 Most of you are still here. That's good. After Tuesday, so. Well, I'm in the middle of it right now. How is it so far? Okay. Okay. I graded the first problem, and I was surprised at how many issues there were with stating the first and second principle of induction. Oh, you haven't... Oh, what does that mean? It means that I haven't seen it. Oh, right. Okay. I appreciate your honesty. You have a man of great integrity. Okay. Well, now you know something about it. Okay. Okay. I won't say much about it then. Well, you'll get your exam back on Tuesday. It'll be done on Tuesday. All I will say is that some of you need to study your definitions a little bit harder. That's all I'm going to say for now, I suppose. So, what's that? It's a lead job and definitions, Joe. Oh, yeah. Yeah. Well, you already knew that it was probably going to be on the exam anyways. Yeah. Yeah. So, okay. So, the one thing I'll say is... Okay. Can you guys, do you mind if you just stop talking for a second? I appreciate it. The thing I want to emphasize, this isn't going to give anything else away, but, you know, some of you, I think, are... You're sort of trying to make these definitions like history dates, and you're not actually understanding what they're saying. For all or there exists, you know, for all acts there exists, some of you really don't know what that is saying, and that's why there are a lot of mistakes on these definitions. I'm confident of it because I've taught this kind of stuff before. You really need to be looking over to the point where you feel like you actually really understand what it's saying, not that you've just memorized a bunch of formulas. Some of you will, instead of saying s an element of with the epsilon, or sorry, instead of saying s is a subset of, right, with the sort of turned u and the bar underneath it, you'll say s is an element of n when you actually mean that s is a subset of n. So, these are the kinds of things. This is just basic mathematical lingo. Some of you still don't quite have that down yet. So, that's important. It's very important to get that down because changing those symbols, you're saying something completely different. And often, by just changing a symbol, what you say doesn't make sense. Yes. Yes, you will. Because the definition is right in your notes. It's right in the book. You guys should have done better on this. I'm sorry, you really should have done better on this. I told you it'd be on there. It's in the notes. It's in your book. All you had to do is memorize it. That's it. So, yes, you are going to get hosed a bit on this. Yes, that's correct. You'll get partial credit. But, you know, there's really no excuse to not get this. And most of you did not get it. Even though I said it many, many, many times. And I even told you, for example, okay, my rant will be over, okay. But I even said, this has nothing to do with properties. I said that in class. The day before, well, it was a week before because of the snow day. But I said, you shouldn't be writing p of n when you're defining the first principle of finite induction. I didn't do it. I didn't do it. I never did this when we talked about it, right? You use this idea when you're actually proving statements using the principle. But the principle, the first principle of finite induction has nothing to do with properties. I saw a lot of p of n prove for p of n plus 1 and all this stuff. No, that's not what it says. It doesn't say that in the notes either. Okay? And I'm just a little disappointed because I told you that specifically. I think I was even yelling, don't say p of n. If I asked you this on the test, no p of ns. I still saw, like, half of you writing p of ns. Which, you know, just... No, that's bad. That's not good. Okay. So, all right, I'm done scolding you for now. I'll scold you more next Tuesday. Okay. So, okay, and this is... Yeah, this is not good. Now, I have no more edge lines to make things neat anymore. Okay. This section is actually, I think, called the Euclidean algorithm. We're not going to do the Euclidean algorithm. We're going to do some other stuff in this section. If you knew... Well, maybe some of you do know this, but you really should be thanking me for not doing the Euclidean algorithm in class. It's kind of messy, and it's just... Yeah. But, I mean, to actually write it out rigorously, you have all these... You have all these remainders and things, and then it just... No. No. No. So, well... Yeah. Well, I... Here's the thing, though. In the homework, there's very little on the Euclidean algorithm in the homework anyway. So, really what I'm going to do is I'm going to actually do the stuff that's going to help you more with the homework problems. So, I think, really, this is the best way to go, honestly. So, basically what we're going to do is we're going to just keep talking about the GCD for a bit, and then we're going to talk about the least common multiple, and we're going to try to relate these things together. So, what I'm going to do is just kind of go straight into a theorem. This is still talking about the GCD, and so we're going to suppose that A and B are integers, right? That's what this means. A and B are in Z, and they're not both zero. If you remember, we talked about the GCD defining it. We don't want both A and B to be zero, because then there is no greatest common divisor, because everything divides zero. So, we kind of imposed this in order for the GCD to exist in a natural way. And the