(a)OR(b)

If N denotes the set of all natural numbers and R is the relation on $\displaystyle \mathrm{N} \times \mathrm{N}$ defined by $\displaystyle (\mathrm{a}, \mathrm{b}) \mathrm{R}(\mathrm{c}, \mathrm{d})$, if $\displaystyle \mathrm{ad}(\mathrm{b}+\mathrm{c})=\mathrm{bc}(\mathrm{a}+\mathrm{d})$. Show that R is an equivalence relation.

Reflexive :Here, $\displaystyle (\mathrm{a}, \mathrm{b}) \mathrm{R}(\mathrm{a}, \mathrm{b}) \forall(\mathrm{a}, \mathrm{b}) \in \mathrm{N} \mathrm{X} \mathrm{N}$ since $\displaystyle \mathrm{a} \mathrm{b}(\mathrm{b}+\mathrm{a})=\mathrm{b} \mathrm{a}(\mathrm{a}+$ b) is always true.

Symmetric: Let $\displaystyle (\mathrm{a}, \mathrm{b}) \mathrm{R}(\mathrm{c}, \mathrm{d}) \forall(\mathrm{a}, \mathrm{b}),(\mathrm{c}, \mathrm{d}) \in \mathrm{N} \mathrm{X}$ N. Then, $\displaystyle \mathrm{ad}(\mathrm{b}+\mathrm{c})=\mathrm{bc}(\mathrm{a}+\mathrm{d})$

$$\Rightarrow \mathrm{bc}(\mathrm{a}+\mathrm{d})=\mathrm{ad}(\mathrm{~b}+\mathrm{c})

$$

Transitive: Let $\displaystyle (\mathrm{a}, \mathrm{b}) \mathrm{R}(\mathrm{c}, \mathrm{d})$ and $\displaystyle (\mathrm{c}, \mathrm{d}) \mathrm{R}(\mathrm{e}, \mathrm{f}) \forall(\mathrm{a}, \mathrm{b}),(\mathrm{c}, \mathrm{d}),(\mathrm{e}, \mathrm{f}) \in \mathrm{NXN}$. Then $\displaystyle \mathrm{ad}(\mathrm{b}+\mathrm{c})=\mathrm{bc}(\mathrm{a}+\mathrm{d})$ and $\displaystyle \mathrm{cf}(\mathrm{d}+\mathrm{e})=\mathrm{de}(\mathrm{c}+\mathrm{f}) \Rightarrow \frac{\mathrm{b}+\mathrm{c}}{\mathrm{b} \mathrm{c}}=\frac{\mathrm{a}+\mathrm{d}}{\mathrm{a} \mathrm{d}}$ and $\displaystyle \frac{\mathrm{d}+\mathrm{e}}{\mathrm{d} \mathrm{e}}=\frac{\mathrm{c}+\mathrm{f}}{\mathrm{c} \mathrm{f}} \frac{1}{\mathrm{c}}+\frac{1}{\mathrm{b}}=\frac{1}{\mathrm{d}}+\frac{1}{\mathrm{a}} \Rightarrow \frac{1}{\mathrm{e}}+\frac{1}{\mathrm{d}}=\frac{1}{\mathrm{f}}+\frac{1}{\mathrm{c}}$

Adding, we get

$$\frac{1}{\mathrm{c}}+\frac{1}{\mathrm{b}}+\frac{1}{\mathrm{e}}+\frac{1}{\mathrm{d}}=\frac{1}{\mathrm{d}}+\frac{1}{\mathrm{a}}+\frac{1}{\mathrm{f}}+\frac{1}{\mathrm{c}}

$$$$ \Rightarrow \frac{1}{\mathrm{b}}+\frac{1}{\mathrm{e}}=\frac{1}{\mathrm{a}}+\frac{1}{\mathrm{f}}

$$$$ \Rightarrow \frac{\mathrm{e}+\mathrm{b}}{\mathrm{b} \mathrm{e}}=\frac{\mathrm{f}+\mathrm{a}}{\mathrm{a} \mathrm{f}}