Question

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

Answer 1

f : W → W is defined as \[f(x)=\begin{cases} x - 1, \text { if x is odd }\\ x + 1, \text { if x is even}\end{cases}\] 

First, we need to show that f is one-one.

Let \[f\left( x_1 \right) = f\left( x_2 \right)\]

Case I : x₁ is even and x₂ is odd
Then,

\[f\left( x_1 \right) = f\left( x_2 \right)\]

⇒ x₁ + 1 = x₂ − 1
⇒ x₂ − x₁ = 2, which is impossible.
Case II : x₁ is odd and x₂ is even
Then,

\[f\left( x_1 \right) = f\left( x_2 \right)\]

⇒ x₁ − 1 = x₂ − 1
⇒ x₁ = x₂
Case IV : x₁ and x₂ are even
Then,

\[f\left( x_1 \right) = f\left( x_2 \right)\]

⇒ x₁ + 1 = x₂ + 1
⇒ x₁ = x₂
Thus, we can see that f is one-one.
Now, we need to show that f is onto.
Any odd number 2y + 1, in the co-domain W, is the image of 2y in the domain W. Also, any even number 2y, in the co-domain W, is the image of 2y − 1 in the domain W.
Thus, every element in W (co-domain) has its pre-image in W (domain). So, f is onto.
Therefore, f is a bijection.
So, it is invertible.
Now, let x, y

\[\in\] W, such that

f (x) = y
⇒ x − 1 = y, if x is odd
x + 1 = y, if x is even
⇒ \[x = \begin{cases}{y + 1, \text { if y is even }}\\y - 1, {\text { if y is odd }}\end{cases}\]

⇒ \[f^{- 1}(y)= \begin{cases} y + 1,{ \text { if y is even }} \\y - 1, {\text { if y is odd }}\end{cases}\]

Hence,

\[f^{- 1}(x) = \begin{cases}{x - 1, {\text { if x is odd}}}\\x + 1,{ \text { if x is even}}\end{cases}\]

Clearly,f = f ⁻¹.