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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
asked in Set Theory & Algebra by Veteran (59.6k points) | 1.8k views

3 Answers

+17 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 $R2.$
answered by Veteran (55.4k 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?
+2
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

??
+2

@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. 

+3 votes

R1 and R2 both are equivalence realtion so R1∩R2 is also an equivalence relation becoz ∩ 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 realtion
R2={(a,a),(b,b),(c,b),(b,c)} is equivalence realtion
R1∪R2={(a,a),(b,b),(a,b),(b,a),(c,b),(b,c)} is NOT equivalence realtion becoz transitive pair (a,c) isnt include in it .
assertions (i) is not true

Ans is C
 

answered by Active (2.4k points)
0
when R2 is defined over (a,b) how have you taken c ?
0
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 vote
Ans: C
answered by Loyal (7.4k points)


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