Question

Let A={1,2,3}. Then find the number of relations containing (1,2) and (1,3), which are reflexive and symmetric but not transitive is:
(A). 1
(B). 2
(C). 3
(D). 4

Answer 1

Hint: Think of the basic definition of the types of relations given in the figure and start with each the relation set (1,2) and (1,3), and keep adding other relations to it till it becomes an equivalence relation.

Complete step-by-step solution -
Before starting with the solution, let us discuss the different types of relations. There are a total of 8 types of relations that we study, out of which the major ones are reflexive, symmetric, transitive, and equivalence relation.
Reflexive relations are those in which each and every element is mapped to itself, i.e., $\left( a,a \right)\in R$ . While the symmetric relations are those for which, if \[\left( a,b \right)\in R\text{ }\] then $\left( b,a \right)$ must also belong to R. This can be represented as $aRb\Rightarrow bRa$ . Now, transitive relations are those for which, if $\left( a,b \right)\text{ and }\left( b,c \right)\in R$ then $\left( a,c \right)$ must also belong to R, i.e., $\left( a,b \right)\text{ and }\left( b,c \right)\in R\Rightarrow \left( a,c \right)\in R$ .
Now, if there exists a relation, which is reflexive, symmetric, and transitive at the same time, then the relation is said to be an equivalence relation. For example: let us consider a set A=(1,2). Then the relation {(1,2),(2,1),(1,1),(2,2)} is an equivalence relation.
Now let us start with the solution to the above question. See, the set given to us is A={1,2,3} and asked the number of relations that can be formed, which contains (1,2) and (1,3).
Therefore, the relation to be reflexive must contain (1,1), (2,2), (3,3). Next, for the relation to be symmetric and contain (1,2) and (1,3), the relation must contain (2,1) and (3,1) as well. So, the compulsory condition we get is that the relation must contain (1,1), (2,2), (3,3), (1,2), (2,1), (3,1) and (1,3).
Therefore, the possible relations are:
${{R}_{1}}=\{(1,1),(1,2),(2,2),(2,1),(3,3),(1,3),(3,1)\}$
No other relation exists which would be reflexive and symmetric but not transitive. For examples:
${{R}_{2}}=\{(1,1),(1,2),(2,2),(2,1),(3,3),(1,3),(2,3),(3,1)\}$
${{R}_{3}}=\{(1,1),(1,2),(2,2),(2,1),(3,3),(1,3),(3,2),(3,1)\}$
${{R}_{2}}\text{ and }{{R}_{3}}$ are reflexive but not symmetric.
So, only the above three are the possible cases. Therefore, the number of relations containing (1,2) and (1,3), which are reflexive and symmetric but not transitive is 1.
Hence, the answer to the above question is option (a).

Note: Remember a relation can also be called a transitive relation if there exists $aRb$ , but there doesn’t exist any relation $bRc$ . Also, most of the questions as above are either solved by using statements based on observation or taking examples, as we did in the above question.