[Math] Prove or disprove that there is a rational number $x$ and an irrational number $y$ such that $x^y$ is irrational

Question

If $x$ is a rational number and $y$ is an irrational number then $x^y$ is irrational.

I have tried to prove it,

let $x= 2$ and $y=\sqrt{2}$ then $x^y = 2^{\sqrt{2}}$,

if it is an rational number then let $x=2^{\sqrt{2}}$ and $y=\sqrt{2}$, then, $x^y= 2^{\sqrt{2} *\sqrt{2}}=4$ which is rational,

therefore $x^y$ is rational. Is it correct?

Answer 1

very close!

$\color{white}{\text{I like the try though +1}}$

Let $x=2$ and $y=\sqrt{1/2}$.

Either $2^{\sqrt{1/2}}$ is irrational, or $\left(2^{\sqrt{1/2}}\right)^{\sqrt{1/2}}$ is irrational