Question

Let $f:W\to W$ be defined as

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

Show that $f$ is invertible and find the inverse of $f$. Here, $W$ is the set of all whole numbers.

Answer 1

Let $n_1$and $n_2$ be two distinct elements of $W$.

To Prove : $f$ is one-one.

Case I. When $n_1$ and $n_2$ are both odd

Here, $n_1\ne n_2$

$\Rightarrow n_1-1\ne n_2-1$

$\Rightarrow f\left(n_1\right)\ne f\left(n_2\right)$

Case II, When $n_1$ and $n_2$ are both even

Here, $n_1\ne n_2$

$n_1+1\ne n_2+1$

$\Rightarrow f\left(n_1\right)\ne f\left(n_2\right)$

Case III. When $n_1$ is odd and $n_2$ is even

Here, $f\left(n_1\right)=n_1-1$ is even

$f\left(n_2\right)=n_2+1$ is odd

$\Rightarrow f\left(n_1\right)\ne f\left(n_2\right)$

Case IV. When $n_1$ is even and $n_2$ is odd

Here, $f\left(n_1\right)=n_1+1$ is odd

$f\left(n_2\right)=n_2-1$ is even

$\Rightarrow f\left(n_1\right)\ne f\left(n_2\right)$

Thus, in all cases $n_1\ne n_2\Rightarrow f\left(n_1\right)\ne f\left(n_2\right)$

$\Rightarrow f$ is one-one

To Prove : $f$ is onto

If $n\in W$ is any element, then

$f(n-1)=n$, if $n$ is odd

and $f(n+1)=n$, if $n$ is even

$\therefore$ Each element of $W$ is $f$ image of some element of $W$.

Thus $f$ is onto

Hence, $f$ is invertible

Now, $f(n-1)=n$, if $n$ is odd

$f(n+1)=n$, if $n$ is even

$\Rightarrow n-1=f^{-1}\left(n\right)$, if $n$ is odd

$n+1=f^{-1}\left(n\right)$, if $n$ is even

Thus, $f^{-1}\left(n\right)=\{\begin{array}{ccc}n-1,& if& nisodd\\ n+1,& if& niseven\end{array}$.

Hence, $f^{-1}=f$