 One is that I signed a problem from the book that's actually wrong. I don't mean I signed one, I didn't mean to assign it to me and the statement is incorrect. So I want to talk about that really quickly. It's a very subtle thing, but some of you probably would have noticed that somebody already pointed this out to me. So number three in section 2.4. And it's very easy fix. I still want you to do the problem in just the way it's stated is not correct. So this says that if D is a common divisor of A and B, then D is the GCD of A and B if and only if the GCD of A over D and B over D is one. Well, the problem is if anyone has their book open, you can follow along here. So the problem is that if D happens to not be positive, then D certainly can't be the GCD of A and B because by definition the GCD is positive. So if D happens to be a negative divisor of A and B, then the statement is false. For example, D is minus one? Not true. So that's the easiest example. Minus one is certainly a common divisor of A and B. Minus one divides everything, but minus one can't be the GCD because it's negative. So how do you fix it? The assumption now is that D is a common positive divisor of A and B. And it's true, but if it's not positive, the statement is false. You guys clear on this? So let me just restate this. Two four, number three, insert the word positive between common and divisor. That's the best way I can say it. So the, yeah, is everybody okay with this? But you do have to insert that word. If not, it is not true. It's definitely false. Right, but what I'm saying is it's possible for the GCD of, for example, suppose A, B were both one and D was minus one, then the GCD of A over D and B over D would be one, but minus one is not the GCD of A and B. So it's true that the GCD has to be positive, but the other implication, if the GCD of A over D and B over D is one, that does not imply that D is the GCD of A and B. That implication is false if D is not assumed to be positive. So anyways, yeah, you have to, you have to fix that. But just with that in there, everything works out fine. Okay, second thing. So I have good news and bad news. The bad news, I don't think most of you probably won't really care that much about, but bad news is that I don't have the homework done. The good news, the good news will outweigh the bad news. The good news is that I'm going to extend your due date for the next assignment to Thursday. Okay, so, yeah. Both of them. Both of them, both of them, yeah. And the reason is that I have to go up to Boulder to give a talk on Tuesday, so I'm not going to be here now. You do have class, okay? You do have class, sorry. But you do have class. But I, yeah. James Parmenter, do you know who that is? Yeah, he's pretty good. You'll like him more than me, for sure. He's got, he's got, he looks like he has lamb chops on the side of his face. He's got big, he kind of looks like Elvis a little bit. I call him Lamb Boy. So I'm sure he would be happy if you all addressed him as Lamb Boy. So, yeah. But it's, but you do have class on Tuesday. Yes. Oh, yeah, it is. And I'm going to keep the scheduling of the next test the same, which actually is going to work out in your favor, really, because there's going to be less material on it. So, yeah, the next exam is coming up here before 2-1. The next exam is in two weeks, actually, from today. It's right before spring break. So you can get nice and depressed before you go to Florida. Yes, it's the 21st. Yeah, it's the 21st. You sort of have to keep, I don't want to push it back after spring break. If I put it to the Tuesday after spring break, that, I mean, that could be good for some people, but disastrous for other people. So I'm going to keep it on that date. Yeah, I think it's best to keep it that way. Okay. And the last thing I want to say before we get started, I'm wasting a lot of time as usual here, but last thing I want to say is just a couple things about the exam. Of course, the solutions are posted online. Some of you maybe have checked that out. You can go to Google to look at the solutions online. They're up there. They're on the web page. It's on the syllabus. And if you lost it, I mean, all you have to do is go on Google. Just Google my name, if you remember my name, Greg Oman. Go on the Math Department web page, and you just click, click, click. You'll find it. I was looking on Blackboard. Yeah, it's definitely not on Blackboard. I have my own web page aside from Blackboard. Blackboard crashes. Every three semesters it crashes, and I just don't want to risk it anymore. But yeah, it's up there. You'll find it. You'll find it. Two things I want to make comments about real quick. One, this is still happening a lot. I know I wrote this on some of your exams. A lot of you are still, this isn't a hugely bad mistake, but it's something you should be paying attention to. A lot of you, when you're writing down your variables, you're not specifying where they live. So some of you might say, okay, well, what's the GCD of A and B? Therefore, 1 is equal to xA plus yB. Really, you should be saying 1 is equal to xA plus yB for some integers, x and y. It's, again, the same thing. Remember what I said before about the pronoun? If I came up to you and said it is cold, I didn't tell you what it was. Okay, that's what you're doing when you do this. You're not saying what they are. You should say, again, you should be writing this clearly for the benefit of somebody who doesn't know what's going on. I know probably that you think that they're integers, but, hang on one second. Some of you, depending, here's the other problem. Here's the other reason why I like to see it, is that some of you are not sure what these things actually are. So if you don't specify, I don't know if you know or not for sure. I think some of you are saying things like, oh, x divides y because I know that x times some real number is equal to y. Therefore, x divides y. We'll know if that real number is not an integer. It doesn't count. That's why I'm also looking for this, because I need to know if you know what you're doing or not. If you don't tell me, I don't know if you know. Yeah. Okay, I've got a question. I am listening to you, by the way. Yes, definitely. Yes. Yes. Yes. Well, if you say something like that, you've already said, okay, s is a subset of n and k is an s. Well, then I know that it's an integer, for sure. If you have a question about your exam, we can talk about it after class. Okay. All right. So everything's still the same. The homework is just due on Thursday. And like I said, you will have class on Tuesday. Here is what we're going to do. So we've proved a bunch of these lemmas and such about the primes, and that's sort of a setup to prove this main theorem of the section, which I want to get through today. It's called, again, the