Let R be a relation on N × N defined by (a, b)R (c, d) if and only...

Given, $\left(a, b\right)R \left(c, d\right) ⇒ ad$\left(b - c\right) = bc \left(a - d\right)

$⇒ \left(a, b\right)$R\left(c, d\right)$⇔$\dfrac{b - c}{bc} = \dfrac{a - d}{ad}$</p><p>$⇒ \dfrac{1}{c} - \dfrac{1}{b} = \dfrac{1}{d} - \dfrac{1}{a}$</p><p>$⇒ \dfrac{1}{a} - \dfrac{1}{b} = \dfrac{1}{d} - \dfrac{1}{c}

Reflexive:

\left(a, b\right)R\left(a, b\right) ⇒ \dfrac{1}{a} - \dfrac{1}{b} = \dfrac{1}{b} - \dfrac{1}{a}$</p><p>It is false. So, given relation is not reflexive.</p><p><strong>Symmetric:</strong></p><p>$\left(a, b\right)$R\left(c, d\right)$ ⇒ $\dfrac{1}{a} - \dfrac{1}{b} = \dfrac{1}{d} - \dfrac{1}{c}</p><p>$⇒ $\dfrac{1}{c} - \dfrac{1}{d} = \dfrac{1}{b} - \dfrac{1}{a}</p><p>$∴ \left(c, d\right)$R\left(a, b\right)$

It is symmetric.

Transitive:

\left(a, b\right)$R\left(c, d\right)$ ⇒ $\dfrac{1}{a} - \dfrac{1}{b} = \dfrac{1}{d} - \dfrac{1}{c}</p><p>\left(c, d\right)R\left(e, f\right) = \dfrac{1}{c} - \dfrac{1}{d} = \dfrac{1}{e} - \dfrac{1}{f}</p><p>\dfrac{1}{a} - \dfrac{1}{b} = \dfrac{1}{e} - \dfrac{1}{f}$

$⇒ \left(a, b\right) R \left(e, f\right)</p><p>So, not transitive.</p>$