GATE CSE 1998 | Question: 1.7

Question

Let $R_1$ and $R_2$ be two equivalence relations on a set. Consider the following assertions:

1. $R_1 \cup R_2$ is an equivalence relation
2. $R_1 \cap R_2$ is an equivalence relation

Which of the following is correct?

• A. Both assertions are true
• B. Assertions (i) is true but assertions (ii) is not true
• C. Assertions (ii) is true but assertions (i) is not true
• D. Neither (i) nor (ii) is true

Comments

Question no 4 in this link

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Question : 

Let R1 and R2 be two equivalence relations on a set. Consider the following assertions:
i. R 1 ∪ R 2 is an equivalence relation
ii. R 1 ∩ R 2 is an equivalence relation

 It means they can’t be null because in the worst case it is satisfying reflexive.

E.g; If R contains {a,b,c} then R1 and R2 both contains at least {(a,a),(b,b),(c,c)}.  So it is symmetric and transitive too, consequently an equivalence relation.

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R1 U R2 is reflexive and symmetric but may not be transitive. We can prove it-- suppose (a,b)∈R1 U R2

then (a,b)∈R1 or (a,b)∈R2.

Similarly suppose (b,c)∈R1 U R2

then (b,c)∈R1 or (b,c)∈R2.

So it is possible that (a,b)∈R1 and (b,c)∈R2.

So a and c are not related and (a,c) does not belongs to R1 U R2 . So it is not transitive.

Similarly we can prove that R1∩R2 is an equivalence relation

Answer 1

Answer: $C$

$R_1$ intersection $R_2$ is equivalence relation..
$R_1$ union $R_2$ is not equivalence relation because transitivity needn't hold. For example, $(a, b)$ can be in $R_1$ and $(b, c)$ be in $R_2$ and $(a, c)$ not in either $R_1$ or $R_2.$

Comments

yes if there is empty set  then also it follows transitive same in case of symmetry

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But empty set does not satisfy reflexive property

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But empty set does not satisfy reflexive property only satisfy transitive and symmetric

Answer 2

$R1$ and $R2$ both are equivalence relation so $R1∩R2$ is also an equivalence relation because $\cap$ include only those pairs which are in both $R1$ and $ R2$.
Assertions (ii) is true 

$R1∪R2$ is NOT an equivalence relation 
see counter example over $(a,b)$ : 
$R1=\{(a,a),(b,b),(a,b),(b,a)\}$ is equivalence relation
$R2=\{(a,a),(b,b),(c,b),(b,c)\}$ is equivalence relation
$R1∪R2=\{(a,a),(b,b),(a,b),(b,a),(c,b),(b,c)\}$ is NOT equivalence relation because transitive pair $(a,c)$ isn't include in it .
assertions (i) is not true

Ans is C
 

Comments

when R2 is defined over (a,b) how have you taken c ?

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Let relation is defined on set s={a,b,c}

Let R1={(a,a),(b,b),(c,c),(a,b),(b,a)}

And let. R2={(a,a),(b,b),(c,c),(b,c),(c,b)}

Here both are equivalence relation

But R1UR2={(a,a),(b,b),(c,c),(a,b),(b,a),(b,c),(c,a)}

Now here it can be seen that (a,c) is not pre sent in R1UR2 as it should be to make it transitive since (a,b) and (b,c) is present.

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R2 must have (c,c) to be equivalence relation

Answer 3

learn this property and solve with in second

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