# Is $x^x$ rational for $x=\sqrt{2}^\sqrt{2}$

This might be naive. Is $x^x$ a rational number for $x=\sqrt{2}^\sqrt{2}$ ?

I remember reading somewhere a long time ago that such $x^x$ is a rational number, as an example of issues with non-constructive math and excluded middle. (I think the question was whether there exists irrational $x$, such that $x^x$ is rational.) But I didn't give it much thought or figured out why. I dismissed the issue as I mistakenly thought exponentiation is associative, and assumed $x^x = \sqrt{2}^2=2$.

Can someone help show constructively whether this is a rational number?

Related:

Can an irrational number raised to an irrational power be rational?

irrationality of $\sqrt{2}^{\sqrt{2}}$.

• When $x$ and $y$ are algebraic numbers then we can use the Gelfond-Schneider theorem to tell us if $x^y$ is trancendental irrational. When $x$ and $y$ are trancendental it becomes a much harder problem (note that $\sqrt{2}^{\sqrt{2}}$ is trancendental by GS). Your number is almost surely irrational (and trancendental), but it might be hard to prove this. – Winther Jul 15 '16 at 8:33
• @Winther: You probably mean "When $x$ and $y$ are irrational algebraic", right? – barak manos Jul 15 '16 at 10:36