Consider the relations $\mathrm{R}_{1}$ and $\mathrm{R}_{2}$ defined as $\mathrm{aR}_{1} \mathrm{~b} \Leftrightarrow \mathrm{a}^{2}+\mathrm{b}^{2}=1$ for all $\mathrm{a}, \mathrm{b}, \in \mathrm{R}$ and $(\mathrm{a}, \mathrm{b}) \mathrm{R}_{2}(\mathrm{c}, \mathrm{d}) \Leftrightarrow \mathrm{a}+\mathrm{d}=\mathrm{b}+\mathrm{c}$ for all $(\mathrm{a}, \mathrm{b}),(\mathrm{c}, \mathrm{d}) \in \mathrm{N} \times \mathrm{N}$. Then

(B) Only $R_{2}$ is an equivalence relation

(A) Only $R_{1}$ is an equivalence relation

(C) $R_{1}$ and $R_{2}$ both are equivalence relations

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

JEE Main · 2024 · 01 Feb 2024 · Shift 2 · Sets Relations and Functions

Consider the relations

R_{1}

and

R_{2}

defined as

aR_{1} b \Leftrightarrow a^{2} + b^{2} = 1

for all

a, b, \in R

and

(a, b) R_{2} (c, d) \Leftrightarrow a + d = b + c

for all

(a, b), (c, d) \in N \times N

. Then

Answer: (B)

R_{1}

a R_{1} b ⟺ a^{2} + b^{2} = 1

for

a, b \in R

. Reflexivity would require

a^{2} + a^{2} = 1

for every real

a

, which is false (e.g.

a = 2

R_{1}

fails reflexivity, so it is not an equivalence relation.

R_{2}

(a, b) R_{2} (c, d) ⟺ a + d = b + c

N \times N

.

Reflexive:

(a, b) R_{2} (a, b) ⟺ a + b = b + a

, always true.

Symmetric: if

a + d = b + c

, rearranging gives

c + b = d + a

, which is exactly the condition for

(c, d) R_{2} (a, b)

. True.

Transitive: suppose

(a, b) R_{2} (c, d)

and

(c, d) R_{2} (e, f)

, i.e.

a + d = b + c

and

c + f = d + e

. Adding these,

a + d + c + f = b + c + d + e ⟹ a + f = b + e,

which is exactly

(a, b) R_{2} (e, f)

. True.

R_{2}

is reflexive, symmetric and transitive, so only

R_{2}

is an equivalence relation.

JEE Main · 2024 · 01 Feb 2024 · Shift 2 · Single correct option