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+11 votes
1.6k views

Let $R$ be a symmetric and transitive relation on a set $A$. Then

  1. $R$ is reflexive and hence an equivalence relation
  2. $R$ is reflexive and hence a partial order
  3. $R$ is reflexive and hence not an equivalence relation
  4. None of the above
asked in Set Theory & Algebra by Veteran (59.5k points) | 1.6k views

4 Answers

+12 votes
Best answer
Answer is $D$.

Let $A={1,2,3}$ and relation $R={(1,2),(2,1),(1,1),(2,2)}$. $R$ is symmetric and transitive but not reflexive.( $(3,3)$ is not there.)
answered by Loyal (5.9k points)
edited by
+7
Also, the Empty relation is symmetric and transitive by default but not reflexive.

Hence ans-(d)
+10 votes
Answer: D

Let A = {(1,2),(2,1),(1,1)}

A is symmetric and transitive but not reflexive as (2,2) is not there.
answered by Boss (34k points)
+3
According to the example you have assumed, thought the answer remains correct, but you must include (2,2) into the relation as well because of transitivity. and may be you can change the set A to , A={1,2,3}
0
simply, take (1,1) as relation  which is symmetric and transitive, but for reflexive it should have other pair ( b,b) &(c,c) if i consider set {1,2,3}

hence option D
0
The explanation is not right!
0
@rajshree you must include (2,2) in your explanation
+7 votes
here ans should be D

explanation:

here the relation is symmetric and transitive. if relation is symmetric and transitive then it need not necessariy be reflexive;i.e. it may or may not be reflexive. therefore ans is D
answered by Loyal (8.2k points)
+6 votes
We can take an empty set { } which is both symmetric and and transitive but not reflexive because diagonal elememts are not present in the set so not reflexive.
answered by Active (4.1k points)


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