 We're not going to do much new stuff, so of course last time we talked about the first principle of finite induction, I didn't do any problems. I was trying to squeeze everything in because we missed, you know, Tuesday. So what I'm going to do is try to do some problems today, and then I'm going to talk about the second principle of finite induction, which if you've started the homework, you know, maybe you know, number 14, I believe, uses this second principle, which I didn't get to just yet. But we'll talk about that, and basically the only thing new that I'm going to do is talk about that principle, and then we're just going to try to get through a few problems, try to, you know, get you prepared for the homework. Induction, of course, as I said before, this is a very, you know, I mean, this is a very common proof technique, not just in number three, but in mathematics in general. Most of you have done this before, but I just want to, you know, kind of go slowly through the first part just to make sure that everybody's sort of up to speed on this stuff. And then we'll probably pick up the pace a little bit, but initially, you know, I think it's best to err on the side of caution and go more slowly than more quickly in the beginning. So that's the plan. So what I'm going to do is just sort of remind you again what this first principle of finite induction is, and again, I'm aware that most of you have seen this before. Then I'm going to do a problem using induction, and then like I said, we'll move on to the second principle. So what I'm going to do, for the most part, I'm just going to start the labeling over again. But this is just a continuation of what we did last Thursday. Okay, so again, let's just, hopefully this is going to get better here soon. Let's recall theorem one, which was the first principle, like I said. And again, this is called just the first principle of, I'm just going to call it induction. I don't have anything to practice on, unfortunately. This doesn't belong, as far as I know, this doesn't belong to the math department, so we don't have any of these gadgets. And I don't have an iPad or any of these things, so I just, I don't have any way of practicing, unfortunately. Okay, I'm thinking about writing in cursive, and I think that might make this a little bit easier. Okay, so we've got our set S that's contained in N. We talked about this before. This N denotes the set of positive integers. Okay, so the first condition was that one is an element of S. Of course, if you have this in your notes, there's not necessarily any good reason to write this down again, unless you just kind of want to reinforce this and you think it'll help you. But the second condition is that natural numbers N, or in other words, for all little N and capital N, if, and this is a conditional statement, if N is in S, then the next positive integer is in S. And the conclusion is that this subset S is actually equal to the set of all natural numbers. And I try to give you kind of a heuristic argument as to why that was true. I said I would give you a formal proof using the well-ordering principle, but because of time, I don't think I'm going to do that. I think time would be better spent just talking about some problems and kind of showing you what I'm expecting in terms of work. So, of course, we're going to get to plenty of the formal stuff here when we start actually, you know, proving some of these exercises. So I don't think I'm going to say anything else about this for now. So what I'm going to do is, let's see if I can, hopefully I can squeeze this in here. I'm going to start, again, I'm going to sort of start slowly here. And again, those of you that are old hands at this, I apologize if you're bored, but I just want to make sure that everybody is on the same page here. So let's do an example. And so let's look at the following. The book, this is actually a little bit easier than the example, I think the examples that the book starts with, but I want to start with something that's easy to digest. Okay, so for every positive integer n, let's let, sorry, let's let p sub n be the statement. I'm going to put this in quotes. Let's see, I'll tell you what, I'll tell you what, sorry. Let me, let me change the notation slightly here. Oops, wrong thing. Okay, p of n, instead of sub n, okay. And it's going to be a statement. And it's going to be 1 plus 2 plus 3 all the way down to n equals n times n plus 1 over 2. Is this homework? No, it's okay. Is this actually one of your homework problems? Okay, well, that's fine. I don't care. Let's see. I thought, I thought why they said we're a little more complicated than this, but okay, well, that's, that's fine. That's okay. I'll, I'll still talk about it. It's okay. Okay, okay. Okay, well, that's, okay, good. That's even better, actually. So, yes, I'm not used to this. I'm not used to this. So, yeah, I'm going to try not to treat you like your Calc 1 students, but this is all I've been teaching since I've been here. So, I'm very glad about that. Okay, so I want to be clear about this too. This is a, this is a proposition. Okay, we're going to end up proving this, that this is actually true for every positive integer n, but for now you just treat this as an assertion, which may be true or false. Okay, it is true, but in general I don't want you to think of it this