Functions Question 549
Question: Let $ A=Z\cup {\sqrt{2}}. $ Define a relation R in A by aRb if and only if $ a+b\in Z. $ The relation R is
Options:
A) Reflexive
B) Symmetric and transitive
C) Only transitive relation
D) None of these
Show Answer
Answer:
Correct Answer: B
Solution:
[b] we have, $ R={(a,b):a+b\in Z,a,b\in Z\cup {\sqrt{2}}}. $ $ \sqrt{2}\in A $ and $ \sqrt{2}+\sqrt{2}=2\sqrt{2}\notin Z $
$ \therefore (\sqrt{2},\sqrt{2})\notin R $
$ \therefore $ R is not reflexive. Let $ (a,b)\in R $
$ \therefore a+b\in \mathbb{Z} $
$ \Rightarrow b+a\in \mathbb{Z}\Rightarrow (b,a)\in \mathbb{R} $
$ \Rightarrow $ R is transitive. Let $ (a,b),(b,c)\in R $
$ \therefore a+b,,b+c\in \mathbb{Z} $
$ \therefore $ None of a, b, c, is equal to $ \sqrt{2} $
$ \therefore a,b,c\in Z $
$ \therefore a+c\in Z $
$ \Rightarrow (a,c)\in R \Rightarrow (a,b)\in R \land (b,c)\in R \Rightarrow R $ is transitive.
$ \therefore $ R is not an equivalence relation.
BETA
