show a conclusion from the homomorphism $\phi : \Bbb R _{>0} \to \Bbb R$ such that $\phi (r) = \log(r)$ I need two show homomorphism and get a conclusion from iso1 in the following: 
a) I have $\phi : \Bbb R _{>0} \to \Bbb R$
$\phi (r) = \log(r)$
I assume here that $\Bbb R _{>0}$ is with multiplication and $\Bbb R$ is with addition??  Is that correct to assume ? 
because I don't have this information in the question. if it is with addition then I don't know how to make homomorphism. 
so i get $\ker (\phi) = 1 =
 e$
so my conclusion  from iso1 is that 
$\Bbb R _{>0} \cong \Bbb R$
which dosen't make sense to me..
b) I have $\phi : \Bbb R^* \to \Bbb R$
$\phi (r) = log(|r|)$
again I assume that $\Bbb R$ is with addition 
and I get to the conclusion that
$\Bbb R^*/Sign \cong \Bbb R$
which again dosen't make sense. 
any help will be very appreciated
 A: They ask you to show there is a homomorphism in both cases. It is left for you to make the additional necessary assumptions to do so. Your guess for this is correct, as $\log(xy)=log(x)+log(y)$
As for your surprise about the correct conclusion that you draw, remember you are dealing with the continuum: the whole real line is contained within any connected open subset of $\mathbb{R}$. For example, $f(x)=\arctan(x)$ is a bijection between $(-\pi/2,\pi/2)$ and $\mathbb{R}$.
Mind you, a similar bijection occur for, say the natural numbers, where there are as many even natural numbers as the whole set of natural numbers.
A: Yes, your assumptions are correct. 
To show that $\phi$ is a homomorphism you must show: 
$$\phi(xy)=\phi(x)\phi(y)$$ 
Define: $\phi(x)=log(x), s.t. \phi: \mathbb{R_{>0}} \to \mathbb{R}$ 
Consider: 
$$\phi(xy)=log(xy)=log(x)+log(y)=\phi(x)+\phi(y)$$ 
So we have that $\phi(x)$ is a homomorphism. 
Here, $ker(\phi(x))=1$ since $log(1)=0$ where 1,0 are identity elements of the domain and codomain respectively. So you have that the identity of your domain maps to the identity of your codomain. 
Thus, $\phi(x)$ mapping between these two groups as defined is an isomorphism and these two groups are isomorphic.
So, why is your observation true? Infinite sets can map 1-1 onto subsets of themselves. Cantor used this to describe the different sizes of infinite sets to show that the cardinality of (0,1) is the same as the cardinality of $\mathbb{R}$. 
Hope this helps a bit. 
A: You are correct in your assumptions--after all, $\Bbb R_{>0}$ isn't a group under addition, and $\Bbb R$ isn't a group under multiplication.
The thing to keep in mind is that, unlike what we see in finite sets, infinite sets can be in one-to-one correspondence with proper subsets of themselves! In fact $\Bbb R$ can be put in one-to-one correspondence with $(a,b)$ for any two $a,b\in\Bbb R$ with $a<b,$ but that is not particularly relevant.
You have in fact proved that $r\mapsto\log(r)$ is a one-to-one homomorphism $\Bbb R_{>0}\to\Bbb R.$ If you can prove it is surjective, then you have proved that it is actually a bijection and so an isomorphism.
From there, since $\Bbb R^*/\{1,-1\}\cong\Bbb R_{>0}$ readily, then the first result gives you another way to verify that $\Bbb R^*/\{1,-1\}\cong\Bbb R$.
