Question

Let f : W → W be defined as f ( n ) = n − 1, if 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

The given function f:W→W is defined by,

f( n )={ n−1,   if n is odd n+1,   if n is even

Consider that f( n )=f( m ).

If n is odd and m is even, then,

n−1=m+1 n−m=2

This cannot be possible.

If n is even and m is odd, then,

n+1=m−1 m−n=2

This shows that both n and m should be either even or odd.

If both n and m are odd, then,

n−1=m−1 n=m

If both n and m are even, then,

n+1=m+1 n=m

Thus, f is one-one.

It can be observed that any odd number 2k+1 in the co-domain N is the image of 2k in domain N. Also, any even number 2k in the co-domain N is the image of 2k+1 in domain N.

Thus, f is onto.

Therefore, f is an invertible function.

Now, assume that the function g:W→W is defined as,

g( r )={ r+1,  if r is even r−1,  if r is odd

When n is odd,

gοf( n )=g( f( n ) ) =g( n−1 ) =n−1+1 =n

When n is even,

gοf( n )=g( f( n ) ) =g( n+1 ) =n+1−1 =n

When r is odd,

fοg( r )=f( g( r ) ) =f( r−1 ) =r−1+1 =r

When r is even,

fοg( r )=f( g( r ) ) =f( r+1 ) =r+1−1 =r

Thus, gοf=I and fοg=I.

This shows that f is invertible and the inverse of f is f −1 =g, that is same as f.

Therefore, the inverse of f is f itself.