 probably. I guess I'm good. Alright, so we're going to start chapter four and this is we're going to start talking about something called congruences. Some of you use this notation on some of your homework and unfortunately I had to stick to my guns and say that you know you need to just use what we talked about in class otherwise it puts other students kind of at an unfair advantage. But now you can use this because we're going to talk about it today. Okay, so some of you have seen this before, you know ends can grow into three mod five this kind of stuff. We're going to just define all this and kind of go through the basics today. I think you'll find I think you will find this to be maybe a little more intuitive given what we've already done. I think you might find that the homework questions are a little more, well, maybe, maybe not, but you might find that they're a little bit easier than some of the ones that we've done so far. So I definitely think they're a little more interesting at least so you can really kind of do some some neat stuff with with this this concept. So let's go ahead and get started. Okay, I'm just going to call this this is not it's called basic properties of congruence but I don't want to write all that out. But yeah, but then no one would know what that was if I just abbreviated it without saying. Okay, so this idea of congruence is actually very simple. It's not it's not complicated at all really doesn't. So what I'm going to do is I'm just going to go straight into the definition of what a congruence is. And then we'll just do a couple of easy examples. You will all under I mean, at least just at the kind of basic level, all of you will understand this. There's no question about it. Okay, so I just want to emphasize a couple of things here. Let N be a positive integer. Okay, so I want to underline this. Of course, this is just another way of saying natural number, right? Okay, so two integers A and B are congruent modulo N. Here's how you denote this. Okay, I'm going to make it just a little bit of a fuss here. Okay, hopefully I won't screw this up. That was okay, that wasn't really very good. But at least, well, at least you can see that there are three lines here, right? Yes, yes, I am a little bit. Okay, now am I going to get really upset if you use equals? No, no, I'm not a little bit, not that upset. But really, this convention is to use these bars. This is something, here's the reason why, well, one of the reasons why I really want you to use this is some of you are going to go on to take abstract algebra and maybe take some other higher courses. This is a sort of standard symbol that denotes something which I won't talk about now. But it's just good to get in the habit of using it now. Okay, because then later on it's really going to be even worse if you use equals. So I would rather you just do it now. Okay, and then the definition is just, it's very simple, provided n divides a minus b. That's it. Because, right, this term really comes from abstract algebra, which I know you've already taken. It's something called a congruence relation, which I probably won't say anything about right now. But I mean, that's a good question and there is a good reason for it. But really, the answer is in abstract algebra and most people aren't going to know what I'm talking about. But I can certainly, before after class, I can tell you if you want. Okay, so that's the definition. And we're just going to do a couple of examples to make sure that you kind of have an idea of what this is saying. So example one, and what I really would encourage you to do is just think about the definition. If you haven't seen this before, think about it for a second. I'm going to do three examples here for those of you that haven't seen it. But it's not that hard. So the point is this is the way that you want to think about it. I mean, it's kind of weird, new notation. But you just think, okay, whatever this is, the mod part, it just divides the difference a minus b, left minus right. That's all you have to think about. So in this case, of course, the a in this problem is seven and the b, or in this example, the seven, the b is two. And so being congruent just means that five, in this case, five is the n divides a minus b divides seven minus two, which is, of course, true. Five divides five. So that's all you're asking yourself. Take the left-hand guy, subtract the right-hand guy, and ask yourself if the number next to the mod divides it. That's it. That's it. Okay. So I don't think that this is even going to show up in your book, but I'm just to be on the safe side. I'm going to say it. Okay. Is there anything, do we have any issues with this? It's not positive. So this is undefined. This is undefined. Okay. Okay, sorry. So example one was seven is congruent to two, mod five. Example two, 12 congruent to two, mod minus 10, which is undefined because the modulus part is negative, not positive. Okay. Of course, you might say, well, why, why, we could define it. Just define it the same way. Does minus 10 divide 12 minus two? Yes, it does. It's just that more or less there's just no reason to define it for negative integers, which I may see more about later. But the point is that this just doesn't really have any meaning. Okay. Now, I do want to make this clear, though. Notice that A and B are integers. I didn't make any more additional restrictions on A and B, right? I never said they were positive. So the congruence, the only number, the only integer that actually has to be positive is the one next to mod on the right. That's the only one that actually has to be positive. Otherwise, everything's fine. So it's the same question. Does seven divide minus 31 minus 11? Think about this. Minus 31 