten imply that it's a multiple of a hundred? No. Does n being a multiple of ten imply that n squared is a multiple of a hundred? Yes. So the three conditions that are necessary for the number n to be a multiple of ten are condition one, condition three, condition five. Okay. Let's move on to part two. For this time we have to ask the question does the statement imply n is a multiple of ten? Okay. Well does this statement imply that n is a multiple of ten? No it doesn't. Five is itself a multiple of five. But five is not a multiple of ten. Look at number two. Is it the case that if n is a multiple of twenty then it has to be a multiple of ten? Does this imply that? The answer is yes. Does n being even and a multiple of five imply that it's a multiple of ten? Yes. If n is a hundred, does that imply that it's a multiple of ten? Yes. If n squared is a multiple of a hundred, does that imply it's a multiple of ten? Yes. So in this case two, three, four and five are the correct answers. They're all sufficient for n being a multiple of ten. Okay. Let's move on to part three. For this one we have to compare our answers for the two previous questions. The first question was necessity and the second question was sufficiency. For necessity we had one, we had three and we had five. For sufficiency we had two, three, four and five. This asks for necessity and sufficiency. There we've just got necessity. There we've just got sufficiency. There we have both of them. There we just have sufficiency. There we have both of them. So the ones that are necessary and sufficient are this one and that one. Okay. Let's move on to part four. For this one the question we have to ask ourselves is what does the implying? In number one this does the implying. So that's the antecedent. In question two of statement two, well even though it's written the opposite way around it's essentially the same statement. It's the alarm ringing that does the implying. What about number three? Keith cycles only if the sun shines. What's doing the implying? Keith's cycling. If you see me cycling you can conclude that the sun's shining because I cycle only if the sun shines. Incidentally I was brought up in England so that's not the case. I'm quite happy to, I'm not happy to ride in the rain but I do ride in the rain. But it's a good example. Number four, what does the implying? Well Amy arrives to see implying. So far I've distinguished between genuine implication and equivalents and their formal counterparts, the conditional and the biconditional. In their daily work however mathematicians are rarely that particular. For instance we often use the arrow symbol as an abbreviation for implies and the double headed arrow as an abbreviation for is equivalent to. Although this is invariably confusing to beginners it's simply the way mathematical practice has evolved and there's no getting around it. In fact once you get used to the notions it's not at all as confusing as it might seem at first. And here's why. The conditional and biconditional only differ from implication and equivalents in situations that do not arise in the course of normal mathematical practice. In any real mathematical context the conditional effectively is implication and the biconditional effectively is equivalents. So having made note of where the formal notions differ from the everyday ones. Mathematicians simply move on and turn their attention to other things. The very act of formulating formal definitions creates an understanding of implication and equivalents that allows us to use the everyday notions safely. Of course computer programmers and people who develop aircraft control systems don't have such freedom. They have to make sure all the notions in their programs are defined and give answers in all circumstances. Okay that's the end of lecture four. As I said at the start it's been a fairly short lecture. My reason for keeping the lecture short is that the upcoming assignment is much longer than the others. It has to be. Implication and equivalents are at the heart of mathematics. Mastery of those concepts and of the terminology associated with them is fundamental to mathematical thinking. You simply have to master implication and equivalents before you can go much further. And there's only one way to achieve mastery right? Remember the story of the elderly lady who approached a New York City policeman and asked officer how do I get to Carnegie Hall? The officer smiled and said lady there's only one way. Practice, practice, practice. So I suggest you carve out some time, grab some food and drink and head off somewhere quiet to complete as much of assignment for as you possibly can. Well question one picks up the same theme as question 11 in the previous assignment. So let's see how it works. It's actually much more difficult I think than question 11 of assignment two. Some of these really are tricky. Okay so let's take them one by one. New trade agreement will lead to strong currencies in both countries. So there's a trade agreement that leads to, that results in, so I think that's implication, the fact that both currencies are strong. A dollar strong and a yen strong. Okay that one's fairly, fairly straight forward. Strong dollar means a weak yen or strong dollar implies that the yen is weak. Well again, I think that one's fairly straight forward. A dollar strong, it follows that the yen is weak. Trade agreement fails on news of weak dollar. So we hear that there's a weak dollar and that implies that the trade agreement fails. Okay and I think that one's, I'm fairly confident in writing that one down. I think that's, I can't think of another one that would really capture part C. Okay part D. If a trade agreement is signed, well this one is signalling very loudly implications for the trade agreement signed. Then they can't both remain strong. So it's not the case that the dollar strong and the yen strong. Okay now with part E, I think we're into an example where, you know, your marriage might be different from mine. Okay, the question really is what is a but? And I think the but means and. Following however, I think is, it could mean one of two things. It could mean the following. The dollar's weak and the yen is strong and there's a trade agreement. In other ways, there's a trade agreement and after there's that trade agreement, the dollar's weak and the yen strong. In other words, it's all conjunction. So you could interpret following is just meaning that happens first and then those two happen. However, you could also say, well, it was a new trade agreement that led to these currency strengths and weaknesses. So you could, I think, quite legitimately say the following. There's a trade agreement and as a result, the dollar's weak and the yen is strong. I think that's a, that's perfectly legitimate. I think there's two interpretations depending on whether you mean following, just one thing follows after another in time or it follows because of causality. Okay, what about number five? Well, this one is again signal in implication. So if the trade agreement is signed, then a rise in the yen will result in a fall in the dollar. You could actually write that a different way. You could say that there's really two assumptions here, two things that tend to happen. If there's a trade agreement signed and then if the yen rises, it follows that the dollar falls. Now, those are actually equivalent as you'd be able to demonstrate using truth tables if you want. But there are both ways of interpreting that. You could say that you've got an implication of an implication or you could say you've basically got an implication with two assumptions. I think this is more of a literal interpretation and this one is more sort of semantic interpretation where you look at what these things mean. Okay, this one, the way I would be, I think I would write this one, was to say if there's a trade agreement signed, then these are linked which is basically equivalent, dollar equivalent to the yen. Okay? Or if you want to spell it out, you would say trade agreement implies that the dollar rise leads to the yen rise and the yen rise leads to the dollar rise. Okay, they're both the same thing. That actually is an abbreviation for that. The buy conditional abbreviates the two conditionals. Let's try and clean that one up a little bit, okay? Okay, new trade agreement will be good for one side but no one knows which. Okay, this is, what we're really doing here is talking about an exclusive all. So that was something that caused lots of use and grief in the, in an earlier example, in an earlier assignment. Okay, new trade agreement will be good for one side. So either the dollar strong or the yen strong but and but is conjunction. We've already noted that. It's not the case that they both hold. Okay, this really forces the practice of saying an exclusive all. One of them holds but not both of them. Well, that was question one and as I mentioned, some of these you may, you may come up with different answers. I'm pretty confident about these that when I look at these sentences, these are what these things mean to me. But we are taking something in everyday language, in natural language. In fact, we're taking stereotypical newspaper headlines, which are abbreviated natural language. And we're trying to interpret them in the formal language of logic. So we're taking something that's imprecise and depends upon context and knowledge of culture. And we're trying to express it in a logical formalism. And that's, that's, that's, do I have interpretation differences? Your marriage may differ from mine on, on one or two of these things. As I say, I could defend each of these and I, I tried to as I went through, but you might come up with an answer that you could defend. And, and that's fine. Okay, let's move on. Well, for number two, I've put my answers in bold. If that doesn't show up on your screen, these are the ones that have been bold. Okay, I started with five and I negated it so two becomes false, two becomes false, false becomes true, false becomes true. I have Psi. We've already worked out the truth table for, for the conditional and we've seen it's T F T T. Now we can take Phi, now we can take not Phi and Psi, we can combine these two columns with a disjunction to get this guy. So this will be true whenever one of these is true, which means one of them's true here, neither is true here, so we get a false. They're both true here, we get a true. One of them's true here, we get a true. So this column comes from combining these two with disjunction. And now we simply compare these two, and this is really getting to number three. We can compare these two and observe that every entry is the same. And since the truth values are the same, the conclusion we can draw is that Phi U Psi is equivalent to not Phi O Psi. And that takes care of two and of three. We're turning to number four now. We've got four columns to fill in and again I've written them in boldface but you may not be able to see the boldface on, on your screen. But these are the four entries that I put. I began with the, the two columns were given the standard range of, of properties for T and F for Phi and Psi. I negated the Psi values to give me not Psi, so T became an F, and F became a T, T to F, F to T. Then I wrote down the values for Phi U Psi, which we know we've already worked that out, which T, F, T, T. Then I negated those to give me Phi does not yield Psi, so a T becomes an F, F becomes T, T to F, T to F. And then I combined the first column with the third column using conjunction to give me this one, and with conjunction you have to have both of them true to get a true. Well we've got a T and an F, and that F means we're going to have an F there. Here we've got a T and a T, so I get a T there. Here I've got two Fs, gives me an F, and here I've got one F, so I've also got an F. Well that fills in all of the columns. And then to answer question three, would you notice that these guys are all the same? And hence we may conclude that Phi does not yield Psi is equivalent to Phi and not Psi. And in fact what I would like to do now is recall the discussion we had to obtain the truth table for Phi yields Psi. If you remember to do that, we had to look at Phi does not yield Psi. In order to find the truth values in the case where Phi was false, we ended up looking to look at Phi does not yield Psi. And this truth table here sort of shines a light on what was going on. It illustrates or it explains to us why looking at that enabled us to work out the two problematic truth values for the truth table for yields, for implication when Phi was false. Because it's this equivalence that we were capitalizing on. That Phi does not yield Psi if Phi is true and nevertheless Psi is false. Okay, okay, well that's questions four and five for you. How did you do? Okay, let me finish this tutorial with this little puzzle I'll leave you with. A woman was driving in her car along a black road. She did not have her car lights on. There was no moon and no light from the stars. A black dog was asleep in the middle of the road. As the woman approached the dog, she swerved to avoid it and the animal slept on. How did the woman know to swerve around the dog? And let me just give you one clue. Remember the focus in this part of the course is on being precise about the use of language and being very careful about the information that language conveys. And with that clue, I'll leave you to puzzle this one out on your own. Bye bye for now. Well, assignment four was pretty long, but I would hope that by now, working with other students in the class, you were able to get through most of the ones that you attempted. I'm just going to focus on three of them, I think. I'll look at number six, ten and twelve. Most of the other ones were just straightforward truth table calculations and I'm pretty sure that by now you should have done those and if not then do go back and talk with friends and other students and try and sort it out. And you may not have had too much difficulty with six, ten and twelve as well, but let me just play it safe and work through these. There's at least one here that's a little bit tricky. It's not the first one. Incidentally, the first one, it's a true statement. This number is a prime number, but that's not what the question is about. The question is what's the negation or the denial and then the simplest way to just say, well, 34,159 is not a prime number. Or else you might say 34,159 is a composite number. Either way, I think that's fairly straightforward. This one is straightforward too, but there's, I can't think of any nice way of writing the answer that sounds, that sounds like good English. You would have to say, roses are not red or violets are not blue. Which is the kind of sentence that you will only see or hear in a class on logic as we're looking at now. You know, mathematical uses of language, precision in language. So it's correct, but it just is not an elegant sentence. You wouldn't normally say something like that. This one is tricky actually. And it's tricky because there's a negation floating around in here. Remember, when you've got a conditional, faillu sai, when you deny it, what you get is you get the antecedent conjoined with the negation of the consequence. In this case, the antecedent is that there are hamburgers. Sorry. In this case, the antecedent is that there are no hamburgers. There we are. I almost slipped into that mistake myself. You've got to be very careful with this. The antecedent is there are no hamburgers. So let's just write that down. There are no hamburgers conjoined with the fact that I won't have a hot dog. Well, I think in this case it's more natural to write the conjunction using but rather than and and say, but I won't have a hot dog. Okay? So we've got the antecedent that there are no hamburgers conjoined, in this case written with a but, because I think it's more natural to say that there's a but, but I won't have a hot dog. Okay? Well, think about that one for a little bit. In my experience, students often have trouble with this because of that negation in there. It throws you off, so you have to be a little bit careful. Fred will go, but he will not play. Again, the but is really just the same as an and, and so when we negate that, we'll get there. Fred won't go. The and becomes an or, and we've got or, he'll play. In fact, I think in English we're probably more likely to write it differently. We're probably more likely to say, Fred will play or he won't go. And I think we'd do that because we'd, we'd, we'd read into this some causality, some connection between the two. Now that's going beyond just doing the straight negation, but I think that's typically how we would understand it. So to me that seems more natural way of saying it. Just, just a cause with the kind of thing people, people normally do. But in terms of the negation, we just took the negation of the two parts together with the negation of the but, which becomes a disjunction, which is an or, so got Fred won't go or he will play. Or more naturally, Fred will play or he won't go. This one is actually messy if we do it in English, but it's much easier to do it symbolically. So let's do it symbolically first. The original statement is that x is negative or x is greater than 10. If we take that statement and negate it, then the x less than zero becomes an x greater than or equal to zero. The or becomes an and x less than, x greater than 10 becomes x less than or equal to 10. So that just becomes zero less than or equal to x less than or equal to 10. If you do it in English, you're going to have to say something like the number x is non-negative. You can't say positive because the negation of negative is not positive because of zero. You've got to say non-negative. The or becomes an and, and the negation of greater than 10 is not less than 10. It's less than or equal to 10. You have to say less than or equal to 10. And I don't see any way of doing it in English other than with a, with a, with an awkward sounding sentence like this. Symbols provide a much more efficient way of expressing this idea and indeed for dealing with the negation. Okay. But anyway, we've, we've found an answer to the question. The last one part F, we'll win the first game. The second, the negation is that we'll lose the first game and we'll lose the second game. And the simplest way to write that is to say we lose the first two games. Remember the question asked for a denial. Now that's just not the strict negation. It's, it's a natural, not necessarily natural, but it's a, an equivalent version of the, of the negation. And for some of these we're able to find pretty good answers. In others it was, it was kind of messy. But there were a couple of trickery ones floated around in there. So I think you need to be a little bit careful about these kind of things. Dealing with language precisely is actually not easy because our mind jumps ahead. How many of you jumped from this one to something like, if there are hamburgers, I won't have a hot dog? Students often do that one. You've got to be careful about the precision of language. Human beings are very smart with using language in everyday terms. But the cost is that we, we, we drop precision. We think in terms of meanings rather than what the, we think in terms of the way we understand the meanings rather than what the literal meanings are. The whole point of this analysis of language we're doing now is to be very, very precise. And then examples like this remind us that precision is not easy. You have to work at it. Okay. Well, that was number six. What was the next one I was going to do? I was going to do number ten now. So let's look at number ten. Well for number ten we have to refer back to number nine. I'm not going to do number nine. I would hope that by now you're able to do a truth table verification of these two equivalents. Work out the truth table of that. Work out the truth table of that and observe that they're the same. So how do we go about, about proving the, the same equivalents? But by means of a logical argument. Well there's two directions to prove. I'm going to begin by proving left to right. Okay. I'm going to assume this and conclude that. Okay. So I'm going to assume phi yields psi and theta. Okay. That means we can deduce psi and theta, that formula, from phi. But we know that we can deduce psi from psi and theta. And we know that we can deduce theta from psi and theta. Because from any conjunction we can conclude either of the two conjuncts. If you know psi and theta then you know psi and you know theta. Okay. Hence by chaining two things together, by chaining the deduction of this from that, and then those two from that, it means we can deduce psi from phi and we can deduce theta from phi. Okay. With me? This means we know if we can deduce theta, if we can deduce psi from phi, that means we know phi yields psi and we also know phi yields theta. Hence we know if we know that, and we know that, we know their conjunction. So assuming this, which is the left hand side, we have concluded that, which is the right hand side. Okay. Let's go the other direction. So let's assume phi yields psi and phi yields theta. Then we may conclude the first conjunct, phi yields psi, and we may conclude phi yields theta. Okay. That is, we can deduce psi from phi and we can deduce, whoops, what did I do here? That was meant to be a theta. I hope you spotted that. Okay. One of the problems with doing this over a video is that you can't yell out when I do it, but at least I spotted it at that time. Phi yields theta. Sorry about that mess, that's a theta. Okay. We can deduce psi from phi and theta from phi. No, I'm really making a mess of this, aren't I? We can deduce theta from phi. Okay. Let's take this from the top. Okay. We can assume, we can assume this. That means we can assume the first conjunct, then we can conclude the second conjunct. Right? If we're assuming that, then we know the first conjunct and we know the second conjunct. If you know a conjunction, you know the two conjunct. So that's the first conjunct and the second one. The first one means we can deduce psi from phi. The second one means we can deduce theta from phi. Well, if we can deduce psi from phi and we can deduce theta from phi, that means we can deduce psi and theta from phi. If you can deduce psi from it and you can deduce theta from it, then you can deduce a conjunct from it. That means we know yields psi and theta. So now, assuming the right-hand side, I've concluded the left-hand side. Now, this is going to look a bit circular, right? To figure out what's going on, to follow what's going on, you have to sort of separate out specific formulas from what they mean. So this is a single formula. And what we're doing is we're pulling it apart into what it tells us. And then we're arguing with what it tells us. And then at the end, we put it back into a formula again. So we start with a formula, we end with a formula. In the middle, we're arguing with what the formula tells us. And ditto here. Now since this really corresponds in all real cases to implication, and since that is conjunction, we end up using words like can deduce, it follows from, and we use the word and a lot. So there is a sort of forced circularity in that this actually does say this. That's the point. So if this seems as though there's a sort of a bonus to it, indeed there is in that this is specifically, these formulas are specifically set up to correspond to these things. But if you really look into it, this isn't totally trivial. We have gone from formulas to what the formulas tell us. Okay, well enough of that. Let's move on to the next one. Okay, and that next one I want to look at is question 12. And that is talking about the contrapositive. And these are the four statements that we have to write down the contrapositive zone. So if two rectangles are congruent, they have the same area. So the contrapositive would be if two rectangles do not have the same area, they're not congruent. Number two. Contrapositive would be if in a triangle which sides A, B, C, with C being the largest, which is the case that A squared plus B squared is not equal to C squared, then the triangle is not right angle. So if in a triangle which sides A, B, C, C the largest, A plus B squared is not equal to C squared, then the triangle is not right angle. Contrapositive to this one, if N is not prime, then two to the N minus one is not prime. Or if you like, if N is composite, then two to the N minus one is composite. And then for part D, if the dollar does not fall, then the Yann won't rise. Well actually writing these down was fairly straightforward. It's just a case of looking what the contrapositive does. It flips the order of the things and puts negations in front of them. So actually writing them down was not what the challenge was. The point of this exercise was to give you an example to sort of help cement the fact that the contrapositive is actually logically equivalent. Because is this one true? Yes. Congruent rectangles have the same area. If rectangles don't have the same area, they're not congruent. Those are both true. If a triangle is right angles, so we've got Pythagoras theorem. If a triangle is added. Each one of these is clear that one is equivalent to the other. If this thing happened to be true, then that thing would happen to be true. If that thing was true, then that would be true. So I think these four examples will illustrate the fact that contrapositives are logically equivalent to the original statement. Okay, I said I was just going to do six, ten and twelve, but as I was working through this one, I thought it would be a good idea to look at question 14, which is very similar. So let me do question 14 as well. Okay, and number 14 is about converses. Where again, as with a contrapositive, you flip the order round, but in this case you don't inject any negation signs. It's simply the implication in the opposite direction. Okay? Which means actually these four are going to be very straightforward because we're just doing the order. So this one would be if two triangles have the same area, then they're congruent. Now already we've shown by an example now that the converse is not necessarily logically equivalent to the original statement. Because the original statement is true, but the converse is false. Now that's not always the case. Let's look at this one. The converse of this one is the following. If in a triangle with sides A, B and C, with C the largest, it's the case that A squared plus B squared equals C squared, then the triangle is right-angled. Now this is true. It's Pythagoras' theorem. And the converse of Pythagoras' theorem is true. So these are both true. But what's going on is that we've actually started out with something that's not just an implication. When it actually is an equivalence, then when you've got an equivalence, then you take the converse. You've just got another part of it. Okay, the statements and the converse are both the two halves of an equivalent. So you can get the same true values, but only when there's an equivalence already there. In the case of this one, the converse would be if n is prime, then 2 to the n minus 1 is prime. And the converse of this one is if the dollar falls, the yarn arise. And in this case, there's no reason to assume that one necessarily follows from the other, unless in some way the other two conferences are linked. Okay, well, I hope you managed to get the four converses written down correctly. But as I indicated a moment ago, that wasn't really the point of this. The point was to sort of give us examples of the fact that converses don't necessarily yield equivalent statements. When you take a converse, you don't get something that's equivalent. Whereas with the contrapositive that we looked at in the previous example, in example 12, when you take the contrapositive, you do get something that's equivalent. Okay, okay. Well, that should, I think, take care of assignment number four. Okay, let's see what we've got with the problem set two. By the way, this part of the course, we've been looking at taking expressions in everyday language or problems that might arise from the real world in a certain sense. And try to make them precise. And that of course means that for people whose language, whose native language is not English, this is even more complicated because we have to deal with the complexities of the English language and try to eliminate the ambiguity from that and make things precise. When it comes to these issues about necessary and sufficency, I actually don't think it makes a lot of difference if you're a native speaker in English. This is tricky when you first meet it. And it's easy to make a slip. So this is one of those occasions where I don't think there's an advantage to being a native English speaker. We native English speakers find this difficult too. Okay, in the case of necessity, a condition is necessary if it follows from this. So what we're asking for is, is it the case that if six divides n, then the condition x holds. That's what we're asking. Does the condition x follow from n being divisible by six? Well, let's see. Is it the case, for example, that if six divides n, does it follow that three divides n? Well, the answer is yes. If six divides it, then three divides it. Okay? So that one's necessary. We'll hold this one. Well, notice that n equals six itself satisfies this condition. Okay, if n equals six, then six divides n. But it's six divisible by nine. In the answer now. Again, n equals six satisfies the condition of being divisible by six. But does it then follow that n's divisible by 12? In other words, it's six divisible by 12. No, it's not. What about n equals 24? Well, again, n equals six gives us an example of a number that's divisible by six. But that number's not necessarily 24. Right? Because it's six. So it's not that one. Okay. N squared divisible by six. N squared divisible by three. Well, let's just see. Six divides n certainly implies three divides n. If six divides it, then three divides it. And if three divides n, then three divides n squared. So that one's okay. Again, if six divides n, then, of course, two certainly divides n because six does. And if six divides n, then three divides n. So two divides n and three divides n. In other words, n is even and divisible by three. So it's that one. So we've got parts a, e and f. And this is the condition. This is the way it caches out in terms of implication. And in the case of the counter examples that we use to prove some of these false, the simplest counter example was n equals six itself. Okay? Well, that takes care of number one. Let's move on to number two. Well, in the case of sufficency, we have to look to see if the condition implies that n is divisible by six. Okay? Sufficency means implies n is divisible by six. So for each of these statements, we have to ask ourselves does it imply that n is divisible by six? Well, if n is divisible by three, in order to, if we, if we think the answer is it's not sufficient, then we have to find an example, we have to find a counter example of a number that's divisible by three that's not divisible by six. Well, why don't we check n equals three? Three is divisible by three, but three is not divisible by six. So that's a counter example. So that one's not sufficient for being divisible by six. That doesn't imply x is divisible by six. Because in particular, three is divisible by three, but not divisible by six. What about part b? Let's take n equals nine. That's a counter example. n is divisible by nine, but n is not, but, but, but if we take n equals nine, then the n is divisible by nine, but nine is not divisible by six. So that's a counter example. n is divisible by 12. Is it the case that if 12 divides into n, does it follow that six divides into n? The answer is yes. So that one's okay. What about n equals 24? Is it the case that if n equals 24, does it follow that six divides in? In other words, does six divide 24? Again, the answer is yes. So that one's true. n squared is divisible by three. Does it automatically follow that n is divisible by six? Well, why don't we take the same counter example we did in part a? Let's take n equals three, then certainly three divides n squared. Okay. But does six divide n? Does six divide three? The answer is no. Six does not divide n. So three divides n squared. Six does not divide n. So it can't be that one. Finally, n is even divisible by three. I could write that as saying n is even is to say that two divides n and it's divisible by three. So I can just rewrite this statement to say that two divides n and three divides n, but if two divides n and three divides n, then it's the case that six divides n. Okay? So that one's true. And when we come to question three in a minute, it's really going to be a matter of combining questions one and two. So we've now got all the information we need to answer the next one. Well, question three asks us for necessary and sufficient. So let's just see what we did in question one without the necessity. In in question two without the sufficiency. Let's remind ourselves what we had. Let me see now, we got necessity that was true in A, wasn't it? It was true in E, it was true in F. Remember correctly? Yeah. Okay? Actually, I'm not remembering. I'm working these out as I go through because I don't have the one in front of me anymore. Sufficiency, let's see, that was C, it was D, and it was F. Okay? By the way, if I'm making this look fluent, it's because I've been doing this for many years. Although you can still catch me out with a well, with a puzzling, with a question that's described in a puzzling way, even the experts can be caught out with this kind of thing. Okay? So we just have to look for the ones which have an X in both rows, in both columns, and the only one that does is this one. In all of the other ones, you've either got necessity or sufficiency or neither in the case of Part B. Okay? So you've got a one necessity, you've got a neither, you've got a sufficiency, you've got a sufficiency, you've got a necessity. The only one that's both is Part F. And that takes care of number three. Well, we're moving along rapidly now. Let's go ahead. Question four starts out simply enough because it's a very straightforward if-then statement. And when you've got an if-then statement, identifying the antecedent is pretty straightforward because it's the part that goes with the if. Okay? So it's this one. If the apples are red, then they're ready to eat. Okay? So this was an easy start, but these things are going to get a little bit more tricky as we move forward. Let's have a look at the Part B. Okay? Well, we're talking about sufficiency and sufficiency is the thing that does the implying. Okay? So what does the implying here? Well, F been differentiable. That's the one that does the implying because the differential property of a function implies that it's continuous. Okay? Sufficiency does the implying. So sufficiency is the thing we're looking for. So the antecedent is a sufficiency, which is differentiability. Well, I said these become a little bit tricky as soon as you get into them. Even though in one sense this is straightforward, my experience is that many students, including myself, when I was a student and the kids today, if I don't really put my mind to it, I have to think a little bit to just flesh these out. Alrighty? Let's move on to Part B. Okay? Remember, we're still looking for the antecedent, the thing that does the implying. This case, again, this actually is fairly straightforward because it doesn't matter whether you put the if clause first or second. It's the thing that goes with the if, so long as it's an if and not an only if. The thing that just goes with a naked if is the thing that's the antecedent. So in this case, it's this guy. It's F is integral. That's the thing that does the implying. And whenever actually, it's sort of the same as if. It's another way of using, a way of saying if. This tells us the condition under which something happens. This is bounded under the conditions that F is convergent, whenever F is convergent or if F is convergent. So it's that guy. Okay? So this is the antecedent. This is the antecedent. Let's move on to the party. When the case of necessity. Necessity is the thing that follows. So our necessary condition is the thing that follows. Okay? So we have to ourselves ask ourselves, what is it that's following in this case? Okay. Any answer is? This thing. Okay. So this is the antecedent because n being prime is necessary. So n being prime is the thing that's the consequence of the antecedent. So it's this that implies that. So this is the antecedent. All ready? Let's move on to part F. How did you do with that one? Well, this is where we start combining words like if and when with an only. And because it says only when Carl is playing, this guy is the consequence. The team wins only when Carl is playing. So if you know that the team wins, you can conclude that Carl is playing. Because the only win when Carl is playing. So that's just another way of saying that Carl is playing is a consequence of the team winning. Okay? So this is the antecedent. And as I mentioned at the beginning of this discussion, even if you're a native English speaker, these typically cause people a lot of trouble. Just the way the human mind works. It's not to do with the native languages to do with the way the mind works. Well, these two are a little bit more straightforward than the last one or two because the when is really almost the same as if. That just tells you the condition on which something is. So the thing that just goes with a when without an only combined with it is the antecedent. Okay? So the antecedent in this case is that Carl is playing because it's when he's playing that the team wins. If you know that Carl is playing, then you can conclude that the team is going to win on the basis of this statement. So that's the antecedent there. And it doesn't matter whether the when clause comes first or second, as was the case with an if clause, it can come first or it can come second. That's still the antecedent. Okay? Being the antecedent is not directly related to whether you're the first clause or the second clause in a sentence. It gets to go with which word you're combined with. If it's an if or if it's a when, then that combines with the antecedent. If it's an only if or an only when, then that flips it round. And then you're dealing with the consequence. Okay? Well that's takes care of those kind of examples. And the last two parts, or the last three parts of this question of this problem were a little bit different. Okay? Well in this case, it's certainly true that if m and n are even, then mn is even. We know that. So the question is, is it the case that if m, n is even, then m and n are even? So that's what this boils down to. This, this implication. Okay? We know what, there's two implications here. It's an if and only if. So that means the equivalence. It means the implication holds in both directions. And one equivalent, and one implication is certainly true. If m and n are even, then mn is even. And so it boils down to the question, if the product is even, then are the two numbers necessarily even? And once you get it down to that stage, all you need to do is just observe. There are many counter examples. We could take m equals two, n equals three, then mn equals six. So here we've got a product that's even, but it's not the case that both numbers are even. So this is a counter example. So the answer to the question is no. This is a statement about any pairs of integers. And if we found one pair of integers that, that makes it fail, then the whole statement fails. So it's a counter example that we need to find. And we found one, m equals two, n equals three. The product is even, but it's not the case that both numbers are even. Okay? Now let's move on to number six. So number six asks us, is it the case that mn is odd, if and only if m and n are odd? Well, I'm sure you all realise that the answer is yes. And you won't necessarily realise the reason, you know the reason why, because there were two facts. We know that odd times odd, I better say is not equals, is odd. And we know this even times anything, any number odd or even is even. So if you take two odd numbers and multiply them together, you get an odd number. If you take a number, a pair of numbers, any one of which is even, and multiply them together, you get even. And when you combine those two, this falls out of it. And if you don't see that, I mean, I'll leave it to you actually. If you want to sort of give a little bit more detail and express it as an implication in both directions, that's fine. You could discuss this, and I'm sure you could discuss this endlessly on the forums and that would be a good idea if you want to. But I'm just going to leave it with the observation that it's really just these two facts that give you this result. Okay? This tells you how the parity even and odd works. And once you know that, you know that. But have fun with this one and discuss it amongst yourselves and settle to your own satisfaction what constitutes a rigorous proof of this thing. Remember, there's actually no sort of gold standard of what is or is not a rigorous proof. It depends on the experience of the audience. A proof in many ways involves audience design. You've got to cast that proof at the level of detail and precision that matches the audience. You know, typically a mathematician, a professional mathematician would simply say, and you see this actually in books and in papers, a mathematician might very well say, this is trivial. Okay? And that would be the proof. In advanced works on mathematics, you often see remarks like the proof is trivial. It has to be said that a beginner might take several days to see why something's trivial. When mathematicians use that kind of expression, they're doing it with a particular audience in mind, namely other professional mathematicians. So, you know, if you read that and it doesn't seem trivial to you, it doesn't mean you're stupid, it just means you haven't spent many years working as a professional mathematician. It's just a way we classify things. And proofs involve a lot of audience design when you write and formulate proofs. Okay? Well, one way to do all of these is by truth tables. And if you work out the truth tables, you find that A is true, that B is true, C is not true, D is true, D is true, and F is true. So simply by using truth tables, you could answer this. This one we've already seen in the lectures and discussions. We've sort of looked at these, this equivalence. This one, and where is it? That one. Those are examples of what's known as De Morgan's Laws after a mathematician Augustus De Morgan. That if you take a negation with a disjunction, you end up with a conjunction of the two negations. And if you take a negation of a conjunction, you end up with a disjunction of the two negations. Okay? So that was a basic fact about implication that a conditional is true if either the antecedent is false or the consequence is true. So that was the truth table that we worked out for the conditional. Those are De Morgan's Laws. That one's not the case anyway. This one, you could probably reason this one out in terms of implication. Just think in terms of when implication. Under which circumstances can you start with an assumption P and deduce two conclusions? And then see when that doesn't happen. Okay? And you should end up with this. So it's possible to reason this one out. Just in terms of implication. And the same is true for this one. You could reason it out. This one I think is easier to reason out than that one because of fewer symbols to deal with. This says that if you have an assumption and then you have another assumption, you can make a conclusion from it. Well, do you do it in two steps? Do you assume P? And then on the basis of that show that if Q is true, then R is true? Or do you simply assume that both P and Q is true and deduce R? Those two things would be equivalents. This is sort of doing it in two steps, and this is combining the two assumptions. They both really tell us that there are two assumptions. There's one assumption, then there's another assumption. And in this case, we've explicitly said there are two assumptions. And in both cases, it's the R that's following from them. Okay, so the question is how and when do you get from P and Q to a conclusion R? So this is, you could reason it out. And if you feel uneasy about that, you could just work out the truth table. Okay, but since we've spent a lot of time on truth tables and the assignments, my recommendation would be that you would go through these and actually try to reason them out in terms of what they mean. You know, truth tables are good if you're a computer. But people are not computers. They're much more interesting creatures than that. And I think we have the power of reasoning. And so I would suggest you go through these and try to reason them out in terms of what these things mean. Because that, after all, is really what this entire course is about. We're in question eight. We make use of the course evaluation rubric. This is the first of many questions that you'll be asked to do, making use of this rubric in order to evaluate purported mathematical proofs. Well, here's the first one. The claim is that for any two propositions P and Q, not P, conjoined with not Q, is equivalent to not P and Q. The argument is fairly short. In fact, all of the ones I'm going to be using as examples are short. That's why they're good as examples. In particular, I want to be able to give the solutions on a single slide, as I'm going to do here. So in a way, these are not typical. In fact, if you do test flight at the end of the course, you will almost certainly see proofs presented by other students, which are much more complicated and much longer and perhaps have all sorts of mistakes in them. The arguments I'm giving you, they are, they all began as arguments that were produced by students over the years when I've taught this material. But what I've done is I've picked particular aspects of proofs where students typically go wrong. And so the examples will have one or two common mistakes embedded in them, just so you get used to looking at proofs from the different perspectives, as captured by the rubric, and seeing how they work and don't work. As I mentioned in the, in the description of the use of the rubric on the course website, the way it will work is that because I'm using short examples, in each particular example, some of these, these, these, these features won't really apply. In which case you have to sort of default and give four, or zero, depending on how the student handles it. Typically you would end up giving full marks because it just doesn't really apply. Okay. Well, let's, let's take a look at how the, how this one was done. And so we'll, we'll, I'll, I'll pretend that it was done by a single student, even though this is a composite of the kind of things that students have done over the years. Okay. Well, it's an equivalence. So we have to prove it in two directions. We have to prove that this implies that and that implies that. So there are two equivalences to prove here. So let's just check the left-right implication. So suppose that not p and not q is true, then a conjunction is true if and only if the two conjuncts are true. So if the conjunction is true, then both not p and not q are true. Okay. That's correct. That's what conjunction means. If not p is true, that means p is false. That's what negation means. If not q is true, that means q is false. So the truth of these two negations means the falsity of p and q. But if p and q are both false, then again because of the way a conjunction works, p and q is false. Hence not p and q is true again because of the way negation works. So that's absolutely okay. Okay. The left-to-right was proved great. What about right-to-left? Well in this case the student makes an attempt to be fairly sophisticated by saying the other argument, the argument in the other direction works the other way. If it does work the other way, it's absolutely okay to simply say that. There's no requirement that you would have to repeat something that's, that's obviously the case. But we'd better check that it is obviously the case because we're actually evaluating whether this is the case. Okay. So let's, let's spell out what this person didn't do. In other words, we'll try to do the same as here going to the direction. So we're going to assume it's not the case that p and q. Okay. Let me just say we'll assume that's true. And that's sort of redundant but I want to talk about truth and falsity. So we'll assume that not p and q is true. That means that p and q is false because of the negation. If the negation is true, it means that p and q is false. Well