theorem says, for any natural number K, the GCD of K, A, and K, B is equal, so basically this is saying you can factor out the K. This is equal to K times the GCD of A and B. Okay, this isn't that surprising. You have the largest number that divides both A and B. It stands to reason that if you want to find the largest number that divides both K, A and K, B, it's probably going to just be K times that number, probably. The book goes through this and does a proof using the Euclidean algorithm, which there's nothing wrong with it, but it's a little bit overkill in a way. There's kind of a nicer way to do this, which I'm going to show you now. So what we're going to do is we're going to let D, B, the GCD of K, A and K, B. Okay, so again, it would do you some good just to try to follow the argument here. So what we're going to do is we're going to prove that D is equal to K times the GCD of A and B. So I'm going to kind of go through this slowly here. We want to prove this, right? So I'm just calling the left side D. That's all I'm doing. So proving this then is equivalent to proving that D is equal to K times the GCD of A and B. There's nothing deep going on yet. It's easier just to use one symbol than a bunch of them when you're doing the proof. So I'm just going to call this D. Okay. So here's what we're going to do. We're going to let D prime be equal to the GCD of A and B. Thank you. Thank you for noticing. I appreciate that. I'm really glad about that. I'm going to sleep better tonight knowing that. So D prime is going to be the GCD of A and B. I could have done this all in one statement, but I'm just going to do this gradually. What is it that we have to show then? So we're going to show that D is equal to K times D prime. D is K times the GCD of A and B. And so I'm just letting D prime be the GCD of A and B. So we're trying to prove that D is equal to K times the GCD of A and B. And so I'm just going to call D prime the GCD of A and B. So now we just need to prove that D is equal to KD prime. All I'm doing, if you go back up to what we're trying to prove, we're going to go to the left-hand side and D prime B, this guy. Just for notational convenience. That's all. So now we just have to show that D is equal to KD prime. So here's what we're going to do. So the first thing is that D prime divides A and some of these that just are right from the definition, I'm not going to write out explicitly just for space, but I would really encourage you to try to follow along here. Why does D prime divide A? Well, D prime is the GCD of A and B. I'm just reminding you what the definition is. The GCD of A and B in particular is a divisor of A and B. It's the greatest one. So since D prime is the divisor of A, D prime divides A. It's just from the definition of GCD. Okay. Thus KD prime divides K times A. See if you believe that. I could write all this out, but like I said, I'm not just to save space. How would you get this? If you know that D prime divides A, how do you know that KD prime divides KA? How would you prove that? Well, this just means D prime times X is equal to A. Just multiply through by K. Then you get the second statement. That's all you have to do. Does that make sense? Just multiply it through. If A divides B, then C A divides C B. It's probably part of the theorem, too, somewhere. That would be my guess. I don't remember off the top of my head, but probably. Okay. Similarly, KD prime also divides KB. For the same reason. It's the same argument. It's just with the B replaced. D prime is the GCD of A and B, and you just go through the same argument. It's exactly the same. Just with B in place of the A. What do we know then? KD prime divides KA and KB. There's something I have to think about now for a second. Whatever this is, it's an integer that divides both KA and KB. Notice that D is the GCD of KA and KB. It's the greatest common divisor of KA and KB. Since KD prime is a divisor of KA and KB, therefore KD prime has to be less than or equal to D, because D is the biggest one that divides both. Yes, yes. That's what I'm saying, because KD prime is a common factor of KA and KB. D, by definition, is the greatest common factor of KA and KB. Since KD prime is a factor of KA and KB, it has to be less than or equal to the greatest common factor. That's incorrect. It's just a little bit. It just doesn't really follow the standard. I don't know that you lose points for it, but I prefer that you write it out. Any questions about this last part? KD prime being less than or equal to D? Here's the strategy. We're trying to prove that D is equal to KD prime. We know that KD prime is less than or equal to D. If I can prove that D is also less then they have to be equal to each other. You can't have one thing less than another thing and that thing less than the first thing. They're going to have to be equal. Let's see here. Can I hide this somehow? Yeah, right here. Up here? Yeah, I got it. There we go. Thank you. I'll give you some extra credit for that. I'm on video too. I'm not going to dig myself any deeper than I already am. The second thing we want to observe is let's note that there's not much to say here. K certainly divides KA and K divides KB. You guys by that? How do you know that K divides? Why is this true? It definitely matters. It definitely matters. A. I'm just trying to get you to think about the definitions. Think about what this means. What does this mean? This means that there's some integer you multiply KA to