It is given that A = N × N and ∗ be the binary operation on A defined by

(a, b) ∗ (c, d) = (a + c, b + d)

Let (a, b), (c, d) ϵ A

Then, a, b, c, d ϵ N

Now, we have,

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

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

⇒ (a, b) * (c, d) = (c, d) * (a, b)

⇒ The operation * is commutative.

Now, (a, b), (c, d), (e, f) ϵ A

Then, a, b, c, d, e, f ϵ N

Now, we have,

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

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

Then, ((a, b) * (c, d)) * (e, f) = (a, b)* ((c, d) * (e, f))

Therefore, the operation * is associative.

Now, an element e = (e₁, e₂) ϵ A will be an identity for the operation *

if a * e = a = e * a a = (a₁, a₂) ϵ A,

(a₁ + e₁, a₂ + e₂) = (a₁, a₂) = (e₁ + a₁, e₂ + a₂)

which is not true for any element in A.

Therefore, the operation * does not have any identity element.