Deepak Scored 45->99%ile with Bounce Back Crack Course. You can do it too!

Mark the tick against the correct answer in the following:

Question:

Mark the tick against the correct answer in the following:

Let $R$ be a relation on the set $N$ of all natural numbers, defined by $a R b \Leftrightarrow a$ is a factor of $b$. 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 $N=\{1,2,3,4 \ldots \ldots\}$

And $R=\{(a, b): a, b \in N$ and $a$ is a factor of $b\}$

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 , $(\mathrm{a}, \mathrm{a})$

$\mathrm{a}$ is a factor of a

$(2,2),(3,3) \ldots(a, a)$ where $a \in N$

Therefore , R is reflexive ……. (1)

Check for symmetric

$a R b \Rightarrow a$ is factor of $b$

b $\mathrm{R} \mathrm{a} \Rightarrow \mathrm{b}$ is factor of a well

$E x_{-}(2,6) \in R$

But $(6,2) \notin \mathrm{R}$

Therefore , R is not symmetric ……. (2)

Check for transitive

$a \mathrm{R} b \Rightarrow a$ is factor of $b$

b $R c \Rightarrow b$ is a factor of $c$

$\mathrm{a} \mathrm{R} \mathrm{c} \Rightarrow \mathrm{b}$ is a factor of $\mathrm{c}$ also

Ex $(2,6),(6,18)$

$\therefore(2,18) \in \mathrm{R}$

Therefore , R is transitive ……. (3)

Now, according to the equations (1), (2), (3)

Correct option will be (B)