minus 11 is minus 42. Seven times minus six is minus 42, so it does, right? So I'll just put this in parentheses. Seven divides minus 31 minus 11, okay? Seven times minus six is equal to that. Okay, so for those of you that haven't seen this before, really you just have to kind of get in your mind just what it is. When you see this mod notation with this congruence symbol, these three bars, you just need to translate that into something that you sort of just translate this into an assertion about divisibility, which we've been talking about now for quite a while. So basically what we can do is this is going to end up being sort of a shorthand. It's going to kind of give us a shortcut for solving problems where before we might use the division algorithm, now we can really save a lot of time. So this is kind of, this isn't a really great example, but in some sense this is sort of like what you did with derivatives when you learned the derivative computing it by the the limit definition, right? You got to write everything out. Now we can kind of get a shortcut and we can actually solve these problems a lot more quickly. Did you have a question? Oh, yeah, I mean we're gonna get to that here in a little bit, but yeah I know what you're thinking and yeah, so you can write 11. 11 is the same thing as four, mod seven. But yeah, we'll talk about that in a second, yeah. Okay, so what is this really saying? Let's see, let me, whoops, okay, I did this again, all right, sorry, okay, where do I, where, where, purple, oops, sorry, yep, thanks. Okay, so let's see, let me, everybody's got this down, I've been blabbing now for a few minutes, right? Okay, all right, I'd like to just get to the top anyways. So here's the first theorem. This theorem is not very, very complicated. This kind of just lets you know what direction we're going, so you might say well okay, great, why did you define this? Who cares? What's the point here? So here's the point. We're gonna let n be a natural number. Instead of saying positive integer, I'm just gonna write it this way. So you should all, you should all remember what capital N is, right, the positive integers. Sorry, make this a comma. And let a and b be integers, then a is congruent to b, modulo n, or mod n, if and only if a and b the same remainder upon division by n. Okay, and you're gonna see that this really just reduces down to things that we've already done. There's a longer theorem I'm gonna give you which I'm not gonna do all of, I probably won't prove every part of it, but I'm gonna do some of these proofs, well of course I normally do that, but some of the homework problems you're gonna have to kind of get these techniques down, these kind of ideas down. So you'll see, I think you'll see if you follow along here that this is not really that bad. Okay, so we're gonna let n, a and b just b as stated in the theorem. So let's suppose this is an if and only a statement, right? So we're gonna suppose that a is congruent to b modulo n. Okay, so we want to show that a and b have the same remainder when we divide by n. Well, this kind of implicitly invokes the division algorithm, excuse me, the division algorithm, right? So we know we can divide by n, we get a remainder that's between zero and n, right? We've proved this already. So I'll just say it this way, by the division algorithm, a equals nq plus r, right? We're dividing by n, q is the quotient, r is the remainder for some integer q and some integer r with, mostly we're pretty good with this on the exam, with zero less than or equal to r less than n, right? That's what the division algorithm says. Okay, so what we're trying to get to is that b has, okay, let me say it this way first. r is the remainder upon division of a by n, right? r is the remainder. So we want to show that the same r is also the remainder for b when we divide by n. So somehow we're going to want to use this fact. And so what you want to do, this is just to give you kind of a tip here, how you would proceed with the proof, is say, okay, well, we want r to be the remainder when we divide b by n. We've got this piece of information that we're assuming, so let's use this somehow. So the best thing to do is just go back to the definition, go back to what it means. It means that n divides a minus b. And using that, hopefully we'll be able to get to what we want, right? Since a is congruent to b mod n, I'll just remind you what this means, n divides a minus b. Okay, right, that's just the definition. So nx equals a minus b sum manager x, right? Okay. Here, why don't, just to make this a little bit clearer. I should have done this before, I apologize. Let me put an asterisk next to this equation up here, that a is nq plus r. So by this, by star we have nx equals a minus b, right? That's just coming right from here. And then we can replace the a with nq plus r. You see that? Okay, so everybody see what I did? A is nq plus r, so I'm just replacing it right here. And from this, okay, let me kind of go slowly through this. So what we have then is, we just jumped from the first term to the last part of the equation. We have nx equals nq plus r minus b, right? nx is equal to this. So what I'm going to do, okay, so now I want to say where we're going with this, we want r to be the remainder when we divide b by n. So what I'm going to do is just solve this equation for b. And you're going to see that we have two n terms. So I'll be able to get b equals n times something plus r, so it has the same remainder. r is the remainder in this case as well. Okay, b equals nq minus nx plus r. All I'm doing is shuffling the algebra here, right? I'm just bringing the b over here, subtracting the nx and bringing that over here. Nothing too complicated. Sorry I didn't mean to put