Proof: √2 is irrational

What I want to do in this video is prove to you that the square root of 2 is irrational. And I'm going to do this through a proof by contradiction. And the proof by contradiction is set up by assuming the opposite. So this is our goal, but for the sake of our proof, let's assume the opposite. Let's assume that square root of 2 is rational. And then we'll see if we lead to a contradiction, that this actually cannot be the case. And if it cannot be the case that is rational, if we get to a contradiction by assuming the square root of 2 is rational, then we have to deduce that the square root of 2 must be irrational. So let's assume the opposite. Square root of 2 is rational. Well, if the square root of 2 is rational, that means that we can write the square root of 2 as the ratio of two integers, a and b. And we can also assume that these have no factors in common. Let's say that they did have some factors in common. If we divided the numerator and the denominator by those same factors, then you're getting into a situation where they have no factors in common. Or another way of saying is that a and b are co-prime. Or another way of saying it is we could write this as a ratio of two integers where this is irreducible, where these no longer share any factors. If you can write anything as the ratio of two integers, then you could obviously simplify it further, factor out any common factors to get it to a point where it is irreducible. So I'm going to assume that my a and b, that this fraction right over here, is irreducible. And this is important for setting up our contradiction. So I'm going to assume that this right over here is irreducible. a and b have no factors in common. Let me write that down because that's so important for this proof. a and-- want to do that same color-- a and b have no factors in common, other than, I guess, 1. So this is irreducible. These two numbers are co-prime. So what does that do for us? Well, let's just try to manipulate this a little bit. Let's square both sides of this equation. So if you square the principal root of 2, you're going to get 2. And that's going to be equal to a squared over b squared. And that just comes from a over b squared is the same thing as a squared over b squared. And now we can multiply both sides of this by b squared. And so we get 2 times b squared is equal to a squared. Now, what does this tell us about a squared? Well, a squared is some number, b squared times 2. So anything times 2 is going-- this is going to be an integer. We assumed b is an integer, so b squared must be an integer, and so you have an integer times 2. Well, that must give you an even number. That must give you an even integer. So this right over here, a squared, must be-- so this tells us that a squared must be even. Now, why is that interesting? Well, a squared is the product of two numbers or is the product of the same number. It's a times a. So this is another way of saying that a times a is even. So what does that tell us about a? Let's just remind ourselves. a is either going to be-- we're assuming a is an integer-- a is either going to be even or odd. We just have to remind ourselves if we multiply an even times an even, we get an even number. If we multiply an odd times an odd, we get an odd number. So we have a number times itself. We got an even number. Well, the only way to get that is if that number is even. So this tells us that a is even. And another way of saying that a is even is to say that a can be represented as the product of 2 times some integer. So let's say some integer k. So where is all of this going? Well, as you'll see, we can then use this to show that b must also be even. So let's think about that a little bit. So let's go back to this step right over here. If we say that a can be represented as two times the product of some integer, and that comes out of the fact that a is even. Then we can rewrite this expression right over here as 2-- I'll do it over here-- 2 times b squared is equal to 2k squared. Instead of a squared, I could write 2k squared. We're claiming, or we're deducing, that, assuming everything we've just assumed, that a is even. So if a is even, it can be represented as a product of 2 and some integer. And then we can write that 2 times b squared is equal to 4k squared. And then you divide both sides by 2. You get b squared is equal to 2k squared. And this tells us that, well, k squared is going to be an integer. You take any integer times 2 you're going to get an even value. So this tells us that b squared is even. So that tells us that b squared is even. Well, if b squared is even, by the same logic we just used, that tells us that b is even. So here's our contradiction. We assumed, in the beginning, that a and b have no common factors other than 1. We assumed that this fraction right over here, a/b, is irreducible. But from that and the fact that a/b must be equal to the square root of 2, we were able to deduce that a is even and b is even. Well, if a is even and b is even, and they both have 2 as a factor, and then this isn't irreducible. You could divide the numerator and the denominator by 2. a and b have a common factor of 2. So let me write this down. So this is just to make it clear. So from this and this, we have a and b have common factor of 2, which means that a over b is reducible. And so that's the contradiction. So you assume that square root of 2 can be represented as an irreducible fraction a/b, irreducible because you can say ratio of two integers right over here, that leads you to the contradiction that, no, it actually can be reducible. So, therefore, you cannot make this assumption. It leads to a contradiction. Square root of 2 must be irrational.