Let R1 and R2 be equivalence relations on a set A- then R1- R2 may or may not be-

The correct option is C transitiveIf R1 and R2 are transitive on a set A, then R1 #8746;R2 may or may not be transitive. As an example, con ...

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Standard XII

Mathematics

Definition of Relations

Let R1 and ...

Question

Let $R_{1}$ and $R_{2}$ be equivalence relations on a set A, then $R_{1} \cup R_{2}$ may or may not be:

reflexive

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symmetric

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transitive

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anti-symmetric

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Solution

The correct option is C transitive
If $R 1$ and $R 2$ are transitive on a set A, then $R 1 \cup R 2$ may or may not be transitive. As an example, consider the set $A = 1, 2, 3$
Let $R 1 = (1, 1), (2, 2), (3, 3), (1, 2), (2, 1)$
and $R 2 = (1, 1), (2, 2), (3, 3), (1, 3), (3, 1)$ then the $R 1$ and $R 2$ are equivalence relations on A. However,
$R 1 \cup R 2 = (1, 1), (2, 2), (3, 3), (1, 2), (2, 1) (1, 3), (3, 1)$ is not transitive as $(2, 1)$ and $(1, 3) \in R 1 \cup R 2$ but $(2, 3) \notin R 1 \cup R 2$

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Similar questions

Q. Let

R_{1}

and

R_{2}

be two relations defined on a non empty set A. Which of the following statements is false? Give reasons in support of your answer.
(a) If

R_{1}

and

R_{2}

are reflexive, then so is

R_{1} \cap R_{2}

(b) If

R_{1}

and

R_{2}

are symmetric, then so is

R_{1} \cup R_{2}

R_{1}

and

R_{2}

are transitive, then so is

R_{1} \cup R_{2}

(d) If

R_{1}

and

R_{2}

are transitive, then so is

R_{1} \cap R_{2}

(e) If

R_{1}

and

R_{2}

are symmetric, then so is

R_{1} \cap R_{2}

Q. Let

R_{1}

and

R_{2}

be two equivalence relations on a set. Consider the following assertions:
(i)

R_{1} \cup R_{2}

is an equivalence relation
(ii)

R_{1} \cap R_{2}

is an equivalence relation
Which of the following is correct ?

Q. Let A = {1, 2, 3}, and let R₁ = {(1, 1), (1, 3), (3, 1), (2, 2), (2, 1), (3, 3)}, R₂ = {(2, 2), (3, 1), (1, 3)}, R₃ = {(1, 3), (3, 3)}. Find whether or not each of the relations R₁, R₂, R₃ on A is (i) reflexive (ii) symmetric (iii) transitive.

Q. Let

R_{1}, R_{2}

are relation defined on

Z

such that

a R_{1} b ⟺ (a - b)

is divisible by

3

and

a R_{2} b ⟺ (a - b)

is divisible by

4

. Then which of the two relation

(R_{1} \cup R_{2}), (R_{1} \cap R_{2})

is an equivalence relation?

Q. Let

R_{1}

and

R_{2}

be two relations defined as follows:

R_{1} = {(a, b) \in R^{2} : a^{2} + b^{2} \in Q}

and

R_{2} = {(a, b) \in R^{2} : a^{2} + b^{2} \notin Q} .

Then

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Let R1 and R2 be equivalence relations on a set A, then R1∪R2 may or may not be:

Let $R_{1}$ and $R_{2}$ be equivalence relations on a set A, then $R_{1} \cup R_{2}$ may or may not be: