 No. No class on Tuesday. At least I think that's how it works here. I think finals week there's no class. I hope so. Okay. So yeah, there'll be no class. I mean, you have a week just to, you know, go over stuff. And then if there's a real big problem with getting the solutions to the third test post, I'll just probably just put it in an email and just send it out to everybody as a PDF. Okay. But I'm hoping it won't have to go that far. All right. So why don't we continue talking about 5.2 here for a minute. I'm going to mention a few things about some of your problems. This I could just have you do yourself. And I'm going to talk about this a little bit. I did last time, but if you remember number 2B, and I only did part of this, but part of this problem, okay, this is just a part of the problem, not the entire problem, essentially amounts to you proving that 8 divides 8 to the 6th minus 1. We talked about this a little bit last time. So I guess the original problem was to show that 8 to the 6th is congruent to 1 module of some bigger number. But the whole point is that you break the modulus down into its pairwise relatively prime pieces, right, by factoring, and then you just have to do it, you have to prove this result for each of those pieces and then put everything together at the end. Okay. And I gave you a hint. I kind of just did this off the top of my head. It's a good hint because it works out pretty well. This, I'm going to tell you this just because I think this is actually a really pretty cool solution. Okay, so 8 to the 6th minus 1 is the same thing as, I may have written this for you yesterday, A cubed squared minus 1 squared, right, that's for sure, definitely true. This is a difference of squares, so you can factor this as A cubed plus 1 times A cubed minus 1. Yep, and I know I did this before. And I told you, some of you may have forgotten this, but I told you that both of these can be factored because both of these are, well, the first one is a sum of cubes and the second is a difference of cubes, right, because one is the same thing as one to the third. You can factor both of these and if you forgot how to do it, nice thing is we have the internet now. You can easily just go on Google and say factoring sum of cubes or difference of cubes. You can get all these formulas in, you know, three seconds, right. So this actually factors as, the first part factors as 8 plus 1 times A squared minus A plus 1. Let's see if I can squeeze this in here. And the second factor is as A minus 1 times A squared plus A plus 1. Okay. Okay, so what was the assumption in the problem, which I didn't write this down, but the assumption, somebody can tell me if I screw this up because my book is closed, is that in the beginning of the problem is that the GCD of A and 42 is equal to 1, right? I get that right. Now, and I think I mentioned this last time, but if A and 42 are relatively prime, that means that A can't be even because then 2 would divide A and 2, of course, divides 42. So the GCD couldn't be 1. It'd have to be at least 2. So A is odd, right? And again, I'm just jotting down notes here. So A is odd. Now, here's the idea. This is cool. All right. You want to prove that A divides A to the 6 minus 1. And we have A to the 6 minus 1 factored like this. See that? Well, if A is odd, then A plus 1 is certainly even, right? And A minus 1 is even. And now think about this. I want you to think about this for a second. What do we know about A plus 1 and A minus 1? Well, A plus 1 is exactly 2 bigger than A minus 1. So these are consecutive even integers. Think about pairs of consecutive even integers. 0, 2, 6, 8, 14, 16, 22, 24. One of them is divisible by 4. You can prove that. And you should if you want to do it this way. If you just doodle some examples, you should convince yourself pretty quickly that if you have a pair of consecutive even integers, 1 is divisible by 4. And you guys actually should all be able to prove that. If you have basically what it boils down to is just if the integers even, start with the A minus 1, the smaller one. It's even. It's of the form 2k. Well, that k is either even or odd. If k is even, then it's of the form 4k. And then it's divisible by 4. If k is odd, then you add 2. And you get that it's divisible by 4. So one of them is divisible by 4. And again, this is something you guys should be able to prove. This is not that hard. So think about this. And I'm actually just, I'm not going to write anything else down. Think about this. They're both even. Both of these guys are even. And I'm telling you one of them, at least, well, it's not going to be both of course. Exactly one of them is divisible by 4. But the other one is still even. So you can extract an 8 from this expression. You see that? Slick. This is slick proof. So that's how you can do it. Oh, I don't know. This is probably, I mean, I don't know. I mean, I can't say that this is the best possible proof. But I can't imagine it gets much easier than this. This way you avoid using the division algorithm dividing by 8 and all that kind of nonsense. This way you can just