get KA. Well, obvious, A. K times A is KA. And same thing here. That's why K divides KB. I don't want you to just sort of memorize this vertical dash. I want you to, when you see it go, what does that mean? We're saying K divides KB. B. I want you to try to think about this. Some of you really need to have practice, really associating meaning to these things. You're making it overly complicated. You don't even need to introduce this third auxiliary variable. It's just simply this. K times A equals KA. So by definition, K divides KA. You don't need a C. It's sort of like saying, let me call this X. Then X is messy. I don't need to do that. I can just say a rag that's messy. You see what I'm saying? Yeah. That's also, I'm glad that you brought that up though. I still see a lot of you do this. A lot of you are throwing in as many variables as you can. You're doing an induction proof. Let N equal K. And then let's prove it for K plus 1. Let's call it K prime now. There's no reason to do that. Just stick with 1. It just makes it so much easier. I've kind of felt the author for this in part because he does this a lot. He'll change variables 8,000 times and approve for no reason. But you really don't need to do that. Yeah. Well, they suck. There's no reason to do that. It's just confusing for no reason whatsoever. Especially at this level. It's really bad for them to do this. Yeah. Right. Well, it is what it is unfortunately. But okay. Here, I want you to think about this too. I'm going through this kind of slowly. But I do want you to think about this. By the way, I should point this out. Your next homework will be do a week from this coming Tuesday. Okay. So you don't have any homework due next week. Okay. So why is this true? Can anybody tell me why this is true? Guys, can you try not to do that? I would appreciate that. Okay. Why is this true? It comes from the definition that we talked about. This is in your notes. I'm just going to remind you. The GCD of A and B has the property that number divides the GCD. That's actually in your notes. That's kind of how we define the GCD. And D, remember, D by definition is the GCD of KA and KB. So since K divides KA and K divides KB, it's a common divisor of KA and KB. It has to divide the GCD. And that's where that comes from. Okay. Yeah. Well, I mean, yeah, I mean, we just, I guess you'll wait and see how it all ties together. I have no question in asking at the end. Okay. So what do we know? Hence, k times alpha equals D for some integer alpha, right? That's what it means to divide D. Okay. So I'm going to, underneath this, I'm going to put an asterisk down here because we're going to use this here in a second. Okay. And since D is the GCD of we know that D definitely divides KA. I'm doing this. This proof may not be the most streamlined, but I'm doing this because I just want to remind you of all of these things. Okay. Definition, right? It's the greatest common divisor. In particular, it is a divisor of both of them. That's why D divides KA. Okay. Okay. But, remember that by star so k alpha divides KA, right? And that's just coming back from over here. Sorry, I should move this a little bit. Bless you. D divides KA, but remember that D is equal to k alpha, so k alpha divides KA. What can we conclude from this? Thus, if we know that k alpha divides KA, is there some obvious conclusion we can make from this? I mean, this is vague, but what can I say as a result of this? That has to do with alpha and A. Alpha divides A because you can just cancel the k's out. If k alpha divides KA, then that means k alpha x equals KA for some integer x. You can cancel the k's. You know k's not 0, right? Because we're already assuming that for any k and n. Remember what n starts. 1, 2, 3, 4, 5. k's not 0, so we can cancel it out. Okay. So alpha divides A and you can do the same kind of thing. You can play the same game to conclude that alpha also divides B, right? Because D divides both KA and KB. So if we do the same argument with D dividing KB, you're going to get that alpha divides B. So we're almost done. So we're running out of room here, but so if we know that alpha divides A and alpha divides B, what can we say let's see, what do we do? We let D prime be the GCD of A and B. Right? It's the greatest common divisor of A and B. Alpha is a common divisor of A and B. D prime is the greatest common divisor of A and B. So alpha is less than or equal to D prime. Okay? Alright, let's see if you believe this. Hence k times alpha is less than or equal to k times D prime. Do you buy that? Because k's positive. That's true by K, for sure. Okay, and now by this starred equation k alpha is D, right? So we get that D is less than or equal to kD prime. See that? I'm just replacing the k alpha because we know that it's equal to D. So what do we have? We're basically done now. Okay? What do we prove? We prove that kD prime is less than or equal to D and we also prove that D is less than or equal to kD prime. So they have to be equal. And that's exactly what it is that we're trying to show. Okay? If one number is less than or equal to the other and then the reverse, they have to be the same number. Do you guys understand that? Does that make sense? Okay? So that's it. One and two, give it to us. Yeah, let me say something more about that here in a second. Okay, I'm going to put a little box here because this is not part of the proof. Just sort of to address this. You don't even need to write this down. Okay? But