 Hi, I'm Zor. Welcome to a new Zor Education. I would like to talk about one very important theorem in calculus. It's called Intermediate Value Theorem. Well, the problem is that this theorem is really formulated quite simply. However, as everything else is very obvious and intuitively easy to understand statements, it's very difficult to prove because it's very close to the axiomatic foundation of real numbers. Alright, anyway, this lecture is part of the Unizor.com website which is dedicated to presentation of advanced math course for teenagers and high school students. I suggest you to watch this lecture from this website because it has very detailed notes for each lecture. It has exams, so basically it's very good for self-study or some kind of flipped classroom study. Okay, so, first of all, why actually this theorem is needed? There is actually some practical reason for this to exist. Let's just consider you are solving some equations. Let's say 2 to the power of x plus x cubed equals to 0. Can you solve this equation? Well, the answer is in simple terms, no. I mean, we know how to solve quadratic equations, sometimes a little bit different ones, but in any case, this is something which is completely outside of the regular scope of equations which we can solve analytically. So, what do we do? Here is a very important consideration. Let's consider that we have two points and at these two points of argument x, our expression on the left takes two different signs of values. For instance, if x is equal to minus 1, 2 to the power of minus 1 is what? It's 1 half plus minus 1 to the power of 3 it's minus 1 and it's equal to minus 1 half. If you put the value 1, you will have 2 to the power of 1 which is 2 plus 1 to the power of 3 which is 1 which is 3. So, look at it this way. This is some kind of a function, whatever that function is. Now, at minus 1, it has negative value. Let's say this is minus 1 half. On the plus 1, it has positive value 3. So, function actually is moving as we are moving from minus 1 to 1 of its argument. The function is moving from minus 1 half somehow to 3. It is obvious, again the word obvious is probably should be understood in quotes, that it should cross the x line. And wherever it crosses the x line, the value of the function is equal to 0 and this particular value is actually the root of this equation because at this particular value of x, let's call it A, function 2.A plus A to the power of 3 is really equal to 0. That's the solution. So, what we definitely know that between minus 1 and 1, there is some point which represents the solution. What do we do next? Well, let's divide this segment in half. So, the midpoint is 0. And let's analyze what is the value of this function at the midpoint. Is it negative or positive? Well, put 0 and that will have to the power of 0 which is 1 plus 0 to the power of 3 which is 1. So, this is 1. So, it's still positive. Now, we are actually having a narrower interval from minus 1 to 0 where we know that our root of this equation is located. So, we have narrowed it a little bit better. Can we do better? Of course. Let's divide the game by half. From minus 1 to 0, this is minus 1 half. Now, we can actually calculate the value minus 1 half, what would be 2 to the power of minus 1 half is 1 over 2 to the power of 1 half which is 1 over square root of 2 which is something like 1 divided by 1.4 something. Now, minus 1 half to the power of 3 is minus 1 eighths, right? It's minus 1 eighths. Okay. Now, this is definitely smaller than this one so it's definitely greater than 0. So, what I can have, I know that at this particular value my function is still positive. Graph is not this. My graph should be somewhere here. So, my solution to the equation should be between minus 1 and minus 1 half. Now, this process I can continue any number of times obviously to get closer and closer each time I'm basically reducing the length of the interval where my solution is located in half, right? In like 4, 5, 7, whatever, 100 steps I can get any precision of where exactly my solution is located which is a good way to approximate the solution since we don't know really the exact solution. It's irrational number anyway. So, any kind of a... So, the process which gives you any kind of degree of precision is good. And in many cases actually we have computer programs that do exactly this. They don't know how to solve this but they can really do this type of a process. Programmatically it's very easy to program a computer to perform all these calculations and eventually you will get to the solution. I mean, there are other ways to do it but this is as good as anyone else. Now, this is all practical aspects of this but what exactly did we use as far as the theory, as far as the certain foundational statements what exactly did we use to approach this problem in this particular way? We used a very simple statement that if some function is negative on one end of the segment and positive on another end of the segment then it should cross zero line somewhere in between left and right boundaries of this segment. This is actually the theorem which is attributed to Bolzano and it's not such a simple thing. I mean, it seems to be obvious but it's not always obvious. So, first of all, we have to make a very concrete statement. We have to formulate this theorem really mathematically correct because not every function which is negative here and positive here cross the x-axis. For instance, you can have the function which is something like this. It's equal to minus one here and equal to three here. At zero it can be, let's say, minus one or whatever, minus one. Half doesn't really matter. So this function doesn't