get it really quickly. Because you're trying to prove that 8 divides 8 of the 6 minus 1. Don't forget what we're trying to prove. One of these guys is divisible by 4. The other by 2. So we've got a 4k and say a 2l. So now we can get an 8 to be pulled out. So 8 divides the whole thing. Does that make sense? So fill in the details here. And yeah, the other part, I think, is just essentially just for Moz Theorem to get the other two factors. And then this is the one where you just kind of have to play with it a little bit. So you see kind of how to proceed. But I just wanted to show this because I thought this was kind of a neat way to do it. Okay. Any questions about this one? Okay. Now, I'm also going to talk about a couple more of these. And again, I apologize again for kind of screwing things up on Tuesday. But actually these problems all follow essentially the same format. And for the most part, there's not a whole lot going on with these. You just have to really notice one thing and then apply for Moz Theorem. And that'll solve almost every one of these. Almost every one of them. I mean, here's an example where you actually had to use some trick or you could get it to work out. But most of them just involve doing what I said. So let's see. Everybody have this down now? We good? Okay. So let me talk about, I think maybe I feel like I started talking about this on Tuesday. 4D, show that A to the ninth is congruent to A mod 30, right? For every integer A. Okay. Two parts to this. And what I'm going to do here really will apply to more than just this problem. We talked about this a little bit before. But first thing you want to do, and I know this is a little bit vague, but you want to take your congruence and you want to split it up into, okay, I'm just going to be informal here, into pieces by factoring the modulus. Okay. So what do I mean by that? I just mean, right. I think I said this to you guys on Tuesday. I'm not mistaken. This is an if and only if. 1, A to the ninth is congruent to A mod 2. 2, A to the ninth is congruent to A mod 3. And 3, A to the ninth is congruent to A mod 5. And I'm going to pause here because I want you guys to, I really encourage you to listen to what I'm going to say here in a second. This is almost as fundamental to doing these problems as the actual for mod's theorem is, really. Okay. And actually, I think it was you, Matt, that asked about this sort of idea on Tuesday. So think about this. Just go back to the definition. This is nothing deep, really. It isn't. Just go back to the definition of congruences. What does this mean? It means that 30 divides A to the ninth minus A. That's what that means. Well, 30 divides something. Sorry. Hang on. Okay. So if 30 divides A to the ninth minus nine, sorry, A to the ninth minus A, then all the factors are going to divide it as well. Two, three, and five, okay? And conversely, if two, three, and five divide A to the ninth minus A, then by a theorem that we have talked about before, their product, because they're pairwise relatively prime, we can squish them all together and the product's going to divide it, right? Okay. So all we have to do now is establish these three different conditions, okay? So what about number one? How do we do that? Well, these now you just apply for mod theorem two. That's really what it boils down to. Okay, so the first one or the corollary, right? A squared is congruent to A, right? Mod two. This is, actually it's a corollary, but this is for mod, for mod theorem. So when you have this, remember this, for any prime P in any integer A, A to the P's congruent to A mod P, right? So once you've got this, well, what can you say? I'm not going to go through all of this, but you can raise both sides to the fourth power, right? So what do we get? We get A to the eighth is congruent to A to the fourth mod two, right? Okay, but from here, right? This is where we started. We know that A to the fourth, if we score both sides, right? A to the fourth is congruent to A squared mod two. Does that make sense? So we know that A to the eighth is congruent to A to the fourth, which is congruent to A squared, which is congruent to A mod two. And so I'm going to stop with this one, but you can see how this is going to play out probably. Once you know that A to the eighth is congruent to A mod two, then A to the ninth is congruent to A squared, and A squared is congruent to A. Basically, the point is once you know that A squared is A, then A to everything is A. Basically, that's what it pulls down to. Yeah? It's easier to understand what was going on by instead doing if A squared is congruent to A, using the proper two E, we can multiply both of them by A, to get A to the third is congruent to A squared. I'm going to repeat that process, and the fourth is congruent to A to the fourth, which is A to the ninth. Yeah, I mean, it's, yeah, I mean, either one's about more or less equivalent. So, yeah, you could do that too, for sure. Yeah. Okay, so I'm going to sort of leave you to finish this yourself, but how do you do the mod three and the mod five? It's the same thing. You