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Prove that Q under addition is not isomorphic to R+

  1. Oct 20, 2004 #1
    How do I prove that Q under addition is not isomorphic to R+ under multiplication?
     
  2. jcsd
  3. Oct 20, 2004 #2

    matt grime

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    They cannot be isomorphic as groups because they are not even in bijective correspondence as <insert one word to get the answer>
     
  4. Oct 20, 2004 #3
    Isn't it f(x) = exp(x) a bijection between Q and R+?
     
  5. Oct 20, 2004 #4

    jcsd

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    No an isomorphism must be onto.
     
  6. Oct 20, 2004 #5
    Why f:Q -> R+, f(x) = exp(x) is not onto?
    For all r of R+, there exists r' = lnr in Q such that r = exp(lnr) = exp(r') = f(r'). Where do I go wrong?
     
  7. Oct 20, 2004 #6

    jcsd

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    If r is irrational is r in Q? is er in R+?
     
  8. Oct 20, 2004 #7
    You're totally right. Then is there any way to show that Q and R+ are not isomorphic?
     
  9. Oct 20, 2004 #8
    I think I know the answer. If I say that any map between Q and R+ is not onto, is that enough?
     
  10. Oct 20, 2004 #9

    jcsd

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    Yes, you just need to look at the two sets Q and R+ to see that the two groups cannot be isomorphic (as Matt grime indicated).
     
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