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

It is given that:

f: W → W is defined as f(n) = `{(n -1, "if n is odd"),(n+1, "if n is even"):}`

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:

g(m) =  `{(m+1,"if m is even"),(m-1,"if m is odd"):}`

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.