Question

Let $f:W\to W$ 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.

Answer 1

It is given that
$f:W\to W$ is defined as $f\left(n\right)=\{\begin{array}{c}n-1,ifnisodd\\ n+1,ifniseven\end{array}$
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 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
$f\left(n\right)=f\left(m\right)\Rightarrow n-1=m-1\Rightarrow n=m$
Again, if both n and m are even, then we have
$f\left(n\right)=f\left(m\right)\Rightarrow n+1=m+1\Rightarrow n=m$
Therefore, f is one-one.
For onto.
It is clear that any odd number 2r +1 in co-domain W is the image of 2r in domain W and any even number 2r in co-domain W is the image of 2r +1 in domain W.
Therefore, f is onto Hence, f is an invertible function.

Let us define $g:W\to$ as $g\left(m\right)=\{\begin{array}{c}m+1,ifmiseven\\ m-1,ifmisodd\end{array}$
Now, when n is odd gof(n)=g(f(n))=g(n-1)=n-1+1=n
$(\because$ when n is odd, n-1 is even).
and when n is even gof(n)=g(f(n))=g(n+1)=n+1 -1 =n
$(\because$ when n is even, n +1 is odd).
Similarly, when m is odd fog(m)=f(g(m))=f(m-1)=m-1+1=m
when m is even fog(m)=f(g(m))=f(m+1)=m+1-1=m
$\therefore gof=I_w$ and $fog=I_W$
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 itself.