Let A = RxR 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 binary element for * on A, if any.

A = R X R

For the ‘ * ’ to be commutative, p * q = p * q must be true for all p, q belong to A. Let’s check.
Note: Here p, q A represent the ordered pairs (a, b) and (c, d) respectively.

p * q = (a, b) * (c, d) = (a + c, b + d)
q * p = (c, d) * (a, b) = (c + a, d + b) = (a + c, b + d)
⇒ p * q = q * p (binary operation * is commutative)

For the ‘ * ’ to be associative, a * (b * c) = (a * b) * c must hold for every a, b, c ∈ A. Here r = (e, f)

p * (q * r) = (a, b) * ((c, d) * (e, f)) = (a, b) * (c + e, d + f) = (a + c + e, b + d + f)
(p * q) * r = ((a, b) * (c, d)) * (e, f) = (a + c, b + d) * (e, f) = (a + c + e, b + d + f)
⇒ p * (q * r) = (p * q) * r (Associative)

Binary elements:

Identity Element: Given a binary operation*: A X A → A, an element e ∈A, if it exists, is called an identity of the operation*, if p*e = p = e*p ∀ p ∈A.

Here p = (a, b) and e = (x, y). For the element e to exist,

(a, b) * (x, y) = (a, b)
⇒ (a + x, b + y) = (a, b)
⇒ a + x = a, b + y = b
Since, ordered pairs are only equal when both the first and second terms are equal
⇒ x = 0, y = 0
Hence the identity element e = (x, y) = (0, 0)

Given a binary operation with the identity element e in A, an element a A is said to be invertible with respect to the operation, if there exists an element b in A such that a * b = e = b * a and b is called the inverse of a and is denoted by a^–1.

Let i = (r, s) be the inverse of p = (a, b) in A.

Therefore, (r, s) * (a, b) = e = (0, 0)
⇒ (r + a, s + b) = (0, 0)
∴r + a = 0, r + b = 0
r = – a, s = – b

Therefore, (r, s) = ( – a, – b) is the inverse pair.