The super square root of $n$ is the solution/solutions to $x^x=n$. Is the super square root of $2$ irrational?



Is the super square root of $2$ irrational?
Yes.


If $x$ is rational, say $x=p/q$ with coprime integers $p,q$. We know $x$ cannot be an integer, so $q\neq 1$. Then $(p/q)^{p/q}=2$, or $(p/q)^p=2^q$. But the left-hand side cannot be an integer because its simplest fraction representation is $p^p/q^p$ with denominator $q^p\neq 1$. (Note that $p^p$ and $q^p$ are also coprime.)


Extension:
In fact, $x$ is transcendental. [Gelfond–Schneider theorem](https://en.wikipedia.org/wiki/Gelfond%E2%80%93Schneider_theorem) says that the only cases where $a, b$ and $a^b$ are all algebraic numbers are (1) $b$ is rational, or (2) $a$ is $0$ or $1$.


So if $x^x=2$ and $x$ is algebraic, then it must be case (1), so $x$ is rational, a contradiction to what we have proved.


