Let $f: W \rightarrow W$ be defined as $f(n)=n-1,$ if 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.

It is given that:

$f: W \rightarrow W$ is defined as $f(n)=\left\{\begin{array}{ll}n-1 & \text { If } n \text { is odd } \\ n+1 & \text { If } n \text { is even }\end{array}\right.$

For one-one

Let $f(n)=f(m)$

It can be observed that if $n$ is odd and $m$ is even, then we will have $n-1=m+1$

$\Rightarrow n-m=2$

However, this is impossible. Similarly, the possibility of $n$ being even and $m$ being odd can also be ignored under a similar argument.

$\therefore$ Both $n$ and $m$ must be either odd or even. Now, if both $n$ and $m$ are odd, 

Then, we have $f(n)=f(m)$

$\Rightarrow n-1=m-1$

$\Rightarrow n=m$

Again, if both $n$ and $m$ are even,

Then, we have

$f ( n )= f ( m )$

$\Rightarrow n +1= m +1$

$\Rightarrow n=m$

$\therefore f$ is one-one.

For onto

It is clear that any odd number $2 r+1$ in co-domain $N$ is the image of $2 r$ in domain $N$ and any even number $2 r$ in co-domain $N$ is the image of $2 r +1$ in domain $N$.

$\therefore f$ is onto.

Hence, $f$ is an invertible function.

Let us define   $g: W \rightarrow W$ as $(m)=\left\{\begin{array}{ll}m+1 & \text { If } m \text { is even } \\ m-1 & \text { If } m \text { is odd }\end{array}\right.$

Now, when $n$ is odd

$g o(n)=g(f(n))=g(n-1)=n-1+1=n$ and

When $n$ is even

$g o(n)=g(f(n))=g(n+1)=n+1-1=n$

Similarly,

When $m$ is odd

$f o(m)=f(g(m))=f(m-1)=m-1+1=m$ and

When $m$ is even $f o(m)=f(g(m))=f(m+1)=m+1-1=m$

$\therefore $ $g o f= I w$ and $f o g= Iw$

Thus, $f$ is invertible and the inverse of $f$ is given by $f^{1}=g$, which is the same as $f$ .

Hence, the inverse of $f$ is $f$ itself.