If $x$ and $y$ are rational then is $x^y$ also rational?

Question

I can think of the counter example $x = 2$ and $y = 1/2$ but how would a proof to disprove this look like?

Answer 1

Providing a single counterexample suffices to prove that a conjecture is false, i.e.,providing a counterexample is all that's needed to disprove a conjecture.

You've done just that!

Answer 2

$\underline{Theorem:}$ If $x, y \in \mathbb{Q}$ then not necessarily $x^y \in \mathbb{Q}$.

$\underline{Proof:}$ Suppose $x^y \in \mathbb{Q}$. Let $x = 2$ and $y = \frac{1}{2}$. Then $\sqrt{2} \in \mathbb{Q}$, but that is a contradiction. RAA