Given $A:N \times N$ with binary operator $\;\ast \;$defined\;by $(a,b)*(c,d) =(a+c,c+d)$

$\textbf {Step 1: Checking if the operation is commutative}$:

An operation $\ast$ on $A$ is commutative if $a\ast b = b \ast a\; \forall \; a, b \in A$

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

Similarly, $(c,d)*(a,b) =((c+a),(d+c)) = ((a+c),(c+d))$ as addition is commutative in $N$.

$\Rightarrow $(a,b)*(c,d) =(c,d)*(a,b)$ \rightarrow$ the operation $\ast$ is commutative

$\textbf {Step 1: Checking if the operation is associative}$:

An operation $\ast$ on $A$ is associative if $a\ast ( b \ast c) = (a \ast b) \ast c\; \forall \; a, b, c \in A$

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

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

$\Rightarrow$ $((a,b) * (c,d))*(e,f)=(a,b)*((c,d*(e,f)) \rightarrow $ the operation $\ast$ is associative

$\textbf {Step 2: Checking if the operation has an identity}$:

We know that the element $e \in N $ is an identify element for operation * if $a*e=e*a$ for all $a \in N$

Let $e=(e_1e_2) \in A \qquad a=(a_1,a_2) \in A$

Therefore, $ a*e=(a_1,a_2)*(e_1*e_2)$ $=(a_1+e_1,a_2+e_2)$

However, this is not equal to $a = (a_1,a_2)$, which for example would imply that $a_1 = a_1+e_1 \rightarrow e_1 = 0$, which is not possible.

Hence no identify element $(e_1e_2)$ exists in N.for the operation *.