just use, for the three, just use the fact that A cubed is congruent to A by four months there. And for five A to the fifth is congruent to A. And then you just, you just, you know, kind of like this, you just kind of mess with it and it just falls out. And once you get all those three parts, then you get what you're looking for. Is this okay? Any questions here about this? About this particular problem? Yeah, okay. It's possible. Yeah, it's possible. Okay, are we okay then with this? Okay, so this is what you're doing for a lot of these problems is you, whatever your modulus is, you want to break it up, in fact, into powers of products of primes and then just attack each of those pieces individually using from us there. That applies to the vast majority of these. Let's see. So, for example, I'm going to do another one here. Six B is to show that A to the fifth and A have the same units digit. And this is going to follow the same, basically the same thing. I actually have to deal with something real quick. I'll be right back. Okay. So, and I think, yeah, there's no assumption on A here. This can be anything. Show that A to the fifth and A have the same units digit. So, this goes something like this. You have the same units digit, and we talked about this a little bit last time. If and only if. Okay, so remember what I said about the units digit? And it's just what this is congruent to mod 10, reduced between zero to nine, right? So, this just means if A to the fifth and A have the same units digit, that means A to the fifth is congruent to, say, C, right? Mod 10. And A is also congruent to C, mod 10. And here, C is just some integer between zero and nine. Okay, so remember, just think about this for a second. If we call C the units digit of A to the fifth, if we just call it C, right? It's some integer between zero and nine. That's all it is, right? Then A to the fifth is congruent to C, mod 10. And if A has the same units digit, then A is congruent to C, mod 10. So, what can we say about this? What can we conclude? A to the fifth, sorry, there's really, I really mean an if and only arrow here. If and only if. A to the fifth is congruent to A, mod 10. That's really what this problem boils down to now. And so, if A to the fifth and A are the same module of 10, that means they have the same units digit. So really, this is, now this is what you have to prove right here. A to the fifth is congruent to A, mod 10. Then you're done. And so, you go about this the same way we talked about with the other problem, right? You factor 10, two times five. So, you just want to show that A to the fifth is congruent to A, mod two. And A to the fifth is congruent to A, mod five. Okay, and again, remember, and this is really easy. A to the fifth congruent to A, mod five is just for mass theorem, the correlator for mass theorem, right? So, we've grown to A, mod 10 eventually? Yes. So, we've broken into two congruences by factoring the modules. Yes. And then we have to show that that's been wrote to both of those. You have to show that A to the fifth is congruent to A, mod two and A to the fifth is congruent to A, mod five. Those are the two things you have to show. I apologize. No, it's okay. That's the idea, yeah. And mod two is easy because A squared is congruent to A, mod two by the correlator for mass theorem. And again, we remember when A squared is the same as A, then A to any power is the same as A. So, that's easy. The other part is just for mass theorem. That's easy. You're done. So, there's not a whole lot going on here. Once you see how to, you know, word the, I mean, how, you know, how to attack it the appropriate way using these congruences, there's not much to do here, really. Okay. Is this okay? Any questions? No? Okay. I'm going to talk about one more, and then I'm going to, I think I'm going to leave the rest to you to finish. But again, since this, you're not going to have solutions to this before the final, I thought it would be good to go through a few more than I normally would. So, 10A says that you're trying to prove something. So, it says to prove the following thing. Prove if A to the p is congruent to B to the p, mod p, then A is congruent to b, mod p. Okay. Sorry, this thing is, I've got to straighten this out. Okay. So, this is kind of interesting. It says that you can cancel the p out in some sense. P's a prime, of course, in this case. So, this is really all it boils down to. There's not a whole lot going on here. You are not, some of you might think that you want to just use the definition and go back to the definition of the congruence and say, okay, p divides A to the p minus B to the p, then p divides A minus B. Just do it directly, or try to do it just from the definition of congruence. You don't want to do that. You want to use the stuff that you have at your disposal from this section to do this. And so, what you want to note is that A to the p is congruent to A, right? Mod p. And B to the p is congruent to B, mod p. And this is just from Pramod's theorem, right? Prove that if A to the p is congruent to B to the p, mod p, then A is