 In this video, I wanna show two more examples of how a typical induction proof looks like. When it comes to an induction proof, there's always three parts. You have a base case where you're gonna prove that the statement is true for basically the smallest natural number that's appropriate. Then you're gonna do the inductive hypothesis, which is just a statement of an assumption. You're gonna assume it's true for an arbitrary case. And then you'll have the inductive application for which then you'll prove that the next number also has the property because of the inductive hypothesis. And those three ingredients together give us a typical induction proof. So with this one right here, let's do a divisibility type argument. Maybe you'd do something like this in number theory. Prove that every integer of the form 10 to the n plus one power plus three times 10 to the nth power plus five is always divisible at nine for any natural number n. And so the proof right here, it's usually good when you start a proof by induction to say that what you're gonna do, right? So we'll say something like we proceed by induction. Now, when you're watching these videos, I wanna make you aware that when I write these proofs in real time, this isn't the best polished proof. You might wanna go to the lecture notes, which are linked to this video to actually see a better looking version of the proof. The reason I'm doing this in real time, as opposed to just reading a proof here, is that this is meant to be instructional. As a professor, I'm trying to teach you how to write proofs, right? And so I wanna write it in real time. So you know what I think about when I would write in a proof here? Different styles, different structures of the proof. Although this is not necessarily a final draft of a proof, I'm trying to show you the structure of how one does these type of proofs. So you should start off by telling people what you're gonna do, right? We're gonna proceed by induction. You don't wanna just kind of randomly walk through a proof and leave the audience unaware of what's going on here. It's like, what you should tell them we proceed by induction or something equivalent to that. And if you wanna actually enumerate the parts, you can do that. I don't necessarily do that myself when I write the proof, but as your professor here, I'm gonna indicate exactly what we're doing here. So the base case. So we wanna consider what's the smallest natural number for which this property is gonna be true. And although n equals one makes sense, it turns out we can be a little bit better, n equals zero. When you're improving stuff by induction, it's always a good idea to think about what n equals zero makes sense. And if it does make sense, I would actually include that as the base case. So let n equals zero. And so then consider in that situation, your number, right? You have 10 to the n plus one plus three times 10 to the n plus five. That would look like 10 to the first plus three times 10 to the zero plus five. 10 to the first will of course be 10. 10 to the zero is a one. So you're gonna get a three plus five. This adds up to be 18. 18 is two times nine. And so therefore it holds for the base case. All right, so that's kind of like our base case. So we finished it, so now we move on. Great. So then we go to our inductive hypothesis in which case we're going to assume assume that nine divides the number 10 to the k plus one plus three times 10 to the k plus five for sum number k, which is great equal to zero, right? That's one way you could phrase it or another way you could say it like, so there exists some integer, what do I wanna call it? Let's call it say m. There exists some integer m such that 10 to the k plus one plus three times 10 to the n, 10 to the k, excuse me, be consistent in your variable there plus five equals nine m. That's what we'll do. So we'll actually have the multiple specific here. Now to the inductive case, right? So now using the inductive hypothesis we wanna consider when we take n to be k plus one. So now consider the case when n equals k plus one. So we're gonna take this number. We're gonna take 10 to the k plus two plus three times 10 to the k plus one plus five. We wanna show that nine divides this number in some way or another, okay? And so this is why proofwriting is a creative process because the next step might not be obvious why we take it as one explores these problems and just kind of see natural to do it here. It's like, okay, what's the next step? Well, I have to somehow find the number 10 k plus one plus three times k and plus a five, right? And I kind of have that here, but I have an extra 10. Maybe we should factor out that 10. So for example, if you factor out the 10 from the first two numbers there, you're gonna get 10 times 10 k plus one plus three times 10 to the k, like so. And then you have the plus five still. So it's kind of what you have there. It's like, well, you know, I almost have what I need except I need a plus five right here. So what if I just add a plus five, you know? No one's looking, you sneak it in there. Well, we want it to be true, right? We need equality here. And so if we add a five, we should just subtract a five. But we really didn't add five. We took 10 times five, right? So this 10 would distribute onto each of these terms right here, right? So this is kind of like when we complete the square. So we're gonna actually subtract five times 10 to balance out the five we added there. The reason why this is useful is that this number right here is then by our inductive hypothesis gonna equal nine M. And so this is how we're using the inductive hypothesis when we use it right here. So continuing on there, we're gonna get 10 times nine M plus five minus 50. Like so, simplifying this, we get nine times 10 M minus 45. Oh, looky there, 45 is a multiple of nine, isn't it? It is nine times five. So we get nine times five. If you factor out the nine, you're gonna get nine times 10 M minus five. And therefore we see that nine divides the number in question here. And so this is exactly what we were trying to show. And so that then proves the inductive case. And so then we conclude with something like therefore the result follows by induction. And that is how our induction proof will work. We had our base case, our induction hypothesis and our inductive case. That's how it finishes. Let's look at one more example of this. This time, we're actually gonna prove it in equality using induction. Again, this is gonna be a little bit more tricky than in an equation, but the principles of induction are gonna work the same way here. As we try to prove it, this is gonna be approved by induction. So we proceed by induction. And so we're gonna first consider the situation. What's the right natural number to start here? Oftentimes they should tell you, I'm gonna start with n equals zero because n equals zero, the statement's true. So notice when n equals zero here, you're gonna get that two to the zero is equal to one, which is greater than zero, like so, great. And so this, whoops, this proves the base case. So we have some base case. I usually like to start with the base case. So then next, we'll make our induction hypothesis. Assume that two to the k is greater than just k for some k greater than or equal to zero. This is our inductive hypothesis. Now, consider the case where n equals k plus one. So you're gonna have some inequalities here, right? You're gonna have two to the k plus one. We have to somehow relate this to our inductive hypothesis. So I'm gonna probably do that by factoring, two to the k plus one is two times two to the k, which then by our inductive hypothesis, we know that two to the k is greater than k. So this will be two times k right here. And again, it's a good idea to be explicit here. Our inductive hypothesis was used on this inequality right here. And so now, well, how do we connect this to what we're looking for? Two k right here, well, two k, this is equal to k plus k, like so. And k is actually gonna be greater than or equal to one, k plus one here. And so we actually, since we already took care of the base case, we could actually, which is zero, we can actually say that k is strictly greater than zero, right? I guess actually, I have to be a little bit careful here. We want k plus k to be great equal to one. So the thing is, this is where it gets a little bit fishy, right? It turns out that the argument doesn't quite work when k equals zero. We actually need that this thing is greater than or equal to one to make the following argument work. So that if k is great equal to one, then this k is great equal to one. And then we get the statement we want right there. And so in this situation, you might actually be justified in doing a second base case. Sometimes you need to do that. Let's do n equals zero or n equals one. If you take n equals one, that's equal to two, which is greater than one, it's not a big deal. If you have to do an extra base case or two, that's perfectly fine. Do as many as necessary so we get this thing to work, right? It's true for zero, but for this argument to follow, it kind of needs to be greater than equal to one. So we make that statement right there and then we're done. Therefore, the result follows my induction. And so this gives us two more examples of a typical induction argument. And in the homework, you'll have plenty of opportunity to practice this principle some more.