the period in. So this is equal to n times q minus x. We've backed out the n and then plus r. Okay, so now I know that r is the remainder upon division by n. All right, now there's a uniqueness part of the division algorithm that said they're unique integers q and r that satisfy, you know, n equals, sorry, b equals nq plus r where r satisfies 0 less than or equal to r less than n. Okay, it's unique. So because we know that b is equal to n times something plus something that's strictly between, sorry, it's between 0 and strictly less than n, we know that r has to be the remainder because it's unique. That was part of the division algorithm is that these things are unique. So we automatically get for free that r is the remainder. Okay, what I'm going to do for the sake of time is I'm going to omit the other direction. This is an if and only of proof. So technically we need to prove that if a and b have the same remainder, then a is congruent to b mod n. That's, we've only done half of it, but I'm just, I'm not going to do that because we've got a lot to do today. So it's in the book also. Page 64 in the book, yeah. Actually, I can just kind of tell you what it is really quickly because it's really easy. Suppose just, just say that a and b both have remainder 5 when you divide by n. Well then what form does a have? a has the form nq plus 5 and b has the form nq prime plus 5. So when you subtract, the remainder is going to cancel and n's going to divide it. Okay, I mean it's, there's really nothing to it. So this was the part that required a little more work. All right, so we have this down. Yeah, am I good? All right. So this, this is important also. We're going to use this later. The book goes into something that they call a complete set of residues, modulo n, and it's, I'm going to avoid using that terminology just, just because I don't want you to get confused with this kind of weird lingo. What I'm giving you is the corollaries, essentially, when they're saying complete set of residues, this is what it, what it means. Okay, I'm not going to use that terminology in this class. Okay, n's a natural number, a's an integer. Okay, so I'm writing out a little bit more than I normally would, but I think this will make it a little bit clearer. So hopefully this is clear. I didn't have room to keep doing this, but what I mean is, of course, the pattern continues. So a's can go into zero mod n, a's can go into one mod n, a's can go into two mod n, a's can go into three mod n, all the way down to n minus one mod n. Exactly one of these holds. Okay, in the, in the book, you mean? Yeah, well, it's sort of, it basically is, except they don't use, they use this complete set of residues. It's, yeah, it's in here somewhere. No, it's okay. It's, let me see. No, you know what it looks like is that the book just sort of mentions this, and they kind of give a, a proof, but they don't actually list it as a certain theorem or, or a corollary. They just kind of mentioned it on page 64. So I'm going to, I want to outline this specifically as a corollary because I think it's important. Okay, again, we're just going to let n and a be as stated. And I'll tell you, before I go into this, really what this is, kind of what this is saying. Really, this is, this is the content of this corollary, even though it's in this kind of funky notation. All it's saying, really, is that when you divide an integer by n, you have n minus one possible rematers. That's all it's saying. It really is. So this is nothing more than the division algorithm more or less. So I'm not going to say a whole lot here, but in fact, I'm not even going to write all the, all these symbols out. Let me, I'll tell you what I will do though, just underneath these, I'm just going to label this zero, one, and then all the way down to n minus one, because I'm going to refer to these ingruences here in a minute. Say that a has a remainder of r upon division by n. Yes. What would we know about r? Then, then zero is less than or equal to r. And I'm going to write this a little bit differently now. Right? I mean, normally what I write is zero is less than or equal to r, which is less than n, but less than n is the same thing as less than or equal to n minus one. r is an integer. Right? Okay. Well, I'm going to, there's a subtle point here that the book kind of glosses over, which I want to mention. But just to make this concrete, I'll just say it this way. If you divide one by five, the number one by the number five, what's the remainder? One, one. Sorry. What if you divide two by five? What's the remainder? Two, right? Two is five times zero plus two. Right? What if you divide three by five? What's the remainder? What's the remainder when you divide four by five? Four. Four. No, it's four. Right? Okay. You divide four by five. It's five times zero plus four. And four is zero is less than or equal to four, which is less than five. So that's the remainder. What's my point? My point is when you divide all of these integers here, zero, one's remain minus one. These are all less than n. So the quotient is zero, and each of these numbers, they're their own remainder. Okay? That's what I'm trying to get at here. It's kind of a subtle point, but I want to make sure to mention this. So what can we say if a has a remainder of r upon division by n and r is between zero and n minus one, r is less than n, then what's the remainder of r upon division by n? So a and r have the same remainder upon division by n. They're both r. Therefore, by the previous theorem, a is congruent to r mod n. That's what the theorem says in your notes. If they have the same remainder, then they're congruent mod n. Upon division by n, they're congruent mod n. Since we're just saying a has a remainder of