Consider the following relations R={x,y/x,y are real numbers and x=wy for some rational number w}, S={mn,pq/m,n,p and q integers such that, n,p≠0 and qm=pm} then

We have x,x∈R for w=1 in plying that R is reflexive for a≠0,(a,0)∉R for any W but 0,a∈R. Then R is not symmeric Hence R is not an equivalence relation Asmn,mn∈s ∵mn=mn, S is reflexive mn,pq∈s⇒qm=pn ⇒np=mq ⇒pq,mn∈s, So S is symmetric again mn,pq∈s and pq,ab∈s means qm=pn and bp=aq ⇒mn=pqand pq=ab⇒mn=ab Somn,ab∈S This means S is transistive

Consider the following relations R={x,y/x,y are real numbers and x=wy for some rational number w}, S={mn,pq/m,n,p and q integers such that, n,p≠0 and qm=pm} then

We have x,x∈R for w=1 in plying that R is reflexive for a≠0,(a,0)∉R for any W but 0,a∈R. Then R is not symmeric Hence R is not an equivalence relation Asmn,mn∈s ∵mn=mn, S is reflexive mn,pq∈s⇒qm=pn ⇒np=mq ⇒pq,mn∈s, So S is symmetric again mn,pq∈s and pq,ab∈s means qm=pn and bp=aq ⇒mn=pqand pq=ab⇒mn=ab Somn,ab∈S This means S is transistive

AI Mentor

Check Your IQ

Free Expert Demo

Try Test

Courses

Dropper NEET Course Dropper JEE Course Class - 12 NEET Course Class - 12 JEE Course Class - 11 NEET Course Class - 11 JEE Course Class - 10 Foundation NEET Course Class - 10 Foundation JEE Course Class - 10 CBSE Course Class - 9 Foundation NEET Course Class - 9 Foundation JEE Course Class -9 CBSE Course Class - 8 CBSE Course Class - 7 CBSE Course Class - 6 CBSE Course

Offline Centres

Consider the following relations $R = {(x, y) / x, y$ are real numbers and x=wy for some rational number w}, $S = {(\frac{m}{n}, \frac{p}{q}) / m, n, p and q$ integers such that, $n, p \neq 0 and qm = pm}$ then

see full answer

Your Exam Success, Personally Taken Care Of

1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.

An Intiative by Sri Chaitanya

S is an equivalence relation but R is not an equivalence relation

R and S both are equivalence relations

Neither R nor S is an equivalence relation

R is an equivalence relation but S is not an equivalence relation

answer is C.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

$We have (x, x) \in R for w = 1 in plying that R is reflexive for a \neq 0, (a, 0) \notin R for any W but (0, a) \in R . Then R is not symmeric Hence R is not an equivalence relation As (\frac{m}{n}, \frac{m}{n}) \in s ∵ mn = mn, S is reflexive (\frac{m}{n}, \frac{p}{q}) \in s \Rightarrow qm = pn \Rightarrow np = mq \Rightarrow (\frac{p}{q}, \frac{m}{n}) \in s, So S is symmetric again (\frac{m}{n}, \frac{p}{q}) \in s and (\frac{p}{q}, \frac{a}{b}) \in s means qm = pn and bp = aq \Rightarrow \frac{m}{n} = \frac{p}{q} and \frac{p}{q} = \frac{a}{b} \Rightarrow \frac{m}{n} = \frac{a}{b} So (\frac{m}{n}, \frac{a}{b}) \in S This means S is transistive$