what does that mean? That means that at least one of p and q is false. Okay. Because you only need one of them false to make the conjunction false. Ah, but the other one, whichever it is, could be true. It doesn't have to be. But there's nothing to rule out the fact that the other one's true. So one of p and q is true. That means one of not p and not q could be false. At least one of them is false. So the other one could be true. That means one of those could actually be false. One of these guys, not p and q could be false. That means if one of them could be false, it means not p and not q could be false. That's not ruled out. In other words, that implication does not work. Just because not p and q is true, it does not necessarily follow that not p and not q is true. It could be false. We're not saying it is, we're saying it could be. So there isn't an implication. There is no implication from right to left because you could have that without having that. That thing could in fact be false. In other words, the original claim is a false claim. It's simply not true. So this statement, though it was a nice attempt to be somewhat sophisticated, it didn't work because in fact the argument does not work the other way. I mean it is essentially the same kind of argument and if it had been correct, that would have been fine. You wouldn't need to give this. But it's not correct. So now we're going to have to sort of put some numbers in here that capture what we've just said. OK? Well what am I going to say? There are four marks available for logical correctness. The left to right part was absolutely correct. This is logically correct. So I'm just going to give half marks for that. I'm going to say two captures the fact that left to right was correct. But remember logical correctness is just one aspect of proof. There are other factors too. And that's what the rubric items are doing. So let's look at what we've got. Clarity. This is absolutely clear. Even here it was clear. It was wrong, but it was clear. So in terms of clarity this is very clear. So I'm going to give full marks for clarity. Is there an opening? Insofar as there's any reason to give an opening here, yes it begins by stating that we're going to go from left to right. And so I'm going to give four for that. Now you could say maybe the person should have begun by saying to prove an equivalence you have to prove implications from left to right and right to left. Well you know it's a judgement call. I would say that for someone at this stage where they're producing arguments like this, and this is a nice argument. This is perfectly correct. For someone that's producing this, I think it's okay to say, yeah it's obvious by the way they're doing it that they're doing that. But remember even though the rubric does break a difficult task into smaller pieces, they're still not easy. So these are going to be judgement calls. I would say that there's enough demonstration of knowing the person knowing what he or she is doing to just say I'm not going to insist that they say proving an equivalence is enough to prove left to right and right to left. I think it's clear from the way they're doing it. But they do start out by saying assume one and prove the other one. Okay? What about stating a conclusion? Absolutely, it was stated we have implication in both directions. This just justifies what I've just said. So the person has now said we have implication in both directions. So just emphasising the fact that he or she knew exactly what they were doing. Remember also that we're really trying to do formative assessment rather than summative. The process is one of seeing what the person has done well and rewarding what they did well. It's not about trying to take marks off. When we do this kind of thing we should always be looking for reasons to give marks, to acknowledge what's been done correct and then to point out things to improve. So we begin by adopting a positive giving marks attitude. That's the best way to do this kind of thing. So I'm always going to be looking for reasons to give marks. On the other hand, if there's any, you can't give marks if they're not justified. Okay. What about reasons? I mean reasons were given but it was really the same as in the first case. The reasons for the other part that should have been given weren't because the person thought it was the same and it wasn't. So some of the reasons are here. They were given as a person was going along. So I'm going to have to give two for that one because it was only half a proof. I mean the proof was listed. The result's false. So this is absolutely false. So the most I could give here is two because even though reasons were given, that's not a reason. That would have actually been a reason if it was correct. If this kind of argument was correct that would have been okay but it's not. And then overall, again I'm just going to have to give two because it's basically just half a result. Okay. That's really what it's coming down to. The person has laid it out well so he or she is going to get a lot of marks for laying it out correctly. But they're going to be losing some marks because they were only half of the thing. I'm just saying there's only half available and so I'm giving half the marks. So the total is going to be 18. It's maybe a little bit generous actually. After all, this thing is false. In fact, if this was a mathematics course and I was sort of grading people for doing mathematics work, I think I'd have been much more harsh. I'd have probably come down at 10, 12, maybe even less. But this is about much more than mathematics. It's about mathematical thinking. It's about communication. It's about understanding proofs. In reality, the mark I've given the person is two. I've given them two out of the four. So essentially, I've said this is worth half marks in terms of the overall evaluation because half the proof is okay. But because we are also rewarding or looking to reward proof structure, communication, all of those other aspects of proofs, which are important, I'm going to give credit for the fact that the person has got the general idea. So I've got 18. As I say, maybe we can add it as a little bit generous. I'm inclined to think it's a little bit generous myself. On the hand, this is a 75%. Okay. Which means 25% has been docked, if you like, or not given. And let's look at what happened. This was essentially just one mistake. The person looked at it and didn't look close enough, but it comes down to just one error. It's one error in logic. The person who could produce that argument almost certainly could produce the correct argument in either direction to show that it was false. So it really comes down to one slip. And frankly, it seems to me that if you docked more than 25% or maybe 30% for just one error, then that's kind of harsh. We are after all trying to turn people into better mathematicians to make them better mathematical thinkers. So let's look at what they do right, and then give credit for that. And this person has done a lot of things right. There was just one mistake. And it was a mistake that almost certainly this person shouldn't have made. They had the ability not to make the mistake. But people do make mistakes. So we're not going to give Markzware for free. But having hesitated a little bit on the 18, I've actually now talked myself back into saying 18 is actually a pretty good grade. Now everybody, well, that's the end of that problem set. How did you get on with that massive assignment four? To give you a chance to complete it, this lecture is also fairly short. And this time, the assignment is down to a single page. So keep working on that last assignment. It really is crucial to master those ideas and that terminology. Knowing when and why one statement implies another and being able to distinguish between necessity and sufficiency are crucial abilities in today's world. There are two more language constructions that are fundamental to expressing and proving mathematical facts and which mathematicians therefore have to be precise about. The two quantifiers, there exists and for all. When we've made these terms precise and are sure we know how to use them properly, our analysis of language will be over. The word quantifier is used in a very idiosyncratic fashion here. In normal use, it means specifying the number or amount of something. In mathematics, it's used to refer to the two extremes. There is at least one and for all. These are all we need to look at because of the special nature of mathematical truths. When mathematics is viewed as a subject in its own right, as opposed to being a set of tools used in other disciplines and walks of life, the core of the subject is the theorem. And the majority of mathematical theorems have one of two forms. There is an object X having property P. For all objects X property P holds. I'll take these one at a time. We're looking at statements of the form, there's an object X having property P. For example, the equation X squared plus two X plus one equals zero has a real root. We can emphasise that this is an existence statement by writing it in the following form. There is a real number X such that X squared plus two X plus one equals zero. Or we could write it this way. There exists a real number X such that X squared plus two X plus one equals zero. Doesn't matter whether we write it with an is or an exists. It's an existence statement. The symbolic abbreviation that mathematicians use for exists is a back to front E. E for exists of course. So a mathematician would write this as follows. There exists an X such that X squared plus two X plus one equals zero. This symbol is called the existential quantifier. The simplest way to prove an existential statement of this form is to find an actual X that satisfies the property. And with this example it's easy. Take X equals negative one and X squared plus two X plus one equals negative one squared which is one plus two times negative one which is negative two. Then we've got the plus one. One minus two plus one equals zero which solves it. Well that was a simple example. Sometimes we can prove an existential statement without finding an actual object that satisfies the property. Let me give you an example of that. I'm going to prove this statement. There exists an X such that X cubed plus three X plus one equals zero. I'm not going to find an actual X that solves this cubic equation but I am going to show that there is a solution. I'm going to start by looking at the function Y equals X cubed plus three X plus one. This is a continuous function. Continuity of functions is actually a fairly deep topic. I will touch on it briefly at the end of the course but from our present purpose all I need to know is that a continuous function is one whose graph is a smooth curve with no breaks and jumps. And you should be sufficiently familiar with cubic equations by now to know that the cubic equation has a nice smooth curve that looks something like that in many cases. Up and down two curves. The point is it doesn't have any jumps. If X equals negative one this curve has value or this function has value. Negative one cube which is negative one plus three times negative one which is negative three plus one which is negative three. If X equals plus one the curve or the function has value Y equals one plus three plus one which equals five. So the curve lies below the X axis for X equals negative one and above the X axis for X equals plus one. So it looks something like this. Put the origin here with a look at negative one. We're going to look at plus one. It's going to lie somewhere down here at negative one. It's going to lie somewhere up here at plus one. And somewhere it's going to cross the axis. It looks like that, that, who knows. The point is it's going to have to cross the axis somewhere or other rather to the left of the origin or to the right of the origin. It doesn't matter where. The important point is somewhere between negative one and plus one. This curve crosses the axis and when it crosses the axis the Y value is zero. So we haven't said which X solves the equation. We haven't even said whether the X is negative or positive. But we do know there is such an X. We've shown that the X exists that satisfies the equation without finding such an X. This is an example of an indirect proof. We haven't proved it directly by finding an X that solves the equation. We've simply shown that there is an X that solves the equation. A lot of mathematical proofs are of this nature. We show that there is a solution to some equation or that there is an object that satisfies some property or other without finding such an object. Pretty cool, huh? Time for a quiz. A simple modification of that last argument use a more specific result. Which of the following is it? Is it one? There is an X less than zero, such that X cubed plus three X plus one equals zero. Or is it two? There's an X greater than zero, such that X cubed plus three X plus one equals zero. Okay, which one do you think it is? Well, the answer is one. And here's how we can get that. In the previous argument, I looked at the curve at X equals negative one and X equals plus one. And I showed it looked something like this. At negative one, the curve was below the X axis. At plus one, the curve was above the X axis. And so it has to cross the X axis somewhere between negative one and plus one. But instead of going between negative one and plus one, I could go between negative one and the origin itself. Well, note that if X equals zero, then Y equals X cubed plus three X plus one is one. So at zero, it's somewhere up here. Which means it's negative at negative one. It's positive at zero. So it's somewhere between negative one and zero that it crosses the axis. Since the curve lies below the X axis at X equals negative one, and above the X axis at X equals zero, it must cross the X axis between negative one and zero. Or I could prove it a different way. I could simply observe that if X is greater than or equal to zero, then X cubed plus three X plus one is greater than zero. And in particular it is not equal to zero. So two is false. And that just leaves one. We know that one of these two has to be true as a result of the previous argument. And so if we can eliminate two, as I've just done here, then I can conclude one. So either way, either by refining the previous arguments or by deducing it from the previous result by this observation, I end up showing that the equation has a root somewhere between negative one and zero. I still haven't found that root, but I have sure that there is a root. It's an indirect proof. Incidentally, in giving that last proof, I was careful to distinguish between the terms negative one and minus one. Chances are your middle or high school teacher made a big deal of that. Like many professional mathematicians, maybe even most of us, I'm generally far less careful. I tend to use negative and minus interchangeably. It's not that we don't know the distinction. It's just that at that university level, we focus on other issues. And since this is a university course, I'll assume you can use your knowledge of arithmetic to determine the intended meaning, just as we do with everyday language. In this course, a similar issue arises with implication and the conditional. And since that distinction is likely new to you, I'm going to try to always use the correct term. But when I'm talking with my professional colleagues, we just talk about implication to cover both cases and rely on our shared knowledge of the concepts to disambiguate. In both cases, the distinctions are important, which is why we emphasise them in our teaching. But once we're sure everyone has fully grasped the issues, we adopt a more relaxed attitude, confident that our understanding will keep us out of trouble. The same kind of argument I just used to show that a certain cubic equation has a real root can be used to prove the wobbly table theorem. Suppose you're sitting in a restaurant at a perfectly square table with four identical legs one at each corner. Because the floor is uneven, the table wobbles. One solution is to fold a small piece of paper and insert it under one leg until the table is stable. But there's another solution. Simply by rotating the table, you'll be able to position it so that it doesn't wobble. You might enjoy trying to prove this. The solution is simple, but it can take a lot of effort before you find it. It's very much a thinking outside the box question. It would be an unfair question on a timed exam. And I'm not giving it as a course assignment. But it's a great puzzle to keep thinking about until you hit upon the right idea. I'll leave you with it. Even if you can't think of a mathematical solution, you could go off in search of a square wobbly table and confirm the theorem experimentally. Meanwhile, let's get back to our main theme. Sometimes it's not immediately obvious that a statement is an existence assertion. In fact, many mathematical statements that do not look like an existence statement on the surface turn out to be precisely that when you work out what they mean. For example, a statement that's a number x is rational is an existence statement. And here's why. Let's take the specific statement square root of 2 is rational. As it happens, this is a false statement. But we're just using it as an example of a statement. So on the face of it, this doesn't look like an existence statement. But it actually is because we can write it as there exists natural numbers p and q such that square root of 2 equals p divided by q. Or using our symbolic abbreviation, exist p, exist q such that square root of 2 is p over q. Incidentally, when we do come to prove that the square root of 2 is not rational, we'll do it by assuring that there are no values of p and q that satisfy this equation. So we'll prove a non-existence statement. Well, this expression is fine. So long as we know in advance that p and q denote natural numbers, I mean here I said it explicitly. Here I didn't. So how do you know that p and q denote natural numbers? I mean maybe they should denote real numbers or complex numbers. The answer is you make it more specific. And you do that by writing it in the following way. There exists a p in n and there exists a q in n such that root 2 equals p over q, where n denotes a set of natural numbers. Incidentally, another font that's often used to denote the set of natural numbers is something like this. An n with a double line for the diagonal. In this course, I'm assuming that you're familiar with basic set theory and you know what the membership symbol means. If you're not, then there is a course supplement and I'm just going to leave it to you to read that supplement. And of course there's a course textbook if you acquire the textbook. That's my own book, Introduction to Mathematical Thinking. In any case, I'm going to assume that you are familiar with set theory and I'm not going to talk about it in the course itself. You sometimes see mathematicians write it in an even more abbreviated fashion. There exists a pq in n such that square root of 2 is p over q. That's fine if you're confident about the material and you're familiar with the notations. But we're going to be looking at examples where there are many different types of quantifiers coming in together and then if you start putting them together in this fashion, there's a possibility of getting confused and going off course. So I would say at the early stages, keep things distinct as they are here and try to avoid that for now. As I say, professional mathematicians write this kind of thing all the time. But just as when we're beginning to learn mathematics, it's important that we distinguish things like negative and minus and then when we master them, we tend to forget the distinctions or we don't forget them but we don't make them explicit. So here when we're beginning to look at quantifiers, I think it's important to be very explicit and then once you really understand it, then you can do what we do in the business. We just write things in the simplest fashion confident that we know what's going on and that the way we write it won't mislead us. Incidentally, see if you can prove that the square root of 2 is not rational. We'll do that in class later but see if you can do it now. What that amounts to is proving that there do not exist p and q in the natural numbers such that the square root of 2 equals p over q or that 2 equals p squared over q squared. This is the statement you actually prove in order to show that the square root of 2 is irrational. It's not a difficult argument but it's rather clever and ingenious. It's fairly short and the chances are that you're not going to come up with it but it's worth giving it a shot. You know, spend a half an hour or so thinking about it to see if you can prove that statement. Good luck on that one but we will come back to that in class. How are you doing? Do try to show that the square root of 2 is irrational. Not so much to find the proof but to get used to what it's like to do university level mathematics. One feature you need to get used to in mastering college mathematics or more generally what I'm calling mathematical thinking is the length of time you may need to spend puzzling about a problem or even one particular detail. For the most part without seeming to be making progress. High school mathematics courses particularly in the US are generally put together so that most problems can be done in a few minutes with the goal of covering an extensive curriculum. At college there's far less material to cover but the aim is to cover it in more depth. That means you have to adjust to the slow pace with a lot more thinking and less doing. At first this comes hard since thinking without seeming to be making progress is initially frustrating but it's much like learning to ride a bike. For a long time you keep falling or relying on training wheels and it seems you'll never get it. Then suddenly one day you find you can do it and you can't understand why it took so long to get there but that long period of repeated falling was essential to your body learning how to do it. Training your mind to think mathematically about various kinds of problems is very much like that. OK, semen over. The one remaining piece of language we need to examine and make sure we fully comprehend is the universal quantifier which asserts that something holds for all x. This notation means for all x it's the case that something other. The symbol here is an upside down letter A which means for all. For example if I wanted to say the square of any real number is greater than or equal to zero I could write it like this. For all x, x squared greater than or equal to zero. So the short way of actually saying that is for all x, x squared greater than or equal to zero. Just as we would write things like there exists an x such that x squared equals zero. Notice that I didn't use the word all in this sentence. I use the word any. The same thing happened with exist. There are different words we can use to express an existence assertion and there are different words we can use to express a for all assertion. This is the universal quantifier. How do I know what the x means? Well I have to specify what it means. If I want to be explicit I would have to write something like for all x in the set of real numbers x squared greater than or equal to zero. Okay let's take a look at combinations of quantifiers. Most statements in mathematics involve two or more quantifiers combined. For example if I want to say there's no largest natural number. What I would write is this. For all m in the set of natural numbers there is an n in the set of natural numbers such that n is bigger than m. For all natural numbers m there is a natural number n such that n is bigger than m. That clearly says there's no largest natural number. Note that the order of the quantifiers is important. If I swap those quantifiers round and write exists an n in n such that for all m in n n is bigger than m. The result says there is a natural number n which has the property that for all natural numbers m n is bigger than m. In other words this says there is a natural number bigger than all natural numbers which is false. All I've done is swapped the quantifiers round but in so doing I've turned a true sentence into a false sentence. Remember that example from the American Melanoma Foundation? In their fact sheet they wrote one American dies of melanoma almost every hour. Using our quantifiers we could write that like this. There exists an American such that for every hour a dies in our age. There is an American such that every hour a dies in our age. Poor guy. I mean quite amazing. What the writer obviously meant was the following. For every hour there is an American such that a dies in our age. It will be different Americans for different hours. Now as I mentioned earlier in the case of everyday English these are almost never a problem. Everyone understands the context. We know what's meant. We know that this one is not meant because this is clearly wrong. We know that that one's the one that's meant. And with a simple mathematical example like this one arguably it's not important either because it's everybody knows what's meant. We know that there is a there is no largest natural number. We know it must be this one and not this one. So in these examples there's probably no problem. But we want to use mathematical notation in complicated situations and to talk about things we don't yet understand. That means we can't disambiguate. It's really important that we say things in the right order. OK. I think it's time for a quiz. So how did you get on? Where the literal meaning of this statement is captured by number one. That's the literal meaning. It says there is a licence for which there are two distinct states that that licence comes from. So this is the literal meaning but it's false. Licences are issued by states. You can't get a licence that's issued by two separate states. So the correct answer to the question is number one. But it's not what the person who wrote that sentence in the form in the driver's licence application meant to say. Let's look at number two. That says there are two different licences, they're different, there are two licences which are issued by one state. Well I guess it's possible that you could own two licences that are both issued by the same state if you had two identities or something like that. So this is a possibly true statement but it's not what the sentence means. Let's look at number three. That says there were two licences and two states, different states, different licences. There were two licences and two states, one licence comes from one state, the other licence comes from the other state. This is the sentence that the driver's licence application meant to say. So the literal meaning is number one which is false. Number two may be true but it's not what the sentence is about. Number three is clearly the intended meaning. It's a meaning that we would all understand from this because we know about licences and about states and so forth. So this is the one we understand but it's not the literal meaning. It doesn't matter when you're applying for a driver's licence unless I guess if you get into trouble you get your lawyer might try to get you off by arguing on the basis of the mathematical meaning. Good luck if you want to try that. But in mathematics we can't allow this kind of thing to happen. We can do all sorts of problems and in mathematics we can't hire a lawyer to get it out of difficulty. Okay let's look at the next question in this quiz. Well the correct answer is number three. Let's see why. What does the first one say? It says there is a licence and there are two different states such that that licence is valid in one state and valid in the other state. So it says there's a licence that's valid in two states. Well okay, it's true because in America when you have a licence valid in one state it's valid in any state you can use it anywhere. So it doesn't say anything particularly interesting. What about number two? That says for every licence and for every pair of states which could even be the same state, the licence is valid in one state or it's valid in another state. Well is it actually false? There's a statement because there can be invalid licences. If you've got an invalid licence that means the whole thing is wrong. This one is the one that captures that sentence. Let's just read it. It says for every licence, any licence, if there is a state in which that licence is valid then that licence is valid in all states. So three is the one that captures this sentence. One is actually a true statement but of very little relevance and two is actually a false statement. It's false remember because you can have invalid licences. So it's not the case that for all licences and for all states the licence is valid in one state or another state. Okay, how did you do on that? You might have had difficulty reading the formulas in that last quiz. If so, check out the supplementary video tutorial called How to Read Mathematical Formulas. You'll find it in the same place you access all of regular tutorial videos. In this video, I'll explain how you read mathematical formulas. Just as you're probably familiar with an arithmetic and algebra, you need to know the conventions regarding the order in which logical operators apply. Well let me begin by just quickly summarising the precedence order for applying the logical connectives. The ones that bind the most tightly, the ones that are sort of the strongest if you want, that have told something close together, are the quantifiers for all and exist. And a quantifier applies to whatever comes adjacent to it. Okay? Now typically what comes adjacent to it involves various other things like hands and ores and knots so you would put them in parentheses or brackets, square brackets or whatever. So very often in fact I almost always make a habit of putting whatever I want next to it in parentheses because so for all then applies to everything that comes there. Okay? It binds tightly to this, applies to everything, the same with existing here. Now if there's only something very simple coming next, for example supposing I wanted to say all the balls are red, I could say for all balls red B, if red is a predicate that applies to balls. So I could say for all B red B and that would apply to the red balls and if there was something else here, you know if there was an da da da then the for all would not apply to that. If I wanted to for all to apply to that I'd have to put parentheses around there. I didn't put parentheses here, I could do. Again especially when I'm giving introductory level courses I usually put parentheses around quantifiers but if you look at some of my research work and advanced courses you'll find I often don't do that. That's fairly consistent among instructors. You know the golden rule is if there's any doubt whatsoever and if you're beginning on this material there certainly will be doubts. If there's any doubts put parentheses in. You know you can't have too many, well that's not quite true if you have too many parentheses it gets hard to read it. So you have to strike a balance but always if there's going to be any ambiguity put the parentheses in and mix parentheses along square brackets or even spaces. I'll try to remember to give an example in a minute with a space because you can sometimes use spaces to disambiguate but the golden rule must be you want to avoid someone being left unclear as to what the meaning is. Negation is about the same strength as, well is the same strength as the quantifiers. So the negation applies to whatever's immediate next to it and since we usually want a whole bunch of things to be negated then the negation is followed by parentheses and then it applies to everything between there. Let me give you the following example. Suppose you wanted to say not the case that three is greater than zero and three is less than zero. Okay, well is three bigger than zero? Yes. Is three less than zero? No. So here I've got a conjunction of something that's true and something that's false. So this conjunction is false so its negation is true. So this guy is true but supposing I wrote it this way, not the case three greater than zero and three less than zero. Supposing the parentheses didn't include both of those, they just included one of them. In that case, what have I got? This guy is false. This guy, three is greater than zero is true so that guy's false. So here I've got a conjunction of false things so I've got something that's false. So these clearly aren't the same because this one's true and that one's false. Here the negation applies to everything in between, which makes it true. Here the negation only applies to the thing next to it. Now I could have gone back here and put parentheses here and I'll mention that in a moment. It actually comes up in the next priority. That wouldn't have changed things. That wouldn't have changed things because the negation would apply to what was in the next parentheses. Negation applies to whatever comes next and what comes next is the whole thing because the parentheses includes the whole thing. So simply putting parentheses inside doesn't change anything. It makes it maybe a little bit clearer. Although this is one of those cases where adding parentheses arguably makes things a little bit less clear. But in terms of the logic, the issue between these two wasn't whether there were parentheses around the three greater than or less than zero. The issue was whether the parentheses governed everything that was next or just the one thing that was next. So these are not the same. The next one is conjunction. Let me now just pick up that thing I mentioned before. When I did that the first time, I wrote this. I said, three is bigger than zero and three is less than zero. Now I in fact left a space. If you watch what I did, I left a space. And I realised at the time I was doing it that that's what I was doing, which is why I decided to pick it up now. So this says that three is bigger than zero and three is less than zero. You actually don't strictly speaking need parentheses around here because this is an atomic formula as we sometimes call it. This is a basic building block out of which we're building more complex formulas. This simply states a fact, three greater than zero, an atomic fact, a single fact. This states another atomic fact, three less than zero. So when you have basic facts about arithmetic or whatever, they stand on their own. The conjunctions and the quantifiers are what combine these basic facts. So you strictly speaking don't need to put parentheses around these. This is a case where I typically would just leave a bit of extra space in here to sort of make it clear that this is a unit and that's a unit. On the other hand, if you want to be safe and it's always wise to be safe if you're at all unsure, you could put those things in. And I did this here. I went back and put them in just to make it clear. Okay? Then some mathematicians would say that conjunction, disjunction are more or less the same or conjunction, disjunction implication. We're getting down to a sort of a general grouping now where everything is roughly the same strength. There are actually some arguments that say that conjunction should be tighter than disjunction, but it's not particularly strong. In any case, I think in, this is a case where one should always use parentheses to disambiguate. The point is that the conjunction applies to whatever's to the left of it and whatever's to the right of it. And if you want it to apply to a whole bunch of things, you put them inside parentheses. Likewise here, you would have a whole bunch of things. And the same is true for disjunction implications. So regardless of whether you think that's stronger than those, the issue should never really arise because you should always put things in parentheses to just say it's this guy or this guy. And in here there could be a whole bunch of things. And that whole bunch of things will be disjoint with this. And likewise here, if you have an implication or a conditional, this whole thing would be the antecedent and this will be the consequence. Now in here there may be all sorts of conjunctions and disjunctions and stuff. There may be quantifiers in here. There may be quantifiers in here. There could be all sorts of stuff in here. Negation signs inside. This whole thing would imply that whole thing. So whenever you're looking at, I mean the basic thing with all of these is when you've got a quantifier or a negation symbol or a conjunction or a disjunction or an implication or equivalence actually. I didn't talk about equivalence but equivalence is just a conjunction of two implications, the by conditional. So we could put that one in here as well. Strictly speaking, if you did assume that that had more tight binding than this, you could write something like this. You could write A and B or C and D. And if you put space in there, that I would think is fairly clear that it meant to be this or that. So this is, I would say, at least the way I was brought up. Let's put it that way. As a mathematician, I was brought up to us to say that that actually is okay and it's not ambiguous. But I would almost certainly now, I think I've cured myself of that childhood sin, I would always put in parentheses and say it's A and B or C and D. I mean you just have to be very careful about not making sure that things are nicely not ambiguous. Okay? So golden rule, put parentheses in. And then everything applies to whatever comes between the next parentheses. It's kind of optional as to whether you have parentheses around the quantifiers. But the parentheses that follow providing you always do it, there shouldn't be any problem. Okay? Now let me illustrate those points with examples that are all related to that quiz at the end of lecture five. Okay? So let's take these one at a time. This was actually from that particular quiz. I've got a for all and for all applies to everything that's adjacent to it. Now that parentheses there teams up with that parentheses there. And I've actually written them not as parentheses but as square brackets to make it absolutely clear. That is the thing that the for all applies to. Okay? For all applies to all of this. In particular, what it's going to say is that pick any license and then it's going to say something about the licenses. Now the license L is going to appear on both sides of this conditional and it will be the same L because the L has been picked here and once you pick the L it'll apply to everything here. So that L is bound by that quantifier. Okay? So I would say L is bound by the quantifier for all L. Okay? Quantifier binding. Okay? So let's read it now in English. It says for any license L if there is a state in which L is valid if L is valid in some state at least one state then L is valid in every state. Okay? For any license L if L is valid in some state the exists here simply applies to this. The exists binds what's next to it so the exist simply binds this thing. Likewise the for all binds what's next to it which is this thing. So what this says is for any license or for every license L something happens. What happens? If that license L is valid in some state then that same license is valid in every state. So this is the one that actually says a license that's valid in one state is valid in every state. Which is true in the United States by the way. Okay? That's the first one. Let's look at the second one. What's the difference? Let's see what we've got for all applies to something in the middle. So for all applies to everything here because I've got the parentheses. The only difference is that instead of having a conditional or an implication I've got a conjunction. So let's see how we would read that. Okay? I mean the binding is the same. The for all applies to everything here. This exists applies to this thing. This for all applies to this thing. And the L is the same L here as here. Because once you've said the for all L within this expression here the L is determined by that. The L is still bound. So as was the case there, the L inside here is bound. Okay? But how would we read it? We'd say for every license L there is a state in which the license is valid and the license is valid in all states. So let me just write that down. For any license L there is a state in which L is valid and L is valid in every state. Now it doesn't mean the same as the previous one. Let's think about this. For any license L for any license there is a state in which that license is valid and L is in fact valid in every state. But this is actually false. This precludes the fact that there are invalid licenses. For example, if you go to California and you drive with too much alcohol in your bloodstream you will find yourself with an invalid license. Not every license is valid. So this is actually a false statement. The one above was true. This is true in the United States. This is false. This means something different. In fact really it's the first thing that was the problem. For every license there is a state in which it's valid. Well that's simply not the case. Already the first conjunt makes it invalid. Didn't arise in the first one because in the first one the part that said it's valid in a state was the antecedent. If it's valid in a state then it's valid in all states. So that said for every license if it's valid in a state. This says for every license it is valid in a state. Well that's not the case. Not all licenses are valid. Okay. So there is a distinction between these two and in fact the distinction is a meaningful one in terms of validity of licenses and so forth. Let's look at this one. Well here we don't have these brackets. So let's see what's going on. This means for all L for every license there is a state that the license is valid in that state. Well is that true? Is it true? Remember this is a unit. The for all and the exist apply to whatever's next. So there's a line here. The for all and the exist don't apply to that. They apply to what's next and there was no bracket so it doesn't get included here. So what this actually says is that for every license there is a state in which that license is valid. So what this really says is that all licenses are valid somewhere. Well okay that's not true and the statement is if that's the case then for all S2 that would say L. Well haha. This would say that L is valid in all states. Well now we've got all sorts of things gone wrong. As a conditional this guy would look on the face of it as if it was going to be true because the antecedent is false. It's not the case that all licenses are valid somewhere. There can be valid licenses. So this is a false antecedent. Now you might be tempted to say since it's a false antecedent the conditional is true. But not quite because this guy isn't even defined. This isn't defined. This is meaningless. What's L? What is that L? It's not governed by that quantifier. This is just an orphan. It's just sitting there. We don't know where it comes from. We don't know what it means. It's just a letter that has no internal meaning to this formula. So it's not the case that this is a valid conditional. It's actually undefined. This is meaningless unless you know what L is. If you know what L is you can assign meaning. And once you know what L is then if L replied to my license, if that L there was my license then we would have a meaningful and in fact a true conditional. But as it stands you've just got a completely unbound variable. So the L there is what we might sometimes call just an unbound or free variable. Unbound variable, free variable. Until you assign a value to it it doesn't have any real meaning. And then let's finally look at this one for all L1. What's the difference here? It's somewhat similar to the one up here but not quite. Let's just read it through. It says for every license and for all pairs of states the license is valid in one state and the license is valid in two states. Well that really just means all licenses are valid in all states. Again this is not the case in the United States because you can have invalid licenses. So we've got something that's actually false. And it's over. I mean there's redundancy here. Because the second S adds nothing new. It simply says for all licenses and for all states the license is valid in that state and it's valid in the other state. So we could just scrap that and scrap that and we'd have the meaning without any of that stuff. So there's nothing actually wrong with this. It's just, I mean it's a false statement but it's got redundant clauses. The second clause says adds nothing that the first one didn't already state. OK. Finally let me just try to distinguish between four cases that beginners typically find to be very confusing. They're actually really very distinct and if you find this confusion between these four that's a sure sign that you haven't yet mastered the notations and what they mean. OK. Let me just write down a transcription of what it means and then let's just ask ourselves exactly what that signifies. So in English that would say for every X if P of X then Q of X. OK. For every X if P of X then Q of X. This is very common. OK. For every number, for every real number, if that number is non-negative then it has a square root, et cetera, et cetera, et cetera. So this occurs a lot in mathematics, this kind of statement for every X if P of X then Q of X. OK. Very meaningful. And it's the same X here notice. Once you've got that for all the X here is the X here. So whatever X providing the X satisfies P then it satisfies Q. So this establishes a relationship between P and Q because if you've got an X that satisfies P then that X will definitely satisfy Q. So this is a very strong and very common statement to make. This is also pretty common. This says for every X P of X and Q of X. It says that every X satisfies P and Q. It's kind of strong. I mean it doesn't occur terribly frequently because that's really the same. I mean you could equally, you could just as equally say for all X P of X and for all X Q of X because you're basically saying everything satisfies P and Q and that's equivalent to everything satisfies P and everything satisfies Q. Notice by the way that this is non-ambiguous because of the binding. The for all binds what's next to it so the for all can only bind that. The for all binds what's next to it and so I don't need the parentheses here because in this case the for all absolutely can't be confused with that. So here's a case where you don't need extra parentheses. I didn't even write the parentheses here. You don't need them. This is totally clear in this case. Okay. And it's equivalent to that. So you don't see this very often because it really is just saying everything satisfies P and everything satisfies Q. But it's okay if that arises. Don't worry about it. It might in a context it might be sensible to write that down. But it doesn't have the same sort of logical force that this does. This has real logical force. It establishes that there's a relationship between P's and Q's. Okay what about this one? This says there is an X for which P of X and Q of X. This is again pretty common in mathematics. This is quite a strong statement. It says you can find a single X which satisfies P and satisfies Q. So you can find an X which has the property P and which has the property Q. So this is a strong statement. That one was strong. That one's strong. I mean this one's strong but it's only strong because each part is strong. So there's almost a redundancy in the way it's written. So this is, maybe I would put strong in quotation marks to sort of say well yes it is strong but it's not strong because of the logical structure. It's strong simply because it's making a statement about P and Q both been satisfied by all X's. Okay what about this last guy? This is one that people often write down and this really means nothing in any real sense. It says what there is an X such that if P of X then Q of X. Okay there is an X such that if P of X then Q of X. It's pretty rare to ever need to say that actually. This really doesn't arise particularly frequently that you would need to say something like this. If you see yourself writing and exist with an implication chances are very high that you've sort of got confused. Let me just put it this way. Let me just say that this is weak. Okay it's not on the same strength as these guys because this says for every X if it satisfies P then it satisfies Q. That's making a strong statement. For every X there's an implication. This simply says there's one X for which there's an implication. Well in a sense the implication is almost vacuous then. I mean one thing to say for example is that if you can find an X that does not satisfy P. In other words you can find an X for which P of X is