r, r also has a remainder of r because it's smaller than n. So therefore, a and r both congruent mod n by theorem one. Okay. So that means that, well, since r is between zero and n minus one, it's certainly one of these congruences. This congruence right here is certainly one of these above, right? Certainly is. Okay? So one of those congruences help. Well, why can't you have more than one of these holding? Well, I mean, okay, this is almost silly, but I'm going to say it anyways. In fact, I'm going to make this rest of this proof a little bit informal, just because I'm trying to get you to think about it a little bit. Why can't we have, say, a congruent to i mod n and a congruent to j mod n with i not equal to j? Why is this? Okay. So notice, remember, we're not quite done with this. Exactly one of those holds. I just showed you that one of them holds. We still need to show that two of them cannot hold, just one. Okay? Well, let me just say it first. This is almost so easy that it'll go over your head if you think too hard. Really, I mean that. It's just, really, there's nothing, not much to this. What did theorem one say? It said that if two integers are congruent mod n, they have the same remainder when you divide by n. They have the same remainder. What's the remainder? Now, remember, i and j are among these numbers between 0 and n minus 1. What's the remainder when we divide j by n? J. J, because j is small, right? j is less than n. So the remainder is j. So therefore, a has a remainder of j when divided by n. But, I don't know if that's but, same thing has to be true here, right? What's the remainder of i when you divide by n? i. So therefore, because they're congruent, a has a remainder of i when divided by n. Well, can't have two different remainders. It only has one. So this is not possible. That's the gist of it. Because by theorem one, a would have a remainder of i and j upon division by n. And of course, that's not possible. So here's kind of the point. We're going to see this. We may not get to this as much today, but we will on Thursday for sure. We'll finish the section. But the main idea here is that in some sense, when you look at this mod n relation, there are exactly n numbers that an integer can be congruent to. And it's just the remainder when you divide by n. So for example, for example, I'm not going to write this down, but I'm just going to say this. So if you give me any integer, I can tell you what that is mod 4 and mod 4 being either 0, 1, 2, or 3. So for example, if you give me 27, well, what is that modulo 4? It's going to be 0, 1, 2, or 3. And the number it is, is it the remainder upon dividing 27 by 4, which is 3. Divide 27 by 4. What's the remainder? 6 times 4 plus 3. So it's 3. It's equal to 3 modulo 4. And everything modulo 4 is going to be either 0, 1, 2, or 3. Exactly one of those. And the number is just simply the remainder upon division by 4. That's it. So basically what this does, in the past, we used a division algorithm, for example, on your exam. And on some of the homework problems, you said, okay, division algorithm. And we've got this divide by 5. It's 5q plus 0, 1, 2, 3, or 4. And then you had to multiply everything out. And it got kind of messy. This is going to give us sort of a shorthand way of doing this without going through all the calculations. And you'll see how that applies here in a minute. But this is the main idea, is that modulo n, there's essentially n different classes. And that's it. There's no more than that. Yes? Do applications ever use the percent sign for modulos? No. No. Yeah, it is. I've seen it done there. But the, yeah, it's just an unfortunate, you know, yeah. Well, you know, I would, I know what it means. I would, yeah, I mean, I would, I would rather see the, you know, and we're talking about a difference of like 0.01 seconds to change the, you know, to write the, between writing these things. So what's that? Exactly. Yeah, that's true. You can lose a lot of points in 0.01 seconds. Just kidding. So, no, I wouldn't, I wouldn't, I wouldn't do that to you, but what's that? Okay. Oh yeah, yeah. Yeah. That'll come up too. Okay. So here's, let me see. Yeah. This is the last theorem that I'm going to talk about. I'm going to prove some of these, but I'm not going to do all of them because I'm going to run out of time here, but we will definitely finish up. This is 4.2. Yeah. 4.1 is something about Gauss, or Gauss, or Riemann, or Euler, one of these, one of these projities that, that, yeah, they all died of pneumonia. Like, yeah, he was, well, yeah, he went, he, he went blind, like later in his life, I think. He mathed himself. Yeah. Oh, he did work after he was, yeah, that's true. No, that's true. Yeah, he definitely did. I know a few things about it. What's that? Euler? Not that I'm aware of. Pythagoras, yeah, yeah, yeah, yeah. He was, well, people were just jealous of him. Yeah, that happened, that, although that does happen on occasion, that happens less than you might think, though. Okay. Okay, so n bigger than 1 is, is fixed. I, I didn't say this, but it's a, it's an integer. Okay. And it's an integer. A, B, C, and D are, are integers. There's no restriction on A, B, C, and D. They don't want it to be positive. Okay. Then, and here's what I'm going to do is just prove these as we go. And some of them, I like said, I'm going to admit, I'm going to do some of the easier ones. Oh, yeah, I will. Right. I will definitely do that. Okay. So the first part, part A is, well, it's unfortunately, we have an A here again, but is that A is congruent to A mod n. Okay. So this is a little bit more abstract now, but this, if you, if you really think about the definition, this is not hard. This is not hard. Remember what I said, what does