# Let A = N × N and

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.

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