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.
