Functions Question 499
Question: Let R be a relation on $ N\times N $ defined by $ (a,b)R(c,d)\Rightarrow ad(b+c)=bc(a+d)R $ is
Options:
A) A partial order relation
B) An equivalence relation
C) An identity relation
D) None of these
Show Answer
Answer:
Correct Answer: B
Solution:
[b] We observe the following properties: Reflexivity: Let (a, b) be an arbitrary element of $ N\times N $ . Then, $ (a,b)\in N\times N\Rightarrow a,b\in N $
$ \Rightarrow ab(b+a)=ba(a+b)\Rightarrow (a,b)R(a,b) $ Thus, $ (a,b)R(a,b) $ for all $ (a,b)\in N\times N. $ So, R is reflexive on $ N\times N $ . Symmetry: Let $ (a,b),(c,d)\in N\times N $ be such that $ (a,b)R(c,d). $ Then, $ (a,b)R(c,d) $
$ \Rightarrow ad(b+c)=bc(a+d) $
$ \Rightarrow ,cb(d+a)=da(c+b)\Rightarrow (c,d)R(a,b) $ Thus, $ (a,b)R(c,d)\Rightarrow (c,d)R(a,b) $ for all $ (a,b),(c,d)\in N\times N. $ So, R is symmetric on $ N\times N. $ Transitivity: Let $ (a,b),(c,d),(e,f)\in N\times N $ such that $ (a,b)R(c,d),and(c,d)R(e,f). $ Then, $ (a,b)R(c,d)\Rightarrow ad(b+c)=bc(a+d) $
$ \Rightarrow \frac{b+c}{bc}=\frac{a+d}{ad}\Rightarrow \frac{1}{b}+\frac{1}{c}=\frac{1}{a}+\frac{1}{d} $ …….. (i) And $ (c,d)R(e,f)\Rightarrow cf(d+e)=de(c+f) $
$ \Rightarrow \frac{d+e}{de}=\frac{c+f}{cf}\Rightarrow \frac{1}{d}+\frac{1}{e}=\frac{1}{c}+\frac{1}{f} $ …….. (ii) Adding (i) and (ii), we get $ ( \frac{1}{b}+\frac{1}{c} )+( \frac{1}{d}+\frac{1}{e} )=( \frac{1}{a}+\frac{1}{d} )+( \frac{1}{c}+\frac{1}{f} ) $
$ \Rightarrow \frac{1}{b}+\frac{1}{e}=\frac{1}{a}+\frac{1}{f}\Rightarrow \frac{b+e}{be}=\frac{a+f}{af} $
$ \Rightarrow af(b+e)=be(a,f)\Rightarrow (a,b)R(e,f) $ Thus, $ (a,b)R(c,d) $ and $ (c,d)R(e,f) $
$ \Rightarrow (a,b)R(e,f) $ For all $ (a,b),(c,d),(e,f)\in N\times N. $ So, R is transitive on $ N\times N $ . Hence. R being reflexive, symmetric and transitive, is an equivalence relation on $ N\times N $ .
BETA
