Existence of Rational Powers of Irrational Numbers/Proof 2

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

We consider the two cases.


 * $(1): \quad$ If $q$ is rational then $a = \sqrt 2$ and $b = \sqrt 2$ are the desired irrational numbers.
 * $(2): \quad$ If $q$ is irrational then $q^{\sqrt 2} = \paren {\sqrt 2^{\sqrt 2} }^{\sqrt 2} = \sqrt 2^{\paren {\sqrt 2} \paren {\sqrt 2} } = \sqrt 2^2 = 2$ is rational, so $a = q = \sqrt2^{\sqrt 2}$ and $b = \sqrt 2$ are the desired irrational numbers.