 In lecture 19 of our lecture series, we introduced the notion of mathematical induction. We called it the first principle of mathematical induction as the name suggests there's a second principle of mathematical induction that we are gonna introduce now. Now the first principle of mathematical induction that we introduced earlier is typically just referred to as the principle of mathematical induction or the principle of induction or just induction for short. The second principle of induction that we're introducing right now is commonly referred to as strong induction. Some people refer to this as the first principle by weak induction, but that's not a very common term, but strong induction is commonly used here. And when you compare the principles, the first principle and the second principle of induction, you'll see why this adjective strong is used here. So let's state what the second principle induction says. Let S of N be a statement about integers where you have N is a natural number. And suppose that S of N sub zero here, well, so just so you'll be clear here, N sub zero could be any integer. It's indexed by the natural number zero in this case. So suppose that we know it's true for S of N sub zero. So this is our base case. We have the base case established. If for all integers K, so that K is greater than or equal to N sub zero, we know that S of N zero is true. S of N zero plus one is true. S of N zero plus two is true. All the way up to SK is true. If that implies that S of K plus one is true, then SN is true for all integers greater than equal to N sub zero. Now on the surface, if you're not careful about how we read this, we might not see what's the difference between the first principle and second principles in induction. For the first principle of induction, the difference comes down to your inductive hypothesis. So you have your base case. That part's exactly the same. For the, for the, for regular induction, your inductive hypothesis only assumes, only assumes that SK is a true statement. And then you use SK to show that SK plus one is true. With strong induction, you actually assume that all of the statements up until SK are true. And then of course, if that implies that SK plus one is true, then it's true for all of the integers greater than N zero. And so that's what makes it strong. It has a stronger assumption. That is to say that the, there's a stronger inductive hypothesis. You have more assumptions about the predecessors. For weak induction, you only assume that the immediate predecessor has the property. But for strong induction, you assume that all of the predecessors have that property right there. So if I were to rewrite these principles in induction, you're gonna get what I call weak induction. Again, this is the first principle induction and we'll abbreviate this as I for short. I for induction. We can then actually write this as a statement. I'm gonna use logical forms right here. So instead of the symbol S, let's recall it P for a moment. The statements, because I wanna use S for strong induction in just a moment. So if we really relabel all of these statements as P, right? So we have this sequence of statements P sub N. So this is a statement. A statement about N, okay? So then we get things like the following. So for induction, we have a base case. And for simplicity sake, I'll just call P sub zero. It should be N sub zero. It doesn't have to start at zero. But again, for simplicity, we're just gonna call it zero. Up to relabeling, if the statement is about N, we could make it a, and if the statement is about N sub zero, we could make it about zero itself. That's just semantics at that point. So you assume that the base case is true. You also assume, so there's an N statement here, for all K, for all K, which are natural numbers, because we're starting with zero here, that P of K implies P of K plus one. Okay? So we know that P of zero is true. And if P of K is true, then P of K plus one is true. That then implies the statement for all N, which are natural numbers, we have, let me rewrite that natural number, it looked a little bit better, we have that P of N holds. Now I could fit that in there. P of N holds. Little cramp, but we can do it. So what this means here is that you read this conditional statement. We have the base case as a hypothesis. And we also have that, this is our inductive step here. If P K holds, then P K plus one holds. If this is true for every natural number, then these imply this right here, that the statement holds. This is, if you write it in symbolic logic, this is what the principle, the first principle of mathematical induction tells us. Now let's then change our perspective here into strong induction. How does that change things? So we'll call this one S for strong induction. Well, we both assume, both of them assume the base case, but what's different now is that we have our inductive step is very, very different. It starts off the same for all K inside of the natural numbers. We have this big old A N statement where I ranges from zero to K, P of I. So we assume all of the previous steps are true. That then implies P of K plus one is true. And so then if you have these hypotheses, this implies that for all N inside of the natural numbers, P of N holds, like so. So the difference I'm trying to highlight here is that in this situation, we only assume P sub K. And in this case, we assume everything up until P sub K. So this gives us the first, this gives us the first implication pretty quickly, that I implies S. We get that one very immediately, that if I is a true statement, then S is a true statement because S has more hypotheses. And so if I can prove this to be true with fewer hypotheses, then adding more hypotheses is perfectly fine. It might be superfluous to add the extra hypotheses, but we get that direction. That's the easy direction. Whoops. This is the easy direction. The other direction is a little bit more challenging to get here, but it's actually not so ridiculous when you actually try to unravel it a little bit here because this is how induction works. Honestly, if you think of it like a recursive sequence, what this is basically telling you is that if P sub zero is true, then that implies because we know these implications here, we know that, so what we're trying to do here is that we're trying to, we assume S to be true. So we assume the principle of