Question

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

Answer 1

Given that $A=N\times N$ and $\ast$ is a binary operation on A and is defined by
$(a,b)\ast (c,d)=(a+c,b+d)$
Let $(a,b),(c,d)\in A$
Then, $a,b,c,d\in N$
we have $(a,b)\ast (c,d)=(a+c,b+d)$
and $(c,d)\ast (a,b)=(a,b)=(c+a,d+b)=(a+c,b+d)$
[Addition is commutative in the set of natural numbers]
Therefore, $(a,b)\ast (c,d)=(c,d)\ast (a,b)$
Therefore, the operation $\ast$ is commutative.
Now, let $(a,b),(c,d),(e,f)\in A$
Then, $a,b,c,d,e\in N.$
We have $(a,b)\ast (c,d)\ast (e,f)=(a+c,b+d)\ast (e,f)=(a+c+e,b+d+f)$
and $(a,b)\ast (c,d)\ast (e,f)=(a,b)\ast (c+e,d+f)=(a+c+e,b+d+f)$
$\therefore \left(\right(a,b)\ast (c,d)\ast (e,f)=(a,b)\ast (a,b)\ast )(e,f))$
Therefore, the operation $\ast$ is associative.
An element $e=(e_1,e_2)\in A$ will be an identity element for the operation $\ast$ if
$a\ast e=a=e\ast a\forall a=(a_1,a_2)i.e.,(a_1+e_1,a_2+e_2)=(a_1,a_2)=(e_1+a_1,e_2+a_2)$
which is not true for any element in A.
Note that a+e =a for e =0 but 0 does not belong to N.
Therefore, the operation $\ast$ does not have any identity element.