congruent to B mod p. So, and this is all there is to it. And you're invoking some theorems, an earlier theorem on congruences that basically say you can kind of change the order around certain ways. But A is congruent to A to the p, right? That's a property we know. We can flip the order. And by assumption, A to the p is congruent to B to the p, right? That's our assumption in the problem. We know that B to the p, over on the right-hand side, this is congruent to B mod p. And there you go. You're done. Use the transitive property of congruence and you get that A is congruent to B mod p. I have a question. Yes? What's wrong, but wasn't there also something in the theorem that you gave us that said A to the k is congruent to B to the k mod p? You could raise it to a power. You could raise both of them to a power. Yes. So couldn't you just use that? No. You're confusing with the converse. If A is congruent to B mod n, then A to the k is congruent to B to the k mod n. But it does not say that if A to the k is congruent to B to the k mod n, then A is congruent to B mod n. It's an if-then statement. It's not an if-and-only-if statement. Right? In fact, there was a question on the third exam that said true or false. If A squared is congruent to B squared mod n, then A is congruent to B mod n. False. Not true. Choose A to B4 and B to B2. Doesn't work. Yes? Right. No, no, I think that's just for the other part of the problem. Here you don't need it. You don't need it in this part. Because, yeah, what we used, A to the p can congruent to A and B to the p can congruent to B are true for any values of A and B. But the last part, I'm guessing, is where that comes into play. Probably. Okay. So really, you've got to, I think I've showed you all the tools, really, that you need to finish up the rest of the homework here. So I'm not going to do the rest of these for you. But it's really, these problems are not that bad once you just see what it is you're supposed to do. Okay. So, anyone want to ask any questions about some earlier stuff or would like to see a problem from, I don't know, maybe an induction problem or something like that. I mean, this is now, this is up to you guys now. Now, I've done my piece here. If you want to ask about stuff, feel free to. Yeah. Right. Yeah, I see what you're saying. I don't think that that is even going, I know what you're worried about here, but I don't think that's even going to come up on the exam. So the problems that you're going to have to use, I mean, really, these are just things that are going to follow from, you know, basic theorems in the section and such. And so, if there's any possible confusion, I'll make some note about it in the directions. But yeah, I know what you're asking, but it's not really going to come up. Yes. Do you just do, I don't know, like chapter one to number 10? It's an induction problem. Okay. Or less than what you're looking for. Okay. Let's see. What is the homework problem? Chapter one, number 10. Okay. Is it 1.1? Yeah, okay. Okay. How many pages is that? Yeah, I see it. Okay. So this is, yeah, page seven of your, your text. Okay. 10A. Yeah. So there are several different kinds of induction problems. You should be familiar with all of them, really. Okay. So it's less than or equal to 2 minus 1 over N. All right. Okay. Well, this is actually a good one, because this, what I'm going to do when I write the solution out here is, this is not going to be totally polished, but I'm going to do this in the way that I think your, the train of thought should sort of progress as you go through this problem. Okay. You're trying to prove this is true for every positive integer N. So remember what you want to do is, there are two steps that you need to follow here. So the first thing you need to, to establish is so-called base case, right? And that is you want to prove that, or verify that if you will. 1 over 1 squared is less than or equal to 2 minus 1 over 1. Right? That's the base case. N is 1, so there's only one term on the left-hand side, and so it just reduces to this. And what is this? This is just 1 is less than or equal to 1, which is, which is true. Okay. So this, this is really just, this first part really is obvious. And the second is the inductive step. Okay. The inductive step then is when you assume that whatever it is you're trying to prove is true for some positive integer N and you want to establish the same assertion for N plus 1. Right? Assume for some natural number N that 1 over 1 squared plus 1 over 2 squared plus on down to 1 over N squared is less than or equal to 2 minus 1 over N. Right? And you want to prove that this same inequality is true with N replaced with N plus 1, essentially. Right? So let's just, this is our inductive hypothesis. That's what this is called. Put a little asterisk next to this. And what we have to prove, this is what we're allowed to assume. What we have to prove then, yes. Well you never write assumed for every N and N because then you've assumed exactly what it is you're trying to prove. You never would write assumed for every N and N that this is true. Because it's like saying, prove for Maas last theorem. I'm assuming for Maas last theorem, therefore it's true. You don't do that. Okay. So what we have to prove is we just keep going. Right? Keep going. And instead of stopping at N, we stop at N plus 1 instead. Right? 