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Does there exist an $$0<x<1$$ such that $$\forall q \in \mathbb{Q^+}$$ $$q^x \in \mathbb{Q^+}$$

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What can easily be seen is that the set $X\subseteq \mathbb{R}$ of all $x$ satisfying $\forall q\in\mathbb{Q}^+:q^x\in\mathbb{Q}^+$ is a subring of $\mathbb{R}$; in particular $\mathbb{Z}\subseteq X$. It would be nice if one could somehow show that $X=\mathbb{Z}$; unfortunately, I can't find a way to do it. – jpvee Apr 15 '14 at 14:02
up vote 4 down vote accepted

There cannot be. First, if $x$ is rational, say $a/b$ where $a,b$ are relatively prime, let $q=p$ be a prime number. Then $q^x$ is irrational, because the $b$-th root of an integer is rational iff it is itself an integer.

However, $x$ cannot be irrational, since it follows from the six exponentials theorem that if $p,q,r$ are three distinct primes, and $p^x,q^x,r^x$ are all rational, then indeed $x$ is an integer.

(I expect there is an elementary proof under the much stronger assumption in your question. Also, to avoid unwanted detours, you may want to restrict $q$ to the positive rationals.)

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