# Q. 2 Let $f : W \rightarrow W$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.

$f : W \rightarrow W$

$f (n) = n -1$    if n is odd

$f (n) = n +1$    if n is even.

For one-one:

Taking x as odd number and y as even number.

$f(x)=f(y)$

$x-1=y+1$

$x-y=2$

Now,   Taking y as odd number and x  as even number.

$f(x)=f(y)$

$x+1=y-1$

$x-y = - 2$

This is also impossible.

If both x and y are odd :

$f(x)=f(y)$

$x-1=y-1$

$x=y$

If both  x and  y are even :

$f(x)=f(y)$

$x+1=y+1$

$x=y$

$\therefore$  f is one-one.

Onto:

Any odd number    2r+1  in co domain of N is image of 2r in domain N and any even number 2r in codomain N is image of  2r+1 in domain N.

Thus, f is onto.

Hence, f is one-one and onto i.e. it is invertible.

Sice, f is invertible.

Let  $g : W \rightarrow W$   as   $m+1$  if m is even   and   $m-1$ if  m is odd.

When x is odd.

$gof(x)=g(f(x))= g(n-1) =n-1+1=n$

When x is even

$gof(x)=g(f(x))= g(n+1) =n+1-1=n$

Similarly, m is odd

$fog(x)=f(g(x))= f(m-1) =m-1+1=m$

m is even ,

$fog(x)=f(g(x))= f(m+1) =m-1+1=m$

$\therefore$             $gof(x)=I_W$      and   $fog(x)=I_W$

Hence, f is inertible  and inverse of f is g i.e. $f^{-1}=g$, which is same as f.

Hence, inverse of f is f itself.

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