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Why is this function injective?

  1. May 5, 2013 #1
    1. The problem statement, all variables and given/known data
    The function from R to R satisfies x + f(x) = f(f(x)) Find all Solutions of the equation f(f(x)) = 0.

    Part of the problem solution says that if f(x) = f(y), then "obviously" x = y. I understand the rest of the solution, but why does f(x) = f(y) imply that x = y?
     
  2. jcsd
  3. May 5, 2013 #2
    Have you tried comparing f(f(x)) and f(f(y))?
     
  4. May 5, 2013 #3
    sorry im new to this stuff, but do you mean f(f(x)) = 0 implies f(f(y)) = 0 which implies f(f(x)) = f(f(y)) ? If yes then how does this prove f(x) = f(y)?
     
  5. May 5, 2013 #4

    HallsofIvy

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    No, he meant nothing of the sort. And f(f(x))= 0 does NOT imply f(f(y))= 0.
     
  6. May 5, 2013 #5
    If f(f(x)) = 0, independently of the argument f(x), then doesn't substituting f(y) for f(x) give f(f(y)) = 0?
     
  7. May 5, 2013 #6

    Office_Shredder

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    If f(x) = f(y), then f(f(x)) = f(f(y)). Use this to prove that x=y. This has nothing to do with anything equaling zero
     
  8. May 5, 2013 #7
    If f(y) = f(x) then substituting f(y) into the original equation gives f(f(y) = f(x) + y = f(x) + x, then subtracting the f(x) from the last equation gives x = y, is this correct? Thanks for the help.
     
  9. May 5, 2013 #8
    That is right.
     
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