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

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    Proving that real with multiplaication is not isomorphic to real with addition

    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 $\displaystyle y\in \mathbb{R}$...)
    Last edited by Stormey; Dec 21st 2013 at 04:59 AM.
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    Re: Proving that real with multiplaication is not isomorphic to real with addition

    Quote Originally Posted by Stormey View Post
    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.
    Suppose that $\displaystyle \varphi :\mathbb{R}^* \rightarrow \mathbb{R}$ is an isomorphism (I'll try to derive a contradiction).
    Then it must hold that $\displaystyle \varphi (0)=1$
    If you have posted what you actually meant then your mapping is mistaken.

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

    Did you mean $\displaystyle \varphi :\mathbb{R} \rightarrow \mathbb{R}^*~?$
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    Re: Proving that real with multiplaication is not isomorphic to real with addition

    yes of course. thank you.
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    Re: Proving that real with multiplaication is not isomorphic to real with addition

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

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

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

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

    Contradiction -- $\displaystyle 1\neq-1$

    Aside - the group of positive reals under multiplication is isomorphic to the additive group of the reals; any logarithm function shows this.
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    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 $\displaystyle \varphi$ can't be one-to-one.
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