In solving the following problem: Prove or disprove that there is a rational number $x$ and an irrational number $y$ such that $x^y$ is irrational. I let $x=2$ and $y = \sqrt 2$, so that $x^y = 2^\sqrt 2$.
Why is this not enough? How come I have to go through the case whether $x^y = 2^\sqrt 2$ is rational?
If $2^\sqrt 2$ is rational then let $x = 2^\sqrt 2$, and $y = \sqrt 2 / 4$
$x^y = (2^\sqrt 2)^{\sqrt 2 /4} = 2^{(\sqrt 2*\sqrt 2) /4} = 2^{2/4} = 2^{1/2} = \sqrt 2$ (previous value for y that was established as irrational.
Best Answer
Let $\mathbb{I}$ denote the set of irrational numbers.
Function $f : \mathbb{I} \rightarrow \mathbb{R} : x \mapsto 2^x$ is an injection.
$\mathbb{I}$ is uncountable $\Rightarrow$ image of $f$ is uncountable.
The set of rational numbers is countable $\Rightarrow$ image of $f$ contains something more than rationals.
There exist such irrational $x$ that $2^x$ is irrational.