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

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

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

So, the correct answer is “Option A”.

Note: Here in this question for $\ast$ to be commutative, we have to show that$(a,b)\ast (c,d)=(c,d)\ast (a,b)$. 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.