fundamental theorem of arithmetic. So before we do that, we're going to kind of do half of it now. There we go. That's a little better. Shut up. I'll tell you what. I'm just going to close this. I think it'll be okay. If it gets really hot, then we'll open it up. If you get really hot, feel free to open it back up. I think it's just people waiting for the class to start. Wow, that was really scary. Actually, I kind of like you to do that. This is something that you all know. Any integer n bigger than 1 is either prime or a product of primes. This is actually just part of it. I'm going to state the whole thing here in a minute. This is sort of half of it, if you will. The other part of it is actually a little harder, but we'll talk about that here in a second. I don't know. Again, I'm old. Maybe you don't do this anymore. But maybe way back in sixth, seventh, eighth grade, something like that, you had problems that said, here's the number 168, write the prime factorization of this. You have a little tree where you break it into these things and then wherever you can't go any further, those are the primes and then you multiply everything together. You all sort of have experience that suggests that this is probably true. And it is. Of course, we're not going to do that. We're going to actually try to write a formal proof of this. As it turns out, it's really not too bad. Let's suppose that this statement is false. Let's suppose not. What we're going to do is we're going to let S be the set of all positive integers n bigger than 1, such that n is not prime and n is not a product of primes. Okay. So what can we say? Well, we're trying to prove that every integer bigger than 1 is a prime or a product of primes. We're supposing that's false. We're going to try to get a contradiction. If it's not the case, then that means there has to be an integer that's not prime and also not a product of primes. If it's not true, that means there's a counter example. There's something that's not prime and not a product of primes. So that means that S is non-empty, by our assumption. We're trying to get a contradiction. By our assumption that S is non-empty. And the reason why S isn't empty again is just because we're assuming... Essentially, we're just assuming it's not empty and we're going to try to find a contradiction. So what can we do? Well, if we have a non-empty set, everything in S is certainly a natural number, right? I didn't say, okay, I committed this sin that I was yelling at you about before, but this is a positive integer. I just got lazy. When I get lazy, I'll tell you I'm getting lazy. And it's bigger than 1. It's not a prime and it's not a product of primes. So everything in S is a natural number. Everything in S is a positive integer, right? So what can we say? There's a property we learned about the positive integers here. We have a non-empty set of positive integers. What do we know about it? Has a least element, right? Well-ordering property, right? We'll call it N, say, okay? Well, here's my question. So N is an element of S. Look at the definition of S, okay? It's a set of all integers, N bigger than 1 with two properties, right? Is N prime? Can N be prime? It's an element of S, right? So it means it has to satisfy that whatever condition is imposed on S. So N is certainly not prime, right? The members of S are the elements that are not prime and not products of primes. So N is certainly not prime. Just by definition of S. Okay. So what can we say about N? And this is why I proved this silly lemon in the beginning, Joe. Well, ultimately, that's true. That is ultimately where we're going to go with this, yes. But we need to get a contradiction at some point that proves that N has to be 1. But yes, you're right. You are right. But the point is, though, that we can't say that just yet. We need to get a contradiction that forces that on us. You'll see here in a second. Okay, so if you go back to Lemma 1, what would Lemma 1 say? If we have a number, an integer N bigger than 1 that's not prime, right? I showed you that it has to be a product of two things that are strictly between 1 and itself, right? That's what we did in the very beginning of class, I think, right, on Tuesday. Okay, so N is RS. For some integers, R and S satisfying 1 less than R less than N and 1 less than S less than N. Again, it's just straight from Lemma 1. Okay, so N was the least element of S, right? R and S are both less than N. So R and S cannot be an S because N was the least one and they're smaller. So they're not an S, right? Does this make sense? Okay, again, try to follow this. Some of these ideas are going to come up a little bit in your homework and certainly in problems that are going to come up later. So really, it's going to help you just to try to follow the argument. So what can we say about R and S? Thus, okay, this goes back to some logic. Maybe you did this in discreet. If R is definitely bigger than 1, right? So if it's not an S, then this condition applied to R can't be true, right? If it were, it would be an S. So what can we say about R? If R is bigger than 1 but not an S. Anybody tell me, Joe? Exactly, yes, exactly right. The members of S are both prime, sorry, are not prime and not a product of prime. So if that's not the case, you negate that. It becomes an element that's not an S is either prime or a product of prime. It's one of the two. It's got to be prime or a product of prime. So if it's neither, then that's exactly what S is saying, right? If it's neither. R is prime or a product of primes. Can we say the same thing about S? Yeah, because S is also not in capital, little S, I should say. It's not in capital S. So the same thing is true of little S, right? Okay, now here's the punch line. Remember what N was, right? N is equal to R times S. What can we say about N? So R is either prime or a product of primes. S is either prime or a product of primes. And N is R times S. So N, what can we say about N? Well, you can say even something more specific than that. N is a product of primes. Because it's a product of two things. If those guys were both prime, then N is still a product of primes because it's got two of them, right? So N is definitely a product of primes. And that contradicts the fact that it's an S, right? It's not a prime and it's not a product of primes. There's your contradiction. There is proven now. Okay. So this contradiction then proves the proposition. This is how proof by contradiction goes, right? You assume whatever it is you're trying to prove is false and get both statement P and statement not P, which is, of course, impossible. Therefore, you conclude that your original statement had to be true. Okay. Okay. And then there was another technical lemma, which is just more work than it's worth. And I think I'm just going to skip it and I'm just going to