 We don't have a lot to do to finish the section up. I'm not going to go on to a different section. What we're going to do is just, there's this theorem we have to finish up, one more theorem in the corollary. And then we'll, I'm going to give you some hints on how to start some of these problems in homework. OK, so that's the plan. All right, so we were on, let me make sure we are here. OK, so, oops, I got to click this, sorry. OK, yes. OK, so we're just going to finish this up. I think we were on theorem two before, and I think I got through part D, all right? OK, so we're just going to finish this up. OK, so this is just a continuation. OK, so I'm not going to have time to go through and prove both of these, but in fact, I don't know that I'll do either of them, but I'm just going to list these. So E says that if A is congruent to B mod n, OK, so I'm using some shorthand here because I want to get this all in one line, implies, OK, so that just means if then, right? If A is congruent to B mod n, then you can add whatever you want to both sides. A plus C is congruent to B plus C mod n, and you can multiply whatever you want on both sides, too. Oops, sorry, OK, let me just try to fix this. A, C is congruent to B, C mod n, OK, I think I really messed up at the end there, but hopefully you can see that. Mod n, right? OK, so instead of writing a proof of this, this is actually very easy, really. So if you think about this, and I want you to start trying to do this in your head a little bit because it's going to save you a little bit of time on the homework. What does this mean? So just remember what we were talking about before. This means that n divides A minus B. And certainly divides A plus C minus B plus C because the C's just cancel out and you just end up with A minus B again. So there's really not much to this part. And here it's kind of the similar, right? I mean, if you subtract the BC, then this assertion is just that n divides C times A minus B. Well, if n divides A minus B, it's going to divide every multiple of it, right? Sort of, yeah. I mean, so the main thing to take away from this is just that if you have some sort of congruence, you can sort of treat this like an equation. And if you add the same thing and multiply the same thing on both sides, it preserves the congruence. That's really all this is saying, OK? And the last part, which I'm not going to prove, this is an inductive proof. It's going to take a while. So I'm just going to skip that. But if, I'll write this out and we'll be lazy this time. So if A is congruent to B mod n, then A to the k is congruent to B to the k mod n for every positive integer k. Right. So B, I mean, A and B could be any integers. They could be any integers whatsoever. And so, yeah, the restriction is just that k is positive, but A and B can be anything. When you were talking to me, were you saying that B could be anything? Yes. Yes. I know that it has to be. Yes, that's right. Yes, that's right. Yes, so for example, it makes sense to say, you could say 2050 is congruent to minus a billion mod n. There's nothing wrong with that. It's just that if you take any integer mod n, so let me say it better way, what we talked about last time was if you take any integer A, then A is congruent to either 0, 1, 2, 3, 4, all the way up to n minus 1. It's congruent to exactly one of those mod n, right? So you can always sort of replace it this way. We're going to get more into that as we talk about some of the homework problems here in a bit. OK, so that's the end of the theorem. But these six things, you're actually going to use these quite a bit in your homework, and I'll show you how this works here once we get through a few more things. I'll tell you what, just to be on the safe side here, I think what I'm going to do is just do an example, showing you how you use this. You have a couple homework problems that are similar to this. Let's see, where are we at here? Example, I think we're on 4 now, right? Well, I'm pretty sure that's right. I did these three examples in the beginning of the congruence system. If it's wrong, I guess it doesn't really matter. OK, so problem is to show that 41 divides 2 to the 20th minus 1. OK, and so the book, this is an example in the book, and I'm going to basically do this the same way that the book talks about it. This is probably the most reasonable way to do this. OK, so what you're going to do is you're going to use the theory of congruences to sort of transform this into a bunch of, you're going to sort of separate this out into a bunch of pieces that are easier to work with. You don't want to actually try to figure out what 2 to the 20th is, and then use a computer and say, you know, I used a computer and it works. That's not what you're doing here. So what you're really trying to prove is that 2 to the 20th, I want you to think about this for a second, minus 1 is congruent to, sorry, yeah, that's right, so 0 mod 41, OK, must maybe isn't the best word here, but we will show this. And so I want you to just pause for a second, make sure you understand why this is the same thing as this. What does this mean? The congruence, it just means that 41 divides 2 to the 20th minus 1 minus 0, which is exactly that. OK, so what you're doing with these problems you're trying to, the first thing you're trying to do is just translate it into some sort of congruence. Then you're going to use some of these properties, well, some of these parts of theorem 2 to try to cut it up into simpler pieces and you'll see what I mean when I go through this here in a second. So here's what you want to do. So we've got a 2 to the 20th. And what we