Question

Let f : W → W be defined as

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

Show that f is invertible a nd find the inverse of f. Here, W is the set of all whole
numbers.

Answer 1

Let f : W → W be defined as

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

We need to prove that 'f' is invertible.

In order to prove that 'f' is invertible it is sufficient to prove that f is a bijection.
A function f: A→B is a one-one function or an injection, if

f(x)=f(y) ⇒ x=y for all x, y ∈ A

Case i:
If x and y are odd.
Let f(x) = f(y)
⇒x − 1 = y − 1
⇒x = y

Case ii:
If x and y are even,
Let f(x) = f(y)
⇒x + 1 = y + 1
⇒x = y

Thus, in both the cases, we have,
f(x) = f(y) ⇒ x = y for all x, y ∈ W.
Hence f is an injection.

Let n be an arbitrary element of W.
If n is an odd whole number, there exists an even whole number n − 1 ∈ W such that
f(n − 1) = n − 1 + 1 = n.
If n is an even whole number, then there exists an odd whole number n + 1 ∈ W such that f(n + 1) = n + 1 − 1 = n.
Also, f(1) = 0 and f(0) = 1

Thus, every element of W (co-domain) has its pre-image in W (domain).
So f is an onto function.
Thus, it is proved that f is an invertible function.

Thus, a function g: B→A which associates each element y ∈ B to a unique element x ∈  A
such that f(x) = y is called the inverse of f.

That is, f(x) = y ⇔ g(y) = x
The inverse of f is generally denoted by f^-1.

Now let us find the inverse of f.
Let x, y ∈ W such that f(x) = y
⇒x + 1 = y, if x is even

And

x − 1 = y, if x is odd

`=>x={(y-1, " if y is odd"),(y+1, " if y is even"):}`

`=>f^-1 (y)={(y-1," if y is odd"),(y+1, " if y is even") :}`

Interchange, x and y, we have,

`=>f^(-1) (x)={(x-1," if y is odd"),(x+1, " if y is even") :}`

Rewriting the above we have,

`=>f^(-1) (x)={(x+1, " if y is even") ,(x-1," if y is odd") :}`

Thus f^-1(x)=f(x)