I want to show that $(\mathbb{Q}^{+}, \cdot)$ is not cyclic.
Suppose that $0<q<1$ a generator for $(\mathbb{Q}^{+}, \cdot)$. Then, for every $a \in \mathbb{Q}^+$ there is a $k \in \mathbb{Z}$ such that
$$q^k=a$$
We know that $q^k$ is decreasing.
$$q^{k+1}<q^k$$
But, there is a number c in $\mathbb{Q}^+$ such that
$$q^{k+1}<c<q^k$$
This means that,
$$q^k \neq c$$ for every $k \in \mathbb{Z}$. So, q cannot be a generator for $(\mathbb{Q}^{+}, \cdot)$
The $q>1$ case is the same.
Is this right? , I would also like to see other forms of proof.
Thanks
Another approach would be the following. Suppose that $a/b$ is a generator for $(\mathbb Q^+,\cdot)$, where $a,b\in\mathbb N$. We consider a prime $p$ which does not divide neither $a$ nor $b$. Then it must exist some $k\in\mathbb Z\setminus\{0\}$ such that $$(a/b)^k=p,$$ but this implies that $a^k=pb^k$. Hence $a^{|k|}=pb^{|k|}$ or $pa^{|k|}=b^{|k|}$, depending on the sign of $k$. As a consequence, $p\mid a$ or $p\mid b$, which is a contradiction.