An isomorphism between $( \mathbb{R} , + )$ and $ ( P , \cdot )$ Am I incorrect in believing that the following exercise is not possible?

Prove that the additive group $( \mathbb{R}, + )$ of real numbers is isomorphic to the multiplicative group $( P , \cdot )$ of positive reals.

My reasoning is that if we had $\phi \colon \mathbb{R} \to P$ as our isomorphism, then we have
$$\phi(\tfrac{1}{3}) = \tfrac{1}{3}$$ 
and
$$\phi(-3) = \tfrac{1}{3}$$
Am I missing something?
 A: Your reasoning is a little faulty since you are assuming that $\phi ( \frac{1}{3} ) = \frac 13$ (since a priori there is not reason to think that $\frac 13$ must be a fixed point of such an isomorphism). But even if $\phi ( \frac 13 ) = \frac 13$, then this would tell us that $$\begin{align}
\phi ( 3 ) 
&=  \phi (\tfrac 13+\tfrac 13+\tfrac 13+\tfrac 13+\tfrac 13+\tfrac 13+\tfrac 13+\tfrac 13+\tfrac 13) \\ 
&= \phi (\tfrac 13)\cdot\phi(\tfrac 13)\cdot\phi(\tfrac 13)\cdot\phi(\tfrac 13)\cdot\phi(\tfrac 13)\cdot\phi (\tfrac 13)\cdot\phi(\tfrac 13)\cdot\phi(\tfrac 13)\cdot\phi(\tfrac 13) \\ 
&= \tfrac 13 \cdot \tfrac 13 \cdot \tfrac 13 \cdot \tfrac 13 \cdot \tfrac 13 \cdot \tfrac 13 \cdot \tfrac 13 \cdot \tfrac 13 \cdot \tfrac 13 \\ 
&= 3^{-9}
\end{align}$$
and therefore $\phi ( -3 ) = ( \phi ( 3 ) )^{-1} = ( 3^{-9} )^{-1} = 3^9$.

If you recall the following rule of exponentiation: $$a^{x+y} = a^x \cdot a^y$$ you should begin to think that a mapping of the form $x \mapsto a^x$ looks "homomorphism-ish," and it is not too difficult to show that if $a > 0$, then such a mapping is an isomorphism between $( \mathbb{R} , + )$ and $( P , \cdot )$.