false. If you can find an X anywhere for which P of X is false then you'll have a conditional that's necessarily true. So this can be made true by finding an X that doesn't satisfy P. That's all it would take to make this thing true. So if you're trying to make a stronger statement, if you're trying to make an existence statement, if you're using this to say there is an X with a certain property then you could make this guy true simply by finding an X that does not satisfy P because if you can make that part false the conditional becomes true. So that's one of the reasons really why this is weak. I'm sure there will be circumstances where this could have some significance but basically my message for you would be forget that one. It's just if you see yourself writing an exist with an implication after it the chances are very high that you've got confused. Always be prepared to override what I say. All sorts of circumstances can arise but in general these guys are all quite significant. That's particularly significant. That one's very significant. This one is sort of less so because it really just reduces to the two separate things. And this one is really pretty weak. So exists combined with implications. If you see that flag it and say do I really mean what I'm writing that? Okay well I hope that's helped clear up some of the basic issues about reading formulas but like many things at this stage really the only way to get rid of any confusions is to just do a whole bunch of examples for yourself. Okay good luck on sorting these things out as you work through the rest of the course. Okay problem set three. So we've got a variable ranging over doubles tennis matches. Doubles by the way is when you have two people playing against two other people, two teams of two people playing against one team against the other. T is another variable ranging over doubles tennis matches where Rosario partners with Antonio. WX means Rosario and her partner. Whoever that partner is wins the doubles match X and we have to find the English sentences that mean the same as a symbolic formula exists to TWT. Now this kind of thing is always a stretch because we're taking everyday language which is ambiguous and vague and underspecified and can be sort of pummeled and stretched in different directions and we're going to capture it with a formula which is precise and well defined. So when it says mean the same there's going to be some slack here and sometimes there's a little bit of argument so that one could have. I've picked an example where I think it's fairly clear at least from a mathematical perspective it's clear. If you haven't got used to dealing with mathematical language you may find this more difficult and this whole course is aimed at people who haven't got that familiarity so I think you will find it difficult. Okay so let's just take them one at a time. Rosario and Antonio win every match where they are partners. Well that's actually every match where they are partners there's really a hidden for all T in this thing. It's about a for all T. In fact basically what you're saying is in every match where they are partners they win or she wins. I mean one wins they both win, they both win. It's a partnership team. Okay whereas this is existing so it's not that one. Okay let's look at the next one. Rosario sometimes wins a match when she partners with Antonio. So there is a time, there is a game when they partner together when they win. That's this one. So that's correct. Now that really does capture it. There is a time when they play together and as a partner there's a partnership and they win. Now now you can look in some sources to say that sometimes it is used exclusively to mean more than one. I mean I don't think that's the case. I mean you've found people to say that. Language is flexible. In any case in mathematics we always interpret things like some, sometimes at least one. That's a whole point about the way we set things up. We eliminate these ambiguities by being specific. And we are specific to say that whenever you are setting something exists, some time, some game or whatever, you mean at least one. Okay? In which case you've got existential quantifier. That means there is at least one tennis game where they play together and they win. Okay? So that one's okay. Whenever Rosario plays with Antonio, she wins the match. Well that's really, again, that's for all T, W, T. So it's not that one. Rosario and Antonio win exactly one match where they are partners. Well that one isn't going to work, all right? Because that says exactly one match. There's no specification here of exactly one match. If you wanted to do that, there's a notation. This notation exists a unique T such that W, T. You can say it and you can say it other ways. I mean you can, this is just an abbreviation for an expression. We've seen that in the problem sets. Actually one of the assignments. So you can capture it, but this doesn't capture it. This just says there's at least one. It doesn't say there's exactly one where they win. Okay? Rosario and Antonio win at least one match when they are partners. That's it. That's another one. That's fine. At least one match when they are partners. Okay. If Rosario wins the match, she must be partnering with Antonio. Well, first of all there's a universal quantifier floating around here, I think. Yeah. Because it's saying whenever she wins the match, she must be partnering. So there's a universal quantifier here. But it's even worse because the universal quantifier is actually for all x, for all matches. Okay? For all doubles tennis matches where she wins something or other, something else. So there's actually here what's generally known as a scope problem. In this statement, the quantification is actually over something different. It's over all possible double tennis matches, not just the ones where she's partnering. So not only does this not capture it, there's actually another issue. There's a scope issue involved. Okay? Because here, the T ranges only over games where they play together. In this case, we're looking at games where Rosario plays with whoever she's playing with. Okay? So there are a couple of things that prevent this one been the right answer. Okay? Well, I'm, you know, as I said at the beginning, from a mathematical perspective, this is actually very clear. It's definitely B and it's definitely E. And the reason I can be so definitive about that is because I'm familiar with the way we've set up the meaning of this in mathematics to correspond to mean at least one. And we interpret in mathematics, we interpret anything that exerts an existence to mean exists one, at least one. And so things like sometimes, some of these, some of those, they're all interpreted to mean at least one. Okay? Let's look at the next one. Well, same setup as in question one. The only difference is now we're talking about, for all T, W of T. So let's, let's run through this one. Rosario and Antonio win every match where they are partners. So every match where they are partners, that would be for all T, because that's what T captures, T is the doubles tennis matches where they partner, and they win. Well, that is that. So that one's okay. What about this one? I mean, this has really got nothing to do with that, has it? Rosario always has got nothing to do with winning. It's just saying she always plays together. That would essentially say that did X and T have the same variable, in fact. I mean, it's just saying that there's no distinction between X and T. So I think rather than cross that as said, that was wrong, I'll say this isn't even a candidate. I mean, it's got nothing to do with winning. Okay. Let's look at part C. Whenever Rosario partners with Antonio, she wins the match. Whenever she partners, that's for all T, she wins. She wins, they wins. So it's all the same in doubles tennis. So that's okay. What about this one? Sometimes Rosario wins the match. No. I mean, first of all, it's, it's not about T, it's about X. Sometimes she wins with whoever she's playing with. And it's an existential one, so it's essentially of the form Exist X W of X. That's really what it means. Sometimes Rosario is in the winning team. She wins. Well, that's not that. It's a different quantifier, and it's over X, not T. So, not that one. I won't cross it out because at least it does talk about winning, so it's a candidate, but it's not the right candidate. Okay. Rosario wins the match whenever she partners. This is whenever she partners, that's essentially for all T, and whenever that happens, she wins. Okay. Bingo. How much is that? That's correct. And finally, if Rosario wins the match, she must be partnering with Antonio. Well, essentially you've got something like for all X here, et cetera. So, as before, as in question one, we've got a scope issue here. This is actually about all possible matches, not just the ones where she's, she's partnering. And the conclusion is that she's partnering with Antonio. So, so this doesn't, I mean, it does talk about winning, but it's, it doesn't come close because there's, there's a scope problem. The variable doesn't mean different things. Okay. So it's not that one. So in this case, we've got a, we've got c, and we've got e. That's kind of unusual to have one of these multiple choice things where three of them are correct, but there you go. Sometimes, sometimes that happens. Okay. Let's move on to question three. When question three, if you look through these, looking for something that seems to say there's no largest prime, I think you quickly end up looking at this one d, which says that for any number x, there is a number y, which is prime and bigger than x. So that certainly says there's no largest prime. Now the question is, do any of these say the same in a different way? Well, let's just look at them in turn. That says there don't exist any x's and y's for which x is a prime, y is not a prime, and x is less than y. Well, there are plenty of pairs x and y that satisfy that. So this is actually false. So, I mean, we weren't, we weren't asked to say whether things are true or false. But this is false, and we do not, we do actually know there was no largest prime. That's Euclid's theorem that the primes are infinite, and the list of primes goes on forever. So it can't be this because this is actually false. But in any case, it doesn't mean the same. This just means something just nonsensical. Of course, there exist pairs x and y with x prime and y not a prime and x less than y. So to say it's not the case, it's clearly wrong. What does this say? For every x, there is a y, such a x. Well, first of all, that would say for all x, x is prime, that would say every number is a prime. So that's false as well. So that can't possibly be it. That can't be it. That can't be it. What does this one say? For all x and for all y, x is, that says as well, every number is a prime. That's false. Wow. Why are we, are we rattling through these? These are, these are just plain false. They're absurdly false, if you like. That was the one that was okay. What does this say? There is an x such that for all y, whoops, that says all numbers are prime again. Forget that. It simply says all y is a prime. Well, they're not. So that's also absurdly false. Scratch that one. I mean, I'm scratching them because not only don't they actually say that, but they actually say something nonsensical. What does this one say? For all x, there is a y, x, whoo, same thing again. What this actually says is that everything is a prime, which is false and it's absurdly false. So it's not just that A, B, C, E and F don't capture it. They don't capture anything sensible whatsoever. This is one on the other hand does say there's no largest prime. Okay, let's move on to number four. Well, I won't go through these in detail as I did with the previous question because the same kind of considerations apply. The only one that actually captures it is E. Let's just read what it says. There is an x which satisfies phi and for all ys that satisfies phi, x and y have to be equal. In other words, it's impossible to find a y that satisfies phi other than the x that you start with. So there is one. The statement begins by saying there is something that satisfies phi and this part says it's the only thing that satisfies phi because if you look at all the possible ys that could satisfy phi, the only one that does so is y equals x. So this actually captures the busy unique x such that phi of x and these four, they're not only, if you try to figure out what they say, if they say anything vaguely sensible at all, they turn out to be just nonsentrical and they certainly don't capture that. Incidentally, this symbol is a moderately common symbol in mathematics. It's not something I made up for the exercise. The exist with the exclamation point does actually mean there is a unique x. You'll find that quite a bit in mathematics. Often you need to be able to say there is a unique solution to something. Now, this exercise tells you that you don't need to have a separate quantifier to mean there is a unique one because you can actually define it in terms of the standard existential quantifier and the universal quantifier. So this is in fact just an abbreviation but it's a useful abbreviation and so you'll often see it. Okay? Well, for question five, let me observe that this symbol, I mean you do see this symbol in sometimes in computer science, very rarely in mathematics except in a situation like this where I'm using this to refer to some arbitrary but unspecified binary operation. So this isn't a particular operation, I just mean there is some operation which I'll call x up at r or y and we need to be able to express the fact that that's not commutative. So this doesn't have a particular interpretation, I'm just using it to mean any particular, any unspecified binary operation. And so what we need to do is ask ourselves which one of these means that it's not commutative. Well commutative, let's write down what it means to be commutative. Commutative means for all x and for all y, x r o y equals y r o x. So that's what being commutative means. Which one of these negates that? Well, when you negate universal quantifiers you get existential quantifiers and things that are true become false or the equality becomes an inequality. And so if you skip through these you find, yep, there it was, c. That says there is an x, there is a y for which they are unequal. There is an x, there is a y for which they are unequal. So that's certainly a negation. Do either, do any of the other ones fall in as being a negation? Not really, not even close, because when you negate both are for alls become an exist. They don't remain for alls, so it's not that one. They don't remain for alls and there's a for all there. So for a variety of reasons, none of these three qualify for that. So there's no, there's no possibility of having two possible expressions. That's the only one. Okay. Okay, question six, evaluating this proof. And this is very typical of the kind of work you see from students who are beginning to look at proofs. Because what this person's done is they've identified the key idea. This is absolutely the key mathematical idea behind this. The fact is that this, this is not prime. And in fact, what I'll do, let me just give the proof that the person should have done. Okay, and then I'll discuss why there's a problem with writing this down. Okay, so what the person should have done is, is something like the following. You begin by saying the claim is logically equivalent to the following statement for any positive integer n, n squared plus 4n plus 3 is not prime. It's logically equivalent to that. To say there doesn't exist an integer for which that's prime is logically equivalent to saying that for any positive integer, it's not prime. And the person then was going to, should maybe prove this is, is, is true. Okay. We prove this, this statement. So we're proving a logically equivalent statement. Okay. So let n be a positive integer. And I'm doing this one in gory detail because I'm trying to get maximum points for this one. Okay. Then by basic algebra, n squared plus 4n plus 3 equals n plus 1, n plus 2, n plus 3. But n plus 1 and n plus 3 are positive integers, each greater than 1. N plus 1 is at least 2, n plus 3 is at least 4. So these are positive integers greater than 1. So by definition, n squared plus 4n plus 3 is not prime. Because it's a product of two positive integers each greater than 1. Okay. So that's what the person should have done. Now, let's go back to what was here. This is the key algebraic heart of this thing. Alrighty. But it's not a proof. And the reason students often do this kind of thing is they're used from high school, they're used to the fact that algebra is all about algebraic manipulation. And indeed it is. But we're talking about proofs here. And a proof is much more than getting the algebraic manipulation rights. If the algebraic manipulation is not right, you don't have a valid proof. But a proof is all about giving reasons and making a, giving an explanation. It's a story. A story with a beginning and middle and an end. You know, most, you know, there's a, there's a, I mean, there's a joke about a Woody Allen in a Woody Allen movie where Woody Allen in the character, Woody Allen character says he's been reading a book. It's about, it was war and peace. And he summarises it by saying it was about some Russians. Well, this is like saying it's about this. Obviously war and peace is just about some Russians, but there's much more than that. And that's really what we're doing here. We're looking for the full story. This one's actually, I think, going to be fairly difficult to grade. I'm going to put four for logical correctness here because logically this is the heart of it. Once you realise that, that's the logical heart. So that's, that's fine. Okay. Clarity, I'm going to have to give a zero. There's just no clarity about this because there's no explanations, nothing. Okay. This is really nothing, nothing valuable here. Opening. There isn't an opening. Just jump straight in. Let's see. Stating conclusion. Whether person did state the conclusion. So I'm going to get that. You have to form acts. The conclusion is stated. That's, that is how you're supposed to end. Okay. As I did here, that's not prime. Reasons, there are no reasons given. Okay. I mean, that's just going to be a zero. And then, let's see what I've got here. Overall valuation is zero. I mean, as a proof, I can't really give anything for that. Okay. Well, let me think. You know, actually, now I'm going to be a bit generous here. I'm going to give two for that. I think because this is key. I mean, I'm going to give some credit for this. I mean, it is the key part of this. And it was the setting that was wrong. So I'm going to give two for that. So that means I've got four eight. I've got ten for that one. Yeah, a little bit generous maybe as a proof. The person certainly has the algebraic ability. And seeing this is key. I mean, that really is, yeah. Okay. I was, I was, I think that's okay. I think that's actually a good mark to give for that one. Okay. But these are not easy to do. You're making value judgments. You're trying to sort of assess a whole bunch of different things. The way we've structured this course is you'll be seeing a lot of exercises like this. And the intention is that by the time we get to the end of the course, you'll have got the general gist of how to do this. I mean, all instructors develop their own methods. And I'm just giving you mine as an example. But we're something that's sort of essentially qualitative as grading proofs is. And it's like grading essays. You know, people, people end up with difference, you know, there are stricter graders and less strict graders. My goal is always when I'm grading this kind of a context is to look for reasons to give people marks because I want to give the credit for what they're doing, but at the same time points out the things that still need to be done. Okay. All righty. Well, that was the end of problem set three. How did you get on with assignment five? Mastery of quantifiers is the one last ability you need to be able to cope with definitions of mathematical concepts and with mathematical reasoning. I'm going to begin by looking at the way negation affects quantifiers. But before I start, let's see how you're progressing by a way of a quiz. Okay. Let's see what we have. Is it the case that for every real number x, x plus one is greater than or equal to x? Yes it is. Is it true there is a real number x? So it's at x squared plus one equals three. Well that would mean x squared equals two. So we're really saying, is it true that there exists a number x? Whose square is two? Well, if we were talking about the rational numbers, the answer would be no, because the square root of two is irrational. But for the real numbers, it's true. What about this one? Well, for very large negative x, x cubed is an extremely large negative number which will dominate all of these numbers. So for large negative x, this expression, this cubic expression, is negative. So it's not the case that it's always positive. So that one's false. What about this one? Is it true that for all x there is a y? x cubed plus y cubed equals zero. That would mean y cubed is equal to minus x cubed. The answer is yes. Given any x, just let y be negative x. When you cube a number, you cube a positive number, it's positive. You cube a negative number, it's negative. So by taking y equals negative x in all cases, you get zero. So this one's true. What about this one? Is there an x such that whatever y you add to it, you get zero? The answer is no. If the quantifier is with the other way around, the answer would be yes. For every y, there is an x such that x plus y equals zero. Namely, you let x be negative y. But it's not the case there is one x that works for all y. So that one's false. What about this one? Well, is it the case that for all x there is a y such that if x is non-negative, then y squared equals x? Well, let's look. Supposing we took an x. If x was negative, then game over because this conditional would be true. It would have a false antecedent. If that x was non-negative, then there is a y whose square is x, namely the square root of x. So whatever x we take, there's always going to be a y that satisfies this. It's true. You might have to think about this one a little bit because there are two cases involved. Given an x, you can find a y that makes this thing true. If the x you're given is negative, any y will make that true because if the x is negative, the antecedent is false. But if the x you're given satisfies this condition, if the x you're given is non-negative, then the y you pick is the square root of x. See what's going on? If you're given a negative x, there is no y that satisfies y squared equals negative x, not in the real numbers. But that eliminates that case. So this one's quite subtle. I suggest you think about this one. It's the kind of thing that you encounter quite a lot in mathematics. Okay? Well, how did you do? Okay, let's move on. In mathematics and in everyday life, you often find yourself having to negate a statement involving quantifiers. Yeah, of course, you can do it simply by putting a negation symbol in front. But that's not enough. At least often it's not enough. You need to produce a negation, not a negative one. The examples I'll give should make it clear what I mean by positive here. But roughly speaking, a positive statement is one that says what is rather than what is not. In practice, a positive statement is one that contains no negation symbol or else one in which any negation symbols are as far inside the statement as is possible without the resulting expression being unduly cumbersome. Let's look at our first example of negating a quantified statement. Let A of x be some property of x. For example, x is a real root of the equation x squared plus 2x plus 1 equals 0. And I'll show that not for all x A of x is equivalent to, there exists x such that not A of x. For example, it's not the case that all motorists run red lights is equivalent to there is a motorist who does not run red lights. In this case, it's pretty obvious that these two are equivalent. The proof that I'm going to give in the general case is actually the same reasoning you automatically do when you look at this specific case. If it seems hard to follow, the issue is purely one of the abstraction. I'm proving an equivalence, so I'm going to prove an implication from left to right and then from right to left. So we begin with the left to right case. I'm going to assume not for all x A of x. Well, if it's not the case that for all x A of x, then at least one x must fail to satisfy A of x. So, for at least one x, not A of x is true. In symbols, there exists an x such that not A of x is true. Well, that's the left to right implication. I assumed not for all x A of x, and I concluded there exists an x not A of x. Now we want to do the other implication. So, assume exists x not A of x. Well, in this case, there's an x for which A of x is false. That's just expressing this in everyday English. Well, if there's an x for which A of x is false, then A of x cannot be true for all x. In other words, for all x A of x must be false. Expressing that in symbols, it's not the case that for all x A of x. And now I've proved the second implication from exists x not A of x to not for all x A of x. Assume exists x not A of x conclude not for all x A of x. If you found this reasoning had to follow, it's purely because of the abstraction. The logic is exactly the same as in the example. Almost certainly you had no difficulty with this example and that's because you're familiar with this situation. The human brain is very good at doing logical reasoning about familiar situations. When we turn those situations into abstract versions, the brain finds it difficult, at least at first. And in fact one of the things about becoming a mathematician is learning to take familiar everyday reasoning that we don't even think about, we just do it automatically and reproduce it in an abstract situation. And it's the abstraction that's causing your difficulty. You can follow the logic but you find it difficult to follow it in an abstract situation. And that's just because of the way the human brain works. Okay, why don't you try this one? Show that not there exists an x A of x is equivalent to for all x not A of x. And before you start, you might want to look at an everyday example. This is coming up as an example in the assignment for this lecture but you might want to try it now while this example is fresh in your mind. Okay, good luck on that one. With that one under our belts, now we can go back and look at that earlier example of the negation of all domestic cars are badly made. Let's see with a set of all cars, D of x means that x is domestic and M of x means that x is badly made. With this notation, the sentence becomes for all x in C, D of x implies M of x. For all cars, if the car is domestic, then it's badly made. We've already seen what happens when you negate a universal quantifier. The negation becomes there exists an x in C such that dx does not imply M of x. And here I've actually abbreviated the following. Not the case dx implies M of x. Well, it's not much of an abbreviation but instead of writing not dx implies M of x, I've used this simple notation dx does not imply M of x. But in any case, looking at our previous example, the one that I illustrated with the motorist, when you take for all x something and you negate it, you get there exists x with the negation of the thing inside. Just compare what's going on here with the previous example. One comment, why am I not saying exists x not in C. Beginners students often do this kind of thing. And the reason they do it is they are trying to go through a formula symbolically and negate things. And that's not the way to go about these things. You've got to think about what the symbols mean. In this case, it's talking about x's which are in C. The only x's we're interested in are x's which are in C. In other words, the only object we're interested in are cars. But if the only objects we're interested in are cars, the negation will only be talking about cars. So the in C part simply tells us which kinds of objects we're dealing with and we don't negate that. Okay, now let's look at this part. We know that dx does not imply m of x is equivalent to d of x and not m of x. In fact, this was the key part of our reasoning to figure out the truth table for the conditional. d of x does not imply m of x if d of x can hold and nevertheless m of x can fail. So we looked at this when we did truth tables. So the negation of the original sentence, which is this, is equivalent to, there exists x in C, this part is the same as this part. d of x and not m of x. Well, what does this mean in everyday language? It means there is a car which is domestic and is not badly made. And that's it. So if you found it difficult to figure out the negation of this sentence when we first looked at this example, now because we've got this notation and this little method for dealing with these negations, you should be able to follow it through fairly straightforwardly. Moreover, the previous time when we looked at this, even if you got the answer, you might not have been totally confident that you got the answer right. This kind of reasoning makes it clear that this is the right answer. And that's the whole point of introducing this formal notation and this logically precise reasoning. We're no longer left not certain if we've got the exact best correct, logically correct negation. The mathematics leads us to the answer. And that is the correct answer. Okay, let's do another example. Let's look at the claim all prime numbers are odd. Now we all know that this is false. And we know it's false because we know that two is a prime number, which is not odd. But let me tease apart the logic that underlies that conclusion. Because in more complicated situations where we're not familiar with the domain or with the result, all we'll have to use is the logic itself. So I want to make heavy weather of this in order to understand the kind of reasoning that we'd use in situations where we don't know the answer. Okay. So I'm going to let P of X mean X is prime. And I'm going to let O of X mean that X is odd. So the sentence can be written symbolically as for all X if X is a prime, then X is odd. When I negate this, I get, exists X, not PX, gives O of X. Or another way of writing it, there is an X such to P of X and not O of X. The negation of not for all X, P of X implies O of X is there exists an X such to P of X and not O of X. In words, IE, there is a prime that is not odd. So in order to prove that this is false, which is the same as proving that the negation is true, the logic says, find a prime that is not odd. And we can. Two is a prime that's not odd. So if I want to prove conclusively that this statement is false, which is the same as proving that its negation is true, what I need to do is demonstrate that there is a prime that's not odd. And in this case, that's immediate. We just pulled two out of the hat, observed that two is prime and it's not odd. Therefore, the negation is true, the original statement is false. Well, in that simple example, that was making much a do about nothing. But in more complicated situations, going through the underlying logic is often the only way forward. By now, you should have noticed the pattern of behavior that's happening with these symbolic negations. When we negate, a for all becomes exists and the for all sort of jumps inside and when it jumps inside, we can manipulate it and get the result into a positive form where the negation is at the most innermost point. You can actually turn these into formal symbol manipulation rules, which means you can implement them on computer systems and there are computer systems that do this kind of reasoning for you. I'm recommending very strongly that you don't do that because what we're trying to get at is mathematical reasoning and moving symbols around doing symbolic manipulation is not mathematical reasoning. It's connected with mathematical reasoning, it supports mathematical reasoning, we can use it in mathematical reasoning, but the real essence of what we're trying to do is to understand what the statements mean, understand what the symbols represent, and reason with the concepts. It's reasoning with concepts that mathematical thinking is all about, not symbolic manipulation. Okay. We showed that this was false by finding a counter example. In this case, the counter example was two. Supposing I modify it as follows. All prime numbers bigger than two are odd. In symbols, for all x bigger than two, if x is a prime, then x is odd. Okay. Now I'm going to give you a quiz. What is the negation of this statement? Is it one there exists an x less than or equal to two such that p of x are not o of x? Or is it two? There exists an x greater than two such that p of x are not o of x. Okay. Which one of those two do you think it is? Where the correct answer is two? It says there exists a number bigger than two which is prime that is not odd. That's the negation of that statement. Why isn't it number one? Well, because the original statement is about numbers bigger than two. So the negation must be about numbers bigger than two. We don't simply negate every symbol in sight and say, well, in terms of the ordering on the real line, if we negate something being bigger than two, then it means it's less than or equal to two. That's a symbolic negation. That's a sort of a detailed localized negation about the ordering on the real line. But the sentence as a whole means something about all numbers bigger than two. So its negation means something about all numbers bigger than two. Okay. Well, let's move on. Let x denote a person, let p of x mean x players for sports team T, and let h of x mean x is healthy. And let me look at the statement, there is an x such that p of x and not h of x. What does that mean in everyday English? There is an unhealthy player on team T. Now let me negate that. I'm going to find the negation by symbolic manipulation. Now this remember is the method that I'm suggesting that we don't use. I'm doing it that way in order to illustrate the mathematics and the underlying logic. If we negate this, we know that when we negate an exists, it turns into a for all. And then the negation moves inside, if you like. And then when we move that negation inside, we get not p of x or h of x. So the negation moves inside, we get p of x becomes not p of x. When we negate a conjunction, we get a disjunction. And when we double negate something, we get the original thing back. Remember this is not the way I'm recommending you do it. I'm doing this as an illustration of the underlying logic. This thing looks familiar. Remember when we looked at the conditional? p conditional q means the same as not p or q. They have the same truth table. So this guy can be rewritten as p of x plus h of x. In English, all players on team t are healthy. Well, is that the negation of that? There is an unhealthy player on team t. Well, if it's not the case that there's an unhealthy player on team t, then all the players on team t must be healthy. Which is definitely the case. That is definitely the negation of that. And again, let me stress. This is not the way you should do this kind of thing. I'm using this to illustrate how negation behaves with quantification. Next point. Suppose I write down a sentence like the following. For all x, if x is greater than 0, then there is a y, such that xy equals 1. Is that true or is it false? Well, there's no way of knowing. It depends what the x denotes. If x denotes a motorist or an automobile or a healthy team player, it's completely nonsensical. If x denotes a natural number, it makes sense but it's false. If x denotes a rational number, it makes sense and it's true. But the point I want to make is that a quantifier only tells you something if you know what the variable denotes. Associated with any quantifier, we have what's called the domain of quantification. Which tells us what does the x denote. If the domain of quantification is obvious, if we've set it in advance or the context makes it clear what it is, then we can use statements like that. If there's any danger of misunderstanding, if there's any potential ambiguity, we can make the quantifier more explicit by stating the domain of quantification. To make this more explicit, I could write it as, for all x in the set of rational numbers, if x is positive, then there is a y, such that xy equals 1. And now I've got a true statement that's unambiguous. It tells us that for every positive rational, well, wait a minute, wait a minute. I haven't actually been explicit as to what the y is, have I? Now arguably, having made it clear that the x denotes a rational, it's reasonable to assume that the y denotes a rational. But if I want to, I could be even more explicit and say for all x in q, if x is greater than 0, then there is a y in q, such that xy equals 1. I think most mathematicians would agree that if you're explicit about the quantifier at the beginning of a statement, then absent any indication to the contrary, the other quantifiers denote the same thing. But if there's any possibility of misunderstanding or if there's any potential ambiguity, it's better to be explicit everywhere as to what the quantifiers denote. Something else you need to be aware of with regards to quantification is that mathematicians sometimes omit the quantifier. We write things like if x is greater than or equal to 0, then square root of x is greater than or equal to 0. If you see an expression like this, what that means is the following. For all x, perhaps in r, or perhaps in something else, for all x in r, x greater than or equal to 0 implies square root of x greater than or equal to 0. Or something like that, depending on what the x denotes. I've assumed from the context, and this is a reasonable assumption, I've assumed that the variable is meant to range over the real numbers. Even though there's no explicit quantifier here, this would be read as meaning something like that. This is what's known as implicit quantification. The mathematician is simply leaving out of the formal expression any explicit mention of the quantifier. It's left implicit in the way this is written. Now, this is a fairly advanced point. I'm certainly not recommending you do this, but the professionals do this all the time, so you need to be aware of it in case you come across it in something that you read. But please avoid doing this. I realise by saying this that I'm saying do as I say, not as I do, because like all professional mathematicians, I do write things this way all the time, but while you're still learning about these ideas, it's best to avoid doing it because if you don't use this in the right kind of context, all kind of difficulties can arise. And having given you one caution, let me give you another one. This is about combining quantifiers with conjunction and disjunction. Let me just do this by way of an example. Let n, the set of natural numbers, be the domain of quantification. Let e of x mean x is even and let o of x mean x is odd. Look at the formulas for all x, e of x or o of x and for all x, e of x or for all x, o of x. I guess I don't need that last bracket, right? Okay? For all x, e of x or o of x and for all x, e of x or for all x, o of x. Notice that the first one is true. It says for every natural number, it's either even or it's odd. What does the second one say? It says every natural number is even or every natural number is odd. Well, that's false. Well, the point I'm trying to make here is you can't just take a for all and take it inside a bracketed expression. If you do that, you'll like it to end up turning a true statement into a false statement or vice versa. Similarly, if I take there exists x, e of x and o of x, we do it with an an this time and I'll take exists an x, e of x and exists an x, o of x. Again, I've put that extra bracket in there. Can't stop myself doing that. What have I got? This says there is an x which is both even and odd, which is false. What does this one say? There is an x that's even and there is an x that's odd. That's true. So the same thing happens with exists and conjunctions say. You can take an existence statement like this and in this case it's false but if you pair them up in this way, it's true. In other words, be very careful when you're reasoning with quantifiers, conjunctions and disjunctions. You can't simply take things inside the way you sometimes do in arithmetic. If you think about what these things mean, you're not likely to run into that difficulty. But if you start treating these as symbolic expressions to manipulate, then things could go badly wrong. Time for a quiz. Okay, well we've just seen that this expression something like for all x a x or b x is not equivalent to for all x a x or for all x b x. The example we looked at was whether numbers are even or odd. And it was a case that for all natural numbers, the number is either even or odd. But it's not the same as saying that all natural numbers are even or all natural numbers are odd. In the case of even or odd for natural numbers, this is true, but both of these disjunctions are false or the disjunction is false. Okay, what about this? Instead of having disjunction, we've got conjunction. Are these equivalent or not? What do you think? Well the answer is yes. They are equivalent. Now you could argue this in an abstract form, but let's just look at an example that's pretty illustrative. Let's take something like all athletes are big and strong. Okay, so x denotes athletes, a means big, b means strong. So that says all athletes are big and strong. This would say all athletes are big and then we'd have all athletes are strong. It's pretty clear that in the case of the example, that is equivalent to the conjunction of those two. If all athletes are both big and strong, then in particular they're all big and they're all strong. And conversely, if they're all big and they're all strong, then they're all big and strong. In other words, when you've got universal to quantification, if it's combined with a conjunction, then you get the equivalence of these two forms. But if universal quantification is combined with a disjunction, they're not necessarily equivalent. I mean, this isn't a proof. This is just an example. But if you take that example, you should be able to come up with a simple little logical argument showing that that implies that and conversely that implies that. So the answer is yes in this case. Let's look at another variant. Well, again, in this case, when we had even and odd numbers, that showed that these are not necessarily equivalent. But what about this, where we have an existential quantifier combined not with a conjunction, but with a disjunction? Do we have equivalence in this case? What do you think? Again, the answer is yes. These are in fact equivalent. What would be a good example? Oh, let's take something like there is a player. Let's keep with athletic examples. There are player who, let's say, who is a good attacker or a good defender. Okay, so there is a player who is a good attacker or a good defender. Is that the same saying? Let's just simplify it a little bit. There is a good attacker or there is a good defender. Okay. And we're going to disjain those. So if there is a player who is a good attacker or a good defender, one or the other, then whichever one it is will be here. If the player who is referred to here is going to either be a good attacker or a good defender, so one of those two is going to be true, which means that the disjunction is true. So if this is true, then that's true. And conversely, if there's a player who's a good attacker, then that player is a good attacker or a good defender, in fact a good attacker. Or if it's this one, then that's true. So we have an implication that way with a disjunction and we have an implication that way with a disjunction. The equivalent. Again, this isn't a proof. This is just an example to illustrate the fact. But you shouldn't have any difficulty taking that example and turning that into a simple little argument. At least I hope you won't have any difficulty doing that. Turning that into a simple argument to show that that implies that and that implies that. So in the case of existential quantification, if we have it with a conjunction, then we don't get equivalents. But if we have it with a disjunction, then we do have equivalents. And if you think about what's going on here, the fact is that for all is like, and let me put quotes on here because this is a sort of a specialised use. For all is like conjunction. And exists is like disjunction. Because for all says it's something's true for all and conjunction means that all of the conjuncts have to be true. This simply says there's at least one and this says there's at least one. So that's all about all things. And that's about at least one. So we get good behaviour. What do you call it? It's good or not? You get this nice behaviour if we have exists with disjunction because exists is essentially a disjunction sort of thing. And for all is a, that was on the previous slide, for all is really a conjunction, sort of thing. And it's when they're mixed up, when you get the disjunctive thing here and the conjunctive thing here that things fall apart. Okay. Well, there you go. How did you get on with those two parts of the quiz? And this is the last point I want to make about quantifiers. Suppose we're having a discussion about, oh, let's say the real numbers. So if we have variables x, y and z, they're assumed to denote real numbers. The domain of quantification will be the set of real numbers. And in the course of the discussion about the real numbers, we want to talk about rational numbers, or particular rational numbers. And then we might want to mention that there's a rational number x. Well, there's one way we can do that. We can talk about the rational number in this context. We can say there is an x in q. Or we can say for all y y q. Or if I want them to talk about a natural number, I can say for all n in the set of natural numbers. So by introducing explicit domains of quantification in this way, in the course of discussing real numbers, I can restrict attention to particular rationales or natural numbers or whatever. And so in the course of an argument, I can actually use multiple domains of quantification. Well, in examples like this where we're talking about different sets of numbers, this is fine. But in other situations, it really doesn't make sense to introduce multiple domains of quantification. For example, suppose I'm going to be talking about, oh, let's say animals. So suppose the domain of quantification is the set of animals. And then I might want to say something like every leopard has spots. Well, how would I do that? Well, I suppose I could say for all x in the set of lepids, x has spots. But in the next sentence I might want to talk about tigers, and then giraffes, and then who knows what. And very soon I could have a whole range of different domains of quantification floating around. And that's not very clean. It's not very nice, and it's certainly not very sensible. Because the discussion isn't really about lepids, or tigers, or giraffes or whatever. It's about animals. And if the discussion's about animals, then the set of animals should be the domain of quantification. So instead of writing something like that, what I should write is for all animals, if that animal is a leopard, then it has spots. That allows me to say something like there is an animal which is a horse and has spots. There's a spotted horse. Or I could say for all x, if x is a tiger, then x doesn't have spots. The point is that when I'm talking about animals, then when I refer to specific animals, I'm not changing the domain of quantification. I'm still talking about animals. I'm just describing properties of particular kinds of animals. The discussion is still about animals, so the domain of quantification should be animals. It's different here, because the domain of quantification, arguably, is the set of real numbers, or the set of all numbers. And then, because we have these standard subsets, it's okay to introduce things like this. You don't have to. You could actually use something analogous to this. But in this kind of situation, yes, if you're talking just about real numbers, then you really don't want to be introducing particular sub-dimage. You should use the analogue of these kinds of expressions and talk about if it's a rational and this, if it's a natural number than this. But if this is really a discussion about all numbers, not just the real numbers, then these are just particular categories of real numbers