 So I'm just going to sort of go in order here more or less. Okay, so let's just start off with section 1.1. Here are some things that you should know. When I say know, what I mean is you should just be able to define these. If I asked you what any of these are, you should just be able to tell me. That's what I mean. Okay. First is the, let's call the well ordering property. I'll just go ahead and remind you what these are. If you've been coming to class and most of you have, these are all in your notes and of course in the book. The well ordering property for the natural numbers just says that every non-empty subset of the natural numbers has a least element. So if you were to say that, that's exactly it, that you get full credit for that if I asked you that. Okay. Every non-empty subset, non-empty is important. If it's empty, it can't have a least element because it doesn't have any elements at all. So these are the things I expect you to pay attention to the details. Because really, I'm kind of throwing you a bone here. I'm telling you all the kinds of things I could be asking you. All you have to do is remember it. That's it. Okay. The second is the, yeah, huh? The second part. Mm-hm. Well, no, you can write it out. You can say it in words or you can write it out specifically. So what he's saying is you can say the well ordering property like this. So let S be a non-empty subset of the natural numbers. Then there exists some X and S such that X is less than or equal to Y for all Y and S. Or you can just say that every non-empty subset of the natural numbers has a least element. I really don't care. I just want you to know what it is. So that's fine. No, it doesn't, I mean in this context the way I stated it is, it did. But the same thing is true as long as you, I mean if you look at the integers, of course it's not true for the integers. Because if you look at the integers themselves, the integers themselves constitutes a non-empty subset of itself. And there's certainly no least integer, right? You can always go down. But as long as you're taking just whatever your set of integers is, as long as it's sort of what's said to be sort of bounded from below. As long as you're only taking sort of a segment of the negative part and you can just go all the way up to infinity, then that same principle is going to hold, okay? But I don't expect, I mean I'm not going to ask you that in general. It's just going to be applied to the natural numbers. First principle of finite induction. Okay, and I want to be clear on this on these two points too. The way I've written the notes, at least if memory serves here, the first principle of finite induction is not, I'm not talking about any properties per se. I mean you sort of employ this principle when you're working on specific problems. The first principle of finite induction just says that if you take S to be a subset of the natural numbers that satisfies one is an S and two for all natural numbers N, if N is an S, then N plus one is an S. The conclusion is that S is equal to all the natural numbers, the set of all natural numbers. Okay, so this has nothing to do with P of N and you shouldn't be writing P of N. It's a very specific statement about if you know a subset has these two properties and you know that it has to be everything. That's what the first principle says. Okay, and I'm sure that was the case when I wrote it in, I mean I'll look back at the notes but I'm sure that's what I said in class too. Okay, sorry I'm going to be a little lazy here. Second principle, oops, left off the D, sorry, I got really lazy. There we go. Second principle of finite induction, and so you remember that this says that, okay, you strengthen the hypothesis. And so the second principle of finite induction says that if you have a subset of the natural numbers that satisfies S, say one is an S, and for every natural number N, if you know that one through N, or those are all an S, then N plus one is an S, then the conclusion is the same, that S equals N. That's the second principle of finite induction, according to the book, which remember I called strong induction, right? Okay, I will, I mean if this comes up on the exam, I'll probably state, you know, if I ask you this, I'll probably say state the second principle of finite induction, a.k.a. strong induction, so that there's no confusion. You'll see both of them. Okay, so, let's, what I want to do now, and this of course has come up in other sections too, is I want to go ahead and do a proof using induction. And I'm going to do, I'm going to kind of do this a little bit overkill on purpose, just because I want you to see the structure of a, of an argument that flows in the right direction. Some of you guys are going the wrong direction, and I'm, like I said, I'm, I'm kind of doing this on purpose to try to make this point. So let me, let me do this. This is actually one of your homework problems that you, that you, that wasn't graded, plus you. Okay, and it was this one, prove that one over one squared plus one over two squared plus on down to one over n squared, sorry, is less than or equal to two minus one over n for all natural numbers n. Okay, so we are, we're going to use induction. The first principle of finite induction, because of time, I'm, I'm not going to write that out, but there are two things that you need to do. You need to establish the base case, and then you have to do the inductive step. Okay, so, okay, so I'm going to, I'm going to offset these, just because I, I want this to be very clear what the separation is, base case. Some of you are expending too much effort on the base case, okay? What I would like you to do just for clarity, this is a nitpicky comment, but the base case here, so this is an assertion that's being made for all natural numbers n. The base case of the assertion is what, what happens when n is equal to one. It's the specific statement with, with n being one. And so, what's the assertion? What's the left-hand side of the inequality? Well, if n is one, there's