**Question:**

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.

**Solution:**

It is given that:

$f: W \rightarrow W$ is defined as $f(n)= \begin{cases}n-1, & \text { if } n \text { is odd } \\ n+1, & \text { if } n \text { is even }\end{cases}$

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.

∴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*) ⇒ *n* + 1 = *m* + 1 ⇒ *n* = *m*

*∴ f is one-one.*

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

=∴*f* is onto.

Hence, *f *is an invertible function.

Let us define $g: W \rightarrow W$ as:

$g(m)=\left\{\begin{array}{l}m+1, \text { if } m \text { is even } \\ m-1, \text { if } m \text { is odd }\end{array}\right.$

Now, when *n* is odd:

And, when *n* is even:

Similarly, when *m *is odd:

When *m *is even:

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.

Click here to get exam-ready with eSaral

For making your preparation journey smoother of JEE, NEET and Class 8 to 10, grab our app now.