Use proof by contradiction. Assume sqrt(2) is rational. This means that sqrt(2) = p/q for some integers p and q, with q <>0. We assume p and q are in lowest terms. Square both side and we get: 2 = p^2/q^2 p^2 = 2q^2 This means p^2 must be an even number which means p is also even since the square of an odd number is odd. So we have p = 2k for some integer k. From this, it follows that: 2q^2 = p^2 = (2k)^2 = 4k^2 2q^2 = 4k^2 q^2 = 2k^2 q^2 is also even, therefore q must be even. So both p and q are even. This contradicts are assumption that p and q were in lowest terms. So sqrt(2) cannot be rational.
