 Proof by contradiction is one of the more important proof strategies. We make an analogy. Suppose the bus you get on to go home does not cross a river. You get on a bus, and after a while you look out and see you're crossing a river. Based on this observation you can conclude, you shouldn't have gotten on the bus. Or at the very least, this is not the bus you should have gotten on if you want to go home. Proof by contradiction rests on the following idea. An important guideline is the law of the excluded middle. A statement must be true or false, but it can't be both. Suppose we assume a statement x to be true and eventually deduce a statement y. Then we've proven y true. But if we know that y is in fact false, this is a problem. Since letting x be true takes us to a statement y that is true and false, we can't let x be true. So this is like recognizing that we're not supposed to be crossing a river, and so that means we got on the wrong bus. And the law of the excluded middle says that it must be true or false, and if x can't be true, x must be false. Note that we didn't have to assume x to be true. We might have assumed x to be false and eventually deduce a statement y. And again, if we knew y is false, this is a problem. And again, the only option is to decide that we got on the wrong bus. And since the bus we got on was x was false, it follows that x has to be true. The essential feature here is that whatever we assumed, we eventually got at something we knew to be false, which can't happen. Now the classical example of a proof by contradiction is the following. We're going to prove that the square root of 2 is irrational. So again, let's start off by rewriting our statement as a conditional. If a number is the square root of 2, then it's irrational. Again, we can always assume the antecedent of a conditional, so we have a number that's equal to the square root of 2. We want to prove the consequent is true, that the number is irrational, and we'll do that by showing it can't be false. So if it's not irrational, then it must be rational. Definitions are the whole of mathematics. All else is commentary. A rational number is the quotient p divided by q of two integers with q not equal to zero. And so the bus we'll get on is the bus that says, suppose square root of 2 is equal to p divided by q. Now a useful idea here is you can always make additional assumptions provided you make them explicit. And so in this case, when we write a fraction like p divided by q, we might be able to reduce this fraction. Well, reducing the fraction or not doesn't really affect whether or not we have a rational number. So let's assume that p divided by q is a fraction in lowest terms. But definitions are the whole of mathematics. All else is commentary. What does that mean? Remember that a fraction is in lowest terms when the numerator and denominator have no common factors. And so the additional assumption we'll make here is that p and q have no common factors. And we'll do some algebra. Square root of 2 equals p divided by q allows us to rewrite this as definitions are the whole of mathematics. All else is commentary. p squared is 2 times q squared. And so that tells us that p squared is even. Now, we've already shown that if p squared is even, then p has to be even. Definitions are the whole of mathematics. All else is commentary. The fact that p is even says that p is equal to 2k. And so p squared is equal to 4k squared. But then that means that 4k squared is equal to 2q squared. So q squared equals 2k squared. And so q must also be even. Since p and q are both even, so they have a common factor. But we know this is false. We said they did not have a common factor. And that means we got on the wrong bus. Our initial assumption must be false. Since our initial assumption was that the square root of 2 is equal to p divided by q, since this must be false, and so the square root of 2 is not equal to p divided by q, it must be irrational. Now, if you think about this, there is a second way that our assumption can give, which is maybe it's not true that p and q have no common factor. And in fact, this leads to a very important idea. Once you've proven a statement, try to prove it again in a different way. And part of the reason here is that this gives you additional practice in proving statements. So remember that proof is a way of studying mathematics. It's a way of reminding yourself what you should know about mathematics. The other reason that proving something in a different way does is that it may allow you to discover something you hadn't known about mathematics. Proof is a way of creating or discovering more mathematics. So it turns out we don't actually need the requirement that p and q are reduced to lowest terms. Instead, we'll use another strategy, lather, rinse, repeat. So again, we'll start off with the same assumption. Square root of 2 is p divided by q, where in this case, p and q are positive integers. Let's pick up from the point where we show that both p and q are both even. Definitions are the whole of mathematics. All else is commentary. That means that we know that p is 2 times something and q is 2 times something where p is greater than p prime and q is greater than q prime. And if we remove that common factor, we get square root of 2 is p prime divided by q prime. But that takes us right back to our starting point. Square root of 2 equals p prime divided by q prime. And so now we know that p prime and q prime are both even. And so p prime is 2 times something. q prime is 2 times something where our somethings are smaller numbers than we had before. And removing our common factor, we find that square root of 2 is p double prime divided by q double prime. And that takes us right back to our starting point. And there's no way out of this loop. And what happens is this gives us an infinitely long but decreasing sequence of positive integers, which is impossible. And again, this means that we got on the wrong bus. And so square root of 2 cannot be equal to p divided by q for any integers p and q. And if you think about what we said earlier, there's still a way out. We could still have square root of 2 be p divided by q. And we'll leave closing off that last possibility as an exercise for the reader because once you've proven a statement, try to prove it again in a different way.