How did you get on with assignment eight? Though the focus of this class 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, number theory and elementary real analysis. The principle advantage is example domains. Is it everyone has some familiarity with both systems. Yet very likely you won't have been exposed to their mathematical theory. I'll take the integers first. Most people's experience with the whole numbers, is by way of elementary arithmetic. Yet in the mathematical study of the integers, looking beyond mere calculation to the abstract properties those numbers exhibit, goes back to the very beginnings of recognizable 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 fascinating causes they take. Not only is the subject full of tantalizing 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 them 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 relates to two numbers, a quotient and a remainder. For example, if you divide nine by four, you get a quotient of two and a remainder of one. Nine equals four times two plus one. This is a special case of our first form of theorem concerning integers. The division theorem. The division theorem says the following, let a be b integers with b greater than 0. Then the value unique integers q and r, such that a = qb + r And 0 less than or equal to r less than b. Well 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 that I'm going to prove existence first, 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. Well it will orientate the reader, to see 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 entities of the form a-kb. Work here as an integer. Ensure that one of them is less than b. What the theorem says, is that among the integers a- 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 k b or a minus q b 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 that q that I'll take with the k. Well, first of all, I need to show that there are such integers. For example, take k equals negative absolute value of k. Then, since b is greater than or equal to 1. A- kb = a + absolute value of a times b, which is greater than or equal to a + 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's r be the smallest such integer, and let q be the value of k for which it occurs. IE we'll have r = a- 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 zero and b, in this fashion Then I'll have proved the existence. So to complete the proof, we show that r < b. But 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 show that r is less than b. Well, if r is greater than or equal to b then a minus q plus 1 b equals a minus qb minus b which equals r minus b which is greater or equal to 0 by virtue Of this assumption. Thus, a- (q + 1)b is a non-negative integer of the form a- kb. But r is the smallest such And yet a-(q+1)b<a-qb which is r, and that's a contradiction. If r is as small as such, we can't find smaller one. There's a contradiction, and so r is less than b, because we obtain 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 certain looks very complicated. It's actually not complicated or deep, it's just intricate in it's 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 that by 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 minimum 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 integral 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 will 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 turn to uniqueness. To prove uniqueness, we show that if there are two representations of a, a = qb + r, and a=q prime b + r, with 0 less than or equal to r and r prime less than b, then r = r prime, and q = q prime. So the first part of the division theorem, the existence part that we've already proved shows that there are representation of this form. What we're now doing is showing that if there were two such representations, then in fact they're identical. The r is the same and the q is the same. Incidently, you probably realized by now the letter q is going to denote quotient, and the letter r is going to denote remainder. We haven't officially defined quotient and remainder yet. So I haven't written 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 already have about arithmetic. Dividing one integer by another. Okay, so we have an equation here. Qb + r = q prime b + r prime. Let's rearrange it. Rearranging, I get r prime- r = b(q- q prime). And I've labeled that equation 1 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 check absolute values, b remains the same. But -b is less than -r which is less than or equal to 0, and 0 is less than or equal to r prime which is less than b. So -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 2 b times absolute value of q minus q prime is less than b. I can divide by b now to give me, Absolute value of q-q prime is less than 1. 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 their absolute value is zero. Which means q as to equal q i. And then by equation one, r is to equal r i and that proves uniqueness. And with it we've proved the division theorem. 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 within equalities. Take it step by step and you should be able to follow the arguments right there to the end, okay? Good luck with this. If this is the first full blown rigorous proof of a theorem like this 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 pay 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 becomes 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 and it is has infinitely many rooms. As in most hotels the rooms are numbered using the natural numbers one, two, three, etc. 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 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 reject 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 occupants of room one moves into room two, the occupants of room two moves into room three, and so on through out the hotel. In general, the occupants of room n moved 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 reject 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 Hilbert's solution, try the following variants. First variant, the Hilbert's Hotel scenario is as before, but this time two guests arrived 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 am n, for each of the natural numbers n equals 1, 2, 3, etc. 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's 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, which I'll give you now. So, theorem, general division theorem lets a, b be integers with b non-zero. Then there are unique integers q and r switched with a = qb + r and zero 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 power of b is greater than 0, the previous theorem tells us that there are unique integers q prime and r prime. Such that a=q prime times the absolute value of b + r prime and 0 less than or equal to r prime, less than absolute value of b. Well now we simply let q = -q prime, and r = r prime. Then, since absolute value of b In this case, with b negative, is negative b, We get a is qb + 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 results. And with the General Division Theorem now established, we can formally give names to the two numbers q and r, 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 learn in elementary school about dividing numbers, and that there were sometimes remainders, there were quotients and remainders. And now we've done it with some sophistication and we've shown that everything is well defined. How about that?
