Question

If * is a binary operation defined on $A=N\times N$ by (a,b)*(c,d)=(a+c,b+d), then prove that * is both commutative and associative.

Answer 1

Hint: We are given the definition of a binary operation on the set $A=N\times N$ i.e. the set of all ordered pairs in which both the components belong to N, the set of natural numbers. We are asked to prove that it is commutative and associative. Therefore, to prove commutativity, we should interchange the order of the elements while taking the relation and prove that the value remains the same, i.e. we should prove that (a,b)(c,d)= (c,d)(a,b). Similarly, for associativity we should prove that (a,b)((c,d)(e,f))= ((a,b)(c,d))(e,f) i.e. the order in which the relation is taken does not matter and the value remains the same in any order we take the operations.

Complete step-by-step answer:
We are given that the binary operation is defined on the set $A=N\times N$ i.e. the set of all ordered pairs in which both the components belong to N, the set of natural numbers. The definition of the binary operation is
$\left( a,b \right)*\left( c,d \right)=\left( a+c,b+d \right).................(1.1)$
i.e. the first and second component of the final ordered pair is the sum of the first components and second components of the two elements on which the binary operation is taken respectively. We also know that the set of Natural numbers, N is commutative and associative, i.e. for arbitrary elements$p,q,r\in N$ ,
$\begin{aligned} & p+q=q+p.................................(1.2) \\\ & \left( p+q \right)+r=p+\left( q+r \right)................................(1.3) \\\ \end{aligned}$
To prove that the binary operation is commutative, we should prove that for arbitrary elements (a,b) and (c,d) belonging to A, $\left( a,b \right)*\left( c,d \right)=\text{ }\left( c,d \right)*\left( a,b \right)......................(1.4)$
To prove it use the definition of the operation in (1.1) and use (1.2) to write

& \left( a,b \right)*\left( c,d \right)=\left( a+c,b+d \right) \\\ & =\left( c+a,d+b \right)\text{ (using 1}\text{.2)}....................\text{(1}\text{.5)} \\\ \end{aligned}$$ However, from definition of the binary relation in (1.1), we can write $$\left( c,d \right)*\left( a,b \right)=\left( c+a,d+b \right).................(1.6)$$ Therefore, as the RHS of (1.5) and (1.6) are equal we can write $$\left( a,b \right)*\left( c,d \right)=\left( c,d \right)*\left( a,b \right)........................(1.7)$$ Therefore, comparing (1.7) and (1.4), we have proved that the given relation is commutative. Similarly, to prove that the given relation is associative, we should show that for any three elements (a,b), (c,d), (e,f) in A, $$\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)..........(1.8)$$ Form (1.1) and (1.3), we can write $$\begin{aligned} & \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( \left( a+c \right)+e,\left( b+d \right)+f \right) \\\ & =\left( a+\left( c+e \right),b+\left( d+f \right) \right)\text{ (Using 1}\text{.3)} \\\ & \text{=}\left( a,b \right)*\left( \left( c,d \right)*\left( e,f \right) \right)\text{ (Using 1}\text{.1)} \\\ \end{aligned}$$ Therefore, comparing the last line to (1.8), we see that we have proved that the binary relation is associative. Now, if there is an identity element (p, q) in A, then p, q should be natural numbers as A contains ordered pairs of natural numbers………………(1.9) However, as (p,q) is assumed to be identity, for any other ordered pair (r,s) $\begin{aligned} & \left( p,q \right)*\left( r,s \right)=\left( r,s \right) \\\ & \Rightarrow \left( p+r,q+s \right)=\left( r,s \right) \\\ & \Rightarrow p+r=r\text{ and q+s=s} \\\ & \Rightarrow \text{p=0 and q=0}..................\text{(1}\text{.10)} \\\ \end{aligned}$ However, as 0 is not a natural number, (1.9) and (1.10) contradict each other. Hence, there does not exist any identity in A. Note: We should note that in equation (1.1), the elements (a,b) and (c,d) correspond to any arbitrary elements of A, for example, if we take (a,b)=(5,6) and (c,d)=(8,9), equation (1.1) would imply (a,b)*(c,d)=(5,6)*(8,9)=(5+8,6+9)=(13,15). Thus, we should not consider the variables a,b,c,d etc. in this question to be of a specific value, they are just variables and are used to represent a general element in A.