really cross the x-line at all. It's horizontally here and horizontally there. The point zero is included here, for instance. This is not zero. This is approaching from the right, it will be always three. And the function never crosses the x-line. So when exactly this particular approach is correct, well, the answer is, as you, I'm sure, kind of guessed, the function should be continuous. So, we are ready to formulate the theorem. And I will not formulate it using the original Boltzana way that negative here and positive here. Therefore, it intersects the x-axis. I will do it slightly more general, but it's basically exactly the same from the difficulty to prove. Here is the theorem which I would like actually to prove. Let's consider we have a continuous function. Now, the word continuous is absolutely the key point here. We have a continuous function defined on some segment. And let's talk about real segment from A to B, real numbers. There is no infinity here at all, just real numbers. And let's consider this, the function is continuous, right? On this particular segment. Let's also consider that function on the left end is strictly less than function value on the right end. This is basically equivalent to saying that on the left function is negative, on the right function is positive in the Boltzana original theorem. All right. If this is given, so the function is continuous, defined on this segment, and the values between F A and F of B are in this particular way. So, for any real number C, which belongs to interval from F of A to F of B, or if you wish, in this formulation. So, for any C in this segment from F of A to F of B, there is, there exists such point H from this interval AB, where the function is equal to this chosen value. Let me just illustrate it. It's really very simple. It looks much more complex than it really is. So, let's say you have your interval from A to B, and let's say you have your function. Now, this is F of A and this is F of B. So, for any point in between C, we can always find at least one H, where the value of the function equal to this. So, from F of A to F of B, we cross, eventually cross any value. Maybe it's not exactly this, maybe it's like this. So, we cross it twice or three times, whatever. It doesn't really matter, but we definitely cross at least once the line C. Now, it's equivalent to, let's say you are going up the mountains. Now, from the level where you are on the ground, where the mountain starts, to the top of the mountain, you inevitably have to cross every height in between these two levels. From the ground zero to the top of the mountain, every level in between you have to cross. You cannot reach the mountain without crossing any height in between. You might actually go up and down, up and down, but eventually, since you are reaching the mountain, you must cross every level in between. So, it's absolutely obvious statement, of course, for continuous function. Because as you are climbing up the mountain, you don't really skip anything. You don't disappear at one point and appear on the top of the mountain. It happens only in Star Wars or something like this. So, that's the theory. And being so obvious, again, as I was saying many times, the obvious thing is very, very difficult to prove in mathematics. Because if you will ask somebody on the street, they will probably answer, okay, this is an obvious statement. It does not require any proof. Well, it does require any proof. But unfortunately, it's not easy because it goes really to the structure of the real numbers. Because, for instance, if you are talking about function with relative arguments, and let's say relative values of the function, that's not true. Because if you are going from one place to another, sorry, not relative, rational, if you are jumping from one rational number to another, if the function is not defined at all on irrational points, then there is this skip. And if your C level is on irrational point, you might not actually find anything. So, it goes to the heart of the structure of the real numbers. Real numbers are in some way complete. There is no gaps between them. And this is actually the first very important consideration which I would like to talk about, completeness. Okay, so we are talking about completeness. And this is something which, again, goes to the very beginning of what real numbers actually are. And this is something which we call an axiom because otherwise we will not be able to do anything with real numbers. So, real numbers are complete in terms of we accept the completeness axiom. Now, what is a completeness axiom? Alright, so let's consider you have certain set of values. And not necessarily contiguous. Let's say these and these values. Now, completeness axiom starts with the concept of the bounded set. So, bounded from above in this particular case. Now, the upper boundary is any number which is greater than anything within this set. So, if this set is, let's say, s, then for any x which belongs to this set, this a would be greater or equal to x. So, any a which basically has this property is upper boundary. And again, from the intuitive understanding, it's fine. I mean, it's understood that these numbers exist. Whatever these numbers, whatever this set is, doesn't really matter. It might contain even one point, but doesn't really matter. The result is something which is greater if we are talking about this particular set as being upper bounded. Okay, so upper boundary is existence of this particular thing. Now, the completeness theorem is the following. If you consider all the different upper boundaries for this particular set, completeness axiom states that there is something which