Let N denotes the set of all natural numbers. On NxN define R as follows: $(\mathrm{a},\mathrm{b}),(\mathrm{c},\mathrm{d})\in \mathrm{N}\times \mathrm{N}$ (a, b) R (c, d) if ad(b + c) = bc(a + d), then

$\begin{array}{rl} & \mathrm{ad}(\mathrm{b}+\mathrm{c})=\mathrm{bc}(\mathrm{a}+\mathrm{d})\\ & \iff \frac{\mathrm{b}+\mathrm{c}}{\mathrm{bc}}=\frac{\mathrm{a}+\mathrm{d}}{\mathrm{ad}}\\ & \iff \frac{1}{\mathrm{b}}+\frac{1}{\mathrm{c}}=\frac{1}{\mathrm{a}}+\frac{1}{\mathrm{d}}\\ & \iff \frac{1}{\mathrm{a}}-\frac{1}{\mathrm{b}}=\frac{1}{\mathrm{c}}-\frac{1}{\mathrm{d}}\end{array}$

R is reflexive

Let $(\mathrm{a},\mathrm{b})\in \mathbf{N}\times \mathbf{N}$

$\begin{array}{rl} & \frac{1}{\mathrm{a}}-\frac{1}{\mathrm{b}}=\frac{1}{\mathrm{a}}-\frac{1}{\mathrm{b}}\\ \Rightarrow & (\mathrm{a},\mathrm{b})\mathrm{R}(\mathrm{a},\mathrm{b})\end{array}$

R is reflexive. 
R is symmetric. 

Suppose $(\mathrm{a},\mathrm{b}),(\mathrm{c},\mathrm{d})\in \mathbf{N}\times \mathbf{N}\mathbf{ }\text{and }(\mathrm{a},\mathrm{b})\mathrm{R}(\mathrm{c},\mathrm{d})$

$\begin{array}{ll}\iff & \frac{1}{\mathrm{a}}-\frac{1}{\mathrm{b}}=\frac{1}{\mathrm{c}}-\frac{1}{\mathrm{d}}\\ \iff & \frac{1}{\mathrm{c}}-\frac{1}{\mathrm{d}}=\frac{1}{\mathrm{a}}-\frac{1}{\mathrm{b}}\\ \iff & (\mathrm{c},\mathrm{d})\mathrm{R}(\mathrm{a},\mathrm{b})\end{array}$

R is symmetric. 

R is transitive.

Suppose $(\mathrm{a},\mathrm{b}),(\mathrm{c},\mathrm{d}),(\mathrm{e},\mathrm{f})\in \mathbf{N}\times \mathbf{N}$ and (a, b) R (c, d), (c, d) R (e,f)

$\begin{array}{rl} & \Rightarrow \frac{1}{\mathrm{a}}-\frac{1}{\mathrm{b}}=\frac{1}{\mathrm{c}}-\frac{1}{\mathrm{d}}\text{and}\frac{1}{\mathrm{c}}-\frac{1}{\mathrm{d}}=\frac{1}{\mathrm{e}}-\frac{1}{\mathrm{f}}\\ & \Rightarrow \frac{1}{\mathrm{a}}-\frac{1}{\mathrm{b}}=\frac{1}{\mathrm{c}}-\frac{1}{\mathrm{f}}\Rightarrow (\mathrm{a},\mathrm{b})\mathrm{R}(\mathrm{e},\mathrm{f})\end{array}$

R is transitive. 

R is an equivalence relation.