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 $A:N\times N$ with binary operation $\ast$ definrd by $(a,b)\ast (c,d)=(a+c,c+d)$.

Step:1-Checking if the operation is commutative:

An opertion $\ast$ on $A$ is commutative if

$a\ast b=b\ast a\forall a,bϵA$

$\therefore (a,b)\ast (c,d)=\left(\right(a+c),(c+d\left)\right)$

Similarly, $(c,d)\ast (a,b)=\left(\right(c+a),(d+c\left)\right)$

$=\left(\right(a+c),(c+d\left)\right)$ as addition is commutaitive in $N$.

$\Rightarrow (a,b)\ast (c,d)=(c,d)\ast (a,b)$

the operation $\ast$ is commutative.

Checking the operation is associative:

An operation $\ast$ on $A$ is associative if

$a\ast (b\ast c)=(a\ast b)\ast c\forall a,b,cϵA$

$\therefore \left(\right(a,b)\ast (c,d\left)\right)\ast (e,f)=(a+c,b+d)\ast (e,f)$

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

$(a\ast b)\ast \left(\right(c,d)\ast (e,f\left)\right)=(a,b)\ast (c+e,d,f)=(a+c+e,b+d+f)\Rightarrow \left(\right(a,b)\ast (c,d\left)\right)\ast (e,f)=(a\ast b)\ast \left(\right(c,d)\ast (e,f\left)\right)$

the operation $\ast$ is associative.

Step:-2 Checking if the operation has an identity,

We know that the element $eϵN$ an identity element for operation $\ast$

if $a\ast e=e\ast a$ for all $aϵN$

$Lete=(e1,e2)ϵA,a=(a1,a2)ϵA$

$\therefore a\ast e=(a1,a2)\ast (e1\ast e2)=(a1+e1,a2+e2)$

however,this is not equal to $a=(a1,a2)$ which for example would imply that

$a1=a1+e1\to e1=0$,which is not possible.

hence no identity element $(e1,e2)$ exists in $N$ for the operation $\ast$.