Question

Let f: W->W be defined as f(n)=n-1, if n is odd and f(n)=n+1, if n is even. Find the inverse of f.

Answer 1

Looks f is its own inverse ?

Answer 2

How can I prove it?

Answer 3

I have proved, n is odd: n=y+1
and n is even, n=y-1; while proving f(n) is surjective.

Answer 4

I'll give it a try after eating in 1 hour

Answer 5

How can I conclude that f is its own inverse?

Answer 6

It's noticeable that the cases get reverse in case of finding the inverse.

Can I write: when n is odd, f inverse of n= n+1
when n is even, f is inverse of n, n-1?


But here f is not it's own inverse.

Answer 7

so f(2n+1)=2n and f(2n)=2n+1

Answer 8

to prove the map has an inverse, it must be surjective and injective (onto and one-to-one). Surjectivity is fine since all element from the codomain are mapped to either the element after of before. Injectivity is implied by the linearity of the function (each number maps only to the next or previous one). So an inverse exists, and you found it.