Consider the following relations: R = |(x,y)|x,y are real numbers and x = wy for some rational mumber w/; S = \( (m / n),(p / q) )| m,n,p and q are integers such that n,q x 0 and qm = pm\. Then

x=w y Reflexive : (a, a) in R => a=w . a Symmetric: (a, b) in R => (b, a) notin Rnot symmetric\ S=\((m / n), (p / q)) / q m=p n\ (m / n)=(p / q) ((a / b), (a / b)) in S => (a / b)=(a / b)( Reflexive ) ((a / b), (c / d)) in S => a d=b c => ((c / d), (a / b)) in S ( Symetric ) ((a / b), (c / d)) in S,((c / d), (e / f)) in S => ((a / b), (e / f)) in SS is an equivalence relation but R is not an equivalence relation

Consider the following relations: R = |(x,y)|x,y are... | Relation (Mathematics)

Consider the following relations:

R = |(x,y)|x,y

are real numbers and

x =

wy for some rational mumber w/;

S = \{\left( \frac{m}{n},\frac{p}{q} \right)|\ m,n,p

and

q

are integers such that

n,q \times 0

and

qm = pm\}

. 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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Correct Answer:

Explanation

$x=w y\\$

Reflexive : $(a, a) \in R \Rightarrow a=w . a\\$

Symmetric: $(a, b) \in R \Rightarrow(b, a) \notin R$

not symmetric

$\\ S=\left\{\left(\frac{m}{n}, \frac{p}{q}\right) / q m=p n\right\} \\$

$\frac{m}{n}=\frac{p}{q} \\$

$\left(\frac{a}{b}, \frac{a}{b}\right) \in S \Rightarrow \frac{a}{b}=\frac{a}{b}(\text { Reflexive }) \\$

$\left(\frac{a}{b}, \frac{c}{d}\right) \in S \Rightarrow a d=b c \Rightarrow\left(\frac{c}{d}, \frac{a}{b}\right) \in S \quad(\text { Symetric }) \\$

$\left(\frac{a}{b}, \frac{c}{d}\right) \in S,\left(\frac{c}{d}, \frac{e}{f}\right) \in S \\$

$\Rightarrow\left(\frac{a}{b}, \frac{e}{f}\right) \in S$

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

Explanation

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