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Homework Help: Prove no isomorphism from rationals to reals

  1. Jan 25, 2010 #1
    1. The problem statement, all variables and given/known data
    Prove that there is no isomorphism, [tex]\phi[/tex], from Q under addition to R under addition

    2. Relevant equations
    An isomorphism [tex]\phi[/tex]:Q to R is a bijection such that [tex]\phi[/tex](x + y) = [tex]\phi[/tex](x) + [tex]\phi[/tex](y), where x,y are elements of Q

    [tex]\phi[/tex](0) = 0.

    [tex]\phi[/tex](-x) = -[tex]\phi[/tex](x)

    3. The attempt at a solution
    My inclination is to attempt to attempt to show a contradiction from two equal rationals p/q and p'/q' occurs when,

    [tex]\phi[/tex](p/q - p'/q') = [tex]\phi[/tex](0) = 0, for p,q integers.

    So, [tex]\phi[/tex](p/q) - [tex]\phi[/tex](p'/q') = 0,

    However, I cannot arrive at an algebraic contradiction.

    Is there a better way to go about this proof, relying only on group theory?
  2. jcsd
  3. Jan 25, 2010 #2


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    The rationals have the property that for every nonzero x and y there are integers n and m such that n*x+m*y=0 where n and m are nonzero integers. Can you prove that? phi(n*x)=n*phi(x), right? Now find two reals that don't have that property.
    Last edited: Jan 25, 2010
  4. Jan 25, 2010 #3
    Hi Dick,

    Thank you very much for your help. I feel that I can prove the property of rationals that your described.

    However, while I know that there are reals that don't have that property, such as [tex]\pi[/tex] and the sqrt(1), I do not know how to prove that those two numbers do not have that property.

    Thanks in advance,
  5. Jan 25, 2010 #4


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    They don't. But let's pick an easier example, suppose phi(x)=1 and phi(y)=sqrt(2). Can you have n*1+m*sqrt(2)=0?
  6. Jan 26, 2010 #5
    No, because that would imply [tex]\sqrt{2}[/tex] = -n/m, i.e. that [tex]\sqrt{2}[/tex] is rational..

    Thanks a lot, I really appreciate it.
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