way. This is just a statement. Okay, so just to make sure that everybody is on board here, the first thing I will say is, let's just actually look at a specific case of this. P of 4 is, what, it's, okay. So, this statement just says that this, the assertion is that the sum of the first n positive integers is equal to the very last one in the sum times the next guy divided by 2. That's what this is saying. But don't get hung up on the letter n. It could be k or it could be n plus 5. It's just, the way that you interpret it is 1 plus, whatever that last guy is, that sum is that guy times the next guy over 2. That's what it's saying. Okay, so P of 4 then of course is just the statement, and it's 4, right? So this is just 1 plus 2 plus 3 plus 4 equals 4 times 5 divided by 2, right? That's what P of 4 is. And you can check this. Alright, so the left side is 10. The right side is 10. So this assertion is true. So, I was going to do some other ones here, but of course, I assume you guys are, you know, you've done this stuff before. You know, if I said what's P of 3, you could tell me exactly what that is. So, what is, what's P of 1? I'm not going to write this down. What's P of 1? What's that assertion? It's an equation, right? It's not a number, right? So just, yes, that's right. I'll just say it, there's sort of a shortened version. P of 1 then is 1 equals 1 times 2 over 2, right? Because n is 1 in this case. You can write that down if you want to. I'm not, this thing is already still awkward. I don't think I'm going to do that. But okay, so hopefully that's clear, at least what these statements are. Okay, and what we're going to do now is actually prove that this assertion, this P of n assertion that I gave you is actually true for every positive integer using this first principle of induction. Okay. So, let's call this example 2. Now this, I'm going to, some of you may ask this, like, oh, do you expect me to write all this exactly as you write it in the homework? No, I don't necessarily. I'm going to be a little bit more verbose about this just because I'm just trying to explain everything where it's coming from. Before the end of class today, I will try to tell you a little bit more of what I'm expecting as far as streamlining some of the things that I'm saying. But for now, I'm just trying to get the idea across as clearly as I possibly can. So I'm going to say a little more than maybe you might say in your homework. Okay, so the problem is to prove that this holds, as I said, for every positive integer. So 1 plus 2 plus 3 plus dot dot dot plus n equals n times n plus 1 over 2 for all positive integers n. Okay, so as I said, I'm going to take my time on this. Okay, so I already gave you example 1, so this p of n that you're going to see is just going to be the same p of n that I just gave you on the previous page. Okay, so we must prove that p of n is true, all natural numbers n. I'm going to switch back and forth between using natural numbers and positive integers just because I want you to equate these in your mind because they are the same thing. Okay, so going back to this first principle of finite induction, if you remember, if you have a subset of n that has the property that 1 is in there, and whenever something is in there, then the next guy is in there, then that subset has to be all of n. And so if we want to prove that this statement is true for every natural number n, I'm not going to write all of this out, and you don't have to either. But the point is, here's the idea. You're going to let, even though we're not writing this down, we're going to let s be the set of positive integers and for which p of n is true. And we want to show that s is everything. In other words, p of n is true for everything. So using this first principle of finite induction, it's suffice to show that 1 is in s, and whenever n is in s, then n plus 1 is in s. In other words, p of 1 is true, and whenever p of n is true, then p of n plus 1 is true. And then the first principle of induction then gives us that s, in fact, has to be everything, which is exactly what it is we're trying to prove. Okay? So let me just say what it is exactly that we're going to do. So we will show, I'll label these here. So 1, p of 1 is true, and 2, some of you that have, those of you that have taken to the screen may recognize this first, this first task as the inductive step, or sorry, the base case of the induction, and the second, the inductive step. You may have heard that terminology before. Have you heard this? Okay. Whenever p of n is true, so is p of n plus 1. Okay, so then if we can do this, then theorem 1 says that p of n is true for all n. Right? So let's just go ahead and do this. Let's go ahead and verify these two conditions. So the first one, I'll just circle this. I should say, and again, those of you that have done point of induction know this, the first part showing that your statement is true for n equals 1 should generally be very simple. There shouldn't be something you need to do much for. And in general, some of these problems you're just going to say, duh, obvious. There's nothing to say. If p of 1 is 1 equals 1, well, okay, that's true. You don't have to go into a long explanation. Well, by the tautological philosophical principle number 4, don't do that. Things that you know from second grade, you can just assert. That's