# Let A = N × N. Define * on A by

Question:

Let A = N × N. Define * on A by (a, b) * (c, d) = (a + c, b + d).

Show that

(i) $A$ is closed for $*$,

(ii) $*$ is commutative,

(iii) $*$ is associative,

(iv) identity element does not exist in $\mathrm{A}$.

Solution:

(i) $A$ is said to be closed on $*$ if all the elements of $(a, b) *(c, d)=(a+c, b+d)$ belongs to $N \times N$ for $A$ $=\mathrm{N} \times \mathrm{N}$.

Let a = 1, b = 3, c = 8, d = 2

$(1,3) *(8,2)=(1+8,3+2)$

$=(9,5) \in N \times N$

Hence A is closed for *.

(ii) For commutative,

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

As addition is commutative a+ c = c+ a and b+ d = d+ b, hence * is commutative binary operation.

(iii) For associative

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

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

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

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

As $(a, b) *((c, d) *(e, f))=((a, b) *(c, d)) *(e, f)$, hence $*$ is an associative binary operation.

(iv) For identity element $\left(e_{1}, e_{2}\right),(a, b) *\left(e_{1}, e_{2}\right)=\left(e_{1}, e_{2}\right) *(a, b)=(a, b)$ in a binary operation.

$(a, b) *\left(e_{1}, e_{2}\right)=(a, b)$

$\left(a+e_{1}, b+e_{2}\right)=(a, b)$

$\left(e_{1}, e_{2}\right)=(0,0)$

As $(0,0) \notin N \times N$, hence identity element does not exist for $*$.