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Math Help - 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 (\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}...)
    Last edited by Stormey; December 21st 2013 at 04:59 AM.
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  2. #2
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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 (\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}^*~?
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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 \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.
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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 \varphi can't be one-to-one.
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