trust that you'll just believe it. Otherwise, I may not get through all this today. So the book doesn't do it either. The book doesn't do it either. I'm just trying to be as rigorous as possible, but sometimes time is more important than rigor. So I'm going to skip this and I'll just, you will all be convinced that it's true anyways. Okay. And so there's just one other thing we have to talk about and then we'll just go into the last part of the proof. And then again, I want to at least spend about 10 minutes or so talking about homework since I won't be here on Tuesday. Okay. Everybody copy this down? No? Okay. All right. I think, and I left out a couple of things here, so somebody please tell me if this is not right. I think we're on definition three now. So I'm pretty sure I defined prime and composite, and now we're on three. Somebody can tell me if that's wrong. Is definition two that we're on? Okay. Let's see. No, I think it is three, right? Okay. So let n bigger than one be an integer or a natural number, they're equivalent in this case. Okay. And let's say that n can be written as a product of primes in canonical form if n is equal to p1 to the alpha 1 p2 to the alpha 2 on down to say pk to the alpha k where p1 is less than p2, which is less than p3 on down. So basically the primes are listed in increasing order. And each of these exponents, each of these alpha sub i's are natural numbers, positive integers. Okay. And so this is just sort of canonical. Just in mathematics really just sort of means natural. That's the way you interpret this. It just means you're just writing, you're grouping all the primes together and then you're just going up an increasing order. And so I want to make an observation here that I'm not going to prove. And this is just something that we could try to prove, but again the amount of work that we're going to it is just not, it just doesn't justify, let me say it a better way, it's just not justified because this would be transparently clear to you that this is certainly the case. Any integer n bigger than 1 that can be written as a product of primes can be so written in canonical form. And again I'm not going to write a proof of this, but I just want to give you the idea. What's the idea? The idea is just all the primes you see, just group them all together and then just do it so that you're going in increasing order. That's it. So in this case what we want is 2 squared times 3 squared times 5 times 7. The right side is in canonical form. Okay. Do you guys all buy that? You just have a bunch of products of primes, just group them like this. And you can certainly get canonical form very easily. I don't think I need to write a proof, a formal proof of this. This is just obvious that you can do this. Okay. Hopefully this is obvious. This example made it obvious. What's that? Well, yeah, I mean more or less. Well the proof, technically the proof would be by induction on the number of primes in the product. But I'm not going to do it. I just don't want to. So yeah. You say well if there's only one prime, well there's nothing to do. Suppose that's true for n primes, then you have n plus one of them. And then you just basically say, okay well let me just look at all the first n ones. Well we can write that in canonical form and then just the other guy, wherever it is you just slide it into the right spot. I mean that's really all the proof amounts to. So yeah. Okay. I think that's about all I want to say here. Where am I going? Okay. All right. I'm going to do, okay, oh yeah, okay. So this is the last thing that I'm going to do, is just the fundamental theorem of arithmetic. So we'll do that now. And we should definitely have time to talk about a couple of homework problems before we leave. So okay. All right. Everybody have this down now? Okay. Let me go ahead and start a new page here. Okay. So this is the main, this is sort of the main point of the section here. So let me start up here. Fundamental theorem of arithmetic. And this actually has applications. It seems like I learned this in fifth grade, but there are some applications of this that allow you to do things that are not as obvious as you might think. We'll get into some of that later. Okay. And the statement's very simple, but I'm going to need to explain exactly what this means. Every integer and bigger than one can be written uniquely. That's the key. And this unique part is actually important. It really allows you to prove some other theorems about the rationality of square roots and things like that. I'm getting ahead of myself, but written uniquely in only one way in canonical form. Okay. I should have put this canonical form at the end, but that's okay as a product of primes. And there is something I definitely need to clarify here. Some of you might object and say, no, no, no, no. That is not true. And depending on how you interpret this, you're right. And that's why I need to make a clarification. And this is just the way it's done in mathematics. There's a little fudging going on here. But product of primes, you might say, well, what about five? Five's just a prime. It's not a product of primes. Five's an integer bigger than one, but it's just a prime by itself. It's not a product of anything. Well, when we say product of primes, we mean the product could just include one thing. Okay. So you have to be aware of that. I really want to impress that upon you. Product of primes, five by itself, we include as a product of primes. It's sort of a trivial product, if you will, in the sense or degenerate product, in the sense that there's only one guy there. But that's something you should be aware of. Okay. Okay. So here is the proof. And it turns out the proof is actually not that hard. We have to do a few things here, but it's not that bad. I think you should be able to follow this. Okay. So by proposition one, did I label this observation as observation one? Okay. I'll just say, okay. Well, let me just say observation. Every integer n bigger than one can be written as a product of primes in canonical form. Okay. All right. Proposition one said that every integer n bigger than one is a product of primes. And remember, the observation says that if we get a product of primes, we can always put it into canonical form. So if you combine these together, then we know that every integer n bigger than one can be written as a product of primes in canonical form. The only other thing we have to prove is that it's unique. This is going to require a little bit of work, but it's not that complex. It's really not that hard, though. No, actually, we can just do this directly, more or less. You might be able to do that too. Oh. It depends. It really does depend. And certainly it's possible that you could write a very natural proof that way here, too. I'm just not going to do it that