Let f: W$\rightarrow$W be defined as f(n) $=$n$-$1, if n is odd and f (n) $=$ n + 1, if n is even. Show that f is invertible. Find the inverse of f. Here, W is the set of all whole numbers.

f : W$\rightarrow$ W

$f(n) = \left\{ {\begin{array}{llllllllllllllllllll}{n - 1,}&{if}&{{\rm{n is odd}}}\\{n + 1,}&{if}&{{\rm{n is even}}}\end{array}} \right.$

Injectivity

Let n, m be any two odd real whole numbers.

$\therefore$ $f(n) = f(m) \Rightarrow n - 1 = m - 1 \Rightarrow n = m$

Again, let n, m be any two even whole numbers.

$\therefore$ $f(n) = f(m) \Rightarrow n + 1 = m + 1 \Rightarrow n = m$
If n is even and m is odd, then n $\ne$ m.

Also, if f(n) is odd and f(m) is even, then f(n)$\ne$f(m)

Thus, if $n \ne m \Rightarrow f(n) \ne f(m)$ $\therefore$ f is an injective.

Surjectivity :

Let n be an arbitrary whole number.

If n is an odd number, then there exists an even whole number

(n + 1) such that f(n + 1) $=$n + 1$-$1 $=$n

If n is an even number, then there exists an odd whole number,

such that f(n$-$1)$=$(n $-$1) + 1 $=$n

Thus, every $n \in W$ has its pre-image in W.

So, f: W$\rightarrow$W is a surjective.

Thus, f is invertible and ${f^{ - 1}}$ exists.

Now, f(n$-$1) $=$ n, if n is odd and f(n + 1) $=$n, if n is even.

$\Rightarrow$ n$-$1 $=$ ${f^{ - 1}}$(n), if n is odd and n + 1 $=$ ${f^{ - 1}}(n)$, if n is even.

Hence, ${f^{ - 1}}(n) = \left\{ {\begin{array}{llllllllllllllllllll}{n - 1,}&{if}&{{\rm{n is odd}}}\\{n + 1,}&{if}&{{\rm{n is even}}}\end{array}} \right.$. Hence, ${f^{ - 1}} = f$