# How can I prove $\sqrt{\sqrt2}$ to be irrational?

How can I prove $\sqrt{\sqrt2}$ to be irrational?

I know that $\sqrt2$ is an irrational number, it can be proved by contradiction, but I'm not sure how to prove that $\sqrt{\sqrt2} = \sqrt[4]{2}$ is irrational as well.

Suppose $x= \sqrt{ \sqrt 2}$ was rational, then so is its square $x^2=\sqrt 2$ which you have shown is irrational. Contradiction!
• Maybe this? If $\sqrt{\sqrt 2}=\frac{p}{q}$, then $\sqrt 2=\frac{p^2}{q^2}$, a contradiction? I personally think that it is clearer to do simple stuff in words. – RKD Apr 7 '16 at 22:31