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Bijections result when the function is surjective and injective

  1. May 5, 2005 #1
    Bijections result when the function is surjective and injective.

    How do I find a bijection in N and the set of all odd numbers?

    f(x) = 2x+1

    Do I have to prove that this is one-to-one and onto? Am I on the right track?
     
  2. jcsd
  3. May 5, 2005 #2

    honestrosewater

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    Yes. Can you prove that for all n and m in N, if f(n) = f(m), then n = m? (Just plug and play.)
     
  4. May 6, 2005 #3

    HallsofIvy

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    You would also, of course, have to prove that it is a surjection: that is, that if m is an odd number then there exist an integer n such that 2n+ 1= m.
     
  5. May 6, 2005 #4
    You can get the surjection from the definition of odd number.
     
    Last edited: May 6, 2005
  6. May 8, 2005 #5
    Thank you for your help. I think that I need a lot of practice using the definition.
     
  7. May 8, 2005 #6

    honestrosewater

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    Which definition- bijection?
     
  8. May 8, 2005 #7
    Yeah. It's kind of frustrating to prove if a function is one-to-one or onto when one way of proving for one problem doesn't apply to the next. It's all about playing around with the definition and all.
     
  9. May 24, 2005 #8
    What's the difference between finding a bijection between P(N) and P(Z) AND N and Z? P stands for power set (the curly P).
     
  10. May 24, 2005 #9
    You mean a bijection between P(N) and P(Z) compared to a bijection between N and Z?

    I've never done that kind of thing but I think if you have a bijection between N and Z then you automatically have a bijection between P(N) and P(Z) by replacing every element x of N in a given element y of P(N) with f(x) and mapping y to the resulting element z, which will be in P(Z). If you have a bijiection between P(N) and P(Z) then I don't think you necessarily can say you have a bijection between N and Z.
     
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