Question
Question: Let A be the set of all triangles in a plane. Show that the relation \(R=\left\\{ \left( {{\Delta }_...
Let A be the set of all triangles in a plane. Show that the relation R=\left\\{ \left( {{\Delta }_{1}},{{\Delta }_{2}} \right):{{\Delta }_{1}}\sim {{\Delta }_{2}} \right\\} is an equivalence relation on A.
Solution
Hint: In order to solve this question, we will prove that the given relation R is reflexive, symmetric, and transitive because an equivalent relation satisfies all these relations. Also, we have to remember that 2 triangles are said to be similar if their corresponding angles are congruent and corresponding sides are proportional.
Complete step-by-step answer:
In this question, we have been given a relation R, that is, R=\left\\{ \left( {{\Delta }_{1}},{{\Delta }_{2}} \right):{{\Delta }_{1}}\sim {{\Delta }_{2}} \right\\} and we have been asked to prove this as an equivalent relation on set A. So, to prove this, we will prove that the given relation R is reflexive, symmetric, and transitive relation because we know that any relation is equivalence relation only when it is reflexive, symmetric and transitive relation.
Now, let us go with reflexive relation. We know that reflexive relation is a relation which maps for itself. So, for Δ1RΔ2, we will see whether the triangles are similar or not. So, for that, we can say that any triangle that has all the angles congruent to the corresponding angles and all the corresponding sides are in proportion. So, we can say any triangle is always similar to itself. Hence, R=\left\\{ \left( {{\Delta }_{1}},{{\Delta }_{2}} \right):{{\Delta }_{1}}\sim {{\Delta }_{2}} \right\\}.
Therefore, relation R holds reflexive relation.
Now we will consider symmetric relations. Symmetric relation is a relation which also satisfies the converse relation. For example, if aRb is satisfied, then bRa should also be satisfied for symmetric relation. Here we have been given relation R, that is, R=\left\\{ \left( {{\Delta }_{1}},{{\Delta }_{2}} \right):{{\Delta }_{1}}\sim {{\Delta }_{2}} \right\\}. So, if it is true, then we have to prove that, \left\\{ {{\Delta }_{2}},{{\Delta }_{1}} \right\\} also satisfies relation R. For that, we will start from Δ1∼Δ2. We know that similarity of triangles shows commutative property, that is, Δ1∼Δ2 is same as Δ2∼Δ1, which is the condition for \left\\{ {{\Delta }_{2}},{{\Delta }_{1}} \right\\}. Hence, we can say that relation R is true forR=\left\\{ \left( {{\Delta }_{1}},{{\Delta }_{2}} \right):{{\Delta }_{1}}\sim {{\Delta }_{2}} \right\\}.
Therefore, we can say that relation R is a symmetric relation.
Now, let us go with a transitive relation, which states that if aRb and bRc, then aRc. So, to check whether R is transitive or not, we will consider Δ1,Δ2 and Δ3. And we will see that Δ1RΔ2⇔Δ1∼Δ2 and Δ2RΔ3⇔Δ2∼Δ3 and now we will see whether relation Δ1RΔ3 satisfies or not. We know that Δ1∼Δ2 and we can write it as,
Δ2∼Δ1………(i)
And also, we know that, Δ2∼Δ3………(ii)
So, from equation (i) and (ii), we can say thatΔ1,Δ3 are similar to one triangle, that is, Δ2. Therefore we can write Δ1∼Δ3, which is the condition of Δ1RΔ3.Hence, we can say R is transitive relation.
Therefore, we have proved that the given relation R is reflexive, symmetric, and transitive relation. Hence, R is an equivalent relation.
Note: While solving this question, we need to remember that a triangle is always similar to itself. Also, we should know that if Δ1∼Δ2, then Δ2 will be definitely similar to Δ1. Also, to prove this question, we need to remember that the equivalent relation is the one that is a reflexive, symmetric and transitive relation.