Is this an isomorphism possible?

Question

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.

Answer 1

Your map is not a homomorphism, because $\phi(4)=-4$, while $\phi(-2)\phi(-2)=4$, so $\phi((-2)(-2))\neq \phi(-2)\phi(-2)$.

To prove that $\phi(r)>0$ whenever $r>0$, remember that every positive real number is a square. So if $r>0$ then $r=s^2$ for some $s\neq 0$, hence $$ \phi(r)=\phi(s)\phi(s)>0$$