 Oops. Okay, which one of these is the pin? Up here. There we go. Okay. All right. Okay, this should be good enough. Okay. Okay. So here's what we're going to do today. We're not going to go as long as normal because I'm going to pass your exams back and talk about it at the end. So I want to get this done first. Otherwise, I'll just start ranting and I won't have enough time to finish everything. I want to finish today. Okay. So I want to do that at the end. I have to get through what I need to get through today. So we're going to do that first. But I will spend some time to give your exams back and talk about it at the end of class. Okay. All right. Some of you are happy and some of you are not, but I can't please everybody. Okay. So what we're going to do is we're going to start with, okay, we've finished 2-4, but we're going to start by me giving you some help with one of your homework problems. So we're going to talk about number 5A. And I should tell you that a couple of these problems require a little bit of work. And I will just warn you that I think I signed 6-2. You will probably need to give 6 a little bit of thought. Okay. If you think that you did it in two lines, maybe you did, but probably it's not right. You're going to need to do a little bit of work with number 6. Okay. So I want to, if we have time on Thursday, I'll talk more about this, but what I want to do is give you some help on number 5A. Okay. And so 5A just says that if the GCD of A and B is 1, then the GCD of A to the N and B to the N is also 1. Positive integers to N. Okay. Now here's the thing you have to realize. If you think about primes and factorizations and such and things that you learned maybe in middle school, there's a sense in which, somebody said that this seems obvious, so it's probably you, Christine, I'm guessing. So you're kind of a brat. But there is a sense in which that's true. It is in the sense that if you think about it in terms of prime factorizations, GCD of A and B being 1 really means, well, okay, this is assuming A and B are bigger than 1. But they don't have any primes in common. If you were a factor at A and B and a primes, there's nothing in common. Because if there was, then the GCD wouldn't be 1, right? It would be at least the prime that's in common to them, if not more than that, right? So if there's no primes in common to A and B, then when you raise A and B to the Nth power, the prime factors are all the same, right? They just keep getting raised to the Nth power so there's nothing in common there either. So the GCD should be 1. Now we haven't talked about prime factorizations yet. So unfortunately, your life's going to be a little more difficult than it will be later on. But I also want to say something before we go into this. If you have a problem in a given section, you need to use, I expect that you just use the tools that you've learned up to that point, okay? You can't just, you can't say, well, we can use this theorem 5 from, you know, made the 9th, which takes care of this in one line. You can't do that, okay? I'm sorry, can't do that. So you still have to do this kind of the tedious way using just what you've learned so far, okay? And what you want to do is this, and I'm not going to go through the whole thing, but here is the sketch of how you go about this, sketch of the solution. The book actually gives you a hint. So the first thing you want to do is this. First, you want to prove number 20 part A of, I'm guessing this is section 2.3. So we have the book open. Does it say 2.2? But that's not true, because there is no 2.8 and 2.2. The author means 2.3, okay? Yes, you brought that. Thank you for bringing that to my attention. And by the way, just so you know, the book tells you how to do it, basically. It says here, here's 20A, do this. Oh, by the way, here's exactly what the solution is. They essentially do the whole problem for you, okay? So do that first. Just follow what the book is giving you in the hand. It basically just works out exactly the way the book sort of outlines it. So what that is, I'll just write it down here. Let's see, actually, let me just open the book and make sure that I don't miscopy this. Okay, so GCD of A and B is 1, and the GCD of A and C is 1, then the GCD of A and BC is 1. This is number 28. So you should start by establishing this. And then from there, the cleanest way to prove this is to use induction. And I'm just going to give you kind of an outline of how you want to go about this. I'm not going to fill in all the details for you here, but I'm going to give you an idea of what you want to do. So I want to be very clear. This is not, okay, if I end up grading this problem, and you just copy exactly what I have copied, you are not going to get full credit. So you don't come up to me afterwards and say, well, I just wrote down what you wrote down. I'm not telling you this is the polished solution. I'm just giving you a guide as to how you want to go about thinking about the problem. Okay? So what you're