and natural numbers and so forth. And you can have sub-domains. You can have different domains of quantification. This isn't an issue of right or wrong. It's an issue of whether it's sensible and helpful to work in a certain way. And using separate sub-domains of quantification is fine if they are natural domains and it makes sense to talk about it that way. But the idea of a domain of quantification is that tells us what it is we're talking about. If we're talking about animals, then don't introduce different domains of quantification that are not animals. Even if they're sub-domains, it gets complicated. Okay, that's enough about quantifiers for now. I suggest you go away and see how you get on with assignment number six. Well for assignment six, I'm just going to do the last three questions, number seven, eight and nine. I think you should be able to do the first six on your own and check it on your own or with other students. So in number seven, we have to negate statements and put them in positive form. Part A was for all x in n, there is a y in n, x plus y equals one. The negation for that is, and I think you'll get the same answer as I do if you get it right. There is an x in n, such of all y in n, x plus y is not equal to one. You could have written that last part as not x plus y equals one, if you like, and that's equally correct. I'll just use this way of writing it. Okay, and in this one, for all x greater than zero, there is a y less than zero, x plus y equals zero. The negation of that in positive form is there is an x greater than zero and it's greater than zero. Okay? For all y less than zero, these don't change around. They stay the same way because these simply tell us what x is and why as we're looking at. And we get x plus y not equal to zero here. Okay, for part C, there is an x greater, there is an x, such that four greater than zero, negative epsilon less than x less than epsilon. And when you negate that, you get for all x, there is an epsilon greater than zero. And in this case, you're going to have to sort of split this into two. If one of two, if x is not between negative epsilon and plus epsilon, then either x is less than or equal to negative epsilon or x is greater than or equal to epsilon. I didn't put parentheses around either of these. I mean, I could have done if I wanted to be also clear, but this takes precedence over the logical operations, arithmetic expressions or inequalities and so forth. Take precedence over logical connectives because these are connectives. They connect statements about mathematics. And this is a statement about mathematics and that's a statement about mathematics. But if you wanted, you could have put parentheses around that, and you could have put parentheses around that. That would have been another way of doing it. Moving on to the last part, for all x and n, for all y in n is a z in n, x plus y equals z squared. We're not talking about whether these are true or whatever, we're just writing them down and negating them. So you get exist an x in n, exist a y in n, for all z in n, and I've just written this again, x plus y not equal to z squared, you could simply put a negation sign in front of that expression. Okay, well that's, that's, that's number seven. It's fairly straightforward. Let's move on to number eight. Question eight is our old friend, Abraham Lincoln, and the, this famous statement of his, or allegedly famous statement of his. And we need to negate that famous sentence, or let's let fx t mean you can fool person p at time t, and then his statement is this one. You can fool all of the people some of the time. There are some times when you can fool everybody. You can fool some of the people all of the time. So there are some people that can be fooled for all, at all times. But you cannot fool all of the people all of the time. Okay, let's just mechanically go through and negate that, looking at the formalism, and then we'll try to interpret the answer and, and, and express the answer in terms of English. Okay, so negate exists, it becomes for all, when it for all it becomes exists. Fpt becomes not fpt. Okay, conjunction becomes disjunction. Again, I'm not putting these in parentheses. And I'm not going to do it here because this is a, this is a, is a whole. The quantifier is binding. They're, they're very tight binding. And then we have disjunction, conjunction here, and then disjunction is here, they're, they're less tightly binding. So looking at this one, they exist becomes for all, for all becomes exist, we get negation. Conjunction becomes disjunction. And the negation here just disappears. So I've got a positive statement. Okay, so in terms of the formalism, this is fairly routine. In fact, this is such a routine thing that this is essentially algorithmic. I just went through and applied the, the patterns, the rules that I've observed happening. The interesting part of this, I think, and you weren't asked to do it, but let's do this. Let's see how we can express this in English language. And we're not going to get something quite as, as nice as this, I don't think, because we need to try and be, to avoid ambiguity as much as possible. So the first part would be the best I can think of at the moment is, let's see, at any time there is someone you can't fool. And, oops, or, because we've negated, or, okay. Let's have a look at this one. I can't really think of anything better than to say the following. For every person, you can't always fool them. Well, you, you might be able to say, well, I think if you try and swap it round, you run into sort of a American melanoma type problems. So to try and avoid that, I think I would, I would want to write it this way. And then the final clause, but that's easy, of course, that's just say you can fool all the people all the time. Okay, I'm, I'm more really happy with that. You might have different opinions on, on the best way to write them. But number two is a bit ugly, but I, I'm not sure we can make a better job of that one. Okay. In any case, we've, we've, we've negated the thing and, and this was nice and clean. Okay, now, now let's look at number nine. Well, number nine involves one of the most important and most famous, some might say infamous formulas of advanced mathematics, of university level mathematics. It's this definition when students entering math, entering university to study mathematics for a math major, it's this definition that, that usually causes them the most problems during their first year. In fact, I think it's this definition that probably is responsible for more math majors giving up mathematics in their first year at university than anything else. It's, it's a really tricky thing to understand as, as many of you have noticed and I've seen from discussions on the, on the forum. Figuring this out is really hard. This was hundreds of years of effort starting with the invention of the calculus by Newton and Leibniz in the, in the 17th century. It took a long time, you know, 700 years before mathematicians were able to figure out the notion of continuity and come up with this definition. This was late 19th century that this was done and it was a, it was a tricky thing. Negating it is relatively straightforward actually because we've, we've got rules for doing them and so when you negate it, what you get is that for all becomes exists and the, it's still a greater than. We're still talking about positive numbers epsilon. The exist becomes a for all. The for all becomes an exists and then there's a bunch of stuff that's in a bracket here. This is a conditional, an implication and so when you negate it, you get the antecedent conjoined with the negation of the consequence. So here's the antecedent. Again, I haven't put parentheses around that because this is a mathematical statement and that, that takes precedence over the, over the conjunction or disjunction. These are just connectives. They're operators that bring things together. So that's a piece on its own, that guy, conjoined with the negation of this guy. Okay? And the negation of it being less than epsilon, is it greater than or equal to epsilon? Okay, so, so that's the negation and that's relatively mechanical to do that as long as you sort of pay attention to preserving things that need to be preserved, changing for all to exist and negating a conditional. The interesting question is what on earth does this thing mean? You know, you can read it through as I just did for all x greater than zero. There's a delta greater than zero such that for all x yada yada yada. What does it mean? Well, this is capturing in symbolic language in an algebraic formalism. It's capturing something geometric. So let's see what it's capturing. Let's look at the original definition of continuity. It's about functions. So let's look at a function this way. I'll draw the real line vertically. This is a real line. And I'm going to draw the real line vertically here. So this is the real lines instead of writing horizontally as we usually do. I'm going to write it horizontally. And the function f is going to take numbers here to numbers here. Okay? So somewhere here we've got a, f applies to that and it gives me f of a. Okay. Now we try to capture the notion of continuity today. That means when we go slightly to the left or right of a, and left or right is up and down the way I've represented it, then the numbers don't sort of have a discontinuity. And the way to capture that, it turns out, and chances are very, very high that you're not going to follow this the first time. It's going to take you weeks if you need to to master this. But here's what this formula says. It says the epsilons are going to work on here. It says let's take an epsilon here. And let's look at f of a plus epsilon and f of a minus epsilon. So I'm going to take an epsilon interval around f of a. And what this definition says is that given an epsilon and one of these intervals, I can find a delta. So here's a plus delta and a minus delta. So starting with an epsilon, which gives me an interval here. For any one of those, I can find a delta which gives me an interval here such that, now let's look at this. Any x in this region, because this says that x is within delta of a. So any x in this interval gets sent to an image f of x in here. So it's saying in order to make sure that all the values of the function are in this interval, I can find an interval around a such that they're all sent into here. So if I want to hit the target, imagine this as hitting the target, like throwing data to that board. If I want to hit the target within a specified accuracy of a, of f of a, I can always do it by starting out within an interval of a. So to get within a given interval around f of a, I can always find an interval around a that does it. So everything from here gets sent into here. And if you think about it long enough, you'll realise what that means is that the function is continuous at here. There's no, no, no jumps. And to try and understand that, let's look at what this guy means in terms of a diagram. I'll do the same thing again. I'll draw the real line. And I'll draw the real line. And here's a. And here's f of a. Now the negation says there is some epsilon. In, in the previous case, this was happening for all epsilons. You could find a delta. In this case there's a fixed epsilon that we can find. And we look at the interval f of a minus epsilon. Okay. What it says is that there is an epsilon such that no matter what you take here, no matter which one you take here, here we found one of these guys. We said take any interval here, we can find an interval here. Here we're saying there is an interval here for some epsilon such that no matter what delta we take here, no matter how small you make this delta, no matter how close you are here, that was a point that gets sent to that. You can always find a point in here that gets sent outside of there. Maybe sent that way, maybe sent that way. So the sum epsilons here that no matter what you do here, no matter how close you are to a, something gets sent out here. In other words, there are points really, really, really close to a that gets sent outside this region. So there's no way that you can, you can get all points in here. A gets sent to that, but arbitrarily close to a, there are points that get sent away. So there's a discontinuity because only a gets sent close to a. When you get, well some of the points may be, but no matter how close you get to a, you'll find points that are sent further away. So that's a discontinuity. Things jump. There's a sort of jump from there to things here or here. Now I, the chances are high that if this is the first time you've seen this, that explanation is going to be hard to follow. This is really difficult to understand. This is extremely, extremely hard. And the goal of this course actually isn't for you to understand this. The goal of this course is to give you the machinery so that you're now able to take a whole semester course on real analysis during which you should be able to understand that. So I wouldn't worry too much if you can't understand this. The point is if you could understand the first part and understand how to deal with the formalism, you now have the machinery you need to understand this, this very deep conceptual definition from within mathematics. Okay, and to have reached that stage in four weeks, I think this is pretty, pretty remarkable. The hard part is to come, and for those of you that want to go on and study more mathematics sooner or later and hopefully sooner, you're going to need to master this definition of what it means. Okey dokey, well as far as we're concerned, we're done with this one. Well for question one on problem set four, we have to choose which one of the following is equivalent to this expression. Okay, we're negating a universally quantified statement in which there's a conditional. And we're told that there's only one of these, one of these answers is correct. The correct answer, let me just jump to the correct answer and then see what's wrong with some of the others, is part C. The not for all becomes an exist, and the negation goes into this part. When you negate a conditional, you end up with the antecedent together with the negation of the consequence. And when you negate a disjunction, the negations filter in and the disjunction becomes a conjunction. So if you follow the rules about, well they're not rules, they're, I mean they sort of are rules, but we, you know, I recommend you not to think of them as in terms of rules because that's not really getting at what this course is about. But there are certainly patterns of activity that you can get to, get to recognise. Then what happens is universals become exists. You have the truth of the antecedent in place of an implication of a conditional, you have a conjunction. And then when you negate these guys, the negation applies to each one and that becomes a conjunction. Okay, but as I just indicated or I just referred to, really what I want you to do is concentrate always on why it is that you get that behaviour. Why does this give you px and not that? Why does negating this give you a conjunction there? So it's all about understanding. If you simply learn to apply the rules, you really don't have a useful skill, okay? Computers are good at applying rules. That's what they do. That's all they can do. People can go much beyond that. Okay. What's wrong with somebody others? Well, the first one is just hopeless. I mean, there's just nothing remotely like that. If you've got that as your answer, then either you are having a temporary aberration or you really, really, really haven't got the issue of this. The others, there are reasons why people could go wrong and the others were put in because of the mistakes that people frequently make. Okay? In the case of this one, the negation is in the wrong place. When you negate a conditional, the negation doesn't come together with this one. The negation should come in front of here and in front of there. So it's just mixing up where the negation comes in. Okay? But otherwise you apply. So it's like applying, you're basically applying the right sort of, let me call it rule. You're doing the right thing but you've got one step out. It's like getting a negative sign, a minus sign in the wrong place in an equation. Looking at this one, well, everything went fine except you forgot to change disjunction to conjunction. But everything else was fine. And in the case of this one, you forgot to, that that should be conjunction. And everything else was fine. Okay? So in cases B, D and E, there was just one thing wrong. And so it's even possibly just did that by a slip. I mean heaven only knows, you've seen me make slips often enough in the lectures and the tutorials where I write the wrong thing down or whatever. We have to keep a lot in our minds when we're doing these things. And frequently what our hand does isn't what we're thinking it's doing and sometimes what we say isn't what we think we're doing. Mathematics is like that. When you're really focusing on the mathematical concepts and the heart of it, you can make slips with the writing and with the words you use. I do it all the time. And we all do. That's just part of thinking mathematics. It takes a lot of concentration to focus on the mathematics. And the everyday things like writing and using words tend to miss out on that because our mind is focused on the content. Okay, let's go on and look at number two now. So for number two, let me just begin by reading this formula out in everyday English, or at least stilted everyday English. It says for every person P, there's a person Q and a game T such that P beats Q in game T. So more colloquially, everybody beats somebody in some game. Everybody beats somebody in some game. Everybody beats somebody in some game. Well, which one say the same as that? That one certainly does. That means the same as everybody beats somebody in some game. It doesn't mean the same as that. It's not that one. That's not a possibility. For every player there's another player, they beat all the time. That's not quite the same. If there was a for all there, for all T, then whenever they play, P would beat that person, that person Q. So that's not quite. You would need a different quantifier. That's just some game. If there was a for all there, you would have it in all the games. There is a player who loses every game. No, no, that's about losing. That's a player who. Well, actually, while we're on this, what would this look like? If I wanted to put that in a formula, I would put that with the following. I would say there is a Q such that for all P and for all T, W, P, Q, T. In a minute I'll say why I said start because I was looking ahead to what would be wrong with this. But let's just read it so it says there is a player such that for all players and for all times, P beats Q in that game. So it sort of says the player loses every game. The problem is when you take for all P here, one of the P's that are going to crop up is Q him or herself, okay? And a player can't beat themself. So this doesn't quite work because the P could include the Q itself. In fact, the one of the P's will be the Q. Okay? And it's also, well it depends. This says that player is going to, yeah I think that would be okay. You've got to be a little bit careful actually. It depends how you interpret the thing. For all T, T just goes over games of tennis. Whereas we're just talking about games of tennis in which people win. So you'd have to sort of think about it and decide whether you need to put clauses in here to make it clear that you are only looking at T's where they play together. There'll be a similar issue here. Again, you could start with, if you wanted to do this one, you would start with saying there is a P, let's see, for all Q, for all T, W, P, Q, T. The only difference is here we had QP and here we've got PQ. But you'd have the same issue. Among the Q's is the possible P and you'd have to be careful to take account of the fact that the T itself ranges over all possible games and you only want to be making a statement about the games that they play together. And you could have a tremendous, you could put clauses in. There are various ways out of that. If I thought this was a big deal out of thought about it before I started to record this little piece. But I think we've solved the problem in any case. That one came up the first time. That was the one that has that meaning. None of the others do. So put those as no's and we've solved that one. Okay? Let's go on to number three. Now normally when I'm going through examples in classes, I don't make deliberate mistakes. I don't need to because I make plenty of real mistakes when I'm going through mathematics in any case. But this time I'm going to make a deliberate mistake and I'm going to do it to emphasise the issue that's going to come up here. Okay? So what I want to do is go through these three and give you answers to all of them. But one of them is actually going to be a wrong answer and then we'll correct it. So this one, actually if you remember the previous question, this is the one that we saw that means everyone wins a game. Okay? For all players P, there's a player Q and a time T. So it's a P beats Q at that time. What does this say? Everyone beats, well it's almost the same, right? Except instead of playing every player beats one player, you say everyone beats everyone else at some time in some game. Okay? Because it's all pairs of players P and Q. And this one, it's essentially, well actually what's the difference? That was for all P there exists a Q. This is for all Q there exists a P. So this doesn't say everyone wins a game. This one says everyone loses a game. Okay? Seem plausible? Well, one of these is actually not possibly true. Let's just see. Could that be true? We're talking about, whoops, there's a typo there. That should be possibly. Okay, let's correct the typo. I've got it right up there. Okay, let's see which one cannot possibly be true. Okay, that's not the issue is here. Well, that could possibly be true. Okay, that's okay. That could be true. Everyone beats everyone else all at a time, at some time. Well, everyone loses a game. Well, that's certainly possible. That's okay. Come into this one. Can this, I mean, is it possible for this to be true? It's certainly possible for that to be true. It's certainly possible for this to be true. We're talking about the real world here when people are playing tennis. Is this one the one that can't possibly be true? No, there doesn't have to be one. You could have said that none of these are the case. In fact, if you've got this far, you probably did say they're all okay. But actually, this one's not okay. Because when you say for all P and for all Q, that includes the case of Q and P being equal. That would mean everybody has to beat themselves at some time amongst everything else. So a player cannot beat every other player because the problem is no player can beat herself or himself, whichever you want. Okay? No player can beat herself. So that can't possibly be true. So this is the guy that can't possibly be true. And it can't possibly be true because the Q and the P have to be equal. What you would have to say if you wanted to make that something that could be true was you'd have to say for all P, for all Q, if P and Q are different, then there is a T, so it should W, P, Q, T. Then you'd be okay because you'd say for all pairs P and Q providing their difference, then at some time, at some game, P does beat Q in that game. So that would be a way of making it possible. But as it stands, that cannot possibly be true because players cannot beat themselves. Okay? So the issue was whether you can take the formula doesn't correctly tie them into what they say about the real world. It wasn't mathematics that was deciding these. It was the real world. And in the real world, players cannot beat themselves. Well, actually in a figurative way, players beat themselves all the time, but in the sense we're talking about here, that's not the case. Okay? Now let's move on to question four. Okay, now this expression is a colloquial expression, and so when I set this question up, I realised that for non-native speaking English students, this would be, this would be somewhat challenging. So what I did was I only gave you options that you should be able to distinguish between the logical structure. The reason I like to give these kind of examples is because they capture an awful lot of social and cultural knowledge and there's an interesting challenge in capturing that kind of thing in formalisms. But to help you along, I gave you three options that you should be able to sort out just on the basis of logic if you know what being a lover means in this case. So being a lover means you're in a mutual relationship, which means you've got, you love someone and that person loves you. But if you look at this one, this says for this person X, so these are all about a person X, they all talk about some person X. This doesn't say that person's a lover, this part says that person is in a relationship with everybody else. So this part because there's a universal quantifier that says X loves Z and Z loves X. So this person X is in a loving relationship with everybody. Well, that's nonsensical, so we can forget that one. So it comes down to these two because in each of these, it says the person is in a loving relationship. X is in a relationship with some Z and it's mutual. X loves Z and Z loves X. The same clause here. So the choice is between A and C. Well, let's look at what A, what A says. That says for all X and for all Y, if the X is in a loving relationship, then Y loves X. Now, the Y doesn't come in here so that the for all Y is to do with this part. So it says take any person X, if that person is a lover, then every person loves them, if they know them. So that actually is the correct one. Okay? All people love a lover. Of these three, that's the one, you know, you could argue about whether that's the absolute best interpretation, but out of these three, it's certainly a correct one. And so it's the one here. Let's look at this one. This one is a little bit different because it's got for all Y in here. So it says for all X, if X is a lover, then when it doesn't say then it says and for all Y, L, Y, X. So the for all actually applies to this part as well. So what this really, what follows from this is that for all X and for all Y, L, Y, X. In other words, everybody loves everybody. Well, that's not the case because this isn't conditional on being a lover. This just really says that's the case and that's the case. So part of this is saying that for all X and for all Y, L, Y, X. Well, that's not the case. I mean, it's not the case that everybody loves everybody. The world would be a nice place, I guess, if that was true, but it's not true. So it can't be that one. So we've eliminated this one because it doesn't capture, it doesn't use the fact that it's being a lover. And we've eliminated this one because it basically just boils down to saying everybody loves everybody and that leaves this one. And this is definitely one good interpretation of everybody loves a lover. Okay? So we were able to reason that one out and I would hope that even without the detailed understanding of what this means in English, there's only one of these that will stand up to analysis. Then after all, the idea is to sort of look at how the formalisms capture relationships from the real world. Well, for question five, we have to find which statements are false. Okay, so let's just hear what they say. For all X, for all Y, for all Z, if X is less than or equal to Y and Y is less than or equal to Z, then X is less than or equal to Z. That's true. It's actually known as the transitivity of the order relationship. Okay, if X is to the left of Y and Y is to the left of Z, then X is to the left of Z. Okay. So that one's true. What does this one say? For all X, for all Y, if X is less than or equal to Y and Y is less than or equal to X, then X is equal to all. That's true. Okay? The only way you can have X less than or equal to Y and Y less than or equal to X is if they're actually equal. What about part C? For all X, there is a Y. Well, that's certainly the case because given an X, you can take Y equals X, and then you'll have X less than or equal to Y and Y less than Y. So that one's true. It's beginning to look as though they're all true. Let's look at the last part. There is an X such that for all Y, Y is less than X, or X is less than Y. Well, you might be tempted to say, yep, given any X, it's the case that every other Y is either less than X or bigger than X. But wait a minute. Among the Ys governed by a universal quantifier is the X you start with. So if there was an X with that property, how could this happen? Because when you look at all the Ys, among those Ys would be X itself. So you would have X less than X, which is impossible. So that one's false. And it's false because the universal quantifier includes the X itself. Given an X, any X that you find, when you universally quantify over Ys, you include that X. And then that fails, and that fails. So there's the one that's false, and there is a false one. And it fails because universal quantifiers go over everything, and that means the Y, among the Ys that you're looking at, is the X that you start with. Okay? Well, for the last question on problem set four, which is question six, we have to look at this piece of reasoning that a student gave. This was actually when I first gave this course in the fall of 2012. And a student was trying to understand the Euclid's proof that the prime numbers are infinite in number. And there was a key step where you form that product when you add one, and the student wanted to verify that in fact, rain plus one was not divisible by a P, and came up with this argument. So I thought this would be a good answer to look at. It raises a number of interesting issues. So first of all, let's just see what the student does. Suppose N is divisible by P. Well, arguably the student, you could say the student doesn't begin by saying what N is, and doesn't say what P is. But I think in this context there's no need to. It's already been stated that N is an integer. It doesn't matter whether it's positive or negative, by the way. And it's already said that N is prime. So there's, in this context, I don't think there's any need to demand that the student repeats it here. What's key is that the student is writing down the assumption on which the argument is going to be based. Okay. So I'm going to say that I'll come to the various other issues in a minute. The opening I think is fine. Okay. That's good. What does the student do then? Then there's an integer Q. Should I say A equals PQ? Yes. That's a definition of divisibility. I'm not going to demand that the student spells that out because it's fairly clear what's going on. So N plus one is P percent of that's okay. Then dividing through by Q, by P. If I name this one over P. Now the student's now written a lowercase Q there. That's obviously meant to mean uppercase Q. I'm not going to worry about that. That's just a typo inviting it. Ditto here. So I'm not going to deduct anything for that. Those are just typos. I mean just using the wrong lowercase and uppercase confusion. So N plus one is not divisible by P. Okay. Logical correctness. Well in a sense this is okay, right? I mean everything steps fine. So I'm going to give three. Why aren't I giving four? I'll come to that in a minute. What about clarity? I've got to give four. This is absolutely clear what's going on. Stay at the conclusion. Yep. Conclusion stated. That's four for that one. A reasons given. Absolutely reasons are given. Yeah. I mean as I observed at the time the student sort of doesn't say this is by definition of divisibility and so forth. But you know there's a limit to how much you can write down within the context of this class and the intended audience which is the students in this class. Then this is fine. You know, it's fine to put more details in the reasons if you want to own the side of caution. But given the fact that this is absolutely clear, I think there's enough reasons given. Okay. And that brings me to the overall evaluation. And this is where I'm going to put a zero down. And this is also why I didn't give a full four for this because this is logically correct in terms of doing mathematics. But this is an argument about integers. And the integers is a number system in which you can add, subtract and multiply. What you can't do is divide. Division is not an operation on the integers. It's an operation on the natural, on the rational numbers and on the real numbers. But it's not an operation on the integers. So in going into division, this argument goes beyond the realms that we're permitted to use. So I'm giving a, I'm being generous here actually and saying I'll give you three because it's, this is mathematically correct in terms of the rational numbers. I mean I could have been tougher and I could have reduced it to two. I could have even put a zero down. But as I've said before, I'm looking for reasons to give people and take them away. We're trying to make people better thinkers, not to make them feel bad about themselves, okay? So I'm going to give credit where it's due. But strictly speaking it's not right. Here's what the students should have done. Okay, so we've got the first part. We've written n equals pq. Okay, that was okay. Okay. Now what I'm going to do is I'm going to argue by contradiction. I'm going to say suppose n plus one were divisible by p. Then there is an integer, call it r, so I said n plus one equals pr. Then we have n plus one minus n equals pr minus pq, which is p into r minus q. But n plus one minus n is one. So what I've shown is that p times r minus q equals one. That means that p, that means that one is divisible by p. But that's a contradiction. p is a prime number. So it's at least equal to two. So it can't divide into one. One isn't divisible by any, by two or three or anything. So there's a contradiction. Hence the original assumption here was false. Hence n plus one is not divisible. Notice that this is almost the same in a sense. This is equivalent to this. It's not the same but it's equivalent. Because the r here is n plus one over p. So this is the r. So I've said, I mean I'm sort of doing the same thing except here I'm not using division. I'm doing everything in terms of divisibility which means I'm doing it in terms of addition, subtraction and multiplication. All I'm using here is addition, subtraction and multiplication. I'm not using division. I'm getting around it by introducing this r, if you like. And that's a significant difference because we simply don't have division as an operation in the integers. We have divisibility which is a property that may or may not hold between two integers. But we can't divide one integer by another because division is not an operation in the integers. Okay? So I give a reasonable amount of credit for this. I think I was generous with this one because I tend to be generous. And that means the total grade is 19. Okay? Well that was the end of the problem set for. How did you get on with assignment six? If you followed everything so far and managed to do fairly well on all six assignments, you should have a good idea of the kind of linguistic precision required in mathematics. Now we can start to put that precision to use in proving mathematical statements. In the natural sciences, truth is established by empirical means involving observation, measurement and the gold standard experiment. In mathematics, truth is determined by constructing a proof, a logically sound argument that establishes the truth of the statement. The use of the word argument here is, of course, not the more common everyday use to mean a disagreement between two people. But there is a connection in that a good proof will preemptively counter explicitly or implicitly all the objections, the counter arguments that a reader might put forward. When professional mathematicians read a proof, they generally do so in a manner reminiscent of a lawyer cross-examining a witness, constantly probing and looking for flaws. Learning how to prove things forms a major part of college mathematics. It's not something that can be mastered in a few weeks. It takes years. What can be achieved in a short period and what I'm going to try to help you do here is gain some understanding of what it means to prove a mathematical statement and why mathematicians make such a big deal about proofs. First, what is a proof and why do we use them? I'll answer the second question first since their purpose dictates what they are. Proofs are constructed for two main purposes, to establish truth and to communicate to others. Constructing or reading a proof is how we convince ourselves that some statement is true. I might have an intuition that some mathematical statement is true, but until I've proved it or read a proof that convinces me, I can't be sure. But I may also have to convince someone else, and that's the second purpose of a proof. For both purposes, a proof of a statement must explain why that statement is true. In the first case, convincing myself, it's generally enough that my argument is logically sound and I can follow it later. In the second case, where I have to convince someone else, more is acquired. The proof must also provide that explanation in a manner the recipient can understand. Proofs written to convince others have to succeed communicatively as well as be logically sound. There's actually not as much of a distinction here as my words might imply. For complicated proofs, the requirement that a mathematician can follow his or her own proof a few days, weeks, months, or even years later can also be significant. So even proofs written purely for personal use need to succeed communicatively. The requirement that proofs must communicate explanations to intended readers can set a high bar. Some proofs are so deep and complex that only a few experts in the field can understand them. For example, for many centuries most mathematicians believed, or at least held a strong suspicion, that for exponents n greater than or equal to 3, the equation x to the n plus y to the n equals z to the n has no positive whole number solutions for x, y and z. That was conjectured by the French mathematician Pierre de Fermat in the 17th century, but it wasn't finally proved until 1994 when the British mathematician Andrew Wiles constructed a long and extremely deep proof. Over the centuries it became popularly known as Fermat's Last Theorem. Since it was the last of several mathematical statements Fermat announced that remained to be proved. Most mathematicians, myself included, lacked the detailed domain knowledge to follow Wiles's proof ourselves. But it did convince the experts in the field, the field by the way's analytic number theory, and as a result Fermat's ancient conjecture is now regarded as a theorem. Fermat's Last Theorem is an unusual example however. Most proofs in mathematics can be read and understood by all professional mathematicians. Though it can take days, weeks or even months to understand some proofs sufficiently to be convinced by them. I've chosen the examples in this course to be understood by a typical student in a few minutes, or possibly an hour or so. Examples given to college mathematics majors can usually be understood with at most a few hours effort. Proving a mathematical statement is much more than gathering evidence in its favour. To give one famous example, in the mid-18th century the great Swiss mathematician Leonard Euler said he believed that every number beyond two can be expressed as a sum of two primes. This property of even numbers had been suggested to him by Christian Goldbach and became known as the Goldbach conjecture. It's possible to run computer programs to check the statement for many specific even numbers, and to date, 2012, it's been verified for all numbers up to and beyond 1.6 quintillion. Most mathematicians believe it to be true, but it's not yet been proved. All it would take to disprove the conjecture would be to find a single even number n for which it could be shown that no two primes sum to n. Incidentally, mathematicians don't regard the Goldbach conjecture as important. It has no known applications or even any significant consequences within mathematics. It's become famous solely because it's easy to understand, was endorsed by Euler and has resisted all attempts at solution for over 250 years. Whatever you may have been told at school, there's no particular format that an argument has to have in order to count as a proof. The one absolute requirement is that it is a logically sound piece of reasoning that establishes the truth of some statement. An important secondary requirement is that it's expressed sufficiently well that an intended reader can, perhaps with some effort, follow the reasoning. In the case of professional mathematicians, the intended reader is usually another professional with expertise in the same area of mathematics. Proofs written for students or laypersons generally have to supply more explanations. This means that in order to construct a proof, you have to be able to determine what constitutes a logically sound argument that convinces not just yourself, but also an intended reader. Doing that's not something you can reduce to a list of rules. Constructing mathematical proofs is one of the most creative acts of the human mind and relatively few are capable of true original proofs. But with some effort, any reasonably intelligent person can master the basics and that's my goal here. Euclid's proof that there are infinitely many primes, which I gave in the first lecture, is a good example of a proof that requires an unusual insight. Let's look at it again. Here's what we did. The idea was to show that if we list the primes in increasing order, then the list can be continued forever. So we imagine that we've listed the primes. P1 is 2, P2 is 3, P3 is 5, etc. All of it was some stage Pn. And we show that we can always add another prime to the list. And to do that, we look at this number n, which we obtain by multiplying together all the primes in the list so far, and then adding one. Now this number n consists of the product of all those numbers P1 through Pn plus one, so it's certainly bigger than all of those numbers. So n is bigger than all the primes in the list. Well, if n is prime, then we know that there's a prime bigger than Pn, namely n. In which case, we can continue the list. Probably not by adding n itself. n, because it's the product of all these plus one, is going to be a lot bigger than Pn, so n is almost certainly not the next prime. But that doesn't matter. If n is prime, it shows there is a prime bigger than the one at the end of the list, and that means we can continue the list. The alternative is that n is not prime. In which case, there's a prime q less than n that divides it. But none of the primes in the list can divide n. Since if you divide n by any of those primes, you're left with that remainder one. If you try to divide P1 or P2 or any of these primes into this number n, it gets swallowed up by this part, and then there's a remainder of one. Okay? So q has to be bigger than Pn. Those are the first n primes, so if q's not equal to one of those, it must be later on in the list. In which case, we've shown again that there's a prime bigger than Pn, and the list can be continued. Again, this particular q that divides n is not necessarily the next prime, but as before, that doesn't matter, showing that there is another prime is all you need to do because then you can take the next prime, whatever it is, and add it to the list. Either way, either if n is prime or if n is not prime, either way, there's another prime to add to the list. It follows that there are infinitely many primes, and the theorem's proved. There are two creative ideas in this proof. The first one is here to show that if we list the primes in increasing order as P1, P2, etc., then the list can be continued forever. So the first creative idea is to think about listing the primes and showing that the list can always be continued. The second creative idea is this one. Defining this number n in such a way that it guarantees that we can always find another prime. I would say that this idea is one that most mathematicians would come up with sooner or later. It's a fairly obvious one. This one is genius. Okay. This is true genius. Let me give you another example. And this time I'm going to prove that result I promised earlier that the square root of 2 is irrational. And I'm going to write it the way mathematicians typically do when they write up results for publication in professional journals or in books. Namely, we call it out by calling it a theorem. So in mathematics a result that's sufficiently significant or important that it's worth mentioning as such is called a theorem. In this context let me mention there's another word we often use called lemma. And a lemma is a result which is worth calling out for some reason but doesn't quite merit the status of being called a theorem. It's actually if you like a little theorem. Okay, the next thing a mathematician typically does is indicate that we're going to begin the proof. Okay? So this is just part of the way mathematicians lay things out. We specify the theorem and then we say we're going to give the proof. You don't have to do it this way. It's just a convention. The essence of being a proof is what comes next. Proofs are about their logical structure, not the way we write them down. I'm going to begin by assuming on the contrary that the square root of two were rational. Now if you've never seen the proof that the square root of two is a rational before, this first step is going to seem pretty mysterious. Why do I begin by assuming the opposite of what I'm trying to prove? Well by the time I get down here, you'll see why. The reason this is a great example is in about six or seven lines, I can make it clear why I'm doing something right in the first step. Okay? Well in that case, if square root of two were rational then there were natural numbers pq with no common factors such that the square root of two is p over q. Remember a rational number is one that can be expressed as the quotient of two integers. In the case of a positive number it would be two natural numbers. And we can always pick those natural numbers or those integers to have no common factors. In other words, when we write a rational number as a quotient, we can always cancel out any common factors and express it as a quotient where the two numbers themselves have no common factors. Again it might seem a little bit mysterious why I'm being particular about any common factors but as with the first step by the time I get down here it will be clear why I'm doing this. Well squaring that equation gives me two equals p squared over q squared. Rearranging I get two q squared equals p squared. I multiply both sides by q squared that gives me a two q squared on the left and then when I multiply the right by q squared it cancels that q squared and I'm left with a p squared. So p squared is even. It's equal to twice something. Hence p is even. Why? Because the square of an even number is an even number, the square of an odd number is an odd number. So the only way I could get the square of a number p to be even is if the number p itself is even. Even squared is even, odd squared is odd. So p is two r for some r. I'm now going to take this equation p equals two r and use it to substitute back in this equation. So I take this equation, I've got two q squared equals p squared and p equals two r. So I've got two q squared equals two r all squared which is four r squared. Well forget the middle term now. I've got two q squared equals four r squared. I can cancel the two. Well if q squared is two r squared, then q squared is even. But exactly as happened before with p squared, if q squared is even, then q is even. Aha. You see what's happened? I've deduced here that p is even. I've deduced here that q is even. So p and q are both even. But they can't be because we assume p and q had no common factors. If they're both even, then they have two as a common factor. But this is impossible since p and q have no common factors. Well the logical reasoning here, the algebra, the arithmetic, is all sound. That's absolutely, everything's perfectly sound. How can we have arrived at an impossible conclusion by a piece of sound reasoning? Well the only thing that can possibly have gone wrong is we began by making a false assumption. Remember we began with an assumption. The only way we can have reached a false conclusion by a valid argument, and this is valid, if you don't believe me, go and check it for yourself, there's not many steps. See if there's anything