1 over N plus 1 squared. And we have to prove that this is less than or equal to 2 minus 1 over N plus 1. Now the idea behind induction is that you want to take your inductive hypothesis and you want to mess with it until you get what it is you're trying to prove. Okay? So here's the idea really. What you're assuming for some N, this is your starting point and you want to get here. You don't want to start and some of you are doing this before. You're starting with what you want to prove and then doing some stuff to that. No, that's not right really. You start with what you assume and you get to where you need to go. Right? If you start with what you're trying to prove then you've tacitly assumed what it is you're trying to prove to be true. If you're working with it, maybe it's not true, maybe it doesn't even exist. That's what you have to establish. You can't get your hands on what you want to prove until you start with the inductive hypothesis and get there through logical steps. Okay? That's the thing I want to stress about induction. Do not mess with what you're trying to prove. Start with what you assume to get there. Okay? So that's the biggest slip-up I've seen when we were doing this before. So... One minus one. Yeah. Mm-hmm. You said that we were essentially trying to... Sort of. I mean, that's not exactly right, but... Okay, so... You're not... I mean, if you were just literally replacing in with n plus one, then you wouldn't have an n. Then you would have skipped from n minus one to n plus one. So I didn't really mean it that way. I just mean the statement, which is going up to n, instead go up to n plus one. So why are you adding an extra term on the left side if you don't have an extra term on the right side? Because... Okay. Look at the original statement here. Here's our original statement that we're trying to prove. So... If I were to call this inequality p of n. p of n. Well... Notice... Okay, this is probably a better way to say it. This guy right here and this guy right here are the same. So if I'm going to n plus one, it still has to have the same form. So this guy right here and this guy right here are going to have to be the same. I mean, imagine starting... Just imagine that n was four, right? Then the assertion would be one over one squared plus one over two squared plus one over three squared plus one over four squared is less than or equal to two minus one over four, right? So it's... Wherever you end up is less than or equal to two minus one over wherever you end it up. That's exactly what the assertion is saying. Okay? So what do we do? Well, we need to get to this point and this has a one over n plus one squared that this doesn't have. So why don't we just add one over n plus one squared to both sides of this, right? I'm not going to write that down, but that's what I'm doing right now. So I just took our inequality that we're starting with and I added one over n plus one squared to both sides of it. That's where I got... That's how I got from here down to here, okay? Well... Now this is why I said I'm going to do this a little bit more informally. What is it that we'd really like to be true in order to finish the proof here? Here, and this is where... I want you to listen to this for a second. Okay? This is important. This is what we have right now. So what we're trying to prove, we have the left-hand side of the inequality right here, but we do not have the right-hand side. This is different than this, okay? So what I would really like to show now, if possible, and think about this for a second, think about it. I really would like to show that whatever this is, is less than or equal to this. Because if I can do that, then think about it. Then we've got what we want on the left-hand side, less than or equal to this, and if I can show that this is less than or equal to this, then by transitivity, this is less than or equal to that. That's exactly what I want to show. That should be your thought process here at this point. So now your goal should be 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. Yes? Is it redundant or equal if you just say less than because there's no way those terms would be equals? Yeah, I mean, well, I mean, you're not really gaining anything by this, really. In fact, maybe they are the same because you can cancel out the twos, and it's not it. I mean, they're not, but let me put it this way. You're not gaining anything by doing that. You might just end up being wrong. So since you only have to prove less than or equal to anyways, why not just leave it there? You see what I'm saying? Okay. Okay, so this is now what you have to show is this. And now here's where this is what you would do now. This is what you would do on scratch paper, and then you would work your argument backwards in your solution. Okay? Well, what is this equivalent