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.  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?  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.  So n is n plus 1. So given the m, we've already found the n.  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.  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.  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.  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.  Let's rearrange it.  And I've labeled that equation one for later reference. Now I'm going to take absolute values in this equation.  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.  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.  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.  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.  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.  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.  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.  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.  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.  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.  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.  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.  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.  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