fine. So, as I said, so the first thing we're going to do is just verify that p of 1 is true. And as we said before, I'll actually write this down. p of 1 is the assertion that 1 equals 1 times 2 over 2. Right? Okay, so everybody clear on this point? And all you have to say is clearly true. That's it, okay? 1 times 2 over 2 is 2 over 2, which is 1. You don't have to write all that out. Just clearly this is true. Okay, so now the second part, I'll tell you what, let me just wait until everyone's copied this. I think I'm going to go to a new page so I don't get stuck in the middle. Okay, is everybody okay? You want to have this down now? Okay, all right. So the second part is you'll just say, we're going to assume that p of n is true and we must show that p of n plus 1 is true. Something you might find useful as you're writing proofs is to just keep track of what it is that you're assuming and what it is that you want to prove. And it can't hurt just to be very explicit and say it, not only for my benefit but for your benefit so you know exactly what it is that you're trying to do. So again, remember this principle of the second condition, from theorem one, part of the hypothesis is that if p of n is true or if n is an s, then p of n plus 1 is true. So it's a conditional statement. If then, this is an implication. So what you're doing is, if you want to prove an implication, you might say, okay, well if it rains tomorrow, I'm going to stay inside. So you're not really, in some sense, you're not really saying anything about what's going to happen if it doesn't rain. You're saying, but if it actually does happen to rain, then you have a conclusion. So an if-then statement, if you want to prove it directly, what you do is you assume what's called the antecedent or the hypothesis, and then you prove the consequence. That's the direct way of doing this. And so in general with induction, that's what you're going to do. So I'm going to just write it this way, and I'm going to use this language because I think it's a little bit suggestive, which is what I want. So what we have, I'm using the word have, and I want you to think of this like, this is a fact, it's in your hand. You can use this fact, what I'm about to write, you can use to prove that p of n plus one is true. Okay, so we're assuming p of n is true, so that means that, of course, in example one, I defined what this meant, right? So there's no, nothing's really left to the imagination here. One plus two plus three, all the way down to n, that sum is n times n plus one over two, right? That's our assumption. And we have to show, and I'll label this a little bit differently here, let me use an actual star here. Okay, p of n plus one. So we talked about what p of four was, right? Here's p of n. So p of n plus one then is just, instead of the sum from one to n, it's the sum from one to n plus one, right? On the left side. And instead of n times n plus one over two, it's n plus one times n plus one plus one over two, on the right side. Does that make sense? Okay, so that's what it is that we have to show. Okay, so I'm going to condense this slightly. So this just becomes n plus one times n plus two all over two. Okay, so everybody see where I'm getting that? Yeah, okay, good. Okay, so here's what you're going to do. And in general, you may have, there are some problems I know. I think there's one in the homework that deals with an inequality called the Bernoulli inequality. So you may not always be proving the qualities. You may be doing inequalities. So the point is the assumption, this p of n assumption, as I said, that's why I'm using the word have. You can use this. You can do whatever you want to it. For all intents and purposes, it is a fact. It's true. And you use that to get to where you want to go, which is here, okay? So if we know that the sum for one to n is n times n plus one over two, we have to prove that when we go all the way up to n plus one, we get n plus one times n plus two over two. And this is what we have to work with right here. So think about what we need to get on the left-hand side. We need to get one plus two, I left this out, but the term before the n plus one is n, right? Okay, and we have that up here. So what's the only difference between the left sides of these equations? Well, the bottom one has an n plus one, whereas the top one does not. That's the only difference. And I can use this as fact, the top one. So if I want to get down to this, it seemed reasonable then that we might want to try to add n plus one to both sides. Because then at least the left-hand side is exactly what we want. And we just hope that the right-hand side can be made into what we want by factoring or something. That's the idea. You guys with me on this? Okay? All right. So that's exactly, as I said, that's exactly what we're going to do. So we're just going to add n plus one to both sides of this asterisk equation. Okay, so when we do that, what do we get? Well, we get n times n plus one over two plus n plus one, right? Anybody with me on this? Okay, so what is it that we want the left-hand side to be? We want it to be n plus one times n plus two over two. That's what we want to prove. So all I have to do is try to make this into that. Okay? You're just going to fiddle with this using some basic algebra and it'll fall out without a whole