Question

Let f : W \(\to\) 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

It is given that: 
f: W → W is defined as  
\(f(n) =   \begin{cases}     n-1      & \quad \text{if } n \text{ is odd}\\     n+1  & \quad \text{if } n \text{ is even}   \end{cases}\)
One-one:
Let f(n) = f(m).
It can be observed that if n is odd and m is even, then we will have n − 1 = m + 1.
⇒ n − m = 2
However, this is impossible.
Similarly, the possibility of n being even and m being odd can also be ignored under a
similar argument.
∴Both n and m must be either odd or even.
Now, if both n and m are odd, then we have:
f(n) = f(m) ⇒ n − 1 = m − 1 ⇒ n = m
Again, if both n and m are even, then we have:
f(n) = f(m) ⇒ n + 1 = m + 1 ⇒ n = m
∴f is one-one.
It is clear that any odd number 2r + 1 in co-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.
∴f is onto.
Hence, f is an invertible function.
Let us define g: W → W as: 
\(f(n) =   \begin{cases}     m+1      & \quad \text{if } m \text{ is even}\\     m-1  & \quad \text{if } m \text{ is odd}   \end{cases}\)
Now, when n is odd:
gof(n)=g(f(n))=g(n-1)=n-1+1=n
And, when n is even:
gof(n)=g(f(n))=g(n+1)=n+1-1=n.
Similarly, when m is odd: fog(m)=f(g(m))=f(m-1)=m-1+1=m
When m is even:fog(m)=f(g(m))=f(m+1)=m-1+1=m.
∴ \(gof=I_w \,and \,fog=I_w.\)
Thus, f is invertible and the inverse of f is given by \(f^{-1}=g\) which is the same as f.

Hence, the inverse of f is f itself.