Question

Let $A$ be the set of all triangles in a plane and let $R$ be a relation in $A$ , defined by R=\left\\{ \left( {{\Delta }_{1}},{{\Delta }_{2}} \right):{{\Delta }_{1}}\cong {{\Delta }_{2}} \right\\} . Show that $R$ is an equivalence relation in $A$ .

Answer 1

Hint:For solving this question first we will see an important property of congruence of triangles that if ${{\Delta }_{1}}\cong {{\Delta }_{2}}$ and ${{\Delta }_{2}}\cong {{\Delta }_{3}}$ then, ${{\Delta }_{1}}\cong {{\Delta }_{2}}\cong {{\Delta }_{3}}$ . After that, we will prove that the given relation is reflexive, symmetric, and transitive then the relation will be automatically an equivalence relation.

Complete step-by-step answer:
Given:
It is given that there is a set $A$ of all triangles in a plane and let $R$ be a relation in $A$ , defined by R=\left\\{ \left( {{\Delta }_{1}},{{\Delta }_{2}} \right):{{\Delta }_{1}}\cong {{\Delta }_{2}} \right\\} . And we have to prove that $R$ is an equivalence relation in $A$ .
Now, before we proceed we should know one of the very basic properties of congruency that if triangle 1 $\left( {{\Delta }_{1}} \right)$ is congruent to triangle 2 $\left( {{\Delta }_{2}} \right)$ and triangle 2 $\left( {{\Delta }_{2}} \right)$ is congruent to triangle 3 $\left( {{\Delta }_{3}} \right)$ then, automatically triangle 1 $\left( {{\Delta }_{1}} \right)$ will be congruent to triangle 3 $\left( {{\Delta }_{3}} \right)$ . Then,${{\Delta }_{1}}\cong {{\Delta }_{2}}\cong {{\Delta }_{3}}$
Now, we will be using the above result for proving the desired result in this question.
Now, we will prove that the given relation is reflexive, symmetric, and transitive one by one.
1. If ${{\Delta }_{1}}$ and ${{\Delta }_{2}}$ are the same triangles in the set $A$ then, we can say that ${{\Delta }_{1}}\cong {{\Delta }_{1}}$ always that’s why $\left( {{\Delta }_{1}},{{\Delta }_{1}} \right)\in R$ . Thus, the given relation is reflexive.

2. If ${{\Delta }_{1}}$ and ${{\Delta }_{2}}$ are two different triangles in the set $A$ such that ${{\Delta }_{1}}\cong {{\Delta }_{2}}$ which means $\left( {{\Delta }_{1}},{{\Delta }_{2}} \right)\in R$ . Now, we can also write that ${{\Delta }_{2}}\cong {{\Delta }_{1}}$ which means that $\left( {{\Delta }_{2}},{{\Delta }_{1}} \right)\in R$ . Thus, now $\left( {{\Delta }_{1}},{{\Delta }_{2}} \right)\in R$ and also $\left( {{\Delta }_{2}},{{\Delta }_{1}} \right)\in R$ . So, the given relation will be symmetric also.

3. If ${{\Delta }_{1}}$ , ${{\Delta }_{2}}$ and ${{\Delta }_{3}}$ are three different triangles in the set $A$ such that ${{\Delta }_{1}}\cong {{\Delta }_{2}}$ , ${{\Delta }_{2}}\cong {{\Delta }_{3}}$ which means $\left( {{\Delta }_{1}},{{\Delta }_{2}} \right)\in R$ and $\left( {{\Delta }_{2}},{{\Delta }_{3}} \right)\in R$ . Now, we can also write that if ${{\Delta }_{1}}\cong {{\Delta }_{2}}$ and ${{\Delta }_{2}}\cong {{\Delta }_{3}}$ then, ${{\Delta }_{1}}\cong {{\Delta }_{3}}$ always that’s why $\left( {{\Delta }_{1}},{{\Delta }_{3}} \right)\in R$ . Thus, now $\left( {{\Delta }_{1}},{{\Delta }_{2}} \right)\in R$ and $\left( {{\Delta }_{2}},{{\Delta }_{3}} \right)\in R$ then $\left( {{\Delta }_{1}},{{\Delta }_{3}} \right)\in R$ . So, the given relation will be transitive also.
Now, as we have proved that the given relation is reflexive, symmetric, and transitive. Thus, the relation will be an equivalence relation.
Hence, proved.

Note: Here, the student must try to understand the relation given in the problem then prove that the relation is reflexive, symmetric, and transitive separately and don’t mix up their conditions with each other to avoid the confusion. Moreover, the student should apply the basic concept of congruence of triangles with clarity.