The super square root of $n$ is the solution/solutions to $x^x=n$. Is the super square root of $2$ irrational?
-
1$\begingroup$ If the super square root is $p/q$ with $p$ and $q$ coprime (note that obviously $p,q\neq0$), then $p^p=2^qq^p$, and taking the 2-adic valuation, $p\nu_2(p)=q+p\nu_2(q)$. Since at least one of $\nu_2(p)$ and $\nu_2(q)$ has to be zero, this is a contradiction. $\endgroup$– Pierre PCAug 27, 2020 at 17:57
-
1$\begingroup$ And then, by Gelfond-Schneider, it's also trascendental $\endgroup$– Pietro MajerAug 27, 2020 at 18:03
1 Answer
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 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.
-
1
-
1$\begingroup$ Found it! Gelfond–Schneider theorem. I am going to edit it. $\endgroup$ Aug 27, 2020 at 18:40