Let $a = \dfrac{7}{12}$ and $b = \dfrac{5}{6}$. Express both a and b in the form $\dfrac{k$1}{m} and $\dfrac{k2}{m}$ where $k$1, $k2$ and m are integers and $k$2 - k1 > 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 - k1 > n + 1 is necessary to find n such rational numbers between the two rational numbers a and b using this method.

Given,

⇒ $a = \dfrac{7}{12}$ and $b = \dfrac{5}{6}$.

L.C.M. of 12 and 6 is 12.

$\Rightarrow a = \dfrac{7}{12} \\[1em] \Rightarrow b = \dfrac{5}{6} = \dfrac{10}{12}.$

Here, k₂ - k₁ = 10 - 7 = 3, which is not greater than 6.

To make k₂ - k₁ > 6, multiply numerator and denominator of both fractions by a sufficiently large integer. Take m = 12 × 3 = 36 :

$\Rightarrow a = \dfrac{7}{12} = \dfrac{7 \times 3}{12 \times 3} = \dfrac{21}{36} \\[1em] \Rightarrow b = \dfrac{5}{6} = \dfrac{5 \times 6}{6 \times 6} = \dfrac{30}{36}.$

Now, k₁ = 21, k₂ = 30 and m = 36.

⇒ k₂ - k₁ = 30 - 21 = 9, which is greater than 6.

The 5 distinct rational numbers between $\dfrac{21}{36}$ and $\dfrac{30}{36}$ with integer numerators are :

$\Rightarrow \dfrac{22}{36}, \dfrac{23}{36}, \dfrac{24}{36}, \dfrac{25}{36}, \dfrac{26}{36}.$

Why the condition k₂ - k₁ > n + 1 gives enough integer numerators :

The integers strictly between k₁ and k₂ are k₁ + 1, k₁ + 2, …, k₂ - 1.

Number of such integers = k₂ - k₁ - 1.

To pick n distinct integer numerators strictly between k₁ and k₂, we need :

⇒ k₂ - k₁ - 1 ≥ n

⇒ k₂ - k₁ ≥ n + 1

If we want to ensure more than enough space, we take

k₂ - k₁ > n + 1

Hence, this condition ensures that there are enough integer numerators between k₁ and k₂ to write n rational numbers using the same denominator.

For n = 5, we needed k₂ - k₁ > 6, which gave us the buffer needed to easily select 5 numbers between a and b.

Hence, with $a = \dfrac{21}{36}$ and $b = \dfrac{30}{36}$ (m = 36),

the 5 rational numbers between a and b are $\dfrac{22}{36}, \dfrac{23}{36}, \dfrac{24}{36}, \dfrac{25}{36}$ and $\dfrac{26}{36}$.

The condition k₂ - k₁ > n + 1 ensures that there are at least n integer numerators strictly between k₁ and k₂.