I'm trying to prove that $\displaystyle (\mathbb{R}^*,\times)$ is not isomorphic to $\displaystyle (\mathbb{R},+)$ (the notation $\displaystyle \mathbb{R}^*$ means $\displaystyle \mathbb{R}\setminus \left\{ 0 \right\}$) but I'm having difficulties when I get to the final conclusion.

This is my work so far (from now on, I'll drop the $\displaystyle \times$ and $\displaystyle +$ notations for conveniency):

Suppose that $\displaystyle \varphi :\mathbb{R}^* \rightarrow \mathbb{R}$ is an isomorphism (I'll try to derive a contradiction).

Then it must hold that $\displaystyle \varphi (0)=1$.

Let $\displaystyle x\neq0 \in\mathbb{R}$, then it has its inverse $\displaystyle x^{-1}=(-x)$.

So:

$\displaystyle 1=\varphi (0)=\varphi (x-x)=\varphi (x+x^{-1})=\varphi (x)\varphi (x^{-1})=\varphi (x)\varphi (x)^{-1}=\varphi (x)^{-2}$

Finally:

$\displaystyle \varphi (x)^{2}=1$ (properties of multiplication in $\displaystyle \mathbb{R}^*$)

Now, since we know that $\displaystyle \varphi (x)$ can be either $\displaystyle 1$ or $\displaystyle -1$... what does that tell me? that $\displaystyle \varphi$ is not well defined (and therefor cannot be isomorphism)? ($\displaystyle x$ is sent to two different $\displaystyle y\in \mathbb{R}$...)