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I was trying to prove that $\sqrt{2}, \sqrt[3]{2}, ... $ are linearly independent using elementary knowledge of rational numbers. But I could not come up with any proof using simple arguments. How to prove that the set $$\{\sqrt[n]{2}\; :\; n=2,3,4,...\}$$ is linearly independent over the field of $\mathbb{Q}$?

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suppose, for $q_k \in \mathbb{Q}$ and $1 \lt s_1 \lt \cdots \lt s_n \in \mathbb{N}$, that $$ \sum_{k=1}^n q_k \sqrt[s_k]{2} = 0 $$ let $s$ be the least common multiple of the $\{s_k\}$, set $\rho=\sqrt[s]{2}$ and $t_k=\frac{s}{s_k}$ so that we have: $$ \sum_{k=1}^n q_k \rho^{t_k} = 0 $$ but this has degree less than $s$, whereas the minimal polynomial for $\rho$ is $$ x^s - 2 = 0 $$

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  • $\begingroup$ How do we know $x^s-2$ is irreducible? $\endgroup$ – Akiva Weinberger Nov 20 '15 at 13:54
  • $\begingroup$ For instance by the Eisenstein criterion. $\endgroup$ – MooS Nov 20 '15 at 16:40

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