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Let R and S be two equivalence relations on a set A.
Then,
Then,
- a)R ∪ S is an equivalence relation on A
- b)R ∩ S in an equivalence relation on A
- c)R - S is an equivalence relation on A
- d)None of the above
Correct answer is option 'B'. Can you explain this answer?
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Verified Answer
Let R and S be two equivalence relations on a set A.Then,a)R ∪ S i...
Given, R and S are relations on set A.
Then R ε A x A and S ≤ A x A
=> R ∩ C ≤ A x A
implies R ∩ S is also a relation on A.
Reflexivity: Let a be an arbitrary element of A. Then, a
A implies (a, a ) 
R and (a, a) 
S,
[Since, R and S are reflexive]
implies(a, a) ε R ∩ S
Thus, (a , a ) ε R ∩ S for all a ε A.
So, R ∩ S is a reflexive relation on A.
Symmetry: Let a, b ε A such that (a, b) ε R ∩ S.
Then, (a, b) ε R ∩ S
implies (a, b) ε R and (a, b) ε S
implies (b, a) ε R and (b , a ) ε S
[Since R and S are symmetric]
implies (b, a) ε R ∩ S
Thus, (a, b) ε R ∩ S implies
(b, a) ε R ∩ S for all (a, b ) ε R ∩ S .
So, R ∩ S is symmetric on A.
So, R ∩ S is transitive on A.
Hence, R is an equivalence relation on A.
Then R ε A x A and S ≤ A x A
=> R ∩ C ≤ A x A
implies R ∩ S is also a relation on A.
Reflexivity: Let a be an arbitrary element of A. Then, a
A implies (a, a ) R and (a, a) S,[Since, R and S are reflexive]
implies(a, a) ε R ∩ S
Thus, (a , a ) ε R ∩ S for all a ε A.
So, R ∩ S is a reflexive relation on A.
Symmetry: Let a, b ε A such that (a, b) ε R ∩ S.
Then, (a, b) ε R ∩ S
implies (a, b) ε R and (a, b) ε S
implies (b, a) ε R and (b , a ) ε S
[Since R and S are symmetric]
implies (b, a) ε R ∩ S
Thus, (a, b) ε R ∩ S implies
(b, a) ε R ∩ S for all (a, b ) ε R ∩ S .
So, R ∩ S is symmetric on A.
So, R ∩ S is transitive on A.
Hence, R is an equivalence relation on A.
Most Upvoted Answer
Let R and S be two equivalence relations on a set A.Then,a)R ∪ S i...
Let R and S be two equivalence relations on a set A.
a) R ∩ S is an equivalence relation on A:
To show that R ∩ S is an equivalence relation, we need to prove three properties: reflexivity, symmetry, and transitivity.
1) Reflexivity: For any element a ∈ A, (a, a) must be in R ∩ S. Since R and S are both equivalence relations, (a, a) ∈ R and (a, a) ∈ S. Therefore, (a, a) ∈ R ∩ S, and R ∩ S is reflexive.
2) Symmetry: For any elements a, b ∈ A, if (a, b) ∈ R ∩ S, then (a, b) ∈ R and (a, b) ∈ S. Since R and S are both equivalence relations, (b, a) ∈ R and (b, a) ∈ S. Therefore, (b, a) ∈ R ∩ S, and R ∩ S is symmetric.
3) Transitivity: For any elements a, b, c ∈ A, if (a, b) ∈ R ∩ S and (b, c) ∈ R ∩ S, then (a, b) ∈ R and (a, b) ∈ S, and (b, c) ∈ R and (b, c) ∈ S. Since R and S are both equivalence relations, (a, c) ∈ R and (a, c) ∈ S. Therefore, (a, c) ∈ R ∩ S, and R ∩ S is transitive.
Since R ∩ S is reflexive, symmetric, and transitive, it is an equivalence relation on A.
b) R ∪ S is not necessarily an equivalence relation on A:
To show that R ∪ S is not necessarily an equivalence relation, we need to find a counterexample.
Let's consider a set A = {1, 2, 3}, and two equivalence relations R = {(1, 1), (2, 2), (3, 3)} and S = {(2, 2), (3, 3)}. The union of R and S is R ∪ S = {(1, 1), (2, 2), (3, 3)}.
Although R ∪ S is reflexive and symmetric, it is not transitive. For example, (1, 1) ∈ R ∪ S and (1, 1) ∈ R ∪ S, but (1, 1) ∉ R ∪ S. Therefore, R ∪ S is not an equivalence relation on A.
In conclusion, R ∩ S is an equivalence relation on A, while R ∪ S is not necessarily an equivalence relation on A.
a) R ∩ S is an equivalence relation on A:
To show that R ∩ S is an equivalence relation, we need to prove three properties: reflexivity, symmetry, and transitivity.
1) Reflexivity: For any element a ∈ A, (a, a) must be in R ∩ S. Since R and S are both equivalence relations, (a, a) ∈ R and (a, a) ∈ S. Therefore, (a, a) ∈ R ∩ S, and R ∩ S is reflexive.
2) Symmetry: For any elements a, b ∈ A, if (a, b) ∈ R ∩ S, then (a, b) ∈ R and (a, b) ∈ S. Since R and S are both equivalence relations, (b, a) ∈ R and (b, a) ∈ S. Therefore, (b, a) ∈ R ∩ S, and R ∩ S is symmetric.
3) Transitivity: For any elements a, b, c ∈ A, if (a, b) ∈ R ∩ S and (b, c) ∈ R ∩ S, then (a, b) ∈ R and (a, b) ∈ S, and (b, c) ∈ R and (b, c) ∈ S. Since R and S are both equivalence relations, (a, c) ∈ R and (a, c) ∈ S. Therefore, (a, c) ∈ R ∩ S, and R ∩ S is transitive.
Since R ∩ S is reflexive, symmetric, and transitive, it is an equivalence relation on A.
b) R ∪ S is not necessarily an equivalence relation on A:
To show that R ∪ S is not necessarily an equivalence relation, we need to find a counterexample.
Let's consider a set A = {1, 2, 3}, and two equivalence relations R = {(1, 1), (2, 2), (3, 3)} and S = {(2, 2), (3, 3)}. The union of R and S is R ∪ S = {(1, 1), (2, 2), (3, 3)}.
Although R ∪ S is reflexive and symmetric, it is not transitive. For example, (1, 1) ∈ R ∪ S and (1, 1) ∈ R ∪ S, but (1, 1) ∉ R ∪ S. Therefore, R ∪ S is not an equivalence relation on A.
In conclusion, R ∩ S is an equivalence relation on A, while R ∪ S is not necessarily an equivalence relation on A.
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Let R and S be two equivalence relations on a set A.Then,a)R ∪ S is an equivalence relation on Ab)R ∩ S in an equivalence relation on Ac)R - S is an equivalence relation on Ad)None of the aboveCorrect answer is option 'B'. Can you explain this answer?
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