Let R be the real line. Consider the following subsets of the plane $\mathrm{R} \times \mathrm{R}$. $\mathrm{S}=\{(\mathrm{x}, \mathrm{y}): \mathrm{y}=\mathrm{x}+1$ and $0<\mathrm{x}<2\}, \mathrm{T}=\{(\mathrm{x}, \mathrm{y}): \mathrm{x}-\mathrm{y}$ is an integer $\}$. Which one of the following is true?

(D) $T$ is an equivalence relation on $R$ but $S$ is not

(A) neither S nor T is an equivalence relation on R

(B) both S and T are equivalence relations on R

(C) S is an equivalence relation on R but T is not

JEE Main · 2008 · 27 Apr 2008 · Offline · Sets Relations and Functions

Let R be the real line. Consider the following subsets of the plane

R \times R

S = {(x, y) : y = x + 1

and

0 < x < 2}, T = {(x, y) : x - y

is an integer

}

. Which one of the following is true?

Answer: (D)

For

T = {(x, y) : x - y \in Z}

, reflexivity holds since

x - x = 0

, symmetry since an integer

x - y

makes

y - x

an integer too, and transitivity since the sum of two integers is an integer, so

T

is an equivalence relation. For

S = {(x, y) : y = x + 1, 0 < x < 2}

, even reflexivity fails because

x = x + 1

is impossible, so

S

is not an equivalence relation. Thus

T

is one but

S

is not.

JEE Main · 2008 · 27 Apr 2008 · Offline · Single correct option