Lemma: The square of an odd number is odd.

Proof. Let p = (2n - 1), then p^2 = (2n - 1)(2n - 1) = 4n^2 - 2n -2n + 1 = 4n^2 - 4n + 1 = 2(2n^2 - 2n + 1) - 1 which is of the form 2j - 1 where j = 2n^2 - 2n + 1. Thus, p^2 is odd.

Corollary: If p^2 is even, then p is even.

Proof. Assume p is not even. Then by the prior Lemma, p^2 is not even, a contradiction. Thus, p is even.

Theorem: The root of 2 is Irrational

Proof. Suppose there exists a rational number r such that r^2 = 2. Then by definition, r = (a/b) where a and b are integers. Factor our any common divisors k_j such that a = (k_1 * k_2 * ... * k_n) * p and b = (k_1 * k_2 * ... * k_n) * q, so that a/b = p/q and that p and q do not share common factors. Since p^2/q^2 = 2, then p^2 = 2*q^2 so that p^2 is an even number. By the above corollary, we know that p is even. Now since p and q share no common factors, we know 2 is not a divisor of q, which means that q is odd.

Also, since p is even, there exists an integer m such that p = 2m. Thus, p^2 = 4m^2. Recalling that p^2/q^2 = 2 by hypothesis, and that this means p^2 = 2q^2, we have that 4m^2 = 2q^2, which simplifies to 2m^2 = q^2. This of course means that q^2 is even, and by the above corollary, we have demonstrated that q is even.

Since we can argue from the same premise that q is both even and odd, we know the premise must be at fault. Thus, there is no rational number whose square is 2.