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Question: Let R be the relation over the set \(N\times N\)and is defined by \(\left( a,b \right)R\left( c,d \r...

Let R be the relation over the set N×NN\times Nand is defined by (a,b)R(c,d)a+d=b+c\left( a,b \right)R\left( c,d \right)\Rightarrow a+d=b+c. Then R is,
A. Reflexive only
B. Symmetric only
C. Transitive only
D. An equivalence relation

Explanation

Solution

Hint: We will be using the concepts of functions and relations to solve the problem. We will be using the definitions of reflexive relation, symmetric relations and transitive relations to verify if each relation holds or not and hence deduce the answer.

Complete step-by-step answer:

Now, we have been given a relation and we have to find whether the relation is reflexive, symmetric, transitive or a combination of these.
Now, we know that reflexive relations are those in which every element is mapped to itself i.e. (a,a)R\left( a,a \right)\in R while symmetric relations are those for which if a R b then b R a. Also, holds and transitive are those relations in which if a R b and b R c then a R c must be held.
Now, we know different types of relations, we will check the given relation for these.
Now, we have been given a relation R over the set N×NN\times Nas,
(a,b)R(c,d)a+d=b+c\left( a,b \right)R\left( c,d \right)\Rightarrow a+d=b+c
Now, for the relation to be reflexive we have,
(a,b)R(a,b)a+b=b+a =a+b=a+b \begin{aligned} & \left( a,b \right)R\left( a,b \right)\Rightarrow a+b=b+a \\\ & =a+b=a+b \\\ \end{aligned}
Which is true always, so this means that,
(a,b)R(a,b)\left( a,b \right)R\left( a,b \right) and hence, the function is reflexive.
Now, for symmetric we have,
(a,b)R(c,d)a+d=b+c..........(1) Now, (c,d)R(a,b)c+b=d+a d+a=c+b a+d=b+c.........(2) \begin{aligned} & \left( a,b \right)R\left( c,d \right)\Rightarrow a+d=b+c..........\left( 1 \right) \\\ & Now,\ \left( c,d \right)R\left( a,b \right)\Rightarrow c+b=d+a \\\ & \Rightarrow d+a=c+b \\\ & a+d=b+c.........\left( 2 \right) \\\ \end{aligned}
So, we have from (1) and (2),
(a,b)R(c,d)(c,d)R(a,b)\left( a,b \right)R\left( c,d \right)\Rightarrow \left( c,d \right)R\left( a,b \right)
Hence, the relation is symmetric also.
Now, for transitive we have if (a,b)R(c,d) and (c,d)R(e,f)\left( a,b \right)R\left( c,d \right)\ and\ \left( c,d \right)R\left( e,f \right) that is we have if,
a+d=b+c and c+f=d+ea+d=b+c\ and\ c+f=d+e
Now, adding both we have,
a+c+f+d=b+c+d+e a+f=b+e \begin{aligned} & a+c+f+d=b+c+d+e \\\ & a+f=b+e \\\ \end{aligned}
Which implies that,
(a,b)R(e,f)R\left( a,b \right)R\left( e,f \right)\in R
Hence, the relation is transitive also. Now, we know that if a relation is reflexive, symmetric and transitive then the relation is an equivalence relation.
Hence, the correct option is (D).

Note: To solve these types of questions it is important to note that a R b means that a is related to b by a relation R. Also these types of questions are solved easily by giving examples and counterexamples.