 back to being due on, well, starting to be due on Tuesday. They never have been because I was sick. So what's going to happen is this. The assignment that was going to originally be due on Thursday will be due next Tuesday, one week from today. And in addition, though, of course, we're going to be doing new stuff. So it's going to be that along with another assignment that we're going to get into today. But you will at least have a lot more time to finish the last one. So what I'm going to do this week is, here's the plan for the week. What we're going to do is finish up the binomial theorem stuff. I'm going to talk about what, specifically, I think I'm going to do one of your homework problems for you in the binomial theorem section. And we give you an idea of what you should be thinking about when you work on the other problems. And then we're going to move on to section 2.2. And we're going to actually start talking about algebraic properties of integers. So we're really going to start digging into the number theory here. And then on Thursday, I'll finish that up. We'll talk about some problems in the next section on Thursday. And then that assignment then will be due in addition, as I said, to the one previously that was going to be due on Thursday. That new assignment in 2.2 will also be due on Tuesday. OK, so as I said, I kind of got stuck in the middle of the binomial theorem last time. So let me just get caught up here and kind of remind you where we were before. Let's see, am I recording? Yes, let's see here. Let me get, I got to click on this pen here, OK. OK, so we're just continuing section 1.2. And so you want to recall that we were trying to prove the binomial theorem. So this, of course, you have this in your notes already if you were here and you were paying attention. So what we were trying to prove is this. OK, so if I remember, I had established the base case of the induction here. So I'm just going to kind of pick up where we left off. I don't want to waste a lot of more time writing down things I've already written down. So prove the base case. And now we have to assume that the binomial theorem is true for some natural number n. And we'll prove that it's true for n plus 1, right? So let me just go straight into that. So now we're going to assume that this equation holds. Yeah, sure. So it's just the binomial theorem. So a plus b to the n equals the sum from k equals 0 to n. N choose k, a to the n minus k, times b to the k, OK? Oh, sure. I understand that, yeah. OK, I'll try to remember. If I forget to read something, something's not clear. Just ask me and I'll clarify it. OK, so we will assume that this holds for some natural number n. I'm not just to say space here. I'm not going to actually write down what it is that we're going to try to prove. Because I'm already going to be writing down a decent amount here. But of course, what we're trying to establish, and I'll go back to this here in a second, once we do it, what we're trying to prove is that this holds for n plus 1, right? So we're trying to prove then that a plus b to the n plus 1 equals the sum from k equals 0 to n plus 1, n plus 1 choose k, a to the n plus 1, minus k times b to the k. That's what we're trying to establish. Is everybody with me on this? OK? So let's just go ahead and do that then. So I'm just going to, this is going to take a little bit of work, but we're just going to look at a plus b to the n plus 1 and see how to figure this out. a plus b to the n plus 1, here's what I would encourage you to do. Don't worry so much about where these steps are coming from. A lot of times, things that seem to come out of nowhere that seem really clever in mathematics are a result of just tinkering around with something and eventually seeing a pattern. You see the finished product. You don't see the elbow grease that went into it. But that's OK. You're not going to have to do something like this in the homework. But it would be good for you just to understand this and understand the technique. So what I want to encourage you to do is just try to understand all the steps that I'm going to take here, not so much where they came from. OK. So I saw this a little bit in the homework. By the way, you'll get your homework back on Thursday. I'm spending a lot of time on the grading because I want to write a lot of comments to get you kind of squared away and going in the right direction here. But a plus b to the n plus 1, OK. This is a plus b to the n times a plus b, right? I don't think that's too hard. You add the exponents, right? You have the same base. You add the exponents. A plus b is a plus b to the first. I did see a couple of you in here apparently believe. I'm not trying to insult you. I just want to get you squared away here. Believe that a plus b to the n plus 1 is a plus b to the n plus a plus b. That is definitely not true. And that's a very bad error, OK. So 2 cubed is 2 times 2 times 2, which is 2 times 2 times 2, right? You add the powers, OK. So this is true. Now let me, OK, I'm going to go slowly. I want to go too slowly here. Let's just distribute this. So this is the same thing as a plus b to the n times a plus a plus b to the n times b, right? You see what I did here? I just distributed, just treat a plus b to the n as a single term. Call it z, right? Then that's z a plus z b. And so now what I'm going to do is bring the a and the b over to the other side. So this is a times, oops, sorry, a plus b to the n plus b times a plus b to the n, OK? Also, pretty easy. Multiplication is commutative. We can definitely do this. No, no. I mean, really the step you would skip here would be this one and just go straight to that. I'm just doing it for clarity. Well, I'm just doing this to dot every i and cross every t so that I don't lose anyone. That's all. But yeah, in general, no, you wouldn't list all of it. I mean, a mathematician would know that you