How do I prove that the square root of 2 is irrational?

Proof by contradiction. Assume sqrt(2) is rational, so sqrt(2) = p/q where p and q are integers with no common factors (lowest terms). Squaring: 2 = p^2/q^2, so p^2 = 2q^2. This means p^2 is even, so p must be even (since odd^2 is odd). Write p = 2k. Then (2k)^2 = 2q^2, giving 4k^2 = 2q^2, so q^2 = 2k^2. This means q^2 is even, so q is also even. But if both p and q are even, they share factor 2, contradicting our assumption that p/q is in lowest terms. Contradiction! Therefore sqrt(2) is irrational.