Answer :

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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