TITLE: Linear independence of $\sqrt{2}$, $\sqrt[3]{2}$, $\sqrt[4]{2}, \dots$ over the rationals
QUESTION [9 upvotes]: 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 $\mathbb{Q}$?

REPLY [13 votes]: 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 
$$
x^s - 2 = 0
$$