this mean? It means you take the left hand guy, subtract the right hand guy, and ask yourself, does the number next to the mod divide it? Yes. Which is true, because A minus A is zero, and n times zero is zero. So this is very simple, right? Okay. So proof is, it's almost kind of silly, but n times zero equals zero, which is also equal to A minus A. So n divides A minus A. Put that on the test. I might actually do, I might, because I, yeah, I could be a, I would like to just put this one question alone on the test and you either get a hundred percent or zero. Put that right there on the test. That's fine. That's, I mean, what I just said, is not going to actually happen, but it'd be easy grading for me, though. That's true. That is for sure. Okay. Okay. So by definition, A is congruent to A mod m, because n divides A minus A. Okay. The second part of this theorem, if A is congruent to B mod n, let's see, I didn't look at this. Maybe I should be careful. Yes. Then B is congruent to A mod n. Okay. And this really gives, I want to do this example, because it's not very hard, but I'm going to kind of talk you through this. You know, for example, this was a homework problem, how you would think your way through the proof. This is an if-then statement. You can, there are lots of ways you can prove an if-then statement. Most of the ways we're going to use in this course are just direct proofs in which you assume what's called the hypothesis, the if part, and then you derive the consequent or the then part. So we're going to assume, and this is the way you start with an if-then proof. If you're going to do it directly, you assume the part that comes after the if and before the comma, right? Assume that A is congruent to B mod n. I'll write this out in a little more detail. Don't forget what it is you're trying to prove. We must prove that B is congruent to A mod n. Okay. Now, here's the way that you want to go about this. Whenever you're trying to prove something, you should, if it's written in some sort of notation that's maybe not quite as clear or intuitive as what you're used to, break it down a little bit. You can just do this in your head. What does it mean for B to B congruent to A mod n? It means we want to prove that n divides B minus A, right? Left hand minus right hand, n divides B minus A. We have that A is congruent to B mod n. So let's write out what the definition is. A is congruent to B mod n means that n divides A minus B, right? So from there, we want to be able to prove that n divides B minus A, and then we're done. So that's the idea, and that's basically what it's going to boil down to. So n times x equals A minus B, right? For sum integer x. I want to say this again, and some of you are losing a point or two on your homework for this. When you introduce a new variable, you should always be seeing where that thing exists in what the universe is. Don't just say x, okay? Say x is an integer. Okay, so we've got this, and what we want to do is show that in fact n is a factor of B minus A, okay? So n times something is B minus A. We've got that n times something is A minus B. So somehow from this, we should be able to manipulate this to get n times something is B minus A. So what you should be thinking about is, is there some algebraic operation you could do to both sides that will get you what you want? Yeah, that's exactly what we're doing. Yes, very good. So all you have to do is, you should have seen this in calculus probably at some point with factoring, right? Or you can think about factoring out of minus 1 if you want to think of it that way. But if you multiply both sides by minus 1, what's going to happen? The minus B will become plus B and the A will become minus A. So then you're going to get B minus A, which is exactly what you want. Okay? Okay. Joe? Well, because we haven't, I know what you're saying, but you're saying the right side of the congruence. You have n is equal to A minus B. And n times x is equal to A minus B. That's exactly why you have to do that. Well, because we haven't proven really any properties of the congruence relation just yet. So I mean, really, right now, we kind of have to just get back down to the definition because that's all we have to work with. I haven't really built up any machinery on what happens with congruences. So for example, if A is congruent to B mod n- Never mind. Never mind. I was thinking of this thing. Yes. Yes. That's why I mess up. Okay. So then you have nx is equal to A minus B? Yep. Okay. So if we multiply through by minus 1, we get n times minus x. I'm going to suppress the basic algebra here. I assume that you guys can do this. I'm multiplying through by minus 1. It's just going to switch everything, right? So n times minus x is B minus A. Okay? As I was saying before. But that means that we just showed that n divides B minus A. So B is congruent to A mod n. Okay? Not too bad. Okay. And we're going to do a couple more of these. Yes. Well, I mean, I definitely don't expect you to do that. I mean, we're basically just using the fact that if x is an integer, then minus x is an integer. So, I mean, really, yeah, you're kind of doing more work than you have to. I mean, there's nothing wrong with it. But this is totally fine. I mean, you can assume that the negative of an integer is an integer. The product of integers is an integer. The sum of integers is an integer. These are things that you can use without proof. And you're not really proving that anyways. All you're really doing is just saying, okay, minus x is an integer. I'm just going to call it something else. But there's really no reason to have to call it something else. So, you can just... Right, n times an integer, yeah, right. But no, you really don't need to do that. You're