Let $$R$$ and $$S$$ be two non-void relations in a set $$A.$$ Which of the following statements is not true ?

Solution :
$$\eqalign{ & \left( {\bf{A}} \right){\text{Let }}\left( {a,\,b} \right),\,\left( {b,\,c} \right)\, \in \,R \cup S \cr & {\text{It is possible that }}\left( {a,\,b} \right)\, \in R - S{\text{ and }}\left( {b,\,c} \right)\, \in S - R \cr & {\text{In such case, we cannot say that }}\left( {a,\,c} \right)\, \in R\,\,{\text{or }}\left( {a,\,c} \right)\, \in S \cr & \therefore \,\left( {a,\,c} \right){\text{ may not be in }}R \cup S \cr & \therefore \,R \cup S{\text{ in not transitive}}{\text{.}} \cr & \left( {\bf{B}} \right){\text{Let }}\left( {a,\,b} \right),\,\left( {b,\,c} \right)\, \in \,R \cap S \cr & \therefore \,\left( {a,\,b} \right),\,\left( {b,\,c} \right)\, \in R{\text{ and }}\left( {a,\,b} \right),\,\left( {b,\,c} \right)\, \in S \cr & \therefore \,\left( {a,\,c} \right)\, \in R\,\,{\text{and }}\left( {a,\,c} \right)\, \in S \cr & \therefore \,\left( {a,\,c} \right){\text{ }} \in \,{\text{ }}R \cap S \cr & \therefore \,R \cap S{\text{ in transitive}}{\text{.}} \cr & \left( {\bf{C}} \right){\text{Let }}\left( {a,\,b} \right)\, \in \,R \cup S \cr & \therefore \,\left( {a,\,b} \right)\, \in R{\text{ or }}\left( {a,\,b} \right)\, \in S \cr & {\text{Now, }}\left( {a,\,b} \right)\, \in R \Rightarrow \left( {b,\,a} \right)\, \in R\,\,\,\left( {\because \,R{\text{ is symmetric}}} \right) \cr & \left( {a,\,b} \right)\, \in \,S \Rightarrow \left( {b,\,a} \right)\, \in \,S\,\,\,\left( {\because \,S{\text{ is symmetric}}} \right) \cr & \therefore \,\left( {b,\,a} \right)\, \in \,R \cup S \cr & \therefore \,R \cup S{\text{ is symmetric}}{\text{.}} \cr & \left( {\bf{D}} \right){\text{ Let }}a\, \in \,A \cr & \therefore \,\left( {a,\,a} \right)\, \in \,R\,{\text{and}}\left( {a,\,a} \right)\, \in S \cr & \therefore \,\left( {a,\,a} \right)\, \in \,R \cap S \cr & \therefore \,R \cap S{\text{ in reflexive}}{\text{.}} \cr} $$