@BlotterFan said in #3:
> Prove non-existance
Suppose there exist two integers p and q such that (p/q) ^2=2. Dividing p and q by 2 as many times as necessary (which doesn't change the value of p/q), we can assume at least one of them is odd. But p^2=2q^2 so p^2 must be even which implies p is even. Say p=2n. Then p^2=4n^2 thus q^2=2n^2. Hence q^2 is even, and so is q, which is a contradiction.
There you go.
