Mark the tick against the correct answer in the following:
Let $S$ be the set of all real numbers and let $R$ be a relation on $S$, defined by $a R b \Leftrightarrow|a-b| \leq 1$. Then, $R$ is
A. reflexive and symmetric but not transitive
B. reflexive and transitive but not symmetric
C. symmetric and transitive but not reflexive
D. an equivalence relation
According to the question,
Given set $S=\{\ldots \ldots,-2,-1,0,1,2 \ldots . .\}$
And $R=\{(a, b): a, b \in S$ and $|a-b| \leq 1\}$
Formula
For a relation $\mathrm{R}$ in set $\mathrm{A}$
Reflexive
The relation is reflexive if $(a, a) \in R$ for every $a \in A$
Symmetric
The relation is Symmetric if $(a, b) \in R$, then $(b, a) \in R$
Transitive
Relation is Transitive if $(a, b) \in R \&(b, c) \in R$, then $(a, c) \in R$
Equivalence
If the relation is reflexive, symmetric and transitive, it is an equivalence relation.
Check for reflexive
Consider, $(a, a)$
$\therefore|\mathrm{a}-\mathrm{a}| \leq 1$ and which is always true.
Ex_if $a=2$
$\therefore|2-2| \leq 1 \Rightarrow 0 \leq 1$ which is true.
Therefore , R is reflexive ……. (1)
Check for symmetric
$a R b \Rightarrow|a-b| \leq 1$
$b R a \Rightarrow|b-a| \leq 1$
Both can be true.
Ex $_{-}$If $\mathrm{a}=2$ and $\mathrm{b}=1$
$\therefore|2-1| \leq 1$ is true and $|1-2| \leq 1$ which is also true.
Therefore , R is symmetric ……. (2)
Check for transitive
$a R b \Rightarrow|a-b| \leq 1$
b R c $\Rightarrow|b-c| \leq 1$
$\therefore|\mathrm{a}-\mathrm{c}| \leq 1$ will not always be true
Ex_a $=-5, b=-6$ and $c=-7$
$\therefore|6-5| \leq 1,|7-6| \leq 1$ are true But $|7-5| \leq 1$ is false.
Therefore , R is not transitive ……. (3)
Now, according to the equations (1), (2), (3)
Correct option will be (A)