lot of work. Okay, so let's see. Well, there are actually multiple ways you could do this. I'm just going to, I think, go with the most natural approach here. Okay, so we want this to become n plus one times n plus two over two. So one thing that might be a reasonable thing to do first off is to get a common denominator here, right? Because then at least we can combine and we'll have a two as our denominator, which is exactly what we want, and we just hope the top will factor. That we're done. That's it. Okay, so n times n plus one over two plus two times n plus one over two. Okay, and as I said, I'll just go through this kind of slowly, just make sure that everyone's with me on this. So we can combine the numerators here, of course. You're not basic algebra students, so I assume you guys can follow this. And so, even so, I'm still going to go through this kind of slowly. So this is n squared plus n plus two n plus two, right? Which, of course, we're going to combine even more, but that's certainly true. Okay? And so what is this? This is n squared plus three n plus two all over two. Now, what we want, right, is to get the right hand side to be n plus one times n plus two over two, so what we want is for this to factor is n plus one times n plus two, and it does, the numerator, right? And that's it. You're done. Okay. Now, really, at this point, you might say, well, what do I need to say at this point? Therefore, you know, by the first principle of finite induction, this p of n is true for all positive integers n, sure, that's a nice way to end it. If you want to just end it like this, that's fine too. You've clearly established what you need to do establish, and any, you know, well, of course, mathematicians don't prove things as simple, really, but would say, oh, well, yeah, of course, that's done. You've done exactly what you wanted to do, so you're good. Yes? No, I don't. I mean, there's a lot of different conventions here. There's a little open square, a filled-in square, QED, you know, ha. I mean, there's lots of things that you can say. Oh, I'm done. LOL, yeah. You suck. Don't say that. That's disrespectful, but, no, you can do whatever you want. I mean, if you have a convention that you like, that's fine. But you do want to end it. Oh, I mean, if you wanted to leave, to finish like this, that's okay. You have done what you needed to do. And that's fine. Yes? I did things a little differently. So I would take the star equation and copy it out there on the left-hand side of it. I would replace the 1 plus 2 block up to n with the right-hand side of the asterisk equation. Right, right, right. And then on the total side of the equation. Yeah, I know what you're saying, except the problem is that you don't want to make assertions being true before you've actually proven them. So the slight issue is saying n times n plus 1 over 2 plus n plus 1 equals n plus 1 times n plus 2 over 2. That's what you're trying to end up proving eventually. So I really would rather you not do that because you are making an assertion that hasn't been verified yet. So the point is don't make claims until they've actually been verified. Here, we haven't done that. We haven't assumed what we're trying to prove yet. I know what you're saying. Then you just simplify so that you get 5 equals 5 or something like that at the end. But that really is not the best way to write the proof because there's a part where you're assuming what you want to prove. Hang on one second. Therefore, I can replace anything that I see. No, but you haven't proven this yet. But if your next step is to write, and maybe I miss understanding what you're saying. I know what you're saying too. We were taught to do it the same way. Okay, do you have it written down maybe? Yeah, and he's plugging in the quality for 1 plus 2. Oh, I see what you're saying. So you're just saying at this point right here. You're saying here, then just replace this with n times n. Yeah, yeah, yeah, that's fine. I thought you were saying something else. And that's what I thought you were saying. No, no, no, sorry. What I thought you were saying is something that a lot of people do. And so that's why I kind of interpreted it that way right away. But yes, that's fine. That's okay. Yes? I think I saw what you thought he was saying. Oh. If that was the only way you could see to do that, wouldn't you have to assume it wasn't true to derive a contradiction? That, I mean that, okay. It wouldn't be pretty. Yes, yes. Logically that would be okay. Although yes, like you said, it would not be a pretty way to go about it. But that's one way you could circumvent that issue. But yeah, what you said is fine. And just to be clear, it is not the case, certainly with proofs, that there's only one correct proof in general. That is not, that's certainly not the case. The only thing I want to, want you not to do is what I said before is assuming what it is you're trying to prove somewhere in your argument. That's the only thing I want you to avoid. Is it too much true therefore it is? Yes. No, no. I mean, you know, really I'm looking for, for these induction proofs, using the first principle of induction, what I'm really looking for is that you, clearly established the base case and the inductive step. As far as actually mentioning it explicitly, you don't really need to do that because, you know, this