# Linear independence of $\sqrt{2}$, $\sqrt[3]{2}$, $\sqrt[4]{2}, \dots$ over the rationals

I was trying to prove that $$\sqrt{2}, \sqrt[3]{2}, ...$$ are linearly independent over the rationals 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 $$\mathbb{Q}$$?

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 $$x^s - 2 = 0$$
• How do we know $x^s-2$ is irreducible? – Akiva Weinberger Nov 20 '15 at 13:54