way. Okay. So the question here, this is kind of the complicated part, is what does that even mean, uniqueness? Okay. What does it mean to be able to be written uniquely in canonical form? And I'll tell you what it means. And if you try to follow this long, I think you'll be okay. So let's suppose that, okay, so there's going to be a lot of symbols here, but p1 to the alpha 1, p2 to the alpha 2, on down to pk to the alpha k is equal to, oops, I can fix this, okay, q1 to the beta 1, q2, hopefully my bad handwriting isn't going to screw this up too badly, to the beta 2, on down to qn to the beta n. Now I'll put an asterisk next to this because we're going to refer to this multiple times. Okay, can you guys all read that? Beta 1, beta 2, if you just want to write b's, that's fine too, but really I'm intending this to be the letter beta. Okay. Seriously, can you guys read this? Is this okay? Okay. Well, let me specify exactly what this means. Where 1, each exponent alpha i and beta i, all of these guys are natural numbers. These are all positive integers. 2, each p sub i and q sub i is prime. 3, p1 is less than p2, which is less than dot, dot, dot, which is less than pk. And 4, same thing is true for the qs. q1 is less than q2, is less than on down to qn. Okay. I'm going to try to squeeze this all on one slide here. Sorry for being a little sloppy here. What is it that we have to prove? I want you guys to think about this. Okay, so before I say, you see what I've got here? What this says is really just that this is a product of primes in canonical form and this is a product of primes in canonical form. Right? That's what all this is saying, really. And if I want every integer n bigger than 1 to have a unique expression as a product of primes in canonical form, I really mean that these should all look exactly the same. And so the point I want to make first is let's try to enunciate that. Let's try to clarify that. What does that mean that those expressions are exactly the same? There are several things, actually. There are three things that we need to know. Okay, so if those expressions, and this is what I want you to think about, if those two sides have to look exactly alike, what has to be true? Can anybody tell me one thing that has to be true? Yes, that's true. That's true. I'm going to write that down in a second. Yeah, that's exactly right. All the exponents have to be the same. And the other thing, this is going to write down first because this is the easiest one to prove. k and n have to be the same. They have to be the same number of primes. So let me just say it all and I'll write it down. n and k have to be the same. The same number of primes on the left is on the right. Each of the corresponding primes have to be the same. p1 and q1 have to be the same. p2 and q2 have to be the same, etc. The corresponding powers have to be the same. Then those two expressions will be identical. So those are the three things we need to prove. You guys with me here? All right, so that's what we're going to do. So the first thing we're going to prove is that k in fact is equal to n. The second thing we'll prove is that these p's and q's, the corresponding p's and q's are the same. pi is equal to qi for any i between 1 and n. And remember, after we've proven 1, k and n are the same. So I can pick either n or k. It's the same thing, right? And the third is that the corresponding powers have to be the same. And then we certainly have the exact same expression on both sides. Okay. So that is our goal. So we're just going to establish these three things. And it's not as hard as you might think. It's really not that bad. Then we'll be done with this section. Okay here. Anyone need more time? All right. So let's do one first. So what we're going to do is we're going to let A be the set of the primes with p, right? So A is going to be p1, p2, on down to pk. And we're going to let B be the q primes, right? q1, q2, down to q sub n. And we're going to show that A is equal to B. These two sets are the same. Okay. So I want to be clear about this. Suppose you know the two sets are the same. Well, then k and n are automatically have to be equal because if the sets are the same, whatever number of elements you have in one set, it has to be the same as the number of elements in the other set because they're the same set, right? So you get k equal to n for free if we do that. We're actually going to get a little more than that too. But that's the idea. So do you remember maybe from Discrete how you show that two sets are equal? What do you do, Joe? Yes, that's right. And that is something, if you took Discrete, if you had a reasonable instructor for this class, you should have been exposed to that before. Okay. So we're just going to do one direction because the other direction follows the same way. And I don't want to spend that time on a class doing exactly the same argument again. But well, I'll show you that A is a subset of B. And then B has to be a subset of A for the same reason. Okay. So how do you do this? So you pick an arbitrary element of A, right? Pick any, we'll just say P sub I, right? P sub I in A. And I'm going to suppress this for a time. But we're going to show that P sub I is in B. That's what we're going to do. Okay. This is in your notes. I'm not going to flip back now. But the star, the star equation was just that those two products of primes were equal, right? Just the P sub I's to these powers equals the Q sub I's to these powers. By star, we know that P sub I divides Q1 to the beta 1 dot, dot, dot on down to Qn to the beta n, right? If you look back in your notes to that star equation, you guys should buy that, right? Because P1 times something is certainly equal to that right side. So P1 divides it. Hopefully that's not too much of a stretch here. You guys all buy that? If you look at the equation, right? P1 times something is the right hand side. So by, let's see, by corollary 1, I'll remind you what this is in a second. Okay. Corollary 1, you can look back in this in your notes, but corollary 1 was, the corollary said it, prime divides a product of a bunch of integers and then that prime has to divide one of the integers in the product, okay? So if this prime divides a product of a bunch of these, here's the way I want you to think about it, okay? Imagine actually writing, instead of Q1 to the beta 1, just write Q1 times itself, beta 1 times. And the same thing with Q2 and the same thing on down. So you have all these primes written out. So since Pi divides that whole product, Pi has to divide one of those primes. Not just one of the primes to a power, it has to actually divide one of those primes if you write them all out. It's a product of a bunch of integers. The prime has to divide one of them. It doesn't mean any one of them, right? No, it doesn't. It just means it