algorithm is very well known. And so you don't, it's not, of course it's not wrong if you're very clear and verbose about things, but it's not something I expect that you would have to do. Any other questions? Yes. For proof by induction, or is it just like, use whatever method you want? So, that's a good question. Because I haven't come close to writing the exam yet, I don't know, honestly. We can definitely talk about that closer to the time of the exam. I mean, you know, if it has any problem, use induction, we don't need that. Yes, yes. Are you using the principle? No, no, no, no. But yeah, but, you know, also if I say, you know, use the first principle of mathematical or finite induction to prove blah, blah, blah, then if it's specifically, I say that, then yeah, I would definitely expect that you would do that, for sure. But no, you wouldn't have to be that explicit. Any other questions about this? No? Okay. Good. All right. So, what we're going to do now is, and this is, like I said, this is the only new, this is the only new thing that we're going to talk about today, is the second principle of finite induction, which by the way, I won't get ahead of myself. Let me just go ahead and write this down first. So, second principle of finite induction. By the way, this, everybody else in the universe calls this strong induction, okay? So, just letting you know, in case you see this in a future math course, strong induction, that's what this is. I'm going to just stay consistent with the book unless the book says something that's actually wrong. So, I'm just going to call it this, because that's what it's going to be called in the homework and such. But yeah, that's what everybody else calls it. Okay. And it just says this. Suppose that S is a subset of N, and it satisfies two conditions. First condition is that one is an element of S, this is just like before. And the second condition is a little bit different for all natural numbers N. If, okay, so, here's the difference. If all the numbers, all the positive integers, less than or equal to N, are in S, then N plus one is in S. And the conclusion then is the same. Then the set S is actually equal to the set of natural numbers. Okay. So, I'm probably not, just for the sake of time here, I'm not going to go into a written explanation of this. But you can see the difference is that for the second condition, the hypothesis is stronger now. Instead of just N being in S, if you remember that was the first principle of finite induction, now everything less than or equal to N is in S. So, why is this true? Well, it's a very similar argument as before. So, why does S have to be N equal to N informally? Well, the first condition says that one is in S, so the smallest natural number is in there. The second condition says that whenever you know that one through a certain number, all of those are in S, then the next thing is in S. So, the second condition says, because we know one is in S, two has to be in S. Now we know that one and two are in S. And by the second condition, three has to be in S. Now we know that one, two, and three are in S. So, by the second condition, four is in S. So, that's the idea. As I said, I want to spend more time actually working problems than going over the kind of the foundation. So, intuitively I think it should be, if you think about it, it should be pretty clear that this is what's going on. So, what's the point? Why do we have these two principles? Well, here's the idea, and this is where the word strong comes from. Very roughly, and this is very informal now, if you can prove something by the first principle, you can prove it by the second. Okay. Why is that? Well, okay, this is kind of a mouthful, but I'm just going to say this. So, if you can prove something by the first principle, what is it that we, what was conditioned to for the first principle? If N is in S, then N plus one is in S. So, basically, if you can prove that N plus one is in S just by knowing N is in S, then you can certainly prove that N plus one is in S knowing that one, two, three, four all the way up to N, or in S. It's a stronger assumption. Okay, so that's the idea. And so now there are situations, and I'm going to do an example for you here in a second, where instead of just, so to prove that P of N plus one is true, it's not enough just to know that P of N is true. You actually need to know that the other guys below N plus one are true as well. And so, again, I'm getting ahead of myself now, but I'm going to say this and I'll repeat myself again, but how do you know the difference? How do you know when to use what? And it's basically just when you have a situation like that. When to prove that P of N plus one is true, you can't do it just from N. You need maybe N and N minus one and N minus two. That's when strong induction, or the second principle of finding induction applies. Okay, so let me, just because, again, I want to get through a couple more examples here. Let me do this example. Actually, what I'm going to try to end up doing is number 13 from your homework. You have to do number 14. I'm going to talk about number 13, which uses strong induction. I will say strong induction because it's just easier to say, but that's what, I mean the second principle, right? Okay, so suppose that the numbers A sub N are defined as follows. And so, again, number 