 Welcome back to our lecture series, Math 3120, Transition to Advanced Mathematics for Students at Southern Utah University. As usual, be a professor today, Dr. Andrew Misildine. Now, for the next several lectures, we're going to explore some important properties of the integers. That is our good old set Z right here. Now, we've already explored the notion of divisibility with regard to integers. We said things like, A divides by B, meet such and such, talk about even numbers, odd numbers, prime numbers, a little bit here or there. But in this current unit, I want to dive deeper into the properties of the integers, but we're going to consider things besides just divisibility. We'll talk about properties, for example, as the ordering of integers, for example. We could talk about A is less than B or A is greater than B or A is equal to B. It turns out if I give you any two integers, one of these three things always happens. There's this trichotomy that's available that either the first number is less than the second or it's greater than the second or it's equal to the second. Now, of course, when I talk about two integers being less than each other, I'm using the ordering of the real numbers for which we say that X is less than Y. These are real numbers, of course. This happens if and only if their difference, Y minus X is a positive number. But if you think that's a circular term there, you can just say like, okay, Y minus X is positive. This is what it means for real numbers to be bigger than each other. Since the integers is a subset of the real numbers, then we can use the ordering of the real numbers to also talk about ordering of integers. That's going to be our goal right here for the next couple lectures. Of course, all of this is circling around the idea of mathematical induction. This is an extremely important proof pattern that we're going to learn about in the very next video for lecture 19 right here. But the method of mathematical induction is very much intrinsic to the study of integers. More specifically, the study of the natural numbers. But a lot of what we can say extends for all integers. So we're going to focus on that for a while. Now, in order to prepare for us to talk about induction, I first want to introduce the well ordering principle, which is a very important property of the natural numbers, for which in order to understand that, let's say a little bit more about this ordering that we've talking about up here. So imagine we have a subset of real numbers, which this could be a set of integers for all we know, but we have some subset of real numbers. Then we say that the set A has a maximum, sometimes called a maximum element, which will typically be denoted as max of A. This is an element X such that X belongs to the set, and if we take any other element of the set, it turns out that Y is less than or equal to X. So this is an important thing to describe the maximum of a set. You do take something in the set, and then everyone else in the set is less than or equal to it. Now, I mentioned that we call this A maximum, but we really could talk about the maximum, for which we can improve that if there's two maxima to a set, they're actually equal to each other. I'll leave that for now as an exercise to the viewer because we'll do some more conversations about uniqueness proofs in the future. So for the moment, we do allow the possibility, ooh, there could be more than one maximum, but not really. Similarly, we define the notion of a minimum of the set, a minimum which will denote min of A. This is an element Z such that it actually belongs to the set, and for all other, this includes Z as well, but for any element Y in the set, we have that Y is greater than or equal to Z. So it's just the dual notion of a maximum here. And again, we could talk about the minimum. You can prove that if a minimum or maximum exists, they have to be unique, but it's also important to mention that I have to sort of qualify that if a maximum exists, if a minimum exists, it turns out that not all sets of real numbers have a maximum or a minimum. Take, for example, the open interval zero to one. I claim, this of course is a subset of the real numbers there, so that's one we can consider here, but this set has no maximum or minimum, which of course I should also mention before we go farther into this example here that if you take the sets like the integers or the real numbers, these sets likewise have no maximum minimum element. There's no maximum real number. There's no maximum integer. There's no minimum real number. There's no minimum integer. But that's because these sets are unbounded. They kind of grow towards infinity or decrease towards negative infinity. That's a very big difference when it comes to this set zero to one. It actually is an interval of finite length. There's only one unit long. It is definitely bounded. There are elements that don't belong to the set that are bigger than everything there. Like if I take, for example, the number two, if I take all elements of that interval zero to one, then in fact, you do have that two is greater than equal to all of these Xs. So that gives us what we call an upper bound, but it's not a maximum element. The notion of a bound is not the same thing here as a maximum or minimum. And I don't wanna go into the details of partially ordered sets at this moment, but be aware that these sets, of course, don't have any maximum or minimum because they're unbounded. On the other hand, this set is very much bounded, like we mentioned a moment ago, you could do this for lower bounds as well. But there is no maximum element of the set. That is, is there an element of the set for which