Question

Let f : W → W be defined as f(n) = n -1, if n is odd and f(n) = n + 1, if n is even. Show that/is invertible. Find the inverse of f. Here, W is the set of all whole numbers.

Answer 1

We are given, f : W → W, defined as

\(\left\{\begin{array}{ll} n-1, & n \text { is odd } \\ n+1, & n \text { is even } \end{array}\right.\)

Let f(n1) =f(n2). If n1 is odd and n2 is even,

then n1 – 1 = n2 + 1, i.e., n1 – n2 = 2, which is not possible as n1 is odd and n2 is even.

When n1 and n2 both are odd, then

f(n1) =f(n2) ⇒ n1 – 1 = n2 – 1 ⇒ n1 = n2

When n1 and n2 both are even

f(n1) = f(n2) ⇒ n1 + 1 = n2 + 1 ⇒ n1 = n2.

This shows f is one-one.

Any odd number 2r + 1 in codomain is the image of 2r and any even number 2r in the codomain W is the image of 2r + 1.

⇒ f is one-one and onto.

∴ f is invertible.

Further, when n is odd, y = n – 1, n = y + 1, y is even.

When n is even, y = n + l, n = y y is odd.

Hence, f-1(y) = g(y) defined as

g : W → W, such that g(y) = \(\left\{\begin{array}{ll}

y-1, & y \text { is odd } \\

y+1, & y \text { is even }

\end{array}\right.\)

⇒ The inverse of f is f itself.