$\sqrt[n]{2^k}$ is irrational for $k=1,2,...,n-1$

Question

We all know that $\sqrt{2}$ is irrational. We can use the same proof to show that $n \in \mathbb{N}: \sqrt[n]{2}$ is also irrational.

But how to show that

$$ \sqrt[n]{2^k} \quad \text{is irrational for} \; k = 1, 2, ..., n-1 $$

?

Answer 1

Write $\sqrt[n]{2^k} = \dfrac{a}{b}$ with $a,b \in \mathbb Z$. Then $a^n = 2^k b^n$.

Write $a=2^\alpha a'$ and $b=2^\beta b'$, with $a',b'$ odd.

Then $n \alpha = k + n \beta$ and so $k$ is a multiple of $n$. Since $ 0 < k < n$, this is impossible.

Answer 2

For another proof, note that $x^n-2 \in \mathbb Q[x]$ is irreducible by Eisenstein's criterion.

Actually this shows that $1,\sqrt[n]{2}, \dotsc, \sqrt[n]{2^{n-1}}$ are $\mathbb Q$-linear independent, which is an even stronger result.