can go from here straight to here. OK, so now we're going to use the inductive hypothesis, which I'm not going to specifically write down, but I told you up here that we're assuming that this holds, right? So a plus b to the n is equal to that. That's our inductive hypothesis. So I'm just going to replace the a plus b to the n, both of those occurrences with that, with the summation. Yes, sorry. Yes, with the summation. OK, so this is a times the sum from 0 to n, as k ranges from 0 to n. n choose k, a to the n minus k, b to the k, plus b times the same thing, right? The sum, as k ranges from 0 to n, n choose k, a to the n minus k, b to the k. Now, a is just a fixed constant. And I say constant, I mean, it could be a real number. It could be a variable, but it's fixed. It's not moving. So this a, we can actually pull inside the summation. And you should have learned about these kinds of things in calculus. You should, hopefully, you've all seen these things before. But the constant's on the outside you can throw inside, just like with derivatives, right? The derivative of a constant times a function is just the same thing as the constant times the derivative of the function. So this property happens all the time in mathematics. And this is one of the instances where we can just throw them inside. So this is equal to the sum as k ranges from 0 to n. So what do we get when we do that? Well, we get n choose k. We might as well just write this as nicely as possible. Once we throw the a inside, we have a times a to the n minus k, which is a to the n minus k, plus 1, right? Times b to the k plus. And then we'll do the same thing with the b, right? Sorry, this is a little bit messy. OK, does everybody see what I did? I threw the a's and the b's inside. That just increased the powers of the a's and b's by 1. OK? All right, so let me make sure everybody has this down first. Unfortunately with this, I have to go to a completely blank page here. But anybody still writing? I definitely don't want to go until you've got all this down. Does everybody have this down now? Yes? OK. OK, so what are we going to do? Well, what we're going to do is we're going to use this Pascal's rule, which I don't have time to go back and write down. But I'll tell you exactly when we're going to use it. So you might wonder why I'm doing what I'm doing in this next step. What I'm doing is I'm going to rearrange this just slightly so that we can apply Pascal's rule. And then that will enable us to finish the induction. OK, so if you look, and then now you're going to have to just look at your notes here, suppose that, I'll tell you what, let me go ahead and go back for a second. Look at this first term here. Some from k equals 0 to n, n choose k, a to the n minus k plus 1 times b to the k. What happens if I plug in 0 for k everywhere inside here? The question is, what does this become? OK, so what's n choose? OK, let's just do it. I'm not going to write it down, but just think about it. What's n choose 0? It's n factorial over 0 factorial times n minus 0 factorial. So it's n factorial over 0 factorial times n factorial, which is 1. What does this become when I plug in 0? This becomes a to the n plus 1. What does this become? 1. So this whole thing, when I plug in 0 for k, this expression, this term right here, is just a to the n plus 1, right? Yes, definitely. And we're actually going to get to that here in a minute. Except I have to do something, I have to tweak with it a little first, but yes, you can definitely do that. OK, now, the second question, let me, I'll tell you what, let me just go through this. Let me write this down first. I don't want to get ahead of myself. OK, so this is equal to a to the n plus 1 plus the sum as k ranges from 1 to n. And choose k a to the n minus k plus 1, the speed of the k. OK, plus, and I'll explain where this comes from in a second. I'm going to do the same thing here. I'm going to take off the k equals 0. And we're going to just see what that additional term is, and we're going to add it on at the end. I'll explain this again as we get closer to it. OK, now this is where it's unfortunate that I had to cut off the last page, but this may look a little mysterious to you at first. I squeezed it in here. All right, now here's where I'm going to have to say something. The rest of the proof is actually very, very straightforward. This is the part that you have to think about for a second. If you understand this part, then the rest of it, in fact, we're almost done. So I'll wait till you have that copied down, and I want to explain where this is coming from. And I'll use the previous slide here in a second, but you're going to have to look at your notes once you copy this down, because it's going to be confusing to toggle between the two every three seconds. So we have this. You guys got this? So now I'm just going to have you look at your notes for what we're about to do. So here is where we, OK, I know this is a little bit messy here, but here's where we were before. So what has changed? So here is our last line, right here. So I told you that when we plugged in 0 for k here, this became just a to the n plus 1, right? So now, OK, I actually lied to you. I'm going to keep doing this now just to confuse you as much as possible. So you see what I did here? Just look at the first two. OK, this is annoying. OK, just look at the first two expressions. I put these in parentheses for a reason, which I'll tell you later. But see what I did? See how this has changed? k equals 1 to n instead of k equals 0 to n. So we're missing the k equals 0 term. But the k equals 0 term, as I just said, is a to the n plus 1. So this taken together is just the sum from k equals 0 to n, which is what we had before. You see that? I just took the first one out. And then I just took the sum then from 1 to n. OK, now what about this? OK, now again, now I'm not going to go