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.
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Suppose $x= \sqrt{ \sqrt 2}$ was rational, then so is its square $x^2=\sqrt 2$ which you have shown is irrational. Contradiction!
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$\begingroup$ Thanks! Just curious, Is there more of a formal way to write this? @Jake $\endgroup$ – Peanutcalota Apr 7 '16 at 22:18
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4$\begingroup$ 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. $\endgroup$ – RKD Apr 7 '16 at 22:31
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