wrong in any of these steps, there isn't. If we reach a false argument by a logical argument, then we must have started with a false assumption. Hence the original assumption that the square root of two were rational must be false. Hence, square root of two must be irrational. And when I was at school, teachers used to insist that we write QED at the end of a proof. Quadirat demonstrandum, which is Latin. It is actually not a bad idea to indicate when a proof ends. And mathematicians have different ways of doing it. Sometimes they write a little box at the end. How you express it, how you lay it out on the page is not that critical. I mean, the idea is to lay it out in a way that can be followed. What makes a proof a proof isn't the fact that you call it a proof, isn't the fact that you end in a QED. It's the logical flow of the steps. It has to be logically precise and you have to be able to follow it. But there's the proof. The reason this is a good example to give is it's short, it's concise, and all of the steps are simple arithmetical steps. It's very easy to follow every step, and yet when you follow this small number of simple steps, you've proved a significant result. In fact, this result when it was first discovered by one of the Pythagoreans in Ancient Greece was dramatic. It changed the course of Greek mathematics because until then they felt that quotients of integers were sufficient to measure any length. But square root of two is the length of the diagonal of a right angle triangle whose sides measure one. And when this result was proved by one of the Pythagorean mathematicians, it showed that quotients of integers were not sufficient to measure all lengths in geometric figures. And that changed the course of Greek mathematics and subsequently the rest of mathematics. It was extremely dramatic. Incidentally, there's a story you'll read about in books and on websites that say that this was discovered by a young mathematician and that the Greek mathematicians of the Pythagoreans were so annoyed and so scared that this would kill their career and their profession that they threw him overboard. There's absolutely no evidence whatsoever that that was the case. It's a great story, but like many stories it's probably not true. Okay, but in any case we've now shown that the square root of two is irrational. And there we go. I mean this is really a remarkable result, very short, very elegant. Incidentally, when mathematicians talk about aesthetics, when they talk about an elegant proof or a beautiful proof, this is the kind of thing they have in mind. Not that it looks beautiful the way it's written out, in fact I've just worked through it and it doesn't look particularly beautiful to look at, but the logical structure is beautiful. Every step counts, the result is established and every step can be understood fairly straightforwardly. The complexity doesn't come because there's deep results involved, deep facts, deep concepts. The proof works because of the structure, the logical structure. Okay, so now you know why root two is irrational. Let me say a little more about that proof. It's an example of what mathematicians call proof by contradiction. And proof by contradiction is a general method that works as follows. You want to prove some statement phi. You start by assuming not phi. You reason until you reach a conclusion that's false. Often by deducing both psi and not psi for some statement psi. For example, in the proof that the square root of two was irrational, we proved that P and Q have no common factors, and yet we knew that P and Q were both even. So we have a statement and it's negation. Or not strictly speaking it's negation, but close enough for these purposes. This is not an factor rendering of psi and not psi. But the point is we've deduced two contradictory things. It cannot be the case that they have no common factors and that they're both even. But a true assumption cannot lead to a false conclusion. Hence the assumption not phi must be false. Remember, we began by assuming not phi. We reach a false conclusion, a contradiction, a true assumption can't lead to a false conclusion, so the assumption must be false. In other words, phi must be true. If not phi is false, then phi is true. Well we begin by wanting to prove some statement phi, and we're going to end up with phi. And in the middle, we reason by assuming the contrary. So here's the contradiction proof going on in here. There's phi, there's phi. In here, we're reasoning with not phi. We start by assuming not phi and we reach a conclusion, a false conclusion, and conclude that that's false. Let me say a little bit more about this. We can look at proof by a contradiction in terms of truth tables. What can we conclude from a proof of theta yields psi where psi is false? In the case of the proof of root two being rational, the theta was the statement square root of two is rational, and the psi was the false statement that we never actually wrote out in full, that P and Q are both even and have no common factors, which is false, because there's no such, there are no pairs of integers which are both even and have no common factors. So in terms of the proofs by contradiction, the theta is the assumption you make at the start of the proof and the psi is the false conclusion you reach at the end. So I'm going to use truth tables to understand how it is that starting with an assumption and reaching a false conclusion leads to concluding that the assumption was false. And since the assumption was counter or contrary to the thing we're trying to prove, that would amount to proving a theorem. Okay. So I'm going to write down the truth table for theta yields psi or theta conditional psi. Right, the truth values down T, T, T, F, F, T, F, F. And we know what the table looks like here, it's T, F, T, T. Well if we've carried out a proof of this, that means this thing is true. So we can forget that line. We're only interested in what happens when this thing is true. So we've carried out a proof that this implication, this conditional is true. So it's one of these three. In this case, however, the psi is false. There's the psi false. Here psi is true. Here psi is true. Here psi is false. So this is the only line in the truth table which fits this. We have this being true, which means it's one of these three, and we have psi being false. That's the only possibility. In other words, theta is false. So when in a proof by contradiction, you make an assumption, a counter assumption, an assumption counter to what you're trying to prove, and you carry out some reasoning to deduce a false conclusion, a contradictory conclusion, or a contradiction. So you've established in your argument that this thing is true. This says that the argument is valid. Then the conclusion from having reached a false conclusion, the big conclusion, the global conclusion, is that your original assumption is false. Okay, it's a little difficult to explain this in words because we have to keep talking about truth and falsity and assumptions and conclusions and assumptions and sub assumptions and so forth. But if you go through the original proof that square root of two is irrational once more, and then think about it in terms of what I just said a moment ago about proof by contradiction in general, and then think about it in terms of this truth table analysis, and hopefully you should be able to understand why proof by contradiction work. It's a very clever idea, and they used a lot in mathematics, and I've just noticed I've misspelled truth tables, so let me put the L in there. Okay, there we go. That was a fairly lengthy discussion of such a short argument, but I know from many years of experience the beginners find the root to proof hard to really understand. You may think you understand it, but do you really? Let's see if you can produce a similar one. Try to prove that the square root of three is irrational. You should definitely try to do this exercise, but be prepared to spend some time at it. Go on, give it a try. I'll say it again. This course is about the process of thinking, not about getting results. You can use the thinking abilities you develop in a course like this to get results in other courses and in other situations in your life. It doesn't matter if you don't get the root three proof, you'll have benefited from trying. Truths by contradiction, which we use in the root two theorem, are a common approach because they have a clear starting point. To obtain a direct proof of some statement phi, you have to generate an argument that culminates in phi, but where do you start? The only way to proceed is to try to argue successively backwards to see what chain of steps ends with phi. There are many possible starting points, but just one goal, and you have to end up at that goal. That can be difficult, but with proof by contradiction, there is a clear starting point, and the proof is complete once you have deduced a contradiction. Any contradiction with such a wide target area, that's often a much easier task. The proof by contradiction approach is particularly suited to establishing that a certain object doesn't exist. For example, that a particular kind of equation does not have a solution. You begin by assuming that such an object does exist, and then you use that assumed object to deduce a false consequence of a pair of contradictory statements. The irrationality of the square root of two is a good example, since that states the non-existence of two numbers, p and q, whose ratio is equal to root two. Even though there is no cookie-cutter template approach to constructing proofs, there are some guidelines. We just met two. Proof by contradiction is often a good approach when there's no obvious place to start. And proof by contradiction is a useful method to prove non-existence statements. Of course, you still have to construct a proof. You've simply replaced a narrow goalpost with an unclear starting point by a much wider one with a known starting point. But like Robert Frost's fork in the trail, that choice can make all the difference. There are a number of other guidelines. I'll tell you some, but do bear in mind that these are not templates. As long as you continue to look for templates to construct proofs, you're going to encounter significant difficulties. You have to start each new problem by analysing the statement that you want to prove. What exactly does it say? What kind of argument might establish that claim? Let's look at another guideline. How might you go about proving a conditional statement? One of the form A implies B. We want to prove a conditional phi yield psi. When we know this is true if phi is false, so we can assume phi is true. Why do we know it's true if phi is false? Well, that was part of the definition of the conditional that we developed using truth tables. So to prove it, we assume phi and deduce psi. This of course confirms the point I made earlier that despite its strange definition, its counterintuitive definition perhaps, the conditional really does capture genuine implication because in all actual practical examples, when you try to establish a conditional, what you do is you assume phi and you deduce psi. And this is genuine implication. This says that psi is following from phi. For example, let X and Y be variables for real numbers and prove the following. If X and Y are rational, then X plus Y is rational. Okay, this is not a surprising result. This is nothing deep. I'm focusing not on the result but on the method I use to prove it. And so I've deliberately chosen an example that's extremely simple so that we can look at the process of reasoning that's involved in proving a conditional. So step one, assume X and Y are rational. In that case, there are integers P, Q, N and M such that X is P divided by M and Y is Q divided by N. Where in that case, X plus Y is P divided by M plus Q divided by N, which equals PN plus QM divided by MN. Hence X plus Y is rational. Okay, as I mentioned a moment ago, there's nothing surprising here. It's almost not a proof at all. It's just really adding two things together. But it actually is a proof because it has the right structure of a proof. Here's what we did. We began by assuming X and Y are rational. We concluded that the sum is rational. And in the middle, there was an argument to demonstrate that fact. The argument was actually fairly typical in what we first did was take the assumption and then unpack the assumption in terms of some useful information. And once we'd done that, we reasoned with that information to get a conclusion that in fact was the thing we were aiming for. So we write down the assumption, we carry out some reasoning and we reach the conclusion. All three steps are important. Declare the assumption, carry out the reasoning in a clear, understandable fashion, and then state the conclusion when you've reached it. Remember that proofs have two purposes. One is to convince yourself, and two is to convince other people. And you may know what you're doing if you don't mention what your assumption is, or what your conclusion is. Or I'll guarantee from experience that a week from now, you'll forget exactly what you were doing. So it really is good, even for your own purposes, to write down what your assumptions are. But certainly from a communicative angle, it's important to begin by stating the assumption, then to lay out the reasoning in a simple understandable fashion, and then to state the conclusion that you've reached. Okay? Well, that's the most basic method of proving a conditional. Let me give you a quiz. Let r and s be irrational numbers. See which of the following are necessarily irrational. And let me stress that word necessarily. Number one, r plus three, number two, five times r, number three, r plus s, number four, r times s, and number five, square root of r. No, this is a quiz format, where I'm just asking you to select the ones that you think are necessarily irrational. But unlike most of the quizzes, where I expect you to be able to answer very quickly, I'd like you to think a little bit before you answer each of these. Because the focus isn't really on getting the answer right. I mean, obviously I want you to get the answer right. I'm sure you do too. But that's not the focus. The focus is on the reasoning that you need to carry out in order to get those answers. In each case, you're probably going to have to carry out one or two lines of simple reasoning in order to answer the question. That's the focus of this particular quiz. In fact, in the assignment that's coming up in assignment seven, I'm going to ask you to write out proofs of each of the five answers. So let me stress that the focus is on the logical reasoning. Okay, see how you do. Well, the ones that are necessarily irrational are one, two, and five. In each case, you use the fact that a rational number is one that can be expressed as a quotient of two integers and an irrational number is one that cannot be so expressed. And then you carry out a couple of lines of reasoning to show that r plus three has to be irrational. If r plus three was rational, then r would be rational. Likewise, you would carry out a couple of lines of reasoning to show that if five r was rational and could be expressed as a quotient of two integers, then so could r. And similarly, you carry out a similar argument for the square root of r. The two that are necessarily irrational are three and four. In order to show that r plus s is not necessarily irrational, what you would need to do are find examples of irrationals r and s for which the sum is rational. Well, how about taking r is a square root of two and s equal to, let's say, ten minus the square root of two. r plus s is equal to ten, which of course is rational. It's an integer. We know that the square root of two is irrational, and a very simple argument shows that ten minus the square root of two must be irrational. In fact, the argument you show that this number is irrational is a combination of the two arguments you used in parts one and two. Okay, so r and s are irrational, but their sum is rational. In the case of this guy, we could take r equals square root of two and s equals square root of two. r and s are both irrational, and yet r times s is two, which is rational. Okay, how did you do? Remember that then in assignment seven, I'm going to ask you to give proofs of each of the five answers. Well, conditionals involving quantifiers are sometimes best handled by proving the contrapositive. What's the contrapositive? Well, we met that in assignment four. So to prove a conditional phi yield psi, what you can do is prove not psi yields not phi. That's the contrapositive of phi yields psi. You reverse the phi in the psi, and you put negations in front of them. And in assignment four, if you look back, we prove that those two are equivalent. We prove that using truth tables. So let's look at an example of how this might work. And the example I'll take is this one. I want to prove that if the sine of an angle theta is not equal to zero, then for all n in the natural numbers, theta is not equal to n pi. Okay, there's a conditional. And I'm in the middle of some argument, we'll imagine, and I want to prove that if the sine of theta is not zero, then theta is not a whole number, multiple of pi. Where the statement's equivalent to not the case that for all n in n theta not equal to n pi implies not that sine theta is not zero. But there's lots of knots in here, so let's clear them all out and put this into a positive form. So in positive form, this is not for all n in n. We're going to get the exists n in n, and the knot is going to move inside, and it's going to negate this knot. It's going to wipe that one out, and then I'm going to have theta equals n pi. Yields, knot that sine theta is not zero means that sine theta is zero. So this is the contrapositive of this. And that's the thing we're trying to prove. Well, we know this. We know that whenever you've got a whole number multiple of pi, its sine is zero. So this proves the desired result. Obviously, this is a highly contrived manufactured example. Again, the focus is not what I'm actually proving. It's the method I'm proving. I'm picking simple examples so that we don't have to worry about the mathematical content. We can just look at the logical reasoning. And in this case, we start out. We want to prove something. We replace it by the contrapositive, and then we prove the contrapositive. In this case, the proof of the contrapositive, we just pull on well-known knowledge about the sine function. That the sine crosses the x-axis whenever you're at a whole multiple of pi. Okay, let me tell you about one more thing involving proofs of conditionals. To prove a biconditional, phi equivalent to psi, we generally construct two proofs. One of phi yield psi, the other of psi yields phi. And since the biconditional is just a conjunction of the two conditionals, that clearly amounts to a proof of the biconditional. Occasionally, it's easier to prove the two conditionals phi yield psi and not phi yields not psi. And I'm going to leave you to find out why this is enough. Why does this work? If you look back at the assignments, you should find a clue as to why it's enough to prove these two in order to prove the biconditional. Okay, now I'd like you to complete assignment seven. As usual, completing an assignment means at least you should attempt all of the questions. You may not be able to get them all out, at least not at first. But remember, the benefit from working on the assignments in this course is actually trying them. Whether you get them out or not is not that important. Remember my favorite example of learning to ride a bike? The fact that you don't succeed in riding a bike for many days or weeks when you're learning doesn't mean you're not making progress. The same with learning to swim. You actually get the benefit during the process when you're trying and failing, and then suddenly you find you can either ride the bike or you can swim. And it's the same with this kind of material in this course. Okay, good luck on assignment seven. Well, I wonder all of the questions in assignment seven, but there's still a few of them. Starting with number one, prove or disprove the statements, all bears can fly. Well, we'll find a counter example. I mean, that's false. And to show it's false, what I need to do is to find a counter example. And an obvious one is the ostrich. That's a bird that can't fly. So we've got a counter example. Okay. Well, now let's look at number two. Prove or disprove the claim for all x in r, for all x and y in r, x minus y squared is greater than zero. That's also false. And again, it's approved for something that's a universal quantifier. I just like the one above it, it's a universal quantifier, or in this case two universal quantifiers. And it's approved that a universally quantified statement is false. What you need to do is find a counter example. Well, that's one way of doing it at least. And the most obvious one here, well, anything that takes x and y equal will do it. So let's give a specific counter example. Let's take x equals y equals one. And in that case, x minus y squared equals zero, and zero is not strictly greater than zero. Okay. It came close, right? If we'd excluded x and y being zero, then we'd have a positive result. But it says it's true for all x, y in r, and a single counter example is all it takes. And in this case, any pair of equal numbers gives us a counter example. Okay, number three. Prove that between any two and equal rationals there's a third rational. So let's let x and y be rationals, x less than y. Okay. Then, because they're rationals, x is p over q. We can write y is r over s where p, q, r and s are integers. And we have to show there's a number between. Well, the obvious thing is to take the mean. Let's just take x plus y over two. And if that's rational, then, then we'll prove the result. Well, here's the proof that it's rational. X plus y over two is equal to p over q plus r over s over two. Okay. Which is p s plus q r over q s all over two. Which is p s plus q r over two. What was it, q s? Can't see underneath my hand here. Which is rational because it's a quotient of two integers. But of course, x is less than x plus y over two is less than y. And we're done. Okay, that's one, two and three knocked off very, very, very quickly. Let's go on to do number seven, the one about square root of three. Well, just as we did with the square root of two, we're going to prove it by contradiction. So, I'm going to assume square root of three were rational. Then, I could write three root three is p over q where p and q are natural numbers. And I can always assume that they have no common factors. Okay? Because if you pick a pair of integers, a pair of natural numbers that do have a common factor, you can cancel it out. So, you can always express a rational number in that form. Then if we square that, I get three is p squared over q squared. So, I can multiply across by the q squared and get three q squared equals p squared. So, three divides p squared. But three is prime. And if a prime divides p squared, that means three divides p. The prime divides a pair of numbers multiplied together, it divides one of the numbers. So, three divides p. So, that means p is of the form three, let's call it three r. Okay? So, that means p squared equals nine r squared. So, I can take p squared and substituting it back in here to get three q squared equals p squared equals nine r squared. So, if I forget the middle term now, three q squared equals nine r squared. Let me factor the three out. I've got q squared equals three r squared. So, that means three divides q squared. But if three divides q squared, then just as before with p, that means that three divides q. And now I've got a contradiction. Since p and q have no common factors, then yet we've just shown that three is a common factor. So, there's a contradiction. Almost exactly the same as the one for square root of two. So, if we compare the two, cf, the proof for square root of two. In the case of the square root of two, we talked about numbers being even. But if we talk about something like p being even, that's just another way of saying two divides p. So, really all I've done here is I've taken the free risk proof, the one for root two. And instead of talking about it in terms of even and odd, I could recast it in terms of whether it's divisible by two or not. And then the facts, the fact about two that we use there, we said if two divides p squared, then it divides p, it used the fact that two was prime. So, I might just as well use three. So, it's exactly the same proof as before, except instead of talking about divisible by two, I'm talking about being divisible by three. I didn't express it that way before. I talked in terms of even and odd. But even just means it's a multiple of two. Okay, well, we've moved on. Let me decide which one to do next. I think I'll do number eight. That shouldn't take too long. So, we'll do number eight next. Okay, so the converse of an implication, of a conditional, is the implication in the opposite direction where you swap around the antecedent and the consequence, but everything remains more or less, otherwise unchanged. So, the converse of this is if the yarn rises, the dollar falls. We just swap them around. Same here. If negative y less than negative x, then x is less than y. And in the case of c, if two triangles have the same area, then they are congruence. Now, in terms of what we're doing, this is totally trivial. This is just swapping things around without doing anything. What I'm really trying to get at with this exercise is in part to contrast it with the contrapositive. And also to observe that truth and falsity can change. In this case, we start out with something that's true. This is a true implication. That's a true implication too. So sometimes the converse of a true implication is a true implication. Let's look at this one. If two triangles are congruent, they have the same area. That's true. But this one, if two triangles have the same area, they're congruent, that one's false. So sometimes the converse of a true statement is true, and sometimes the converse of a true statement is false. Now, that's a different situation from what we find with the contrapositive. But when you swap around the order, you can sometimes get truth going to false, and you can sometimes preserve truth. Okay, well, that's really all of this to that. It was just to sort of give an opportunity to reflect on these things and to recognise that truth and falsity plays its own game when you're dealing with converses. Okay, let's go on and do number 11. Okay, that was that one about the rational numbers. We've already looked at some of these in the lecture where we've observed that some of them are rational. Let me see we observe that this one can be rational. We observe that this one can be rational. And there's an issue about this one being rational. Okay, these were easy. We did those in the lecture. That was a later issue. That leaves one, two, and five. They're the ones where they're irrational. And so we have to prove them this time. Okay, and so starting with number one, say, what we're trying to show is that it is irrational. So yes, it's irrational. Okay, and let's see. Suppose, suppose that r plus three were rational. Okay, that's a plus sign there. Then r plus three would be of the form p over q, where p and q are integers, where then r would equal p over q minus three, which is p minus three q over q, which is rational. And that's a contradiction, because r is assumed to be irrational. And we've shown that if r plus three were rational, then r would have to be rational. And the others are similar. You simply assume the contrary, you express it in terms of p over q, and then you do a tiny little amount of manipulation, and you end up showing that the number r is rational here, and that square root of r is, and that r is rational in the square root example. Okay, very straightforward. Nothing really more to be said about this one. Okay, the only one left now to do on assignment seven is question 12, so let me just quickly go through question 12. Well, the key facts that you use in dealing with all of these are that if n is even, then, and actually if and only if, then n is 2k for some k, some integer k. Okay? And that n is odd if and only if n equals 2k plus 1 for some k. Okay, the even numbers are the ones that are multiples of two, and the odd ones are the ones that are one more than the multiple of two. Okay, they've been between the multiples of two. So for example, if we do part a, if m and n are even, then we would have m equals 2k. For example, we'd have n equals 2l, so m plus n would equal 2k plus 2l, which is 2 into k plus l, which means it's even. It's 2 times something else. Do I need to do one more? Well, let me just do one more anyway, just because I've, I mean, I'm on a roll now. Let me do a number d, part d. If one of them is even, it's 2k, and the other one's odd, it's 2l plus 1, then m plus n equals 2k plus 2l plus 1, which equals 2 into k plus l plus 1. Okay? So it's two, it's twice something plus 1, et cetera, for all the others. Okay? Nothing terribly deep about, about these things, but in terms of providing out a proof, it's basically a case of just capitalising evens and odds in this way, and doing a tiny amount of algebra. Okay, that was a very easy assignment, I think. Well, it was meant to be an easy assignment, it's easy for me to say that, right? I've been doing this for years. If you haven't met this kind of thing before, then it probably isn't an easy assignment. I fell into the trap of mathematicians there of of saying something's easy in the same way we talk about trivial. That's the sort of a term of art that we use in the game. So everything's easy when you know how to do it, right? Okay, but I was looking ahead to assignment 8, which was decidedly difficult for me and for you guys. Okay, assignment 8 coming next is difficult. Question 1 in problem set 5 asks us to judge whether this is a valid proof or not. Well, certainly the algebra looks correct, right? If M and N are odd, then by definition an odd number is simply one that's not even and an even number is one that's a multiple of 2. So an odd number has to be of the form 2P plus 1, so that's okay for some P and for some Q. So if the odd introduces that, in which case multiplying them together, you get that, the algebra's correct, so M and N is odd. Everything's so far to this part because it doesn't complete the proof. What we've shown is if. We've shown that M and N is odd if M and N are odd. In other words, we've proved the implication that way, okay? If M and N are odd, then M and N is odd. So in this statement, we've proved the right-left implication and we need to show the other part. We need to prove the only if part. In other words, the implication from left to right. It's if and only if. So we've proved the if part from right to left. We need to prove the only if part from left to right. Ie, we need to show, let's write it out in full. If it's not the case that M N are odd, then M N is odd. Is not odd. Yeah, is not odd. Okay, there are various words you can talk about this, okay? Ie, but if it's not the case that M N are odd, that should have been a comma there. Okay. If it's not the case that M and N are odd, then M times N is not odd. Okay? I got the tense correct, the plurality, that was I and that is. Okay. If at least one of M and N is even, then M times N is even. If M equals two K, it doesn't matter which one we assume is going to be even because multiplication is going to be commutative. If M equals two K, then M N equals two K times N, which is even. So this was hard, this is hard to going to set the world on fire, right? I mean, there's nothing terribly deep about this. It's extremely elementary. And it's not that you don't know this and understand it. The question is, can you write it down in a proof? And the point here is that if M equals two K, then M N equals two K times N, which is odd. And the point here is that we have to start with the definitions. And the definitions are, N is even if and only if there's an integer K, so it's that N equals two K. And N is odd if it is not even. So starting from these two definitions, this is the proof we give. The first part that was actually given to us, that was fine from those definitions. We started with the definitions. Going the other direction, we took the definition. So the question is simply, all this really involves is reducing this to the definition. And then the argument itself is very simple. That was a slightly more complicated bit of algebra, but not much more. So it's not about the complexity of the algebraic manipulations. You've been able to do that for years, I know that. The question is, can you reduce the statement to something coming from the definitions? Of course, in many cases you simply would. This would be so simple compared with most proofs in mathematics. You wouldn't go into these details, but the focus here isn't on the, at this stage, isn't on the complexity of the argument. The focus is on the structure of the argument. Are you able to get the logical structure right? And experience tells me that it takes most of us, and it certainly took me a long time to get the sense of what do I have to prove in order to prove something? That's not something that's not cookie cutter. You've got to build up experience in what constitutes a proof and be able to judge whether a proof is correct or not. Okay, let's go ahead and look at number two now. Let me jump first of all straight to the correct answer. And the correct answer is A. This says there are people and there are times at which you can fool those people. So you can fool some of the people, some of the time. Remember, existence quantifiers are what we use in mathematics to capture the word some or at least one. A little bit atypical in terms of what the word some means in everyday language. Okay, and here the way I've expressed it is there is a person and there is a time such that you cannot fool that person at that time. Okay, there's a person and a time such that you cannot fool that person at that time. Which is actually equivalent to saying you can't fool all the people all the time. Formally you could take that part and you could rewrite it as it's not the case that for all x and for all t, fxt. It's not the case that you can fool all the people all of the time. Okay, but I wrote it this way because we're only asking for the equivalence. I'm not saying which is the closest way of capturing it. I'm just saying which is one's equivalence. Okay, let's look at part B. What does this, well does the first part look the same in all of these three? So the distinction is in the second part. So let's see what this one says. Can we express that in English? Well this is tricky to express in English because of these American Melanoma Foundation type issues. I mean you could say something like, oh let's say you can't fool everyone. Let's try something like this. That's some time or other. For everyone it's the case that you can't fool them at some time or other. I mean really what it says is it's not the case for every person. There's a time when you can fool them. Okay, so you can't fool everyone at some time or other. Yeah, that's sort of, I can't think of a way of saying this that really makes, that reads well in English. And yet which really captures this. But hopefully it's clear what the thing means. It's not the case for every person. There's a time when you can fool them. Okay. To me that says the same but you may interpret that one different. Because this is one of these things like American Melanoma Foundation that it's ambiguous as natural language softness. Okay, let's see again in the case of this one, what does this one say? It's not the case that there is an X and there is a T so that you should conform. This one I think is easy to say in English. Okay, it basically says you can never fool anyone. Okay, it's not, you cannot find a single person in a single time, such that you can fool that person at that time. You can never fool anyone. So this one I'm very happy with. Okay, that one really captures it. And the smiley doesn't indicate that I think that's true. It indicates that I think that's a really clean interpretation of what that one means. And number four doesn't arise because we've already found something correct. Well questions three, four and five were just truth table arguments. That was essentially revision material. So I won't go through those here. Let's just take a look at question six, which when you first meet it looks as though it might be terribly deep, especially since we've seen this now after we've looked at things like square root of root two and square root of root three. But actually this one, when you step back and think about what it says, turns out to be very simple. We simply observe that whenever n is a perfect square, in other words, it's the square of some integer, any integer, no infinitely many integers, so there are infinitely many perfect squares. Then of course the square root of n is just k, which is rational. And that's all that is to that one. I mean not only is it rational, it's a whole number. The point is there are infinitely many numbers n for which the square root of n is a whole number and hence rational, namely all the perfect squares. Okay, that's all that is to that. You don't really need to say any more. That's it. That shows that it's true and it tells you why it's true. And that's the proof. Of course you only asked really to sort of say whether it was true or false. But it was this that I was interested in. Can you construct a proof of its negation, of it or its negation? That was really what it's about. It's all about proving things here. Okay. Well finally, question seven is one of these fallacious proofs. There's a whole range of these things. Quite amusing. Usually they depend upon dividing by zero in some disguised form. This one's different, okay? So let's follow it through. Because normally the point is just to find the mistake. We're clear that there's going to be a mistake because one doesn't equal two. And indeed one of the things I asked you to do was to find the mistake. But the focus here isn't so much on finding the mistake. It's on how do you grade this as a proof? And remember when we're grading proofs, logical correctness is certainly part of it so there's going to be some losses of grade here. But also it's about communication. Okay. So let's go through this one and just see how we would grade the thing. Okay. Well for example, is it clear? It's absolutely clear. So we're going to have to look at four marks for clarity. Is it very clear? There's a good strong opening. We start with the identity. One minus three equals four minus six. Okay. Both sides are equal to two. So full marks for giving the idea of the opening. The conclusion is stated. So it's four marks for stating the conclusion. And reasons given, yes. Absolutely. Adding nine plus four, nine over four to both sides. This is completing the square to give you a perfect squares. Then we're seeing factors, which it does. That's why we added nine over four to both sides. Taking square root of both sides. Absolutely. Reasons, reasons, reasons. So in terms of the structure, this is wonderful. Now overall I'm going to have to give you zero, right? I mean that's clear because the thing is plain false. The question is, you know, what do I give for logical correctness? Well, first of all we need to identify the identity, where the errors are occurring, right? And the errors here, it's in this line. Because when you take square roots, then of course there are positive and negative square roots that we had. And in fact the correct solution is to say the minus left hand side, the negative root on the left hand side minus one minus three over two. And it's the positive root you would take on the right hand side when you take square roots. So that's what you should have. In other words, three over two minus one equals two minus three over two. In other words, a half equals a half. Okay, so no big deal there. I mean that's nice. The world still exists. A half does equal a half even though all doesn't equal two. So the issue here was when you're taking square roots, there are two possible signs. And the one that makes it valid is the negative on the left and the positive on the right. Okay. So this is where the mistake is. The only question is, do we give partial credit for the fact that there's logical correctness everywhere else? This is a judgment call. I would say that these steps are so simple. It's just a very elementary algebra arithmetic that at this level, I'm not going to give particular credit for getting these bits right. And I'm going to say this is a big mistake. This is huge. Okay, when you take square roots, you have to take a plus for a minus. That's a mathematical mistake. Realising that. Not recognising that you need to worry about the signs when you take square roots. So that's a bigger of a mathematical point of view. So this grading that I've given it, which means I'm actually giving 16, means I'm giving no points at all on the mathematical correctness. I'm giving lots of credit on everything else. Now, if this was a mathematics course, I wouldn't be using this rubric this way. I would take account of these, but they would have less weight. I mean, these have all got the same weight at maximum of four. And that's because the focus of this course is on proofs and reasoning and communication. And these are important parts of that. You know, the assumption in this course is that you already can do some mathematics. And so I'm not really grading you on that. I'm grading you on mathematical thinking and mathematical communication. So there's zero marks here on the mathematics, but there's four marks and everything else. And this is the reason why I did this one. This is where everything is well laid out, but the thing is playing wrong. So it distinguishes between mathematical correctness and the mathematical thinking and the communication and then writing a proof out correctly. Having said that, in this case, it's blatantly obvious that something's gone wrong. But very often in mathematics, mathematicians, professional mathematicians make mistakes, buried in proofs. And because they're professionals, they can usually write things correctly. We know how to, we learn as part of becoming mathematicians, we learn how to write things out, we give reasons. So in fact, this, although it looks absurd here, you often find that because mathematicians know how to lay things out well, results get published even though they're absolutely wrong. And whenever result gets published, basically what that means is that the referee of the journal where it's published has graded it and said, this is, this is correct, this is absolutely correct. So, you know, this is not an unrealistic situation in principle when false results get proved and when false results get published, that is, of course they never really proved. The false results get published when what's going on then is that the referee has gone through it and thought everything's okay. Okay. Here it's dramatic because the answer's absurd and it's blatantly false. But this is actually not an unrealistic scenario. And so giving a high grade isn't, it's not a bad thing. I'm doing this to emphasise the distinction between the various things we're looking at here. And this one, I think, makes it clear that there are other issues involved other than, other than logical correctness. But as I say, if this was a regular mathematics course, I wouldn't be giving 16 out of 24. In fact, I'm not even sure I'd give any marks for this if a student came up for this because it really, this is really a big mistake. Okay. But within this context, within the context of this course, this is the kind of thing we're looking for in looking at mathematical thinking and mathematical proofs and communication. Okay. There we go. Well, that's the end of the problem set questions. But let me leave you with one more tantalising little puzzle. We are probably familiar with the story of Archimedes, lived in Greece about 250 BCE or thereabouts, who was asked by the king to determine whether a crown he'd been given was actually made of pure gold or not and that involved calculating the density. And he knew how to calculate to find out the, the, the mass of the, the weight of the crown. But the question was how do you calculate its volume? No, no, Archimedes knew lots of mathematics for calculating volumes. Indeed, he'd invented a lot of that mathematics. He was able to calculate areas of circles and volumes of spheres and various other shapes like boxes and rectangles and pyramids and so forth. He knew all of that stuff. So he had a lot of mathematics at his disposal that he could have applied. But it didn't seem to work for something irregular, like a crown or at least not easily. But then one day when he's taking a bath, this is the story, the story goes, when he's taking a bath, he has this, this amazing insight. He says to himself, if I immerse the crown in water, it will displace some water. In fact, the amount of water it will displace is exactly equal to the volume of the crown. So if I collect the water when it spills out of the bath and when I put a crown inside it into the water, then I'll be able to just measure the volume of the water in a standard way and I'll know the volume of the crown. And, and as that story goes, he was so impressed and tickled by his solution that he jumped out of the bath and ran stark naked through the streets crying out, Eureka, Eureka, which is Greek for I found it, I found it. Now, I've known lots of mathematicians, I certainly haven't known Archimedes, but I doubt if even a mathematician deep in the throes of solving a problem would run naked through the streets. However, I can imagine him being extremely pleased with himself and having a great adrenaline rush when he had that insight, because that's a great example of thinking outside the box. He knew lots of techniques for calculating volumes. He invented many of them, but on this occasion he thought outside the box and found a really elegant solution that was different. And the puzzle I'm going to give you is very much along those lines and it actually is about taking a bath. And here it is. If it takes half an hour for the cold water faucet to fill your bathtub and an hour for the hot water faucet to fill it, how long will it take to fill the tub if you run both faucets together? Now this looks like one of those frustrating little word problems you get in high school, okay, where you end up, you sort of say that the rate of flow of the cold water be F and then T and you write down some equations and you figure something out, okay. And I'm sure you know how to do that. You can apply a standard technique for doing this kind of thing involving rates of change and you get the answer. But you don't need to do any of that. You don't need to do any calculations at all in order to solve this one. If you think outside the box, you can answer it without doing any of those calculations based on the rates of flow or anything like that. You simply have to think of the problem a different way. If it helps, if you're inspired by the Archimedes story, you might want to run a bath and get in the bath and see if you can come up with the idea then. I must admit from personal experience that a lot of my best ideas in solving mathematical problems and improving mathematical theorems throughout my career have actually been obtained when I've been sitting in a bath or shortly after taking a bath. There's something so relaxing about getting in a bath that it frees the mind to come up with these out of the box solutions. So this isn't really about do you know the standard methods? You probably do. This is about can you think about this in a different way that allows you to solve it without really doing any of this music whatsoever. Okay? Well, with that hint, I'll leave you to it. Enjoy. How did you get on with assignment seven? In this lecture, I'll say something about how to prove statements that involve quantifiers, which is true for almost all mathematical theorems. Universally quantified statements of the form for all natural numbers N, some property A N holds, where the quantification is over all natural numbers are often proved by a method known as induction. I'll explain the method of induction with some examples from number theory. Number theory is one of the most important branches of mathematics. It studies the properties of the natural numbers, one, two, three, etc. We look at some elementary parts of the subject in later lectures, but for now it provides good examples of induction proofs. But first, let's look at how we might prove an existence