Originally Posted by
jzellt 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