certainly not going to lose points for it, but it's okay. Okay, so the next one is a little more complicated. If A is congruent to B, mod n, and B is congruent to C mod n. I'm going to be able to guess what the conclusion is here. Then A is congruent to C, right, mod n. Okay. I don't know if any of you have taken some other math courses, maybe even discreet. So this three-bar relation here, this congruence relation, has these three properties. Does anybody know what a relation with these two properties is called? Is that a curiosity? No. Good guess, though. What's that? Well, that's what this property is, yes. So anyway, it doesn't matter. You may see this later. This is called an equivalence relation. Maybe some of you have heard that before. It's a reflective symmetric and transitive, yeah. Okay, that's... But I was just curious. Okay. I'm not going to... I'm not going to prove this part. It's actually really straightforward. I will say something about this. In fact, if you just listen, you probably can see the proof right away. Anyways, what does this mean? This means that n divides A minus B. This means that n divides B minus C. So n's going to divide the sum, too. This is an old theorem from a long time ago. So A minus B plus B minus C is A minus C, and divides A minus C, and that's exactly that. So that's it. You just add them together and you're done. Yeah, I just did the proof. I just didn't write it down. So that's how that one goes. Okay. Everybody have this now? Okay, so let's see. Part D. A is congruent to B. Mod n. And C is congruent to D mod n. Then A plus C is congruent to B plus D mod n. A, C is congruent to B, D mod n. Okay. There are two more parts which I'm not going to prove. I'm going to write the proof of this. I'll give you the last two and then we'll stop. Okay. So we're going to suppose that A is congruent to B mod n and C is congruent to D mod n. We need to get that A plus C is congruent to B plus D mod n. All right. So again, what are we going to do? We're going to translate these congruences into divisibility statements, right? Okay. Well, now I'm going to suppress going back to basic principles every time. There's a theorem, I think it's, you know, theorem two in section 2.2 or something like that that says that if, say, n divides x and y, then n also divides all linear combinations of x and y, right? So if n divides A minus B and n divides C minus, and so I'm just going to use that just, I'm assuming you guys are with me on this now, n divides A minus B and n divides C minus D, then of course it divides the sum, right? You can write this out directly, nx equals A minus B, ny equals C minus D, then n times the quantity x plus y equals A minus B plus C minus D, right? Okay. And I'm just going to rearrange the form slightly. That is, n divides A plus C minus the quantity B plus D. I want you to think about this for a second, convince yourself that this is the same thing as what we have above. This quantity and this quantity are the same. I'll go ahead and put this in parentheses. Of course we really don't need the parentheses around the additive part here, but do you guys buy that? What is this? This is A plus C minus B minus D. A plus C minus B minus D. Same thing. Why did I write it this way? Because this is exactly what it means for A plus C to be congruent to B plus D mod n. It means the left-hand guy minus the right-hand guy is divisible by n, okay? And this is just, we don't have to say more. It's just definition now. A plus C now is congruent to B plus D mod n. Okay. Now, for the last part, we're going to actually, I'm going to go back to basic principles. We know that nx equals A minus B and ny equals C minus D for some integers x and y, right? Okay. Why am I writing this? Just from above, right? We have n divides A minus B, n divides C minus D. That's where this is coming from. Just a definition of divides. Okay. Well, what do we have to prove? What's the last thing we have to prove? And like I said, this is the last proof I'll do. I'll just give you the last two parts of the statement. We have to prove that n divides A C minus B D. So somehow I want to go from here. These are my assumptions. I want to mess with this somehow so that I can get that n times something is A C minus B D. Okay. Well, here, let me do this again. I'm going to call this equation 1, and I'm going to call this second ny equals C minus D equation 2. So first equation is nx equals A minus B. The second is ny equals C minus D. I want to show that A C is congruent to B D mod n. So from nx equals A minus B and ny equals C minus D, I want to show that n divides A C minus B D. That's what I'm trying to prove. Okay. Well, here's how you do it. I mean, you can't do this just by adding, right, because you're not going to have an A C come out just by adding. So you're going to have to do some sort of multiplication here. So we want, again, remember, we want n to divide A C minus B D. So look at this first equation. Here's the idea. It's a little bit of a trick, but not much of a trick. We need to get an A C to come out somehow. So let's just multiply everything through here by C. So then we're going to have A C minus B C on the right side. So at least we have the A C part, but we don't want the B C. We want B D. So we need to get rid of the minus B C somehow. Well, then if we multiply this through by B, we're going to, when we add the B C is going to cancel and we'll be left with the minus B D, which is what we want. So what I'm doing is I'm multiplying the first equation by C and I'm multiplying the second equation by D and then I'm going to add them together and then n will divide A C minus B D. Okay. So once we do that, we get nxC equals A C minus B C. Multiply the second by D. Let's