Let S be the set of all real numbers. For a,b∈S, relation R is defined by aRb iff |a−b|<1, then R is

|a−a|=0<1 ∴aRa ∀a∈R⇒R is reflexive aRb⇒|a−b|<1⇒|b−a|<1⇒bRa ∴R is symmetricfurther (1,1.5)∈R and (1.5,2)∈R⇒(1,2)∉R∴R is not transitive

Let S be the set of all real numbers. For a,b∈S, relation R is defined by aRb iff |a−b|<1, then R is

|a−a|=0<1 ∴aRa ∀a∈R⇒R is reflexive aRb⇒|a−b|<1⇒|b−a|<1⇒bRa ∴R is symmetricfurther (1,1.5)∈R and (1.5,2)∈R⇒(1,2)∉R∴R is not transitive

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Let $S$ be the set of all real numbers. For $a, b \in S,$ relation $R$ is defined by $a R b$ iff $| a - b | < 1,$ then $R$ is

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only reflexive

only symmetric

only transitive

reflexive & symmetric

answer is D.

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Detailed Solution

$| a - a | = 0 < 1 ∴ a R a \forall a \in R \Rightarrow R$ is reflexive
$a R b \Rightarrow | a - b | < 1 \Rightarrow | b - a | < 1 \Rightarrow b R a$
$∴ R$ is symmetric
further $(1, 1.5) \in R a n d (1.5, 2) \in R \Rightarrow (1, 2) \notin R$
$∴ R$ is not transitive