has to divide some of them. Yes, it has to divide some, one of them, yeah. So by corollary 1, P sub i divides, say, Q sub j, right? Or some j with 1 less than or equal to j less than or equal to n. Okay. Any questions about this? Do you guys see where I'm saying this? It's not just that it'll divide some QI to a power. Well, if it divides that to a power, then just write them all out. Then it has to divide that one guy too, okay? So then what can we say about this? Well, there's also a lemma I proved specifically for this because I knew it was coming. I think it was lemma 2. Lemma 2 said that if a prime divides another prime, they have to be equal. Okay, that's something else we did on Tuesday. So we know that P sub i has to actually be equal to Q sub j because they're prime. Thus, tell me if you believe this, thus P sub i is an element of B, right? It's equal to something in B. It's got to be in B. And of course the exact same sort of reflected argument applies to show that B is a subset of A. It's exactly the same argument, right? Yeah. Are we assuming that j is an integer? Yes, yes. So again, when I'm doing this, I will get, I will be just slightly lazy from time to time because time, well, yeah. But yeah, j's definitely an integer. Similar, right? Okay, so A is equal to B. And if A and B are equal, then they have the same number of elements and therefore K has to be equal to N. That was the first thing we needed to show. Okay, the second case, there's really not much to say here. So the second thing we need to show is that each of these P i's and Q i's are the same. Okay, this is the one I was going to prove with lemma before this that I decided not to. Actually, I'll tell you what. Let me go back here for a second. Okay, and this is the thing that I'm hoping you guys will just take on faith because of time. The P's are listed in increasing order. That's the assumption. The Q's are in increasing order. But if the set of the P's and the set of the Q's are the same and they're listed in increasing order, then the corresponding ones have to match up. They have to, right? The smallest one has to correspond to the smallest one. If it's off at all, then you can't, well, basically you're going to get a contradiction if it's skewed in any way. Since the sets are the same and they're in increasing order, it all has to line up. So all the P's and Q's have to be equal to each other. If the corresponding P's and Q's have to be the same. Like I said, I was going to prove a lemma that proved this rigorously, but I just don't have time to do that. So P I equals Q I, that's an I, sorry, for any I between one and N by the proof of one. In general, if two sets are equal and you have them written this way, they don't have to match up. Because, for example, the set one, two, and the set two, the set one, two, and the set two, there's no order built into the sets. But one and two aren't equal. But if they're listed both in increasing order, then they have to be the same. That's the point. Okay, so we've got two things done. We just have one more thing to do. Okay, so the last thing we have to prove is that the corresponding exponents have to be the same. Okay, so that's what we're going to try to prove. So we now have a one and two. Let me put a star next to this. P one to the alpha one, P two to the alpha two, P N to the alpha N is equal to P one to the beta one, P two to the beta two, on down to P N to the beta N. Okay, and why am I doing this? Well, this just goes back to the original equation that I put an asterisk next to. We've already proven that the number of primes on either side are the same. So we can just use N on both of them, because N and K are the same, and we know that the corresponding primes are also the same. That's why on the right side, I can replace each QI with PI, because we know that they all are the same. We just proved that. So what is it that we have to prove? We need to know that alpha one equals beta one, alpha two equals beta two, et cetera, and then we know that they look exactly the same at that point. Okay, so this is the last thing we're going to do. And again, for the sake of time, I'm just going to prove that, and also for simplicity, I'm just going to show you that alpha one is equal to beta one. And then it's just automatic that all the other powers have to be the same. Why is that? Because the order doesn't matter. You can, I mean, well, okay, at this point the order doesn't matter. You can just, the P two to the alpha two and the P two to the beta two, well then just, because it's multiple agents, just flip those to the front and then just run the same argument again. Okay, so then you can just get it for everything for free after that. Okay, and again, I'm doing this now just because I don't want to say, we're going to prove alpha i equals beta i and then it just gets messier. So let's just do it for the first one and you'll see the idea. Okay. Now here's where, okay, we actually are going to use a contradiction here, but let's suppose not. Then if they're not the same, then either alpha one is less than beta one or beta one is less than alpha one, right? Let's suppose that alpha one is less than beta one. The same proof is going to work the other way. I mean, if beta one is less than alpha one, it's the same exact proof. All we're going to do is divide, oops, sorry, I'm just getting sick of erasing stuff. Okay, so this is in your notes if you're copying it. I just put this equation star. You can look in your notes to see this equation. We're going to divide everything, both sides by p1 to the alpha one. That's what we're going to get. That's what we're going to do. So then what happens on the left side when we divide by p1 to the alpha one? Well, it just cancels out, right? It's gone. And then we just have p2 to the alpha 2, et cetera, on the left-hand side. You guys see this from your notes? Yeah? Okay. p2 to the alpha 2 times p3 to the alpha 3 on down to pn to the alpha n equals, what do we get on the right side? What's the first, what is p1, what is the new power on p1 on the right side? Beta 1 minus alpha 1, right? You guys see that? Dividing through by p1 to the alpha 1, we have, well, when you have the same base, you subtract the power. So we just subtract the power from it and then we get everything else being the same. And now we're basically done. Here's what we have, okay? So we're just going to reuse the proof of 1 to get a contradiction here. And I'll just say what it is first and then you can write it down if you want if you don't want to wait for me to write on the screen. What do we prove in the first, for the first part of this uniqueness proof? We prove that if you have a product of primes written in canonical form equal to another product of primes in canonical form, then the set of primes are the same. We prove that. That's