 Welcome back to another screencast where we're going to see another example of proof by mathematical induction. This time one involving an integer division property, this proposition right here. We're going to prove that for all positive integers n, 3 divides n cubed minus n. We've actually seen this proposition before. I believe it's in the Sunstrom textbook proven in a different method. So this will be an interesting situation where you can prove the same result in multiple quite different ways. We're going to do this by induction this time, so let's identify what the parts are. First of all, notice this is a good candidate for proof by induction because we have a predicate. This guy right here, I'll just call it p and n. That is a predicate because the statement is neither true or false as is, but once I put in a value for n, a specific number value for n, I can definitely determine whether it's true or false. What we're claiming here is that this predicate is true for all positive integers n. So this sounds like mathematical induction. That's why we're going this route. So as with all proofs by mathematical induction, we have two stages to complete. We need to complete the base case first where we're going to show that p of 1 is true, and then we're going to work with the inductive step. So we'll get to the inductive step later, let the base case now. So what we're going to show here is that p of 1 is true. Let's write that in a little bit more informative way. We want to show that 3 divides 1 cubed minus 1. That's what p of 1 literally is. So let's just kind of think our way through this. Well, what is 1 cubed minus 1? Well, that's a simply 0. And we just sort of note that definitely 3 divides 0 in any case. And so that establishes the base case because 3 does indeed divide 0. Base cases usually are quite often very, very simple to do. It's not much even to say, just something we notice. So the base case is true onto the inductive step. And this is where things get a little tricky here. We have to be careful about what we're going to assume and what we're going to prove. Up here in the blue margin, I'm going to rewrite my proposition. p of n, the proposition we're trying to show is true for all positive integers n, is the statement that 3 divides n cubed minus n. So we're going to assume one thing and prove something related. We're going to assume that for some positive integer k, positive integer is the same thing as a natural number. So I'll just write it like this. For some positive integer k, 3 divides k cubed minus k. This is an assumption we are making. Like in the story about the castle and the staircase, we're going to assume, optimistically, that we've made it up to a certain level on the staircase, up to k. And then what we're going to prove is that 3 divides, I'm going to replace the n here. You see up here, following my laser pointer, with a k plus 1. I'm just going to literally do that first. So k plus 1 cubed minus k plus 1. And maybe I'll just leave it at that. That's what I want to show. That is p of k plus 1. Again, that's what the statement is. So I'm assuming that p of k is true. I'm assuming that this statement works automatically. And I'm going to prove that this statement is true, based on that assumption. So let's make a new slide here and start doing the math. What we're trying to prove here is that 3 divides a certain quantity. So what I'm going to do is that quantity. Okay, so let's look at the quantity. I'll change my color to blue here. Let's look at the quantity k plus 1 cubed minus k plus 1. And let's just expand this out. I mean, it seems like the logical thing to do. Let's just do some valid mathematical steps until something becomes clear. So to cube this out, just use your FOIL method, or for all intents and purposes, just use Wolfram and Alford or something to expand this out. That's k cubed plus 3k squared plus 3k plus 1. And then I'm subtracting off k plus 1. I'm going to go ahead and subtract all the terms in that sum. So again, this four-term polynomial came from cubing this. The subtraction is here and then k plus 1 is this. Remember, we're subtracting everything in the group. So now let's do a little cancellation. I can cancel the ones here. So those are gone. And what I'm going to do next is regroup the remaining terms in a somewhat clever way. I could, I know that I could just simply subtract the like terms here, but I'm going to do something a little different. I'm going to take the k, the minus k right here, and move it over next to this. Now, why do that? Well, let me do it first and maybe you'll see why I'm going to go that route. So I have k cubed minus k, and then I have the other stuff left over 3k squared plus 3k. Now, why did I do that? Why did I move this k over to here, just now here? Well, it's because if you look carefully at this term, I know something about it. Remember, the inductive hypothesis said that we can assume that 3 divides k cubed minus k. Well, there is k cubed minus k right there looking at me. So I know that 3 divides that. And so I know that there exists an integer, sorry, let's call it q, such that I could write that as 3 times q. And again, just through that logic again, I move the k over to here to make it look like something in the inductive hypothesis. I know that this expression right here is divisible by 3, so there exists an integer q such that this stuff in the orange is equal to 3q. Now, what about the leftovers? Well, here I have 3k squared plus 3k. And look at all those factors of 3. I'm going to pull that factor 3 out and I have all this stuff left over. So let me just verbally go through an argument here. By closure, this thing here is an integer. And so what I have, if you read all the way from the top to the stack to the bottom is that the k plus 1 cubed minus k plus 1, that's the thing I wanted to show was divisible by 3 is equal to this, which is equal to this, which is equal to this, which is equal to this, and this is 3 times an integer. And so what that allows me to conclude is the very thing that I want to conclude. And that is that 3, I don't have a lot of room to write this, but I can conclude it, 3 divides k plus 1 cubed minus k plus 1. That's exactly what I wanted to show. And so the proof is done because I have established the base case. That was easy. That came early on. And if I assume the inductive hypothesis, this expression up in here, what I get is that the next step works out and the proposition holds for that k plus 1 step as well. And so there's no limit to the heights that I can reach on this staircase. This predicate is true for all positive integers n. So that's the magic and the power and everything else of mathematical induction. Keep watching if you want more examples because we have more to come.