nothing, we're not going, we're not going anywhere, we're ending where we start, right? So it's just one over one squared is less than or equal to two minus one over one. That's exactly what the statement is, right? Okay, now here's where things get a little bit murky maybe because you, you might say, well, I don't know what you want me to prove. This is the base case. Here's, okay, you are all in this class, you certainly have enough sophistication to look at this and say, the left side is one, the right side is one. You don't have to spell it out, you really don't. You don't say, left side is equal to one over one squared versus one. Right side is two minus one over one, one over one equals one. Therefore, two minus one over one is two minus one, which a lot of you are doing that kind of stuff. I, something like this that I, that I would expect a fourth grader to see immediately is you do not have to justify. I'm not saying it's wrong if you do it, and I know some of you don't know. Maybe you think I'll take out points if you don't do it. I'm not blaming you for it, I'm just letting you know. This is obvious, you don't have to say anything else. You should write it out, because it is important that you know that it's true, that you, you check the box. That's all you have to do here is just state what the, what the base case is, and then just, if it's obvious, that's, it's fine. Yes. Yeah, it is. Just, no, but what I'm expecting is that you would write it like you would an English paper. Don't write check marks in, in Booyah, and that kind of stuff. No, some, a couple of people did that. That's all I'm saying is write it out. Like you wouldn't put a check mark in an, in an English paper. That, that's all I'm saying. Okay, form of, you know, obvious is, is subjective, right? But, but, you know, and so formal is subjective too, to a degree. But it's, this is not a big deal. I actually didn't take off points for that specific thing. But, but yeah, I mean, write it, write it like you would. I mean, so for example, in the book, you're not going to see that. You're not going to see, oh, check mark. That's not in the book. So what I mean is write it as if you're presenting it formally. That's all, that's all I'm saying. Okay, now, yeah, huh? Yeah, oh, oh, over here. I thought, sorry, I thought you did, okay. Well, okay, I'll, I'll try to address that here in a second. Okay, yeah, you should, well, you should finish. I mean, it should at least be a sentence and it should be. Yeah, I would, I would, I would like you to write that. It just, it just kind of gives it a nice flow. You know, I mean, if you just say, we must check that this, but you didn't check that this, you didn't say anything. You just say, we must check it and then you move on. It's not, you know, mm-hmm, mm-hmm, right. Yeah, I mean, the base case is really just, though this is an assertion. I mean, you, it would be better just to say that it's true, just so it doesn't leave the reader wondering. Oh, is he just saying the proposition? Or is he, does he actually going to nip it in the button and tell me that it's actually correct? It's just better to do it because it only takes a few words. Yeah, this is, this is a nitpicky, just to be, this is very, I'm being very nitpicky right now, but I'm just kind of letting you guys know kind of what, yeah, I know, I know, that's true. But I'm just kind of letting you know where I kind of want you guys to go. So the inductive step, okay, here's where there's some issues. Okay, in fact, because of time, I may not, I may not go through the entire proof, but I'm at least going to get this started. This, I want you guys to listen to this, is some of you are doing this. When you're doing induction, the inductive step is, consists of two parts. The first part, and a lot of you are missing this, is to state your inductive hypothesis. If you don't do that again, it's just, it's not clear to the reader. It's clear again, I know how this works, but if you don't, if you leave out the inductive hypothesis, then your reader's going, where is this coming from? Where, where is this coming? It should all be tied together neatly so that there's no question where you're pulling stuff out of. The inductive step consists of, assume for some natural number N that, that holds. That should be in your proof. We must prove, and you don't necessarily have to write this, although it's nice for yourself and for me, so I know where you're going. We must prove that, that's true with N being replaced with N plus one. Okay, but stating your inductive hypothesis is an essential component of the proof, okay? If you don't do it, you're leaving things out, then you're pulling things out of hats that you've never introduced before, okay? So you really have to write that out, and some of you are not doing that. Write out the inductive hypothesis very, very clearly, and it's, the inductive hypothesis is always just, let N be a natural number, or let N be arbitrary or whatnot, and it's just, it's just what you're trying to prove. That's it, that's all the inductive hypothesis is. Yes? Mm-hmm. No, no, no, but you're not assuming it holds for all N. That's assuming what you want to prove. You're assuming that it holds for some N, and then you want to prove that it's true for N plus one. You would never assume what you're trying to prove. You're just assuming that it's true for some N. What you're trying to establish is this. The second part of the induction principle is a conditional statement. It's, if it's true for N, then it's true for N plus one. So to prove a conditional statement directly, you might remember this from discreet. You assume that it's true for N, and then you show that it's true for N plus one. But you definitely do not assume it's true for all N. No, you're assuming that there exists, that N is arbitrary, and it is true for that N, you want to prove it's true for the next one. If you assume it's true for all N, then you don't need to show it's true for N plus one. You've already assumed it's true for