Existence of Rational Powers of Irrational Numbers

Theorem
There exist irrational $$a$$ and $$b$$ such that $$a^b$$ is rational.

Proof

 * $\sqrt{2}$ is irrational;
 * $$2$$ is trivially rational, as $$2 = \frac{2}{1}$$.

Consider the number $$q = \sqrt{2}^{\sqrt{2}}$$, which is irrational by the Gelfond-Schneider Theorem.

Thus $$q^{\sqrt{2}} = (\sqrt{2}^{\sqrt{2}})^{\sqrt{2}} = \sqrt{2}^{(\sqrt{2})(\sqrt{2})} = \sqrt{2}^2 = 2$$ is rational, so $$a = q = \sqrt{2}^{\sqrt{2}}$$ and $$b = \sqrt{2}$$ are the desired irrationals.

Alternative Proof
Given that $$2$$ is rational and $$\sqrt{2}$$ is irrational, consider the number $$q = \sqrt{2}^{\sqrt{2}}$$.

We can prove the proposition using a similar argument to the first proof without the Gelfond-Schneider Theorem by considering two cases.


 * 1) If $$q$$ is rational then $$a = \sqrt{2}$$ and $$b = \sqrt{2}$$ are the desired irrational numbers.
 * 2) If $$q$$ is irrational then $$q^{\sqrt{2}} = (\sqrt{2}^{\sqrt{2}})^{\sqrt{2}} = \sqrt{2}^{(\sqrt{2})(\sqrt{2})} = \sqrt{2}^2 = 2$$ is rational, so $$a = q = \sqrt{2}^{\sqrt{2}}$$ and $$b = \sqrt{2}$$ are the desired irrationals.