really want is for 2 to the 20th to be 1 mod 41, in other words, have a remainder of 1, because then when we subtract the 1 we get 0. That's the idea, OK? So let's try to figure out what 2 to the 20th is mod 41. So I'm going to kind of write this more informally now just so that you can see kind of what the strategy is, OK? So what we're going to do is we're going to look at 2 to the 20th and, OK, I'm just going to write it this way. This is, like I said, a little bit informal. Break it up, if you will, into smaller pieces. And I'll show you what I mean. I know this is vague, but OK, so here's what you're going to do. So 2 to the 20th, OK, what we're looking for is we want to see what the remainder is when we divide 2 to the 20th by 41. It's a big number. So we want to break it down into a smaller number, maybe to a power, and see what we get. Well, here's what we can do. So 2 to the 5th. Now you may say, well, why am I choosing 2 to the 5th here? And how am I getting this? Once I do a couple of these, this will make a little bit more sense. And I'll kind of tie this in, and hopefully at the end of the problem. Well, OK, so here's what you're trying to do. You're trying to get a power of 2 that's close to 41. So that mod 41, it becomes a small number. That's the idea. And what we really want, you're going to see this at the end. I know this may seem mysterious right now. We want this power to be something that's going to divide the bigger power. And you'll see why in a second here. So what we want is write 2 to a power that divides the bigger power that we're interested in, and pick that number so that it's close to the modulus number, so that modulo, whatever that number is, it's relatively small, OK? So what is 2 to the 5th? OK, well, 32. OK, so what is that congruent to mod 41? Well, you might say, here's the point. OK, you've got to be a little, we'll cover here. So 32 mod 41, 32 is, well, what's the remainder when you divide 32 by 41? Well, it's 32. But we can actually rewrite this. And it is 32, but it's also congruent to, I'll let you think about this for a second, minus 9. Why am I getting the minus 9? Or why am I doing this? Well, OK, 32 by itself, if you just write 32, it's congruent to 32 mod 41. That doesn't really tell you anything. Every number is congruent to itself, mod anything. It doesn't matter. The minus 9, though, now we're getting down to smaller numbers. And again, all you have to do is just think about this. This is 32. What does it mean for this to be true? It just means that 41 divides 32 minus 9. 32 minus minus 9 is 32 plus 9, which is 41. So once you've got this, and I'll try to be clear where these are coming from. So by f, again, remember what we're interested in right now is 2 to the 20th. That's what we're interested in. 2 to the 5th to the 4th is congruent to minus 9 to the 4th. Mod 41, right? OK, so what's the point here? So 2 to the 20th, then, is congruent to minus 9 to the 4th, which is the same thing as 81 times 81, mod 41. You see that? Minus 9 to the 4th is minus 9 squared times minus 9 squared. Both of those are 81. So we're breaking this down into numbers that we can manage. That's the main idea here. So now, just note that. OK, so let's look at 81. Let's see if we can figure out what this is. Mod 41. So here's what you want to be thinking about. So we've got 2 to the 20th being congruent to 81 times 81, mod 41. But let's see if we can get 81, mod 41, to be something smaller, mod 41. So the point is, 41 almost divides 81, but it doesn't quite. It divides 82. So if I write a minus 1 here, this is certainly true. Just think about what this means. It just means that 41 divides 81 minus minus 1, which is 82. So that's true. So again, by f, again, above, and I mean theorem tube, I'm running out of room here. 81 squared is congruent to, OK, I'm just going to write this. Hopefully you guys can see where this is coming from. I'm just squaring both sides. By part f, I can take either side to whatever power I want. Minus 1 squared is 1. So I get this, right? OK, we're almost done, in fact. So let's see what we've got. So we've got 2 to the 20th is congruent to, so by this. So this is congruent to 81 squared, which we just found is congruent to 1, mod 41. And I'm going to quote this theorem just so that you can see where this is coming from. I can find it. We talked about this last time. V is congruent to b mod n, and b is congruent to c mod n, and a is congruent to c mod n. This is part c of theorem 2, OK? So by part c, 2 to the 20th is congruent to 1 mod 41, OK? So really, I should have probably written this in the beginning. This is really all you need. This is really all you need. You don't need to get the 0. I really should have just had the 1 on this side. But so what's the point, though? What's the point once we've got this? Is that by definition, 41 divides to the 20th minus 1. That's exactly what we wanted to get. OK, so that's it. So 41, you might want to write there for 41 divides 2 to the 20th minus 1. Now we're done. OK, so you can kind of get a feel for, well, it might take a little practice. But what you're doing is you're breaking these things down into simpler pieces, OK? And then, well, if I have time, I'll do another one at the end of class today. OK, so there's another theorem here, which I don't know. We'll see how much of the proof I go into. I want to save some time for homework, because I think this modulus stuff, modular arithmetic, is really important to get to. This comes up a lot. And you're going to see this. Those of you who are taking abstract algebra, you're going to see this kind of idea come up. So I want to make sure that we spend