$a$ transcendental $\implies a^a$ is transcendental?

by user349289   Last Updated January 14, 2018 13:20 PM

Suppose $a\in \mathbb{C}$ is not a algebraic number. Then is $a^{a}$ also transcendental number ?

I've not idea about how to do it. I got motivation for asking this question from the fact that $e^{i\pi}$ is rational while $e,i\pi$ both are transcendental.

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Let us take $x=\exp\left(W(\log 2)\right)$, i.e. a solution of $x^x=2$.

Step 1. $x\not\in\mathbb{Q}$.
Assuming $x=\frac{p}{q}$ with $\gcd(p,q)=1$, we have $p^p=2^q\cdot q^p$, absurd.

Step 2. $x$ is not an algebraic number.
Assuming that $x$ is in algebraic number, the Gelfond-Schneider theorem gives that $2$ is a transcendental number. It is not, so:

$\color{red}{\text{your claim does not hold.}}$

Jack D'Aurizio
June 21, 2016 16:48 PM

$$x^x=2$$ if $x=a/b$ then $$x^a=2^b$$ therefore $x$ is irrational. According to Gelfond Scheninder theorem $$a^b$$ is transcendental for $a b$ algebraic, $a\neq0,1,\,b$ irrational . $$x^x= 2\implies x \text{ transcendental.}$$

Rajarshi maiti
January 14, 2018 12:26 PM