# Proof that there exists a rational^irrational=irrational

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 $$\sqrt{2}$$, $$\sqrt{2}$$, $$\sqrt{3}$$, and $$log_{2}3$$, but I could prove the irrationality of some other number if I needed it to prove this.

• How about $2^\sqrt{2}$? Sep 26 '20 at 17:35
• @molarmass how would you prove $2^{\sqrt 2}$ is irrational? Sep 26 '20 at 17:47
• By the Gelfond–Schneider theorem, $2^\sqrt{2}$ is transcendental and thus it is irrational. In fact, this number is known as the Gelfond–Schneider constant. Sep 28 '20 at 18:28

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