14 is similar to this. Okay, so the first three are just explicitly specified. A sub one is one. A sub two is two. A sub three is three. Okay, and after that we have a recursive definition. A sub N is the sum of the three preceding guys. A sub N plus one plus A sub N, sorry, A sub N minus one plus A sub N minus two plus A sub N minus three, if N is bigger than or equal to four. So then, okay, I assume most of you have seen this. I'm just going to do a couple of these real quick just to make sure everybody understands this definition. The author calls this inductive definition. It's really, that's not really right. The word is recursive. Induction is something that you use to, I mean, informally to prove theorems. There's something, you may have learned, if you remember implicit differentiation, maybe from COC-1, that's a special case of a more general theorem called the implicit function theorem. And in this case, these definitions come from a very general abstract theorem called the recursion theorem. So really, when the author says inductively defined, that's not really the right word. It should be recursively, and that's the word that I'm going to use. Okay, so what is A4? Well, okay, so now we have to use this recursive definition, right? A sub four then is, if we plug in four for N, it's A sub three plus A sub two plus A sub one. Right? Okay, so what do we want here? Anyone? Six. What's A sub five? Anyone? Eleven. Are you guys getting this? You guys see this? Okay. So the second condition just, again, just says that for N bigger than or equal to four, A sub N is the sum of the three previous ones. That's it. So as you go along, you can just keep knocking them off as you go down, right? So now I could easily find A sub six, right? It's A sub five plus A sub four plus A sub three. Okay. Does anybody have any questions about this? Okay, I just assumed that you've probably seen this before, so I wasn't going to spend a lot of time on this. Okay, so, et cetera. Okay, so now what we're going to do is we're going to prove something and we're going to use the principle of strong induction. In fact, what I'm going to do is incorporate another problem that I think I assigned to you. I'm not going to have time to actually do it, but I think I gave you, I think I assigned this. I think I gave you number two in the homework, right? Okay. And I'll tell you what, again, because of time, I was going to write this number two down because we're going to use it in the next problem, but I don't think I'm going to have time to do that. So I'm going to refer to it here in a bit. So let me show you how this goes. So I'm going to, we're going to do another problem now using the strong induction. Okay. And this is going to be a notation as in the previous example. We're going to prove that A sub n is less than two to the n, positive integer n. So here's what we're going to do. I'm going to try to get, I'm sure, yeah. I'll have plenty of time to get through this one. If I have time, I'm going to talk about this Bernoulli inequality problem that you have to do in your homework. I'll help you get started on that. So what we're going to do is we're going to use theorem two. You find the same way as what you were already talking about. Yes. Okay. Yes. So again, second principle of finite induction. It's much quicker just to say strong induction because that's what everyone else calls it. So what is it that we're doing? Well, it's very similar to the problem that I did before with the first principle. So the first thing I'm going to do is just verify the assertion is true for n equals one. And then instead of assuming it's true for n and proving it's true for n plus one, you assume it's true for one through n and then you prove it's true for n plus one. That's the only difference. The strategy in general is the same. It's just that now you have more to get your hands on because your assumption is stronger. So you have more to work with now. Okay. And instead of writing this p of n, I'm not going to write that now. So I'm just going to say, well, if we let p of n be a sub n is less than two to the n, p of one then would just be a sub one is less than two to the first. So that's our base case for the induction in this case. You guys with me on this? You with me? Okay. So again, we're going to... I'm going to label it the same way that I labeled the other one. So one. Okay. And I'm going to be very clear as to what it is that we have to show. We must prove or we must show. I use this word show very loosely here with the base case. A sub one is less than two to the first. All right. That's our base case for the induction here. Well, okay. What's the definition? A sub one is one. So A sub one being less than two then is obvious. Okay. So that's the first part. That's it. That's all you have to say. And as for the second part, okay. Now we're going to let n be a natural number. There are a couple of ways, of course, to do this. Again, I want to be very clear as I said before. I'm not saying that this is the only way to do it. I have to do something, though. So I have to choose one of the ways. This is one way to do it. If you're not sure about your way, if you want to ask as I'm going through this, that's fine, too. And assume, let's assume that the assertion is true. In other words, A sub n is less than two to the n, or A