suppose you have real numbers A and B and suppose you know that A is less than or equal to B and B is less than or equal to A. Okay? If, suppose they weren't equal. So if A wasn't equal to B, then it's not, so if A is not B then, okay, we know A is less than or equal to B. If A weren't equal to B, then they would have to be less than B or equal. Right? Then you'd have A is less than B and B is less than A. Right? Because of this. Okay? If you're assuming this, I want to show you that A and B have to be equal. Well, if they're not equal because A is less than or equal to B and they're not equal, it has to be less than B. And the same thing here. So then you get this, but by transitivity A is less than itself and that's not possible. No, I mean I'm sort of explaining where it comes from. You know, there's always this question of, well, how much do you have to actually show? You know, how much do you have to prove? And there is no exact answer I can give you. It just doesn't exist. I can't say you have to show, if any of these two million things show up, you have to show these, but these five billion things you don't. I can't do that. So this is something that I wouldn't expect you to show. I also wouldn't expect you to know that you wouldn't have to show it. I'm just letting you know that you don't have to. Okay? Okay, so that's it. So we've proved this theorem. All right, any questions here? No? Everybody's asleep? Okay. All right, that's fine. Okay, so you guys have this down then? Okay. All right. So I'm going to prove now a lemma. This is actually, I think the rest of the proofs are not as kind of tedious as this one. And this is sort of a corollary, but a and b are integers, which are not both zero. Usually, there's not a big difference. Usually lemmas are things that you prove that are not as complicated or important as a theorem, and then you use them to prove a theorem. So it's sort of like, it's sort of like a piece. I mean, what usually will happen if you want to prove a theorem that's really complicated, you'll break it down into little pieces. And so you might have several lemmas and then you'll put them together to prove the theorem. Instead of having a 10-page proof, you'll break it up into pieces. It just makes it easier to follow that way. Okay, and this, I don't think you're going to have a hard time buying this. The GCD of a and b is equal to the GCD of minus a and minus b. Not really a whole lot to this, really. But what we're going to do is we're just going to do something similar to what we did before. Don't worry, this is not very long. D is going to be the GCD of a and b and we're going to let D prime be the GCD of minus a and minus b. In fact, I'm not even going to do the whole proof. I'm just going to do part of it. D divides a and D divides b thus and I'm not going to write out the details. If somebody wants me to, I will. But if D divides a, that just means D times some other integer is equal to a, then D definitely also divides minus a because if Dx equals a, then D times minus x equals minus a. Oh, yeah, I believe. Okay, yeah. Well, we're going to just incorporate that. Actually, that's good. Okay, because now we can just reinforce this. So you've got experience with this. So I guess I won't need to say much else about this. So D divides minus a and D divides minus b. So what can we say? Think about the idea that we had before when we were trying to prove that the GCDs were the same. What did I do? I showed that one was less than equal to the other and the other one was less than equal to the first thing on the other side. So, because it's the same thing as before. Since D is a divisor of minus a and minus b, and D prime is the largest divisor of minus a and minus b. That means that D has to be less than or equal to D prime because it's the largest one. And D is one of them. So if you do the same thing the other way, you get that D prime is less than or equal to D and you're done. And that's it. Oh, that D was really bad. I'm regressing now. Okay. So that's it. You show that they're both less than or equal to the other one and then you conclude that they have to be equal. All right. Yeah. Oh, no, because that's what I want though. Because I mean it's true that D prime divides minus a and minus b. But what I really want is that D divides minus a and minus b. So, because D divides a, and as I said, whatever you have to multiply x by to get a, multiply minus x by to get minus a. And so D is definitely going to divide minus a and minus b. The reason why we want D to divide minus a and minus b is that then we can conclude that D is less than or equal to D prime. Because D prime is the largest thing that divides minus a and minus b. That allows us to conclude that the first one is less than or equal to the second one. That's why we want the D there. So we can get that it's less than or equal to. And we flip the argument around so we get D prime is less than or equal to D and then we conclude that they're the same. What you're saying is definitely correct. It's just that we want the D there so we can get the less than or equal to and approve. Is this okay? Okay. So now I'm going to introduce another thing that I don't know if I've ever used this before. Maybe I have, maybe I haven't. Okay. Why am I doing this instead of calling a theorem? Well, it doesn't matter really. It's just, usually in mathematics, in, you know, research papers, a proposition is sort of a theorem that's not that great. A lemma