Proving any automorphism of the group $\Bbb R^\times$

Question

Let$\Bbb R^\times$ denote the non-zero real numbers. Prove that any automorphism of the group $\Bbb R^\times$ under multiplication maps positive to positive numbers and negative numbers to negative numbers.

Answer 1

Positive numbers $a$ can be characterized by the fact that there exists a number $b$ such that $a=b^2$. This property survives an automorphism.

Answer 2

Let $x > 0$, and choose $y$ so that $y^2 = x$. Then $$\phi(x) = \phi(y\cdot y) = \phi^2(y) > 0.$$ If $x < 0$, then $-x > 0$, so $$\phi(-x) = \phi(-1)\phi(x) > 0.$$ This implies $\phi(x)$ and $\phi(-1)$ have the same signs. From $$\phi(1) = \phi^2(-1),$$ either $\phi(-1) = 1$ or $\phi(-1) = -1$. Since the former would imply $\phi$ is not injective, $\phi(-1) < 0$, and $\phi(x) < 0$.