Let $A = N \times N$ and $^*$ be the binary operation on $A$ defined by $(a, b) \,^*\, (c, d) = (a + c, b + d)$. Determine whether the operation $^*$ is commutative,associative,and has an identity element.

(A) Given $A = N \times N$ and the binary operation $^*$ defined by $(a, b) \,^*\, (c, d) = (a + c, b + d)$.
$1$. Commutativity:
For any $(a, b), (c, d) \in A$,we have:
$(a, b) \,^*\, (c, d) = (a + c, b + d)$
$(c, d) \,^*\, (a, b) = (c + a, d + b) = (a + c, b + d)$
Since addition is commutative in $N$,$(a, b) \,^*\, (c, d) = (c, d) \,^*\, (a, b)$. Thus,$^*$ is commutative.
$2$. Associativity:
For any $(a, b), (c, d), (e, f) \in A$,we have:
$[(a, b) \,^*\, (c, d)] \,^*\, (e, f) = (a + c, b + d) \,^*\, (e, f) = (a + c + e, b + d + f)$
$(a, b) \,^*\, [(c, d) \,^*\, (e, f)] = (a, b) \,^*\, (c + e, d + f) = (a + c + e, b + d + f)$
Since addition is associative in $N$,the operation $^*$ is associative.
$3$. Identity Element:
Let $e = (e_1, e_2) \in A$ be the identity element. Then $(a, b) \,^*\, (e_1, e_2) = (a, b)$,which implies $(a + e_1, b + e_2) = (a, b)$.
This requires $a + e_1 = a$ and $b + e_2 = b$,so $e_1 = 0$ and $e_2 = 0$.
Since $0 \notin N$,there is no identity element in $A$.