Multiplicative Group of Positive Rationals is Non-Cyclic

Theorem
Let $\left({\Q_{>0}, \times}\right)$ be the multiplicative group of positive rational numbers.

Then $\left({\Q_{>0}, \times}\right)$ is not a cyclic group.

Proof
Aiming for a contradiction, suppose $\left({\Q_{>0}, \times}\right)$ is cyclic.

Then $\left({\Q_{>0}, \times}\right)$ has a generator $x$ such that $x > 1$.

It would follow that:
 * $\Q = \left\{{\ldots, x^{-2}, x^{-1}, 1, x, x^2, \ldots}\right\}$

... where the elements are arranged in ascending order.

But then consider $y \in \Q: y = \dfrac {1 + x} 2$.

So $1 < y < x$ and so $y \notin \Q$.

From this contradiction it is concluded that there can be no such single generator of $\left({\Q_{>0}, \times}\right)$.

Therefore, by definition, $\left({\Q_{>0}, \times}\right)$ is not a cyclic group.