exactly what we proved. Now you see what we have? These are both in canonical form, right? They're still increasing even though we're missing p1, p2 to pn is still increasing. So we have a product of primes in canonical form on the left equal to a product of primes in canonical form on the right, but the sets of primes are not the same because p1's missing now. That contradicts the first thing we proved, that the set of primes is the same, okay? And therefore alpha 1 and beta 1 have to be equal. And then you apply the same argument to all the powers. It's the same thing. Hopefully you guys are with me on this, okay? Can't have this. This is what we've, the first thing we proved is that the set of primes is the same. If you have two primes, products of primes in canonical form, they have to be the same. We have two primes equal in canonical form. The sets of primes are the same, but they're not. There's a contradiction. This is false. And if you want to, just to totally nip this in the bud here, why is it, why are these not the same? Because p1 is in this set, but it's not in that set. They can't be the same because they don't have the same members. Okay? So the conclusion is that our assumption, our contradictory assumption is actually, it leads to a contradiction. And thus these powers actually do have to be the same. And similarly, powers all match up. Alpha i equals beta i for all i. Satisfying one less than or equal to i less than or equal to n. And that's it. That's the proof. To establish the existence of the product, of an integer bigger than one is a product of primes of canonical form. And that any two such expressions have to be the same. Okay. And I told you, and this is good, and now we actually have a little more time this time. So I didn't babble on as long as I usually do. Definitely want to take this time. This will be good for you to talk about a couple of Homer problems. Okay? I can tell you right now because I just, I think I have a pretty good sense of what people find tricky and what they don't. 2.4 number six is going to cause some trouble. Okay? I'm going to give you a, I'm actually going to give you more than a hint. I'm going to give you kind of a roadmap as to how you can go about trying to do number six. Okay? And I'll write it down for you. Okay, so this is it. This is the end of the section. And so we are done now. So homework. I don't know how, if any of you have actually looked at number six yet, maybe it's not due to, well, now it's a week from today. Okay, good. That tells me that you have actually some idea of what you're doing because you just wrecked the whole class by saying that now. Now everyone's going on Google to get their homework solutions. Yeah, but it doesn't make any sense. Okay. They can all go. Yeah, well, for, yeah, that's true. The question is like. Okay. Well, okay. My tea guy did. So there's, there's, there are several ways you can go about doing this. I'm going to do it. I'm going to, okay. In general, there's not just one way to do a proof through multiple ways to do a proof. Some of them, and sometimes you might have a proof that short, but kind of masks the idea. So you kind of left it going, how did you even come up with this? And some of them are a little bit longer, but they're clear to understand this one is going to be maybe a little bit longer, although it's actually not that long. But I think it's, it's much less. It's actually a little bit less messy maybe than what you're talking about. I'm not sure exactly. I have an idea of probably where he was going with it, but here's, okay. So this is, let me just write down the problem first. Okay. Two, four, number six says to prove it. I'm going to abbreviate this. You really shouldn't do this in your proofs. I'm just doing this now to save some time to actually talk about this problem. If the GCD of A and B is one, then prove that the GCD of A plus B and A times B is equal to one. If, so the problem says, basically it says assume the GCD of A and B is one. So in other words, A and B are relatively prime. Yeah. Okay. So then what we're going to try to, you need to prove is that the GCD of A plus B and A B is one. Now I can tell you this. If you just try to go through it by the definition, or if you, okay. So let me just give you a couple of ideas first. There's a theorem that said if you know that some integer linear combinations of two, linear combination of two integers is one, then those two integers have GCD equal to one, then they're relatively prime. So you might think, okay, well I know that, that, you know, XA plus YB equals one for some integers X and Y. And I'll try to use that to somehow mess with that to get something times A plus B plus something times A B is one, and then I'm done. Then I'm done. You, that is something you should probably be able to do. Although I'm not going to go in that direction, actually. I haven't even really thought about it. It's going to probably require a little bit of cleverness and manipulation and such. And this does too a little bit, but I think maybe not as much. So it's not as mysterious where the solution's coming from. So I'm going to go, I'm going to kind of talk about it this way. This is a suggestion. This is not saying you have to do it. If you want to try to mess with this and do what I was just saying, and it's right, that's fine. I'm not saying this is the only way that you can do it. Okay? Here's what I'm going to suggest here. And I'm just going to write it exactly that way. Suggestion. Okay? Not have to or only way. I'm not saying that. Let D be the GCD of A plus B and AB. And I also think this has the advantage of kind of getting you more familiar with some of the tools that are going to come up. You're going to, this is going to come up too a little bit in a homework problem I gave you in 3-1, which I'm also going to talk about here in a minute. Okay. Try to do this. And again, remember, I'm just giving you a kind of a roadmap. I'm not going to tell you exactly how to do these, but I'm going to break it down into a bunch of smaller pieces that should be more manageable. Okay? First thing, one, prove that, and you might want to do this on scratch paper first. Make sure you have all the pieces together and then write a nice coherent argument in your homework. Prove that D divides A squared minus B squared. I'll even give you a hint as to how you go about doing this. This may seem a little mysterious for now. Can you guys see how you might be able to pull this off? Anyone? You know that D divides A plus B for sure, right? This is a GCD. How can you show that it divides A squared minus B squared? Think about factoring, but so what is A squared minus B squared? It's A plus B times A minus B. What's that? Okay, so wait a minute. A squared minus B squared is A plus B times A minus B. I'm just going