some time on this today. OK, everybody have this? Does anybody have any questions here about this? Yeah. OK, right here? Yeah, I was just curious. Oh, OK. Yeah, I know roughly what you're getting at. And I mean, yeah, I mean, there's a sense. I know what you're thinking. There is an, this is kind of an analog of that, in the sense that if you have a congruence, you can add whatever you want to both sides, and it preserves it. And you can multiply by a scalar, and it preserves it. So yeah, I mean, I'm not really going to go into the vector space analogy, because congruence modulo of prime actually gives you a vector space over a certain. But this is going to get really into abstract algebra, which I kind of want to steer clear of right now. But yes, there's sort of that idea. This is sort of an analog of that. So yes, I mean, in a sense, that's right. Although there's no real vector space going on here, but the idea is similar. Yeah. OK. So unlike a homework problem, if we worked it out, similar to that, that would be acceptable. For sure. Yeah. OK. So here's the last theorem that we're going to do. I told you that before. I think. You know, you just worked it too well. OK. So this just says, suppose that CA is congruent to CB mod n, then you can sort of cancel the C in some sense here, but not exactly. So A is congruent to B mod n divided by D. Yes. Where D is equal to the GCD of C in it. OK. I think that, yeah, I don't. I have that. Well, no, I do. But I was going to write out the proof. But the proof is there's a lot of letters and it's kind of messy. So I think I'm going to not write out the proof of this. OK. Of course, proof's in the book. But, you know, and normally, of course, I pretty throw with writing out proof. But me not writing out proof is not going to hurt you in the homework at all. OK. So I will tell you this, though. I will give you a corollary. And the corollary is the one that's going to be used, I think, most often. So the moral of this theorem is that if you have a congruence, we have the same C on both sides. You can cancel it out, but you have to change the modulus to n over D, where D is the GCD of C in it. So whatever you're canceling, you have to take n divided by that, and that's what you have to change, the GCD. So the corollary, I'll tell you what. Here's what I'm going to do, actually, because the corollary is, like I said, going to be the most useful of the two. And of course, corollary, this just means it's a special case of this theorem. And I'm actually going to prove this, because it's very easy. It more or less follows right away from some of the stuff we've done before. So suppose that CA is congruent to CB mod n, and the GCD of C and n is 1. Then in fact, you can just cancel the C, and you don't have to change the n at all. So then A is congruent to B mod n. OK. Well, this really does follow right away from the previous theorem, because the theorem says that you can cancel out the C, but the modulus has to change from n to n over d, where d is the GCD of C and n. Well, if the GCD of C and n is 1, then you haven't changed it, right? You're just dividing by 1. So it's the same thing, right? So this may actually help you out in one of the homework problems. So I'm just going to write this really quickly. This just comes from stuff we did already. It's very short. So let's just assume that you notice that I'm not writing C, A, B, or N, Z, and n is an n. I'm just not writing that now. It's assumed that, remember, it's assumed that this guy, whatever this guy is, is a natural number. It can't be negative. But everything else can be an integer whatsoever. OK. So assume C, A is congruent to C, B mod n, and that the GCD of C and n is 1. OK. Well, we're going to show that n divides A minus B. So what does it mean for C, A to B congruent to C, B mod n? It means that n divides C, A minus C, B, right? OK. IE n divides, well, we can pull out a C, right? C times A minus B. Shouldn't be too hard, right? That's just definition of mod n. It means it divides the left-hand guy minus the right-hand guy. Just factor out the C. OK. And now this goes back to a theorem that we learned a while ago. Hopefully most of you remember this. This is just saying that C and n are relatively prime. And this is a corollary from chapter 2, I think, or a theorem from chapter 2, is that if you know that a number divides another number times something else, and if these two are relatively prime, then you know that this number divides that number. Right. Maybe you remember this from a while ago. I don't remember the theorem at number or whatever. But this is something we did in chapter 2. C and n. OK. So let me just restate this. Well, you would probably know better than me. So yeah, if c and n are relatively prime and n divide c times something, then because c and n are relatively prime n has to divide that other guy. So in other words, let me use other letters. That's lemon. OK. So this is just page 24. OK. Yeah. OK. So yeah, this is Euclid's lemon, page 24. Yeah. So this is one I actually remember now. I gave this to you, but I don't think I wrote the proof out. But yeah, it's Euclid's lemon, page 24. OK. So since the GCD of c and n is 1, and I'll just go ahead and use that here, it follows from Euclid's lemon that n divides a minus b. OK. So a is congruent to b mod n. That's what it means. And so the moral of this is that if you have a congruence and you have a scalar on both sides that's relatively prime to n, then you can just cancel it. So any questions here? OK. So here's what I want to do. I'm going to just stop. This is the end of the lecture on this stuff. So what I want to do now