James' Proof
Statement: Using the numbers $\sqrt{2}$ and $\sqrt{2}$$\sqrt{2}$ prove that there exist irrational numbers a,b such that ab is rational.
Proof: We must first identify $\sqrt{2}$ as an irrational number. We then consider two cases: 1) $\sqrt{2}$$\sqrt{2}$ is rational, or 2) $\sqrt{2}$$\sqrt{2}$ is irrational. For the first case, let $\sqrt{2}$$\sqrt{2}$ = cd when c = $\sqrt{2}$ and d = $\sqrt{2}$. Since $\sqrt{2}$ is an irrational number and $\sqrt{2}$$\sqrt{2}$ is rational, the statement is satisfied. Let us consider the second case. When a = $\sqrt{2}$$\sqrt{2}$ and b = $\sqrt{2}$, ab = ($\sqrt{2}$$\sqrt{2}$)$\sqrt{2}$ which is rational. Since $\sqrt{2}$ is an irrational number and $\sqrt{2}$$\sqrt{2}$ is an irrational number, the statement is also satisfied.