Let R1={(a,b)∈N×N:|a−b|⩽13} and R2={(a,b)∈N×N:|a−b|≠13}. Then on N :

R1={(a,b)∈N×N:|a−b|≤13}R2={(a,b)∈N×N:|a−b|≠13}.For R1 :i) Reflexive relation (a,a)∈N×N:|a−a|≤13 R1 is reflexiveii) Symmetric relation (a,b)∈R1 then |a−b|≤13⇒|b−a|≤13 ⇒(b,a)∈R1iii) Transitive relation (a,b)∈R1,(b,c)∈R1⇒(a,c)∈R1:(1,3)∈R1,(3,16)∈R1 but (1,16)∉R1For R2 :i) Reflexive relation (a,a)∈N×N:|a−a|≠13ii) Symmetric relation (a,b)∈N×N:|a-b|≠13⇒|b-a|≠13⇒(b,a)∈N×Niii) Transitive relation (a,b)∈R2, (b,c)∈R2, (a,c)∈R2(1,3)∈R2, (3,14)∈R2, but (1,14)∉R2

Let R1={(a,b)∈N×N:|a−b|⩽13} and R2={(a,b)∈N×N:|a−b|≠13}. Then on N :

R1={(a,b)∈N×N:|a−b|≤13}R2={(a,b)∈N×N:|a−b|≠13}.For R1 :i) Reflexive relation (a,a)∈N×N:|a−a|≤13 R1 is reflexiveii) Symmetric relation (a,b)∈R1 then |a−b|≤13⇒|b−a|≤13 ⇒(b,a)∈R1iii) Transitive relation (a,b)∈R1,(b,c)∈R1⇒(a,c)∈R1:(1,3)∈R1,(3,16)∈R1 but (1,16)∉R1For R2 :i) Reflexive relation (a,a)∈N×N:|a−a|≠13ii) Symmetric relation (a,b)∈N×N:|a-b|≠13⇒|b-a|≠13⇒(b,a)∈N×Niii) Transitive relation (a,b)∈R2, (b,c)∈R2, (a,c)∈R2(1,3)∈R2, (3,14)∈R2, but (1,14)∉R2

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

Let $R_{1} = {(a, b) \in N \times N : | a - b | ⩽ 13}$ and $R_{2} = {(a, b) \in N \times N : | a - b | \neq 13}$ . Then on N :

see full answer

High-Paying Jobs That Even AI Can’t Replace — Through JEE/NEET

🎯 Hear from the experts why preparing for JEE/NEET today sets you up for future-proof, high-income careers tomorrow.

An Intiative by Sri Chaitanya

Both R₁ and R₂ are equivalence relations

Neither R₁ and R₂ is an equivalence relation

R₁ is an equivalence relation but R₂ is not

R₂ is an equivalence relation but R₁ is not

answer is B.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

$\begin{array}{l} R_{1} = {(a, b) \in N \times N : | a - b | \leq 13} \\ R_{2} = {(a, b) \in N \times N : | a - b | \neq 13} . \end{array}$

For R₁ :

i) Reflexive relation

$(a, a) \in N \times N : | a - a | \leq 13$

$R_{1}$ is reflexive

ii) Symmetric relation

$(a, b) \in R_{1} t h e n | a - b | \leq 13 \Rightarrow | b - a | \leq 13 \Rightarrow (b, a) \in R_{1}$

iii) Transitive relation

$\begin{array}{l} (a, b) \in R_{1}, (b, c) \in R_{1} \Rightarrow (a, c) \in R_{1} : \\ (1, 3) \in R_{1,} (3, 16) \in R_{1} but (1, 16) \notin R_{1} \end{array}$