everything, right? No, just say something like this. I mean, this is all you have to do. Assume for some N and N that, okay, and let me make this clear, I could have done this above, but like I said, I'm glad you guys are asking these questions. Now I'm probably not gonna have time to go through the whole thing in detail. So assume that one over one squared plus one over two squared all the way down to one over N squared is less than or equal to two minus one over N. Everybody clear on this? Okay, it's the inductive hypothesis. All right, and let me know when I can, if anyone's still writing, of course, I want you to get this down here. Okay, see a couple of you, are you both okay? I'm gonna give you some leeway because you made me some sweet donuts. Are you okay if you got this mostly down there? Are you good? Okay, all right, you're okay. Okay, well I wasn't sure, I was just saying, maybe you're thinking about baking you more donuts and stuff and you're not paying attention, that's fine with me. What can I say? Okay, so there was the inductive hypothesis, right? So now what do we have to do? So we have to prove this. And it's just nice for the benefit of your reader to write down what it is you're gonna prove before you do it just so that whoever's reading it can see what your goal is. Okay, and yeah, I'd hope to go through all this in detail but I just don't think I'm gonna have time. Okay, so we have to prove that one over one squared plus one over two squared plus one over three squared on the way down to one over n squared plus one over n plus one squared is less than or equal to two minus one over n plus one. And so this is, again, just what we get by replacing the inductive hypothesis with n plus one. We have to establish this. And so remember, the whole point of induction is that inductive hypothesis, that's in your hands. You can use it as if it were fact and you use it to extract this out somehow. That's the whole idea. The other thing I wanna say, and I did an example of this before and some of you are still doing this, overall I'm actually pretty happy with your work just in case I seem like I'm really angry all the time. I'm not really. I'm pretty happy overall with your work. Most of you really seem like you have at least a good hand on the mathematics of what's going on. Don't start with what you want to prove. That's very important. And I know I did this silly example before but some of you are just assuming that it's true and then you just wanna keep filling with it until you get something like 20 equals 20. And then you say, ah, now it works. Well, maybe. But then reverse, your proof should go in the opposite order. You start with what you know and you get to what you wanna prove. You do not start with what you wanna prove. You'd never do that, okay? That's just not a proper argument. You start with what you know and then you follow an argument until you get to where you need to go, okay? So for those of you that are still doing that, don't stop doing that. Try to stop doing that. And a lot of you, all you have to do is reverse your arguments. All you have to do is just take your proof and go like this, okay? But it's just not proper flow to do it this way. Yes. So for some of you, that's what's happening here. So what you wanna do is, and this is, unfortunately, I'm gonna not have time to go through this, but. But you're not. What do you mean? Oh, no, no, no, no, I'm not. So I'm gonna, I'll tell you what, yeah. Okay. If you go back up here to the inductive step, okay? So most of you, when you're doing your proofs, most of you, this applies to, you should be doing some stuff on scratch work first until you can see how everything fits together and then construct your proof. It's not, again, it's not like, you know, calculus where you wanna find the slope of the tangent line to y equals e to the x plus one at zero two. It's not like that. You can just start right away because you know what you're doing. Oh, plug this in, plug and shug, you know. Some of these problems are not gonna be that way. Sometimes you're gonna start and you're gonna go, oh, well, now I'm gonna need this fact in the middle of my proof. I didn't know I was gonna need, so let me prove that first, and then I'll refer to it later. The flow is much, much better than when you say, digression, here's another digression. You know, state those things first and then state your argument. It just flows much better that way. Yes. I thought you wanted to do this pretty quick. Well, on the exam, you know, well, first of all, I'm going to give you scratch paper for the exam. Second of all, the problems are not gonna be impossibly difficult on the exam either, okay? So, yeah, I don't think that's gonna be a major issue on the test. It's a good question, but I don't think it's gonna be a problem. Okay. Well, I mean, you've got several sections, and so, I mean, the first section is induction. And of course, as you know from doing the homework, there are later sections where induction also comes up. You will almost certainly see some induction problem on the test. That is, I can tell you right now you will. You will definitely see that, okay? So, you should sort of know how to proceed with constructing these things. So, let me just say that, and I wish, again, I could go through this, but what do we do? What's the idea? The idea is you don't start here. This is where you want to get, okay? This is the main point I wanted to make anyways. You want to start with this, okay? Or what you can say is, okay, there are a couple things you can do. You say we want to prove that this is less than or equal to two minus one over n plus one. So, what you could do, of course, is say by the inductive hypothesis, the left-hand side is less than or equal to two minus one over n plus one over n plus one squared, right? Okay, so I mean, the more natural way to do this is just to say, okay, let's just start with the inductive hypothesis. What do we do? We want to get this, right? So, let's add one over n plus one squared to both