everything else is smaller than that? No, no. Unlike with this set, you could take any number, right? Take the number X, for example, inside of our set. It's not the smallest element because you could also then pull out X over two, which would still be in that set because it would be positive, but it would be smaller. So you could shrink, shrink, shrink, shrink, shrink. You never find a minimum element. You also could do the same sort of a similar trick, a little bit more complicated, but you could always get bigger numbers, bigger numbers, bigger numbers. You never, ever, ever are going to get a maximum or minimum. And the reason I chose this example, it's because it's an open interval. The elements that should be the maximum or the minimum don't belong to the set. Like if you take the interval zero to one, that does actually have a maximum element. It's one. It does have a minimum element, which is zero, but they don't belong to the set. And so there is a problem going forward with that. And again, I don't necessarily want to get into the technicalities of partially ordered sets or calculus issues right here with like supremum, infamum or anything like that. But I just want you to be aware that this is what a maximum and minimum element is, and it does have to belong to the set to count. All right, so with that clarification then mentioned, we can talk about the so-called well-ordering principle. The well-ordering principle tells us that all non-empty subsets of the natural number always contain a minimum element. And so this is a very curious thing, because you take, for example, this interval, it's an interval of real numbers, it didn't contain a minimum element. You kind of want that to be zero, but it doesn't belong to the set. It didn't have a minimum element. But for natural numbers, every set of natural numbers will always have a minimum element. There's no guarantee that it has a maximum element. Like if you take the set of natural numbers itself, there is no maximum, but it does have a minimum element that's zero. And if you take any collection of natural numbers, you're gonna get guaranteed a minimum element. And so I mentioned the well-ordering principle in comparison to these other examples we just considered, because for typical sets of real numbers, minimum elements are not guaranteed, but for a set of natural numbers, you always have a minimum element. It's guaranteed because whatever set you take, there's always the smallest natural number inside of that. And so this is an extremely important property about the natural numbers, which then as the natural numbers is sort of like the most important subset of the integers, a lot of information we can learn about the integers can be drawn from this well-ordering principle. And it turns out that the well-ordering principle we just talked about is logically equivalent to the induction principles that we're gonna start talking about later on in this unit. And that's why we wanted to begin with it. Now, what I next wanna do is use the well-ordering principle in a proof. And so the proof that we're gonna do here is one of the most important properties of the integers, the so-called division algorithm. The division algorithm tells us that if we take any two integers A and B, such that B is a positive number, A doesn't have to be, but B is a positive number, then there's gonna exist unique integers Q and R, such that A equals QB plus R. So we think of Q as the quotient if you were to divide A by B, and then R is the remainder for which the remainder satisfies the condition that the remainder is always smaller than the divisor, which would be B in that situation. And, but it could, it's also gonna always be non-negative. So the remainder could be zero up to but not included B. If you got R equals zero, that actually means that B divided A because if R was equal to zero, you'd have that gone and you get A equals Q divided by B. So B divides A in that situation. But most of the times when you divide two random integers, you will have a remainder. So this is division with remainder. And it turns out this is always, always possible to do. And this is an application of the well-ordering principle that I want to illustrate here because the well-ordering principle tells us that if we have a set of natural numbers, and I should mention that it's a non-empty set of natural numbers because clearly the empty set contains no minimum element because it contains nothing. So that is one possible exception, but it's the only exception. Every non-empty set of natural numbers contains a minimal element. So if we come up with a set of natural numbers, then we guarantee it has a minimum element that will then exist by the well-ordering principle. And then oftentimes we can make an argument that that minimum element is the thing we're looking for if we construct the set of natural numbers well enough. And then oftentimes there's a statement about uniqueness, like these are the unique numbers that do it. We can use the minimality here to help us with the cyberproof. So this division algorithm is a perfect example of the well-ordering principle in practice. And so that's what I wanted to do. So like I said, we're gonna start off with a set of natural numbers. So we're gonna take the set S to equal the set of numbers of the form A minus BK, where K of course is any integer and A minus BK is greater than or is equal to zero. So where do these numbers come from A minus BK? The idea is if you take this equation and you rewrite it as A minus QB, this equals R, we're basically trying to look for the set of potential remainders. But we of course don't have this condition in play here. We could give