back now. I'm just going to have you look at your notes. But what's different now between this and the second term on the last line of the previous slide? Well, again, I'm starting with k equals 1 instead of k equals 0. You see that? OK, but what I had on the previous slide was k equals 0 to n, the sum from k equals 0 to n, n choose k. But k was starting at 0. Now, because we're starting at 1, we've subtracted 1 to compensate. You see that? So now it kind of tells you a reply. Yes, that's right. That's exactly right. So I had to subtract 1 to compensate for the fact that we're going up by 1. And that's why we have everything change slightly, because we've shifted down by 1. All the k's now become k minus 1 from the previous line to compensate for that. You guys follow what I'm saying here? OK, since we were supposed to start at 0, so now we have to take the k's down by 1 to get the same thing we had before. But notice, we are missing here. We are missing a term. The original expression was the sum from k equals 0 to n, n choose k, blah, blah, blah. But if we're starting from k equals 1 to n, n choose k minus 1, what are we missing? We're missing the n choose n term. If you look at your notes, it was the sum from k equals 0 to n, n choose k. That was the first part of it. But by changing from k equals 1 to n, we've lost a term. From 0 to n, there's n plus 1 numbers between 0 and n. From 1 to n, there's only n of them. We lost one of the terms. Which term did we lose? We lost the n choose n term. Notice that, right? No matter what I plug in for k, this will never be n choose n. You see that? We lost that. So we have to, like we did with the other expression, we have to add that on at the end so that we don't lose it. Does this make sense? Sort of? Well, I hope it does. That's where the b to the n plus 1 comes from. If you go back, I'll go back one slide here. Let me actually do that now. You see this? If you go back down here, what happens when you plug in n for k? That's one of the terms in the summation, right? It's n choose n, which is 1. You can check that. It's 1 times a to the n minus n, which is also 1 times b to the n plus 1. You see that? So we get b to the n plus 1 when we plug in n for k. And that's exactly what we lost when we changed, oops. What happened? Did it not say that or what's going on here? Oh, there we go. There we go. OK. So that's the term that we lost when we changed the index. We have one last term, so we have to add that one that we lost at the end. That's all I'm going to say. If you didn't get it yet, you're probably not going to for now. You can go back to the notes later and look at it. But n choose, yeah. So at the end, it's actually the term that we lost was n choose n a to the n minus n b to the n plus 1, which is just b to the n plus 1. That's the term that we lost by shifting the index. Yes? Uh-huh. On the very last slide, do you mean the one after this? The next one. OK. Yeah, oops. OK. Right. Yes, because here's where it came from. You're just talking about the a part, right? The a to the n minus k plus 1. So remember, from the previous slide, since we're changing from k equals 0 to k equals 1, we have to subtract 1 from k and all the occurrences of k. So instead of a to the n minus k, it's a to the n minus the quantity k minus 1. Which becomes n minus k plus 1 when you distribute the negative. OK, good question. Is this any other questions I can help with? No? OK. OK. So now, as Joe pointed out, we can actually use Pascal's rule here to get where we need to go. I have a question about that. Pascal's rule? Well, I have a question. When I said that you could use it, I thought it was easier to get the pattern at the end with anything, but now it's n choose k minus 1. So now I'm not sure that I see why you use it. Right, gotcha. OK. No, that's fine. I will try to make that clear once we get to that point. OK. So now what we can do, notice that we have the same bounds on the sigma, right? k equals 1 to n, k equals 1 to n. So as was suggested earlier, we can squeeze these into just one sum, right? OK, so this just becomes the sum from 1 to n. And notice also, you see the terms next to the binomial coefficients. On the first one, it's a to the n minus k plus 1 b to the k. On the second one, it's also a to the n minus k plus 1 b to the k. So we can actually just factor that out, basically, right? And then we end up with, on the inside, this just becomes n choose k plus n choose k minus 1. Put this in brackets here. Times a to the n minus k plus 1 times b to the k. OK, I'm going to put all of this in big parentheses here as well. You'll see why in a second. So we've got that. We've condensed that. And then we still have the plus b to the n plus 1 on the outside, right? OK, so all I did was combine, right? These two sums, they have the same indices. And so I can just add what's inside together and form a single sum. And so now this is equal to a to the n plus 1 plus, OK. So I don't think I'll write the parentheses anymore. But the sum k equals 1 to n. And now I want you to look at this, what I have blocked off in these square brackets. If you go back to your notes from before, I gave you this lemma called Pascal's Rule. If you might want to look that up now and see how this applies here. Pascal's Rule said that if k and n are integers, it's important to check the hypothesis here. If k and n are integers and 1 is less than or equal to k is less than or equal to n, then n choose k plus n choose k minus 1, which is what we have right here, is equal to n plus 1 choose k. So does the hypothesis hold? Is it the case that k is between 1 and n? Well, yes it is. Because that's part of the reason why we dumped off these other terms on the outside is to make k be bigger than 0. So now the hypothesis for Pascal's Rule is satisfied so we can apply the conclusion now. Yes, you did have it. Yes, in other words, yeah. OK, so