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Having a hard time seeing how to approach this.

Given $\phi : \Bbb R^x \rightarrow \Bbb R^x$ is an automorphism of $\Bbb R^x$ (the multiplicative group of nonzero real numbers), and $P$ = {$x \in \Bbb R : x > 0$} (the set of all positive real numbers) and $N$ = {$x \in \Bbb R : x < 0$} (the set of all negative real numbers)

How can I prove $\phi(P) = P$ and $\phi(N) = N$ given $\forall a \in P, a = \sqrt{a}\sqrt{a}$ and since $\phi$ is an automorphism, $\phi^{-1}$ exists and is also an automorphism.

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    $\begingroup$ x for $x$, \times for $\times$. Also, you are not proving a map is equal to itself. $\endgroup$
    – anon
    Apr 14, 2015 at 0:22
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    $\begingroup$ Why are you describing the set $N$? It doesn't appear to come up elsewhere in the problem. $\endgroup$
    – Newb
    Apr 14, 2015 at 0:23

1 Answer 1

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Hint: apply $\phi$ to the equation $a=\sqrt{a}\sqrt{a}$. What happens?

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  • $\begingroup$ I'm not sure how to apply it? That's my problem? I'm having a hard time figuring this out. Typically when I've run across these maps they define the function. I'm not sure what $\phi$ outputs in this case? $\endgroup$
    – shparkison
    Apr 14, 2015 at 19:01
  • $\begingroup$ @shparkison You get $\phi(a)=\phi(\sqrt{a}\sqrt{a})$ when you apply $\phi$ to $a=\sqrt{a}\sqrt{a}$. You're thinking too hard, take baby steps. Now, how can $\phi(\sqrt{a}\sqrt{a})$ be rewritten using the fact that $\phi$ is a homomorphism? $\endgroup$
    – anon
    Apr 15, 2015 at 1:20
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    $\begingroup$ So would it be correct to say since $\phi$ is a homomorphism then $\phi (\sqrt{a}\sqrt{a}) = \phi (\sqrt{a}) \phi (\sqrt{a})$ and since $\phi$ is also an automorphism with $\phi^{-1}$ then this can be written as $\phi (\phi^{-1} (\sqrt{a})) \phi (\phi^{-1}(\sqrt{a}))$ which is equal to $\sqrt{a}\sqrt{a} = |a|$ which is a positive real $\forall a \in P$ $\endgroup$
    – shparkison
    Apr 15, 2015 at 14:25
  • $\begingroup$ @shparkison That's true but I'm not sure what you think you're getting out of that. What I want to you to see is that $\phi(a)=\phi(\sqrt{a})^2$ is positive, and thus conclude $\phi(P)\subseteq P$. $\endgroup$
    – anon
    Apr 15, 2015 at 21:34

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