going to do is just assume, of course, that the GCD of A and B is 1. And what is it that you're proving? Of course, you're trying to prove that the GCD of A to the N and B to the N is 1 for every natural number N. So the base case, I'm not going to write this down. The case of the induction, of course, is just that the GCD of A to the first and B to the first is 1. But of course, that's true because that's just the GCD of A and B, which we're already assuming is 1. So you should write that down, but the base case is just trivial. Okay? So to be very clear what we're trying to do, we're trying to prove this for every positive integer N. We're assuming that GCD of A and B is 1. Okay? So now what you're going to do is we're going to assume that the GCD of A to the N and B to the N is equal to 1 for some natural number N. And you have to prove that the GCD of A to the N plus 1 and B to the N plus 1 is equal to 1 as well. That's the inductive step. Okay, so here's one way you can do this. So the first thing we're going to do, this will actually go through in a little bit of detail. First claim is that the GCD of A to the N and B is equal to 1. Okay? So, and the reason I'm going to go through this is because this idea is going to come up a little bit later on in your argument if you kind of follow it the way that I kind of want you to. Not that you have to do it this way. Again, in general there are multiple ways to prove a theorem. So as long as it's correct, it's fine. Why is this true? Well, we're assuming that the GCD of A to the N and B to the N is equal to 1. And let's see if we can see why that forces this to be true as well. What if the GCD is not 1? Then that means that there is some integer. So the GCD is always a natural number. If the GCD is not 1, it has to be bigger than 1, right? And there's, oops, sorry. I'll just, okay. I'm too lazy to go erase this. All right. So then there's some integer D bigger than 1, right? Such that D divides A to the N and D divides B. Okay. Right? You believe that? The GCD is not 1. It's got to be bigger than 1. So we get this. How does that, can you see how that is going to contradict this? Anyone see why that might be true? If D divides B, okay, this is, I'm not going to write all this down. This is where it's up to you to fill in the steps here. And it's just some positive integer. If D divides B, then D certainly divides B to the N, right? DX equals B, then DX times B to the N minus 1 is equal to B times B to the N minus 1, which is B to the N. So if you know an integer divides something, then that integer is going to divide every positive power of that thing as well, right? So by this you would get that D divides B to the N, but then D divides A to the N. D divides B to the N is bigger than 1. Contradicts that. See that? Okay. So D divides A to the N. D divides B to the N. D is bigger than 1. And this contradicts our inductive hypothesis, right? Okay. Yeah, I mean, you can, you can do that. Are you talking about, oh, you're talking about here? Yeah. Or just the proof in general, just. Of course you never, you said larger numbers and it wasn't a prime one. That's prime two. No, I mean, I know what you're saying, but what you're, you could, you could possibly make that work out, but it's just going to end up being a lot messier than it needs to be, because then you're just going to be representing A and B as products of primes which we haven't proven yet, by the way. So I'd rather you not do it that way. Okay. So we've got this now, right? Here's what I'll say, and then I'm probably going to stop here in a second. Let me use two asterisks so it's not confusing. So what can we do? Well, by number 20 part A, which I just told you what that was up above here, and I'm going to explain where this comes from. The GCD of A to the N and B to the N plus 1 equals 1 as well. So here's the idea, right? 20A says that if the GCD of two guys is equal to 1, and then the GCD of two other guys is equal to 1, but the left hand number is the same in both cases, then the GCD of A and Bc is equal to 1. So as long as you have the same first number, and those GCDs are both 1, then the GCD of that first number and the product of the two second numbers is also 1. That's what this is saying. You want to think of it abstractly. Don't get hung up on the A and the B, okay? So what do we have in this case? We've got the GCD of A to the N, B is equal to 1, right? Think of the A to the N as the A in 20A now. And the GCD of A to the N, B to the N is equal to 1. So in particular those first numbers are the same. So 20A says then that the GCD of A to the N, B to the N plus 1, right? Multiply the second components is equal to 1 as well. You guys see this? Okay? So, oops, that's not good. This has been launched, what happened? Oh, I hit something I didn't want to hit. Okay, yeah. All right. Thank you. Okay. Let's see, yeah. How do I do that? Okay. Up here? Oh, just up here. Okay, there we go. Yeah, sorry. I do know how to do a few things on a computer. I feel