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A relation $R$ is said to be circular if $a\text{R}b$ and $b\text{R}c$ together imply $c\text{R}a$.

Which of the following options is/are correct?

  1. If a relation $S$ is reflexive and symmetric, then $S$ is an equivalence relation.
  2. If a relation $S$ is circular and symmetric, then $S$ is an equivalence relation.
  3. If a relation $S$ is reflexive and circular, then $S$ is an equivalence relation.
  4. If a relation $S$ is transitive and circular, then $S$ is an equivalence relation.
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Option (A) If a relation $S$ is reflexive and symmetric, then $S$ is an equivalence relation.

For a relation to be equivalence relation, it should be reflexive, symmetric and transitive. So for $S$ to be equivalence relation , we need to show that :

If a relation is reflexive and symmetric then it is also transitive.

But if the see the example below:

It is reflexive and symmetric. But it is not transitive [There is $(2,1)$ and $(1,3)$ but $(2,3)$ is not there.] So from the above counter-example we see that: option $A$ is wrong.


option (B) : If a relation $S$ is circular and symmetric, then $S$ is an equivalence relation.

So for $S$ to be equivalence relation , we need to show that :

If a relation is symmetric and circular then it is reflexive and transitive.

But the empty relation $\{\}$ is both symmetric and circular.

Definition of symmetric relation is $aRb \implies bRa$, but since the relation has no such $aRb$ so the relation is symmetric.

Definition of circular relation is $aRb \land bRc\implies cRa$, but since the relation has no such $aRb$ pr $bRc$ so the relation is circular.

But the empty relation is not reflexive, since it does not have $aSa$ $\forall a \in A$, where $A$ is the set on which the relation $S$ is defined.

So option $B$ is wrong .


option D: If a relation $S$ is transitive and circular, then $S$ is an equivalence relation.

So for $S$ to be equivalence relation , we need to show that :

If a relation is transitive and circular then it is reflexive and symmetric.

Now the empty relation $\{\}$ is both transitive and circular.

Definition of transitive relation is $aRb \land bRc \implies aRc$, but since the relation has no such $aRb$ or $bRc$ so the relation is transitive.

Definition of circular relation is $aRb \land bRc\implies cRa$, but since the relation has no such $aRb$ pr $bRc$ so the relation is circular.

But the empty relation is not reflexive, since it does not have $aSa$ $\forall a \in A$, where $A$ is the set on which the relation $S$ is defined.

So option $D$ is wrong .


Inspite of being an MSQ, since 3 options are eliminated and at least one option has to be true, so option $C$ is the correct answer. But let us see why.

option (C): If a relation $S$ is reflexive and circular, then $S$ is an equivalence relation.

So for $S$ to be equivalence relation , we need to show that :

If a relation is reflexive and circular then it is transitive and symmetric.

Since $S$ is reflexive we have tuples of the form $(a,a)$ in $S$  $\forall a\in A$ where $A$ is the set on which $S$ is defined.

Now suppose $(p,q) \in S$, We also have $(q,q)$ as $S$ is symmetric. But $S$ is circular. So $(p,q)\land (q,q) \implies (q,p)$, due to the circular property. So $S$ is also symmetric as $(p,q)\in S \implies (q,p)\in S$

Now suppose $(p,q), (q,r) \in S$. Since $S$ is symmetric $(q,p), (r,q) \in S$. But $S$ is circular so $(r,q),(q,p) \in S \implies (p,r) \in S$. So $S$ is transitive, since  $(p,q), (q,r) \in S \implies (p,r) \in S$

So $S$ is equivalence relation.

Answer:

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