is talk about a few of your homework problems, give you some guidance as to how to begin some of these. OK. So number, I think the first problem I sent was number three. All right. So we have this copied now? Yeah. OK. OK. So this is definitely going to be important to pay attention to. All right. So let's see where we are here. 4.2, I guess, number three. Yeah. So this just says if a is congruent to b mod n, then prove that the GCD of a and n is equal to the GCD of b and n. I'm going to give you some advice here. I'm not going to do the whole problem, but I will tell you the strategy here. So for some of you, you see this, you want to prove that the GCD of a and n is equal to the GCD of b and n. You think, well, I don't know exactly how to go about doing this. Well, the first thing to do just for simplicity's sake to economize is to just, instead of writing GCD of a and n, just call it, for example, d1. And call the GCD of b and n d2, for example, or d and d prime, whatever you want to do. But just it makes. Yes, because they, well, yes, yes. But you don't want to call them the same thing outright, because then you're saying they're the same, but before you've even proven that they're the same. So the point is in the beginning, you don't know that they're the same. You don't want to use the same letter. But then you prove that those two things have to be equal at the end. OK, so here's the first thing that you want to do. So, and again, you can use whatever you want, whatever letters you want, but you definitely want to make them different. So let's let d1 equal the GCD of a and n. And let's let d2 be the GCD of b and n. Again, this is, I'm not writing out the whole proof, so you're going to have to write this out in a nice way. So we use complete sentences and all that kind of stuff. Here's the idea. OK, now this is something I know some of you have been struggling with, and I understand why. This is a little bit tricky. So the GCD, and I want you guys to listen to this, is going to help you, I think. How do you work with the GCD? How do you prove that d1 is equal to d2? Well, there are a lot of facts about the GCD. So let's just look at a and n for now. So what do we know about d1? Well, d1 divides a, d1 divides n. And we also know that d1 is a linear combination of a and n. We also know that any number x, say, that divides a and n, also divides d1. So there's lots of facts floating around here about the GCD. So the question is, how do you know which of these to use? So here's what I will say, first of all. And this is just a general rule of thumb. I'm not saying it will never apply, but generally speaking, this is what you want to do. A lot of you guys are using this linear combination fact. And maybe from linear algebra, you just feel like, oh yeah, I remember that. I know about linear combinations. Well, for most of these problems, this linear combination idea is just not going to be relevant. It's just going to make things a lot messier. And you're going to end up with 18 variables. And then your heads are going to start spinning. Generally speaking, the main time when you're going to use this linear combination fact is when the GCD is 1. When the GCD is 1. If it's not 1, don't try that first. It's probably not going to help you. So I really want to make sure everyone's clear on this, because I'm seeing a lot of this in the homework. And then when you start doing that, like I said, once it gets all messy, your brain shuts down after a while. So you're not going to use the linear combination fact here. That's not going to come up. So here's what you want to do. So the first thing you want to do is you want to show that. And this is not hard. I can tell you, this does not require much work. Show that d1 divides b. And OK, the second part is a little silly, actually. But and d1, that sucked. OK, that's a d, divides n. Right, that's why I said it's a little silly. But yes, you don't have to show the second part. You already know that, because d1 is the GCD of a and n. So it automatically divides n. This is the part that you need to do a little bit of work on. It's not hard. Use the fact that a is congruent to b mod n, n divides a minus b. It doesn't require much work. It really doesn't. So what can you conclude from this? OK, now this is where you are using something about the GCD that we know. If d1 divides b and d1 divides n, then it divides the GCD of b and n. That's something that maybe some of you forgot about. But anything that divides the two guys divides their GCD. That is in your notes. We've talked about that before. So the conclusion then is that d1 has to divide d2. Any questions about this part? You guys understand what I'm saying here? Show that d1 divides the two guys. And again, just from the theorem that you already know about the GCD, it divides their GCD. That's where this last part is coming from. Yes, d1 divides b? Yes, yes, yes. You'll use the fact that n divides a minus b and then it's not hard. It just takes one or two lines and you can get it very, very quickly, very quickly. And then you might guess what the next strategy is. You do the same thing the other way, right? Show that d2 divides a and d2 divides n. OK, again, the d2 dividing n is just automatic from the definition, right? You don't have to do any work here. Then conclude that, conversely, that d2 divides d1, right? If d2 is dividing both of these guys, it divides their GCD just like before. So now what can you say? Well, OK, I'm going to leave it to you to finish this. But what do we know about d1 and d2? They're positive. They're both positive. Think about positive numbers, just positive integers. If