statement. We want to prove there exists x such as A of x. The obvious way is to find an object A for which A of A. For example, to show that there's an irrational number, just prove that square root of two is irrational, which we did. Unfortunately, this doesn't always work. Sometimes we use indirect proofs. For example, I worked through that problem about the cubic equation, where we showed that there was a point on the x axis where the curve crossed it, and hence that the cubic had a solution. We never found the solution, we just showed that one existed. Let me give you a really fascinating example of an indirect proof. I'm going to prove that there are irrationals, R and S, such that R to the power S is rational. That actually came up in assignment seven, if you remember, and I promised then that I would do this later. Well, now is that later. The proof involves considering two cases. Case one, if the square root of two raised to the power square root of two happens to be rational, we can take R and S both equal to square root of two. Then, by the assumption that square root of two to the square root of two is rational, we've shown that there are numbers R and S which are irrational with R to the power S rational. Well, if case one isn't the case, then we must be in case two, which is where root two to the root two is irrational. In which case, take R equals square root of two to the square root of two, and S equals square root of two. And if we do that, then R to the power S is root two to the root two, all to the root two, which by the way that exponents work is root two to the power root two times root two, which is root two squared, which is two, and two is rational. So, again, we've proved that there are irrationals R and S such that R to the power S is rational. Taken together, case one and case two prove the theorem. I really like this example. It's not just clever, but it's cute, and it's cute because of the way it works. We really don't know whether this number root two to the root two is rational or irrational. It's probably irrational actually, but the point is we don't need to answer that question. We just look at the two alternatives. If it does happen to be rational, seems weird, but if it does happen to be rational, then we can find irrationals R and S without the S rational this way. On the other hand, if root two to the root two is irrational, which probably is the case, but again we don't need to know that, then we take R to be this number, S to be that number, and then R to the S is two, which is rational. So we don't know officially which one of these is the case, and more to the point, we don't need to prove which one of these is the case. We simply look at the two alternatives, and depending on which alternative we're in, we take different rationals, we take different irrationals R and S with R to the S being rational. So the choices of R and S are different in the two cases, but because these two cases exhaust all possibilities, this number is after all either rational or irrational, then this proves the result. By the way, this is known as the method of proof by cases, and it's a method that's used a lot in advanced mathematics. Pretty neat, eh? Well now let's take a look at how we might prove a statement involving a universal quantifier. Suppose we wanted to prove a statement for all x, a of x. One way is to take an arbitrary x and show that it satisfies a of x. For example, to prove that for all n, there is an m, m bigger than n squared, where the variables m and n range over the natural numbers, but this is a pretty trivial example. But I want to focus on the method that you would use to handle the two quantifiers in a statement like this, where in particular the first quantifier is a universal quantifier. Let n be an arbitrary natural number. Set m equal to n squared plus 1. Then m is bigger than n squared. And this proves the statement, there is an m, m bigger than n squared. And it follows that the statement for all n, there is an m, m bigger than n squared is true. Okay, well as I said, this is pretty trivial. Let's just see what's going on in terms of the logical reasoning. You want to prove a statement involving two quantifiers of the form for all n, there is an m, m bigger than n squared. What we're going to do is eliminate one quantifier by replacing it by an arbitrary natural number. Now by arbitrary, I mean we make no assumptions at all other than the fact that it is a natural number. The argument will have to hold for any natural number whatsoever because we're trying to show that this is true for all n. So we pick an arbitrary one. Then we carry out some reasoning to verify the rest of the formula. Okay, when we eliminate that for all n, we're trying to prove that there is an m, m bigger than n squared, which is this statement. Now in this case, it's a trivial argument. We just square n, add one, and let n be that number. So we explicitly find an m that satisfies this property. In practice, you're probably going to have to do a lot of work here with a more significant example. There could be several lines or even pages of arguments involved at this stage. But the point is you eliminate the first quantifier by picking an arbitrary n, then you carry out some reasoning to prove the rest of the formula, which is this thing here. And when you've done that, because the argument works for an arbitrary natural number, it follows that you've proved it in this form. The for all n has been handled by reasoning with an arbitrary natural number. So we've taken away one of the quantifiers, carried out some reasoning, and proved the result. But let me stress that this works because the n that we pick is arbitrary. Well another approach would be to use a method of contradiction. To prove for all x of x, assume not for all x of x. This is equivalent to exist x, not A of x. Well, let c be an object such that not A of c. Notice that this isn't an arbitrary object. It's a specific object, admittedly an object whose identity we may not know, that is guaranteed to exist by this. The point is to prove this universal statement we began by assuming it's negation. It's negation gives us an existential statement, and that tells us that there will be some object, a specific object, for which not A of c. Now we've got a starting point to carry out a proof to derive a contradiction. Now a reason with c, together with the fact that not A of c, to derive a contradiction. Now the exact reasoning you use will depend upon what the statement a says. But the general pattern is this. You want to prove for all x of x, and you're going to use contradiction. So you assume not for all x of x. That's equivalent to exist x, not A of x. And that tells you you can find an object, or at least you know that an object exists for which not A of c. And then if you can, you reason with that object using the fact that not A of c, to derive a contradiction. Okay, again, we'll see examples of this later on. I'm just giving you the general overview of the way these arguments work. So, as in the previous case, if this looks a little bit mysterious, it'll become clearer, I hope, when we look at some specific examples later. In fact at this stage you might want to look back at assignment 7 and see what was going on there in the light of one I've just said here. And before I let you go, let me give you a little quiz. Well, the correct answer is no. This is not a valid proof. The point is, picking a positive rational p arbitrarily is not the same as letting p be arbitrary. The choice of p may indeed be arbitrary, but once you've made it, you've got a specific p. We've even said what it is. That's actually different from saying let p be arbitrary. An arbitrary choice of p leads to a specific p, which you may have chosen arbitrarily, but you might have been unfortunate in your choice. You might have chosen a p that just happens to make something happen without there being any specific reason. It's a subtle point I know, but it's an important point, so you really need to sort this one out. Letting p be arbitrary is not the same as picking in an arbitrary fashion a specific p. The point is that p is specific. Its choice may have been arbitrary, but once you've made it, it's specific. In contrast when you argue with an arbitrary p, you don't know anything about the p. You certainly wouldn't know that it was equal to 0.001 or whatever. You may need to think about that one a little bit longer, or talk it over with some of your colleagues. Now let's look at proof by induction. This is the special case where we're proving universal quantified statements where the quantifier ranges over the natural numbers. For example, suppose I want to prove that 1 plus 2 plus et cetera et cetera et cetera plus n equals a half n n plus 1. The sum of the first n natural numbers is a half times n times n plus 1. Well, if you're meeting this problem for the first time, a smart thing to do is to begin by checking the first few cases. First of all to see if it's true for the first few cases, and to see if there's a sense of what's going on. So if n equals 1, what do we get? Well then we've just got 1 here, and putting n equals 1 into the formula, we have half times 1, times 1 plus 1, which is a half times 1 times 2, which is 1, and indeed 1 equals 1. So it's true in the first case. In the second case, n equals 2, what do we have? We have 1 plus 2 is the left-hand side equals, put n equals 2 here, we get a half times 2 times 2 plus 1, which is a half times 2 times 3, a half to itself we've got 1 plus 3, and 1 plus 2 does equal 3. So if the formula is correct for n equals 2, we might want to try n equals 3. What do we get then? We get 1 plus 2 plus 3 equals a half, 3, 3 plus 1, which is a half, 3, 4, which is 6, which is correct. So the formula checks out for the first three cases, and we might want to do a couple more cases. But at least that gives us some sense of what's going on. By doing this we understand what the formula, we understand what the equation is saying. Now so far this isn't a proof. This is just checking the first three cases. In fact you've got to be very careful about jumping to conclusions based on the first three or four cases. So let me just stress that. This is not a proof. Okay, let me stress it again. Beware of jumping to conclusions. And to drive this point home, let me give you an example where if you jump to conclusions you'll come to the wrong conclusion. Consider the formula pn equals n squared plus n plus 41. If you start working out values for this formula, you find something rather interesting. You find that all values of pn for n equals 1, 2, 3, et cetera, are prime numbers. Well, not quite. That's true until you reach n equals 41. And when you get to 41, you find that p of 41 works out at 1681, which is 41 squared. So the first series of values for n equals 1, 2, 3, et cetera, all the way up to 41, which is well beyond the number of cases most people would try out by hand. You find you've got a prime number. But if that leads you to believe that this formula generates primes for every n, then you're going to be wrong. Because when you get to n equals 41 it's no longer prime. This example, by the way, is due to the famous Swiss mathematician Lennard Euler in 1772. And it's a cautionary tale telling us not to jump to conclusions based on numeric evidence. Nevertheless, it's always a good idea, it's often a good idea to begin looking at a problem by working out a few cases. Not only do you get an understanding of what the formula is about, but you can sometimes, when you work out a few cases, you can find a pattern. In this case, there actually isn't much of a pattern to find. You work out the values and you just get numbers, and it turns out that the first 40 of them are primes. But in this case, if you look at what's going on after, even if you did maybe a couple more, n equals 4, n equals 5, you'd begin to see a pattern of what's going on, how the two parts of the formula are combining what's happening to the half. The half is dividing into the second one here, the first one here, the second one there. And after you've done a few cases, you can often find a way of capitalising on the pattern that's there in order to produce the proof that I'll show in a minute. So induction proofs can often be derived by looking at the first few cases, seeing what the pattern is when you start plugging the numbers in, and then actually building a correct rigorous proof. Okay, so the next step now is to begin with this numerical evidence, and our understanding of how the formula works in order to come up with a rigorous proof of the fact that this formula holds for every n. The method of induction depends on a principle known as a principle of mathematical induction. Okay, and here's what it says. Suppose you want to prove a statement for all n, a of n. Establish the following two statements. One, a of one. That's known as the initial case, or the initial step. Two, prove the following statement. For all n, for all natural numbers n, if a of n is true, then a of n plus one is true. That's known as the induction step. Now intuitively, this gives the statement for all n, a of n as follows, by the initial step, step one, a of one. If we now apply the induction step in the special case where n equals one, we have a of one by the initial step, and that yields a of n plus one, which is a of two. So by two, or by step two, a of two, a of one yields a of two. So from a of one, we can conclude a of two. Okay, we begin by proving the initial step. So if we've proved a of one, then we know that one's true. We can now apply the induction step in the special case where n equals one, and having proved this, if we have a of one, we can conclude a of two. Then we can do the same thing again. Once we've got a of two, we can put the two in here and conclude a of three. So by a of two and the induction step, we can conclude a of three, et cetera, all the way through the natural numbers. So it's a little bit like taking a row of dominoes, and you knock the first one down, and providing the dominoes are close enough so that when one gets knocked down, the next one gets knocked down, then once you start to topple a row of dominoes, the first one not goes down, then the first one knocks down the second one, the second one knocks down the third one, the third one knocks down the fourth one, the fourth one knocks down the fifth one, and so on and so on and so on, all through the natural numbers. So you get a row of dominoes falling down. So this basically says knock the first domino down and put the dominoes close enough together so that when one falls, the next one falls. Okay, that intuitive explanation is fine as far as it goes, but that's certainly not a rigorous demonstration of this method. The fact that proving these two together does yield that universally quantified statement, that's actually quite a deep result of mathematics. And let me just stress that. You need an axiom or a principle to make this work called the principle of mathematical induction. The principle of mathematical induction is what tells you that one and two above, steps one and two above, yield for all n, a of n. Okay, now we have the method of induction laid out before us. I'll apply it in order to prove that identity that we looked at a moment ago about the sum of the first n natural numbers. So I'm going to prove a theorem now that for any natural number n, one plus two plus three plus yada, yada, yada, plus n is a half n, n plus one. I'm calling it a theorem because I'm going to give a rigorous proof. I'm going to use a method of mathematical induction. And since one of the purposes of a proof is to explain why something is true, it's usually good form to indicate to the reader the method you're going to use. It's orientates the reader to what's going to come next. And I'm using a standard method here, namely mathematical induction, so it's good form to indicate that that's the method I'm going to use. Well, I have to do two things. I have to show that this is true when n equals one, and then I have to show that if it's true at n, it follows at n plus one. So the first step is to say that for n equals one, the identity reduces to, well, n equals one on the left hand side, I've just called one, put n equals one on the right hand side, I've got a half times one times one plus one, which is a half times one times two. And indeed the left hand side does equal the right hand side. So that's true since both sides equal one. So we've proved the first step. Incidentally, you may have noticed that I referred to this not as an equation, but as an identity. And the reason is there's an equation is something that you solve to find the value of a variable, either x in the case of a real number equation or n in the case of a natural number equation. But this isn't an equation that you solve for n. This is an identity. It's something, it tells you that the two sides are always equal. And we use the word identity to refer to an expression like this that claims or that states that two things are equal for all, for all values of n. So these two sides are identically equal. So the correct terms are an identity for something like this, an equation for something that you solve. Having said that, like many mathematicians in practice, I'm sometimes not particular in the words I use. If you caught me talking to my colleagues, you might find me using the word equation where I mean identity. Possibly even using the word identity where I mean equation, all that one's less likely. But strictly speaking, an equation is something you solve and identity is something that states that two things are always equal. Okay, let's move ahead. We've proved that the result is true at n equals one. The next step is the induction step. So I'm going to assume the identity holds for n. Ie I'm going to assume one plus two plus n equals a half n, n plus one. And let me note now and I'll put this in brackets. I want to deduce one plus two plus n plus one equals a half n plus one, n plus one plus one. Okay, that's just a note for myself to indicate the target I'm aiming for. And the challenge now is to start with what I know and deduce where I want to get. Well, how can I get this from this? Well, one way would be to add n plus one to both sides of that identity, which means that I'd get the left hand side here and then hope that when I work out the right hand side, having added n plus one, I end up with this expression. Okay, so let's add n plus one to both sides of, let me call it star. So star is the identity I'm assuming. Let me add n plus one to both sides of that. That gives me one plus two plus plus n plus n plus one is a half n, n plus one plus n plus one. So I've now got the left hand side of what I want to prove. I need to show that the right hand side is the right hand side of the thing I want to prove. So I need to show that that is the same as that. Well, let's just see what happens. I can take the half out as a common factor. So I've got a half into n, n plus one plus two into n plus one. And that equals a half n squared plus n plus two n plus two, which is a half n squared plus three n plus two. And this factors. This is a half n plus one n plus two. But that's the same as that. I can write it explicitly if you like as a half n plus one n plus one plus one, which is the identity with n plus one in place of n. Hence, by the principle of mathematical induction, the identity holds, and we're done. We've proved the initial step, a simple matter of observation, and we've proved the induction step by a fairly simple algebraic argument. And that's it. That's mathematical induction in practice, where this screen came up about a quarter of an hour into the lecture. Did you spot anything wrong with it at the time? Can you see anything wrong with it now? Well, the problem's in here. This isn't the polynomial I meant to write down. What I meant to write down here was the polynomial p of n equals n squared minus n plus 41. I got the sign wrong. And for this, it is the case that for all values of the polynomial for n equals one, all the way up through n equals 40, it's prime. Okay, all the way up. And then p of 41. If you put that in, it's 41 squared minus 41 plus 41, which is 41 squared. Which is this number here? 1681. Let me call that p sub minus to indicate there's a minus sign here. Because the polynomial I wrote down, I'll call p plus of n, was this one, plus n, plus 41. That one is actually prime for n equals one through 39. And if I put 40 in there, let's put the minus in there to distinguish it. If I put 40 in there, I get what? I get 40. N squared plus n is n times n plus one. So it's 40 times 40 plus one plus 41. Which is 40 times 41 plus 41. So I've got 40 times 41. And then another 41. That means I've got 41 lots of 41, which again is 41 squared equals 1681. These are sort of the same polynomial when you look at it in terms of the behaviour here. If you just see what's going on. They were both actually due to, well actually, I'm not sure historically whether, or I wrote them both down, but clearly once this has been discovered, you're just manipulating them. And what I did when I was giving the lecture, this is spontaneous. By the way, almost all of these lectures are spontaneous. They're not meant to be polished lectures. I mean, I'm familiar with the material. So it sometimes looks as though I'm just sort of reading it from a book. But if I am reading it from a book, it's a book in my head. I'm having taught it for many years. Basically what I'm recording is spontaneous. And this just, I just pulled this off from my memory and I confused two things. I knew that it didn't really matter what the sign was, but it does make a difference as to where it stops being primed. If you pick the negative sign, it's primed all the way up to 40. If you pick the positive sign, it's primed all the way up to 39. So it was a technical error, but in principle it was correct. And it certainly, whichever one you pick, gives a wonderful example of the fact that just because you have a repeating pattern, it doesn't mean to say something's true for all n. And remember that I introduced this as an illustration of caution when I was talking about induction proofs. OK, well, what happened was the first time I gave this course, many students found the mistake and I thought it was good to leave it in the second time I gave the course so that students would have a further chance of finding the mistake themselves. It's always fun to find mistakes. This time I decided to leave it in, but put in a quiz at the end just to prompt anyone that hadn't noticed the mistake that there was actually a mistake. OK, so now you know what's going on here and you've seen me make a wonderful mistake in public, which is something that all good mathematicians should do. OK, well, that brings us to the end of 8.0, lecture 8.0. Let's move on to lecture 8.0 where the correct answers are this one, this one, and this one. If you check back at the description I gave of the method of induction, you shouldn't have any difficulty realizing why this is the correct answer, why this is the correct answer, and this is the correct answer. So if you got any of these wrong, you definitely need to go back and look at the lecture again and make sure you really understand induction. It's a very important technique in mathematics. Well, let me give you another example of an induction proof now, and this time I'll lay out the proof in a way that makes explicit the connection between the argument I give and the description of induction I gave a moment ago. And the example I want to give is the following theorem, well, calling it a theorem is actually a bit of a stretch, but here's what I'm going to show, that if x is a positive real number, then for any natural number n, 1 plus x to the n plus 1 is bigger than 1 plus n plus 1x. Okay, this result is not going to set the mathematical world on fire, but it is a good example to illustrate the way mathematical induction works, and in particular to connect the arguments that I'm going to use to the early description I gave for mathematical induction. So here's the proof. Well, I'm going to begin by stating the method I'm going to use to mathematical induction, since one of the purposes of proofs is to explain why something is true. It's always a good idea to begin by stating the method you use that orientates a reader to the argument that's about to come, and to connect it to the description I gave of induction before, I'm going to let n, let a of n be the statement 1 plus x to the n plus 1, bigger than 1 plus n plus 1x. So by induction, I'm going to prove for all n, a of n. Well, the first step is to prove a of 1, and a of 1 is the statement 1 plus x squared is bigger than 1 plus 2x. We put n equals 1 here, we get 1 plus x squared greater than 1 plus 2x. Is this true? Well, the answer is yes. It's true by the binomial theorem, which tells us that 1 plus x squared is equal to 1 plus 2x plus x squared, which is bigger than 1 plus 2x, because x squared is strictly positive. Notice that x is positive, so in particular x is non-zero, hence x squared is strictly positive, hence we have a strictly positive term here, hence that expression is strictly bigger than 1 plus 2x. So a of 1 is true. The induction step involves proving for all n, a of n implies a of n plus 1. Or how do we typically prove a statement like this? We pick an arbitrary n and prove a of n implies a of n plus 1. And how do we prove an arbitrary n like this? We assume the antecedent and deduce the consequence. Well, a of n is the statement 1 plus x to the n plus 1 greater than 1 plus n plus 1x. And a of n plus 1 is the statement 1 plus x to the n plus 2 is greater than 1 plus n plus 2x. So the induction step boils down to assuming this and deducing that from it. So I'm going to focus on this now, and I'm going to begin by looking at the first expression, 1 plus x to the n plus 2. In order to use the induction hypothesis, I'm going to pull out one of these terms and write that as 1 plus x times 1 plus x to the n plus 1. And by the induction hypothesis, 1 plus x to the n plus 1 is bigger than 1 plus n plus 1x. And that equals 1 plus n plus 1x plus x plus n plus 1x squared, which equals 1 plus I've got n plus 1x and another x. So altogether I've got n plus 2x plus n plus 1x squared. And this is positive. So that expression is strictly bigger than 1 plus n plus 2x. And that part of it, the part here connected with this is this. In other words, this is proves a n plus 1. And the theorem follows by induction. Okay, I've laid this out in gory detail here, but my purpose was to connect the kind of argument that you give when you do an induction proof with the description of induction that I gave a moment ago in terms of predicates a n. Okay, so this is, so what I've been doing here is really emphasising how my earlier description of induction connects to an actual proof. If you take out all of the pedagogic stuff that I've put in here, you've really got a two or three line argument. I mean it really boils down to that step and then this step here. Those are the two key steps of the, of the induction proof. Okay, well there we are. Let me summarise how induction works. You want to prove that some statement a n is valid for all natural numbers n. You first prove a 1. That's usually a matter of simple observation. The next step is give an algebraic argument to establish the conditional a of n yields a of n plus 1. Usually you assume a of n and then you deduce a of n plus 1 from it. And typically you do that by looking at the form of a of n plus 1, manipulating it into such a form that you can apply a of n in order to carry out an argument. In other words, reduce a of n plus 1 to a form where you can use a of n. Conclusion, by the principle of mathematical induction, this proves for all n a of n. Occasionally you might have to work pretty hard to establish the first case. Generally this step is easy and this one is where you have to put in the effort. But neither of these two steps on their own constitutes an induction proof. It's the whole package that constitutes induction. My illustration of induction as, as knocking over a row of dominos, while it's intuitively helpful, obscures the fact that this principle of mathematical induction is actually a pretty deep principle of mathematics. The reason it's deep, by the way, is because this is talking about an infinite collection. It's talking about all of the natural numbers and there are infinitely many of them. And the moment you start talking about infinity in mathematics there are subtleties and complexities lurking on the wings. Incidentally, that's why we usually complete a proof by writing a sentence something like, by the principle of mathematical induction, the result follows. Because this is heavyweight. And it's usually a good practice and a shrewd move to mention a heavyweight if there's a heavyweight in the move. Okay, well that's induction. I'd better tell you about a fairly common variant of induction. We sometimes need to prove a statement in the form for all n greater than or equal to n zero a of n, where n zero is some fixed number, five or ten or 20 or a million or something. In this case, the first step is to verify a of n zero. A of one may not be true. In fact, it's usually the case that when we are trying to prove something of this form, the reason we have an a of n zero is to verify a of n zero is to verify a of n zero . The reason we have an n zero as the starting point is precisely because the result doesn't hold at the beginning. We have to go out somewhere through the natural numbers until the result starts to hold. General induction starts at one, this variant starts at some point beyond one. But other than that, the argument is essentially the same. In particular, the induction step is to prove for all n greater than or equal to n zero, a of n implies a of n plus one. In fact, the example of induction I want to give you next uses this variant of induction. And that example is part of a famous result of mathematics called the fundamental theorem of arithmetic. Before I do, let me give you a quiz. Is one a prime number? Okay, I know this isn't the focus of the course, but there's a lot of confusion about this. And if we don't clear it out of the way now, it's going to keep haunting us as we go through the material that's coming up. Now, you know, I could of course simply give you the answer. And in previous times when I've given this kind of course, both in a MOOC and elsewhere, I've given students the answer. But it's one of these things that people just don't sort of really pay attention to. And then it comes up again when there are homework exercises and discussions. So what I'm going to do is give you this as a quiz. So I want you to answer whether this is a prime number or not, whether one is a prime number or not. Okay? Well, the answer is one is not a prime number. And the reason is the definition. Here's the official definition of a prime number. It's a positive integer n greater than one whose only exact divisor is one and n. That's all there is to it. It's a definition. So from now on, please be very careful that you really know the definition of a prime number. One is not a prime number. Incidentally, whilst on the topic of getting definitions cleared out, let me make another remark. Zero is not a natural number. Again, this is by definition. Zero is an integer, but it's not a natural number. Historically, the natural numbers are the counting numbers and people historically started counting at one, two, three, et cetera. Zero was very much a Johnny Com lately. Okay? So these come around all the time. They cause confusion. It's some unnecessary confusion, so I thought I would get it out of the way now and I would draw your attention to this definition by putting it in as a quiz. Okay? Now let's go on to what I was planning to do. Here's the part of the fundamental theorem of arithmetic that I want to prove. And it's the following theorem. Every natural number greater than one is either prime or a product of primes. Now we have to use that variant of induction for this one because it's not true at n equals one. The theorem only holds for n equal to two or more. In fact, one is the only number for which it doesn't hold. So we're only just in the variant. We've just missed one point. But we have to start at n equals two. In other words, the first step of the induction is going to involve looking at n equals two, not n equals one. But with that one change, it's going to mean induction proof. So proof would be by induction. Well, the induction statement n is, well, you might think it's going to be n is either prime or a product of primes. Okay? Based on the examples we've seen so far, you might imagine that a of n, the induction statement, is that n is either prime or product of primes. Well, that turns out not to be the case. That just doesn't work. We have to pick something else. And this is what we pick. For all m, if two is less than or equal to m is less than or equal to n, then m is either a prime or a product of primes. So the induction statement a of n talks about all the numbers between two and n being prime or products of primes. So it's a sort of cumulative statement, if you like. As we go up through the natural numbers, we iterate on the cumulative fact of things being primes or product of primes. And the induction proof establishes that a n is true for all n bigger than one. So we have to start with the first step, namely n equals two. Here of two says what? Well, if n equals two, there only is one m between two and n. So the universal quantify here is essentially vacuous. I mean, it's just a special case. It's a generous case. There is just n equals two. So this statement when n equals two simply says two is either prime or product of primes. Which is true. Two is a prime. Now we're going to assume a of n and deduce a of n plus one. Well, a of n plus one is going to talk about all numbers m between two and n plus one. So let's look at one of those m's. Let m be a natural number, two less than or equal to m less than or equal to n plus one. Well, if m is less than or equal to n plus one, m is either less than or equal to n or m is equal to n plus one. If m is less than or equal to n, then by a of n, m is either a prime or a product of primes. The only other possibility is that m equals n plus one. Well, if n plus one is prime, then m is prime. So what happens if m equals n plus one and n plus one is not prime? Then a natural number is pq such that one less than pq less than n plus one and n plus one equals pq. If n plus one is not prime, then it can be obtained as a product of two smaller numbers. Not being prime means you can find two smaller numbers whose product equals the number. Two smaller numbers other than one that is. So this is just a definition of not being prime. Well, since two less than or equal to p and q less than or equal to n, by a n, p and q are either primes or products of primes. These two numbers p and q are less than n plus one, so they're less than or equal to n. That means these numbers are m's that fit into the induction hypothesis. They're less than or equal to n, so by a of n, each of these is either a prime or a product of primes. But n plus one is a product of them. So n plus one is a product of two numbers, each of which is either a prime or a product of primes. Hence n plus one is a product of primes. The theorem follows by induction. If you look back through this proof now, you'll see why we took our induction statement to be this somewhat more complicated looking statement. We wanted the induction statement to talk about all the numbers up to an including n being either prime or products of primes. So that when we went through the proof and we came down to a pair of numbers like this, we could immediately apply the induction hypothesis a of n in order to conclude that each of those two were either prime or products of primes. And therefore that the n plus one was either a prime or a product of primes. Incidentally, formulating induction statements that are cumulative in this fashion is quite common. It's often the case that in induction proofs, you end up dropping down to some number or some pair or group of numbers smaller than the one you've got and needing to apply the induction hypothesis to those smaller numbers as we did here with P and Q. Okay, how did you like that one? Okay, so now you know about induction. And with induction, we're starting to get into some serious university level mathematics. Though the general idea is simple enough, if you can always go one more step, then you will eventually reach every natural number. In practice, it can require a lot of effort and ingenuity to construct an induction proof. See how you get on with assignment eight. Well assignment eight was really quite difficult I think. On the other hand, looking at the discussions on the forums, I think many people found it difficult because it hadn't yet made this transition to doing mathematical thinking. They were trying to do something that wasn't really required. This was very definite in the case with this one. Remember that the things I keep repeating are that one of the essences of mathematical thinking is that you first of all, you ask yourself, what is this asking me? What do I know? What does it tell me? What kind of object is it talking about? And what do I need to do in terms of what's my target? So you have to stop. You pause. You reflect. You think about it. What you don't do, at least at the beginning, is say, does this remind me of a problem I've solved before that I can just instantly apply that previous technique? Because that can lead you in a completely the wrong direction. For example, there are various words here. In high school, a good strategy was look for key words and try to map them into techniques. But it really isn't a good strategy in terms of advanced mathematics. Okay? There are many questions, many problems we have about perfect squares that are about the natural numbers. Okay? So what? This isn't about the natural numbers. This is about the integers, a completely different set of numbers. So, well, not a completely different set, it's a larger set of numbers. So this is about the integers and the fact that some of the words are often used in discussions about the natural numbers, well, so be it. I mean, words get useful source of context. Okay? So this is a question about the integers. And if it's about the integers, then why don't we just take m equals n equals 0. Then m squared plus mn plus n squared equals 0, which is 0 squared. And we are done. Folks, this is not a trick question. This is a question that gets at the heart of mathematical thinking. And the heart of mathematical thinking isn't knowing a lot of techniques from mathematics. The heart of mathematical thinking is, what do I know and what do I need to do with what I know? And what am I talking about? Stop, slow down, reflect, think. Mathematical thinking at this level is not a sprint. All of the techniques you taught in high, taught in high school were very good for succeeding in high school. And they certainly leave you with a lot of valuable skills. But we've been there, done that. You know, if we'd be on to high school, if we've mastered that, then we're trying to do something else. High school sort of teaches you how to lay bricks. What we're doing now is saying, how can I take all of those bricks, which are valuable things to have, and use them to build a house? We're now going from being brick layers to architects. We certainly need all of those bricks. You know, I am definitely not knocking at high school, right? We use this stuff all the time. But that was providing us with the basic tools. Now we're learning how to make use of those tools. Okay, that's what mathematical thinking's about. And sometimes we need a lot of tools and sometimes we can get by with something very simple because we just ask ourselves, what am I trying to do now? And we're not getting seduced into thinking that this is just another variant of something we've met before. Think of every problem as a new problem. Then you'll find that things get much more doable because you'll be focusing on thinking, not applying techniques. Okay, now that's the end of my sermon. Well, it's not really the end of my sermon because that's what the whole course is about. But let me just move on now to question two. And in this case, you probably haven't seen anything quite like this before. So you have to start by asking yourself, how could it happen that the answer was a perfect square? I mean, just how could that come about? What kind of thing must happen? Well, let's just write it this way. How can we have mn plus 1 equal to p squared for some p? Well, at this point, one of those bricks that I was taught in high school becomes really useful because if mn plus 1 equals p squared, that means mn equals p squared minus 1. And one of the bricks I had drilled into me in high school was that p squared minus 1 was p minus 1, p plus 1. So what we're about to do is take something that was drilled into it in high school and make clever use of it, okay? Because if mn is p minus 1 p plus 1, then we could have m equals p minus 1 and n equals p plus 1. In that case, p would be m plus 1. Remember, m is the number we're given. We're trying to find an n. And I'm just using this to say what could that n be, what must that n look like? So given an m, we'd be able to take p equals m plus 1 and then from this one, n equals p plus 2. And now we've found the n. That is n plus 2. Whoops, n plus 2. Okay? p is n plus 1 and n is p plus 1. So n is n plus 1. So given the m, we've already found the n. So given n, let's just summarize it now, given m, take n equals m plus 2, then mn plus 1 equals m, m plus 2 plus 1, which is m squared plus 2m plus 1, which is m plus 1 squared. So we've found, we've answered the question, we've said that given an m, take n equals m plus 2, then mn plus 1 is m plus 1 squared, which is a perfect square. Now, if I hadn't gone through this, it would have appeared that this was a rabbit out of a hat trick. And unfortunately, it's a consequence of the way that mathematicians often write their papers, that they don't include all of the reasoning, they just give you the conclusions. This would be a relatively simple example. But if I had simply said, given m, why do we take n equals m plus 2? You would have said, how on earth did he come up with that? What made him think of that? You know, and I look as though I've got some kind of magical ability. No, I don't, I just went in and said, well, how could I possibly get to that answer? And then it was just something I learned in high school. Okay? So you know, one of the techniques is just say, how could I possibly get the answer that I'm asked to find? Now I'm couching this in terms of classroom questions, but the same, the same issue arises when you're dealing with mathematical problems in the real world. Look at the problem. What does it tell you? What do you have to do? How could I get the answer that I'm, that I'm going for? So this might be a simple classroomy type example, but it has many of the elements of good mathematical thinking. And in this case it was, how can that answer possibly arise? And as soon as I'll sort of start to look at that, it just drops out. Okay? Recognising that was the entire key to the thing. Once you've done that, straightforward. Okay, let's move on and look at the next one. Question three. Question two is another one of these things that, when you first meet it, you think, wow, this is asking me for something really deep. How could I possibly come up with a quadratic, all of whose values are composite? How could I guarantee that all of the values are composite? Sounds like it's going to be really deep, but now if you sort of just take a breath, sit back and say, you know, how could it happen that those values are composite? Well, if they're composite, they have to be the product of two or two other numbers. Okay? Wait a minute. Why don't we just take, if that's going to always be the product of two whole numbers, let's make it the product of two whole numbers. And lo and behold, that is indeed of the form n squared plus bn plus c, where b and c are positive. That's all there is to it. We just wrote one down instantly. We didn't have to prove that one exists. We did prove it. We did it by just writing one down. And all we had to say to ourselves, oh, this sounds complicated, but no, all it means is that the values are products. And quadratics are products of things. So we just write it. We explicitly make it a product of two things. The value will always be the product of two numbers, n plus one and n plus two. And it's for all positive integers. So we'll have maybe two and three or whatever, so these are always greater than one. So this thing is composite, n squared plus three and n plus two is always composite. That's it. Didn't involve any advanced machinery, just thinking about what the problem asks us to do. Very similar with this one. You know, it sounds deep, but it's about the goalback conjecture, a problem that's been around for hundreds of years, hasn't been solved. How on earth can we do anything with the goalback conjecture? Well, the answer is just look at what it tells us and ask us what we can do with it. And first of all, we observe that this is talking about numbers n bigger than five, just so odd. Well, if n is an odd number bigger than five, then n is of the form 2k plus three, where k is bigger than one. I mean, normally we think of odd numbers as being of the form 2k plus one, but because I'm looking for prime numbers here, I'm going to write it as 2k plus three. And I can do that because n is bigger than five. So if a number is bigger than five, any odd number bigger than five is of the form 2k plus three, where k is bigger than one. Well, in that case, since 2k is bigger than two, because k is bigger than one, 2k is an even number bigger than two. So by goalback's conjecture, 2k equals p plus q, where p and q are primes. In which case, n, which is 2k plus three, is p plus q plus three, the sum of three primes. We've done. Actually it wasn't difficult at all. Once we sort of thought about what it says, all of the complexity is in this unsolved problem, goalback's conjecture. If we assume goalback's conjecture, which we were allowed to, because it says if that's true, then every even number bigger than five is a sum of three primes. One of which, in the case of our proof, is three. That was it, okay? Actually it wasn't difficult at all. It just looked as though it might be when we first met it. So another lesson we can learn is don't get put off because something looks complicated. Until you think about it, you don't really know whether it's complicated or not. Okay. A couple more on this assignment sheet, and then we're done with it. Questions five and question six, the last two questions, they're both induction proofs. So what I'm going to do is I'm going to do example five, and then I'll leave you to do number six. If you've already tried it and failed and you came here looking for an answer, you're not going to find it explicitly, but hopefully by watching me go through another example, namely number five, you'll be able to go back and do number six. Because these things are all very similar, at least the ones I'm giving you are all very similar. Not all induction proofs are similar, but these, the ones I'm giving you from number theory are all very similar. Okay. Now what the first thing you have to do is express this as some kind of an equation, you know, with a formula in it, because otherwise we're going to deal with this expression as sum of the first n odd numbers. So we're going to have to write down a formula for the sum of the first n odd numbers. So what it asks us to do is show that one plus three plus five plus, and then the nth one is two n minus one, and we have to show that equals n squared. Okay. Arguably the hardest part of this whole thing is figuring out what the last term is. Okay, so that's what we have to prove, and we're going to do it by induction, and we're told to use by induction. So we have to begin by looking at the first case for n equals one. This becomes just one equals one squared, which is true. So it's true for n equals one. Okay. So now let's assume the result. Let me call that star. So let me now assume star. And now I need to prove the same, the corresponding result with an n plus one in place of n. So why don't I just add the next term to both sides? The next term will be two n plus one to both sides of star. In which case I get one plus three plus five plus two n minus one plus two n plus one. This is the first n plus one. Okay, that's the first n plus one odd numbers. This is the next case in the induction. And now when I add two n plus one to the right-hand side, it becomes that. Wait a minute. I can now use another of those bricks that I learnt in high school. Thank you to my high school math teachers because now I can just get this one straight out. This was drilled into me. Boy, it's good to have these things at my fingertips. That's n plus one squared. We're done. It's the same formula with n plus one in place of n. Wasn't difficult at all. In fact, all the machinery I needed to solve this one, I learnt in high school. The one thing I didn't learn in high school, at least not very well, was how to strategise about a novel problem. And that's what this course is about. This is how do you take all of that great stuff you learnt in high school and strategise and use it and get new results and think about new problems. But before you do that, you have to ask yourself, what is the problem I'm given? What do I know about it and what am I having to prove? And if you misread a sentence, you're going to end up doing the wrong problem. There were a lot of people writing on the forums where what they were really saying was, I demonstrated all my high school skills and great virtuosity and solved a problem that I wasn't asked to solve. In this course, there's very little penalty for doing that except you feel bad. If this wasn't a course, if you were working for a large corporation, you might be out of a job or demoted or moved to a less interesting job. So this is a low penalty experience for how to think mathematically, which is why these MOOCs are great. OK? Well, as I say, I'm going to leave you to do question six. And having seen another example, I hope you're not put off by the formulas and the complexity in question six. It's really just logical thinking. OK? Mathematical thinking. OK. So much for assignment eight. Seemed hard at the time. Actually was hard. But hopefully it doesn't seem quite so hard now. It's not hard if you become a good mathematical thinker. And that's achieving that is what the whole thing's about. OK? How did you get on with assignment eight? Though the focus of this course is a particular kind of thinking, rather than any specific mathematics, the integers and the real numbers provide convenient mathematical domains to illustrate mathematical proofs. Those domains are number theory and elementary real analysis. The principal advantage is example domains is that everyone has some familiarity with both systems. Yet very likely you won't have been exposed to their mathematical theories. I'll take the integers first. Most people's experience with the whole numbers is by way of elementary arithmetic. Yet the mathematical study of the integers, looking beyond mere calculation to the abstract properties those numbers exhibit, goes back to the very beginnings of recognisable modern mathematics around 700 BCE. That study has gone into one of the most important branches of pure mathematics, number theory. Most college mathematics majors find that number theory is one of the most fascinating courses they take. Not only is the subject full of tantalising problems that are easy to state, but require great ingenuity to solve if indeed they have been solved and many haven't. But some of the results turn out to have applications crucial to modern life. Internet security being arguably one of the most important. Unfortunately we'll barely scratch the surface of number theory in this course. But if anything you see in this lecture arouses your curiosity, I would recommend that you look further. You're unlikely to be disappointed. The mathematical interest in the integers lies not in their use in counting, but in their arithmetical system. Given any two integers you can add them, subtract one from the other, or multiply them together and the result will always be another integer. Division is not so straightforward and that's where things get particularly interesting. For some pairs of integers, say 5 and 15, division is possible. 