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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.

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Solution

The given function f:WW is defined by,

f( n )={ n1,ifnisodd n+1,ifniseven

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

If n is odd and m is even, then,

n1=m+1 nm=2

This cannot be possible.

If n is even and m is odd, then,

n+1=m1 mn=2

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

If both n and m are odd, then,

n1=m1 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:WW is defined as,

g( r )={ r+1,ifriseven r1,ifrisodd

When n is odd,

gοf( n )=g( f( n ) ) =g( n1 ) =n1+1 =n

When n is even,

gοf( n )=g( f( n ) ) =g( n+1 ) =n+11 =n

When r is odd,

fοg( r )=f( g( r ) ) =f( r1 ) =r1+1 =r

When r is even,

fοg( r )=f( g( r ) ) =f( r+1 ) =r+11 =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.


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