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Prove the irrationality of $2^{\frac{1}{n}}$ for $n > 2$

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 a correct proof?

What do you think about this proof?

  • 2
    Can Fermat be proven without this result?2012-10-28

2 Answers 2

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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.2012-10-28