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Is R under addition isomorphic to R\{0} under multiplication?

  1. Dec 27, 2009 #1
    1. The problem statement, all variables and given/known data
    Is [tex]\mathbb{R}[/tex] under addition isomorphic to [tex]\mathbb{R}\backslash{0}[/tex] under multiplication?

    3. The attempt at a solution
    It is true that [tex]\mathbb{R}[/tex] under addition is isomorphic to [tex]\mathbb{R}_{>0}[/tex] under multiplication, by using the bijection [tex]\phi : \mathbb{R}\rightarrow\mathbb{R}_{>0}[/tex] with [tex]\phi\left(x\right)=e^x[/tex]. But we cannot use this isomorphism for [tex]\mathbb{R}\backslash{0}[/tex] since the exponential is positive everywhere, and if we define it to be negative for [tex]x<0[/tex] then we cannot map to the interval [tex](0,1)[/tex]. Intuitively this gives a hint that there is no isomorphism possible... Can I do something with the fact that we know that if there is an isomorphism, it must map inverses to inverses, therefore [tex]\phi(0)=1[/tex]? I really need a hint to be able to find the answer... Thanks for any help!
    Last edited: Dec 27, 2009
  2. jcsd
  3. Dec 27, 2009 #2
    Let [itex]\varphi : (\mathbb{R},+) \to (\mathbb{R}\setminus \{0\},\times)[/itex] be such an isomorphism. For an arbitrary element [itex]x \in \mathbb{R}[/itex] we have:
    [tex]\varphi(x) = \varphi(x/2 + x/2) = \varphi(x/2)^2 \geq 0[/tex]
  4. Dec 27, 2009 #3
    Beautiful argument! Didn't think of that... thanks!
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