Question

Let $A=R\times R$ and $*$ be a binary operation on $A$ defined by
$\left( a,b \right)*\left( c,d \right)=\left( a+c,b+d \right)$. Show that $*$ is commutative and associative. Find the identity element for $*$ on $A$ .Also find the inverse of every element $\left( a,b \right)\in A$.

Answer 1

Hint: Here in this question we will try to show that $*$ is commutative and associative binary operation, by using the definitions for $*$ to be commutative it should obey that $\left( a,b \right)*(c,d)=(c,d)*(a,b)$ and for $*$ to be associative it should obey $\left( \left( a,b \right)*\left( c,d \right) \right)*\left( e,f \right)=\left( a,b \right)*\left( \left( c,d \right)*\left( e,f \right) \right)$.
For identity element and inverse of each element $\left( a,b \right)\in A$ for $*$ binary operation on $A$, we have to consider $\left( a,b \right)*\left( x,y \right)=\left( a,b \right)$for $\left( x,y \right)$to be identity element and $\left( a,b \right)*\left( x,y \right)=\left( 0,0 \right)$ for $\left( x,y \right)$to be inverse of each element $\left( a,b \right)\in A$ for $*$ binary operation on $A$.

Complete step by step answer:
Here in this question we have $A=R\times R$ and $*$ be a binary operation on $A$ defined by
$\left( a,b \right)*\left( c,d \right)=\left( a+c,b+d \right)$.
For $*$ to be commutative, we have to show that $\left( a,b \right)*(c,d)=(c,d)*(a,b)$.
$\left( a,b \right)*\left( c,d \right)=\left( a+c,b+d \right)$
And from the definition we can say that
$\left( c,d \right)*(a,b)=(c+a,d+b)$.
Here, we observe that$\left( a,b \right)*(c,d)=(c,d)*(a,b)$.
Hence, we can say that $*$ is commutative binary operation on $A$.
For $*$ to be associative, we have to show that $\left( \left( a,b \right)*\left( c,d \right) \right)*\left( e,f \right)=\left( a,b \right)*\left( \left( c,d \right)*\left( e,f \right) \right)$.
We have,
$\left( a,b \right)*\left( c,d \right)=\left( a+c,b+d \right)$.
And from the definition we can say that
$\begin{align}
  & \left( \left( a,b \right)*\left( c,d \right) \right)*\left( e,f \right)=\left( a+c,b+d \right)*\left( e,f \right) \\
 & =\left( a+c+e,b+d+f \right)
\end{align}$
And from the definition we can also say that
$\begin{align}
  & \left( a,b \right)*\left( \left( c,d \right)*\left( e,f \right) \right)=\left( a,b \right)*\left( c+e,d+f \right) \\
 & \Rightarrow \left( a+c+e,b+d+f \right)
\end{align}$
Here, we observe that $\left( \left( a,b \right)*\left( c,d \right) \right)*\left( e,f \right)=\left( a,b \right)*\left( \left( c,d \right)*\left( e,f \right) \right)$.
Hence, we can say that $*$ is associative binary operation on $A$.
From, the definition of identity element for $*$ on$A$, we can say that the identity element for $*$ on$A$ is $\left( x,y \right)$, if there exists $\left( x,y \right)$ such that $\left( a,b \right)*\left( x,y \right)=\left( a,b \right)$.
Let us simplify the $\left( a,b \right)*\left( x,y \right)=\left( a,b \right)$
$\begin{align}
  & \left( a,b \right)*\left( x,y \right)=\left( a+x,b+y \right) \\
 & \Rightarrow (a,b)
\end{align}$
This occurs only when $x=0,y=0$. So we can say that $\left( 0,0 \right)$ is the identity element for $*$ binary operation on $A$.
From, the definition of inverse of every element $\left( a,b \right)\in A$ for $*$ binary operation on $A$, we can say that the inverse element for$*$ on$A$is$\left( x,y \right)$, if there exists $\left( x,y \right)$ such that $\left( a,b \right)*\left( x,y \right)=\left( 0,0 \right)$.
Let us simplify the $\left( a,b \right)*\left( x,y \right)=\left( 0,0 \right)$
$\begin{align}
  & \left( a,b \right)*\left( x,y \right)=\left( a+x,b+y \right) \\
 & \Rightarrow (0,0)
\end{align}$
This occurs only when$x=-a,y=-b$. So we can say that $\left( -a,-b \right)$ is the inverse of every element $\left( a,b \right)\in A$ for$*$ binary operation on $A$.

So, the correct answer is “Option A”.

Note: Here in this question for $*$ to be commutative, we have to show that$\left( a,b \right)*(c,d)=(c,d)*(a,b)$. If we take “+” as not a commutative operation it will lead us to a completely different answer, so here we should be clear that “+” is a commutative operation.