you know one divides another and the other divides the first, right? So that's the idea here. So, yes, I see what you're saying. Yeah, at this point, really, you can just say it. There is actually a theorem back in chapter 2 where if a divides b and b divides a, then a is plus or minus b. This is actually a true-false question I give you on the first test. But because they're both positive, they actually have to be equal. You can't have one being the negative of the other because they're both positive. But I don't expect you to quote that at this point. Are you guys with me? Are you guys with me on this? OK, OK. So I want to just stress to you, and maybe a lot of you are hopefully, some of you are looking at the website and seeing the solutions to the homework, the test, and such. And maybe you're having the sense, oh, it's simpler than I thought it was. Or I wrote all this stuff out, and this solution was a paragraph. It's so much easier. Things, the hard part about this is just sort of knowing what to do. But once you have an idea where to go, most of these problems are not as hard as they might seem at first. So I'm trying to just give you an idea of how to strategize here. Yes? Is it safe to, like she was asking, not do the in-detail from chapters previous and only use the more in-depth proofs for the front? If it's, yeah. I mean, if you. Knowing where to stop, like I just assumed that. Well, yeah, I mean, this is something that I can't give a completely precise answer to. But in general, yes. I mean, if you say something that follows from stuff that we've already done, ideally, I would like you to mention it. But if you don't, I'm not going to take offense. I think you're a homework, but you don't, like, do you like theorem? No, no, no, no, no. Definitely not on the exam. No. But yeah, something like this, I'm not going to worry about. I'm not going to get too nitpicky about. You know? OK, so why don't we, let's see, what else do we want to talk about? Oh, yeah, yeah, sorry. On 16, it says use the theory of congruences. Is that to use mama? OK, to verify. Yeah, so this is very similar to the example I did. Just use mama. Yes. I looked for what is the theory of congruences, and so I just make sure there wasn't a particular thing I was supposed to use. No, yeah, I mean, so remember the example that I did. Was that 41 divides 2 to the 20th minus 1? It's like that. It's like that. Yeah, so you're just having to have to figure out how to sort of break it down. I'm going to talk about another one of these here, too. Probably not that one, but I'm going to do one that's maybe even a little harder than that one. OK, so because I think it'd be good for you to see more than one of these, because when you're first starting this, this can be a little tricky to see what to do. OK, so here's what I'm going to do for you. Um, I'm going to, so number five actually is two parts here. So I'm going to talk about the second part of number five just again to give you an idea of how to think about this the right way. I'm going to leave the first part to you to figure out. But so this is this, I want to be very clear here. OK, this is just part of number five that I'm doing. So don't forget to look back on it. There's more than just this in number five, so don't forget that. So the second part is to prove that 111 to the 333rd power, plus, and then you're just switching this, plus, 333 to the 111 is divisible by seven. OK, OK, so again, I'm going to give you the idea. And again, I want to be clear here. I'm just kind of being informal now, just jotting down the ideas here, your actual solution should be more polished than this, OK? So we've got two parts here. And so what I can do is just kind of look at these individually mod seven and see what I get. So for example, for example, if I know that 111 to the 333 is, say, I don't know, minus one mod seven, and 333 to the 111 is positive one mod seven, then when I add them together, I get zero mod seven. And that means that it's divisible by seven. It leaves a remainder of zero upon division by seven. So we're going to look at these two pieces separately mod seven, and then we're going to put it together at the end. And I'll quote the appropriate theorem so that you can see where it's coming from. OK, so first thing to do is to, and you can just, what I'd suggest that you do, if you're not doing this already, have a piece of scratch paper and just play around with it for a while. Don't think that you're going to go right to the solution. And you're going to erase nine times if you do that. Just put it together. You're going to erase nine times if you do that. Just play around with it until you can see what's going on. And then once you have the pieces put together, then write it up. Because I see some of your, I can see that you've erased 15 times. The lines are all gone. And you're going to save yourself some time if you do it that way. OK, so consider. Yeah, and that too, yes. Thank you. All right, so this is going to require, let's see. I wrote this down. A little bit of work. So here's what you're doing. What you want, for example, is kind of think about what we did before. It would be really nice if 111 was 1 mod 7. Because then 111 to the 333 is 1 to the 333, which is 1. So it makes it very, very simple. Well, it's not 1, but it's close to it. So what you want to do is just think about dividing 111 by 7. That's what you want to do. And see what the remainder is. That will give you an idea. It's certainly going to break down a lot more, right? 