Let $f:\mathbb{R^\times }\rightarrow\mathbb{R^\times}$ be an isomorphism. Show that $f$… [closed]

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.

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

closed as off-topic by Namaste, rogerl, Arnaud D., Shailesh, A. GoodierMar 12 '18 at 13:50

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• Could you let us know what you mean by $\mathbb R*$. Is that the multiplicative group on $\mathbb R\setminus \{0\}$? Or the multiplicative group on $\mathbb R^+$, the positive real numbers? In any case, please state explicitly what group you are referring to. – Namaste Mar 12 '18 at 0:58
• Sorry, $R^*$ means $R$ with respect to multiplication. – user539807 Mar 12 '18 at 1:06
• $\mathbb R^\times$ is a better notation for the non-zero reals with multiplication. – paul garrett Mar 12 '18 at 1:13
• If I may ask, why am I being downvoted? Is my question off topic or is it because my question wasn't asked properly? – user539807 Mar 12 '18 at 1:44

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 $x>0, f(-x)=f(-1)f(x)=-f(x)<0$.
• I'm not sure why you said $f(-1)=-1$. If you use specific numbers like that, is there loss of generality? – user539807 Mar 12 '18 at 1:35
• I show that $f(-1)=-1$ to prove that if $x>0, f(-x)=f((-1)x)=f(-1)f(x)=-f(x)$ since I know already that $f(x)>0$ if $x>0$ I deduce that $f(-x)<0$. – Tsemo Aristide Mar 12 '18 at 1:41
• Is there any way I can use $f^{-1}$? I get your way but I don't quite get how I can use this fact. – user539807 Mar 12 '18 at 2:13
• I use that $f$ is bijective to show that $f(-1)=-1$ since $f(-1)^2=1$, $f(-1)=-1$ or $f(-1)=1$, but $f(-1)\neq f(1)=1$. – Tsemo Aristide Mar 12 '18 at 2:15