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\begin{aligned}r_{i}\in \mathbb{Q}\quad r_{1},r_{2},\ldots ,r_{n}\in\left( 0,1\right) \\ Proof:2^{r_{1}},\ldots ,2^{r_{n}}\ \text{is linearly independent on } \mathbb{Q} \end{aligned} Some hints are welcomed.

Jean Marie
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liyushu
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$x^n-2$ is an Eisenstein polynomial so is irreducible. Thus $\mathbb Q(\sqrt[n]2)=\mathbb Q[x]/(x^n-2)$ is a degree $n$ extension of $\mathbb Q$. In particular, $1,\sqrt[n]2,\sqrt[n]2^2,\dots,\sqrt[n]2^{n-1}$ are $\mathbb Q$-linearly independent.

Kenta S
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