So does **Circular** imply **Symmetric** and **Transitive**?

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15 votes

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?

- 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.

16 votes

Best answer

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 does not follow the definition of equivalence relation,

so only option left is C

and **C indeed is the correct answer as proved below**

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

**case 1: x has only diagonal elements ( **$a\text{S}a$** 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 **$a\text{S}b$ ** exists but ** $b\text{S}c$** does not exist**

Both $a\text{S}a, b\text{S}b$ also exist (it is given reflexive)

$(a\text{S}a$ and $a\text{S}b) \rightarrow b\text{S}a $ (from circular property),

so, $a\text{S}b \rightarrow b\text{S}a $, hence it is symmetric

And transitive as well (because i assumed if $a\text{S}b$ exists then no $b\text{S}c$ exists for any three different elements $a,b,c$)

Hence this case will also be equivalence

**case 3: let for some 3 different elements a,b,c both **$a\text{S}b,\ b\text{S}c$** exists**

$a\text{S}a, b\text{S}b, c\text{S}c$ (exists from reflexive property)

$b\text{S}b$ and $b\text{S}c \rightarrow c\text{S}b$ (from circular property) **(Hence it is symmetric)**

$a\text{S}b$ and $b\text{S}c\rightarrow c\text{S}a$ (from circular property)

$c\text{S}a \rightarrow a\text{S}c$ (from symmetric property) (already proved symmetric 2 steps above)

so, we can conclude,

$a\text{S}b$ and $b\text{S}c\rightarrow a\text{S}c$ ** (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.**

edited
Dec 10, 2021
by tusharb

For Option A, if we take B={1,2,3} and the relation = {(1,1) (2,2,) (3,3)}, then option A holds true. Because it is reflexive and symmetric and by default it becomes transitive, so the equivalence relation does hold. Aren’t I right?

Edit: Found the counter example : {(1,1) (2,2,) (3,3),(1,3),(3,1),(2,3)(3,2)} which is not transitive

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18 votes

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

Proof : Assume $a\text{R}b.$ Then since $R$ is reflexive, we have $b\text{R}b.$ Since $R$ is circular, so, $a\text{R}b, b\text{R}b$ will mean that we have $b\text{R}a.$ So, $R$ is symmetric.

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

Proof : Assume $a\text{R}b, b\text{R}c.$ Since $R$ is circular, so, $c\text{R}a,$ and since $R$ is symmetric(we proved above) so $a\text{R}c$ 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 $(a, b) \in R$ 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.

3. 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.

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.

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1 vote

- 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.

Sir in option A it is given the relation is reflexive and transitive,that doesn’t mean it is not transitive right?

For example let us take a simple set S={1,2,3}

the relation {(1,1),(2,2),(3,3)} As mentioned in the option it is both reflexive and symmetric and additionally it is also Transitive.Can’t we concluse that it equivalent relation?

For example let us take a simple set S={1,2,3}

the relation {(1,1),(2,2),(3,3)} As mentioned in the option it is both reflexive and symmetric and additionally it is also Transitive.Can’t we concluse that it equivalent relation?

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Sir in option A it is given the relation is reflexive and transitive,that doesn’t mean it is not transitive right?

For example let us take a simple set S={1,2,3}

the relation {(1,1),(2,2),(3,3)} As mentioned in the option it is both reflexive and symmetric and additionally it is also Transitive.Can’t we conclude that it equivalent relation?

For example let us take a simple set S={1,2,3}

the relation {(1,1),(2,2),(3,3)} As mentioned in the option it is both reflexive and symmetric and additionally it is also Transitive.Can’t we conclude that it equivalent relation?

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