1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    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,
    Doug
     
  5. Jan 25, 2010 #4

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    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.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook