Must an automorphism on the group of real numbers under multiplication maintain sign? Suppose we have an automorphism $\phi$ under the group $(\mathbb{R}^{\#},\,\cdot)$.  I need to show that $\phi$ preserves the sign of the numbers, or that $\phi(\mathbb{R}^+)=\mathbb{R}^+$ and $\phi(\mathbb{R}^-)=\mathbb{R}^-$.
I've had a bit of success.  I determined fairly painlessly that $\phi(r)>0\iff\phi(r^{-1})>0$ for any $r\in\mathbb{R}^\#$.  But I'm having difficulty making any other sort of progress.  My guess is that there's something trivial that I'm missing that I'm currently just too blind to see, but either way, I'm a bit frustrated with my stagnancy here.  Any help or hints or suggestions would really help.
For reference, the only thing I really know about automorphisms is the fact that they're bijective and that it preserves the identity (in this case) $\phi(a) \cdot \phi(b)=\phi(a \cdot b)$ for all $a,b\in\mathbb{R}^\#$.

Can you think of a property, in terms of multiplication only, that positive numbers have but negative ones don't? (HINT: think about $x^2$ . . .)