111 mod 7 is either 0, 1, 2, 3, 4, 5, or 6, for sure. So we're hoping it's 1 or minus 1, because then the powers are easy to deal with. That's the idea. So here's, and this is just after you look at the remainder, this isn't very hard to, I don't think it's that hard to get on your own here. So this is, 111 is congruent to minus 1 mod 7. OK, so here's the thought process that goes into this. If you divide 111 by 7, OK, here, why don't I? And remember, this is just informal now, so I'm not saying this needs to go in your solution, which it certainly doesn't. But OK. So what? We got 1. OK. What's that? I think 7 times 15 is 105, right? OK, so yeah, well, I just want to show you where this comes from here first. So what do we get? We get 41, right? You guys learn this at some point, I hope? OK. I don't know what they teach anymore. I'm old. I don't know. OK, so 15, 5 times, OK, so we get 35. They don't. Oh, OK, yeah. I guess this is too complicated now. Well, that's OK. The point is that the remainder is 6. That's the idea. So here's the sort of clever part here. So the remainder, when we divide 111 by 7, is 6. So we do know that 111 can grow into 6 mod 7. But when we try to take 111 to 333 power, if we write 6 here instead, we're going to still be stuck, because it's not really going to work out very nice. So the key to remember, though, is that 111, well, there's a remainder of 6. So in some sense, it's still 1 away from a multiple of 7. OK, that's the idea. So I mean, think about it, 7, 14, 21, 28. Well, 27 has a remainder of 6 when divided by 7. But it's only 1 away from another multiple of 7. And so that 1 away, that's how we encode it. We just write minus 1 here, because then when we add it, it's divisible by 7, right? Yeah? So can't we just say 6 minus 7 is just negative 1 in a similar manner? Say that again? So can't we just say, well, it's a remainder of 6? Because we're playing in 7 world at the moment. We can just do 6 minus 7. Oh, that's also minus 1. Yeah, sure. Yeah. So the point is that 6, here's the idea. 6 and minus 1 are the same modulo of 7. Yes. But what I want to stress right now, though, is just that, OK, what is this really saying? This is just saying that 7 divides 112, OK? So what you're trying to do, the general strategy is, if you're trying to make these things simpler, sorry, if you're trying to make these things simpler, then you want to reduce this down to something small, and minus 1, at least an absolute value, is smaller than 6. So we want to use minus 1 here. Then what the modulus is? Yeah, it's going to be smaller than that. Oh, OK. So depends. I mean, if we got a remainder of 1 instead of minus 1, then instead of 6, rather, then we could just put the 1 in and we'd be set. But the point here is that once we know this, 111 to the 333, then is congruent to minus 1 to the 333, mod 7, right? This is by, what was it? Theorem 2? Was it Theorem 2? OK. Well, this is nice, right? Because now we know that 111 to the 333, then, is congruent to, well, this isn't hard, right? Minus 1 to the 333 is minus 1. It's an odd power, OK? So this is congruent to minus 1, mod 7. OK, so that's nice. And so this takes care of the first part. Now, what we want to show is that the sum is divisible by 7. In other words, we want to show that this sum is congruent to 0, mod 7, right? Congruent to 0, mod 7. Well, we know that this first guy is congruent to minus 1, mod 7. So what do we want, really? We want this to be congruent to 1, mod 7. Because then when we add them, we get 0. And that's exactly what we want to have, right? Does that make sense? So that's what you're trying to show. I'm going to leave that to you. That's what you're trying to show. You're trying to show that this is congruent to 1, mod 7. And you're just going to have to break it down again and do some work. Yes? I want to make sure. Somebody might have already learned this. No, that's OK. When you pick the first one, whatever it is, like the number, mod 7, this example, you just want the opposite sign for the other one. It could be anything and then less than whatever. Yeah, right. So that's right. So in general, if you want a certain number to divide something, say you have two pieces added, if you get the first piece being, say, 5, then you'd want the other one to be minus 5. So you can add to get 0. Yeah. So now what you want to do is show that 3, and it actually is true. 333 to the 111 is congruent to 1, mod 7. OK, now let me make sure I'm going to, again, I want you to kind of see how all this up fits together. So I would like you to quote theorem 2 when applicable. So then by, let's see, what is it? I think D, theorem 2 D. If you look back to theorem 2 D, it says that if you have the same modulus, a's congruent to b mod n, c's congruent to d mod n, then you can add on both sides. You can just add them like it's an equation, basically. So part D says then once you've got this, that you can just add these together, and this sum is congruent to the sum of these two guys mod 7, which is 0. And that's exactly what you want to get it to be divisible by 7. So whenever you have two congruences, as long as it's the same modulus, you can just add on the left and the right side, and you get another congruence. And so since this is minus 1 and this is 1, we get 0. And of course, that just means that 7 divides this. Is that OK? Yes. So then once we add everything together, once you, yeah. I mean, so as long as you have something congruent to 0 mod whatever, it just means that that modulus divides whatever's on the left-hand side. Because when you subtract 0, you don't change it. Do you think your own 4.2f kind of just knock the powers off anyway, or kind of like this? Well, no, no. You're thinking the converse of f. f says that if a is congruent