If R and S are relations on a set A, then prove that

Solution:

(i) R and S are symmetric relations on the set A.

$\Rightarrow R \subset A \times A$ and $S \subset A \times A$

$\Rightarrow R \cap S \subset A \times A$

Thus, $R \cap S$ is a relation on $A$.

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 $(b, a) \in S$                                     [Since $R$ 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 $a, b \in A$

So, $R \cap S$ is symmetric on A.

Also,

Let $a, b \in A$ such that $(a, b) \in R \cup S$

$\Rightarrow(a, b) \in R$ or $(a, b) \in S$

$\Rightarrow(b, a) \in R$ or $(b, a) \in S$                                         [Since $R$ and $S$ are symmetric]

$\Rightarrow(b, a) \in R \cup S$

So, $R \cup S$ is symmetric on $A$.

(ii) R is reflexive and S is any relation.

Suppose $a \in A$. Then,

$(a, a) \in R$                                            [Since $R$ is reflexive]

$\Rightarrow(a, a) \in R \cup S$

$\Rightarrow R \cup S$ is reflexive on $A$