# Thread: Proving that real with multiplaication is not isomorphic to real with addition

1. ## Proving that real with multiplaication is not isomorphic to real with addition

I'm trying to prove that $(\mathbb{R}^*,\times)$ is not isomorphic to $(\mathbb{R},+)$ (the notation $\mathbb{R}^*$ means $\mathbb{R}\setminus \left\{ 0 \right\}$) 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 $\times$ and $+$ notations for conveniency):

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

Then it must hold that $\varphi (0)=1$.
Let $x\neq0 \in\mathbb{R}$, then it has its inverse $x^{-1}=(-x)$.
So:
$1=\varphi (0)=\varphi (x-x)=\varphi (x+x^{-1})=\varphi (x)\varphi (x^{-1})=\varphi (x)\varphi (x)^{-1}=\varphi (x)^{-2}$
Finally:
$\varphi (x)^{2}=1$ (properties of multiplication in $\mathbb{R}^*$)
Now, since we know that $\varphi (x)$ can be either $1$ or $-1$... what does that tell me? that $\varphi$ is not well defined (and therefor cannot be isomorphism)? ( $x$ is sent to two different $y\in \mathbb{R}$...)

2. ## Re: Proving that real with multiplaication is not isomorphic to real with addition

Originally Posted by Stormey
I'm trying to prove that $(\mathbb{R}^*,\times)$ is not isomorphic to $(\mathbb{R},+)$ (the notation $\mathbb{R}^*$ means $\mathbb{R}\setminus \left\{ 0 \right\}$) but I'm having difficulties when I get to the final conclusion.
Suppose that $\varphi :\mathbb{R}^* \rightarrow \mathbb{R}$ is an isomorphism (I'll try to derive a contradiction).
Then it must hold that $\varphi (0)=1$
If you have posted what you actually meant then your mapping is mistaken.

$0\notin\mathbb{R}^*$ so what you have is it must hold that $\varphi (1)=0$.

Did you mean $\varphi :\mathbb{R} \rightarrow \mathbb{R}^*~?$

3. ## Re: Proving that real with multiplaication is not isomorphic to real with addition

yes of course. thank you.

4. ## Re: Proving that real with multiplaication is not isomorphic to real with addition

Stormey,
I'm afraid your computations are totally garbled.

Since $\phi$ is onto, there is $a\in\mathbb{R}\text{ with }\phi(a)=-1$

So $\phi(2a)=\phi(a+a)=\phi(a)\phi(a)=(-1)(-1)=1$

As you pointed out, $\phi(0)=1$ and since $\phi$ is one to one, 2a=0 or a=0

Contradiction -- $1\neq-1$

Aside - the group of positive reals under multiplication is isomorphic to the additive group of the reals; any logarithm function shows this.

5. ## Re: Proving that real with multiplaication is not isomorphic to real with addition

Thank you.
Actually that was the contradiction I was looking for in the first place - showing that $\varphi$ can't be one-to-one.