First slide

Relations XII

Question

Let R and S be two equivalence relations on a set A. Then

Moderate

A

$R \cup S$ is an equivalence relation on A
B

$R \cap S$ is an equivalence relation on A
C

$R - S$ is an equivalence relation on A
D

None of these

Solution

Given, R and S are relations on set A.
$∴ R \subseteq A \times A$ and $S \subseteq A \times A \Rightarrow R \cap C \subseteq A \times A$
$\Rightarrow R \cap S$ is also a relation on A.
Reflexivity: Let a be an arbitrary element of A. Then,
$a \in A \Rightarrow (a, a) \in R and (a, a) \in S,$
$[∵ R and S are reflexive]$
$\Rightarrow (a, a) \in R \cap S$
Thus, (a, a) $\in R \cap S$ for all $a \in A .$
So, $R \cap S$ is a reflexive relation on A.
Symmetry: Let $a, b \in A$ such that $(a, b) \in R \cap S .$
Then, $(a, b) \in R \cap S \Rightarrow (a, b) \in R and (a, b) \in S$
$\Rightarrow (b, a) \in R and (b, a) \in S,$
$[∵ R and S are symmetric]$
$\Rightarrow (b, a) \in R \cap S$
Thus, $(a, b) \in R \cap S$
$\Rightarrow (b, a) \in R \cap S$ for all $(a, b) \in R \cap S .$
So, $R \cap S$ is symmetric on A.
Transitivity: Let $a, b, c \in A$ such that $(a, b) \in R \cap S$
and $(b, c) \in R \cap S$ . Then, $(a, b) \in R \cap S$ and
$(b, c) \in R \cap S$
$\Rightarrow {((a, b) \in R and (a, b) \in S)}$
and $\{((b, c) \in R and (b, c) \in S)\}$
$\Rightarrow \{(a, b) \in R, (b, c) \in R\} and \{(a, b) \in S, (b, c) \in S\}$
$\Rightarrow (a, c) \in R and (a, c) \in S$
$∵ R and S are transitive So (a, b) \in R and (b, c) \in R \Rightarrow (a, c) \in R (a, b) \in S and (b, c) \in S \Rightarrow (a, c) \in S$
$\Rightarrow (a, c) \in R \cap S$
Thus, $(a, b) \in R \cap S and (b, c) \in R \cap S \Rightarrow (a, c) \in R \cap S,$
So, $R \cap S$ is transitive on A.
Hence, R is an equivalence relation on A.

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