see. Sorry. I want to multiply it by B. Sorry about that. I want to multiply the second equation by B. Yes. Yeah. I meant to write B. Okay. So we get, put two stars here. So we get n y B equals B C minus B D. Okay. So if we add these equations together, we get n times xC plus yB. That's the left-hand side. After adding these two equations, the left-hand side just factoring out the n here. And what happens when we add the right together? This is just what I said before. The minus B C and the B C will cancel and we'll just be left with A C minus B D, which is what we wanted. So that means we don't have what you want. You show us n divides that. Yep. So yeah, that's right. So what we know then is that, I mean, again, we know this is an integer, right? It's a product sum of integers. It's an integer. n divides A C minus B D. Therefore, by definition, A C is congruent to B D mod n. That's what we wanted to show. Yeah. You don't have to mention that. That's okay. Okay. I'll tell you what I'm going to do. I'm going to make a note to myself here, but just for the sake of time now, we're getting low on time. So this is not the end of theorem two. There are two more parts, but I'm just going to do those on Thursday. I'm going to finish up the rest of the section. I'm going to give you your homework assignment. This, along with some of the examples in the book, really should be able to get started on the homework. I'll tell you that this homework will be due next Tuesday. You've got a week. Thursday, we'll finish the section up, and I'll give you some hints and such about how to start a couple of the homework problems. Okay. So I'm going to give you the homework, and I'll give you your exams back here. Okay. So just to remind you again, this is section 4.2, 3, 5, 9, 12, C, 16, and 17. Okay. So six problems. The other ones that don't have numbers, there's only one part to these. So none of these, like 70, doesn't have five parts. They're all just one part. So it's not that bad. We'll finish this up on Thursday, so you have the weekend to work on it. Okay. Did you have the due date? The due date for the assignment? One week from today. One week from today. Okay. All right. So let me get your exams passed back. You can, I'll tell you what, if you want to leave when you get your exam, that's fine. It's going to be a few minutes, though. I will probably say a few things about the exam if you want to stick around. Okay. I will tell you that the solutions to this test should be posted in the next hour and a half or so online. All right. So again, what I encourage you to do, I really encourage you guys to do this, look online, see where you made your mistakes, try to understand the solutions online. Then if you have any questions, come and talk to me, and I'll try to make things a little bit clearer. Okay. The only other thing I'll say before I pass these back is that some of you, I noticed that your scores went down quite, some of you, your scores went down quite a bit from the homework. So I know, you know, some of you are working together, that's fine. I don't really care. I mean, if you want to work together in groups, that is fine with me. No big deal. Although I think some of you are using the internet as more of a resource than you should and are just sort of kind of copying down things that you find online and then not really understanding what the solution is saying. And then when you get to the exam, you're having a lot of trouble. I think, I don't, I'm not saying I have any particular person in mind, but I, I just, from the scores, I think that that is happening for some of you. And it is hurting you on the exams. So I really want to say, try not to do that. That is just not helping you really. I mean, it's helping your homework score. But if your exam score is failing as a result of that in the end, you're not really helping yourself too much. You really should be trying to understand the problems. Come and talk to me. If you're having trouble, just come and talk to me. I will help you. A lot of you are doing this. And you know, I give you hints and stuff in, you know, in my office. That's really what you should be doing. Okay. Okay. David Harris, some of you are not here. We were scared. Micaela, Chris, Jessica is not here. I don't think, okay. You are, you, Jessica is like, carry your pigeon. Okay. Yeah. No, she already told me. So that's fine. Uh, Sarah, I think calling you a pigeon is better than, is pigeon better than Dota? I don't know. Okay. That's true. Okay. Brianna, Josh, Christina, Chad, Eric, Joe, Douglas, Josiah, Andrew, Andrew, not here. Nope. Tori, not here. Okay. Nick, Daniel, okay. Michael. Okay. John, uh, Matt and Talina. Okay. Let me just say, let me say a couple things here for those of you that are interested in sticking around. Okay. I'm not going to write much here. The first problem, of course, was the same as the first problem on the, uh, first test. Okay. For the most part, two was, was pretty good, except M is the least common multiple. So that means it's the smallest one. Some of you flipped the inequality around. So that means that any common multiple of A and B, M is less than positive, right? Common multiple of A and B, M is less than or equal to it, not bigger than or equal to it. Okay. That's what we had with the GCD. The GCD is bigger than or equal to, right? Number two, most people got that one. Sorry. 