TITLE: Exponentials of rational numbers
QUESTION [5 upvotes]: Does there exist an $$0<x<1$$ such that $$\forall q \in \mathbb{Q^+}$$ $$q^x \in \mathbb{Q^+}$$

REPLY [4 votes]: 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.)