Question

If $f:W\to W$ is defined as f(x)=x-1 if x is odd and f(x)=x+1 if x is even. Show that f is invertible. find the inverse of f, where W is a set of whole numbers.

Answer 1

It is given that

$P:w\to w$ is defined as

$\Rightarrow f\left(x\right)=x-1$ if $x$ is odd

$=x+1$ if $x$ is even

For one - one

Let $f\left(x\right)=f\left(y\right)$

It can be observed that if $x$ is odd and $y$ is even then we have

$\Rightarrow x-1=y+1$

$\Rightarrow x-y=2$

This is impossible

Similarly the possibility of $x$ being even and $y$ being odd can also be ignored under similar argument.

So both $x$ and $y$ must be either odd or even.

If both are odd then ,

$\Rightarrow f\left(x\right)=f\left(y\right)$

$\Rightarrow x-1=y-1$

$\Rightarrow x=y$

If $x$ and $y$ both even then

$\Rightarrow f\left(x\right)=f\left(y\right)$

$\Rightarrow x+1=y+1$

$\Rightarrow x=y$

so $f$ is one one

For on to ;

It is clear that any odd number $2r+1$ in c0-domain N is the image of $2r$ in domain N and any even number $2r$ in co-domain N is the image of $2r+1$ in domain N.

$\therefore f$ is on to

Hence $f$ is invertible

Let us define $g:w\to w$ as

$\Rightarrow g\left(x\right)=x+1$ if $x$ is even

$=x-1$ if $x$ is odd

Now when $n$ is odd

$\Rightarrow g$ of $\left(n\right)=g\left(f\left(n\right)\right)=g(n-1)=n-1+1=n$

when $n$ is even $g$ of $\left(n\right)=g(n+1)=n+1-1=n$

similarly,

when $m$ is odd

$\Rightarrow f_{0}g\left(m\right)=f\left(g\left(m\right)\right)=f(m-1)=m-1+1=m$

and when $m$ is even

$\Rightarrow f_{0}g\left(m\right)=f\left(g\left(m\right)\right)=f(m+1)=m+1-1=m$

$\therefore g_{0}f=I_w$ and $f_{0}g=I_w$

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