Q. 2 Let f : W \rightarrow Wbe 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.

Answers (1)

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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