 # Prove that every automorphism of R*, the group of nonzero real numbers under multiplication, maps positive numbers to positive numbers and negative numbers to negative numbers Tahmid Knox 2020-10-28 Answered
Prove that every automorphism of R*, the group of nonzero real numbers under multiplication, maps positive numbers to positive numbers and negative numbers to negative numbers
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$\varphi :R\cdot \to R\cdot$ be an automorphism
Concept used:
An isomorphism from a group G oto itself is called a automorphism of G*
Note that R* is the of nonzero real numbers.
Since $\varphi \left(x\right)\in R\cdot$ then obtain $\varphi \left(x\right)\ne 0$ for every $x\in R\cdot$
Let x>0 then obtain $\sqrt{x}$ make sense.
Recall the theoram, properties if isomorphism acting on elements. Suppose that $\varphi$ is in isomorphism from a group G ont $\stackrel{―}{G}.$
For every integer n and for every froup of elemnt $a\in G\varphi \left({a}^{n}\right)=\left(\varphi {\left(a\right)}^{n}\right)$
$\varphi \left(x\right)=\varphi \left({\left(\sqrt{x}\right)}^{2}\right)={\left(\varphi \left(\sqrt{x}\right)\right)}^{2}>0$
Let x>0 then obtain $\sqrt{-}x$ make sense
$\varphi \left(x\right)=\varphi \left({\left(-\sqrt{x}\right)}^{2}\right)={\left(\varphi \left(-\sqrt{x}\right)\right)}^{2}<0$
Thus, the funtion phi maps positive numers to positive numbers and negative to negative numbers.