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Math Help - bijection help

  1. #1
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    bijection help

    If f: S--->S is bijective, and if a e Z, show that f^a is also a bijective function S--->S. Hint: use Def. 15

    Def. of f^a:

    Supose f: S--->S is a bijection. If a = n is a positive integer, then f^n is the nth iterate of f. If a = -n is a negative integer, then f^a = f^(-n) is the nth iterate of the inverse f^(-1). If a = 0, then f^a is the identity function.

    Def. 15:

    Let f: S--->S be a bijection. SUppose a e Z and that a = [m,n]. Then f^a = f^m o (f^-1)^n

    o = composition
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  2. #2
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    Quote Originally Posted by jzellt View Post
    If f: S--->S is bijective, and if a e Z, show that f^a is also a bijective function S--->S. Hint: use Def. 15

    Def. of f^a:

    Supose f: S--->S is a bijection. If a = n is a positive integer, then f^n is the nth iterate of f. If a = -n is a negative integer, then f^a = f^(-n) is the nth iterate of the inverse f^(-1). If a = 0, then f^a is the identity function.

    Def. 15:

    Let f: S--->S be a bijection. SUppose a e Z and that a = [m,n]. Then f^a = f^m o (f^-1)^n

    o = composition
    use the fact that composition of bijections is bijective
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