Let A= R × R and * be a binary operation on A defined by
(a, b) * (c, d) = (a+c, b+d)
Show that * is commutative and associative. Find the identity element for *
on A. Also find the inverse of every element (a, b) ∈ A.

(a, b) * (c, d) = (a + c, b + d)

(i) Commutative
(a, b) * (c, d) = (a+c, b+d)
(c, d) * (a, b) = (c+a, d+b)
for all, a, b, c, d ∈ R
* is commulative on A

(ii) Associative : ______
(a, b), (c, d), (e, f) ∈A
{ (a, b) * (c, d) } * (e, f)
= (a + c, b+d) * (e, f)
= ((a + c) + e, (b + d) + f)
= (a + (c + e), b + (d + f))
= (a*b) * ( c+d, d+f)
= (a*b) {(c, d) * (e, f)}
is associative on A

Let (x, y) be the identity element in A.

then,

(a, b) * (x, y) = (a, b) for all (a,b) ∈ A
(a + x, b+y) = (a, b) for all (a, b) ∈ A
a + x = a, b + y = b for all (a, b) ∈ A
x = 0, y = 0
(0, 0) ∈ A
(0, 0) is the identity element in A.

Let (a, b) be an invertible element of A.
(a, b) * (c, d) = (0, 0) = (c, d) * (a, b)
(a+c, b+d) = (0, 0) = (c+a, d+b)

a + c = 0 b + d = 0
a = - c b = - d
c = - a d = - b

(a, b) is an invertible element of A, in such a case the inverse of (a, b) is (-a, -b).

Let A = Q x Q, where Q is the set of rational numbers.
Given that * is the binary operation on A defined by (a, b) * (c, d) = (ac, b + ad) for 
(a, b), (c, d) ∈ A.
(i)
We need to find the identity element of the operation * in A.
Let (x, y) be the identity element in A.
Thus,
(a, b) * (x, y) = (x, y) * (a, b) = (a, b), for all (a, b) ∈ A
⇒(ax, b + ay) = (a, b)
⇒ ax = a and b + ay =b
⇒ y = 0 and x = 1
Therefore, (1, 0) ∈ A is the identity element in A with respect to the operation *.

(ii) We need to find the invertible elements of A.
Let (p, q) be the inverse of the element (a, b)
Thus,

Let f: W→W be defined as

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

The set of natural numbers, N, = {1, 2, 3, 4, 5, 6.....}
The relation is given as

Given that 
Let us find fog:


Therefore,