strong induction. Now we wanna prove this principle of weak induction here. Well, induction itself is a conditional statement. So we assume this to be true and then we need to prove that this is true. So what we're gonna do here, so we have that Pk implies Pk plus one and we have the base case, okay? So if P zero works, that means P one works because P zero implies P one, but then also means by this P two works and by this P three works and P four works and P five works until we get up to Pk. And then once we have all of them, P zero, P one, P two, P three, P four up to Pk, we've then satisfied the conditions of strong induction. So P zero plus the strong assumption then gives us Pn for all natural numbers. That's then the conclusion of weak induction. And so sure enough, there's a little bit of more of an argument here, but strong induction implies weak induction, but also weak induction implies strong induction. So from a logical perspective, these two principles of mathematical induction are actually logically equivalent to each other. The domino effect is the same in both of them. And so anything that you could prove using induction, you can also prove using strong induction and vice versa. Now, when it comes to strong induction, I would say that it has the advantage that your inductive hypothesis is stronger because you have it true for all of the previous cases, not just the immediate predecessor. That can be extremely useful in proofs when the regular induction doesn't necessarily have that. So some people say like, I will only use strong induction when I must use it, which to be fair, you never need it because it's equivalent to weak induction. Now I confess that there's, it really comes down to writing, right? It comes to writing style that for me personally, if I'm trying to prove something, do I use strong induction or do I want, do I use weak induction? Which one do I use? Well, it kind of comes down to that if I only need the immediate predecessor to prove my inductive step, I'm going to use weak induction and that's how I'm going to do it now. But if I do want to use more predecessors than just the immediate predecessor, then I'll use strong induction right there. It's never wrong to use strong induction always. That is, you just always take the most assumptions that you can, even if you don't use them, there's nothing wrong with that. Some people shy away from it because they're like, well, that's you're using extra assumptions, you don't need shame on you, but it's like they're there, they're true. But so that's how I break it down. If I'm only going to use the immediate predecessor, then I will write it as an induction proof. And that's how I kind of do it by default. But then as I'm writing the proof, if I realize actually I want to use not necessarily the immediate predecessor, but I want to use any predecessor, I'll then rewrite it in revision to a strong induction proof and go forward from there. So let me give you some examples of proofs by strong induction because these would be two good examples on why you might want a predecessor besides just the immediate predecessor. Because with our previous induction arguments, we'd only needed the immediate predecessor. Why might you want a generic predecessor? Well, it turns out when you make arguments about divisibility, about divisors of an integer, the immediate number in front of it is not a divisor. Like if you have N, the number N, you can actually show that N minus one does not divide it. But at the same time, the divisors take N to be a positive integer here, the positive divisors of N aren't gonna be smaller numbers than N, but not the immediately first smaller number there. So they're gonna be somewhere smaller for which, so this is why strong induction can be very useful because if you know the statement's true for all smaller numbers and the divisor shows up here as some smaller number, it can be quite useful. So let's demonstrate that. Two proofs about divisors. So let's take this one here. Let N be a natural number such that N is greater than one. So I don't wanna use zero here and I don't wanna use one because when it comes to divisibility, N and zero are a little bit different, right? Zero has this property that anything times zero is equal to zero and one has the property anything times one is equal to N. One is the multiplicative identity and zero is the multiplicative absorber. And so because of that, their divisibility traits are a little bit different. So we'll just focus on positive integers greater than one, so like two, three, four, five, et cetera. It turns out that every natural number greater than one has a prime divisor and we can prove this by strong induction. To do this, we start off with a base case. So the smallest number in play here is two, right? N equals two, two does have a prime divisor because two itself is a prime number. Two divides two and that's our prime divisor. So then for the sake of induction, we're gonna assume that all natural numbers from two to K minus one have a prime divisor. So every number from two to K minus one, let's assume we have shown that they have a prime divisor. Now let's consider the K, the K number. So when N equals K, can we show that K has a prime divisor? Well, if K is itself a prime number, then K divides K just like it did with two divides two and we're done, all right? So if K, we're doing this by cases here. If K is prime, we have an argument. If K is not prime, then K is necessarily composite and there exists positive integers A and B such that K equals A and B. Now this gets to the fact that I was mentioning earlier. If K is a positive number and A and B are positive divisors of it, I can then argue that A and B are numbers between one and K, all right? So I can assume since this is a composite number that A is not K itself. But I also know that A is not equal to one because if A was one, that would imply that B was K, but I also can assume that B is a proper divisor of K. So these two numbers, A and B are divisors that sits somewhere between one and K. And as such, the inductive hypothesis applies to these numbers. A and B, neither of them are gonna be K minus one, but there's some numbers between two and K minus one. So the inductive hypothesis applies. Therefore, the number A has a prime divisor, A equals P times C, where P is a prime and C is just any integer, I don't care what it is. And