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

Correct answer: D. T is an equivalence relation on R but S is not T=\ (x, y): x-y ∈ l\ as 0 ∈ I T is a reflexive relation. If x-y ∈ 1 y-x ∈ I T is symmetrical also If x-y=l_1 and y-z=l_2 Then x-z=(x-y)+(y-z)=l_1+I_2 ∈ I T is also transitive. Hence T is an equivalence relation. Clearly x ≠ x+1 (x, x) S S is not reflexive.

JEE Main 2008

Exam: JEE Main
Year: 2008
Subject: Mathematics
Chapter: Sets and Relations

JEE Main 2008 · Mathematics · Sets and Relations

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

Let

R

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

R \times R

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

T=\{(x, y): x-y \text{ is an integer}\}

. Which one of the following is true?

Options

Aneither $S$ nor $T$ is an equivalence relation on $R$
Bboth $S$ and $T$ are equivalence relations on $R$
C $S$ is an equivalence relation on $R$ but $T$ is not
D $\mathrm{T}$ is an equivalence relation on $\mathrm{R}$ but $S$ is not

Correct answer

D. $\mathrm{T}$ is an equivalence relation on $\mathrm{R}$ but $S$ is not

Step-by-step solution

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

0 \in

\mathrm{I}

\mathrm{T}

is a reflexive relation. If

x-y \in 1 \Rightarrow y-x \in I\therefore \mathrm{T}

is symmetrical also If

x-y=l_1

and

y-z=l_2

Then

x-z=(x-y)+(y-z)=l_1+I_2 \in I\therefore \mathrm{T}

is also transitive. Hence

T

is an equivalence relation. Clearly

x \neq x+1 \Rightarrow(x, x) \notin S\therefore S

is not reflexive.

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