# Irrationality of $2^{\frac{1}{n}}$ for $n > 2$

What do you guys think about this proof? So, we suppose $$2^{\frac{1}{n}}$$ is rational ( = $\frac{p}{q}$). Therefore, $$2 = \frac{p^n}{q^n} \Rightarrow q^n + q^n = p^n$$ and this contradicts fermat's last theorem. Is this correct proof?

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mathoverflow.net/questions/42512/… –  wj32 Oct 28 '12 at 4:01
@wj32 Nice and thanks for the link. –  user17762 Oct 28 '12 at 4:01
Can Fermat be proven without this result? –  Baby Dragon Oct 28 '12 at 5:30

You still need to prove it for $n=2$ though. The proof is right for $n >2$. But this proof falls under the category of mosquito nuking proofs.

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Well if $2=p^n/q^n$, then $2\cdot q^n=p^n$. This implies that 2 divides $p$; thus $2^n$ divides $p^n$. This in turn implies that $2^n$ divides $2\cdot q^n$, but since $n>1$, 2 must divide $q$ as well. This is a contradiction.

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I think the OP, and the rest of us, are well aware of this proof, @one, yet the interesting twist presented by the OP is the use of Fermat's Last Theorem. –  DonAntonio Oct 28 '12 at 4:10