to b, then you can raise them to powers. But it doesn't say that if you have the powers, you can cancel the powers out. No, it's just one direction. Yeah. OK, is this all right? OK, so what I want to do, and the other one, the 333 to the 111 will require just a little bit more work, but you're still kind of doing the same kind of idea. What I will say about the second part is, here's my, OK, can I go to the next page now? Is that OK? Here's where I would start. Not that it has to be done this way, but I would start by, OK, so you want to know what 333 to the 111, or sorry, what is it that you're doing? Yeah, 333 to the 111. You want that to be 1 mod 7, right? So what you want to do is see what 333 is mod 7. So you can do this, you can easily just check this out, just by doing the division. This is congruent to 4 mod 7. So this is another way of writing in. This is also 2 squared, right, mod 7. Now if you raise both sides to the 111 power, you're going to get some big power of 2. But OK, here's the last thing I'm going to say. Here's the last thing I'm going to give you. If you go through it this way, this will come up. 2 to the third is 1 mod 7. 2 cubed is 8, right? 8 is congruent to 1 mod 7. So there's where you're 1. And now you can raise the powers, and then you can get it that way. Does that make sense? Just, OK? Can you say that? 2 cubed is congruent to, I'm not giving you all the details here, but this is congruent to 1 mod 7. And again, if you guys are new at this, just think about the definition. What does this mean? It means that 7 divides 2 cubed minus 1. That's true, because it's 8 minus 1, 7 divides 7. And now you can use that result that says that we can raise both sides to the positive power we want. It's preserved. Of course, you need the 111 power, so you're going to have to mess with it a little bit. But that's the idea, OK? OK, so I've done most of this for you now. Here's probably the last thing I'm going to talk about. Let's see here is number 17 says that suppose that there's quite a few pieces to this. A, B is congruent to C, D mod n. And B is congruent to D mod n. OK, there's one more. And the GCD of B and n is 1. Prove that A is congruent to C mod n. Again, I'll give you the idea here. I will tell you, first of all, just a hint. You do not need to change this into divisibility notation. In other words, you shouldn't start the proof by saying, OK, n divides AB minus CD. n divides B minus D. You don't need to do that. You can just use theorems that we've done. You don't have to change any of this around at all. In fact, you can prove this in just a few lines just by using some of the theorems that we've seen so far. Here's the hint I'll give you. Here's what you want to do. And you're going to use theorem 2. So what you're going to do instead is you're going to prove that AB is congruent to, well, let's see. I guess it doesn't. So I'll just put it on the other side. BC doesn't matter, right? It's multiplication is commutative. It's just nice to go in alphabetical order here. But this is what you're going to prove. This is what you should be trying to prove. Now look at the assumption that the GCD of B and n is 1. So if you look back to the previous corollary, corollary says that if you've got the same guy on both sides and it's relatively prime to the modulus, then you can cancel it. That's the corollary that I gave you before. It's in your notes. I think it's just the previous corollary that we talked about, right? So let's see. Here, are you OK? OK, OK, you sure? Let's see if I call this corollary 1 or corollary 2. I don't think I did, did I? I just called it corollary. OK, and so here, you're just going to use these two congruences, basically, and use theorem 2. There's really not much that you need to do here to prove that A, B is congruent to B, C mod n. Sure, I have this right. There's really not much to do here. I mean, really, you're just doing one little thing and applying something in theorem 2, and you're done. It's not a lot of work. It's very, very little work. If you're writing a lot of stuff here, you're doing way too much. Theorem 2 basically takes care of it for you right away after you just have to do one small, little, tiny thing, and then you're done. OK, let's see. I guess we'll do the end here. Does anybody have any questions they'd like to ask? What's an integer? Well, I don't know. What's an integer? Well, that's, I could answer that question, but it depends on what axioms you take for mathematics. And integers are a group, but what actually is it? It depends on how you construct it. Yeah, well, I'm not going to go into that. What's a ring? I can't say what a ring is. Oh, you should be able to tell us. You should be able to answer that. A ring? Is this a tiny thing? Yeah. I don't know. It's a group, second operation. Isn't it a group that's like a two groups or a bijection or something? It's a group. No, it doesn't really have anything to do with that. But if you take abstract algebra, you'll learn about it. Nobody wants to listen to the definition of that right now, so I can tell you. You can look it up on Wikipedia. I mean, you know. I was going to do my abstract book. But yeah, it's just sort of an abstraction from the integers with addition of multiplication. Except instead of adding and multiplying numbers, you're adding and multiplying things that you don't necessarily know what they are. But they have certain properties. That's right. Yes. No, no, you can't do that. Did you pack away your scarves for the summer? No, I still have them. It was warm today, so I just, you know. I just had the idea, and now I'm not saying that. It's a little