I want to check if my proof needs any modification:
Let $q$ belong to $\mathbb Q$
Then $\langle q\rangle =e,q,q^2,…$
But there exist no $q$ such that $\langle q\rangle=\mathbb Q\;$ since between every power of $q$ there is a rational $p$ that is not in $\langle q\rangle$
=> no element generates $\mathbb Q$
=> it's not cyclic
Best Answer
Assume that $q\in\mathbb{Q}^+$ generates the group that is $$\langle q\rangle=\{q^n: n\in\mathbb{Z}\}=\mathbb{Q}^+.$$ Without loss of generality we may assume that $q>1$ (note that $\langle q\rangle=\langle q^{-1}\rangle$ and $\langle 1\rangle=\{1\}$).
Then $r=\frac{1+q}{2}\in \mathbb{Q}^+$, but for any positive integer $n$, $$0<q^{-n}<1<r<q^{n}$$ which means that $r\not\in \langle q\rangle$. Contradiction.