How does the proof that the square root of 2 is irrational use contradiction?

"One of the most famous indirect proofs dates from Euclid’s Elements in 300 B.C. It proves that the square root of 2 is an irrational number, a great surprise to mathematicians at the time, most of whom doubted the very existence of irrational numbers. Remember that an irrational number is one that cannot be expressed as the ratio of two integers, no matter what the integers are. Proving this directly seems pretty hard, since how do you prove that there aren’t any two integers whose ratio is √2, no matter how hard you looked? I mean, 534,927 and 378,250 are pretty dang close: (534,927/378,250)² = 2.000005. How could we possibly prove that no matter how hard we look, we can never find a pair that will give it to us exactly? One way is to assume that √2 is a rational number, and then prove that down that path lies madness. It goes like this. Suppose √2 is rational, after all. That means that there must be two integers, call them a and b, whose ratio is exactly equal to √2: a/b = √2. This, then, is the starting point for our indirect proof. We’re going to proceed under this assumption and see where it leads us. By the way, it’s clear that we could always reduce this fraction to lowest terms in case it’s not already. For instance, if a = 6 and b = 4, then our fraction would be 6/4, which is the same as 3/2, so we could just say a = 3 and b = 2 and start over. Bottom line: if √2 is rational, then we can find two integers a and b that have no common factor. If they do have a common factor, we’ll just divide it out of both of them and go with the new numbers. Okay then. But now look what happens. Suppose we square both sides of the equation, a/b = √2, a perfectly legal thing to do: (a/b)² = (√2)², a²/b² = 2, and a² = 2b². Now if a² equals 2 times something, then a² is an even number. But a² can’t be even unless a itself is even. This proves, then, that a is even. Very well. It must be equal to twice some other integer. Let’s call that c. We know that a = 2c, where c is another integer. Substitute that into the last equation and we get: (2c)² = 2b², 4c² = 2b², and 2c² = b². So it looks like b² must be an even number as well, since it’s equal to 2 times something, and therefore b is also even. But wait a minute. We started by saying that a and b had no common factor. And now we’ve determined that they’re both even numbers! This means they both have a factor of 2, which contradicts what we started with. The only thing we introduced that was questionable was the notion that there are two integers a and b whose ratio was equal to √2 to begin with. That must be the part that’s faulty then. Therefore, √2 is not a rational number. Q.E.D."