Prove or disprove: For every irrational number x, there exists an irrational number y such that x^y is a rational number

Question

I just thought of this statement on a bus on my way back from my Math class. I gave it a lot of thought, but we have not covered how to prove nested quantifier statements, and I have no clue how to approach this. I asked my professor and she said she didn't know how to approach this problem right away either. But this is keeping me up at night for some reason. I don't even know if this statement is true or false. My intuition says it is false, but then again I am unable to find a counterexample of x and prove that for all irrational exponents, the answer is never rational. If someone could point me in the right direction, that would be really really appreciated!

Answer 1

This raises the matter of defining $x^y$ for $x<0$ and $y$ not an integer. For $x\in(0,\infty)\setminus\{1\}$, you can use a trick like the following. Observe that $\log_23$ is irrational: in fact, $2^m\ne 3^n$ for any $n\ne0$. Then, either $\log_x2$ is rational or it isn't. If it is rational, then $\log_x2\cdot \log_23=\log_x3$ is irrational.