 Sometimes, we want to prove an infinite number of cases. For example, the sum of any number of even numbers is even, is really the sum of two even numbers is even, the sum of three even numbers is even, the sum of four even numbers is even, and so on. We'll call this an infinite-ordered sequence of statements. There's an infinite number of statements to prove, but there's an orderly way to list all of them. To prove something like this, we can use mathematical induction. Mathematical induction works sort of like a set of dominoes. If we set things up correctly, then knocking over the first domino will knock over the second, which will knock over the third, and so on, and all of the dominoes will eventually get knocked over. And the proof by mathematical induction works as follows. We'll prove the first statement is true, then we'll prove that if the kth statement is true, then the k plus first statement must also be true. This allows us to prove all the statements after the first. So, for example, we have proven the first statement is true, and since the first statement is true, and we've proved that if the kth statement is true, then the k plus first statement must also be true, then we know that the second statement must also be true. But wait, there's more. Since the second statement is true, and we know that if the kth statement is true, then the k plus first statement is also true, then we know that the third statement must be true. But wait, there's more. Since we've already proven that if the kth statement is true, then the k plus first statement is true, then since the third statement is true, the fourth statement must also be true, and so on. For example, suppose we want to prove that the sum of any number of even numbers is even. And as an infinite ordered sequence of statements, the sum of two even numbers is even, the sum of three even numbers is even, the sum of four even numbers is even, and so on. So this is a candidate for a proof by induction. We'll prove the first case, the sum of two even numbers is even, rewriting this as a conditional, and remember that when trying to prove a conditional, you may always assume the antecedent true. So let n be the sum of two even numbers, and we'll put our destination, n, is even with a little bit of space between the two statements to provide the proof. Definitions are the whole of mathematics, all else is commentary. When we say that it's the sum of two even numbers, remember that an even number is the product of two and an integer. And again, remember, always write definitions as two conditionals. And so our definition says that if a number is even, then it is the product of two and some integer, and also if a number is the product of two and some integer, then it's even. So we know that n is two times something, plus two times something. Similarly, if we want to conclude that n is even, then our previous line has to say that n is two times something. And we can do a little algebra to join the beginning of the bridge to the end of the bridge. And again, we can summarize our proof by rewriting it as a conditional if n is the sum of two even numbers, then it is even. Now we prove the induction step. If the sum of k even numbers is even, then the sum of k plus one even numbers is even. So again, when trying to prove a conditional, you may always assume the antecedent is true. In this case, the sum of k even numbers is even. We want our conclusion to be that the sum of k plus one even numbers is even. And so we might say something like m is the sum of k plus one even numbers. Well, that's really just the sum of k even numbers, plus one more even number. But remember, we're assuming that the sum of k even numbers is even. So this sum is an even number. And so it's two times something. And one more useful strategy, use what you've proved. In our base step, we prove that the sum of two even numbers is even. Well, right here we have the sum of two even numbers. And so we know that the sum of k plus one even numbers is even. And we can summarize our proof by writing it as a conditional if the sum of k even numbers is even, then the sum of k plus one even numbers is even. And we can summarize our work. The sum of two even numbers is even, that's the base step. Then if the sum of k even numbers is even, then the sum of k plus one even numbers is even, that's our induction step. Consequently, since the sum of two even numbers is even, then the sum of three even numbers is even. But since the sum of three even numbers is even, then the sum of four even numbers is even. And since the sum of four even numbers is even, then the sum of five even numbers is even. And so on, which proves our statement.