If R and S be two non-void relations on ...

If R and S be two non-void relations on a set A, which of the following statements if false?

R and S are transitive `rarrRuuS` is transitive

R and S are transitive `rarrRnnS` is transitive

R and S are symmetric `rarrRuuS` is symmetric

None of these

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The correct Answer is:

To determine which statement is false regarding the relations R and S on the set A, we will analyze the properties of these relations step by step. ### Step 1: Define the Set and Relations Let the set \( A = \{ A, B, C \} \). We define the relations as follows: - Relation \( R = \{ (A, A), (B, B), (A, B) \} \) - Relation \( S = \{ (B, B), (C, C), (B, C) \} \) ### Step 2: Check Transitivity of R and S A relation is transitive if whenever \( (x, y) \) and \( (y, z) \) are in the relation, then \( (x, z) \) must also be in the relation. **For Relation R:** - We have \( (A, B) \) and \( (B, B) \) (which is in R), but we do not have \( (A, B) \) leading to any new pairs. - There are no other combinations that violate transitivity. - Thus, \( R \) is transitive. **For Relation S:** - We have \( (B, C) \) and \( (C, C) \), but we do not have \( (B, C) \) leading to any new pairs. - There are no other combinations that violate transitivity. - Thus, \( S \) is also transitive. ### Step 3: Find R Union S Now, we find the union of the two relations \( R \) and \( S \): \[ R \cup S = \{ (A, A), (B, B), (A, B), (C, C), (B, C) \} \] ### Step 4: Check Transitivity of R Union S To check if \( R \cup S \) is transitive: - We see \( (A, B) \) and \( (B, C) \) are in \( R \cup S \), but \( (A, C) \) is not in \( R \cup S \). - Therefore, \( R \cup S \) is **not transitive**. ### Step 5: Find R Intersection S Next, we find the intersection of the two relations \( R \) and \( S \): \[ R \cap S = \{ (B, B) \} \] ### Step 6: Check Transitivity of R Intersection S The intersection \( R \cap S \) contains only one pair \( (B, B) \), which is trivially transitive since there are no other elements to violate transitivity. ### Conclusion From our analysis: - \( R \) is transitive. - \( S \) is transitive. - \( R \cup S \) is **not transitive**. - \( R \cap S \) is transitive. Thus, the false statement is related to the transitivity of \( R \cup S \). ### Final Answer The false statement is that \( R \cup S \) is transitive. ---

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