therefore, when you look at K, K equals A times B. A, I can then factor as PC. And so I can write that as K equals P times CB. Therefore, P divides K and K then has a prime divisor. And hence, every positive integer greater than one has a prime divisor by strong induction. So you saw in this argument here that I used my inductive hypothesis on the number A. I know A sits somewhere between one and K, but I don't know where it is. And this is why it was extremely useful to have my hypothesis cover all numbers between two and K minus one, because I don't know where A landed and that's okay. Strong induction gives that to us. Let's look at one more example of strong induction here. For all positive integers N and primes P, there exists a natural number K, such that P to the M divides N, but P to the M plus one does not divide M, N here. This is sort of like a, this natural number here should be an M, not a K. Sorry about that. So we have, for every positive integer N and every prime, there's this maximum number of times that the prime divides into it. This is sort of like a precursor to the fundamental theorem of arithmetic, which tells us that every natural number has a prime factorization, I mean, other than zero and one. We excluded those for the previous proof for that reason here. So there's this maximum number of times a prime divides it. Like for example, if I take the number 12, 12 is equal to two squared, two squared times three. And so two squares, the maximum number of times that two divides 12. On the other hand, you can take something like say eight, this is two cubed times nothing else. So two, three is the maximum number of times two divides eight. We can take like 15, prime factorization is three times five, and actually two divides that no times. So you get two to the zero in that situation. It is a natural number. So that's what we wanna argue right here. I'm gonna again prove this by strong induction. So we're gonna start off with N equals zero. At this time, I am doing pause integers. So N equals zero, and I don't actually care which prime this is. This is just a generic prime here. When you take N equals one, excuse me, P to the zero, which is equal to one, divides one, one divides one. But on the other hand, P to the first does not divide one. No primes divide one. So this gives us the base case. For the number one, your base case is in fact equal to zero. Now also be clear as we go through this argument, the prime P is fixed throughout the whole thing. It's arbitrary, but it's fixed. It's not a different prime, even though I don't care which prime it is. So for our inductive hypothesis, we're gonna assume the statement holds for all numbers between one and up to K, not including K at that moment. We know it's true for one, and we're gonna assume that every number two, three, four, five, et cetera, up to K, it's also been proven to be true. So let's consider N equals K. Well, we have two possibilities. If P does not divide K, that could happen, then we set M equals zero and we're done. Cause again, M is a natural number. That's the natural number we're gonna choose right there. On the other hand, if P does divide K, then that means we can factor K so that it looks like P times A. Now, like we talked about in the previous proof, A is a positive divisor of K, so it's gonna be less than K, but it's also gonna be greater than or equal to one. So the inductive hypothesis applies in the strong sense. So I don't suspect that A is equal to K minus one. In fact, it can't happen because it's not gonna be divisor of K in that situation. Of course, there is a notable exception, like a K equals two, then of course, one is equal to K minus one. It is a divisor in that situation. But the important thing is I'm not supposing that A equals K minus one. It's probably not. But because a strong induction, as long as I'm any number smaller than K, then the inductive hypothesis applies. So therefore, with regard to the prime P, there is some number M such that P to the M divides A and P to the M plus one doesn't divide A, which actually would suggest that P doesn't divide B in this situation here. So we have this factorization, A equals P to the M times B and the prime P does not divide B, otherwise you'd have a larger power of P dividing A. And so we're gonna insert this into the previous equation here. So K equals PA, which equals P times PM times B, which then gives us PM plus one times B. Now, like we said earlier, P does not divide B. Therefore, it turns out that M plus one is the number we're looking for. P plus two doesn't divide K because that would then force us to have that P divides B and we don't have that situation. This is a consequence of Euclid's lemma, which we'll talk about actually in the very next, later on in this unit about induction. So don't worry about that right now. And therefore the proof then follows by strong induction, just like we saw before. This is again, this example of using divisors is really a good place to emphasize strong induction because I need to know information about a generic predecessor as opposed to the immediate predecessor, strong induction has a huge advantage. And try not to get dog down into this debate about whether I should use strong induction or not. If you write every induction proof using strong induction, there is nothing logically wrong with that. It comes only down to stylistic reasons or cultural reasons, like everyone shuns you for using strong induction when you didn't need to. But using weak induction is logically equivalent to strong induction. So from a logical point of view, it doesn't matter which one you use. It honestly comes down to the composition of the proof. These two examples that we've seen in this video were so much easier to write the proof using strong induction over weak induction, therefore we use strong induction. But I could rewrite the proof using weak induction, that's perfectly fine. On the other hand though, if I'm writing a proof and I only need the previous term, I can write it as a strong induction proof, but you're making assumptions you don't need and that doesn't lead to great mathematical writing. So I would then suggest weak induction in that regard. So in the end, it doesn't matter which one you use, it's only for the sake of composition that you're gonna decide whether you use the strong induction or the regular induction principles.