15 divides by 5 to give the integer result 3. For other pairs, say 7 and 15, division is not possible unless you're prepared to allow fractional results, which takes you outside the integers. If you restrict arithmetic to the integers, division actually leads to two numbers, a quotient and a remainder. For example, if you divide 9 by 4, you get a quotient of 2 and a remainder of 1. 9 equals 4 times 2 plus 1. This is a special case of our first form of theorem concerning integers, the division theorem. The division theorem says the following. Let's A, B, B integers with B greater than 0. Then there are unique integers Q and R such that A equals QB plus R and 0 less than or equal to R less than B. Where there are two parts to this theorem, there's an existence part, there are integers with this property and there's a uniqueness part. Those integers are unique with this property. In proving the theorem, we're going to take those one at a time. I'll first prove existence and then I'll prove uniqueness. So proof. As always, it's a good idea to begin a proof by stating the method you're going to adopt. In this case, what I'll say is I'm going to prove existence first and then uniqueness. I'm not using any standard method, so I'm not able to state that I'm using something like induction or whatever. But it will orientate the reader to say that the first part of the proof will deal with existence and the second part will deal with uniqueness. Okay, so we're going to handle existence. Well, look at all non-negative integers of the form A minus KB where K is an integer and show that one of them is less than B. What the theorem says is that among the integers A minus QB, namely the integers R, there is one with R between 0 and B. So I'm going to look at all possible candidates for A minus KB or A minus QB if you like. And I'm going to show that among those candidates, one of them satisfies that condition. And the K for which it satisfies that condition will be the Q that I'll take for the theorem. Well, first of all, I need to show that there are such integers. For example, take K equals negative absolute value of A. Then, since B is greater than or equal to 1, A minus KB equals A plus absolute value of A times B, which is greater than or equal to A plus absolute value of A, which in turn is greater than or equal to 0. Well, having shown that there are such integers by producing a K that gives me such an integer, I can choose the smallest one. Let R be the smallest such integer and let Q be the value of K for which it occurs, i.e. we'll have R equals A minus QB. So with R and Q defined in this way, I will indeed have A equals QB plus R. If I can show that R lies between 0 and B in this fashion, then I'll approve the existence. So to complete the proof, we show that R is less than B. Well, suppose on the contrary that R is greater than or equal to B. In other words, I'm now going to use proof by contradiction to use less than B. Well, if R is greater than or equal to B, then A minus Q plus 1B equals A minus QB minus B, which equals R minus B, which is greater than or equal to 0 by virtue of this assumption. Thus, A minus Q plus 1B is a non-negative integer of the form A minus KB. But R is the smallest such, and yet A minus Q plus 1B is less than A minus QB, which is R. And that's a contradiction. If R is the smallest such, we can't find a smaller one. There's a contradiction, hence R is less than B, because we obtained that contradiction on the assumption that R is greater than or equal to B. And that proves existence. If you haven't seen a proof like this before, it almost certainly looks very complicated. It's actually not complicated or deep. It's just intricate in its structure. The idea is to look at all of the possible candidates to make this true. Well, first of all, we have to show there is a possible candidate, and we actually do the exhibiting one. And then to get a candidate which satisfies this additional requirement, we pick the one where the R is the smallest. It's the minimal value of R that gives us one of these things. And then we use the fact that it's minimal in order to show that it has to satisfy this requirement. Each individual step is just elementary algebra. It's just somewhat intricate to push the argument through. There's a lot of arguments in mathematics like this. There's no deep mathematics involved. There's nothing more than elementary arithmetic and a little bit of algebra. But there's a bit of intricate structure to make the proof work. So if you haven't seen this kind of argument before, it'll probably take you several reads through. You need to look at it several times before it begins to make sense. Okay? Well, that proves existence. Then we need to look at it to uniqueness. To prove uniqueness, we show that if there are two representations of A, A equals QB plus R, and A equals Q prime B plus R prime, with 0 less than or equal to R and R prime less than B, then R equals R prime and Q equals Q prime. So the first part of the division theorem, the existence part that we've already proved, shows that there are representations of this form. What we're now doing is showing that if there are two such representations, then in fact they're identical. The R is the same and the Q is the same. Incidentally, you probably realize by now that the letter Q is going to denote quotients and the letter R is going to denote remainder. We haven't officially defined quotients and remained the X. I haven't been anything like that down, but I'm using the familiar notation. Because what we're doing is we're giving the mathematical underpinnings of the familiar knowledge that you probably have about, about arithmetic, dividing one into your binary. Okay, so we have an equation here. QB plus R equals Q prime B plus R prime. Let's rearrange it. Rearranging, I get R prime minus R equals B into Q minus Q prime. And I've labeled that equation one for later reference. Now I'm going to take absolute values in this equation. And when I do that, I get absolute value R prime minus R equals B times absolute value of Q minus Q prime, which I've labeled equation two. B is positive, remember? So when you take absolute values, B remains the same. But negative B is less than negative R, which is less than or equal to zero. And zero is less than or equal to R prime, which is less than B. So negative B is less than R prime minus R is less than B. Well, if R prime minus R is between negative B and plus B, that means that the absolute value of R prime minus R is less than B. So by two, B times absolute value of Q minus Q prime is less than B. And divide by B now to give me absolute value of Q minus Q prime is less than one. But everything here is an integer. And the only way you can have a pair of integers, the absolute value of the difference between them being less than one is that absolute value is zero, which means Q has to equal Q prime. And then by equation one, R has to equal R prime. And that proves uniqueness. And with it, we've proved the division theorem. Well, again, if you haven't seen arguments like this before, it's going to seem pretty daunting. But when you go through it step by step, this is really just very basic arithmetic arguments with inequalities. Take it step by step, and you should be able to follow the arguments to the end. Okay? Good luck with that. If this is the first full blown, rigorous proof of a theorem like this that you've encountered, you'll probably need to spend some time going over it. The result itself isn't deep. It's something we're all familiar with. Our focus here is on the method we use to prove conclusively that the division property is true for all pairs of integers. Time spent now making sure you understand how this proof works, why every step was critical will pair dividends later on when you encounter more difficult proofs. By gaining experience with mathematical proofs of simple results like this one, which is obvious, mathematicians become confident in the method of proof and can accept results that are not at all obvious. For an example of a result that's not obvious, in the late 19th century, the famous German mathematician David Hilbert described a hypothetical hotel that has a strange property. Hilbert's hotel, as it's become known, is the ultimate hotel in which it has infinitely many rooms. As in most hotels, the rooms are numbered using the natural numbers, one, two, three, et cetera. One night, all rooms are occupied when an additional guest turns up. I'm sorry, says the desk clerk, all our rooms are occupied. You'll have to go somewhere else. The guest, a mathematician, thinks for a while before saying, there is a way you can give me a room without having to eject any of your existing guests. Before I proceed with this story, you might like to stop the video for a moment and see if you can see the solution the mathematician guest has seen. The clerk is skeptical, but he asks the mathematician to explain how he can free up a room without ejecting anybody already in the hotel. It's simple, the mathematician begins. You move everybody into the next room. So the occupant of room one moves into room two, the occupant of room two moves into room three, and so on throughout the hotel. In general, the occupant of room N moves into room N plus one. When you have done that, room one is empty. You put me in that room. The clerk thinks about it for a moment and then has to agree that the method will work. It is indeed possible to accommodate an additional guest in a completely full hotel without having to eject anyone. The mathematician's reasoning is totally sound. And so the mathematician gets a room for the night. The key to the Hilbert hotel argument is that the hotel has infinitely many rooms. Indeed, Hilbert formulated the story to illustrate one of several surprising properties of infinity. You should think about the above argument for a while. You won't learn anything new about real world hotels, but you will come to understand infinity a bit better. And the significance of understanding infinity is that it's the key to calculus, the bedrock of modern science and engineering. And when you're satisfied you understand the Hilbert solution, try the following variants. First variant. The Hilbert hotel scenario is as before, but this time two guests arrive at the already full hotel. How can they be accommodated? In separate rooms, I should add, without anyone having to be ejected. Second variant. This time the desk clerk faces an even worse headache. The hotel is full, but an infinite tour group arrives. Each group member wearing a badge that says, hello, I'm N. For each of the natural numbers N equals one, two, three, et cetera. Can the clerk find a way to give all the new guests a room to themselves without having to eject any of the existing guests? And if so, how? Examples like the Hilbert hotel demonstrate the importance of rigorous proofs in mathematics. When used to verify obvious results, like the division theorem, they may seem frivolous. But when the same method is applied to issues we are not familiar with, such as questions that involve infinity, rigorous proofs are the only thing we can rely on. Now back to the division theorem. Well, the way I've stated the division theorem, it only applies to division by a positive integer b. There's a more general version that I'll give you now. So theorem, general division theorem, let a, b, b integers with b non-zero. Then there are unique integers q and r such that a equals q times b plus r and 0 less than or equal to r, less than absolute value of b. So the general division theorem is almost exactly the same as the previous division theorem we proved, except that instead of demanding that b is strictly positive, we're saying that b is non-zero. And then here we have the absolute value of b. And the proof follows fairly straightforwardly from the previous result. Then, since the absolute value of b is greater than 0, the previous theorem tells us there are unique integers q prime and r prime such that a equals q prime times the absolute value of b plus r prime and 0 less than or equal to r prime, less than absolute value of b. Well, now we simply let q equal negative q prime and r equal r prime. Then, since absolute value of b in this case with b negative is negative b, we get a is qb plus r, 0 less than or equal to r, less than absolute value of b. And that's this theorem. So we simply threw it back to the previous result. And with the general division theorem now established, we can formally give names to the two numbers q and r. Okay? And officially we say the number q is called the quotient of a by b and r is called the remainder. And so we've now cycled right back to something that you learned in elementary school about dividing numbers and that there were sometimes remainders, there were quotients and remainders. Okay? And now we've done it with some sophistication and we've shown that everything is well-defined. Now, how about that? Well, now that we have the division theorem available, we can look at the important property of divisibility. If division of a by b produces a remainder r equal to 0, we say a is divisible by b. Hence, a is divisible by b if and only if there's an integer q such that a equals bq. For example, 45 is divisible by 9, but 44 is not divisible by 9. The notation that we use for divisibility is this, b vertical line a denotes a is divisible by b. And let me give you a warning at this point. b divides into a or a is divisible by b is not the same as b divided by a with a slanted line. This guy is a relationship between a and b. It's true or false. This guy denotes a rational number. The result of dividing b by a in the rational numbers. So we have a property, a relationship, which is either true or false. And we have a notation for a rational number. Now, these are quite distinct concepts. This is something to do with two integers. It tells you whether two integers are in a certain relationship. This is a notation for a specific number. The reason I'm issuing this as a warning is that beginners in particular often confuse these two things. They both, after all, do have to do with division. But in this case, it's a property that has nothing to do with fractions, with rational numbers. This, it's a specific number. It's a specific rational number that we're referring to. Now, there wouldn't be as a problem if the notation wasn't very similar. And it gets worse when people with sloppy handwriting, and you've probably noticed that I'm one of them, often don't distinguish clearly between a vertical line and a slanted line. In fact, this one is already almost vertical. Maybe it would have been better if I'd written it like that. Okay, that's better. But because the notation isn't very similar, it's often very easy to get confused between these two things. The context does disambiguate. If you understand what's been discussed, you shouldn't have any problems. But bear in mind, especially when you're meeting this material for the first time, that there is an important distinction here. And in fact I've got a quiz coming up in a moment, which I hope is going to help cement your understanding of this notion so that you'll be able to distinguish it from this one. Okay, and once we've got a quiz about divisibility lined up, we can define the unimportant notion of a prime number. A prime number is an integer p greater than one that is divisible only by one and p. Notice that we explicitly exclude one from the prime numbers. So the first prime is two, the next one is three, then five, then seven, then 11 and so forth. Now for that quiz about divisibility. Okay, how did you do? Well, remember the condition we have to check is this one. B divides A if and only if there is a Q such that A equals BQ. Well, let's see what we have. The correct ones are B, D, F, G and H. Why don't I have A? Well, remember the whole definition of divisibility was under the assumption that B is non-zero. We can't consider divisibility by a zero number. So that one doesn't work and likewise that one doesn't work. And of course the reason this one doesn't work is because seven simply doesn't divide into 44. Okay, do be careful about this though. Division by zero is problematic under any circumstances and so we ruled it out and in this case it was ruled out in the very statements of the divisibility theorem. Okay, how about this one? Well again, the condition we have to check is that B divides A if and only if there is a Q such that A equals BQ. And again, this is under the assumption B non-zero. Everything is an integer here. But the correct ones are that, that, so why are the others false? Well, this one is false because that simply doesn't divide that. However, if you, so we've got out a calculator and actually tried to divide it, then you were missing the obvious point. This is an even number and that's an odd number and we cannot have an even number dividing an odd number. So if you had to do any arithmetic for this, then you were making the mistake that I've been warning you against of acting as if you were in high school. This is a course about mathematical thinking and what I hoped you would have done is to think about this mathematically and realise that you can't have an even number divided into an odd number. So this is a cautionary reminder that this is not about jumping into calculations and applying procedures. This is all about thinking about a problem. Okay? So that's false but I would hope that you would recognise it's false without doing any work whatsoever other than a little bit of thinking. Okay? I occasionally like to slick things like that into quizzes just to keep people on their toes. Well, this one is false, of course, because that's simply not the case at 2n divides n squared. The reason, the others are false, well this is false for various reasons, I mean it's the same. 2n simply doesn't divide n squared for various reasons. The reason this is false is because if n can vary over z, then that would include n equals 0. And so we can't have this because we can get a 0 coming in. That includes n equals 0. Likewise for this one. Okay? Same thing. So this one fails because 0 is included. If I excluded 0 it would be fine. The only difference between these two is that this is the positive integers, 1, 2 and 3 and so on and so forth. This one because it's all of the integers includes 0. Okay? Well, let's move on now. I want to prove a theorem now that gives the basic properties of divisibility theorem. Let A, B, C and D be integers with A non 0. Then the following all hold. 1, A divides 0, and A divides A. 2, A divides 1. If and only here, A equals plus or minus 1, 3. If A divides B, then C divides D, then A, C divides BD. This is under the assumption C is non 0 because we can't have divisibility by 0. The division theorem itself excludes the case of division by 0. 4, if A divides B, then B divides C, then A divides C, and this is under the assumption B non 0. Again we have to exclude the possibility of division by 0. 5, A divides B, then B divides A. Those two hold if and only if A equals plus or minus B. 6, if A divides B, then if B is non 0, then absolute value of A less than or equal to absolute value of B. You can only divide smaller numbers into bigger numbers. Or at least numbers that are no bigger than. You can't divide a number B by something that's bigger than it in absolute value. One more. 7, if A divides B, then A divides C, then A divides Bx plus Cy for any integers xy. You prove all of these by going back to the definition of divisibility. What I'm going to do is I'll prove two of them as examples. Let me just pick number 4. Let's prove 4, that's this one here. All the proofs are essentially the same idea. If A divides B and B divides C, then that means there are integers D and E such that B equals DA and C equals EB, that's the definition of divisibility, in which case C equals D times E times A, which again by the definition of divisibility means A divides C. Let me do one more. Let me do 6. So A divides B, so that means since A divides B, it means there is a D such that B equals DA, in which case taking absolute values, absolute value B equals absolute value D times absolute value A, and since B is non-zero, we know that the absolute value of D will have to be greater than or equal to one. So the absolute value of A is actually equal to the absolute value of B. B is non-zero, so D can't be zero, so its absolute value is greater than or equal to one, and if the absolute value of D is greater than or equal to one, then A has to be less than or equal to B. The other statements in the theorem are proved similarly. You just take it back to the definition of divisibility, and remember the definition is B divides A if and only if there is a Q such that A equals BQ, and in all cases you go back to the definition and then the theme drops out in a couple of lines. OK, well that lists all of the basic properties of divisibility. OK, it's time to prove the fundamental theorem of arithmetic. Theorem, every natural number greater than one is either prime or can be expressed as a product of primes in a way that's unique except for the order in which they are written. For example, 4 is 2 times 2 or 2 squared, 6 is 2 times 3, 8 is 2 cubed, 9 is 3 squared, 10 is 2 times 5, 12 is 2 squared times 3, and so on and so on, 3,366 equals 2 times 3 squared times 11 times 17 and so forth. I worked that one out in advance by the way. OK, so 2 is self is prime, 3 is prime, 4 is a product of primes, 5 is prime, 6 is a product of primes, 7 is prime, 8 is a product of primes, 9 is a product of primes, 10 is a product of primes, 11 is a prime, 12 is a product of primes, and so on and so on and so forth. The expression of a number of a product of primes is called its prime decomposition, and when you know the prime decomposition of of a number. You know an awful lot of information about that number and how it behaves with relation to other numbers. What we proved part of this a little while ago proved the existence part. I'm going to give you another proof of existence in a moment, but the new part is to prove uniqueness. The uniqueness proof will require Euclid's Lemma. That says, if a prime P divides a product AB, then P divides at least one of AB. The proof of Euclid's Lemma is not particularly difficult, but it would take me outside the scope of this course. Remember the focus of this course isn't to teach you number theory and number theoretic techniques, it's to develop mathematical thinking and it would be too much of a detail from that goal in order to prove Euclid's Lemma. But if you just google Euclid's Lemma, you should be able to find some references that would lead you to a proof. Let me give you the existence proof. More precisely, let me give you a new proof of existence. Remember, this is the statement of the theorem. Any natural number greater than one is either prime or can be expressed as a product of primes in a way that's unique except for their order. Then we're going to prove existence. The previous proof of existence that I gave you used a method of mathematical induction and I gave it as an illustration of the method of induction. But I'll give you a different proof now. I'll prove it by contradiction. Suppose there were a composite number, that's a non-prime, that could not be written as a product of primes. Then there must be a smallest such number. Call it n. Since n is not prime, there are numbers a, b strictly between one and n such that n equals a, b. If a and b are primes, then n equals a, b is a prime decomposition of n. And we have a contradiction because n was chosen so as not to have a prime decomposition. It was actually the smallest number that didn't have a prime decomposition. Well, if it's not the case that n be primes, then at least one of them must be composite. But if either of a, b is composite, then because it's less than n, it must be a product of primes. So by replacing one or both of a, b by its prime decomposition in n equals a, b, we get a prime decomposition of n and again we have a contradiction. That proves existence. That's the first part of the proof of the fundamental theorem of arithmetic. Now let me prove uniqueness. So what I have to prove is that the prime decomposition of any natural number n greater than one is unique up to the ordering of the primes. And I'll prove it by contradiction. So assume there is a number n greater than one that has two or more in fact different prime decompositions. Let n be the smallest such number. Let n equal p1 times p2 times da, da, da, times pr equals q1 times q2 times da, da, da, qs be two different prime decompositions of n. So we have n as a product of r primes and we also have n as a product of s primes, some of the primes being different. Well p1 is a prime factor of n. So p1 divides n. So p1 divides this product. Since p1 divides q1 times q2 through qs, and let me stop at this moment and see what I've done, I've said p1 divides n. So p1 divides this and I'll split that into two that way and notice that p1 divides q1 times that. And I've done that in order to apply Euclid's Lemma. So now let me continue that sentence. By Euclid's Lemma either p1 divides q1 or p1 divides the product q2 through qs. Remember Euclid's Lemma says that if I have a prime number p, and if p divides a product a times b, then p divides at least one of a b. So Euclid's Lemma was that if p divides a product a b, it divides at least one of those two numbers. So from here p1 divides this product. I've now written that as a product of two numbers. And I can now apply Euclid's Lemma and say that if p1 divides that times that number, it must divide at least one of them. Either p1 divides q1 or it divides the other parts, or maybe both of them. Okay, hence either p1 equals q1 or else p1 equals qi for some i between two and s. Okay, if p1 divides q1, then they're both primes, so the only possibility is that p1 equals q1. Or else p1 divides this product of primes, which means that p1 actually is one of those primes. So first case, I know that p1 equals q1. Second case, I know that p1 equals qi for one of the i's between two and s. But then we can delete p1 and qi on the two decompositions instead, which gives us a number smaller than n that is two different prime decompositions. Contrary to the choice of n is the smallest such. Okay, having established that p1 is equal to one of these qs. We don't know which one, but it's going to be one of them. I can delete the p1 from here. I can delete the appropriate q from there. And when I delete p1 from there and some q from there, I still have equality. I still have a prime decomposition of a number. But having deleted that common prime, the number that I get is smaller than n. So I get a number smaller than n that has two different prime decompositions. The number smaller than n will be p2 times pr, so that will be the number smaller than n. And I've deleted a number from here, and what's left is a prime decomposition here. So the proof involves simply identifying the fact that, recognizing the fact, that this uses Euclid's lemur, that you're going to have the same prime on both sides. You can delete it and get two different decompositions of a smaller number. And that proves uniqueness. Okay, well, now we've proved the fundamental theorem of arithmetic. Cool, huh? So, did that make sense? The chances are, you're going to have to spend some time trying to come to grips with this. Though I know you're familiar with arithmetic, especially a whole number of arithmetic, this is probably the first time you've tried to really analyze numbers. See how you get on with assignment nine. Well, as I stressed in the lecture, you've got to be careful in distinguishing between these two notions. You have a notation for a relationship, something that's true or false. A is divisible by B, or B divides A. And you've got another notation for a number, the result of dividing A by B. This makes sense when you're talking about the integers. This actually is not defined for the integers. I mean, for some integers you get an answer. But this is, I mean, division. Actual division is an operation, is an operation, not on the integers. It's an operation on the rational numbers or the real numbers. In the integers, all you have is addition, subtraction and multiplication. You don't have division. What you can do is say whether you have divisibility. And divisibility is defined in terms of multiplication, okay? So it's all about the distinction between the two. You've got a property, something that's true or false. And you've got a notation for a number. This is an actual number. Okay. So the actual answer I'm going to end up with is the thing here at the bottom. To the question. There's concisely and accurately as you can the relationship. This is the relationship. B divides into A, or A is divisible by B, if and only if A divided by B happens to be an integer. That's summarising what I just said above there. That's a notation that denotes a rational number. Rational number. That's a different set of numbers. This denotes a relation B divides A. Ie, there is an integer Q such that A equals QB. And whenever you're dealing with divisibility, with this notion, and you're working with the integers, you have to reduce that abbreviation to this. There is an integer such that. That's what that means. Okay. That means that. That nothing means that. Okay. And then to get down to here, in the case where you do have divisibility, then of course the Q that's here is the quotient. I mean we use Q to denote the word, to stand for the word quotient anyway. But notice that this says nothing about division. Division doesn't arise here. It's all about the result of multiplying two numbers. So this makes perfect sense when you're talking about the integers. Okay. So we're not doing sort of arithmetic in the sense of calculating here. You know, obviously everything that's involved here about dividing one whole number by another. You know, you learnt it in elementary school. It is just division of whole numbers. The focus here, however, is on what you're doing within certain systems of numbers. We have two systems of numbers here. We have the integers and we've got the rationals. They're just two separate systems of numbers. In the case of the integers, you can add, subtract and multiply. In the case of the rationals, you can add, subtract, multiply and divide. But they're different systems of numbers. And so the focus here is on what you can do in the integers. That's what number theory is about. And then in the very last lecture, lecture 10, we'll be actually looking at the rationals and the rails. But that's a different system of numbers. You can do different things with it. So we're taking a more sophisticated look at elementary arithmetic, but it still is, after all, elementary arithmetic. Okay, let's look at numbers two and three. Where the issue with all problems like this is you have to express the divisibility property in terms of multiplication. Remember, divisibility is a property of pairs of integers. You can't divide integers. All you can do with integers is add, subtract or multiply them. And divisibility arises when you take this definition. A divides B if and only if there is a Q, an integer. So it's that B equals Q terms A. So the general strategy for dealing with the divisibility, actually it's really the only strategy, is you replace issues like this. You replace a statement like that with a statement about multiplication. Because the point is that there is no operation here. There's nothing that that's not an operation to deal with. It's not an arithmetical operation on the integers. Remember, this is all about the integers. So you have to express it in the language of the integers. And the language of the integers allows you to talk about addition, subtraction and multiplication, but not division. So how do you answer this one? What's the proof? Well, this one is sort of immediate because the very definition of divisibility explicitly excludes A not equal to zero. It excludes A being zero. So it's false, and that's the reason. Nine divides zero. Well, that's definitely true. And to show that it's true, you simply express the definition. You express it in terms of the definition. So you would have to show that there is a Q. Look at the definition. You have to show there is a Q. So it's zero equals Q times nine. Well, of course there is. Zero itself is one of those things. Okay, so it's false. It's true. This one's false. For the same reason A was false. You're not allowed to have A equal to zero in the notion. That includes that requirement. This one is definitely true. And the proof is just right in terms of this. This is basically what you end up having to write in each case. If you look at these, that's what I'm going to end up writing. I'm going to end up reformulating it in terms of the definition. That's really all it involves. Just reformulate the statement in terms of the definition. Sure that the definition is true. Okay, well in this case it's true because Q equals one makes it true. In this case, we know that there's no such Q. I mean, you could argue it just by, since any possible Q would have to be less than, what? Less than seven, say. You could actually explicitly, if you wanted to prove that even more detail, you would just let Q be all of the possibilities that have a chance of being that. Q equals one, two, three, four, five, six, seven. And seven sevens is already 49, so you're out of it. So you actually only need to go to six, of course. So you could explicitly list all of the possible Qs if you wanted to. But that would be so trivial. I think you could just leave it like that. At this level, if we were talking to kids in the elementary school, we would ask them to list all of the possible Qs and make sure that none of them give you the answer 44. But at this level, you can just take that as being obvious. This one's certainly true. You exhibit the Q, namely Q equals minus six, Q equals negative six. Ditto here, you exhibit the Q, and again it's negative seven. Here you exhibit the Q, and the Q is eight. Here you need to show that four n one divides n. Well, that's certainly true. And the reason is that for any n in z, n equals n times one. Right? That's trivial, right? One divides everything. Four n in z, four n in n, n divides zero. That's true, because again for any n in z, zero equals zero times n. And this one, this is what we're going to be careful with, because if we're quantifying over all of the integers, that includes zero itself. And you're not allowed to have zero dividing anything. Okay, that's excluded from the definition. So this is the one you have to be careful with, because it includes zero. It's the only case where it goes wrong, you only need one counter example to make a universally quantified statement false. And that one counter example is all it takes to get rid of that one. Okay, so that one's false. Okay? Now we've done them all. Notice it was just the same pattern, express divisibility in terms of the definition of divisibility. And then in each case it just dropped right out, because this is after all just elementary whole number arithmetic. It's not that there's anything deep going on here, it's just that we're looking at it in a somewhat more sophisticated fashion than you did when you were in the elementary school. Everything you need to know to solve this, you learned in elementary school, it's just that we've now got a little bit more of a sophisticated gloss on it. Okay? Well, as with the previous example, what you have to do is reduce each of these to the definition of divisibility. Remember, divisibility is a property of pairs of integers. This isn't a division. It's obviously related to division. But you don't have division in the integers. What you can do with the integers is you can add them, multiply them, and subtract them. I mean subtraction just being the inverse of addition. But you can't divide them, okay? You can have a property of divisibility, but to discuss divisibility within the integers, you have to reduce it to a discussion of essentially multiplication. Okay? So how would you show that a divides 0, that you've got divisibility of 0 by a? Well, you observe that actually because of the properties of 0, 0 is equal to 0 times a. So in particular 0 satisfies the requirement for divisibility. Okay? There is a q in z, so it says 0 equals qa, namely q equals 0. So by definition, a divides 0. Okay? Similarly, in the case of a dividing a, because of the properties of 1, a equals 1 times a. So again, the definition of divisibility is satisfied, and it's satisfied in this case, by letting q be equal to 1. So by definition, a divides a. Okay? So that was that one. And the rest are essentially the same idea. A divides 1 if and only if a equals plus or minus 1. Okay? We've got two implications to prove. First of all, let's assume that a equals plus or minus 1. Then again, all you have to do is show that there is something such that 1 equals q times a. Well, if a equals plus or minus 1, there certainly is, right? Conversely, ah, that should be if. Little typo there. Conversely, if a divides 1, then for some q, 1 equals qa, the definition of divisibility. But if 1 equals q times a, then the absolute value of 1 is the absolute value of qa, which is the absolute value of q times the absolute value of a. And if 1 equals that, then the only possibility, because these are positive integers now, is that absolute value of q is absolute value of a is 1. That's the only way you can get 1. And so if the absolute value of a equals 1, then a has to be plus or minus 1. Let me just do one more and then let you do all of the rest. If a divides b and c divides d, then a c divides bd. Okay? Well, we know that there are q and a, such that b is qa, definition of divisibility. D equals rc, definition of divisibility. Hence, multiplying the two together, you've got bd is qa times rc, because when you rearrange them, qr times ac. So by definition, ac divides into bd. And the others are essentially the same idea. In each case, you just reduce it to the question of multiplication, through the definition of divisibility. So you never actually do any dividing. You express division in terms of multiplication. And you can do that because division is the inverse of multiplication. Okay? So the whole thing is going to work out. So these proofs are always typically just one or two lines. They're really just a matter of translating what it is you're having to prove into divisibility. So the first line of any of these arguments really is just, is a matter of re-expressing what it is you're having to prove in terms of divisibility, in terms of multiplication, via the definition of divisibility. Okay? Well, that's it. How did you get on with assignment nine? There was a lot there, so you might not have been able to finish it yet. Don't worry, you can finish that later. You won't need any of it for this final lecture, where we're going to apply our mathematical thinking to studying the real numbers. Incidentally, if you're not familiar with elementary set theory, you should look at the special course reader on the subject before you proceed with this lecture. Numbers arose from the formalisation of two different human cognitive conceptions, counting and measurement. Based on fossil records, anthropologists believed that both concepts existed and were used many thousands of years before numbers were introduced. As early as 35,000 years ago, humans put notches into bones, and probably wouldn't sticks as well, but none of those have survived, or at least haven't survived and been found, to record things. Possibly the cycles of the moon or the seasons. And it seems probable they use sticks or lengths of vine to measure lengths. Numbers themselves, however, abstractions that stand for the number of notches on a bone or the length of a measuring device, appear to have first appeared much later, around 10,000 years ago, in the case of counting collections. These activities resulted in two different kinds of number. The discrete counting numbers used for counting, and the continuous real numbers used for measurement. The connection between these two kinds of numbers was not finally put onto a firm footing until the 19th century with the construction of the modern real number system. The reason it took so long is that the issues that had to be overcome were pretty subtle. Though the construction of the real numbers is beyond the scope of this course, I can explain what some of the problems were. The connection between the two conceptions of numbers was made by showing how, starting with the integers, it's possible to define first the rationals and then use the rationals to define the real numbers. Starting with the integers, it's fairly straightforward to define the rational numbers. A rational number is, after all, simply a ratio of two integers. Well, it's actually not entirely trivial to construct the rational numbers from the integers. You want to define a larger system, the rationals, that extends the integers, by having a quotient A over B for every pair AB of integers, where B is non-zero. But how will you go about defining such a system? In particular, how would you respond to the question, what is the quotient A over B? You can't answer in terms of actual quotients, since until the rationals have been defined, you don't have quotients. If you're interested, you can find an account of the construction of the rationals from the integers in many books and on the internet. But I'll mention once again that you should be cautious about mathematics found on the internet. The point is, constructing the rationals from the integers, while it has some subtleties, is fairly straightforward. Constructing the reals from the rationals, however, is a lot more difficult. In this final session, I'll look at some of the issues involved in constructing and using the real numbers. We'll start by looking at some properties of the rationals. With the rationals, you've got a system of numbers that's adequate for all real-world measurements. This is captured by the following property of rational numbers. And I'll give it as a theorem. If r and s are rationals with r less than s, then there's a rational t such that r is less than t is less than s. Before I prove that, let me give you a little note. This property is called density. The rational line is dense. If we have the rational line, what density says is that whenever you have a rational number r and another rational number s, we can find a rational number t in between r and s. OK, let's prove that. Well, why don't we just take t to really mean of r and s? Well, clearly r is less than t is less than s. And the only question is, is t a rational number? Well, it obviously is, but let's prove it. Then r be equal to m over n, s equal to p over q, where m, n, p and q are integers. And t is equal to a half m over n plus p over q, which equals mq plus np over 2nq cos mq plus np and 2nq are integers. So that proves that t using q, which was pretty obvious anyway, but since we are focusing on how we prove things in mathematics, at this stage I thought it was a good idea to actually go through a formal proof that one could give. And density means that the rationales are good for doing measurements because what density tells us is that we can get rational numbers as close as we wish to any particular length. If this is slightly smaller than a particular length and if this is slightly larger, then we can always find another rational that's even closer. On the other hand, density does not mean that there are no holes in the rational line. Let me just emphasise that the proof finished here. So that was the end of the proof. And with this remark, we're on the edge of an abyss of understanding. At least it's an abyss if you haven't seen this before. This is some interesting stuff coming up. It's stuff that wasn't fully worked out until the late 19th and early 20th centuries. There's an example of a hole in the rational line. There's root 2. And let me just make this a little bit more precise. Suppose I define A to be the set of all those rationales X, which are either negative or non-positive, or X squared is less than 2. And let me define B to be the set of all those rationales, such that X is positive and X squared is greater than or equal to 2. Let me do a little diagram. We'll have 0 here. And A is a set that's going to go out to infinity at the left. And it's going to go to some point here. That's going to be A. Let's delineate things here. So A is going to be a set here. B is going to be a set. Also it's going to go out to infinity. And we're going to have every element of A to the left of every element of B. And here is where root 2 would be if root 2 existed in the rational line. This is the rational line, Q. So we've got the rational numbers. And I've split the rational numbers into two sets. A, which is everything negative and anything whose square is strictly less than 2. And B is everything which is positive and whose square is greater than or equal to 2. And