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Show an additive homomorphism f:Q->Q is Q linear

  1. Oct 20, 2013 #1
    1. The problem statement, all variables and given/known data
    Let f(x+y)=f(x)+f(y) where x, y, f(x) are all rational numbers. Show that f(qx)=qf(x) for all q, x rational.


    2. Relevant equations



    3. The attempt at a solution
    It's easy to get to f(nx)=nf(x). Because f(x+...+x)=f(x)+...+f(x)=nf(x). But going from there to rational is hard. I think I've done it before, but I can't get it. I've been stumped.
     
  2. jcsd
  3. Oct 20, 2013 #2

    Dick

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    Start with a simple example. f((1/2)x)+f((1/2)x)=f((1/2)x+f((1/2)x)=f(x). So how is f((1/2)x) related to f(x)? Now generalize. If you've done it before this should click pretty fast.
     
    Last edited: Oct 20, 2013
  4. Oct 20, 2013 #3
    ...

    It was that easy, wasn't it? Thank you. :redface:
     
  5. Oct 20, 2013 #4
    I absolutely remember doing this before. I feel so silly

    f(x)=f(nx/n)=nf(x/n)
    so f(x)/n=f(x/n)

    I appreciate it, that's been stumping me. I should have known it
     
  6. Oct 20, 2013 #5

    Dick

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    You're welome and great, it did click fast.
     
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