Question

Let A={1,2,3}. Then find the number of equivalence relations containing (1,2).

Answer 1

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

Complete step-by-step answer:
Before starting with the solution, let us discuss 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., $\left( a,a \right)\in R$ . 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 equivalence relations that can be formed, which contains (1,2). So, for being an equivalence relation, a relation must be symmetric, reflexive, and transitive at the same time.
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) the relation must contain (2,1) as well.So, the compulsory condition we get is that the relation must contain (1,1), (2,2), (3,3), (1,2) and (2,1).
Therefore, the possible relations are:
${{R}_{1}}=\{(1,1),(1,2),(2,2),(2,1),(3,3)\}$
Now the other possibility is if we add a relation (1,3) to ${{R}_{1}}$ . But as we add (1,3), we are bound to add (3,1) to keep the relation to be symmetric. Now as (3,1) is added and (1,2) is a part of ${{R}_{1}}$, so to keep the relation transitive, we add the term (3,2) and further a term (2,3) to get an equivalence relation.
${{R}_{2}}=\{(1,1),(1,2),(2,2),(2,1),(3,3),(1,3),(2,3),(3,1),(3,2)\}$
So, only the above two are the possible cases. Therefore, the number of equivalence relations containing (1,2) is 2.

Note: Remember a relation can also be called a transitive relation if there exists $aRb$ , but there doesn’t exist any relation $bRc$ as in the above first possible case it is transitive relation as well.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.