is a theorem that's really crappy. A proposition is okay and a theorem should be something that's interesting. Okay. So now we're going to let a and b be a lens of z, z rather, sorry. Again, which are not both zero. And again, the reason why I do this is because otherwise the GCD is not defined. Let's see if I can get this on here. For any non-zero integer k, the GCD of k a and k b is equal to the absolute value of k times the GCD of a and b. Okay. So I'm going to ask you a few questions as we go through this. I'd like you guys to be as involved as possible because I think that this is something that you guys should be able to do. All right. So what we're going to do is we're going to assume that let a and b be integers. We're going to assume that they're not both zero. And now we're going to let k also be non-zero. Okay. So we want to prove this. And what we're going to do is we're just going to consider two cases. k is not zero. It has to be either positive or negative. So we're going to just look at these cases separately. And this will fall out pretty quickly. Case one is that k is positive. And the GCD of k, a and k, b. So how can we proceed with this? We want to show that it's equal to that. And there's hardly anything to do. We've basically already done it. I think back to what we did in the beginning of class today. This is equal to k, right? Right. It's equal to k times the GCD of a and b. This was what? Theorem one, right? Theorem one. Was it theorem one? Yeah. Okay. Because theorem one, remember, the hypothesis was that k is positive in theorem one. And in this case, k is positive. So we can definitely use theorem one. And this is, of course, the same thing just one second. This is the same thing as the absolute value of k times the GCD of a and b because k is positive. So we definitely get this to work out in this case. Yes. Yes. Yes. Absolute value of k. Okay. I'm running out of room here. But if you wanted to write a note beside how I got this, it's just because we're assuming k is positive. Absolute value of k and k are the same thing since k is positive. I'm just running out of room. That's why. Okay. And this is going to get messy probably because I'm going to try to squeeze this all on this one page here. But case two is that k is less than zero. Okay. So this requires just a little bit of work. Hmm. GCD of, we're going to try to do that to prove the same thing. GCD of k and kb is equal to okay. So I did prove this lemma for a reason and it's going to come into play here. Joe. Yeah. Yes. Yeah. And that's exactly what I was going to say. Yeah. It's lemma one. Yeah. We're going to use lemma one here. Okay. And you'll see how this applies in a second. This is equal to the GCD of minus kA and minus kB, right? By lemma one. Which is the same thing as the GCD of minus k times A and minus k times B, right? Same thing. Minus kA, the minus on the outside is the same thing as minus k times A. Same thing. Okay. So now, how can we finish this? So what can we say now? Now what can we do? We can pull out the minus k because k was negative. We're assuming k is negative. So the minus k is now positive. So now we can use theorem one to pull the minus k out. Make sense? Okay. And that's why we have the lemma because now we can flip it to being positive and we can pull it out just like before. But we're not completely done. There's still one other thing we have to know, right? Because we're trying to prove this. So we almost have it, but really what we want though or what we need I should say is that minus k is actually the absolute value of k, right? Why is that true? Because k is negative, right? Remember what the absolute value does to a negative real number. It just erases the negative essentially. It makes it positive. Since k is already negative, that minus will flip it to being positive and that's exactly what the absolute value does. Is that wrong? Well, but you still need, you still sort of need this, right? Because right here, oh, you mean, are you talking about this? Yeah, yeah. But here's the thing though. We can pull out the k though because k is positive and theorem one, k has to be positive in order to pull it out. So since k is negative, we can't just pull it out. We have to use the lemma to make it positive first and then pull it out and then we say because k is negative, minus k is the absolute value of k. You really, I don't think you can really get around and you have that. Okay. Sorry about things getting all squished up here, but this is the end of the proof. So this is the absolute value of k times the GCD of A and B. And again, I'm running out of room. I apologize. If you want to put something in your notes to remind you why this is true, it's because k is less than zero. That's why minus k is the absolute value of k. Is this okay? Of course, you guys learned this a long time ago. I think you guys are probably fine with this. Okay. So this gives you a very nice way if you know the GCD of two numbers scaling through, you can just sort of factor out the absolute value of that scalar and then you're done. So it allows you to sort of know what the GCDs of other things are without having to go through all the computation. Okay. So, the last thing we're going to talk about, if everybody's got this down, okay, are we okay? You guys okay? And this is not going to be nearly as painful as it was before. Sorry, let me move this around here. I'm going to call this a theorem, although there's not much to it