Let R and S be two non-void relations on a se | vedclass

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(A) To determine which statement is false,we analyze each property:$(a)$ Let $A = \{1, 2, 3\}$. Define $R = \{(1, 2)\}$ and $S = \{(2, 3)\}$. Both $R$

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$R$ and $S$ are transitive $\implies$ $R \cup S$ is transitive

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(A) To determine which statement is false,we analyze each property:$(a)$ Let $A = \{1, 2, 3\}$. Define $R = \{(1, 2)\}$ and $S = \{(2, 3)\}$. Both $R$ and $S$ are vacuously transitive. However,$R \cup S = \{(1, 2), (2, 3)\}$. Since $(1, 2) \in R \cup S$ and $(2, 3) \in R \cup S$,but $(1, 3) 
otin R \cup S$,the union $R \cup S$ is not transitive. Thus,statement $(a)$ is false.$(b)$ If $R$ and $S$ are transitive,then for any $(x, y) \in R \cap S$ and $(y, z) \in R \cap S$,we have $(x, y) \in R, (y, z) \in R \implies (x, z) \in R$ and $(x, y) \in S, (y, z) \in S \implies (x, z) \in S$. Thus $(x, z) \in R \cap S$. This is true.$(c)$ If $R$ and $S$ are symmetric,then $(x, y) \in R \cup S \implies (x, y) \in R$ or $(x, y) \in S \implies (y, x) \in R$ or $(y, x) \in S \implies (y, x) \in R \cup S$. This is true.$(d)$ If $R$ and $S$ are reflexive,then for all $x \in A$,$(x, x) \in R$ and $(x, x) \in S$,so $(x, x) \in R \cap S$. This is true.Therefore,the false statement is $(a)$.

Let $R$ and $S$ be two non-void relations on a set $A$. Which of the following statements is false?

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