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Question

The relation

R

\mathbb{R}

defined by

xRy

iff

|x - y| \leq 1

is:

(A)Reflexive and symmetric but not transitive

(B)An equivalence relation

(C)Reflexive and transitive but not symmetric

(D)Only reflexive

Solution Path

Check each property independently. Reflexive (|x-x|=0) and symmetric (|x-y|=|y-x|) hold. Transitivity fails: counterexample x=0, y=0.8, z=1.5 shows |x-y|<=1 and |y-z|<=1 but |x-z|>1.

01Question Setup

1/4

Classify the relation

R

\mathbb{R}

defined by

xRy \iff |x - y| \leq 1

. Check reflexive, symmetric, and transitive properties.

xRy \iff |x - y| \leq 1

02Reflexive and Symmetric Check

2/4

Reflexive:

|x - x| = 0 \leq 1

always holds. Symmetric:

|x - y| = |y - x|

, so the condition is symmetric.

\text{Reflexive: } \checkmark \quad \text{Symmetric: } \checkmark

03Transitivity CounterexampleKEY INSIGHT

3/4

Take

x = 0, y = 0.8, z = 1.5

. Then

|x-y| = 0.8 \leq 1

and

|y-z| = 0.7 \leq 1

, but

|x-z| = 1.5 > 1

. So R is NOT transitive.

\text{NOT transitive!}

04Final Answer

4/4

R is reflexive and symmetric but not transitive. It is NOT an equivalence relation.

\text{Answer: (A)}

Concepts from this question1 concepts unlocked

★

Equivalence Relations and Equivalence Classes

STANDARD

An equivalence relation on a set A is a relation that is reflexive, symmetric, and transitive. Every equivalence relation partitions A into disjoint equivalence classes. The equivalence class of an element a is [a] = {x in A : (a, x) in R}. Two elements are in the same class if and only if they are related. The number of equivalence relations on a set of n elements equals the Bell number B(n).

[a] = \{x \in A : (a, x) \in R\}

Equivalence relations are the most commonly tested relation type in JEE. Questions typically ask you to verify all three properties or to find equivalence classes. The partition interpretation helps in counting problems.

Classify relationsFind equivalence classesCount equivalence relationsPartition problems

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