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If $R_{1}$ and $R_{2}$ are equivalence relations in a set $A$, show that $R_{1} \cap R_{2}$ is also an equivalence relation.
Solution
since $R _{1}$ and $R _{2}$ are equivalence relations, $(a, a) \in R _{1},$ and $(a, a) \in R _{2}$ $ \forall a \in A$ This implies that $(a, a) \in R _{1} \cap R _{2}, \forall a,$ showing $R _{1} \cap R _{2}$ is reflexive. Further, $(a, b) \in R _{1} \cap R _{2} \Rightarrow(a, b) \in R _{1}$ and $(a, b) \in R _{2} \Rightarrow(b, a) \in R _{1}$ and $(b, a) \in R _{2} \Rightarrow$ $(b, a) \in R_1 \cap R_2$ hence, $R _{1} \cap R _{2}$ is symmetric. Similarly, $(a, b) \in R _{1} \cap R _{2}$ and $(b, c) \in R _{1} \cap R _{2} \Rightarrow(a, c) \in R _{1}$ and $(a, c) \in R _{2} \Rightarrow(a, c) \in R _{1} \cap $ $R _{2} .$ This shows that $R _{1} \cap $ $ R _{2}$ is transitive. Thus, $R _{1} \cap $ $R _{2}$ is an equivalence relation.
