Functions Question 530
Question: Let R and S be two non-void relations in a set A. which of the following statements is not true?
Options:
A) R and S transitive $ \Rightarrow $ $ R\cup S $ is transitive
B) R and S transitive $ \Rightarrow R\cap S $ is transitive
C) R and S transitive $ \Rightarrow R\cup S $ is symmetric
D) R and S reflexive $ \Rightarrow R\cap S $ is reflexive
Show Answer
Answer:
Correct Answer: A
Solution:
[a] Let $ (a,b),(b,c)\in R\cup S. $ It is possible that $ (a,b)\in R-S $ and $ (b,c)\in S-R. $ In such a case, we cannot say that $ (a,c)\in R $ or $ (a,c)\in S. $
$ \therefore (a,c) $ may not be in $ R\cup S. $
$ \therefore R\cup S $ is not transitive. [b] Let $ (a,b),(b,c)\in R\cap S $
$ \therefore (a,b),(b,c)\in R $ and $ (a,b),(b,c)\in S $
$ \therefore (a,c)\in R $ and $ (a,c)\in S $
$ \therefore (a,c)\in R\cap S $
$ \therefore R\cap S $ is transitive.
[c] Let $ (a,b)\in R\cup S $
$ \therefore (a,b)\in Ror(a,b)\in S $
Now, $ (a,b)\in R\Rightarrow (b,a)\in R $ ( $ \because $ R is symmetric)
$ (a,b)\in S\Rightarrow (b,a)\in S $ ( $ \because $ S is symmetric)
$ \therefore (b,a)\in R\cup S\therefore R\cup S $ is symmetric. [d] Let $ a\in A.\therefore (a,a)\in R $ and $ (a,a)\in S. $
$ \therefore (a,a)\in R\cap S. $
$ \therefore R\cap S $ is reflexive.
BETA
