James' Proof

Statement: Using the numbers $\sqrt{2}$ and $\sqrt{2}$^{$\sqrt{2}$} prove that there exist irrational numbers a,b such that a^{b} 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 $\sqrt{2}$^{$\sqrt{2}$} = c^{d} 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 = $\sqrt{2}$, a^{b} = ($\sqrt{2}$^{$\sqrt{2}$})^{$\sqrt{2}$} which is rational. Since $\sqrt{2}$ is an irrational number and $\sqrt{2}$^{$\sqrt{2}$} is an irrational number, the statement is also satisfied.