sides. We can certainly add anything we want to an inequality. We have the same thing to both sides. So, then we have exactly what we want up here. We have the left-hand side up here now. So, then what do we want to do? We just want to show that two minus one over n plus one over n plus one squared, right? That's what we added to both sides so that we could get this. It's less than or equal to that. And the rest of your proof should just be, it should be trying to prove that two minus one over n plus one over n plus one squared is less than or equal to two minus one over n plus one. And that's how you would finish off the proof. So, the main point again, start with your inductive hypothesis. Don't start with what you want to prove. You end with what you want to prove. That's how an argument flows, okay? Okay, I'm sorry that this is incomplete. If I had more time, I would do the whole thing, but I just don't, unfortunately. And I think you'd probably rather me go on to more stuff so you know what to be studying for the exam. So, I'm gonna just, I just don't have a choice but to stop at this point. Okay, so, the next section is 1.2. This is the binomial theorem. And I can tell you there are not many things you need to know here. And that's, in fact, I mean, as far as the definitions and the theorems, there's very little. Okay, so, let's see. Yeah, yeah, sorry, I'm sorry. Yes. I'm just thinking about complete sentences. When step n is some equality, then step n plus one is azure-braggly equivalent to it, like you're just showing your work. Right. Should I try and make that into a sentence? Or can I say, you know, sentence, sentence, sentence here and then just. Yeah, I know what you're saying. So, you're talking about, you know, you got an equation, you're just gonna do some basic algebra or something. You should tell me what it is that you're doing, okay? So, you know, if you're gonna cancel something out and then take reciprocals and then do this, don't put all that in one step. You should say what it is that you're doing. You know, you could put it even parenthetically, you know, if you went off to the side, that's fine. But, and some of you got docked a few points for, there was, I think this, you know, n choose k business, there was something where r was one half n times n minus one or something like that. And some of you lost some points because you just weren't clear how you got, where, you know, you were getting your next step from. And again, it's not that I don't think you know, it's the same reason that you should be clear in what you're saying. Okay, especially, especially for example, some of you did this, some of you did this and you have this in your homework. You just started canceling things out and you used the fact that r plus one factorial is r factorial times r plus one, but you never stated it. You just sort of, in your mind you hadn't, then you canceled everything out and then all of a sudden you get to this really simple equation. Those are things you should definitely state. You should state in your argument, okay? Now, again, the benefit of your grandma, okay? Not me, not me. But you have to remember that this is what you're doing. No, I'm serious. I know, I know you're giving me a weird look here, but this is the point. The point is you are writing an argument to convince someone who doesn't already have all the steps in their head. That is the point of this. That's just not for me. That's every mathematician, that's what we do. Okay, that's part of the game, okay? So what do we need to know here? One, know the definition of n choose k, right? You should know when it's defined, right? This is not defined for n equals minus 50, for example. This is not defined for n equals five and k equals 50, either, right? You should know what the, you know, if I could say, what are the restrictions on n and k here? Well, what is it? Well, n has to be an integer that's bigger than or equal to zero, and k has to satisfy zeros less than or equal to k, which is less than or equal to n, okay? You should know that. It doesn't, we haven't defined two choose seven. We haven't talked about that, right? Second thing, and you might be able to guess what the, there's just two more here. You might be able to guess what these things are. The second is binomial theorem. You should know what that is. And you shouldn't just know what it is. You should know how to work with it, too. I mean, I could say, okay, what is, without multiplying all this out, I will look at your scratch paper to see if you're doing it. What's a plus b to the fifth? Use a binomial theorem. Instead of a plus b times a equals b, foil, then foil, then foil, then foil, okay? So, I mean, I could ask you something like that just to make sure you actually know how to use it. Because, like I said, a five-year-old can memorize the symbols. You, I do expect that you can work with it, too. Okay? Let's see. Oops. And then the third thing you should know is Pascal's rule. And, of course, that came up in one of your, actually one of the graded problems, Pascal's rule came up. Most of you saw that. You saw how to apply it the correct way. But that's something you should know. Okay. And so this, for now anyways, I'm not gonna say anything about this as far as specific problems are concerned. But you should certainly be able to work with these things. I mean, I could certainly ask you a proof that involves the binomial theorem not in a really messy, nasty, complicated way. But you had a couple of homework problems that involved the binomial theorem. Certainly one, I think it was 2a or something like that where all you had to do was let a and b be one. And you've got everything to work, maybe it was 3a, but you got everything to work out without hardly any work whatsoever. But you should definitely be familiar with how to use the binomial theorem within a proof of some sort. Okay. So what about section 2.2? Not a whole lot to this section. You should know the division algorithm. Actually, that's really all we did in this