Question

A relation $R$ is said to be circular if $aRb$ and $bRc$ together imply $cRa$.
Which of the following options is/are correct?

• A. If a relation $S$ is reflexive and circular, then $S$ is an equivalence relation.
• B. If a relation $S$ is circular and symmetric, then $S$ is an equivalence relation.
• C. If a relation $S$ is transitive and circular, then $S$ is an equivalence relation.
• D. If a relation $S$ is reflexive and symmetric, then $S$ is an equivalence relation.

Answer 1

The correct option is A If a relation $S$ is reflexive and circular, then $S$ is an equivalence relation.

Let $S$ be reflexive and circular. Let us checking symmetry:
Symmetry:
Let $xSy$
Now since $S$ is reflexive $ySy$ true.
So $xSy$ and $ySy$ is true
Now by circular property we get, $ySx$
So $xSy\Rightarrow ySx$
So $S$ is symmetric.

Transitive:
Let $xSy$ and $ySz$
Now by circular property we get $zSx$ and by symmetry property proved above, we get
$zSx\Rightarrow xSz$
So $xSy$ and $ySz\Rightarrow xSz$
So $S$ is transitive.
So $S$ is reflexive, symmetric and transitive and hence an equivalence relation.
So option (a) is true.

Option (b): Let $S$ be circular and symmetric.
Let $S$ be defined on set $\{1,2,3\}$
Now empty relation is circular and symmetric but not reflexive. So $S$ need not be an equivalence relation.
So option (b) is false.

Option (c): Let $S$ be transitive and circular.
Let $S$ be defined on the set $\{1,2,3\}$.
Now empty relation again satisfies transitive and circular but is not reflexive. So $S$ need not be an equivalence relation.
So option (c) is false.

Option (d): Reflexive and symmetric need not be transitive for example on $\{1,2,3\}$.
$S=\left\{\right(1,1),(2,2),(3,3),(1,2),(2,1),(2,3),(3,2\left)\right\}$ is reflexive and symmetric. But it is not transitive because $(1,2)$ and $(2,3)$ belong to $S$ but $(1,3)$ does not.
So option (d) is false.