Prove or disprove the following: For all rational numbers $x$ and $y$, the number $x^y$ is also rational.

Question

Prove or disprove: For all rational numbers $x$ and $y$, the number $x^y$ is also rational.

I think that the statement is true since I can not come up with a counterexample but I am unsure of where to go from here... If anyone can finish the proof for me that would be greatly appreciated along with any explanations!

Proof: Suppose $x$ and $y$ are rational numbers $x = p/q$ and $y = r/s$.

Is $(p/q)^{(r/s)}$ rational since it can not be expressed as the ratio of two integers?...

Thank you in advance!

Edit: Sorry for my confusion! I now understand how the square root of two makes the statement false.

Answer 1

The statement is false.
Counterexample:
Let $x=5\;\text{and let }y=\frac12$, so $x$ and $y$ are both rational.
Then $x^y=5^{(1/2)}=\sqrt5$, which is irrational.
This works for any $x$ which is not a perfect square.

Answer 2

As rational powers indicate roots and as irrational roots exist, the statement should be immediately seen as obviously false. And $2^{\frac 12} = \sqrt 2$ should be an immediate and obvious counterexample.

Answer 3

Assume $2^{\frac{1}{TREE(3)}}$ is rational.

Then there are coprime integers $z_1$ and $z_2$ such that

$$\frac{z_1}{z_2}=2^{\frac{1}{TREE(3)}} $$

Raise both sides to the $TREE(3)$th power.

$$\frac{{z_1}^{TREE(3)}}{{z_2}^{TREE(3)}}=2$$

Multiply by ${z_2}^{TREE(3)}$

$${z_1}^{TREE(3)}=2{z_2}^{TREE(3)}={z_2}^{TREE(3)}+{z_2}^{TREE(3)}$$

So we have a perfect ($TREE(3)$)th power that can be expressed as a sum of two perfect ($TREE(3)$)th powers.

But this contradicts Fermat's Last Theorem, as proven by Andrew Wiles in 1995, because:

if this was true, there would be an immodular, semistable elliptic curve, proven by Ken Ribet in 1986,

contradicting the proof by Wiles that all semistable elliptic curves as modular.

So there is an irrational $x^y$ for rational$x$ and rational $y$