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Q6. Let R and S be relations on a set A. Assuming A has at least three elements, state whether each of the following statements is true or false. If it is false, give a counterexample on the set A = {1,2,3}.

If R and S are symmetric then R ∩ S is symmetric. If R and S are symmetric the R ∪ S is symmetric. If R and S are reflexive then R ∩ S is reflexive. If R and S are reflexive then R ∪ S is reflexive. If R and S are transitive then R ∪ S is transitive. If R is reflexive then R ∩ R-1 is not empty. If R is symmetric then R ∩ R-1 is not empty.

How to check whether is true or false? and what does it mean by give a counterexample? Please help. Thank you.

veeyikpong
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1 Answers1

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Suppose $R$ and $S$ are symmetric, and let $(a,b) \in R \cap S$. Then because $(a,b) \in R$ and $R$ is symmetric, $(b,a) \in R$. Similarly, $(b,a) \in S$, and so $(b,a) \in R \cap S$. Since we started with the assumption that $(a,b) \in R \cap S$ and discovered that $(b,a) \in R \cap S$, it follows that $R \cap S$ must be symmetric.

You should be able to apply nearly identical reasoning to the case of reflexivity as well. Let me know if this helps you to get the idea to solve the others.

Rookatu
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  • can u check for a false one and provide with an example please? – veeyikpong May 06 '13 at 06:53
  • Let $A = { 1,2,3 }$ and let $R = {(1,2),(2,1),(1,1) }$, $S = {(2,1),(2,3) }$. Then both $R$ and $S$ are transitive. However, $R \cup S$ contains $(1,2)$ and $(2,3)$ but not $(1,3)$, and therefore is not transitive. – Rookatu May 06 '13 at 07:50