Is it possible that $n^{\alpha}$ is rational for all $n \in \mathbb{N}$ with $0 < \alpha <1$?

Question

I think that is impossible, i.e., for each $\alpha \in (0,1)$, there exists $n_0 \in \mathbb{N}$ such that $n_0^{\alpha}$ is irrational. Or equivalently, there is no $\alpha \in (0,1)$ such that $n^{\alpha}$ is rational for each $n \in \mathbb{N}$.

But I can't prove it. Could you help me?

score 1 · Accepted Answer · answered Nov 07 '19 at 08:25

1

This is true, but I don't think there is any elementary proof (here I ask about an elementary proof of an equivalent statement). One can prove this using the six exponentials theorem as explained in this answer. Indeed, it is enough to just consider $2^\alpha,3^\alpha,5^\alpha$ to deduce that they can't all be rational for $0<\alpha<1$.

answered Nov 07 '19 at 08:25

Wojowu

26,600

Thank you so much! With six exponentials theorem, this problem is quite easy! – Hsing Nov 07 '19 at 08:43

Is it possible that $n^{\alpha}$ is rational for all $n \in \mathbb{N}$ with $0 < \alpha <1$?

1 Answers1