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 $\mathbb{Q}$?


1 Answer 1


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 $$ x^s - 2 = 0 $$

  • $\begingroup$ How do we know $x^s-2$ is irreducible? $\endgroup$ Nov 20, 2015 at 13:54
  • $\begingroup$ For instance by the Eisenstein criterion. $\endgroup$
    – MooS
    Nov 20, 2015 at 16:40

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.