this becomes just n plus 1 choose k. I want to go back here, sorry. OK, now I'm going to rewrite this because I wanted you to see that it actually does specifically satisfy the conclusion that we need to prove, which is that the binomial theorem holds for n plus 1. n minus k plus 1 is certainly the same thing as n plus 1 minus k, right? No doubt about that. And actually, OK, I keep going back and I'm going to say, I am going to offset this so there's no confusion that the plus b belongs inside the summation, which it doesn't. So plus b to the n plus 1. And this is by, I'm going to just put this in parentheses. I'm saying it now. I'm running out of room here. Pascal's rule is what allowed us to do this, OK? Now, this will require a little bit of explanation, but because I tend to spend a lot more time than I should on these proofs, I'm just going to go ahead and write it out now. This is the sum as k goes from 0 to n plus 1. n plus 1 choose k. a to the n plus 1 minus k times b to the k. And this is, you can check this, if you go back to the original statement of the binomial theorem, this is exactly what you get when you replace n with n plus 1. So that is exactly what it is we're trying to prove in the inductive step. Why is this true? Well, notice what I, so you see this is different, right? I mean, this was the sum from k equals 1 to n. Now I have the sum from k equals 0 to n plus 1. How did I do that? Why is that true? Well, what's the difference? I mean, this is, and I'm just going to say this, OK, because we're running out of time here. But look at this last expression. What is this? Well, this is just the sum from k equals 1 to n plus what we get when we plug in 0 and what we get when we plug in n plus 1. You see that? What do we get when we plug in 0? This is 1 a to the n plus 1, 1. It's a to the n plus 1. It's exactly that term. What do we get when we plug in n plus 1? 1, 1, b to the n plus 1. It's exactly that term. So these are equal. You can see because the first and the last terms are these guys and then what you have in the middle is exactly this. So that's it. Now the proof is done. OK, and here this is something that's going to be very useful to you in the exercises. So now I'm going to spend a little bit of time talking about what to do with the exercises. I'm sure some of you will appreciate kind of how to think about these. I will tell you, and I didn't look at all the problems, but I think this is true. I hope this is true because if it's false, I'm going to feel really bad later. OK, first of all, the problems in this section in 1.2, it is not the case that you have to use induction for every problem. No. And I think, I'm pretty sure this is true, you're only going to be using induction when the book specifically says to use induction in the directions. Otherwise, you should be able to do it without induction. Yes? So there's two things that we would say. Yes. Yes. Definitely. So what I want to do, does everybody have this down now? So what I want to do now is actually, because we've been using the sigma notation, it kind of tends to maybe mask what's actually happening here. What I want to do is write out the binomial theorem without the sigma notation, actually just writing out all the terms with pluses between them. That will help you. If you've looked at the homework, and hopefully most of you have, you notice that a lot of the problems say prove that some equation holds, and on the left side, it's some binomial coefficient plus, plus, plus, plus, plus, plus, right? So that's just the binomial theorem, really. It's just written out instead of using the sigma notation. So this actually will help you. So note, expanded out, we have a plus b to the n is equal to n choose 0 a to the n plus n choose 1 a to the n minus 1 times b plus n choose 2 a to the n minus 2 b squared. I'm going to have to go over here now plus on down to n choose n minus 1 a b to the n minus 1 plus n choose n b to the n. Is this some deep new theorem? No, it's not. This is just the binomial theorem except remember what the sigma notation means, right? The sum from k equals 0 to n means plug in 0 for k. See what you get. Plug in 1 for k, see what you get. Plug in 2 for k, see what you get, and then add them all up. That's it. That's all you're doing here. So this is just sort of a less compact version of the same theorem. It's the same thing. OK, here's what I will say. I'll probably say a little more about the homework on Thursday since you don't have to turn it in until Tuesday, so I'm not going to say a ton about it now. I will say something about 3a and 3, or sorry. Let's see. Let me say something about 3a and 3b. I think I gave you both of these, right? Oh, I gave you 3a and 3c. So here's what I'm going to do. I'm not going to do 3a. If you understand what's happening with 3b, 3a is even easier. So I'm going to do 3b for you. And then, trust me, if you understand 3b, you will get 3a in about two lines. 3a is easy. It's extremely easy if you just see what you're supposed to do. So let's look at 3b. And again, I know this isn't a hand-in problem, but I'm doing it just to help you with the other ones that are similar. 