weird. OK, well, if it's going to really mess you up and hurt your concentration, I'll wear it next time. I'll wear it. I'll do that for you. You can't do it for you. I have a question. A real one? OK. You're concerned with the division algorithm? What kind of remainders we get left over, use multiplication on two numbers, then we get a number that we add to it afterwards? Do we ever get to any kind of thing where we're doing a similar kind of thing by raising the numbers to exponents? Well, I mean, no, not no. I mean, in general, that's not something that works out very nicely. So I mean, you can, this is a question, actually, that is studied in abstract algebra. Something called finite field theory. But it's kind of beyond the scope of this. Yeah, so it's a good question. It's just, yeah, it's something that's going to take us way kind of far away from where we're at right now. But OK, so hopefully this will help a little bit. Kerry? No, not necessarily. I see what you're saying. Yeah, I mean, so you don't have to, for example, in this problem, you don't have to say, suppose A-V is congruent to C-D mod n and A-V, C-D in integers, n's a natural number, that's getting a little bit unwieldy at that point. What's implied in that incident? I mean, it's sort of implied that we've only defined these things for integers anyways. So yeah, I mean, don't concern yourself too much about that, I'm not going to get too picky about that. But if you start introducing your own variables, then you should say where they are. That's all. So question then, that does not apply if they give you two things, A and B, they're integers. And you say, let D be GCD of A and B. I don't have to go out of my way and say, OK, and D has to be an integer. No, no, no, no. No, that's the definition of GCDs. Yeah, that's automatically given by the, that's implied by the definition, that it's a positive integer. Yes. Yes. Yeah. OK, so 12C, is that the only part I gave you? I think it is, right? The product of any set of n consecutive integers is divisible by n. Here's, OK, maybe we'll finish up with this here. Here's what you guys should be thinking about, OK? Don't go right to the abstract problem where n is just some arbitrary positive integer. Think about it first for small values of n, and then see if you can extrapolate from that to see what happens in general. So when n is 2, it says the product of any two consecutive integers should be divisible by 2, right? OK, well, of course, we've already seen this before, but I'm just going to talk about this, right? x times x plus 1 should be then, in congruence notation, should be congruent to 0 mod 2, right? Well, why is that true? Well, one of them, this problem really boils down to the division algorithm, really, OK? One of these guys has to have a remainder of 0 when you divide by 2. They're consecutive, right? So if x has a remainder of 0, then x is even, and the whole thing is even. If x has a remainder of 1, then x is 2k plus 1, so x plus 1 is 2k plus 2, and then you can factor out a 2 and get that x plus 1 is divisible by 2. And you can play the same game for 3. x times x plus 1, I'm going to answer your question in a second, but x times x plus 1 times x plus 2. Well, OK, one of those has to be divisible by 3 for the same reason, the division algorithm, right? If x is divisible by 3, you're done. If it has a remainder of 1, then x plus 2 is divisible by 3. If it has a remainder of 2, then x plus 1 is divisible by 3. So this is the idea, OK? And because there are only n possible remainders, one of those guys is going to have to be 0. Has to be. That's the idea. So in general, how do you write a product of n consecutive integers? OK. But you have to be careful, though. If you want to write this out the right way, OK. Here, let me just ask you this. So suppose I have this. Well, consecutive, it just means we keep going up by 1, right? x times x plus 1 times x plus 2 times, OK, x plus. Is this the product of n consecutive integers? All the way up to x plus n. Sorry, Joe. There's one more, right? That I don't want, OK? So I have n plus 1 terms here. This is not a product of n consecutive integers. It's a product of n plus 1 consecutive integers. So, well, the easiest thing to do here, really, if you want to represent the product, you don't have to just go up to x plus n minus 1. You can if you want to, but the easiest thing to do is just give it to this guy. Because every integer, the left of the first one, the first x, because every integer can be written as a form of something plus 1, right? x minus 1 plus 1. So this is the probably the most compact way to express the product of n consecutive integers, just like this, OK? You could also just cut off the last guy and just say, you know, x plus n minus 1 if you want to do that, that's fine too. But then just so, again, my advice to you is when you have some of these statements, for example, this problem that you asked about, you know, the product of n consecutive integers is divisible by n. Well, if you don't know how to do it in general, play with it. Say, what happens when n is 2? Why does it work when n is 2? Why does it work when n is 3? Why does it work when n? Then once you see the pattern, then you should be able to get it in general. But that's what mathematicians do. They want to prove something. They don't go at it right away, because you may not know how to do it. Look at special cases that are simple, and then try to extrapolate the pattern to the general case. That's how it's done, really. OK? All right. Well, I think we're going to go ahead and stop there for today then. So like I said, tests, homework, and such.