sub i is less than two to the i. For one, two, all the way down to n. Whatever n happens to be. Well, I'm saying it because I'm writing left to right. So I don't mean it from a mathematical standpoint. So what is it that we have to show? So we must prove that it's true for n plus one. And let me just use the same notation I used before. I'll just use the star again. So the assertion for n plus one, then, is the assertion that A sub n plus one is less than two to the n plus one. Is that okay? Okay, so there are a couple of ways you can go with this. I'm going to do it this way because I think it makes my point a little bit better. So what I'm going to do is what we're going to do is we're going to consider a few cases. Case one, so we don't know what n is, right? Maybe n's one, maybe n's two. Who knows? We don't know what it is. So what I'm going to do is I'm going to take care of the easy cases first, and then we're going to assume that n is big enough so that A sub n is defined in terms of the three previous guys. And then we'll use that strong induction hypothesis to take care of that. But remember that A sub n is A sub n minus one plus A sub n minus two plus A sub n minus three only if n is bigger than or equal to four, right? If we're on A2, we don't have three previous guys yet. So we can't apply that yet. We can only apply it when we know that the subscript is bigger than or equal to four because we need to have at least three previous guys in order to use, right? To use in order to get A sub n. So the first case is n equals one. Well, then we're just done by part one because we just verified that, right? We already did that. So we don't have to do that again. We already did it from the top. So let's see. So we have to show, sorry, maybe I was a long clear. We already know that A sub one is less than two. We already talked about that. So we still have to show that A sub two is less than four now, right? That's our assertion when n is equal to one. You see that? When n is one, the assertion is we have to prove that A sub two is less than four. You guys with me on this? So in this case, we have to verify that A sub two is less than four. But this is clear because by definition, A sub two is two, okay? So you should have that in your notes, right? The first three, right? A sub one was one. A sub two is two. A sub three is three. So this is clear just from the definition, okay? These are sort of the simple cases. Case two, n is equal to two. And then what's the assertion we need? We need four. A sub three to be less than what? Eight, right? So we must check that. A sub three is less than eight. Again, that's clear. This is clear since A sub three is equal to three. Okay, so there's another way to think of this. You could sort of think of this as sort of the base case of the strong induction argument, okay? In a sense, okay? Because we don't know that A sub n, remember, we don't know that A sub n is A sub n minus one plus A sub n minus two plus A sub n minus three. That's only true when n is bigger than or equal to four. So we have to take care of these simple cases first before we know that, okay? Now, the next case is when n is bigger than two, right? Then n is bigger than or equal to three, so n plus one is bigger than or equal to four. And now we can use the recursive definition in this case because we know we're big enough so that we can use it. Okay, so everybody have this down? Okay, so the third case is not going to be that n is equal to three. The third case is just that n is bigger than two. And then that'll take care of everything. So again, I'm going to write down just to be clear. Just to remind you what it is that we have to show, we must prove that A sub n plus one is less than two to the n plus one. Okay, now, by definition, okay, so we have to prove something about A sub n plus one, right? Well, we know what A sub n plus one is in terms of some of the previous terms, right? Again, I don't have this on the board. On the screen, I don't think it's going to help to scroll back. A sub n, this is in your notes, sorry. A sub n is A sub n minus one plus A sub n minus two plus A sub n minus three when n is bigger than or equal to four, right? So again, like I said, if n is bigger than two, and it's a positive integer, if it's bigger than two, then it has to be bigger than or equal to three, right? So n plus one is bigger than or equal to four. So then we can use this recursive definition then, because we know the subscript is bigger than or equal to four. So it's the sum of the three previous terms. What are the three previous ones? A sub n, A sub n minus one, and A sub n minus two. You guys with me here? Now we can use our assumption, and now I'm probably going to scroll back one just to make sure that everybody remembers this. Okay, let me wait until you're done writing this. So remember what, let's see, oops, sorry. Okay, so one more. Okay, there we go. What are we assuming? We're assuming that this assertion, that A sub n is less than two, and we're assuming that that's true for one, two, three, four, five, all the way up through n. Okay? So we're assuming that what it is that we want to prove is actually true for n, n minus one, and minus two. Everything below n plus one. Okay? So since n is less than n plus one, n minus one is less than n plus one, n minus two is less than n plus one.