section. Okay. And so the division algorithm does, I mean, there are several parts to it. And if I ask you the question, what does the division algorithm say? You should say everything very clearly, okay? So this has something to do with dividing by n. Okay, so if you take an integer a, so really the statement is, I may be using different letters here, but for any integer a and for any natural number n, in other words, any positive integer n, there exists unique integers q and r such that one a equals nq plus r and two r satisfies zero is less than equal to r which is less than n. There are a lot of points here, but I expect you to get all of those. Okay, it's not just, we can divide and get a remainder. You will get zero points if you say that. Okay, it's a very specific, very clear rigorous mathematical statement. And I do expect you to state it as such as I did in class. Unique is important too. That's part of the division algorithm, okay? All right, so what I wanna do, I think this is probably the best way to proceed. It's just to tell you the things that you need to be looking over and then whatever time we have left, we can go back and spend on problems, okay? Well, fortunately, we only have one more section to talk about. Okay, 2.3, all right, so let me just try to squeeze all this in from left to right. Several things that we talked about in 2.3. You should know the definition of factor, right? Okay, and when I say this, of course I don't mean factoring a binomial. That's not what I mean. Know what it means for an integer A to be a factor of an integer B, right? Or there are lots of other synonyms for this that I'm not gonna write down, right? A divides B, right? B's a multiple of A, all of these things. You should know what these are. Okay, so, and I'm gonna be very clear about this part. So you should also know the definition, GCD. Okay, and so, there's one part here that given the notes that I gave you, you might have a question about. I gave you, unlike the book, I actually proved a theorem first showing that the GCD existed and was unique, even though I didn't call it the GCD in the theorem, and then I defined the GCD after we had the theorem. So, what do I mean by GCD? Well, the GCD of two integers A and B, remember the assumption is that A and B can't both be zero. Otherwise, there is no greatest common divisor because everything divides zero. It's a positive integer D with a property that D divides A and D divides B and if C is any integer that divides both A and B then C divides D. That is the definition of GCD. There's this other part, right, where it's a linear combination of A and B. Okay, that, and of course, if you say it, I'm not gonna take off any points if I ask that on the test, but that's not really part of the definition of GCD. Okay, we had just done it, so I mentioned it, but the definition has nothing to do with the linear combination part. It's just what I said before. You guys okay with this? Does that make sense? You should know, I don't expect you, of course, to cite theorem numbers, but whatever, if we proved a theorem that you need, you could just state what the result is without saying, buy theorem two from this date, I don't expect that, but you may need to know these, which I did prove for you. I'm gonna abbreviate this a little bit, but the GCD of A and B is equal to one, if and only if XA plus YB equals one. Okay, now here's a good place for me to illustrate a point I was trying to make before. I'm not done writing this yet. This right here is, of course, I realized I was lazy over here and I didn't say that A and B were integers that weren't both zero, but just suppose I did. Formally, this is not what you would say, because someone reading this would say, what's X and what's Y? Is it my cat? Is it an integer? Is it the square root of two? What is it? For some integers, X and Y. That's how you complete this. Never introduce, when you're defining or in your proof, never introduce a pronoun that you haven't defined, okay? Don't do that. No, if I wrote this out in general, it would just be this, it would just be, let A and B be integers which are not both zero. Then the GCD of A and B is one, if and only if XA plus YB equals one for some integers X and Y, or for some X and Y and Z. Okay, it's not the case that it's true for every integer A and B. Certainly not, two and four, for example. You can't write a linear combination of two and four to give you one. But we're just saying it is one, if and only if this property holds, yes. That's fine, you can do that. The upside down A and the backwards E for exists. Yeah, that's fine. Yeah, I mean, so yeah, I would prefer not to use modular arithmetic. Yeah, the problem with that is just that, those of you that know about this stuff and know some of the tricks, if I let you use that, then the other students are already disadvantaged who haven't seen it yet. That's awesome for them, I know it. Yeah, but it's not so good for the other people. So generally speaking, your argument should be constructed. What you're using should be a subset of what we've done in the class and what the book has done so far, okay? All right, so here. And I may even, if I don't ask you all these, but one of these theorems comes up in the course of a proof that I give you, I may say, hint, use this fact in your proof, okay? But these are things that you should be comfortable with at the very least, all right? I'm certainly not gonna say what was theorem three from two point, I'm not gonna do that. Okay, and just because I'm running out of space here, I'm not gonna write everything out formally, but if A divides C and B divides C and the GCD of A and B is one, then, sorry, A, B divides C. We talked about this, I approve this for you. You guys see this? And just so I can kind of move on here, sorry for this being very sloppy, there's one more that I wanna say and I'm just gonna write it up here. So this goes in with all this stuff too. If A divides BC and the GCD of A and B is one, in other words, A and B are relatively prime, then A divides C. This is one where I, this is called Euclid's Lemma, I believe, and I just said, okay, the proof is very similar to the previous, I didn't actually prove this, but it's, well, I mean, it all depends on what you mean by opposite, but I mean, the idea is behind why this works, they're the same, they're basically the same. Okay, so here's what I'm gonna do now, wait till you got this down. Let me do, I think I'm gonna do one of your, one of your homework problems in this section. 