# Is there a rational number x such that sqrt(x) is irrational, but sqrt(x)^sqrt(x) is rational?

## The motivation for this question is to consider: ${\sqrt{2}}^{\sqrt{2}}$ and ${\left({\sqrt{2}}^{\sqrt{2}}\right)}^{\sqrt{2}}$. Note that $\sqrt{2}$ is irrational, and ${\left({\sqrt{2}}^{\sqrt{2}}\right)}^{\sqrt{2}} = {\sqrt{2}}^{2} = 2$. So either ${\sqrt{2}}^{\sqrt{2}}$ is rational and we find that this rational number raised to an irrational power $\sqrt{2}$ is also rational, or ${\sqrt{2}}^{\sqrt{2}}$ is irrational and we find that this irrational number raised to an irrational power is rational. Going back to the question asked, I think there is no such rational number $x$, but I also suspect that it is difficult to prove.

Apr 30, 2017

No. If $\sqrt{x}$ is irrational, then ${\sqrt{x}}^{\sqrt{x}}$ is transcendental by the Gelfond-Schneider theorem.

#### Explanation:

The Gelfond-Schneider theorem says that if $a , b$ are algebraic numbers with $a \ne 0$, $a \ne 1$ and $b$ irrational, then ${a}^{b}$ is transcendental (which implies that it is irrational).

A proof of the Gelfond-Schneider theorem can be found at http://people.math.sc.edu/filaseta/gradcourses/Math785/Math785Notes8.pdf

In our example, $\sqrt{x}$ is irrational, so it will not be equal to $0$ or $1$ and is algebraic, since it's a root of ${t}^{2} - x = 0$ with $x$ rational.

Hence with $a = b = \sqrt{x}$ we can deduce that ${\sqrt{x}}^{\sqrt{x}} = {a}^{b}$ is transcendental, so irrational.