I am working on the following homework problem:
Let $\phi$ be an isomorphism from $\mathbb{R}^*$ to $\mathbb{R}^*$ (nonzero reals under multiplication). Show that if $r>0$, then $\phi(r) > 0$.
I was trying to prove this by contradiction, but I haven't been able to find any problems. I know that we must have $\phi(1) = 1$, and $\phi(-1) = -1$ no matter what, but despite playing around with those values, as well as $\pm r$ and $\pm \frac{1}{r}$, I haven't gotten any contradictions, and I'm starting to think this might be possible.
What if we defined $\phi(x)$ as: $$x\ \text{if}\ x=\pm 1 \\ -x \text{, otherwise}$$
Would this not be an isomorphism?
Any hints for this question would be much appreciated.