is right here, which is the least upper boundary, which in many cases is called supremum of this, supremum of set s. So, the completeness axiom states that there is one particular value which is called the least upper bound. So, if there is one upper bound, there is least upper bound. And this least upper bound is very, very important, because it's kind of a highest point towards which any sequence which belongs to this particular set is approaching. Now, not necessarily this point is part of this set, because, for instance, if your s is all the x which are less than, let's say, 2. Well, if this is 2, then everything less than 2 belongs to the set s, right? But the point 2 does not belong to this s. So, what is the point 2? Well, that's the least upper boundary of this set, or supremum of this set, as we saw. So, from one side, this point is the least upper boundary. From another side, this is a limit of all the sequences which are increasing towards 2, because any limit of the points here which are going from the left to the right can go as close to 2 as possible, not touching the number 2. So, this is the border between the set and upper boundaries of the set. So, existence of this border is an axiom. That's very important. We do not prove it. It's axiom which we, well, you know that we unfortunately have to accept certain axioms, because without them we don't have the theory. We cannot prove anything. So, we need something as a foundation. In this case, this particular completeness axiom is a foundation of the real numbers. And the proof of the intermediate value theorem is based on this axiom. And that's next. Let's go back to our function. This is a. This is b. This is f of a. This is f of b. Somewhere in between we have c. And we have to prove that there is number h between a and b, where our function takes the value c. Okay. Here is what I'm proposing to do. Let's consider the set of all values, x, such that f of x is less than or equal to c. Now, on this graph, now, if this is the level c, where is my function less than c? Here. So, on this interval from a to h, my function is less than c or equal. Now, anything to the right of this h would be upper bound for this particular set s, right? So, let's just think about this set s. First of all, this set is not empty because at least point a. We have agreed, if you remember, that f of a less than c, less than f of b, and f of a and f of b do not equal to each other. f of b is really greater. That was our initial consideration, right? So, at least function f at point a gives you f of a, which is smaller than c or equal, which means our set s contains point a. Or a is contained in set s. So, it's not empty. Now, is it bounded? Yes, of course it's bounded, because we are considering only points from a to b, and point b is definitely upper bound. So, we are talking about a set, a non-empty set, s, which has at least one upper bound. And according to the completeness theorem, it has the least upper bound. So, there is, exists supremum of s, which I call h. So, let's consider h to be really this particular least upper bound. And what I'm planning to prove right now is that this particular h, which exists according to the completeness theorem, it's a point between a and b. It's real number between a and b. So, I would like to prove that f of h is equal to c. This is necessary to prove, which means that I have actually found this particular point h. Well, I proved the existence, let's put it this way. I proved the existence of this point where the function takes exactly the value which we are looking for. Okay, how can I prove that? Well, there are three choices, right? f of h is less than c, f of h is greater than c, and f of h is equal to c, right? So, I will prove that this is wrong and this is wrong. And let's just follow my logic. First of all, how can I prove that f of h less than c is impossible? Well, let's think about this way. What does it mean that f of h is less than c? Well, it means that the h belongs to s, right? Now, on the other hand, h is the least upper bound. So, h is supposed to be greater or equal to anything within s, within my set s, right? h is supposed to be on the border or even further from all the points on the left. But now, let's think about this way. f is a continuous function, right? We have actually said it in the very beginning, so the whole theorem is only about continuous functions. So, if f of h is strictly, strictly less than c, not less than equal, just strictly less, then there is some distance between f of h and c. c minus f of h is positive, right? So, let me just take this and since f of h is a continuous function, what I'm going to do is around point h, so this is h, this is f of h, right? Now, if f of h is less than c, c is somewhere above that, and f of h is a continuous function, that means that around some small neighborhood of h, I will also be less than c. How can I prove it? Well, very simply, for instance, this is equal to, let's say, epsilon. Well, let's take epsilon divided by 2 to be absolutely sure that there is no equality. So, this is this distance. Now, for any epsilon, well, in this case, epsilon divided by 2 doesn't matter. For a continuous function, we can have the delta, so this is delta, this is delta. So, if I am deviating from the value of the argument by less than delta, so if x minus h is less than delta, or equal, then follows f of h minus f of x by absolute value would be less than epsilon, or epsilon divided by 2 doesn't matter, for any epsilon. So, since this is basically a definition of the continuous function, so if function is smaller than certain level c at one point, then I can always find this particular