Prove that there exists a rational number raised to an irrational number that is an irrational number

Question

Prove: There exists $a \in \mathbb{Q}$ and $b \in \mathbb{R}\smallsetminus \mathbb{Q}$ such that $a^b \in \mathbb{R} \smallsetminus \mathbb{Q}$.

I've tried using $\log_23$, $\sqrt 2$, and $\frac{1}{\sqrt 2}$ for the irrational number, but couldn't find a way to prove $a^b$ was irrational.

Is there a way to prove this without using Gelfond–Schneider theorem?

Answer 1

Well, either $2^{\sqrt{2}}$ is irrational and we are done, or $(2^\sqrt{2})^{\sqrt{2}/4}=\sqrt{2}$ is an irrational, which is a rational to an irrational power.

Answer 2

fix $a=7.$ The set of $b$ is uncountable.

Answer 3

Try $a=2$ and $b=\log_2(\sqrt{3})$

Answer 4

Let $p$ be a rational in the form $\frac{a}{b}$ and $q$ be an irrational number, now,
$$p^q=1+\frac{\left(q\ln p\right)}{1!}+\frac{\left(q\ln p\right)^2}{2!}...$$
For all $p \gt 1$, Each term in series is irrational thus $p^q$ must be irrational.