Question

Let $N$ denote the set of natural numbers and $R$ be a relation on $N\times N$ defined by
$(a,b)R(c,d)⟺ad(b+c)=bc(a+d).$ Then on $N\times N,R$ is

• A. An equivalence relation
• B. Reflexive and symmetric relation only
• C. Transitive relation only
• D. Symmetric and transitive relation only

Answer 1

The correct option is A An equivalence relation

Given $(a,b)R(c,d)\iff ad(b+c)=bc(a+d)$ where $(a,b),(c,d)\in N\times N$
$ad(b+c)=bc(a+d)$
$\Rightarrow \frac{b+c}{bc}=\frac{a+d}{ad}$
$\Rightarrow \frac{1}{b}+\frac{1}{c}=\frac{1}{a}+\frac{1}{d}$
$\therefore (a,b)R(c,d)\iff \frac{1}{b}+\frac{1}{c}=\frac{1}{a}+\frac{1}{d}$

As $\frac{1}{b}+\frac{1}{a}=\frac{1}{a}+\frac{1}{b}\forall (a,b)\in N\times N$
So, $(a,b)R(a,b)$
$\therefore R$ is reflexive relation.

If $(a,b)R(c,d)\Rightarrow \frac{1}{b}+\frac{1}{c}=\frac{1}{a}+\frac{1}{d}$
$\Rightarrow \frac{1}{d}+\frac{1}{a}=\frac{1}{c}+\frac{1}{b}$
$\Rightarrow (c,d)R(a,b)$
$\therefore R$ is symmetric relation.

Let $(a,b),(c,d),(e,f)\in N\times N$
Let $(a,b)R(c,d)$ and $(c,d)R(e,f)$ $(a,b)R(c,d)\Rightarrow ad(b+c)=bc(a+d)$
$\Rightarrow \frac{1}{b}+\frac{1}{c}=\frac{1}{a}+\frac{1}{d}\cdots \left(1\right)$
$(c,d)R(e,f)\Rightarrow cf(d+e)=de(c+f)$
$\Rightarrow \frac{1}{d}+\frac{1}{e}=\frac{1}{f}+\frac{1}{c}\cdots \left(2\right)$
$\left(1\right)+\left(2\right)\Rightarrow (\frac{1}{b}+\frac{1}{c}+\frac{1}{d}+\frac{1}{e})=(\frac{1}{a}+\frac{1}{d}+\frac{1}{f}+\frac{1}{c})$
$\frac{1}{b}+\frac{1}{e}=\frac{1}{a}+\frac{1}{f}$
$\Rightarrow (a,b)R(e,f)$
$\therefore R$ is a Transitive relation.
Since $R$ is Reflexive, Symmetric and Transitive relation.
$\therefore R$ is an Equivalence relation.