really old now, but okay. Convert to me. Very, very old. I'm not even going to talk about drinking anymore. Okay. So, you guys have this down? Okay. So, and like usual, I'm taking more time on this than I wanted to, but this is going to help you on the homework anyway. So, what do we know? We know this is going to seem a little weird, but the GCD, I'm doing this now for a specific reason, which I'll tell you in a second. The GCD of B to the N plus 1 and A to the N equals 1. Right? This is what I just wrote before, I think, except I've just flipped the order now. Right? And this is silly. This is the kind of thing I would yell at you guys for writing, but I'm actually doing this for a reason. I think the intuitive gain here is going to help. It's going to warrant this redundancy. Do you believe that? Why did I write it twice? I wrote it twice for a reason, and it's actually good that I did that. Here's why. Because you agree that we have the same first component in both places? And of course, the GCD of A and B and the GCD of B and A, it's the same. It doesn't matter what the order is. Right? That doesn't matter. So, what does problem 28 say? If we have the same first components and those two GCDs are both one, then the GCD of B to the N plus one and A to the 2N is equal to one as well. Multiply the second ones together. Okay? See, I'm just using problem number 20A applied to this particular pair of GCDs. Now, all that matters is that the GCDs are one and I have the same first components, which I do. So, I can apply problem 28 of this. Okay? And I'm going to flip the order back around here, but so this is A to the 2N comma B to the N plus one equals one as well. Right? Number 20 part A. Well, no. I mean, yeah. No, I guess, you know, what I sometimes will say not to write are things like five equals five and ten equals ten. And this is sort of along the lines of blah and blah. You don't need to say it again, you know, that kind of thing. But I'm doing it specifically so you can see how the problem 28 applies here. That's why I did it. Yeah. Well, yeah. I mean, well, I mean, I'm doing it just so that, but the thing is I'm teaching this class, so I'm trying to write it down so everyone understands it. If someone's sophisticated, had sufficiently good sophistication mathematically, wouldn't need to see this because they would just say, oh yeah, 20A, of course. I don't need to write it twice. But I'm trying to teach now. So the rules are slightly different for me than they are for you. Okay? What's that? Yes. Exactly. Well, that's what I'm doing right now, too. Okay. Do you see where this is coming from? I kind of did two steps in one now. I applied problem 28 and I flipped the order around again. Okay? Now, I'm going to leave it to you now from here to deduce that the GCD of A to the n plus 1 and B to the n plus 1 equals 1. And the hint I'm going to give you, you can write this down. In fact, I would suggest that you write this down. Here's the hint on how to finish this. Is that 2, n is a positive integer. So 2n is bigger than or equal to n plus 1. So once you've got a bigger power in the GCD being 1, then you can automatically sort of get it for free for the smaller power. Kind of like what I just did. Okay? That's the idea. This is about the cleanest proof I could think of. It's really nice and elegant. Induction, you don't have to write in a bunch of powers and a bunch of weird stuff. You can just do it very cleanly. So the hint is that 2n is bigger than or equal to n plus 1. Once you know that, you can get it pretty quickly. Okay? So I'll leave that to you to finish those few details. But you don't have a whole lot left to do at this point. Okay. So I should go ahead and move on here. So now what we're going to do, I'm going to just go into 3.1. Did anybody have any questions, by the way, about this last part here? No? Okay. And this section will unfortunately annoy some of you. And there's just nothing I can do about that. This is called the fundamental theorem of arithmetic. So here is the idea. What we're going to be doing in this section is basically what we're going to do is we're going to prove very rigorously all of the things that you learned when you were 12. And you're going to think, well, you already know all this stuff. Well, you maybe know that they're true, but you don't necessarily, you never knew why they were true, really. You might think, well, you had some idea why it was true, but now you're really going to get, well, if you care, you're going to now have a good sense of where this is all coming from. Okay? So that is what we're going to do. And actually, in a way, this is good, though, because you can relax a little bit, because some of the stuff we're going to start talking about, you already know. Prime number. Okay, so most of you already have an idea of what a prime number is. And of course, if you have your book open, you have the definition in front of you. So the first question is one prime? Is number one prime? It's not prime. You