Prove that $\mathbb{R^*}$, the set of all real numbers except $0$, is not a cyclic group 
Prove that $\mathbb{R^*}$ is not a cyclic group. (Here $\mathbb{R^*}$ means all the elements of $\mathbb{R}$ except $0$.) 

I know from the definition of a cyclic group that a group is cyclic if it is generated by a single element. I was thinking of doing a proof by contradiction but then that ended up nowhere.
 A: clark's answer is surely a great and simple one.  Thomas Andrews' hint is another great one.  Here's a more complicated answer that shows more, that there are entire intervals of numbers that would not be generated.
Let $x$ be a generator of the cyclic group $\mathbb{R}^*$.  If $|x| = 1$, then all powers of $x$ satisfy $|x^n| = 1$.  So, $|x| < 1$ or $|x| > 1$.
If $|x| < 1$, then $|x^{-1}| > 1$ and $x^{-1}$ is also a generator.  So, assume $|x| > 1$.
If $|x| > 1$, then $|x| = 1 + \epsilon$ for some $\epsilon > 0$.  Any positive power of $x$ will satisfy $|x|^n = (1 + \epsilon)^n > 1 + \epsilon$.  Any negative power of $x$ will satisfy $|x|^{-n} = (1 + \epsilon)^{-n} < (1+ \epsilon)^{-1}$.
So, there are entire intervals that are never achieved.
A: Say $g$ is the generator. It must be negative. Either $g<-1, g=-1, $ or $g>-1$.  $g=-1$ clearly does not work.  Let $h = \frac12(-1 + g)$.  $h$ lies strictly between $g$ and $-1$. How is $h$ generated?
A: $\mathbb{R}^*$ has infinite order. If it were cyclic, it would have to contain an element that does not have a cube root.
A: Suppose $\mathbb{R}^*$ is cyclic.
Let $a$ be its generator.
Since $-1 \in \mathbb{R}^*$, there exists a nonzero integer $n$ such that $-1 = a^n$.
Then $a^{2n} = 1$.
Hence the order of $a$ is finite.
This is a contradiction.
A: HINT 
$\mathbb{R}$ is uncountable
A: Let $g\in\mathbb R^*$ and $G=\langle g \rangle=\{ g^n : n \in \mathbb Z\}$.
If $|g|=1$ then $G \subseteq \{ \pm 1 \} \neq \mathbb R^*$.
Otherwise, we may assume that $|g|>1$. 
If $g>1$ then $x=(g+1)/2 \notin G$ because $1 < x < g$.
If $g<-1$ then $x=(g-1)/2 \notin G$ because $g < x < -1$.
In both cases, we have $G\neq \mathbb R^*$.
A: Another way to see it: Assume $\Bbb{R}$ is cyclic and $a$ is its generator. Then $1=na$ for some $n\in\Bbb{Z}$ and $\sqrt{2}=ma$ for some $m\in\Bbb{Z}$. These imply that $a=1/n=\sqrt{2}/m$ and hence $\sqrt{2}=n/m$, contradiction.