2.3 really, the problems work out very, very nicely really. There's not a lot of trickery involved in this section. For a lot of them, it's just, you know, you just brute force, you expand out with the algebra and it just falls out right away. So this problem I'm gonna do is, there's really not much to it either, but I'm just, I'm doing it just because, again, I want you to see, now I'm gonna do an entire proof here. You can see what it is that I'm expecting as far as how to write the argument out and such, and some of you are still struggling a little bit with this. So I wanna at least get one whole proof in today. What's that? Let me see. Okay, yeah, let me say something about that. I should have given you both. I just felt bad, because I'd already given you 10 problems and I just decided not to, although I should have. No, I'm not gonna do that. Here's what you can do, okay. First of all, 21B in 2.3. Says to prove that two to the 35th minus one is divisible by 31 and 127. You can use part A to do that in about one line. Now, some of you probably, in fact, most of you probably didn't realize this, but this hint is actually directly a result of a problem that you've already done back in section 1.1. And that is this hint, and this is also partly why I didn't assign it. If you go back to, if you have your book open if you're curious, or if you have your homework, this is just problem two. This is a special case of problem two in 1.1. That's what the hint is. So if you wanna, so there's nearly no point in you proving the hint in a sense, because you've already done it before. You can assume part A. Yes, you can. So you do not need to prove part A. And I'm glad that you brought that up because I intended to make that comment about this problem anyways. Part B, you can just use part A and then if you see what to do, it's super, super easy. Okay? I don't expect you to prove part A. Now if you do it, I'm not gonna slash points because you did it more work, but you don't have to do it. And I'm just telling everyone, you did part A already in your first assignment, except actually a more general version of part A in number two and 1.1. Okay. So that's why I also felt a little better just a second ago. What's that? What does your number three say? Your number three says use the second principle, find an induction to establish, blah, blah, blah, blah, blah. Oh, well, I mean, it really doesn't matter, honestly. I think you can make them both work, but yeah. I said number two just because I assigned number two. I didn't assign number three. So that's why I'm saying you've done it before because it's essentially number two. Okay. So this is the kind of stuff that you should know and what I'm gonna do is let's take a look at, if I can find my notes here. Let's take a look at number 20B. Okay. So everyone have, I assume I was yammering for a while. Everybody's got this down now. All right. So again, this is 2.3 and we'll have plenty of time to finish this up. 20B, and it just says this, if the GCD of A and B is equal to one and C divides A, then the GCD of B and C is also equal to one. Okay, so there are a couple of ways that you could do this, actually. And in fact, because we have the time and I want you to get used to thinking different ways, I'm gonna present both of these arguments. So intuitive, we haven't really talked, we haven't of course talked at all about factorization yet, but the idea here is that, think of the, okay, the GCD of A and B being one, just to give you some intuition. What this really means is, of course A and B, they could be negative here, but roughly you think of it this way. Think of writing the prime factorizations out. There's no prime in common to both. That's intuitively kind of what's going on here, okay. So suppose C is a factor of A. Well, what's the prime factorization of C gonna look like? Well, it's gonna have primes that are appearing in the prime factorization of A, because it's a factor of A, okay. So the GCD of B and C then also has to be one because the prime factors in C are amongst the prime factors in A, okay. And there's nothing in common between A and B. So intuitively, that's kind of why this is gonna work this way. It's not really that mysterious once you think of it like this. But of course, we're not gonna appeal to that, because we haven't talked about that yet. We will, but we haven't gone into that yet. So what you can do is just use the basic stuff that you've done already and the theorems in the section and it falls out pretty easily. Okay, so here's the idea. Okay, so we're going to assume A and B are integers. I'm writing everything out here, which are not both zero. Yeah, that's fine. No, no. It's sort of implied by this GCD notation. Otherwise, this wouldn't even have any meaning. I'm doing it though just to reinforce this, especially considering the test is coming up on Thursday so everybody understands this. So there are two things that we're assuming here. The GCD of A and B is one. And the second is that C divides A for sum integer C. I'm gonna ask you a question that probably none of you would even think to ask yourselves. But if you're gonna be completely rigorous here, this is a question that you should be able to answer. Okay, of course, and I'm glad that you asked that. The GCD is only defined for integers that are not both zero. We've talked about that before. But notice then that the conclusion is that the GCD of B and C is one. So if this had any meaning, we should know that B and C aren't both zero, right? How do we know that B and