really. A and B are non-zero integers. Then there is a least positive integer M such that A divides M and B divides M. Okay. Because of time here, I'm thinking I may not write the proof down. There's really nothing to this proof. It's just if A and B are non-zero integers, then A and B both divide the absolute value of A times the absolute value of B. And that's certainly positive. That's not hard to show. That's very easy to show. And so, just by the well-ordering property, just take the least positive integer that A and B both divide. That's it. That's all there is to it. So, like I said, I think I'm not going to write this out. I want to spend time on more complicated things. So, you don't have to prove this. Just take this as fact. It's very simple to establish this. So, let me go to the definition then. And you've also seen this before. These come in multiple. So, this is denoted LCM. And this is simply the smallest positive integer M such that A divides M and B divides M. Okay. So, it's just the smallest positive integer that's multiple of both A and B. Okay. So, just to refresh your memory here, I don't mean this to be an 8th grade algebra exercise or anything, but just to make sure that everyone's on board here. What's the least common multiple of 8 and 12? 24, right? Smallest number, positive number that 8 and 12 both divide into, right? What about this? Sorry. The least common multiple of minus 6 and 7. Okay. I'm going to call this grade algebra teacher if you don't get this right. Okay. Apparently, you really like your algebra teacher. Nope. It's 42. You and you are both, they're coming to class next week. Third grade teacher, okay. All right. I'll take care of her for you. It's okay. Keep forgetting about this camera here. I'm going to incriminate myself. Okay. Because by definition, the least common multiple by definition is the smallest positive integer. And that's why I did this. So remember, by definition, the least common multiple and the GCD, they're positive. Positive. Yeah. I understand it has to be positive. It doesn't, but that's not the definition. The definition is just simply the smallest positive number that both of them divide into. Minus 6 divides into 42 because minus 6 times minus 7 is 42. We're not multiplying by to be a positive number. It can be any integer. Okay. Are we good with this? No? Is this okay? All right. So, this is the last page. Okay. So this is the last, we're going to do one more theorem and then we're done. Okay. And if we finish early, then we finish a little bit early today. All right. We go to the next page. Yeah. We good? Yeah, Joe. About this LCM thing. In this case, with the 6 and 7, I noticed that it's just the same thing is just multiplying the two of them. Does that work? Does that always work if one of them happens to be a prime number? No. So, yeah, I mean, so if you, well, okay, so I mean, I can give you kind of a simple example to this. If it was 7 and 7, for example, the LCM would be 7. It wouldn't be 49. But actually, you're going to have an answer to that question in the next theorem, I think. So you'll see how this works. So the point is, here's the point. What do you mean to? The least common multiple is equal to the product if and only if the two numbers are relatively prime, as she said. Relatively prime. So they don't share any common. So the GCD is one. GCD is one. So this is going to come from this next theorem. Okay. That's too much work. No, no, no. I'll take care of that myself. It's okay. I appreciate the offer, though. I do appreciate the offer. You would do it and we'd get it on video. That would... I'll think about doing it in the future. I'm technically challenged, so this could be a problem for me. Okay. Here's the last theorem. And we're going to let... Now this time it's even easier. So A and B now are natural numbers. Then the GCD of A and B times the least common multiple of A and B is equal to A and B. So, here's the proof. Like I said, this is the last proof, so muster up some energy to try to pay attention to this and then we'll be done. Okay, so let's just assume A and B are natural numbers. And let's let D be the GCD of A and B. So we know a lot about the GCD. We're going to try to prove something about the least common multiple. So here's what we're going to prove. We're going to show that the least common multiple of A and B equals A, B over D. That is what we're going to...that was bad. Well, that's okay. A, B over D. Least common multiple is A, B over D. So I want you to see that this is the same thing as what we're trying to prove, right? Because we just divide through both sides of the equation by the GCD of A and B. We're calling that D, remember. It's the same thing as asserting that least common multiple of A and B is A and B is A, B over D, right? Just divide through. And then use the fact that we're calling it D. Okay. Here's one thing I want you to note which I'm not going to really justify in detail. But A, B over D is definitely a natural number. It's a positive integer. In order for it to be the least common multiple, by definition it certainly has to be a positive integer, right? That's how we define the least common multiple. In particular, it has to be a positive integer. How do we know that A, B over D is a positive integer? How do we know that? Anybody? Joe? D was the GCD, right? D is the GCD of A and B, yes. Because if D divides A and B divides B and A and B are integers. So if you do A times B, you get an integer