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Let T be the set of all triangles in a plane with R a relation in T given by R = {(T1, T2) : T1 is congruent to T2}. Then R is
- a)non commutative relation
- b)a universal relation
- c)an equivalence relation
- d)an empty relation
Correct answer is option 'C'. Can you explain this answer?
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Verified Answer
Let T be the set of all triangles in a plane with R a relation in T gi...
Let T be the set of all triangles in a plane with R a relation in T given by R = {(T1, T2): T1 is congruent to T2}. (T1T2 ∈ R iff T1 is congruent to T2.
Reflexivity :T1≅T1 ⇒ (T1T1) ∈ R . Symmetry :(T1, T2)∈ R ⇒T1≅T2 ⇒ T2≅T1 ⇒ (T2, T1) ∈ R
Transitivity :(T1,T1) ∈ R and (T2, T3) ∈ R ⇒ T1≅T2 and T2≅T3 ⇒ T_1≅T3 ⇒ (T1, T3) ∈ R .
Therefore, R is an equivalence relation on T.
Reflexivity :T1≅T1 ⇒ (T1T1) ∈ R . Symmetry :(T1, T2)∈ R ⇒T1≅T2 ⇒ T2≅T1 ⇒ (T2, T1) ∈ R
Transitivity :(T1,T1) ∈ R and (T2, T3) ∈ R ⇒ T1≅T2 and T2≅T3 ⇒ T_1≅T3 ⇒ (T1, T3) ∈ R .
Therefore, R is an equivalence relation on T.
Most Upvoted Answer
Let T be the set of all triangles in a plane with R a relation in T gi...
Explanation:
Equivalence Relation:
An equivalence relation is a relation that is reflexive, symmetric, and transitive. Let's check if the relation R defined above satisfies these properties.
Reflexive:
For a relation to be reflexive, every element of the set T should be related to itself. In this case, it means that every triangle is congruent to itself. Since any triangle is congruent to itself, the relation R is reflexive.
Symmetric:
For a relation to be symmetric, if T1 is related to T2, then T2 should also be related to T1. In this case, if triangle T1 is congruent to triangle T2, it implies that triangle T2 is congruent to triangle T1. Therefore, the relation R is symmetric.
Transitive:
For a relation to be transitive, if T1 is related to T2, and T2 is related to T3, then T1 should also be related to T3. In this case, if triangle T1 is congruent to triangle T2, and triangle T2 is congruent to triangle T3, it implies that triangle T1 is congruent to triangle T3. Therefore, the relation R is transitive.
Since the relation R satisfies all three properties of an equivalence relation (reflexive, symmetric, and transitive), we can conclude that R is an equivalence relation.
Therefore, the correct answer is option 'C' - an equivalence relation.
Equivalence Relation:
An equivalence relation is a relation that is reflexive, symmetric, and transitive. Let's check if the relation R defined above satisfies these properties.
Reflexive:
For a relation to be reflexive, every element of the set T should be related to itself. In this case, it means that every triangle is congruent to itself. Since any triangle is congruent to itself, the relation R is reflexive.
Symmetric:
For a relation to be symmetric, if T1 is related to T2, then T2 should also be related to T1. In this case, if triangle T1 is congruent to triangle T2, it implies that triangle T2 is congruent to triangle T1. Therefore, the relation R is symmetric.
Transitive:
For a relation to be transitive, if T1 is related to T2, and T2 is related to T3, then T1 should also be related to T3. In this case, if triangle T1 is congruent to triangle T2, and triangle T2 is congruent to triangle T3, it implies that triangle T1 is congruent to triangle T3. Therefore, the relation R is transitive.
Since the relation R satisfies all three properties of an equivalence relation (reflexive, symmetric, and transitive), we can conclude that R is an equivalence relation.
Therefore, the correct answer is option 'C' - an equivalence relation.
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Let T be the set of all triangles in a plane with R a relation in T given byR={(T1,T2) :T1is congruent toT2}. Then R isa)non commutative relationb)a universal relationc)an equivalence relationd)an empty relationCorrect answer is option 'C'. Can you explain this answer?
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