to? Well, the twos, we can certainly cancel the twos on both sides, right? So this is really equivalent to just minus one over n plus one over n plus one squared less than or equal to minus one over n plus one, right? And from here, basically what you're going to do is you're just going to keep sort of messing with this until it gets down to something that is obviously true, and then you work your argument backwards. Okay? Okay, well, maybe we don't want these negatives. So we can rewrite this as... Okay, sorry. I'm running out of room here. One over n plus one plus one over n plus one squared less than or equal to one over n. Okay, sorry, this is getting really... But you see what I did? I just brought the one over n plus one over to this side, the one over n I brought over to the other side. Well, that's not really going to get you anything here because the reciprocal of sum of two reciprocals, you can't just cancel the reciprocal out. One over one over a plus one over b is not a plus b. Okay, I think that's what you're saying. Just take the reciprocal and then I get n plus one plus n plus one squared less than or equal to n or something like that, but that's not the reciprocal when you have two fractions. No, no. So, I mean, for example, one is less than or equal to two. If you take the reciprocals, then you get... Then is one over one less than or equal to one half? No, it's not. Reciprocals in general are only going to flip things, but when you've got sums and things like this, yeah, you can't just take reciprocals and just hope things kind of work out. They generally don't. Okay, so what can we do with this? I don't know if you guys want me to finish this or not, I don't know. So, what you could do here is combine these into one fraction, right? Get a common denominator, just by multiplying the top and bottom of this guy by n plus one, by one over n plus one by n plus one, and then you can cross multiply from there. Okay, I'll go ahead and finish it since we're so close anyways. Okay, so what we had was, sorry, one over n plus one plus one over n plus one squared. What was it? Less than or equal to... What do we have? One over n, okay. Again, this is what we're ultimately going to try to prove this. Well, if we get a common denominator, this is the same thing as n plus one plus one over n plus one squared, less than or equal to one over n. Okay, see what I did here? I just multiplied the first fraction top and bottom by n plus one. Okay, and now let's just clear the fractions here. Now let's multiply everything through by n times n plus one squared, both sides. We can do that. It's positive. We don't have to worry about flipping inequality signs. Then what do you get? Well, the n plus one squared is going to cancel, and then you're still going to have an n here, and then it'll be n plus one plus one, right? Less than or equal to... On the other side, it'll be n plus one squared left over, because the n will cancel with the other n. Okay, so now you can see how this is going in your mind. Just keep going with this. n squared, okay, so this just becomes n plus two. So this is n squared plus two n, less than or equal to n squared plus two n plus one. See that? That is definitely true. How would a really good proof proceed then? Okay, here. Let me just... I want you guys to keep in mind this is important. I don't want to make sure you didn't forget this. I'm not saying that this is the right way to establish the proof. I'm saying this is what you do on scratch paper, and you say, I got to prove this. Well, this is equivalent. I mean, actually, in this case, you could do this, because these are all if and only ifs. In general, you may not be able to do that. But, well, what's this equivalent to? This really is equivalent to the following simple assertion, zero, less than or equal to one. And that is definitely true. So how would a nice proof proceed through this? We would start here, and we'd go backwards. Now, in this case, and I'm being a little more liberal now with the if and only if arrows, if you want to establish something, and everything you have is literally equivalent to what you're writing down after it, you can just do if and only, if and only, if and only, if and only, if and only. And get down to the end like this. That's fine. That's okay. So the proof really hinged on simply the fact that zero is less than or equal to one. That's it. Okay. Well, this one is a little trickier, maybe than some of the other ones were, but most, I would say, the majority of the induction problems are probably not quite this messy. Other questions? Diane, you look like you wanted to ask something. I was just trying to see one that maybe wouldn't be as visible. Anything from the binomial? Okay. Binomial theorem section? I know. Okay. Okay. Sure. Let's see here. Okay. So I'll just remind you of this real quick. And this is the form the book gives it in. Okay. So yeah, just due to time here, I'm probably not going to do a long problem. I'm going to remind you of the binomial theorem and then kind of the thought process. I'm probably going to do a fairly