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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?

  1. Both assertions are true
  2. Assertions (i) is true but assertions (ii) is not true
  3. Assertions (ii) is true but assertions (i) is not true
  4. Neither (i) nor (ii) is true
in Set Theory & Algebra by Veteran (52.2k points) | 2.6k views
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?
+1
they are defined on the same set
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
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
0
In your example R1 is not reflexive ...... missing (3,3) element

R2 is also not reflexive

They both had to be equivalence relations defined on the set .....

"Let R1 and R2 be two equivalence relations on a set"
0
oh got my mistake thnks  a lot :)

3 Answers

+21 votes
Best answer
Answer: $C$

$R1$ intersection $R2$ is equivalence relation..
$R1$ union $R2$ is not equivalence relation because transitivity needn't hold. For example, $(a, b)$ can be in $R1$ and $(b, c)$ be in $R2$ and $(a, c)$ not in either $R1$ or $R2.$
by Veteran (60.5k points)
edited by
0
I am getting answer as option A.Can you please give R1,R2 where R1 union R2 not an equivalence relation?
+6
For example, (a, b) can be in R1 and (b, c) be in R2 and (a, c) not in either R1 or R2.
0
But sir what if nothing is present in the intersection??
0
if R1 intersection R2= fi

??
+5

@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. 

+9 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
 

by Active (2.7k points)
edited by
0
when R2 is defined over (a,b) how have you taken 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.
+1
R2 must have (c,c) to be equivalence relation
–2 votes
Ans: C
by Loyal (7.2k points)
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