Proof that there exists a rational^irrational=irrational

Question

How do I prove that there exist some $a\in\mathbb{Q}$ and $b \in \mathbb{R} - \mathbb{Q}$ such that $a^b \in\mathbb{R} - \mathbb{Q}$? I dont need to find what it is, just that it exists. The only numbers I know are irrational for the purposes of this proof are $\sqrt{2}$, $\sqrt[3]{2}$, $\sqrt{3}$, and $log_{2}3$, but I could prove the irrationality of some other number if I needed it to prove this.

Answer 1

HINT: How big is the set $\left\{2^b:b\in\Bbb R\setminus\Bbb Q\right\}$? How many rationals are there?

Answer 2

If $a > 0$ and $a \ne 1$ then $a^x$ is injective, one to one. As there are uncountibly meany $x$ there are uncountably many $a^x$. As there are only countably many rationals, not all of the $a^x$ can be rational.

Answer 3

Given the list of values you know to be irrational, we could simply take $$2^{\log_2 \sqrt 3}=\sqrt 3$$

We remark that $\log_2\sqrt 3 =\frac {\log_2 3}2$ so that too must be irrational (given that $\log_2 3$ is irrational).