Consider the following relations :-R={(x,y):x,y are real numbers and x=wy for some rational number w} :S={(nm,qp):m,n,p and q are integers such that n,q=0 and qm=pn}. Then :

R: x=wy for a function to be a reflexive function (a,a)∈R⇒a=ωa w=1 (only) ∴R is not a reflexive function. S:{(nm,qp)⇒nqmq=np=mp⇒nm=qp If (ba,ba)∈S, then S will be reflexive function If S∈(ba,dc) and ∈(dc,ba)⇒S is symmetric for a function to be transitive, S∈(21,42),s∈(a,b),S∈(b,c)∴21=84=21S∈(42,84)⇒S∈(a,c)⇒S∈(21,84)∴ S is transitive Hence, S is an equivalence relation. ⇒ (c) is the correct option

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Consider the following relations :- R={ (x,y):x,y are real nu

Question

Consider the following relations :-
$R = {(x, y) : x, y$ are real numbers and $x = w y$ for some rational number $w}$ :
$S = {(\frac{m}{n}, \frac{p}{q}) : m, n, p$ and $q$ are integers such that $n, q \neq = 0$ and $q m = p n} .$ Then :

R

is an equivalence relation but

S

is not an equivalence relation

Neither

R

nor

S

is an equivalence relation

S

is an equivalence relation but

R

is not an equivalence relation

R

and

S

both are equivalence relations

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Oliver Discussed

Consider the following relations :-

R = {(x, y) : x, y

are real numbers and

x = w y

for some rational number

w}

S = {(\frac{m}{n}, \frac{p}{q}) : m, n, p

and

q

are integers such that

n, q \neq = 0

and

q m = p n} .

Then :

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Text solutionVerified

x = w y

for a function to be a reflexive function

(a, a) \in R \Rightarrow a = ωa

w = 1

(only)

∴ R

is not a reflexive function.

S : {(\frac{m}{n}, \frac{p}{q}) \Rightarrow \frac{q}{n} m q = n p = \frac{p}{m} \Rightarrow \frac{m}{n} = \frac{p}{q}

(\frac{a}{b}, \frac{a}{b}) \in S

, then

S

will be reflexive function If

S \in (\frac{a}{b}, \frac{c}{d})

and

\in (\frac{c}{d}, \frac{a}{b}) \Rightarrow S

is symmetric for a function to be transitive,

S \in (\frac{1}{2}, \frac{2}{4}), s \in (a, b), S \in (b, c) ∴ \frac{1}{2} = \frac{4}{8} = \frac{1}{2} S \in (\frac{2}{4}, \frac{4}{8}) \Rightarrow S \in (a, c) \Rightarrow S \in (\frac{1}{2}, \frac{4}{8}) ∴ S is transitive

Hence, S is an equivalence relation. \Rightarrow (c) is the correct option

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Question Text	Consider the following relations :- $R = {(x, y) : x, y$ are real numbers and $x = w y$ for some rational number $w}$ : $S = {(\frac{m}{n}, \frac{p}{q}) : m, n, p$ and $q$ are integers such that $n, q \neq = 0$ and $q m = p n} .$ Then :
Answer Type	Text solution:1
Upvotes	150

Consider the following relations :-R={(x,y):x,y are real numbers and x=wy for some rational number w} :S={(nm​,qp​):m,n,p and q are integers such that n,q=0 and qm=pn}. Then :

Text solutionVerified

Practice questions from similar books

Consider the following relations :-
$R = {(x, y) : x, y$ are real numbers and $x = w y$ for some rational number $w}$ :
$S = {(\frac{m}{n}, \frac{p}{q}) : m, n, p$ and $q$ are integers such that $n, q \neq = 0$ and $q m = p n} .$ Then :