Let $a = \frac{7}{12}$ and $b = \frac{5}{6}$. Express both $a$ and $b$ in the form $\frac{k_1}{m}$ and $\frac{k_2}{m}$ where $k_1$, $k_2$, and $m$ are integers and $k_2 - k_1 > 6$. Using the same denominator $m$, write exactly five distinct rational numbers lying between $a$ and $b$ keeping an integer numerator. Explain why the condition $k_2 - k_1 > n + 1$ is necessary to find $n$ such rational numbers between the two rational numbers $a$ and $b$ using this method.

Choose $m = 24$, $a = \frac{7}{12} = \frac{14}{24}$ (so $k_1 = 14$),

$$b = \frac{5}{6} = \frac{20}{24} \text{ (so } k_2 = 20\text{).}$$

$k_2 - k_1 = 6$. This is not > 6, so enlarge.

Take $m = 48$.

$$a = \frac{7}{12} = \frac{28}{48} \quad (k_1 = 28); \quad b = \frac{5}{6} = \frac{40}{48} \quad (k_2 = 40).$$

$$k_2 - k_1 = 12 > 6.$$

Five rationals strictly between them: $\frac{29}{48}$, $\frac{30}{48}$, $\frac{31}{48}$, $\frac{32}{48}$, $\frac{33}{48}$.

Reasoning: between $\frac{k_{1}}{m}$ and $\frac{k_{2}}{m}$ we need integer numerators strictly between $k_{1}$ and $k_{2}$. The number of such integers is $k_{2} - k_{1} - 1$.
To get at least $n$ distinct rationals, we require $k_{2} - k_{1} - 1 \geq n$, i.e., $k_{2} - k_{1} > n$, equivalently $k_{2} - k_{1} \geq n + 1$ (so $k_{2} - k_{1} > n + 1$ guarantees strictly more than $n$ options).