Let A = N × N and * be the binary operation on A defined by

Solution:

A = N × N

* is a binary operation on A and is defined by:

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

Let (a, b), (c, d) ∈ A

Then, a, b, c, d ∈ N

We have:

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

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

[Addition is commutative in the set of natural numbers]

∴(a, b) * (c, d) = (c, d) * (a, b)

Therefore, the operation * is commutative.

Now, let (a, b), (c, d), (e, f) ∈A

Then, a, b, c, d, e, f ∈ N

We have:

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

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

$\therefore((a, b) *(c, d)) *(e, f)=(a, b) *((c, d) *(e, f))$

Therefore, the operation * is associative.

An element $e=\left(e_{1}, e_{2}\right) \in$ A will be an identity element for the operation ${ }^{*}$ if

$a^{*} e=a=e^{*} a \forall a=\left(a_{1}, a_{2}\right) \in \mathrm{A}$, i.e., $\left(a_{1}+e_{1}, a_{2}+e_{2}\right)=\left(a_{1}, a_{2}\right)=\left(e_{1}+a_{1}, e_{2}+a_{2}\right)$, which is not true for any element in $\mathrm{A}$.

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