3b says to prove this. n choose 0 minus n choose 1 plus n choose 2 minus dot, dot, dot. So that just means you continue the pattern plus minus 1 to the n times n choose n equals 0. That's the problem, is to prove this. Here's the main idea. Some of you might think, and you might be able to, and actually, if you're skillful, you can do it by induction with just ignoring the binomial theorem and everything that we've already done. But then you're reinventing the wheel. There's just no reason to do that. Everything that's been proved in the section is certainly free game to be used unless it's specifically contradicted in the directions for the problems. So you might look at this and say, oh, I'll do it by induction. No, definitely don't do that. You might look at this and go, ooh, that looks hard. I don't know. This is just impossible. I'm just going to drink some beer and watch how I met your mother. That's not a bad idea, by the way, but not when you're doing the homework. No, it's actually not hard at all. What you do is you just use the binomial theorem. And here's what you do. Here's what you do. If you look at the binomial theorem written out this way, so you want to prove that I'm just trying to give you an idea of how you go about reasoning this through. You want to prove that this is equal to 0. Well, this certainly has the flavor of this expansion on the right side of the binomial theorem. Certainly does look something like that. So maybe we can use the binomial theorem to just get it for free. And then the question is, what do we use for A and B? Well, we want this to be 0. So in particular, A plus B to the n should also be 0. We want everything to be equal to 0. So what are the coefficients A and B? Well, what are some sort of natural choices that we can use for A and B that will give us 0, but will also give us something like this? We want all these things to collapse to 1 or minus 1, right? Yeah? What about 1 and minus 1? 1 and minus 1. 1 and minus 1. Yes, very good. For A and B. That's it. Done. That's the proof. You just use the binomial theorem with A being 1 and B being minus 1, and I'll show you how it works. And I would like you to do something kind of like this. OK. And so some of you are writing out more than you need to in your proofs. And I want you to notice that what I'm going to do is just going to proceed very succinctly from one side to the other. We want to prove that these two things are equal. So certainly, if I can prove that 0 equals this, then this equals 0, of course. The quality is so-called symmetric. If A equals B, then B equals A. So it doesn't matter which side you start on. So just think about this, OK? So I'm not prefacing this by saying, let n be an arbitrary natural number. That's technically what I would do in the beginning. But I'm trying to save a little bit of time here. 0 is certainly equal to 1 plus minus 1 to the n. Certainly true. 1 plus minus 1 is 0. 0 at any power, right? Positive integer power is certainly still 0. No, no, no. But what I'm saying here, what I have in the, let me just say exactly what's in the parentheses. 1 plus minus 1, that whole quantity to the n, is the same thing as 0 to the n. I had negative. Right, I thought, yeah. So it's really just 0 to the n, is all that it ends up being. I'm not contriving that problem. No, no, no, I know, I'm just trying to make it clear. And I'm not, yeah, sometimes I'm not being explicit here. So, OK, so really, and this is not really part of the proof. I'm doing this just to illustrate, using the binomial theorem. Let's let, think of 1 as A and minus 1 as B, right? With the binomial theorem, you see that? OK, so what do we get? We get n choose 0. I'm just going to go ahead and write the whole thing out here, even though you really don't need to do this. 1 to the n plus n choose 1 times 1 to the n minus 1 times B, which is minus 1, right? Plus n choose 2. A is 1, right? So it's this 1 to the n minus 2. B is minus 1, so this becomes B squared is minus 1 squared. Plus on down to n choose n minus 1. A is 1, B is minus 1, so this is minus 1 to the n minus 1. Plus, and the final term is n choose n, B to the n, B is minus 1, so this is minus 1 to the n. OK, sorry with the A here. This looks like this is 2 to the A. It's not. Let me block this off just to be, I hate this thing. This is sitting by itself. So that 2 is minus 1 squared. Well, I know I'm going overkill here, but I just want to make sure everyone just sees where I'm going with this. Well, what do we want 0 to be? We want it to be equal to that, right? So let's just do the obvious rewrite here and say, OK, well, what's n choose 0 times 1 to the n? Well, it doesn't matter what n is, right? This is still going to be n choose 0 for sure, right? We're looking at the first term. This is certainly n choose 0. What about this second one? Well, 1 to the n minus 1 is still 1 times minus 1, so the second term is just minus n choose 1, right? Because we have a minus 1. This becomes minus 1 right here, right? Now what about the next one? Well, OK, 1 to the n minus 2 is still 1. Minus 1 squared is 1, so the next expression is just n choose 2 plus n choose 2, right? So the next one, technically, is minus dot, dot, dot. And then what do we end up with at the very end? n choose n times minus 1 to the n, which, of course, is the same thing as minus 1 to the n, times n choose n, little box. Done. That's it. Oh, sure, sure. You could. You could do that. But what's being asked to be proved is just that it's n choose 0 minus n choose 1 plus n choose 2, dot, dot, dot. So I'm just showing you that we can, in fact, get that. If you wanted to simplify more, you certainly could, but it's just not necessary because you have proven explicitly what it was asked of you. You guys see what I did here? You guys following this? Seriously, you guys following this? Yeah. Doesn't matter, though. But it's irrelevant. Whether n's even or odd just plays no role. I'm just going to pattern all of those different. Yeah, I understand what you're saying, because you might say, well, I don't know if this is positive 1 or minus 1, because I don't know what n is, if it's odd or even. But it doesn't matter, because you're not asked to know. So it's just we've proven exactly what was asked of us, done. That's it. OK? Any other questions about this? All right. So now, if you go back to 3A, it's even easier. I mean, there's nothing to it, really. A plus B to the n equals, now it's the same thing. What do we choose A to be? What do we choose B to be? In 3A, you wanted to get, I think what? You wanted to get 2 to the n on the right side of the equation, I think, right? There's