this splits the rational line into two pieces. Okay? So clearly A union B is the rational line. But, and here's the kicker, A has no greatest member and B has no smallest member. Now it was by considering situations like this that mathematicians in the late 19th and early 20th century were finally able to figure out a rigorous construction of the real number system, a theory of real numbers. But it took a couple of thousand years to get to that stage from the time when the ancient Greeks discovered that the square root of 2 was irrational. What this tells us is that the rationals are inadequate. I can't even spell inadequate now. Inadequate to do mathematics. Why do I say that? Because in Q we cannot solve the equation x squared minus 2 equals 0. So the rationals are fine for measuring things. I mean they're fine for doing carpentry and for doing various kinds of geometrical things and for building things and for tracking the stars and doing astronomy and so forth. That's fine. So long as we can get by with, say, the 10 decimal places of accuracy or whatever we need. But if we want an actual solution of equations, then the rationals aren't good enough. Because even a simple quadratic like x squared minus 2 equals 0 cannot be solved. The way out of this problem, this was the work that was done in the late 19th, 30th, 20th centuries, was to say well we've got several different systems of numbers. We've got the natural numbers, then we've got the integers where we add negative numbers or we had negative versions of the natural numbers. Then we have the rationals where we have quotients of integers. These are good for counting. These are good for doing arithmetic when we want to have negative values like my bank account. These are good if we want to measure things. If we want to do mathematics, we have to go one step further to the rails. By the way, this was, in many ways, it was a historical progression. Not quite because z sort of got in in a different way. But going from the natural numbers to the integers to the rational numbers to the real numbers, that was largely a historical progression. And it was done in order to have greater expressivity. When arithmetic began about 10,000 years ago, it was exclusively the natural numbers and it was introduced to provide a monetary system in some area, in what was known as the Fertile Question Region, present-day Iraq, essentially. Then as bankers came along and they wanted to keep track of people's negative bank accounts, I haven't been a little bit flippant, but that was pretty close to what happened. Negatives were introduced, rationals were introduced to measure things in the world, and then the real numbers are required to do mathematics. To get the real numbers from the rational numbers, what mathematicians in the late 19th century did was find a way to fill in the holes. There were these holes in the rational line and the rails were constructed by filling in the holes. Okay? Fill in the holes in the rational line. Well, I'm going to give you a few more details about how this was done, but I should mention that there were some mind-blowing elements to this. There were some really interesting surprises in store for the mathematicians that made this step. One was the fact that however you count it, there are more holes in the rational line than there are rational numbers. Now, there are infinitely many rational numbers. Nevertheless, the infinitude of the holes in the rational line vastly outweighs on an infinite scale the infinitude of the rational numbers themselves. It was a super infinity of holes. That meant that when those holes were filled in, the number system that was obtained, the real numbers contained uncomparably more numbers than the rational numbers. They're both infinite, but mathematicians had to develop systems for counting infinite collections in order to cope with this. And it turned out that the real numbers, as a set, has infinitely many more numbers than the rational numbers in an infinite sense. Now, making that precise is definitely beyond the scope of this course, but I will be able to get into the beginnings of the considerations that led to those surprising conclusions. Okay, well, that's all coming up within the next few minutes. Well, before I give you any details of how the real numbers were constructed, I need to do a little bit of preliminary work. I need to introduce or reintroduce, if you've seen it before, the notion of intervals of the real line. Let A be real numbers with A less than B. The open interval, A B, written with parentheses this way, is a set X in R such that A less than X less than B. So a simple diagram would be this. We've got a point A, we've got a point B. This is now the real line. And the open interval, A B, is a set of all the numbers strictly bigger than A and strictly less than B. But the interval excludes these two. So it's an interval of the line. It's a segment of the line, but it excludes the two endpoints. The closed interval, A B, written with square brackets like this, is the set X in R, A less than or equal to X less than or equal to B. So the diagram for this would be we've got A, and we've got B. This interval actually includes the two endpoints. So the distinction between these two is that A and B are elements of the closed interval, but A and B are not elements of the open interval. Now, this may seem like splitting hairs. After all, this is a segment of the real line, so there are infinitely many numbers in here. In fact, as I just indicated a moment ago on that previous page, the infinitude of the real numbers in there is mind-bogglingly bigger than the infinitude of the rational numbers in there, and the rational numbers in there is already an infinite set. So we've got infinitely many numbers in here, infinitely many points in this line, and yet I'm splitting hairs between these two points at the end. That sounds like splitting hairs, but take it from me. This is a big, big, big difference of distinction. The distinction between these two turns out to be huge, and it's closely bound up with the reasons why the rationales are inadequate to do with mathematics and the reals are actually good for doing mathematics. We're going to use this notation in order to talk about what had to be done to go from the rational numbers to the real numbers. I'm not going to do the construction, it's way too deep for a course like this. It's first and second year university level mathematics, and it takes most of us a long time to understand it, but I will open the door to such a study. There are some variations on the notation. Let me give you those. I'm not doing any mathematics here, I'm just giving you some notation. This is what's known as a half-open or half-closed intervals. So we have half-open intervals or half-closed at the same thing, really. Such as AB, where that's closed, and that's an open, square brackets, parenthesis. That's a set of all X in R, such that A less than or equal to X less than B, and the other way would be AB. The other way is a set of all X in R, such that A less than X less than or equal to B, and this one is called left-closed, right-open, and this one is called left-open, right-closed. So the word closed means you include the endpoint, and open means you exclude the endpoint. So if it's left-closed, the left endpoint is included. If it's right-open, the right endpoint is excluded. And one more bit of notation. We sometimes include intervals that stretch all the way to infinity, and we write things like negative infinity A is a set of all X in R, such that X is straight left and A. It's everything to the left of A, or we might write something like negative infinity A with a closed bracket here, with a square bracket to denote a closed part of the interval. That would be the set of all X in R, such that X less than or equal to A. Similarly, I could have A going out to plus infinity, would be the set of all X in R, such that X strictly greater than A, or I could have closed A infinity, which is the set of all X in R, such that X greater than or equal to A. And with one final remark, I'm done with this summary of notation. We don't have something together with infinity closed, and we don't have infinity closed together with something. You can't have infinity next to a square bracket, because infinity is not a real number. So there's no possibility of this guy, whatever it is, being an element of the interval. Notice that when I defined these, there was no mention of infinity. I just said all those X's to the left of A. Ditto, ditto, ditto. Okay, with this notation available, we can now move ahead and look at how we go from the rational numbers to the real numbers. We can look at what goes wrong with the rational numbers in a deep precise sense, and I can give you some indication of how we would go about rectifying the deficiencies of the rational line. Okay, coming up next. Where the key property that the real numbers have, that the rationales don't have, is what's known as a completeness property. In a nutshell, the completeness property is what makes the real numbers great for doing mathematics, and the absence of which makes the rationales inadequate for doing mathematics. So it was a formulation of a system of numbers that satisfied the completeness property, and indeed a formulation of the completeness property itself, was one of the crowning glories of late 19th century mathematics that set up the 20th century mathematics upon which most of modern science and technology depends. So this is a big deal, folks. Given the set A of reals, a number B such that for all A in A, A is less than or equal to B, is said to be an upper bound of A. We say B is at least upper bound of A if, in addition, for any upper bound C of A, we have B less than or equal to C. Well, duh! What else would a least upper bound be? It's the least one. It's an upper bound and it's the least one. Why am I making such a big deal of this? Because it was by making a big deal of this that mathematicians were able to formulate this, and it was by formulating this that mathematicians were able to construct the real number system, which meant that after 2,000 years of effort, mathematicians finally had a system of numbers that was adequate for doing modern science, physics, technology, et cetera, et cetera, et cetera. In mathematics, it's often the case that something that seems trivial turns out to be fundamental and have enormous consequences. And this is one of those moments. Okay? So I've been very precise about the definition because mathematicians found that it was only by being precise that they were able to figure out how to proceed. The notation we use for least upper bounds is this, least upper bound of A. So the notation for the least upper bound of a set A is lub A. Let me make a note. We can make the same definitions for N, for Z, and for Q. The completeness property of the real number system says that every non-empty set of rails that has an upper bound has a least upper bound. And let me stress that that least upper bound is in the set of real numbers. This one simple, elegant statement is the key to the real number system and to most of modern mathematics. Before I go any further, you really ought to take a look at assignment 10.1. I say you should at least look at it. Ideally you should try to do it, and there was many of the exercises as you can. You need to be familiar with the background material about upper bounds and least upper bounds before we progress. We're about to meet material that most beginners, beginners at universal level mathematics, find incredibly difficult. To go back to my favourite example about riding a bike, it's one of those transitions in life where it seems impossibly difficult until you get the hang of it and then it suddenly seems remarkably straightforward and you wonder why it took you so long. So it's not that we're facing something terribly difficult. It's just that we need to go through some kind of a shift going from thinking it's impossible to thinking, well, okay, that seems straightforward. It's one of those difficult transitions that when you look back, with retrospect, don't seem to difficult at all. We're facing one of those and I strongly recommend that you take a look at that assignment before you go any further. You might very rapidly find yourself lost in the next few minutes of lecture. Okay? Having said that, I'm going to take a pause now and then I'll come back in the second video connected with lecture 10 and I'll see you then. Welcome back, as they say in the media, where the correct answers are number two and number three. Five is not the maximum element of this interval. In fact, five is not an element of this interval at all. This, remember, is a set of all x in the rails, such that zero is strictly less than x, strictly less than five, so five itself is not a member of there. So not only is it not the maximum element, it's not an element at all. Seven is actually not a member of this interval, for the same reason as five wasn't an element of this interval, but nevertheless seven is still the least upper bound. There is no smaller upper bound of that interval than seven. So it is a least upper bound. So least upper bounds are not the same as maximums. And in this case, zero is a member of that set. That interval is defined, remember, as a set of rails, such that zero less than or equal to x, less than or equal to one. And in this case, the end points zero and one are elements of the interval. So zero is in there, and it's clearly the minimum element of that interval. Where the answer is, the first one is correct. This is what it means to say that the rational line is dense. Between any two rationals, you can find the third one. The second one is actually true, but it doesn't express density. It doesn't express density because whether or not there is an irrational number between two rationals is sort of irrelevant. The question is about the rational line being dense. This is actually making a statement about the real line. So it's true, but actually irrelevant to the notion of density of the rational line. So that one's not true. I mean that one's true, but it's not the answer to the question. And this one, that's actually expressing the notion of completeness. Now, if by least upper bound we mean least upper bound in the rationals, in the rational line q, then that would say that the rational line is complete, which is false. If however we interpreted this to mean every set of rationals that is bounded above has a least upper bound in the real numbers, then that would be an instance of the completeness of the real line. And this issue about the existence of least upper bounds is what distinguishes the reals from the rationals. And it's what makes the reals a very powerful system for doing advanced mathematics and calculus in particular and demonstrating the fact that the rationals is not complete is what demonstrates the impoverished nature of the rationals in terms of mathematics and doing things like calculus. Okay? How did you get on? What I want to do now is, well actually really what I want to do is introduce the beginnings of the subject known as real analysis. Now this isn't real analysis as opposed to fake analysis. Real here is essentially short for real numbers or for the real number system if you like. It's the analysis of the real numbers. And I'm going to begin with a theorem. The rational line is not complete. Now if you've done that assignment, assignment 10.1 that I asked you to do, you should be familiar with what that means but let me remind you in case you decided to play it risky and go ahead without doing that assignment. Well let me just remind you that completeness means if a subset of reals has an upper bound then it has a least upper bound in the set of reals that is. That was completeness as applied to reals. But as I mentioned at the time, these notions also apply to any set. So in terms of the rationals completeness would mean if a is a set of rationals having an upper bound then it has a least upper bound in the rationals. What this theorem says is that this property does not hold for the rational numbers. Remember R stands for real numbers here. But if I replaced R by Q and talked about the rational numbers then this property would not hold. It does however hold for the real numbers. That is the completeness property for the real line. We won't be able to prove that but I'll be able to indicate how it's possible to construct the reals in order to make it possible to prove that. Okay, here's the proof of the theorem. Let A be the set of all rationals R such that R is non-negative and R squared is less than two. Now already you can probably sense what's going on. This is going to hinge around that property that the square root of two is irrational. So let me do a picture. Here's zero. Here's two. A is going to be a set where everything is going to be greater than or equal to zero. And A is going to go up to some point less than two. Well, A only contains rationals less than two whose square is less than two. So those rationals themselves are less than two. So it's going to be something like this. And we all know that lurking in there somewhere is the square root of two. I should stress that throughout this argument, the argument I'm about to give, we're talking purely about the rationals. So I'm not going to be talking about any reals. This is why I sort of put this down here somewhat faintly. This is to help guide our intuition. To motivate what's going to go on. But the entire argument I give is going to be in terms of rational numbers, not real numbers. I deliberately did not write our less than the square root of two because there is no such thing as a square root of two in the rationals. I'm using sets of rationals in this argument. It's an argument about the rational numbers, not about the real numbers. Okay? Well, A is bounded above. For example, two is an upper bound. We only need to find one, and two will do just fine. I will show that A has no least upper bound. That would mean that A is a set of rationals which has an upper bound but no least upper bound, and hence the rational line is not complete. Because completeness would say that any set of rationals with an upper bound in the rationals has a least upper bound in the rationals. Well, how would I show that there's no least upper bound? Well, let x in Q be any upper bound of A and show there's a smaller one. Again, let me stress smaller one in the rationals. Remember we use the lesser Q to denote rationals because Q stands for quotient, and rational numbers are numbers of the quotients of integers. We can't use the letter R for rational because R is used for real numbers. Unfortunate I know but there we are. Well, since we're talking about the rationals, that upper bound x is going to be of the form P over Q, where P and Q are integers. In fact, they can be natural numbers because this set A is a positive integer, or at least non-negative number. This set A is non-negative rationals. It's everything to the right of the origin, so everything's positive. So I don't have to worry about negative numbers here. So these two integers can be chosen positive. And I want to show that there's a smaller upper bound. Well, let's suppose x squared is less than two. It's either less than two or it's greater than or equal to two. It's one of the two. Let's just see what happens if x is less than two. In that case, looking at this equation, two Q squared is bigger than P squared. Now, as n gets larger, n squared divided by two n plus one increases without bound. So we can pick an n in n so large that n squared over two n plus one is bigger than P squared divided by two Q squared minus P squared. Now, you might not see where I'm going with this, but hopefully you can believe everything I've said. Okay? Well, assuming x squared is less than two, we'll actually, in a moment, we'll arrive at a contradiction. The conclusion I'm going to get out of this is that x squared is in fact not less than two. But this is where we're starting. If x squared is less than two, then because of that definition, two Q squared is greater than P squared. Okay? So two Q squared minus P squared is positive. That means this number is a positive number. And what I'm saying is that because we've got an n squared here and a linear term involving n here, the bigger n gets, this gets increasingly large. It gets as large as you want it to be. So I can pick an n big enough so that this number is bigger than that one. And if you rearrange that, you'll find that two n squared Q squared is bigger than n plus one squared P squared. Okay? I'll leave you to do the algebra for getting from there to there. Hence, n plus one over n squared times P squared over Q squared is less than two. Just rearranging that, taking those terms to the other side. Now we'll let y be n plus one over n times P over Q. Notice that y is a rational number. It's a quotient of integers. And y squared is less than two. Because this says that y squared is less than two. By the way, this, by now you should have begun to smell why I, I, I started looking at this term. I was trying to get this number y. Remember I started with an x as an upper bound and I wanted to show that there's a smaller one and I'm going to work towards that. And now I've got, I've introduced this y. Okay. So y's in Q and y squared less than two. So y is an element of that set A. But wait a minute, y is equal to a number slightly bigger than one times x. So that means that y is actually bigger than x. So the number y that I've constructed is in the set A and yet is bigger than x. Well, that's a contradiction since x is an upper bound of A. That supposition must be false. So x squared has to be greater than or equal to two. Okay, so what I've done is I've taken an upper bound of A. I'm going to show there's a smaller one and as a first step towards doing that, I've shown by contradiction that that upper bound has to have its square greater than or equal to two. Now I'm going to go ahead using this extra information to show that there's a smaller upper bound and hence there's no chance of any x being a least upper bound. Let me recap where we've got to. We let A be the set of all rationals that are non-zero and for which r squared is less than two. We let x be an upper bound of A and we had x in the form p over q where p and q are integers. Okay, so we have that. And our goal is to show that A has an upper bound smaller than x. Hence there cannot be a least upper bound which would show that the rationals are not complete. And we just showed that x squared is greater than or equal to two. Hence, since the square root of two is a rational, x squared is strictly bigger than two. X is a rational, x squared can't be equal to two, so it's strictly greater than two. Thus, since x equals to p over q, p squared is bigger than two q squared. I'm going to use this fact to find an upper bound of A smaller than x. To do that, I'm going to pick in an integer so large that the following is true. N squared divided by two n plus one is bigger than two q squared over p squared minus two q squared. Ie rearranging that p squared n squared greater than two q squared times n plus one squared. So you just rearrange this, do a little bit of algebra when you get this. Ie p squared over q squared times n over n plus one squared is greater than two. Again, you just rearrange that and do a little bit of algebra to get that. Let y be n over n plus one times p over q. Then y is an element of q, y is a rational number. It's a quotient of integers. So it's in q. And moreover, y squared is bigger than two. Moreover, since n divided by n plus one is less than one, y is less than x. Because p over q is x, and y is just this guy times x. So it's less than x. But for any a and a, a squared is less than two is less than y squared, so a is less than y. Hence, y is an upper bound of a, which is smaller than x. Thus, a does not have a least upper bound. And this proves the theorem. I guess my mathematics is better than my handwriting. This proves the theorem. Final remark, the construction of r from q can be done in several different ways. But in all cases, the aim is to prevent an argument like the above going through for r. And with that, you're at the very gateway to modern real analysis. For our final topic in this course, I'd like to say a little bit about real number sequences. These are connected with one of the ways of constructing the real numbers from the rationales. And they also give us a technique or a concept for doing an awful lot of work in real analysis. To put it another way, sequences of real numbers are a big deal in modern real analysis, which means they're a big deal in calculus. And anything that's a big deal in calculus is a big deal in science and engineering and technology. So whichever way you cut it, sequences are a big deal. Now what is a sequence? Well, in everyday terms it's a list. One, two, three. Let's put some comms in here of numbers. So we have a number, a number, a number going on to infinity. The way we normally express this and try to capture this infinite extent here is by writing it an where n goes from one to infinity. And this is what's called an infinite sequence. If you look in textbooks, you'll find a more formal definition of the sequences of function from the set of natural numbers into the real numbers. But for the purposes of one of the kind of things I want to talk about here, it's enough to think of it simply as an infinite list of real numbers. For example, the sequence of natural numbers, one, two, three, and so forth, that's an infinite sequence. In terms of our notation, I would just write that as n. For n goes from one to infinity. Or I could have the sequence that consists simply of an infinite sequence of sevens. Seven going on forever. That would be expressed in this way. Or I could have the following sequence. Three, one, four, one, five, nine, et cetera, enumerating the decimal digits of pi. Well, there's no simple formula like this to capture this one. I have to use some expression like this. Or let me give you another example. I could have the sequence consisting of negative one to the n plus one from n equals one to infinity. That's the sequence that consists of plus one, negative one, plus one, negative one, et cetera. That's an example of what's known as an alternating sequence, meaning that the sign alternates as you go through the sequence. Okay, so that's what sequences are. Just infinite lists of numbers. Now let's look at the following example. Look at the sequence consisting of the numbers one over n from n equals one to infinity. Okay, that's the sequence one. A half, a third, a quarter, and so on. The thing to notice about this is that the numbers get closer and closer to zero. In fact, they get arbitrarily close to zero. Or this one. One plus one over two to the n. From n equals one to infinity. That consists of the numbers one and a half, one and a quarter, one and an eighth, one and a sixteenth, et cetera. And these numbers get arbitrarily close to one. And going back to this example here, the sequence three, 3.1, 3.14, 3.141, 3.1415, 3.14159, 3.1415926. That sequence gets arbitrarily close to pi. That's as far as I know the decimal expansion by the way. So these sequences have the property that as you go along them, they get arbitrarily close to a fixed number. One is zero in the first case, one in the second case, and pi in the third case. And there's a general property here that we're going to capture by way of a definition. If the numbers in a sequence a n, n from one to infinity, gets arbitrarily closer to some fixed number a, we say that sequence tends to the limit a and writes a n hours a as n hours infinity. An alternative notation is we sometimes write it this way. Limit as n goes to infinity of a sub n equals a. Not all sequences tend to a limit. Look at this one, for example, this alternating sequence. Plus one, negative one, plus one, negative one. That doesn't approach any particular number. It bounces back and forth between plus one and negative one. This one, in a trivial sense, doesn't tend to a limit. This one tends to the limit seven. It doesn't just tend to it, it never gets away from it. This one doesn't tend to a limit at all. These numbers get bigger and bigger and bigger. We would sometimes say that the sequence tends to infinity, but that's beyond the scope of the limited amount I want to talk about sequences in this course. The point is some sequences don't tend to a limit. Other sequences do tend to a limit. Okay, let's move on. Well, so far everything's been very intuitive. Let's get a little bit more formal now. We've got a sequence a n, n from one to infinity. The intuitive explanation or the intuitive description I gave of this, that the sequence or the members of the sequence tend to a limit a, as n tends to infinity. That sort of corresponds fairly roughly to the fact that the absolute value of a n minus a becomes arbitrarily close to zero. Boy, my writing really is a problem, isn't it close? Now let me give you the formal definition. And the formal definition involves being precise about what that means. a n tends to a, as n tends to infinity, if and only if, for all real numbers epsilon greater than zero, there is a natural number n such that for all m greater than or equal to n, absolute value of a m minus a is less than epsilon. And now, perhaps for the first time, you see why we spent so much effort understanding quantifiers and being particular about the order in which quantifiers appeared. This definition is absolutely crucial in real analysis. It's absolutely crucial to our definition and our understanding of the real numbers and that means it's absolutely crucial to calculus and hence it's absolutely crucial to physics, science, engineering, technology, et cetera, et cetera, et cetera. This is the real stuff, folks. Okay, let's try and understand how that captures this intuition. The geometric intuition we have, I think, is fairly clear that you're going along a sequence and the numbers get closer and closer and closer to some fixed number. How does this, somewhat complicated looking expression, capture that in a precise, formal way? Well, let's peel away this part for a minute. By the way, it's traditional in mathematics to use epsilon in this context. And at the back of your mind, you should think that epsilon is not just positive, but it's very small and positive, like a tenth or a hundredth or a millionth or a zillionth or whatever. As you'll see when I go through this, it's when epsilon becomes small but remains positive that this thing kicks into power. Okay, so let's forget that bit. Let's look at the other part. Exists an n such that for all m greater than or equal to n, absolute value a minus a is less than epsilon. And I've avoided mentioning explicitly the set n now. Notice that in the first place, I didn't mention that m was also a natural number because the context makes that clear. n is a natural number and we're talking about all m's greater than or equal to n. And when we discussed quantifiers earlier on, I pointed out that mathematicians do this kind of simplification all the time. So it makes it easy to read, that's all. Okay, what does this part mean? So we've been given an epsilon, we'll assume. We've given some positive number of epsilon and at the back of your mind think of it as a small positive number. What this means is that from some point onwards, from some point n onwards all the members of the sequence are within a distance epsilon of a. So all the numbers in the sequence a n, n equals 1 to infinity are within a distance of epsilon from a. You still with me? Now let's see what this thing means when we have the whole expression. This says that for any epsilon greater than 0, this thing holds. So for any epsilon 0, there is a point such that from that point onwards all the numbers are within a distance epsilon from a. As I mentioned a moment ago, the intuition here is that we can take epsilon greater than 0 as small as we want. So let me draw a picture now. We've got some number a and there's a sequence bouncing around here somewhere. Maybe it's coming in from the left, from the right or maybe it's bouncing around. But there are these numbers a1, a2 scattered along this line here. Now we're given an epsilon greater than 0. So let's suppose it's down here. So here's a minus epsilon and here is a plus epsilon. I've just got a distance epsilon to the left and a distance epsilon to the right. And the arguments I'm about to give will hold for any epsilon. At least that's what this says. So I'm given an epsilon and I look at this interval. And what the rest of this formula says is that from some point on all the elements of the sequence are in here. If I take a small of epsilon, let's do it say here and here. Let's call it epsilon prime. So it will be a minus epsilon prime and here would be a plus epsilon prime. So I'll take a smaller epsilon, still positive so the formula will hold for the epsilon prime. There will be some point from which all the elements of the sequence beyond that point are within this region. And then I could take an even smaller one, an even smaller one. Notice that the N depends on the epsilon. Remember that example about the American Melanoma Foundation? It wasn't a problem for them getting the quantifiers the wrong way around. If we got the quantifiers the wrong way around here it would be in big trouble. This wouldn't work. The point is given an epsilon you could find an N. For each epsilon you may have to go further out in the sequence until you come to a point where all the elements are within that distance. But what this says is that you can always, by going sufficiently far out in the sequence, reach a stage where all of the numbers in the sequence from then onwards are within a given distance of A. And you can do that for smaller and smaller and smaller A. And that's capturing this intuition. And it does it beautifully and elegantly and with enormous power. Let me give you a couple of examples. First example. Let's look at the sequence 1 over N N goes from 1 to infinity. Now we know that 1 over N turns to 0 as N turns to infinity. Let's prove this rigorously in terms of the definition of limits that I just gave. So I want to prove that fact. What I have to show is that for all epsilon greater than 0, all real numbers epsilon greater than 0, there's an N in the natural numbers, such that for all natural numbers M greater than or equal to N, absolute value 1 over M minus 0 is less than epsilon. Well, let's simplify that. For all real numbers epsilon greater than 0, there is an N and let me just drop that explicit mention of the natural numbers. For all M greater than or equal to N, that just says absolute value of 1 over M less than epsilon, which actually just means 1 over M less than epsilon, because M is a positive integer. So I have to verify this in order to prove that. So how do I verify that? Well, let epsilon greater than 0 be given. Another way of saying it would be to say let epsilon greater than 0 be arbitrary. A different way of saying the same thing essentially, at least in this circumstance. What I need to do is find an N such that for all M greater than or equal to N, 1 over M is less than epsilon. Pick any N such that N is bigger than 1 over epsilon. If epsilon is a very small number, this will be a very large number, so I might have to pick a large N. But pick such an N. Now intuitively, you know that since the natural numbers go on forever, as it were, getting bigger and bigger, that's always possible. That actually uses a principle of mathematics. Call it Archimedean property, so named after Archimedean. C assignments 10.2. For a little bit more about that. So pick an N big enough so that N greater than 1 over epsilon. Then, if M greater than or equal to N, 1 over M is less than or equal to 1 over N, is less than epsilon. And we're done. Notice that the choice of N depended on epsilon. Given an epsilon, I picked my N in terms of the epsilon. Please note American Melanoma Foundation. At least when you apply quantifiers in mathematics, it's a big deal if you get them in the wrong order. The N depends upon the epsilon. The smaller the epsilon is, the bigger the N has to be. Intuitively, for this sequence, the smaller the tolerance you impose on the number being close to zero, the further out in the sequence you have to go, the bigger the N is, before you're within that tolerance. So quantify order matters. The choice of N depended on epsilon. Different epsilon, different N. OK. Let me give you one more example. Let me look at this sequence. N over N plus 1 from N equals 1 to infinity. That's the sequence. A half, two thirds, three quarters, four fifths, and so on. Let me prove in terms of the definition that N over N plus 1 tends to 1 as N tends to infinity. Intuitively, it obviously does. These numbers get closer and closer and closer to 1. But to prove that, what I have to show is the following. For all epsilon greater than zero, there is a natural number N such that for all natural numbers M greater than or equal to N, absolute value of M over M plus 1 minus 1 is less than epsilon. OK. I'm going to prove it the same way as before. I'm going to assume I'm given an epsilon. Then I'm going to find an N dependent on epsilon that makes this thing true. Let epsilon greater than zero be given. We need to find an N such that for all M greater than or equal to N, over M plus 1 minus 1 in absolute value is less than epsilon. Well, I'm going to pick N so large that N is bigger than 1 over epsilon. That's actually the same choice as in the previous example. It just works out that way because I'm using very simple examples. In more complicated examples, the choice is perhaps not quite so simple as straightforward as this. OK, but let's see why that one works in this case. Then for any M greater than or equal to N, I've got the following. M over M plus 1 minus 1 in absolute value equals minus 1 over M plus 1 in absolute value. Just work that out as a single fraction which is just 1 over M plus 1 which is less than 1 over M which is less than or equal to 1 over N because M is greater than or equal to N which is less than epsilon. And we're done. OK, you should now be in a position to attempt the questions on assignment 10.2. Bear in mind that the two examples I've given were particularly simple ones where the choice of the N was very simple. You won't always be so lucky. You might have to work a little bit harder to find the N but my focus here as it's been throughout the course wasn't so much on the details and on the procedures and if you graduated from high school I assume you can do mathematical procedures you can do the algebra. The focus is on the reasoning. In particular in the context of sequences the key thing is how given an epsilon you have to choose an N that depends upon it. It's this quantifier switch that's really crucial. For all epsilon there is an N. If you go on to further studies in mathematics in particular in real analysis you're going to see variants of this kind of definition not just for the limits of sequences but for the definition of continuity and then when you come to look at the formal definitions of integration and so forth. So you're going to come across this kind of a definition a lot. Coming up with this formulation was one of the jewels in the crown of the eight-nineteenth century mathematics. It was an absolutely brilliant piece of work and it gave rise to pretty well all of modern mathematics at least all of modern mathematics has to do with real numbers. Okay, well good luck with assignment 10.2. You made it. All you have to do now is complete the final assignment and you should be well set for taking college level mathematics courses or for simply reasoning more effectively in your job or in your everyday life. Since you've got this far I'm sure you'll complete these last steps so let me offer you my congratulations now. These massive online courses are still hugely experimental so simply sticking with us and finishing the course is an accomplishment all by itself. Future generations of students will surely have it easier in some ways. What they won't find easier however is the material with itself. Learning facts is relatively easy. Learning to think a whole new way is extremely difficult. However well you think you've mastered the material you should feel pleased with completing the entire process under your own motivation. As a lifelong teacher much of my motivation and reward has been getting to know each generation of students and working with them, helping them overcome difficulties and then seeing them succeed. With a MOOC that regular close contacts absent and I miss it. On the other hand being able to reach tens of thousands of students as opposed to 25 brings its own reward. To complete the basic course all you have to do now is work on assignment 10 and submit problem set 8. If you're doing the extended course the next two weeks are devoted to the test flight process. See the website for details. Okay well we're at the end of the course now so in keeping with my basic philosophy behind the course I'm only going to answer well the two of the questions from assignment 10. Since there's no more lectures to come no more material to come in this course there's nothing that's going to depend upon on anything now. You'll only need this stuff for future courses using it in the real world. So I'm going to leave most of the questions on assignment 10 for you to do in your own time and to resolve them with colleagues and so forth. But I just do one or two just to sort of set the ball warning and show good faith if you like. So this one I'll do assignment 10.1 number one the intersection of two intervals is again an interval. It's sort of easy if you think in terms of diagrams I mean what are the possibilities they could sort of overlap or one could be inside one of another they could overlap the other way around they could be disjoint completely separate and I guess that's easy isn't it they could overlap one way they could overlap another way one could be completely inside another one or they could be completely disjoint. But let's just do it in a symbolic fashion let's just let it be well I'll take the case for open intervals an analogous argument will work for closed intervals so let's say A be A B and let's C be the interval C D then by definition A intersection C is the set of all X such that A is less than X is less than B intersected with the set of all X this is X in the rails such that C is less than X is less than D already and because of the way conjunction works that's the set of X such that the maximum of A and C is less than X is less than the minimum of B and D okay I think I'll use the right letter there yeah that's correct okay which is an interval it's in fact the interval open interval maximum of A and C to the minimum of B and D and that's an interval it may be empty you know if it's disjoint it's empty the empty set is an interval it's still a set of numbers between two points okay so that I think will deal with that one and let me just say similarly for closed intervals and for half half open intervals half open half closed okay so that's really it and in the case of unions it's false for unions for example if I take the open interval 0 or 1 and I form the union with the open interval 3 or 4 and that's not an interval okay and that's that one okay which one shall I do next let me do what shall I do next let's do question 5 in 10.1 as a question 5 asks us to verify this alternative definition of least upper bound okay so let's see but first of all A just says that B is an upper bound okay so don't be like that A says B is an upper bound so the issue is does condition B say that it's a least one okay well let's see what that amounts to B is a least upper bound if and only if no C less than B is an upper bound okay that's what the word least means right there isn't a smaller one so that's the original concept there's nothing smaller than a lower bound an upper bound okay well that's true if and only if for any C less than B C is not an upper bound that's just another way of saying the same thing if and only if for any C less than B if it's not an upper bound that means there is an A in A such that it's not the case that A is less than or equal to C okay you can find an A for which C is not bigger than it it's not an upper bound if and only if and I'm really laying this one on with incredible detail here for any C less than B there is an A in A so it's that A if it's not less than or equal to C it's bigger than C now arguably I could have deleted one of these lines I've really been very pedantic about writing everything down but in my experience both when I was a student and from teaching this kind of material for many years even though these are very simple ideas it's very confusing at first dealing with the different variations of definitions of great slow bounds least upper bounds there's something about the way the human brain works that even though we recognize on one level that this has got to be really really trivial it causes us problems so it's just something to do with the way the brain works that makes us a challenge at least for most of us well, that's that one let's see what we're going to do next well, I think I'll do just one more and I'll pick one of those limit questions in number four this is from Assignment 10.2 and I'll pick this one show that the limiters of n over n plus 1 all squared tends to 1 as n goes to infinity and these all follow a standard pattern you start by with a given epsilon greater than 0 and what we need to do is we need to find an integer n such that for any natural number n bigger than or equal to n it is the case that the absolute value of n over n plus 1 of all squared minus 1 is less than epsilon from some point on in the sequence the difference between that term and 1 is less than epsilon ok, so now let's just pull this apart and see what this really wants us to find we need to find an integer n such that n greater than or equal to n implies let me write everything over n plus 1 all squared and I've got n squared minus n squared minus 2n minus 1 that is less than epsilon ie I need to find an n n bigger than or equal to n implies the n squared to disappear I've got a 2n plus 1 over n plus 1 squared less than epsilon ok, so the n squared to disappear I'm left with minus 2n minus 1 over n plus 1 squared then taking the absolute value and so when I do take the absolute value the minus 2n minus 1 just becomes 2n plus 1 and the n plus 1 squared is positive of course so that just reduces to that ok, now we can see what to do what we need to do remember we're looking for a big n so let's pick n so big that n plus 1 squared over 2n plus 1 is bigger than 1 over epsilon we can always do that epsilon's given it's a small number we'll assume we use a symbol epsilon to sort of emphasise the fact that in reality epsilon's very small that's not a logical restriction it's just our intuitions behind what the proof's doing ok, so this is going to be a very large number and in this term, in this quotient the numerator is squared and the denominator is linear in n so eventually the numerator dominates the denominator so we can make this as big as we want and we can certainly make it bigger than 1 over epsilon and when we do that we find of course that if n is bigger than or equal to n then 2n plus 1 over n plus 1 squared is certainly less than or equal to big to big n plus 1 over big n plus 1 squared which is less than epsilon because we picked big n to make that happen ok, and that's it we just worked backwards we looked at the goal and we worked backwards and indeed we can always find an n with this property because the numerator goes faster than the denominator without bound, squared over linear and so we get this expression this is very typical of limit verifications I wouldn't say they were all like this but the vast majority of them follow this pattern you work backwards and then you just ask yourself where you have to look how far you have to go out in order to make something happen and you very often end up saying we'll pick something so big that it's bigger than 1 over epsilon for limits that involve quotients ok, well that's I think that's all I'm going to do I'm not going to do many on assignment 10 I'll leave it with those there are plenty more to work with and I hope you'll enjoy working through those I remember when I learned this stuff for the first time as a student I actually enjoyed it, I thought it was a lot of fun first of all it's dealing with infinity in a rigorous way and that's pretty cool anyway dealing with infinity holding infinity in the palm of your hand that's really cool and it's just an intellectual challenge it's a fun game to play so enjoy it have fun with these things and I hope you do ok in the final exam ok bye for now