I saw this proof in class today to prove the square root of 2 is irrational: 1. Assume that √2 is a rational number. Meaning that there exists an integer a and b so that a / b = √2. 2. Then √2 can be written as an irreducible fraction (the fraction is shortened as much as possible) a / b such that a and b are coprime integers and (a / b)^2 = 2. 3. It follows that a^2 / b^2 = 2 and a^2 = 2 b^2. 4. Therefore a^2 is even because it is equal to 2 b^2 which is obviously even. 5. It follows that a must be even. (Odd numbers have odd squares and even numbers have even squares.) 6. Because a is even, there exists a k that fulfills: a = 2k. 7. We insert the last equation of (3) in (6): 2b^2 = (2k)2 is equivalent to 2b^2 = 4k^2 is equivalent to b2 = 2k^2. 8. Because 2k^2 is even it follows that b^2 is also even which means that b is even because only even numbers have even squares. 9. By (5) and (8) a and b are both even, which contradicts that a / b is irreducible as stated in (2). This was copied from wikipedia by the way. I mean I know this works or it wouldn't have been around for as long as it has. I read this and it makes sense I understand why the square root of 2 is irrational. But it seems to me that you can use any number in there, take 4 for example, and reach the same conclusion when clearly the square root of 4 is rational. Can someone clarify this for me?