Let R₁=\{(a, b) N × N: |a-b| ≤ 13\} and R₂=\{(a, b) N × N: |a-b| ≠ 13\}. Then on N: [28-Jun-2022-Shift-2]

Correct Answer is : Neither {R}1\{R\}₁{R}1 nor {R}2\{R\}₂{R}2 is an equivalence relation

Both \{R\}₁ and \{R\}₂ are equivalence relations

Neither \{R\}₁ nor \{R\}₂ is an equivalence relation

R₁ is an equivalence relation but R₂ is not

R₂ is an equivalence relation but R₁ is not

Correct Answer is : Neither {R}1\{R\}₁{R}1 nor {R}2\{R\}₂{R}2 is an equivalence relation

Test Index

Sets and Relations Part 1

Section: Mathematics

Question : 79 of 116

Marks: +1, -0

Let

R_1=\{(a, b) \in N \times N: |a-b| \leq 13\}

R_2=\{(a, b) \in N \times N: |a-b| \neq 13\}

N:

Solution:

R_1=\{(a, b) \in N \times N: |a-b| \leq 13\}

and

R_2=\{(a, b) \in N \times N: |a-b| \neq 13\}

R_1: \because |2-11|=9 \leq 13

\therefore \;\; (2,11) \in R_1

and

(11,19) \in R_1

but

(2,19) \notin R_1

\therefore \;\; R_1

is not transitive

Hence

R_1

is not equivalence

R_2:(13,3) \in R_2

and

(3,26) \in R_2

but

(13,26) \notin R_2( \because |13-26|=13)

\therefore R_2

is not transitive

Hence

R_2

is not equivalence.

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