C can't both be zero? Well, okay, I'm gonna waste about two minutes until someone tells me. Why can't B and C both be zero? No, no, no, no. No, what I'm saying is we need to prove that the GCD of B and C is one. So in particular, it has to be well-defined. So in particular, B and C can't both be zero. How do we know that? Well, no, no, no, but it's not, no, no, no. It's not by definition of GCD because we're trying to prove something. So there are two things. We need to show that one, we need to know it exists. And two, we need to know that it's equal to one. Okay, well, see, what am I saying? So you're on the right track, okay? But A could be zero. You could have zero and one, right? The GCD of zero and one is one. There's no contradiction there yet. Yeah, zero only divides one thing, zero. Zero times X equals something, that thing has to be zero. Here's why. Okay, I lied, I guess I'm just gonna tell you now. But the reason why B and C can't both be zero is because of this. Suppose, but try to follow this. Suppose B and C were both zero. B and C were both zero. So C is zero. C divides A, so A has to be zero. But then B and A are both zero, right? Because we're assuming B and C are both zero. If B and C are both zero, then because C divides A and C is zero, A is zero. But then B and A are zero, contradiction, because it contradicts the definition of GCD, which we're assuming in the beginning of the problem. Okay? Probably none of you would stop to cross that I, but I'm not gonna write that down. But that technically is part of the argument, it is. If you think it's not, you're wrong. It really is. If you wanted to be completely rigorous, that's something you'd have to check, for sure. Okay. I'm not gonna do it. And I'm not gonna take off points if you didn't do it either. But just for the sake of being thorough. Okay, so what do we know? Okay, I'm just kind of going from memory here and somebody can tell me if I've got this label wrong. But I believe that theorem two was the theorem about the GCD where I proved the existence and uniqueness of this D that satisfied all these conditions. Theorem one I think was that divisibility result. Maybe someone can verify this. Theorem two about there exists a unique integer D such as D divides A, D divides B, blah, blah, blah, blah, all that stuff. Theorem two in this section, in your notes. No, no, no, I don't mean the book is labeling it theorem two. I mean in the notes it's labeled as theorem two. Okay, you, okay. I just wanted to make sure. C divides A and B divides D. And C divides A and B, C divides D, and then therefore, you know, and also diesel in your combination, blah, blah, blah, okay, yep, okay. So what do we know? We know that XA plus YB equals one for some integers X and Y, right? I want you to pause for a second to make sure that you're clear on this point. One is that the GCD of A and B is one. Theorem two says though that the GCD is always a linear combination. So since one's the GCD, it's an integer linear combination of A and B. That's from theorem two. You guys with me so far? Okay, now since, what else do we need to use? Well, the fact that C divides A, what does that mean? That means that C times Z equals A for some integer Z, right? Small Z, big Z. Should've made the big Z bigger, but okay. And again, just for clarity here, let me put an asterisk next to this. You guys by that, right? The second assumption that we have is C is a factor of A. By definition, that means C times some integer is equal to A. Okay, now what we wanna do is we wanna plug CZ in for A in this expression, right? In this expression, right? So I have this, so you see what I'm doing here? Everything is, you can follow every little step. I tell you exactly what I'm gonna do before I do it. And I have everything offset. So there's no ambiguity here at all. Not really. I mean, I'm going through it really carefully here, but yeah, I mean, I could type this up in two minutes. I am pro-status, that's true. But the thing is, I mean, you're right. When you start doing this, that's true. But that's just the laws of physics. That's just the way it is, you know? I never thought, no, I'm not gonna do that. Sorry, I'm not gonna do that. Okay, well, luckily we're almost done though, so yes. Actually, there are very, I mean, there's somewhat verbose in that I want, I really want to give everybody a sense of how an argument flows in English using complete senses. So they're more so than I probably would otherwise write. The main thing that I'm getting at here though, as far as writing the arguments is, what you want to do is make things clear for the reader to follow and try mostly to write in complete senses. That's really what I'm trying to get at. And I'm just showing you that, more or less, that's kind of what we're, how we're proceeding through this. I'm just saying that you can play. Okay, well that, it's probably not long enough. Yeah, I know what you're saying. Yeah, yeah, yeah, you just have a streamlined version of what I have here, it's the same ideas. But what I'm, it is too terse. Yes, I want to be clear here. And this is not just because of me. And of course, some of you in discreet, maybe depending on who you had. In fact, I know who a lot of you had, but you're not going to trust me, but you really should trust me on this. A mathematical proof is an English argument, okay? Now, again, I can see your idea. I can know exactly where you're going to go with it. I can follow it like that. But you're, again, you're writing the proof for your grandma, who I hope she's still like. But you're writing your proof for your grandma. That is what you're doing. So you shouldn't be leaving these things to the imagination. Don't, don't, don't throw out pronouns you haven't introduced, okay? And this does take a little bit of extra time, but that is the point of a proof. Is to be very clear and concise and convince the reader that you know what you're doing. Okay? I can maybe know what you're doing, but I know all this stuff, you know? But, you know, again, if