Question

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

Answer 1

$f:\to w,w$ is the set of all whole number such that,

$f^{-1}\left(x\right)=\{n-1,ifnisoddn+1,ifniseven\}$

$Letn1,n2ϵwandf\left(x1\right)=f\left(x2\right)\left\{bothodd\right\}n1-1=n2-1\Rightarrow n1=n2$

Similarly,

$Letf\left(x3\right)=f\left(x4\right)\left\{bohteven\right\}n3+1=n4+1\Rightarrow n3=n4$

So $f$ is one-one function again every element in co-domain w is the image of element in domain of w.So it is also on-to.

Let,$y=n-1$,n is odd.

and $n=y+1$,y is even.

$\therefore f^{{}^{\prime }}=n+1$,n is even

again let $z=n+1$, n is even

$n=z-1$, z is odd

$\therefore f^{-1}\left(x\right)=n-1$, n is odd.

hence,$f^{-1}\left(x\right)=\{n-1,ifnisoddn-1,ifniseven\}$

So, inverse of the given functon is the function itself.