remainders that are potentially larger than B, but we do require that these potential remainders be greater than or equal to zero. Now I were only including natural numbers. So because A and B were integers by assumption, K of course is just an arbitrary integer right now. So A minus BK is good in fact, B and integer. If we require it to be positive, or non-negative I should say, then that will then make this a subset of the natural numbers. Now before we can invoke the well-ordering principle which applies to subsets of the natural numbers, we have to make sure this is a non-empty set. And how do I know? Because after all, since B could be, or B is positive, but like A could be negative, K could be negative, right? I mean, how do I know anything non-zero ever occurs here? Now if A is itself a positive number, we'll actually say if A is non-zero because that's a possibility as well, then you could just subtract off a zero multiple of B from A and now give you a non-negative number. And so this is an element that belongs to S. So there's something if A is non-negative. Now if A is negative, we're gonna treat it separately. This is just a common proof by cases. If A is negative, then what I want to do is then subtract from A B times two A, okay? And since you have this factor of A here, you can factor out the A like so. And so you're gonna get one minus two times B. Now if you take any integer and times it by two, that'll give you something bigger than one so that since this is negative, this here is gonna be negative as well. And I said integer earlier, it actually meant positive integer. Remember B is a positive integer. So if I take any positive number and times it by two, that gives me something larger than one because I mean that the smallest case B was one, so now you're at two. One minus two B is a negative quantity. Times if I negative, you get something positive and so, or non-negative. I mean, it could be zero, right? And so this quantity, which we'll belong to S because it's greater than equal to zero. Now we observe right here. So in either case, we get that S is non-empty. So now at this moment, we can invoke the well-ordering principle because we have a non-empty subset of the natural numbers. You do have to check that it's non-empty because if you'd set up your set here, how do you know that this definition isn't vacuous? How do we know whether something satisfies that definition or not? You have to check. There's at least one thing that's non-empty. So then we can invoke the well-ordering principle. The well-ordering principle gives us that there's a minimum element inside of this set S and we're gonna call that element R, okay? And so that is then what we're gonna do as our remainder. Now we're gonna let Q be an integer that obtains R because after all, if R is inside of this set, that means R is equal to A minus B times some number K. There's gotta be some integer that obtains it and we're gonna call that number Q, all right? So Q is the companion here. Now be aware that by construction, perhaps there could be a different Q, but don't worry about that one here. If you have A minus B by some other K, right? You could subtract A from both sides. You do end up getting negative B, Q equals negative B, K. Since B is a positive integer, you can divide both sides by negative B. You're gonna end up with Q equals K in that situation. So yeah, there's only one possible integer that will create R and we're so calling that element Q just so you're aware going forward here. So next what we have to show is we have to show that the number R that we've now just constructed by the well-ordering principle satisfies the interval conditions we said before. It has to be non-negative and less than B. Now the fact that it belongs to S gives us that R is gonna be greater than or equal to zero. You couldn't be in S if that wasn't the case. That direction's already taken care of. So what we have to then prove is that R is less than B. This is where the minimal conditions comes into play that R is the smallest element of S, all right? I want you to consider the number A minus B times Q plus one like so for which this number, if you play around with it for a second, this is the same thing as A minus B, Q minus B for which A minus B, Q is R. And as B is a positive number, if you take a, which also zero is a non-negative here, if you take R and you subtract from a positive, you're gonna get something here that is less than R. All right, it's smaller than R. Now since R is the minimal element of S by the well-ordering principle, it's gotta be that this number right here is not in S. It is still of the form A minus B times an integer because Q plus one is an integer. Therefore, the reason it's not in S is because it's now a negative number. So A minus B, Q plus one is less than zero in that situation, which of course, distribute the negative B right here. We end up with A minus B, Q, which is R minus B. This quantity, this is equal to B minus R. This is negative. Now, if you add B to both sides, we get that R is less than B. And so that then shows what we're looking for. We've now demonstrated that given any integer A and any positive integer B, there exists a quotient and a remainder such that A equals B, Q plus R and R will sit between zero and B where it could be zero, but it can't be B. That gives us the first part of our division algorithm. The next part, which is a little bit more tricky, we also wanna show that this element is unique. Now, when one wants to show that you have a unique element here, what you do is you basically suppose that you have two elements that satisfy the conditions and you are either equal to each other, okay? So