you're arguing a case in court, you're not just gonna say, oh, he did this at this time. And there was blood on this. And there was this, this, this, and this. Okay, go and deliberate. You wouldn't do that. It would be good for the jury, but maybe. But it wouldn't be, it wouldn't be good for your client. So it's the same idea, really, okay? Okay, so what if we plug in CZ for A? Then what do we get? We get X, CZ plus YB equals one. Which becomes YB plus XZ times C equals one. Is this absolutely necessary? No, but I'm doing this just again for clarity. That's the reason why I'm doing it. Okay, you do not have to say, because this addition is commutative, multiplication is commutative and associative. You don't have to say that, okay? Just to, and again, some of this, some of this is personal taste. So I can't give you a list of exactly what you're gonna have to know in order to write proofs that I'm gonna like. It just doesn't exist. But when you have the mean idea there and your proof is reasonable, you're not gonna get slash a ton of points. You just aren't going to. You might lose two points, maybe, or something out of 30, something like that. Okay, so what can we say now? How do we conclude the proof? This is the question. We wanna prove that the GCD of B and C is equal to one. So what do we know? Well, yes. Okay, for X, C, you mean? Okay, you can do that if you, I mean, you can do that if you want to. But here is, sorry. What, okay, I'll maybe address that in a second, but how should we finish the proof? That's the question. How generally, you need to say something here. You do need to say something. Because we haven't, we haven't actually written out explicitly that the GCD of B and C is equal to one, right? We just wrote that one is a linear combination of B and C. The GCD of any two integers is always a linear combination of those two integers. But just because you know that something is a linear combination of the two integers doesn't make the GCD, right? Take four and eight, for example, right? Two times four plus one times eight is 16. That's not the, it's a linear combination, but it's not the GCD of four and eight. But there's something special about one, right? And this is something you should say in your argument. There's a theorem that I did for you in class. That shows that we're actually done at this point. But you should quote that. You should say that. You guys follow me here? Just because one's a linear combination in general, okay? There's something special about one, but just because the number's a linear combination of the two does not make it the GCD, does not. The fact that it's one makes it the GCD. But there's a theorem I gave you and I proved this. I believe it was probably theorem three. I'm good, I am good. So it's theorem three, right? Okay, theorem three says that if the linear combination, so this theorem three says that the GCD of A and B is one if and only if one is an integer linear combination of A and B. Because one is an integer linear combination of B and C, the GCD of B and C is one by theorem three. That's how you finish the proof. That is how you finish the proof. Some of you give me dirty looks. Don't give me dirty looks. That's how you do it. That is how it's done, okay? No, I'm not gonna expect you to do that on the exam. If that's all on the homework. You can say it either from the theorem that I did in class. Of course the book has different numbering, but you should make some sort of reference to that result. Well, you know, by theorem 2.3 on page, blah, blah, blah. But you need to say something. Yeah, that's fine. It was shown in, yeah, that's fine. I'm not gonna get that picky about it, okay? You know, but as long as you say something that you understand that there's some specific theorem that guarantees this, that it's not just true for anything, that one is special somehow. Okay, so now all you have to say is something that like thus GCD of BC equals 1 by theorem 3. And this is, of course, from class now. Okay, there's the proof. But you can see that it's not just a bunch of symbols like a calculus derivative here. I'm making an argument and I'm justifying things in English. That's what you guys should be trying to do. Is this okay? Okay. I wanna, so we're gonna run out of time here, but I mean, I covered a decent amount of stuff and certainly the definitions and theorems guys, you guys should be able to get these things without too much trouble. I would strongly encourage you, those of you especially that are losing a decent number of points, especially for your exposition, the solutions to every graded problem that I've passed back so far, these are all posted on the website. Some of you have looked at them now, but I've written them out very carefully and you can see how the flow is and you compare your solutions. You should start to get a sense. Oh, crap. I mean, that will help you. If you have that reaction, that's actually good. This takes some time. It does take time to construct really clear airtight arguments. It does take a little bit of work, okay? But I definitely encourage you guys to look online, look at the solutions, okay? I've put forth some effort to try to help you with this stuff. I'll also tell you that tomorrow I will be around for a little while in the afternoon. I will be around tomorrow from 2.30 to 3.30. So if you wanna stop by, feel free to come by. I'm really, if I seem like a jerking class, I'm really not. I mean, a lot of you have come by already. I mean, I'm a pretty nice guy. So, what's that? I will not have beer. I will have vodka and I'll have some dessert wine. And I'll have chocolate, too. No, I'm kidding. Even if you show up and you're like, wait, where's the vodka? I thought you were gonna have vodka, huh? No, there's no, I can't. I would get bided pretty quickly for that. I think anyone would find out. Hey, I'm not saying I would not enjoy that, but I can't, I can't do it. I'll tell you what, after the semester, after the final, I'll buy you guys some stuff at Clyde. How's that? How's that?