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Injective one to one proof?

  1. May 13, 2012 #1
    The problem statement, all variables and given/known data

    Let E ⊆ R and f : E → R a decreasing function for all x ∈ E. Prove that f is injective.

    The attempt at a solution

    I tried that f were not injective.
    Then, there exist x < y such that f(x) = f(y)
    -This contradicts f being a decreasing function.
    I think this is right, but I am unsure what to do now
     
  2. jcsd
  3. May 13, 2012 #2

    Dick

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    In what way are you unsure that x<y and f(x)=f(y) contradicts f being a decreasing function? What's your definition of 'decreasing function'?
     
  4. May 13, 2012 #3
    For a one to one function that is decreasing i have that x < y so therefore f(x) < f(y). So f(x) is not equal to f(y)).

    or have i got this the wrong way around?
     
  5. May 13, 2012 #4

    Dick

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    So, if f(x)=f(y) for x<y then that would contradict the assumption that f is decreasing, yes? I'm still confused about where you are confused.
     
  6. May 13, 2012 #5
    Yes it would. I thought it would be easier to prove though contradiction my question, but I am unsure whether this is the best way to prove this/right approach.

    Sorry for the confusion
     
  7. May 13, 2012 #6

    Dick

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    If you are assuming f(x)=f(y) for x<y then you ARE doing a proof by contradiction. You've assumed f is not injective. And it is the right way to go. Maybe it would be better if you showed your whole proof step by step.
     
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