Let $A=\mathbf{N} \times \mathbf{N}$ and ${ }^{*}$ be the binary operation on $A$ defined by
$(a, b) *(c, d)=(a+c, b+d)$
Show that ${ }^{*}$ is commutative and associative. Find the identity element for * on $\mathrm{A}$, if any.
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.