Let $f:\Bbb{R^\times}\to\Bbb{R^\times}$ be an isomorphism. Show that $f$ takes $\Bbb R^{>0}$ to $\Bbb R^{>0}$, and $\Bbb R^{<0}$ to $\Bbb R^{<0}$.

Question

Let $f:\mathbb{R^\times}\rightarrow\mathbb{R^\times}$ be an isomorphism. Show that $f$ takes positive numbers to positive numbers and negative to negative.

If $f$ is an isomorphism, then it's bijective and that $f(g_1g_2)=f(g_1)f(g_2)$.

I think that $f^{-1}:\mathbb{R^\times}\rightarrow\mathbb{R^\times}$ is also an isomorphism means I only have to show $f$ takes positive numbers to positive numbers, but I'm not exactly sure why. Thanks for the help.

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Note: $\mathbb R^\times$ is the multiplicative group of non-zero real numbers.

Answer 1

If $x>0$ is positive, $f(x)=f(\sqrt x)^2>0$, $f(1)=1$ since $f((-1)(-1))=f(1)=f(-1)^2=1$ we deduce that $f(-1)=-1$ since $f$ is injective, if $x>0, f(-x)=f(-1)f(x)=-f(x)<0$.