to say it for the sake of time. I'm going to have time to write all this down, but I want you to just pay attention to this. This is A plus B times A minus B. Most of you should know that, right? You already know that D divides A plus B. So D times X equals A plus B for some X. But then DX times A minus B is A plus B A minus B, which is A squared minus B squared. You see that? Did I do that too fast? D times X is A plus B because D divides A plus B. Multiply both sides by A minus B. Then D times something is A plus B A minus B, which is A squared minus B squared. Okay? Got it? Okay. There's the first thing to prove that D divides A squared plus B squared. This is not maybe as obvious. Yeah, it does. But in this case, it's a little trickier, though, because we don't know that D divides A actually, right? Here's the hint here. You can write this down again. I'm not going to write all this down for you, but if you want to write it down, you can't. How could we do this? Anybody see? I'm only going to have a couple more minutes to do this, but anybody see how to do this? This may be a little trickier. Well, think about it. D divides A plus B. Say D times X equals A plus B. Then D times X times A plus B is A. So if D divides A plus B, then D divides A plus B squared, the whole quantity squared. That's for sure, right? Just multiply both sides by A plus B. And you get D divides A plus B squared. What is A plus B quantity squared? A squared plus 2AB plus B squared. But you've got an extra 2AB that you don't want. But you also know that D divides AB. So D is going to divide 2AB. So then you can get that extra term out of the way because you know that D divides that extra term. So then it's going to divide the difference, basically. D divides A squared plus 2AB plus B squared. It also divides AB. So it divides 2AB. So it divides A squared plus 2AB plus B squared minus 2AB, which is exactly that. Write it down. If you want to ruin it, just write it all down. What I'm saying right now. Write it down. Okay? Now, once you've got this, you can prove that D divides 2A squared and D divides 2B squared. There's not much to this, by the way. Well, you'll see. How can you get this? Subtract. If D divides something into something else, it also divides the difference. There's a theorem. I think it was theorem 2 part G from section 2.2 or something like that that says if D divides AB, then it divides all linear combinations of AB. Okay? So if it divides A squared plus B squared and D divides A squared minus B squared, it's going to divide their sum. What's their sum? 2A squared. Right? And you can do the same thing the other way, to get 2B squared. So, what do you know about D? D is going to divide the GCD of 2A squared and 2B squared. Right? Do you buy that? If it divides both of them, it has to divide their greatest common divisor as well. That's part of the definition of GCD, right? It divides both, and anything that divides both divides the GCD. Now you can use some theorems that I proved in class. You can pull the 2 out. I know I'm going to stop here, but you can pull out the 2. That's something we did in class. And if the GCD of A and B is 1, then the GCD of A squared and B squared is 1 by what I just told you yesterday. Remember one of the older problems I said that we talked about yesterday, if GCD of A and B is 1, then the GCD of A to the N and B to the N is 1. You see, now you can get it, you can will it down to know that D has to be 1 or 2. And then you have to rule out 2, and then you're done. I'm warning you guys, you have to think in here. You have to think in here. And sometimes you have to be a little clever. There's just no way around it. That's the nature of the beast. That's what we do in this course. And so hopefully some of you are not doing this, and some of you have your heads down, and right now you're head spinning, and you wish you were dead. But hopefully most of you are thinking that, well, you worry about your grade, but you want to think of these as puzzles. You don't want to think of these out. I'm not saying you should be jumping up and down for joy, and you should spend every Friday and Saturday night doing this instead of going out and having fun. But that's kind of the point of this, is to think of this as a challenge. Think of them as puzzles that you want to figure out. And so some of them get a little tricky. All right. Well, I give you a lot of hints here. So this is one way you can go with this problem. All right. So let me talk about one more problem, then we'll finish up for the day. OK. I go to NewPage now. Is that OK? This is another one I think is going to cause some trouble. 3.1, number 8. Suppose that, I'm going to kind of abbreviate this, but p is bigger than or equal to q, and q is bigger than or equal to 5. p and q are prime here. This is almost kind of a weird problem. You almost think that there's no way this is true. Then it is kind of surprising. Then 24, that's kind of a big number. 24 has to divide p squared minus q squared. Well, yeah. It seems random. It does. It does seem random. But this is where I'm going to, OK. Sorry, trying to collect my thoughts here. We're starting to get to a point in the course where just hammering on things from just using the knowledge you came in to this course with is not going to work anymore. So some of you might think, well, how am I going to do this? I'll divide by 24 using the division algorithm. I'll have 24 cases or something like this. That's not going to work. I'm going to have to be a little craftier than that to do some of these. And this is a perfect example of this. Here's my suggestion. I really want you guys to listen to this because this is going to maybe help you a little bit to just have an idea of where to start. A lot of you are going to see this problem, and it's OK because you just don't have a lot of experience yet. There's nothing wrong with this. I have no idea what the hell to do. I have no idea how to start this. How am I going to prove 20 plus a weird number? How do I prove this? I have no idea. Well, think of it this way, OK. There's certainly two has to. Maybe I can just prove that two divides it. Maybe I can prove that p squared minus q squared actually has to be even somehow. And I'm just going to talk this out first. I really want you guys to listen. This is going to help you. It really is. Well, how do I know? OK, so if this is true, then two has to divide it. p squared minus q squared has to be even. Why? Why is that to be even? Well, p and q are primes. So p and q, because they're bigger than or equal to 5, have to be odd. Because if they're even, two would divide into them. And that would contradict the fact that they're primes. You guys buy this? It can't be even. A prime bigger than two cannot be even. Because it can't have two as a factor. It only has one in