nothing. Well, let's see. Let me open the book up here. OK. Right, so yeah, so for 3A, we want to prove that a sum of these binomial coefficients is equal to 2 to the n. Well, what's 2 to the n? It's 1 plus 1 to the n. Then just apply the binomial theorem again, just like we just did. And it falls out right away. I'm not seeing a lot of expressions here. Seriously, really, I want you to ask me if you have questions here. Is this easy? You guys getting it? OK. So 3A, you should all get 3A now. All of you should get 3A. I'm doing this selfishly in part because it's much more pleasant for me to grade things that are right than things that are wrong. Especially with, and you will see, I am going to write a lot of comments on your homework, which I think you should. Can you give us all the test answers ahead of time so it's easier to grade it? No, no. Although you will see things on the test that you've certainly seen before. That's for sure. No, that's OK. Yeah, what's up? I have a question. OK. I'm 5B. Let's see. An orious one divided by 2. OK, so 5B. I'll probably talk a little bit more about this on Thursday. But the point here is they give you a hint. Here's what I request of you, or require of you, probably, if you want full credit. These problems where you're given hints, that hint, I expect you to actually prove that hint. Don't just take it on faith. So your problem should be, whatever the hint is, prove that first, and then use it in the problem. They're giving you the hint, but that doesn't mean you get it for free. You need to prove it. And it shouldn't be hard to prove, really. So in this case for 5B, well, what's the hint? So m squared is 2 times m choose 2. So what you can do is just replace all of these things. And what you're going to get, let's see here. Yeah, so here's what you're going to do. Let me make sure that I assigned. Yes, 5A. I assigned 5A. You will probably find that you will use 5A in your proof of 5B. Because once you apply the hint, you're going to be able to extract a sum that is of the form of the left side of 5A. That's the. I said, using the proof we found in 5A. Yes, OK. On top of that, there was another piece that he assigned 1 plus 2 plus 2 plus 1 plus 1. Oh, yeah, but I also assigned that, too, in 1, 1. That was the first problem in 1, 1. So here's the thing. Here's what you're allowed to use. And there might be occasions where I'll assign a problem that actually uses the previous problem that I didn't actually assign also in the homework. You are allowed to use anything in the book, anything in the lecture, and any of the previous exercises to establish your proof. No, no, no, no. So if you need to use the fact, so if you guys remember this, you all did this in the homework. I think it was the very first problem I gave you. 1 plus 2 plus 3 on down to n is n times n plus 1 over 2. If you need to use that in your proof, you can certainly just quote it and just say, buy whatever it was, 1A from 1, 1, we know this. You don't have to re-prove it again. Does that answer your question? OK, you guys follow me here? Definitely anything you've done. You don't have to do it again. I'm not going to ask you to do that. All right. I think for now, we'll talk a little more later, but for now I think I'm going to jump on to going on to the next section here. So remember, this isn't due till next Tuesday. Thursday I'll at least have a bit of time that we can talk a little more about this. But at least now you've gotten the section done and you have some idea of how to start with some of these things. OK, so we're going to skip section 2.1, which is just sort of a history lesson, essentially. Not that it's not interesting, but I'm not in general going to spend time on history, because you can just read that yourself if you want to know. That's not really integral to the theory that we're going to be developing here. So we're going to jump ahead to section 2.2 now. OK, so I'm going to try to get through the main theorem here. I think I can do it. This one's going to be about as complicated as the binomial theorem. Although its assertion is much easier, and you may just say, well, the theorem is just true, because duh, of course it's true. I learned this when I was in first grade. But it doesn't quite work that way anymore for us. We're interested in actual rigorous proofs. This is called, and this theorem, even though you learned it a long, long time ago, is extremely useful in elementary number theory, the division algorithm. You all learned, I'm not going to write this down, because you all know this, but you learned a long time ago. Hopefully, unless you started, everyone just uses calculators now, I'm old, so I don't know. But I'm assuming at some point you learned how to do long division, 17 divided by 3. You learned this? I don't know if they still teach this or not. I hope they do. So how do you do this? 17 divided by 3. 3 goes into 17, well, five times, and the remainder is 2. So 17 divided by 3. 3 goes into 17, five times, the remainder is 2. Notice that the remainder is less than what you divided by. Because if it wasn't, if it was bigger, then you would go in at least one more time. So that's the idea behind the division algorithm. It's just formalizing this, and then we're actually going to prove this more rigorously now. And it's going to look, again, a little bit more formal. But you'll remember that this is just stuff that you have already done, maybe 10, well, probably more than that, 15 years ago, depending on how old you are, I guess. So here's what I want to do today, if I can get through this. This is it. This is the section. It is the division algorithm. And it's really just amounts to playing around with basic algebra, except there are a lot of, you can see a lot of letters floating in the proof. So it might make your head spin a little bit. Here's what it says. Given integers a and b with