might say, well, why is that true? Well, because we just don't want it to be. Okay? That's why. Because it screws up. If you allow one to be prime, it screws up certain theorems. Certain theorems become false if you allow one to be prime. And we'll talk about that once we get to those on Thursday. Okay, so one's not prime. So if we're talking about prime numbers, they're automatically bigger than one. Right? And so, and you all know this. Right? A prime number is a number that is bigger than one and then has the only, sorry, the only positive divisors of the number are one in itself. Right? Okay, so easy enough. And you all are familiar with some prime numbers, two, three, five, seven, 11. I don't think I need to write those down. So, yes. It's an if and only if. Yeah. So technically, so in definitions, technically the best way to say this is if and only if. A lot of times just the convention is a lot of times people just get lazy in mathematics and they don't say that. It's just sort of understood when you make a definition, if is if and only if. It always is. But yes, that's true. It isn't if and only if. Okay. Again, natural number bigger than one is composite. Probably seen this as well. If, and, we're going to make it as simple as possible. And is not prime. Okay? That's what a composite number is. Okay. So that's easy enough. That should be an I-T-E. Sorry about the typo there or PEN-O or whatever you want to call it. So, we're going to prove a lemma about composite numbers. This is maybe the most annoying lemma that you'll ever see proved. But we're going to do it. So an integer is composited if and only if n equals rs for some integers r and s. Satisfying one is less than r. It's less than n. And one is less than s. It's less than n. Okay. We're going to use this. Essentially this is the theorem. The definition of composite is not prime. And so we're going to prove that a composite number, a number bigger than one is composited and only if it can be written as a product of two things that are strictly between one and itself. Okay. And this shouldn't be too surprising, really. Okay. So we have two things that we need to prove. So the first thing we're going to do is we're just going to assume that n bigger than one is an integer. Suppose that n is equal to r times s where one is less than r, which is less than n, and one is less than s, which is less than n. And, of course, r and s are both integers. Okay. So I'm claiming that n is composite in this case. That's what we're trying to do. I'm kind of going in the second direction. We're kind of establishing then the first hypothesis. If n is equal to r times s where r and s are strictly between one and n, then, of course, n is not prime. It's not prime because if n were prime, the only factors would be one in itself. And if r is between one and n, r can't be one and it can't be n. So it has a factor other than one in itself. So it's certainly not prime in this case. Yeah. And that's really all you have to say here. Then, by definition, n is composite. So everybody see that? It certainly can't be prime. I mean, I could write that out and say, oh, well, r is a factor that's not equal to one or n, so therefore it's not prime. I'm not going to write that. It should just be clear from this. Right. Okay. So let me wait until you get this down here. And I'll go on to the next page. Are we okay? Anybody need more time? Well, I could always go back. Okay. Conversely, suppose that n is composite. Okay. So we're just going to prove this in gory detail here. So that means that n is not prime. So if we know that n is not prime, we know the definition of prime is that the only positive divisors of n are one and n. If n were prime. So if it's not prime, what can we say? Joe. Yeah. You're missing one small detail. But if it's not composite, that means it's not the case that the only positive divisors are one and n. So there's some positive divisor, positive divisor. That was the thing you were missing. That's not one or n. Right? Okay. Does that make sense? Okay. Well, that's, yeah. Well, yeah. I mean, No, no, no. I mean, that's equivalent to what I'm saying. Yeah. Yeah, exactly. Then there is some positive divisor r of n such that r is not equal to one and r is not equal to n. Right? Hence, so what does that mean? If r is a divisor of n, hence, r times s, right? Equals n for some integer s. Yes, exactly. So that's just by definition of a divisor. Right? If r is a divisor, it means that r s equals n for some integer s. Okay. So now, as r divides n, let me make sure I get the quoted theorem right here. By theorem one, section 2.3, we get that r is less than or equal to n. This is the first theorem that I proved. When we started talking about divisibility, I proved this theorem that had like, you know, seven or eight parts. This is a special case of one of those parts. Okay. But remember that r is not equal to one and r is not equal to n. So what can we say? Well, one has to be less than r and r has to be less than n, right?