Say R1 is defined over {a}, R2 is defined over {b}

R1 = {(a,a)} -- Equivalence relation

R2 = {(b,b)} -- equivalence relation

R1 $\cap$ R2 = $\phi$ .... (not reflexive)

Where am I going wrong?

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+19 votes

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

- $R_1 \cup R_2$ is an equivalence relation
- $R_1 \cap R_2$ is an equivalence relation

Which of the following is correct?

- Both assertions are true
- Assertions (i) is true but assertions (ii) is not true
- Assertions (ii) is true but assertions (i) is not true
- Neither (i) nor (ii) is true

0

I am not getting why R1 $\cap$ R2 is Equivalence relation

Say R1 is defined over {a}, R2 is defined over {b}

R1 = {(a,a)} -- Equivalence relation

R2 = {(b,b)} -- equivalence relation

R1 $\cap$ R2 = $\phi$ .... (not reflexive)

Where am I going wrong?

Say R1 is defined over {a}, R2 is defined over {b}

R1 = {(a,a)} -- Equivalence relation

R2 = {(b,b)} -- equivalence relation

R1 $\cap$ R2 = $\phi$ .... (not reflexive)

Where am I going wrong?

0

they are defined on the same set @bikash what does it mean .We can take any subset of A right to form realtions

0

we cant take R1 defined over {a} and r2 defined over {b} ........ They should be defined over same set of elements.

In my example I took R1 over {a} and R2 over {b}

if we take R1 and R2 defined over {a,b} ...... in my example R1 = {(a,a)} cant be reflexive

In my example I took R1 over {a} and R2 over {b}

if we take R1 and R2 defined over {a,b} ...... in my example R1 = {(a,a)} cant be reflexive

0

M saying the same thing .in question it is not mention that we have to take same domain for both R1 and R2 .According to that b option cant be true

0

"Let R1 and R2 be equivalence relations defined on a set" ..... we need to consider 2 relations defined on a single set

0

yes m doing the same

lets A={1,2,3} we have 8 subsets right.

Assuming R1={(1,1).(2,2),(1,2)(2,1)} is an equivalennce realtion

R2=phi

both R1 and R2 are defined on same set A.

Could you pls look into this .Where I m wrong

lets A={1,2,3} we have 8 subsets right.

Assuming R1={(1,1).(2,2),(1,2)(2,1)} is an equivalennce realtion

R2=phi

both R1 and R2 are defined on same set A.

Could you pls look into this .Where I m wrong

+22 votes

Best answer

0

I am getting answer as option A.Can you please give R1,R2 where R1 union R2 not an equivalence relation?

+6

@yankur9 @Shukrayani intersection won't be null in this case. this property is defined for equivalence relations **over the same set** and that is specified in the question. at least all reflexive pairs will be present in the intersection.

+13 votes

$R1$ 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**

+3

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.

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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