1. ## 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

2. 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
use the fact that composition of bijections is bijective