TITLE: Show that $(\mathbb{Q}^*,\cdot)$ and $(\mathbb{R}^*,\cdot)$ aren't cyclic
QUESTION [5 upvotes]: I'm reading a book about abstract algebra, but I'm having trouble solving this excercise: "Show that $(\mathbb{Q}^*,\cdot)$ and $(\mathbb{R}^*,\cdot)$ aren't cyclic"
Where $(\mathbb{Q}^*,\cdot)$ is the group of nonzero rational numbers under multiplication and $(\mathbb{R}^*,\cdot)$ is the group of nonzero real numbers under multiplication.

Here is my attempt for the first.
Suppose $(\mathbb{Q}^*,\cdot)$ is cyclic, then $\mathbb{Q}^*=\langle\frac{p}{q}\rangle=\{(\frac{p}{q})^n,n\in\mathbb{Z}\}$, where $p$ and $q$ are coprime.
$\frac{2p}{q}$ is also in $\mathbb{Q}^*$ so it must be equal to $(\frac{p}{q})^n$ for some $n\in\mathbb{Z}$.
To solve $\frac{2p}{q}=(\frac{p}{q})^n$, I take a logarithm of both sides and end up with $1+\log_\frac{p}{q}(2)=n$, since $n$ is an integer $\log_\frac{p}{q}(2)$ must be an integer too, but it is possible only when $\frac{p}{q}=2^{\frac{1}{k}}, k\in\mathbb{N}$, (i.e. $\frac{p}{q}$ is a k-th root of $2$), but $k$ must be $1$ for $2^\frac{1}{k}$ to be rational so $\frac{p}{q}=2$ contradicting the hypothesis of $p$ and $q$ being coprime.

However I don't know whether this is a proper proof and the same reasoning cannot be applied to $\mathbb{R}^*$, I'd like you to just give me an hint towards a proof, without telling me the whole proof, if possible.

REPLY [3 votes]: Simpler form of your proof: If $(\mathbb{Q}^\times, \cdot)$ is cyclic, it has a generator $\frac{p}{q}$ in lowest terms. Now $\frac{2 p}{q} \in \mathbb{Q}^\times$, so it there must be $n \in \mathbb{Z}$ such that:
\begin{align}
\frac{2 p}{q} &= \left( \frac{p}{q} \right)^n \\
2 q^{n - 1}   &= p^{n - 1}
\end{align}
(we consider $n > 0$ here, switch around $p$, $q$ so both sides are integers if $n < 0$). So $p$ is even, and $q$ must be odd. The only possibility is to have $n = 2$, and so $q = 1$, and thus $p = 2$, the generator must be $\frac{1}{2}$. But there is no way to get $\frac{1}{3}$ as a power of $\frac{1}{2}$.