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

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(D) $1$. Symmetric: If $R$ and $S$ are symmetric,then $(x, y) \in R \implies (y, x) \in R$ and $(x, y) \in S \implies (y, x) \in S$. If $(x, y) \in R

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All $(a)$,$(b)$,and $(c)$ are true.

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(D) $1$. Symmetric: If $R$ and $S$ are symmetric,then $(x, y) \in R \implies (y, x) \in R$ and $(x, y) \in S \implies (y, x) \in S$. If $(x, y) \in R \cup S$,then $(x, y) \in R$ or $(x, y) \in S$. Thus $(y, x) \in R$ or $(y, x) \in S$,which means $(y, x) \in R \cup S$. So $R \cup S$ is symmetric.$2$. Transitive: If $R$ and $S$ are transitive,let $(x, y) \in R \cap S$ and $(y, z) \in R \cap S$. Then $(x, y) \in R, (y, z) \in R \implies (x, z) \in R$ (since $R$ is transitive). Similarly,$(x, y) \in S, (y, z) \in S \implies (x, z) \in S$ (since $S$ is transitive). Thus $(x, z) \in R \cap S$. So $R \cap S$ is transitive.$3$. Reflexive: If $R$ and $S$ are reflexive on $A$,then for all $a \in A, (a, a) \in R$ and $(a, a) \in S$. Thus $(a, a) \in R \cap S$ for all $a \in A$. So $R \cap S$ is reflexive.Therefore,all statements are true.

Let $R$ and $S$ be two relations on a set $A$. Then which of the following is true?

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