like, let's suppose that A equals B, Q plus R and A equals B, Q prime plus R prime, such that B, R, excuse me, Q, R, Q prime, R prime, they're both integers and the numbers R, R prime satisfy the condition that they're greatly equal to zero and B. Notice, I'm not supposing that these numbers are necessarily distinct. It could be that R and R prime are the same and Q and Q prime are the same. In fact, that's what we're gonna show, but we're just saying, oh, you have a pair of numbers, you have a Q and an R that satisfy the condition, you have a Q prime and an R prime that satisfy the condition. We wanna argue that they're in fact gonna be equal to each other. And this is, of course, the argument that we have to provide right now, all right? Now, whether they're equal or not, we don't know, but what we can say without the loss of generality is that R prime is greater than or equal to R, okay? Because like we had talked about at the beginning of this video, when you compare the integers R and R prime, either R is less than R prime, R is greater than R prime, R they're equal. And so by the loss of generality, we can assume R prime is greater than equal to R. The only possibility I'm not considering is that R prime is less than R. If that was the reality, what we can do is we can play switcheroo right here and we just relabel them so that the R prime is the bigger one. So without the loss of generality, we can suppose that R prime is greater than or equal to R. We can't say that they're not equal because that's actually what we're trying to show there, okay? Now, when you look at these equations right here, notice that the left-hand side is equal to A in both situations. So if we remove A, we can actually set BQ plus R equal to BQ prime plus R prime and get the following equation right here. And we wanna manipulate this equation a little bit to help us. So I'm gonna move the BQ prime to the left-hand side, we'll move the R to the right-hand side. And so we get the following setup so that you can factor out the B on the left-hand side to get B times Q minus Q prime that is equal to R prime minus R. And by assumption here, since R prime is greater than or equal to R, this tells us that R prime minus R is greater than or equal to zero. This is a positive quantity, okay? So examining what we've discovered so far is that since B times Q minus Q prime is equal to R prime minus R, we get that B divides R prime minus R. Now, B is itself a positive integer, R minus R prime is itself a non-negative integer, okay? But I also wanna point out that R prime minus R is less than or equal to R since R is itself a non-negative integer if you subtract it, that only makes the number smaller. And so R prime without the subtraction will be bigger than R prime with the minus R there. But by assumption, R prime is less than or equal to zero. So what we have here, when we look at this number, R prime minus R, B divides R prime minus R, but we also have that R prime minus R is less than B. How is that a possibility here? And then I also should mention that it sits between zero and B like so. How can we resolve these conditionals? How can a positive integer divide another integer? But that integer be smaller than it. Now, the only resolution here is actually that R prime minus R is equal to zero because zero does have this property. B does divide zero since it's just equal to B times zero. But we also have that zero is less than B because B is positive. And we also have that zero is greater than zero because it's equal to zero. Zero is the only number that can resolve this conflict here. And as such, we have to then conclude that R prime minus R is equal to zero. But of course, if you add R to both sides, you end up with R prime equals R. And like we talked about previously in the proof, their correspondence, their associates Q and Q prime would also have to equal each other, Q equals Q prime. So this proves that there's only a unique R and a unique Q given any possible division. So while we are able to use, where will the prove the division algorithm using the well-ordered principle? And this is a classic way of how you use the well-ordered principle to prove various facts about integers and natural numbers. We will use this proof technique in future videos as well. So take a look for those. But despite the fact we are able to prove the division algorithm that these numbers exist, I do wanna point out that this proof does not provide us any clues on what those numbers Q and R actually are. This is an example of what we call a non-constructive proof. The difference between a constructive existence proof and a non-constructive existence proof is something we will delve into in the future. Don't worry about it right now. Now the good news is, and this is the reason why we call it the division algorithm, is that the algorithm for long division that we learned back in grade school is the algorithm you use to construct these numbers R and Q. They provide the numbers by basically just doing like just repeated subtraction. Like if multiplication is just repeated addition that you can kinda treat division as repeated subtraction and you just basically subtract off multiples of your divisor until you have your remainder. And if you find this number, which it does exist, it has to be unique and therefore long division exactly gives us that. But the moral of the story is that the well-ordered principle gives us non-constructively the existence of this minimum element and its existence can be useful to prove powerful facts about natural numbers and integers. The most powerful fact of course is the principle of mathematical induction that we'll see in the very next video.