Question

Let $A=R\times R$ and $\ast$ be a 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. Also find the inverse of every element $(a,b)\in A$.

Answer 1

$A=R\times R$

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

$Commutative:let(a,b),(c,d)\in A$.

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

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

= $(c,d)\ast (a,b)\forall (a,b),(c,d)\in A$

$\ast$ is commutative.

$Associative:let(a,b),(c,d),(e,f)\in A$.

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

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

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

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

$=(a,b)\ast \left(\right(c,d)\ast (e,f\left)\right)\forall (a,b),(c,d),(e,f)\in A$

$\ast$ is associative.

$Identityelement:$

Let $(e_1,e_2)\in A$ is identify element for $\ast$ operation by definition.

$\Rightarrow (a,b)\ast (e_1,e_2)=(a,b)$

$\Rightarrow (a+e_1,b+e_2)=(a,b)$

$a+e_1=a,b+e_2=b$

$\Rightarrow e_1=0,e_2=0$

$\Rightarrow (0,0)\in A$

$\Rightarrow (0,0)$ is identity element for $\ast$.

$Inverse$:

Let $(b_1,b_2)\in A$ is inverse of element $(a,b)\in A$ then by definition.

$(a,b)\ast (b_1,b_2)=(0,0)$

$(a+b_1)=0,b+b_2=0$

$\Rightarrow (-a,-b)\epsilon A$ is inverse of every elemnt $(a,b)\in A.$