Let $\mathrm{R}_{1}=\{(\mathrm{a}, \mathrm{b}) \in \mathrm{N} \times \mathrm{N}:|\mathrm{a}-\mathrm{b}| \leq 13\}$ and \[ \mathrm{R}_{2}=\{(\mathrm{a}, \mathrm{~b}) \in \mathrm{N} \times \mathrm{N}:|\mathrm{a}-\mathrm{b}| \neq 13\} . \text { Then on } \mathrm{N}: \]

(B) Neither $R_{1}$ nor $R_{2}$ is an equivalence relation

(A) Both $R_{1}$ and $R_{2}$ are equivalence relations

(C) $R_{1}$ is an equivalence relation but $R_{2}$ is not

(D) $R_{2}$ is an equivalence relation but $R_{1}$ is not

JEE Main · 2022 · 28 Jun 2022 · Shift 2 · Sets Relations and Functions

Let

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} . Then on N :

Answer: (B)

For a relation to be an equivalence relation, it must be reflexive, symmetric, and transitive.

For

R_{1}

: While

R_{1}

is reflexive (since

∣ a - a ∣ = 0 \leq 13

) and symmetric (since

∣ a - b ∣ = ∣ b - a ∣

), it fails transitivity. For example,

(1, 3) \in R_{1}

and

(3, 16) \in R_{1}

(since

∣1 - 3∣ = 2 \leq 13

and

∣3 - 16∣ = 13 \leq 13

), but

(1, 16) \in / R_{1}

(since

∣1 - 16∣ = 15 > 13

).

For

R_{2}

: Similarly,

R_{2}

is reflexive and symmetric, but not transitive. For example,

(1, 3) \in R_{2}

and

(3, 14) \in R_{2}

(since

∣1 - 3∣ = 2 \neq = 13

and

∣3 - 14∣ = 11 \neq = 13

), but

(1, 14) \in / R_{2}

(since

∣1 - 14∣ = 13

).

Therefore, neither

R_{1}

nor

R_{2}

is an equivalence relation.

JEE Main · 2022 · 28 Jun 2022 · Shift 2 · Single correct option