short one this time. Okay. So binomial theorem says that a plus b to the n is equal to the sum from zero to n of n choose k a to the n minus k b to the k. Okay. So yes. Each term is just that n choose k. That's right. Yeah. I remember because I lost. Yeah. Yeah. That's right. Okay. 3D says to show that this might jog your memory. I may have even done this one for you before. I don't remember. But n choose zero plus... Oh, no. Let me see. Is that what I want to do? No. Sorry. Let's do 3B instead. Okay. Sorry. So minus n choose 1 plus n choose 2. This is really messy. Sorry. Minus. And then you just keep on going. And at the very end, it's plus minus 1 to the n times n choose n. And the point is that this is equal to zero. I guess. Right? Yes. And the point is to prove this by using the binomial theorem. Okay. Well, here's what you do. Here's the idea. Now, this one actually happens to be maybe a little easier than some of the other ones. But the idea here is look at your sum and all you're doing here is you're just trying to choose a and b appropriately so that when you expand out a plus b to the n using the binomial theorem, you get exactly that. Okay. And so, first of all, whatever your expansion is, this needs to be equal to zero. And remember, if you look at the binomial theorem, the coefficients of all these powers are of the form n choose k. Right? And all of these coefficients here are either 1 or minus 1. They're all either 1 or minus 1. Okay. So, yeah. So then, sorry. That's not what I meant to say. I mean, the coefficients actually go up there. I mean, we see these n choose k, these binomial coefficients coming up throughout the expression here. It's just that they also have either a 1 or a minus 1 on the outside. Okay. Any n features? No. I mean, if you look here, look at the n choose k and the binomial theorem, we have all of these things here except the signs are alternating. That's all I'm saying. The signs are alternating. Right? If we want this to be equal to zero, then we want, if we're going to use the binomial theorem and want everything to collapse to zero, then a plus b should be zero. Okay? And notice that we also don't have any a's or b's or anything. It's always just either 1 or minus 1. So the most natural thing to try first, then, is just to choose, for example, a to b1 and b to b minus 1. And then expand it out using the binomial theorem. And then just follow the reasoning here. Zero is certainly, whatever n is, zero is certainly 1 plus minus 1 to the n. You should all buy that. It doesn't matter what n is, right? 1 plus minus 1 is zero. Zero to any power of zero. And now this is where we use the binomial theorem. And now we're just, I'm just going to write all the terms out as we go up. Where do we start in the binomial theorem? Well, k equals zero, right? So it's n choose zero, 1 to the n minus zero times minus 1 to the zero. And what is, I'm not going to write that all down, but if you just follow along here, all you're doing is you're now using the right-hand side of the binomial theorem. The sum from zero to n, n choose k, a to the n minus k, b to the k. Well, we start with zero. a is 1 and b is minus 1. So this, since a and b are both 1 or minus 1, this product here has to be either 1 or minus 1. It has to be, always, okay? And so we're just going to be left with, in this case, because we're raising minus 1 to the zero, it's 1. So the coefficient is just 1 and choose zero. And that's exactly the first term that we want. What happens if we plug in 1 for k? Well, again, the a part, look at the binomial theorem. It's 1. It's always going to be 1. Don't think about that. If we plug in 1 for k for the b term, since b is minus 1, we're going to get minus 1 to the first, right? Which is minus 1. So then the second term will just be minus and choose 1. The minus comes from the fact that minus 1 is being raised to the first power. And then there's just, we just keep doing it, right? And this just continues all the way down to the end. And that's it. So all you're doing, once you identify a and b, you just plug them into the theorem and just write the expression out. Instead of the sigma notation, just write it out. And that's it. Okay, I think I probably have to stop because I have to hand your homework back as well before you go. Okay, so what I've decided to do is this, this homework. Man, I've been really nice to you guys lately because I have all this trauma occurred to me and I haven't had time to do my job as well as I should have. So I gave you a completion grade on this one as well. Okay? Now, before the final, you will have a salute. Some of these, I'm going to pick a few of these to write up solutions to. So you'll still have, so let me say it this way. There was another assignment I gave you a completion grade for, although online I wrote up solutions to three problems. Those are fair game in the graded homework portion of the exam. Okay, even though it wasn't graded, that's still fair game. All of these things that you see online that I have solutions for, those are fair game. Okay? All right. Thank you for joining us.