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One-to-One Functions

  1. Feb 8, 2005 #1
    Hi, if anyone could help me with the following question, that'd be great. Thanks in advance =)

    Show that the composition of two one-to-one functions, f and g, is one-to-one.

    Thank you,

    RW Techs
     
  2. jcsd
  3. Feb 8, 2005 #2
    Hmm, I find it hard to believe you have thought about the question, it is very easy.

    The composition of the functions is f(g(x)). We know for each x there exists a unique u such that g(x) = u. So f(g(x)) is one to one iff f(u) is one to one, which it is, by assumption.

    Transform this in to symbolic logic and you have a proof good enough for an analysis class in which you prove things that are already obviously true.
     
  4. Feb 8, 2005 #3

    mathwonk

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    start with the definition of one one. i.e. h is one one if whenever x and y are different, thenm also h(x) and h(y) are different......
     
  5. Feb 8, 2005 #4

    It appears to me that your proof relies on nothing more than the fact that g is a function. g(x) could be any function, and the "proof" still "works".
     
  6. Feb 9, 2005 #5

    Galileo

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    "We know for each x there exists a unique u such that g(x) = u"

    That won't work for each function.

    Equivalent, but maybe easier to see:
    You are given that:
    [tex]f(x)=f(y) \Rightarrow x=y[/tex]
    and
    [tex]g(u)=g(v) \Rightarrow u=v[/tex]
    for all x,y in the domain of f and u,v in the domain of g.
    You need to show that:

    [tex]f(g(u))=f(g(v)) \Rightarrow u=v[/tex]

    for all u,v in the domain of g.
     
    Last edited: Feb 9, 2005
  7. Feb 9, 2005 #6

    matt grime

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    I think crankfan was pointing out that that answer wasn't very clear in its proof. I for one can't stand the "there is a unique" part of it. If you should be of the desire to quantify "there is unique" in logical symbols as pressed by that answer, then you're on hiding to nothing.

    Your proof using the f(x)=f(y) => x=y is much clearer.
     
  8. Feb 9, 2005 #7
    Maybe I was quibbling. I think it's the quantification that I didn't like.

    An ordinary function has a unique u in the range associated
    with each each x in the domain.

    But yes, Galileo's proof is much cleaner.
     
  9. Feb 9, 2005 #8

    Galileo

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    Whoopsie, that should be the other way around anyway:
    For each u in the range of g there is an unique x, such that g(x)=u.
     
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