2B. Okay. 3A, there was some trouble here. Some of you, and I know you know this. I know you all know this, but some of you in sort of, you feel like the only integers are natural numbers, that the negative numbers don't exist. A lot of you are forgetting about them, the negative numbers. So it's not, that's actually false. Okay. Can you take, so I'll write this down. Actually, let me just write this down real quick. Okay. Okay. Yeah, I keep doing this. Okay. So what, what is, what's 3A saying, really? The question really is, so A and B are non-zero integers. The question is, is it true that the least common multiple of A and B can never be greater than? What does it mean to never be greater than? It means less than or equal to, right? To not be greater than means less than or equal to. Is the least common multiple of A and B always less than or equal to A times B? Well, you might say, well, A, B is a, is a multiple of A, A, B is a multiple of B. So the least common multiple should be less than or equal to that. Well, but there's a, there's something you're missing here. We can take the least common, we don't have to have the integers be positive. We can take the least common multiple of minus 2 and 6, for example. Okay. So what if, what if I take here, what if I take A to B1 and B to B minus 1? What's the least common multiple of 1 and minus 1? So here's where some of you are forgetting the definition too. The least common multiple by definition is positive. Positive, right? So what's the least common multiple of 1 and minus 1? I'm here in some matter. It's definitely 1, right? What's A times B in this case? Is 1 less than or equal to negative 1? No, not true. I put this on there specifically because I had this feeling that a lot of you don't believe in negative numbers and you should. Okay. Okay. Now, okay. Also, also 3B, there were, there were some issues with this. This, and here's something that a lot of you are going to find as you go along in this course that you are, and this is very common. This is what almost everybody does when they learn this stuff. You're writing way too much stuff and that it's actually a lot easier than you might think. Okay. This problem, and I'm not going to be precise now since I'm just kind of talking about this. The GCD of A and A plus P is either 1 or P. Well, this is just a special case of a problem you guys already turned in and it was actually a solution that's online. I don't know if you guys will remember this. Some of you probably do. The GCD of A and A plus N is a factor of N and then the last part of that problem was therefore the GCD of A and A plus 1 is 1. This is a homework problem that you guys did. It was graded. Solution's online. This is just a special case of that. So, the GCD of A and A plus B should divide P. It's got to be 1 or P. I mean, of course you have to write it, write out the argument, but that's it. It's just a special case of the homework problem. Okay. So, I'm not going to write all this out in detail. Okay. The detailed solution will be online, but if we let D be the GCD of A and A plus P, right, then what do we know? Again, don't take this as the formal solution. It's not. I'm just trying to make this quick. We know that by definition of GCD, D divides A and D divides A plus B, right? It's a common divisor. So, if D divides A plus P and A, then D is also going to divide their difference, which is P. Since D's positive and P's prime, D is 1 or P. That's it. That's it. That's all there is to it. Okay. I have to say, I don't feel like this was an unfair problem given the homework problem that you had. I don't think it is. Yeah. No, no, no. But by definition, D is the GCD. So, by definition, D is positive. That's why D has to be 1 or P. Yes, minus 1 divides 1 and P, but D is positive. So, D has to be either 1 or P. Okay. But it comes from the fact that D's already positive by definition of being the GCD. Okay. So, also, a lot of you guys, okay, if anyone want to walk out, that's okay. A lot of you guys are doing this, and this is also some of the definitions too. You're saying that if P is a prime, it's only divisors or 1 and P. This is a subtle point, but that's not true. 5's prime, minus 1 is a divisor of 5. Minus 5 is a divisor of 5. If P's prime, the only positive divisors are 1 and P. Okay. I hate to nick you on this, but this is a class on proof. So, we need to be precise about this stuff. Okay. And then the other thing, the last thing I want to say, and most of this, I'm just not going to, I'm running out of time here, but the number 7, number 7, there was also some trouble with this. This is the last problem, right? P equals a squared minus b squared, and I give you a hint. I said to factor this, right? Okay. So, it's a plus b times a minus b. Here's the part that a lot of you missed. And this is in the, you see, you have to read the problem. I put in parentheses, it's assumed that a and b are positive integers. a and b are positive integers. So, if, okay, here's what I want you to think about. Okay. Think about this for a second. If a and b are positive integers, these are not true questions. Is a plus b positive? Definitely. Okay. So, if a plus b is positive, then a minus b also has to be positive, right? Because b is positive. So, we know that they're both positive. Now, which of these two is bigger? a plus b or a minus b? a plus b is bigger. So, if a prime is written like this, and its only positive divisors are 1 and p, and this one is bigger than this one, then this has to be p and this has to be 1. Right? That's it. That's it. And I could try to guide you a little bit on this, but it's not that hard, really. I mean, you're just using the definition of prime. Well, that's it. Well, so what do you know from the second equation? a is equal to b plus 1, right? Plug that back in here. b plus 1 plus b equals p. And you can solve for b pretty easily now. b is equal to p minus 1 over 2. And once you know that, you know what a is. That's it. That's all there is to it. So, anyways, like I said, I'll have the solution posted this afternoon. I would encourage you to go through and look at these.