Are nonzero rationals under multiplication cyclic?
Here's my thinking:
They are not. The generator must be a rational $q = a/b$, $a$, $b$ integers with no common factors. Assume $a/b$ generates $\mathbb Q\setminus \{0\}$, then $q^n = a^n/b^n$ or $(1/q)^n = b^n/a^n$. This is impossible since we can find a prime (there're infinitely many of them) that doesn't divide either $a$ or $b$, which have a finite number of prime factors. Not sure if it proves it though.
Your proof looks fine to me.
An alternative is this. If $\Bbb Q^\times$ were cyclic, it would be infinite cyclic, so $\simeq \Bbb Z$. But $-1$ has order two in $\Bbb Q^\times$; and there is no element of order two in $\Bbb Z$: every element has infinite order, except for $0$.
Let's follow your comment: assume that $\mathbb Q^{\times}$ is cyclic generated by $a/b$ with $\gcd(a,b)=1$ and take a prime $p$. Then $1/p=a^n/b^n$ or $1/p=b^n/a^n$ with $n\ge 1$, that is, $b^n=pa^n$ or $a^n=pb^n$. So $p\mid b^n$ or $p\mid a^n$ and thus $p\mid b$ or $p\mid a$. If $p$ is chosen large enough (let's say greater than any prime that appear in the decompositions of $a$ and $b$), then it's impossible to have $p\mid a$ or $p\mid b$.
If $q\in\mathbb{Q}\backslash\left\{ 0\right\} $ would generate the group then $q^{n}=-1$ for some $n\in\mathbb{Z}\backslash\left\{ 0\right\}$.
Then $q^{2\left|n\right|}=1$, so $\left\langle q\right\rangle =\left\{ q^{r}\mid r=0,1,\ldots,2\left|n\right|-1\right\} $ is finite and a contradiction is found.
It is in the same line as the (more elegant) proof of Pedro.
The subgroups $\langle \frac{1}{2}\rangle=\{ 2^n\colon n\in\mathbb{Z}\}$ and $\langle \frac{1}{3}\rangle=\{ 3^n\colon n\in\mathbb{Z}\}$ of $\mathbb{Q}^{\times}$ intersect trivially, but any two (non-identity) subgroups of infinite cyclic group always intersect non-trivially.
Yes, you could prove it by your prime number argument, but I think it's easy to see this way:
Suppose the multiplicative group $\mathbb{Q} ^ \times$ is generated by a single element $q$. Then the group can be written as $$\dots,\ q^{-2},\ q^{-1},\ q^0,\ q^1,\ q^2,\ \dots$$
It should be easy to see the generalized sequence $$\dots,\ |q^{-2}|,\ |q^{-1}|,\ |q^0|,\ |q^1|,\ |q^2|,\ \dots$$ is either constant, strictly increasing or strictly decreasing (depending on whether $|q| = 1$, $|q| > 1$ or $|q| < 1$).
In any case, you should see it misses quite a lot rational numbers, so it can't be cyclic (the same argument applies to $\mathbb{R} ^ \times$ and even $\mathbb{C} ^ \times$ as well)
The fundamental theorem of arithmetic exactly says that prime numbers define a free basis of $\mathbb{Q}^{\times}$. Therefore, $\mathbb{Q}^{\times}$ is an abelian free group of rank $\aleph_0$; in particular, it cannot be cyclic.
