Isomorphism from <R,+> to <R+,\times>: Proving 1-1 and Onto Function

[LMKIYHAQ]

Homework Statement

Is there an isomorphism from <R,+> to <R+,$$\times$$> where $$\phi$$(r)=0.5$$^{r}$$ when r $$\in$$ R?

2. Homework Equations
For an isomorphism I know it is necessary to show there is a 1-1 and onto function. I am unsure if I can use the steps I am trying to use to show it is 1-1.

The Attempt at a Solution

For phi(r)=phi(s) I want to show r=s. Am I able to take the ln (or log?) of both sides to get ln(.05$$^{r}$$)=ln(0.5$$^{s}$$)? I am not sure which to use (ln or log) and where these logarithmic functions would be defined since for r=s, r and s are supposed to be real numbers.

Thanks for the help.

 

[morphism]

Maybe try rewriting 0.5^r=0.5^s as 2^r=2^s. This may make things cleaner.

 

[LMKIYHAQ]

I don't know how to make that change?

 

[LMKIYHAQ]

Even if that way does clean it up, is my way of taking ln of both sides wrong?

 

[Dick]

Sure, just use logs. (0.5)^r=(0.5)^s iff r*log(0.5)=s*log(0.5). That shows it 1-1. Is it onto? But 1-1 and onto doesn't make it an isomorphism. You have to prove things like phi(r+s)=phi(r)*phi(s), right?

 

[LMKIYHAQ]

Thanks for the help.