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# Mark the tick against the correct answer in the following:

Question:

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

Solution:

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)