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 $(a, b) \in R$
$ \mathrm{f}(\mathrm{a}, \mathrm{b})=2 \mathrm{a}+3 \mathrm{~b} $
For Reflexive
$ \mathrm{f}(\mathrm{a}, \mathrm{a})=2 \mathrm{a}+3 \mathrm{a}=5 \mathrm{a} \text { i,e divisible by } 5 $
$ \Rightarrow(\mathrm{a}, \mathrm{a}) \in \mathrm{R} $
For symmetric
$\mathrm{f}(\mathrm{b}, \mathrm{a})=2 \mathrm{~b}+3 \mathrm{a}=5 \mathrm{a}+5 \mathrm{~b}-(2 \mathrm{a}+3 \mathrm{~b})$, divisible by 5
$ \mathrm{f}(\mathrm{b}, \mathrm{a}) \text { is divisible by } 5 \Rightarrow(\mathrm{b}, \mathrm{a}) \in \mathrm{R} $
For transitive
$ \begin{aligned} & \mathrm{f}(\mathrm{a}, \mathrm{b})=2 \mathrm{a}+3 \mathrm{~b} \text { is divisible by } 5 \ & \quad \Rightarrow 2 a+3 b=5 \alpha \ & \mathrm{f}(\mathrm{b}, \mathrm{c})=2 \mathrm{~b}+3 \mathrm{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, $2 \mathrm{a}+3 \mathrm{c}$ is divisible by 5
$ \Rightarrow(\mathrm{a}, \mathrm{c}) \in \mathrm{R} $
Hence relation is equivalence relation.