and D has to divide that. Yes, that's right. And because they're all positive, right? Remember our assumption is that A and B are natural numbers. Remember the definition of natural number. They're all positive. So A, B over D is also positive because A and B are positive, right? And D is the common divisor of A and B. D is a factor of A. So this is certainly, it's not a proper fraction. It is actually a whole number, right? So we're just going to prove several things here. So the first thing, what do we need, and this is the reason why I'm doing it this way is because I really want you to think about the definition of least common multiple. So we're claiming that A, B over D is the least common multiple of A and B. So one thing that has to be true, there are three things that have to be true for this to be the case, aside from it being a positive integer, which we already know. One is that A, B over D has to be a multiple of A and a multiple of B. And it also has to be the smallest one that's a multiple of A and B. So there are three things that we have to check. One, we need to know that it's a multiple of A. Two, that it's a multiple of B. And three, that it's the smallest multiple of A and B. So that's what we're going to prove. A is a factor of A, B over D. So in other words, A times some integer is A, B over D. And there's not much to this. Note that A times B over D, of course, this is so simple you shouldn't even be thinking very hard about this. You guys buy that? A times B over D is A, B over D? How do we know that? Does that immediately imply that A is a factor or that A divides A, B over D? No. Because what if B over D was a half? It has to be an integer, right? It has to be an integer. Yes. Yes. I'm just trying to get everyone to think this way. But I want to be very clear on this. This is a common mistake that you might make in your homework. Just because you say you want to prove that A divides B. If we know that A times something is B, that does not mean that A divides B. Within the real numbers, you can always do this, right? A times B over A is always B. So just knowing that you can multiply by something to get B does not mean that it divides. Everything has to be an integer. Everything has to be an integer, right? Because you could prove that 2 divides 3, right? 2 times 3 halves is 3. Oh, therefore 2 divides 3? No, it doesn't. Because 3 halves are not an integer. You need it to be an integer. I've never taught number 3, but I'm sure that that's the case that you're going to make is not really realizing that you need to know that what you're multiplying by has to be an integer. That's very important. Yes. Right, sure. We're actually going to do that here in a little bit. But I'm just going to write it out in words just because I want to emphasize this really in words. And B over D is an integer since D divides B. That's the important part. 2. Okay, I want to make sure I get this done. B also divides A, B over D. And the proof is just the same idea. In this case, you just say that B times A over D is A, B over D and D divides A because D is the GCD of A and B. It's the same argument. I'm just flipping the letters around. Okay, and this is where we actually have to do a little bit of work. Let's suppose now that C is a positive integer and that A divides C and B divides C. So in other words, C is a common multiple of A and B. We want to show though that A B over D is the smallest positive common multiple of A and B. So what we have to prove is that if C is any common multiple of A and B, then A B over D has to be less than or equal to C. In other words, A B over D, that shows that it's actually the least one. It means that it's less than or equal to anything that's a common multiple of A and B. It has to be the least. So what we have to prove, we need to show that A B over D is less than or equal to C. And I want you to pause for just a second here. I'm going to kind of reiterate what I said before. C is positive and C is a common multiple of A and B. We are trying to establish that A B over D is the smallest possible one. So since C, we're assuming is one, we need to show that A B over D is less than or equal to it. That shows that it's the smallest possible one. Okay. So, and it's just going to go like this. Since A divides C, A X equals C for some integer X, right? And so this is just coming from our system here. A and B both divide C, right? Since B divides C, B Y equals C for some integer Y, okay? That's just definition. We're assuming A and B both divide C, so this just follows from definition. Okay. And the last thing that we're going to say before we put all this together is as D is equal to the GCD of A and B, what do we know about D? D is equal to A alpha plus B times beta for some integers alpha and beta. You guys remember this? You know why I'm writing this down? The GCD is a linear combination, right? When you prove that, that's actually, that was actually built into the, sort of built into the definition almost, that I gave you. Okay. Let me move this thing out of the way here. See if I can squeeze all this on here. All right. This is going to get a little messy. I apologize for that, but you guys can all fall all these steps though. So I'm going to take C over A, B over D. I'm doing this for a reason. No, no, no. This is the clearest way to do this. You'll see here at the end why I'm doing it this way. So how do you do this? C divided by a fraction, you just flip and multiply, right? You