Question

Let T be the set of all triangles in a plane with R being a relation in T given by $R=\\{\left( {{T}_{1}},{{T}_{2}} \right):{{T}_{1}}\text{ is similar to }{{\text{T}}_{2}}\\}$ . Then R is:
A. Reflexive only
B. Not transitive
C. An equivalence relation
D. Symmetric only

Answer 1

Think of the basic definition of the types of relations given in the question and try to check whether the relation mentioned in the questions satisfies the condition for any type of relation or not.

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. We know ${{T}_{1}}$ similar to ${{T}_{2}}$ implies that ${{T}_{2}}$ similar to ${{T}_{1}}$ . Hence, the relation $R=\\{\left( {{T}_{1}},{{T}_{2}} \right):{{T}_{1}}\text{ is similar to }{{\text{T}}_{2}}\\}$ is symmetric. Also, a triangle can be said to be similar to itself, so the relation is reflexive too. We can also say that if two triangles are such that they are similar to the same third triangle, then they are similar to each other as well. Hence, they are transitive as well.
As we have shown that the relation is symmetric, reflexive, and transitive, we can say that the relation is an equivalence relation.

So, the correct answer is “Option C”.

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.