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  4. Let S be the set of all real numbers. A relation R has been defined on S by a R b Longleftrightarrow|a-b| ≤ 1, then R is

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Q. Let $S$ be the set of all real numbers. A relation $R$ has been defined on $S$ by $a R b \Longleftrightarrow|a-b| \leq 1$, then $R$ is

KCETKCET 2014Relations and Functions - Part 2

Solution:

Given, $a R b \Leftrightarrow|a-b| \leq 1$
For Reflexive
$a R a=|a-a|=0 \leq 1$
So, it is reflexive.
For Symmetric
$a R b \Leftrightarrow |a-b| \leq 1$
$\Rightarrow |b-a| \leq 1$
i.e., $a R b \Rightarrow bRa$
So, it is symmetric.
For Transitive
Take $a=1,\, b=2$ and $c=3$
Now, $|a-b|=|1-2|=1$
and $|b-c|=|2-3|=1$
But $|a-c|=|1-3|=2>1$, which is not true.
i.e., $a R b,\,\, b R c \neq a R C$
So, it is not transitive.