guys know that, right? So you just multiply C by D over A, B, and you get this. Okay. So this is equal to C times A alpha plus B beta, right? Over A, B. You guys see where I'm getting this now? I'm sorry, I'm running out of room here, but D is A alpha plus B beta, so I'm just replacing the D with this. That's all I'm doing. It's equal to that, so it's just going in right there. Okay. So this is, if you distribute, this is C A alpha plus C B times beta simplify this. Hopefully I can do this in well, if anyone's at all lost here, just let me know. This is C alpha over B plus C beta over A. You guys see how I got that? We can just split the fraction up into two pieces. We can cancel the A out for the first piece, so that's where the C alpha over B comes from. In the second piece, we can cancel the B out to get C beta over A. Yes. Yes. We're not, although we still got a little work to do. Well, all we have right now is that we divided by, that's kind of where we're going. That's roughly the idea. We've divided by A B over D. What we're going to show though is that this is actually an integer. Once we've got that, that really gives it to us. Don't you already know that? Right, but we don't know necessarily because we need to show that these are integers too. This is a fraction. Maybe this isn't an integer. Yes. The answer is yes. I'm just going to write the details out. But we do need to, we just need to say a few things to really nip this in the bottom and make it concrete. But your idea is basically right. Yeah. Okay, so let me, I'll tell you what, just to make this part a little bit easier, I just got boxed in here and I'm running out of space. So I'll call this one, this equation one. I'm going to call this equation two. I'm still going to have to do a new page here, I think. Okay, so what I can do now is, so the last part we had before was C alpha over B plus C beta over A. Right? But look over here. C over B is Y. You see that? C over B is Y. So I'm just replacing this part with Y. That's where I got that. So what's going on up here? C over A is X. C over A is X. So I'm replacing the C over A with X to get X B over, X beta over there. Does that make sense? Okay. Okay, so, yeah, now we're almost done now. My, can I go on here? Almost done. Almost done. Okay. So here's what we get if we put all this together. So A B over D times Y alpha plus X beta equals C. If you look, I can go back to the other page if you want to, but remember what we had before. We had C over A B over D is equal to Y alpha plus X beta. Right? That's what we had before. So all I'm doing now is just dividing both sides by A B over D. Go back to the previous, what you just had before I wrote that, multiply both sides by A B over D. That's all we're doing. And you should see where this is coming from. Is this okay? Just A B over D on both sides. So what can we say then? Well, we don't even need to say that we just know that A B over D certainly divides C, right? These are all integers, right? Y alpha, X and B, these are all integers. So this product in sum is certainly an integer as well. So A B over D definitely divides C. What is it that we wanted to prove? We wanted to prove that A B over D is less than or equal to C. That's what we wanted to show. But because it divides it, everything's positive, right? And this is actually a theorem from 2, 2. A B, D and C, these are all positive numbers. If you know that one positive integer divides another one, then that first one has to be less than or equal to the other one, for sure. Right? For sure. And if you want some justification for this, this is also just theorem one of section 2.2. When I say theorem one, I mean according to my labeling. Part F. So we've actually, I mean, this last part we actually already proved this, essentially. Okay. And don't worry. I'm not proving this. I'm just stating this as a corollary. I'm not going to prove it. A and B are natural numbers. Then, and this is what I was just saying just a few minutes ago. The least common multiple of A and B is equal to A times B if and only if the GCD of A and B is 1. Well, yeah. Yeah. Yeah. So the idea here is just that, you know, the GCD is small as possible, of course, when two integers are relatively prime. But if they don't have anything in common, then that makes the least common multiple as big as possible. If you think of prime factorizations, it sort of makes sense. The prime factorization of A and B, they don't have any primes in common. So the only way for, if you have any common multiple of A and B, it's going to have to be all of those primes together because there's nothing in common. So that's sort of the idea. Okay. So I think I will stop there. To start on this, but let me go ahead and give this to you. Okay. So this is 2.4. Okay. So I'm not giving you a ton of problems here. So this is 3, 4A, 5A, 6, 9 and 10A. I should tell you that you will almost certainly have one more assignment that's going to be due aside from this also next Tuesday. Okay. On Tuesday, that will probably be due as well. So you'll still have another week to finish it. And you'll get your exams just to be clear. You'll get your exams back on Tuesday. You'll get your previous assignment back on Thursday. I'd rather not pass all that back in one day. So the exam will be Tuesday because I assume you want to get that back probably more than your last homework. And then the homework will be Thursday. And then the next Tuesday will be your next assignment. Okay.