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Math Help - Prove inverse of a bijection is increasing

  1. #1
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    Prove inverse of a bijection is increasing

    Let f: A->B be a bijection, where A and B are subsets of the set of all real numbers. Prove that if f is increasing on A, then f's inverse is increasing on B.

    My professor said to use the contrapositive of "f: A->B is increasing" to prove this, but I don't understand how that would help me.

    I know that if f is a bijection and g is the inverse of f, then g is also a bijection and f is the inverse of g.

    I don't even know where to start.
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  2. #2
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    If x\in A,~y\in A~\&~x\le y then f(x)\le f(y).

    Suppose s\in B,~t\in B~\&~s\le t then would happen if
    f^{-1}(s)> f^{-1} (t)?.

    HINT: What is f(f^{-1}(t))=?
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  3. #3
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    Ok I know that f(f-1(t))=t

    So if f-1(s)>f-1(t) then that would mean the inverse function is decreasing on B.

    Am I supposed to use f(f-1(s))<=f(f-1(t))? Can that even be true if the inverse function is decreasing?
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  4. #4
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    Quote Originally Posted by letitbemww View Post
    Ok I know that f(f-1(t))=t

    So if f-1(s)>f-1(t) then that would mean the inverse function is decreasing on B. Am I supposed to use f(f-1(s))<=f(f-1(t))? Can that even be true if the inverse function is decreasing?
    No it does not mean that.

    It means s=f(f^{-1}(s))>t=f(f^{-1}(t)).
    What is wrong with that?
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  5. #5
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    Oh, that can't be true if we already established s<=t.

    Is it just that simple? I guess I was confused on how I could use the contrapositive to prove it. I don't even understand my professor's hint because there is no "if...then" statement there.

    Maybe I'm just not understanding right. It seems like you are proving it by contradiction.
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  6. #6
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    Almost all of this type problem can proved by contradiction.
    That is what was done.
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  7. #7
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    Ok thanks!!
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