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

$ad(b+c)=bc(a+d)$

Dividing both sides by $abcd$,

$\frac{b+c}{bc}=\frac{a+d}{ad}$

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

Rearranging,

$\frac{1}{a}-\frac{1}{b}=\frac{1}{c}-\frac{1}{d}$

So, relation R is based on equality of the same expression.

Therefore, R is reflexive, symmetric and transitive.