itself as a factor. OK. But then an odd number squared is still odd. This is a corollary of what you guys proved. Actually, some of you just proved the square of an odd number was odd, but that's not what I asked on the exam. So an odd number squared is odd minus, an odd number squared is odd, odd minus odd is even. Damn, so now we've got two. So at least you're getting somewhere. This is what I'm saying. If you don't know how to do the full thing, break it into a sub-piece that you can handle. And then just keep breaking it up until you see kind of what's going on, and then put it all together. That's kind of how you're, very roughly, what you're doing here. OK. And I'm going to tell you, here, this is huge. This will help you on this problem, and I think maybe another problem in this section too. This is really big. And from the notes, let me make sure, no, I'm going to use the labeling the book does. For this section, I want you to be very aware of this corollary, because this will help you. Certainly will help you in this problem. I'm going to say it. You should all write this down. Corollary 2 on page 23. Corollary 2 page 23. And I want you to listen to what I'm saying. OK. We only have five more minutes. I know it's been a while now. But this is what corollary 2 says. And I'll write some of this down in a second. But I want you to listen. You guys should listen to this. Corollary 2 says if you know that A divides in us some number C, and B divides some number C, and A and B are relatively prime, then their product divides C. Listen to this. 24 is 8 times 3. 8 and 3 are definitely relatively prime. They don't have anything in common. So if you can prove that 8 divides P squared minus Q squared, and you can prove that 3 divides P squared minus Q squared, then by corollary 2, since they're relatively prime, the product, which is 24, divides P squared minus Q squared. So really, you can break this down into two sub-problems that are easier to manage. Use that corollary to squish it together, and then you get the result. This is what I mean by you guys have to start thinking outside the box a little bit. You've got to start to be clever. Use the tools. Sometimes you have to use the tools, or you're not going to be able to solve the problem. Okay? So the idea is to prove that 1, 8 divides P squared minus Q squared, and 2, 3 divides P squared minus Q squared. Now I'll kind of get it started a little bit on 1, and then we'll stop. But with 3, you're going to do something similar. Again, I'm just sketching the rough idea. I'm not trying to make this polished. Your proof should be a little more polished than this. Well, what do we know about P and Q? Since they're odd primes, they are odd. P and Q are both odd. Again, you need to write down everything clearly. I'm just doing this for the sake of time. We know that P is equal to 2R plus 1, and we know that Q is equal to 2S plus 1 for some integers R and S, right? I'm not going to write all that out, because I just don't have the time, but we know. So what do we know about P squared minus Q squared? Well, how can we get somewhere with this? In this case, you may not actually even need to do this, but I think this is going to serve you well. This is P plus Q times P minus Q, right? Okay? You buy that? We just talked about that before. So what is this? This is 2R plus 1 plus 2S plus 1 times 2R plus 1 minus 2S plus 1, which just becomes minus 2S minus 1, right? You follow this? Okay? And you prove that you should say something like, because P is a prime, it must be odd. You should justify why it can't be even. And what is this? Well, okay, I'm running out of time. You see that I can pull out a 2? 1 plus 1 is 2, so I can just pull that out. And I just have R plus S plus 1 left over, right? What do we have left in the second set of parentheses? This is just 2R minus 2S, right? You see that? Well, now what can we do? We can pull out another 2, right? So now this is 4 times R plus S plus 1 times R minus S. You see that? What do we want to prove? We want to prove that 8 divides it. Well, we're kind of close. We've got the 4 on the outside. Well, we really like to have an 8 factored out, because then we know that 8 divides it, and then we're done, right? Well, what you want to do is look at what you've got left over. Listen to what I'm saying, and we're going to stop here in a minute. If you can prove that at least one of these guys is even, then you can pull a 2 out, and you've got your 8. Okay? How do you do that? We'll just consider cases, right? Suppose that, for example, suppose R and S are both even, then R minus S is even, and you've got your 2. Suppose R and S are both odd, then R minus S is odd, and you've got your 2. You see that? You guys listening to this? Hope you are. This is helping you, hopefully. Well, what's the last case? One's even and one's odd, right? If they're both even or both odd, that's even, and you've got what you want. Otherwise, one's even and one's odd. Well, what can you say? If one's even and one's odd, the sum of an even and an odd number is odd, right? But then the plus one makes it even, so then you've got your even in that case too, right? You see that? You guys got to start. These are going to get a little more challenging. They're not going to be all like the induction proofs that work themselves out. You have to be a little clever. Then you can do something similar with three. What's the remainder? If you divided P by three, what's the remainder? It's got to be what? One or two, right? Can't be zero because it's a prime and it's bigger than or equal to five. You see that? So you can do the same kind of game. You know P and Q have to have a certain form and then just go through this and you should be able to extract a three. And then you get the answer by that corollary. Sorry, I went through this kind of fast, but I didn't really have a choice. In section 2.2, it says the square of any odd integers of the form 8k plus one. If you use that line, it's kind of... Oh, I see what you're saying. A little bit. Yeah, yeah, okay. Yeah, eight. Yeah, so the square of any odd integers of the form 8k plus one. Okay, so yeah, so if you use that, then you can get the... Yeah, you can get the eight to pop out right away. Yeah, right. Is that something that I assigned to you or not? Yeah, I don't think I assigned that particular problem. That's not a problem. Okay, so here are the rules, okay? If you want to use a previous homework problem, that's okay, but if I haven't assigned it, then you should write a proof of it. It's not a problem. It's getting information. Oh, it's in the section. Okay, anything in the section is fair game. Okay. Anything in the section is fair game. Even if I didn't do it in class, but you should cite it. You should say by corollary, blah, blah, blah. Say where it's coming from, okay?