InterviewSolution
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Mark the tick against the correct answer in the following: Let S be the set of all triangles in a plane and let R be a relation on S defined by ∆1 S ∆2⇔ ∆1 ≡ A2. Then, R is A. reflexive and symmetric but not transitive B. reflexive and transitive but not symmetric C. symmetric and transitive but not reflexive D. an equivalence relation |
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Answer» Correct Answer is (D) an equivalence relation Given set S = {…All triangles in plane….} And R = {(∆1 , ∆2) : ∆1 , ∆2∈ S and ∆1 ≡ ∆2} Formula For a relation R in set A Reflexive The relation is reflexive if (a , a) ∈ R for every a ∈ A Symmetric The relation is Symmetric if (a , b) ∈ R , then (b , a) ∈ R Transitive Relation is Transitive if (a , b) ∈ R & (b , c) ∈ R , then (a , c) ∈ R Equivalence If the relation is reflexive , symmetric and transitive , it is an equivalence relation. Check for reflexive Consider , (∆1, ∆1) ∴ We know every triangle is congruent to itself. (∆1, ∆1) ∈ R all ∆1 ∈ S Therefore , R is reflexive ……. (1) Check for symmetric (∆1 , ∆2) ∈ R then ∆1 is congruent to ∆2 (∆2 , ∆1) ∈ R then ∆2 is congruent to ∆1 Both the equation are the same and therefore will always be true. Therefore , R is symmetric ……. (2) Check for transitive Let ∆1, ∆2, ∆3 ∈ S such that (∆1, ∆2) ∈ R and (∆2, ∆3) ∈ R Then (∆1, ∆2)∈R and (∆2, ∆3)∈R ⇒∆1 is congruent to ∆2, and ∆2 is congruent to ∆3 ⇒∆1 is congruent to ∆3 ∴(∆1, ∆3) ∈ R Therefore , R is transitive ……. (3) Now , according to the equations (1) , (2) , (3) Correct option will be (D) |
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