### Sets And Relations Question 3

#### Question 3 - 29 January - Shift 2

Let $R$ be a relation defined on $\mathbb{N}$ as aRb is $2 a+3 b$ is a multiple of $5, a, b \in \mathbb{N}$. Then $R$ is

(1) not reflexive

(2) transitive but not symmetric

(3) symmetric but not transitive

(4) an equivalence relation

## Show Answer

#### Answer: (4)

#### Solution:

#### Formula: Reflexive relation (iv), Symmetric relation (v), Transitive relation (vi), Equivalence relation (vii)

Let $(a, b) \in R$

$ f(a, b)=2 a+3 b $

For Reflexive

$ f(a, a)=2 a+3 a=5 a \text { i,e divisible by } 5 $

$ \Rightarrow(a, a) \in R $

For symmetric

$f(b, a)=2 b+3 a=5 a+5 b-(2 a+3 b)$, divisible by 5

$ f(b, a) \text { is divisible by } 5 \Rightarrow(b, a) \in R $

For transitive

$ \begin{aligned} & f(a, b)=2 a+3 b \text { is divisible by } 5 \\ & \quad \Rightarrow 2 a+3 b=5 \alpha \\ & f(b, c)=2 b+3 c \text { is divisible by } 5 \\ & \Rightarrow 2 b+3 c=5 \beta \end{aligned} $

$ 2 a+5 b+3 c=5(\alpha+\beta) \Rightarrow 2 a+3 c=5(\alpha+\beta-b) \Rightarrow \text { a R c } $

So, $2 a+3 c$ is divisible by 5

$ \Rightarrow(a, c) \in R $

Hence relation is equivalence relation.