distance, and for this distance between values of the function and c, I can find distance between arguments, so it's smaller, which means that not only f of h is smaller than c, but also function f of x in an immediate neighborhood of h is also smaller than c. So, all these immediate neighbors also belong to s, to our set of all the points where the level of the function less than or equal to c. But in this case h cannot be upper boundary, because there are certain values to the right of the h, which also belongs to the set s, which means that h is not greater than all of these elements of the s, which means h is not an upper bound. So again, we have proven right now that f of h cannot be less than c, because in this case not only h, but also immediate neighborhood of h would belong to our set s. And that's why h cannot be the upper boundary. So, that's done. Now, how about greater? It's actually very similar. Okay, so let's consider function of h is greater than c. And again, we will come to some kind of contradiction saying that this is impossible. Absolutely similarly to the previous considerations, I can say that f of x is also greater than c, where x belongs to certain neighborhood h minus delta h plus delta. In absolutely the same way as before. So, if the function is greater than c and function is continuous, then in immediate neighborhood of this h, it's also greater than c. Now, what follows from this is that not only h is an upper bound, but also all these values within this particular interval are also upper bound, right? Think about this way. If you have this interval from h, h minus delta h plus delta. Now, we were talking about the fact that h is the least upper bound of the points where the function is less than c. Now, what appears to be that none of these points can be here, because all of these points, that's where f of x is greater than c. So, none of these points can be here, which means they're all to the left. And if they are all to the left, then not only h is upper bound, but also any one of these is also upper bound of this set s, which means that h is no longer the least upper bound, because all these are also upper bounds and they are smaller than h. So, we also came into some kind of contradiction, right? So, this is impossible. This is impossible. So, the only thing which is left is that f of h is equal to c. So, that's how we have proven absolutely obvious statement that if the function takes different values from a to b, then anything in between should be attained at some particular point. And it's a very kind of... I would say it's an obvious fact, but it's not obvious proof. And as I was saying, we need some axiom as a foundation, the completeness axiom of the real numbers. And again, I mentioned it very briefly. It's not true for rational numbers, for instance. The completeness theorem does not take place in rational numbers. So, rational numbers are not complete. So, we cannot say that the function f of x, where x is rational, we cannot say about this function that it has exactly the same properties. And the fact that rational numbers are not complete is very easy to prove. There is a very simple example. Let me just give it to you. So, let's consider a set of numbers s where x is rational and x square is less than 2. It's a set, and it's obviously not an empty set because 0 square is less than 2. And it's obviously bounded because, for instance, any number like 3, for instance, would be a bound for this. If x square is less than 2, then obviously x is less than 3 or 10 or something, so it's bounded. But is the completeness axiom correct? Is there a rational number which is the least upper bound? Well, we all understand that the least upper bound would be square root of 2, but it's not rational, right? So, all these rational points, they are a set, but the rational numbers as a concept, as a unit, as a set is not complete because we have the least upper boundary and non-rational number. Another definition of completeness which might be a little bit more intuitive is the following. If you have, let's say, a set, whether it's a rational number or real numbers, whatever, so if you have some kind of a set here, then if every sequence within this particular set converges to the member of this set, I mean, if every rational, for instance, converges to rational, then rational would be complete. But that's not true, rational can converge to irrational number. But the real numbers, the sequence of real numbers, if it converges, it always converges to the real number. That's the big difference between real numbers and, let's say, rational or integer or anything like that. Okay, so that's all about completeness. Now, yes, Balsana theorem, I was saying before that when the left boundary of the segment is negative and the right boundary is positive, is exactly the particular case of this theorem which I have just proven, intermediate value theorem, where f of a is less than zero and f of b is greater than zero. So obviously, it should cross zero. Other than that, I do suggest you to read the notes for this lecture on Unisor.com. It's very important because the logic is not exactly clear. And again, what we are doing here, we are not really only studying mathematics or facts in mathematics. We are trying to develop the logical thinking, which means that the proofs of the theorems are very, very important. So my presentation is one thing and then there is a written proof which resembles some textbook, if you wish, on the website. I do suggest you to go to the website and read very attentively the description of this intermediate value theorem on the website. Other than that, that's it. Thank you very much and good luck.