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Prove $2^{1/p} - 2^{1/q}$ is irrational for $p, q, \in \mathbb{N}, p \neq q$. Not sure how to begin, as the usual trick of taking $p^{th}$ powers doesn't seem to work very well.

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    $\begingroup$ I'm sure you want to assume $p\neq q$... $\endgroup$ – Wojowu Feb 28 '16 at 19:21
  • $\begingroup$ Try using the rational roots theorem. $\endgroup$ – tmyklebu Feb 28 '16 at 19:26
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Suppose that $p<q$ and $\sqrt[p]{2} -\sqrt[q]{2} =r=\frac{m}{n}\in\mathbb{Q}.$ It is well known that the minimal polynomial of $\sqrt[q]{2}$ is equal to $x^q-2,$ but then we have $W(\sqrt[q]{2} ) =0$ where $$W(x)=-2n^p+\sum_{k=0}^p {p\choose k} m^k n^{p-k} x^{p-k}$$ but this is impossible since $\mbox{deg} (W) =p<q .$

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