In this question, we do not have individuals as elements but ordered pairs as the elements of the Cartesian product R₀ x R. Hence this is the same problem as the above problems but in the form of ordered pairs belonging to the Cartesian product.

A = R₀ x R

i. We are given with the set A which is the Cartesian product of R₀ and R.

Here the function *: A X A → A is given by (a, b)°(c, d) = (ac, bc + d)

So here the operation on two ordered pairs gives us a ordered pair. So the first element of the first ordered pair (a) multiplies with the first element of the second ordered pair (c) to make (ac) in the resulting pair. To make the second element of the final pair, second element of the first ordered pair (b) multiplies with the first element of the second ordered pair (c) to form (bc) and then this is added to the second element of the second ordered pair (d) to make (bc + d). Hence the final ordered pair is (ac, bc + d). So this is the operation basically.

For the ‘⁰’ to be commutative, a * b = b * a must be true for all a, b belong to A. Let’s check.
Note: Here a, b represent the ordered pairs (a, b) and (c, d) respectively.

1. (a, b)°(c, d) = (ac, bc + d)
2. (c, d)°(a, b) = (ca, da + b)
⇒ a * b≠b * a (as shown by 1 and 2)

Hence ‘⁰’ is not commutative on A.

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

3. (a, b)°[(c, d)°(e, f)] = (a, b)°(ce, de + f)

= (ace, bce + de + f)

4. [(a, b)°(c, d)]°(e, f) = (ac, bc + d)°(e, f)

= (ace, [bc + d]e + f)
= (ace, bce + de + f)
⇒ 3. = 4.

Hence ‘⁰’ is associative on A.

ii. 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 in this case, p and e are an ordered pairs.

Let e = (a, b) be the identity element of A and p = (x, y) where p A.

∴, (x, y)°(a, b) = (a, b)
⇒ (xa, ya + b) = (a, b)
⇒ ∴xa = a, ya + b = b

(Since, ordered pairs are only equal when both the first and second terms are equal)
⇒ x = 1 (x gets cancelled out)
⇒ y(1) + b = b

⇒ b = 0

The ordered pair (1, 0) is the identity of the operation ‘⁰’ on A.

iii. 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.

For this example, a, b, e are ordered pairs.

Let us proceed with the solution.

Let (c, d) A be the invertible elements in A of (a, b) here (a, b) A.

∴(a, b)°(c, d) = (1, 0) (We know the identity element from previous)
⇒ (ac, bc + d) = (1, 0)
⇒ ac = 1, bc + d = 0

(Since, ordered pairs are only equal when both the first and second terms are equal)

(Required invertible element)