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Integer-valued function

  1. May 4, 2009 #1
    Is there any function (if any) f: Z -> Z such that
    f(f(n))=-n , for every n belongs to Z(integers) ??


    I think that there is not any function like the one described above but how can we prove it. Any ideas??
    Thanks in Advance
     
  2. jcsd
  3. May 4, 2009 #2

    mathman

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    How about f(n)=in?
     
  4. May 4, 2009 #3

    CRGreathouse

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    Not integer-valued. (I assume if the OP meant Gaussian integers that would have been mentioned, since that's the obvious solution.)

    I've been thinking about this for a few hours now and I can't see any way to do it, but I can't prove that it's impossible.
     
  5. May 5, 2009 #4
    How about

    for n>0
    f(2n-1)=2n
    f(2n)=-2n+1
    f(-2n+1)=-2n
    f(-2n)=2n-1

    f(0)=0
     
  6. May 5, 2009 #5
    Nice, chronon. Nice.
     
  7. May 5, 2009 #6

    matt grime

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    Indeed - it's nice to visualize f as a piecewise permutation

    (0)(-2,-1,2,1)(-4,-3,4,3)...(-2n,-2n+1,2n,2n-1)....

    and recall that (abcd)^2=(ac)(bd)
     
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