b bigger than 0, there exists. I'll explain this more in a minute. Proof theoretic, of course, is sometimes it takes a while to wrap before you can really wrap your mind around what this word really means. Of course, you know what it is, really. Exactly one. But how to apply this in proofs is not maybe as obvious. Unique integers q and r satisfying these two conditions. First condition is that a is equal to q times b plus r. I'll tell you how to think about this. Let these be suggestive. Assign some meaning to these things in your mind. q is going to be the quotient in r as a remainder. That's where these letters are coming from. And the second condition is that r is between 0 and b. Yes, yes. OK, I'm going to split this proof into two parts. So here's the way you, OK, again, I want to give you some framework so that you understand kind of where this is coming from. Think of the 17 divided by 3 example, OK? So a is 17 and b is 3. Just think about dividing, OK? Then the quotient is 5, right? 17 is 3 times 5 plus 2, OK? And notice what the second condition says. It says that the remainder should be less than what you're dividing by, which, again, really does make sense. Because if it's bigger than what you're dividing by, we'll go into it more. So you'll have a different quotient. So all right, this is going to be, I'm just warning you now, this is going to be a little tedious. But this is good for you. Well, I mean, it all depends on what you're allowed to do. I mean, in this course, we're kind of building up everything from ground 0 here from square 1. So we're going to dot every i and cross every t here. This, there are certainly more than one proof here. Even though it's a little long, I think, clarity-wise, I think this is kind of one of the better ones, really. I mean, I really think this is about the best proof. Because we have to start from the beginning. I can't say. Consider a Euclidean domain and a function which is multiplicative, and we're not going to do that. So we have to do this from the very beginning. And I think this is a good proof. OK. So here, like I said, I'm going to split this into two pieces. The first piece, I'm going to call existence. What do I mean by existence? I mean, I'm just going to show you that there exists integers q and r that satisfy 1 and 2. That's all we're going to focus on first. The uniqueness we'll do later. OK, so this may seem a little mysterious to you at first, but you'll see where this comes from in a minute. S is equal to the set a minus xb, where x is an integer, and a minus xb is greater than or equal to 0. OK. a and b, remember, are fixed. These are handed to us. a and b are integers, and b is positive. Those aren't moving around, OK? What's moving here is the integer x. And so s is just going to be the set of all integers of the form a minus xb, where x is an integer, and this expression is bigger than or equal to 0. Yes. These are, yeah, everything inside here is, so x itself can be any integer. But we need a minus xb to be a non-negative integer. OK. So here's the first claim. s is actually not empty. In other words, there's at least something in s. And this, if you think about what s is, this really shouldn't be hard to, you should be able to convince yourself of this pretty easily. I'm going to write a formal proof of this, but just think about what this is. a and b are fixed, and b is positive. OK. So we want, all we're looking for is to prove that there exists some x such that a minus xb is positive. OK. Well, b is bigger than 0. So we can make xb as big in magnitude as we want by just multiplying by a really big integer x. OK. So if we multiply by a really big negative integer, then minus xb will be positive. So here's the idea. Take x to be some huge negative integer. The minus in the front then will make the whole thing positive, because b is positive. So we can choose x big enough so that whatever a happens to b, when we add this, we get something that's bigger than or equal to 0. Is that OK? I don't know. No. I feel like I've lost most of you already. Well, but that's what we're trying to prove, though. We're trying to prove that there exists an x such that a minus xb is bigger than or equal to 0. OK. It's not that hard. It's not that hard, really. OK. OK. All right, I'm going to just say this again. We don't know what a is. You don't know what a is. It's some integer. Do you all believe that if you had it a big enough integer to a, you're going to get something positive? Choose x to be some big negative integer. Then minus x is going to be a big positive integer, multiplied by b, which is also positive. So we can make the minus xb huge as big as we want, and positive. So if I make it big enough and positive, and I add it to a, I can make that whole thing positive. Throughout the monitor, there's a window because you're so excited. Yes. OK, we're back. We're back. OK. Fantastic. OK, sorry for wasting five minutes. That's OK. Well, let's get back into this here. OK. Well, let's see if you believe this. Minus, whatever a is, minus the absolute value of a is definitely an integer, right? Certainly, that's true. Minus the absolute value of a. Yeah. And OK, well, let's just look at this. Minus. OK, now this is going to get a little weird. Minus absolute value of a times b. Let me put a little dot here just to make sure that's clear. OK. This is equal to, see if you believe this, this is equal to a plus absolute value of a times b. You guys by that? Distribute the negative, right? Just becomes a plus. A plus absolute value of a times b. OK, since b is positive, remember that's our assumption, b is positive, what do we know? b is certainly bigger than or equal to 1, right? You by that? Well, it's an integer. And in general, if 1 half is bigger than 0, it's not bigger than or equal to 1. But because it's an integer, it's bigger than or equal to 1. OK. OK, so I'm going to have to go