2 votes

A relation $R$ is said to be circular if $aRb$ and $bRc$ together imply $cRa$.

Which of the following options is/are correct?

- If a relation $S$ is reflexive and symmetric, then $S$ is an equivalence relation.
- If a relation $S$ is circular and symmetric, then $S$ is an equivalence relation.
- If a relation $S$ is reflexive and circular, then $S$ is an equivalence relation.
- If a relation $S$ is transitive and circular, then $S$ is an equivalence relation.

2 votes

let S be an empty relation

empty relation is all **symmetric, transitive and circular **(as all these three are conditional)

but it is not reflexive.

**equivalence relation : reflexive, symmetric and transitive**

S is both circular and symmetric but not reflexive, hence B is false

S is both transitive and circular but not reflexive, hence D is false

option A clearly don’t follow the definition of equivalence relation,

so only option left is C

and **C indeed is the correct ans. proved below**

**Let S be both reflexive and circular,**

**case 1: x only have diagonal elements ( $aSa$ exists for all a)**

then S is all reflexive, symmetric, transitive and circular **(hence equivalence )**

**case 2: let for some 3 different elements a,b,c $aSb$ exists but ** **$bSc$ doesn’t exists**

both $aSa, bSb$ are also exists (it is given reflexive)

$(aSa\ and\ aSb) \rightarrow bSa $ (from circular property),

so, $aSb \rightarrow bSa $, hence it is symmetric

and transitive as well (because i assumed if $aSb$ exists then no $bSc$ exists for any three diffrent elements a,b,c)

hence this case will also be equivalence

**case 3: let for some 3 different elements a,b,c both $aSb,\ bSc$ exists**

$aSa, bSb, cSc$ (exists from reflexive property)

$bSb\ and\ bSc \rightarrow cSb$ (from circular property) **(Hence it is symmetric)**

$aSb\ and\ bSc\rightarrow cSa$ (from circular property)

$cSa \rightarrow aSc$ (from symmetric property) (already proved symmetric 2 steps above)

so, we can conclude,

$aSb\ and\ bSc\rightarrow aSc$ ** (hence it is transitive as well)**

as we just proved it as both symmetric and transitive, it is definitely equivalence relation

In all three cases we proved that if S is both reflexive and circular then it an is equivalence relation.

**C is correct ans.**

1 vote

**Only C the correct answer.**

- Option A is not correct because a relation is equivalence iff it is reflexive, symmetric and transitive.
- Consider the relation {(2,3),(3,2)} on set {1,2,3} as a counter solution to option B.
**Option C is correct.**- Consider the relation {(2,2)(2,3)(3,2)} on set {1,2,3} as counter solution to option D.

1 vote

**If a relation R is reflexive and circular then it is symmetric : True**

Proof :

Assume aRb. Then since R is reflexive, we have bRb. Since R is circular, so, aRb, bRb will mean that we have bRa, so, R is symmetric.

**If R is reflexive and circular then it is transitive : True **

Proof :

Assume aRb, bRc. Since R is circular, so, cRa, and since R is symmetric(we proved above) so aRc so R is transitive.

**So, option C is correct. **

Option B is false. For counter example, take a set A = {a,b,c}, define relation R on A as follows : R = { (a,a) }, R is symmetric and circular but not equivalence relation.

Option A is false. For counter example, take a set A = {a,b,c}, define relation R on A as follows : R = { (a,a), (b,b),(c,c), (a,b),(b,a), (a,c),(c,a) }, R is symmetric and reflexive but not transitive so not equivalence relation.

Option D is false. For counter example, take a set A = {a,b,c}, define relation R on A as follows : R = { (a,a) }, R is transitive and circular but not equivalence relation.

**Some more variations :**

1.

Converse of Statement in option C is also true. i.e.

**Theorem : If R is an equivalence relation then R is reflexive and circular.**

Proof :

Reflexive: As, the relation, R is an equivalence relation. So, reflexivity is the property of an equivalence relation. Hence, R is reflexive.

Circular: Let (a, b) ∈ R and (b, c) ∈ R

⇒ (a, c) ∈ R (∵ R is transitive)

⇒ (c, a) ∈ R (∵ R is symmetric)

Thus, R is Circular.

So, we can say that

**“A relation S is reflexive and circular if and only if S is an equivalence relation.”**

2.

If a relation R is transitive and circular then it is symmetric : False.

If a relation R is transitive and circular then it is reflexive : False.

Counter example(for both above statements) : R = { (a,b) }

PS : Similarly you can try to prove or disprove more similar statements and their converses.

0 votes

- Option A is not correct as a relation has to be reflexive, symmetric and transitive for being equivalence. ( by definition )
- Option B : a relation is symmetric and cyclic implies that it is transitive . But reflexivity is not implied there. In case there is an isolated node in the relation, then a loop to itself is not guaranteed. So reflexivity not guaranteed. Hence, the relation need not be equivalence.
- Option C: a relation is reflexive and circular implies symmetricly and transitivity. Hence it is correct.
- Option D: a circular and transitive relation implies symmetricly but nor reflexivity for the reason stated in point 2.

**So option C is the right answer.**

We can verify this by defining the above relations on a set of four elements and keeping one node isolated.