# Proving that $\sqrt{2}^{\sqrt{3}}$ is irrational

I know how to prove that $\sqrt{2}$ and $\sqrt{3}$ are irrational. But, I am not sure in this case.

$x=\sqrt{2}^{\sqrt{3}}$ gives me equation $\ln {x}-\frac{\sqrt{3}}{2} \ln{2}=0$ which is not polynomial, so I can not use theorem about rational solutions.

If I asume the opposite and show it like rational number $\sqrt{2}^{\sqrt{3}} = \frac{p}{q}$, I am not sure what can I use here.

So, my question is: how to prove or disprove(if it is rational) this statement?

## 1 Answer

Note that both numbers that you're dealing with (I mean, $\sqrt2$ and $\sqrt3$) are algebraic numbers (that is, each of them is a root of a non null polynomial with rational coefficients). Furthermore, $\sqrt2$ is neither $0$ nor $1$ and $\sqrt3$ is irrational. The Gelfond-Schneider theorem tells us then that $\sqrt2^{\sqrt3}$ is transcendental (that is, non-algebraic), and therefore irrational.

If there is a more elementary way of proving that $\sqrt2^{\sqrt3}$ is irrational, I am not aware of it.

• Could you add more justification within your answer (a brief summary of the info at the link), so that your answer isn't merely a link only answer? I might consider upvoting if you did so convincingly, and not just hand-wavingly. – Namaste Feb 2 '18 at 17:26
• @amWhy I was in a hurry when I wrote the answer. It seems to me that it cannot be considered a link-only answer anymore. – José Carlos Santos Feb 2 '18 at 18:17