to another page here. You guys got this down? OK. So now we're going to multiply the previous inequality by the absolute value of a to get absolute value of a times b is bigger than or equal to the absolute value of a, right? So maybe I should have just written this out explicitly. The previous inequality was just that b is bigger than or equal to 1. If you multiply both sides of b bigger than or equal to 1 by the absolute value of a, you get this, right? How do I know I can do that? Why is that a legal move? In general, you can't do that, right? I mean, there's situations where you have to change the inequality sign. Because the absolute value of a is not necessarily positive, but not negative. Non-negative. And this is not a strict inequality. So I mean, if a would happen to be 0, then this is certainly true, because it's just asserts that 0 is bigger than or equal to 0, right? So thus, let's just put all this together now. a minus absolute value of a times b equals a plus the absolute value of a times b, which is bigger than or equal to, OK, now you see what I'm doing here. Look at this previous inequality. I can certainly add whatever I want to both sides of an inequality. It doesn't matter what it is, right? So what if I add a to both sides of this inequality? You see, then this just becomes bigger than or equal to a plus absolute value of a. You see that? And the beauty of this is, well, what can we say? Think about this for a second. What can we say about a plus the absolute value of a? I know it's a very open-ended question, but it's certainly bigger or equal to, bigger than or equal to 0. Yes, Joe? Not necessarily. It could be. Well, OK, so the thing is, if a is negative, suppose a was minus 1, then a plus the absolute value of a is 0. So, but you can say for sure that whatever a is, OK, if it's positive, if a is bigger than or equal to 0, then the absolute value of a is a. Then a plus a is also bigger than or equal to 0. If a is less than 0, what's the absolute value of a? Minus a. You learned this in calculus a long time ago. So it's still greater than or equal to 0, regardless of what a is. This is definitely bigger than or equal to 0. Absolute value of a is always bigger than or equal to a, for sure. And it's positive. So it's bigger than or equal to minus a as well. So this is bigger than or equal to 0, for sure. You can do this by cases, whether a is bigger than or equal to 0 or less than 0. I'm not going to do that, but that's certainly true. OK, so what's the point here? The point, if you look back in your notes, is that if you looked at how we defined s, a minus absolute value of a times b is an element of s. For sure. Sorry, this is a little sloppy. But how do we know that it's in s? Well, look at how s was defined. If you missed this, I'll go back to it here in a second. But look at how s is the set of all a minus xb, where x is an integer, that are positive. a minus xb, that are positive. What do we have? This is a minus, this is an integer, right here, what I'm circling. This is certainly an integer, times b. And it's non-negative, I should say. And it's non-negative. So therefore, by definition, it's in s. It's of the form a minus an integer times b, and it's bigger than or equal to 0. So it's in s. s is non-empty. So what do we know about s? Well, here's the last thing I'm going to get to probably here. So by the well-ordering property, we talked about this a while ago, right? s has a smallest element. Let's call it r. OK, so I'm just going to write two things down that we know about r, and then we'll probably stop, and I'll finish this up on Thursday. And we'll do a few problems. So the rest of the proof isn't going to take long. And that's the rest of the section. Then we'll do problems from this section and problems from the last section. Then you'll have the homework on Tuesday. I think it'll work out pretty well. OK, so here's what we know about r. And this is where we're going to leave off. OK, so let's go back to the definition of s, right? It was everything of this form right here. And we just showed it was non-empty. So there exists some element r in s. So this element r has to look like this, by definition of s. It has to be of that form. So the first thing that we know is that r is equal to a minus qb sum integer q. Just look at the definition of s. That's the definition of s. It's a minus something times b. So since r is in there, it has to be of that. By definition of that set, it's a minus something times b. We're just going to call that something q. What's the second thing we know about r? There's one other thing that we know. OK, sorry. Before I do that, let's rearrange this, because we're going to use this here on Thursday. So if you rearrange this equation, this is the same thing as saying that a is equal to qb plus r. You see that? You all by that? Just by bringing it over to the other side. Notice that that's the quotient remainder form that we're interested in. That'll come into play in a minute. What's the other thing we know about r that we just found out about r? It's non-negative. It's non-negative, exactly. Remember, s has the smallest element called r. Remember the definition of s was everything of that certain form that was bigger than or equal to 0. That's in your notes. It's the things that are bigger than or equal to 0. Since r is in s, r is bigger than or equal to 0. So I think I will stop there. And like I said, we'll do some 1, 2 stuff and we'll do some 2, 2 stuff. The last thing, though, I am going to give you this. I think you would want to maybe get a head start on some of these things. Homework. Yes, yes. Both of them will be due next Tuesday. Yeah. Well, you've got a week. You've got a week. 2, 3a, 3b, 8 and 10. So this is due Tuesday. What is that, the 12th? February 12th? So work hard. And I'll try to get you prepared as well as possible on Thursday. And we'll talk about that.