5. Rational numbers between two rational numbers

Asha wanted to count the numbers between \(-9\) and \(12\). She found that totally \(20\) integers are present, with \(8\) negative integers, \(zero\) and \(11\) positive integers present between \(-9\) and \(12\) (excluding \(-9\) and \(12\)).

 

She also gets to know that there are no integers, present between any two consecutive integers, that is, for example, between \(8\) and \(9\) there is no integer.

 

 

She wondered whether the same would happen in case of rational numbers too?

 

She took two rational numbers $\frac{-3}{2}$ and $\frac{-2}{3}$.

 

She converted them to rational numbers with same denominators, for which the denominator of both the numbers should be converted into LCM.

 

She found the LCM of (\(2\),\(3\)) \(= 6\).

 

Convert both the numbers with denominator as \(6\).

 

$\frac{-3\times 3}{2\times 3}=\frac{-9}{6}$ and $\frac{-2\times 2}{3\times 2}=\frac{-4}{6}$

 

We have,

 

 

(or)

 

 

She could find rational numbers $\frac{-8}{6},\frac{-7}{6},\frac{-6}{6},\frac{-5}{6}$ between $\frac{-3}{2}$.

 

She doubts that are there only \(4\) rational numbers between $\frac{-3}{2}$. She gets to know there are more than \(4\) rational numbers between $\frac{-3}{2}$ because if we find the multiples of denominators, then many more rational numbers can be inserted between $\frac{-3}{2}$.

 

For example,

 

$\frac{-3}{2}=\frac{-3\times 4}{2\times 4}=\frac{-12}{8}=\frac{-3\times 10}{2\times 10}=\frac{-30}{20}$ and $\frac{-2}{3}=\frac{-2\times 4}{3\times 4}=\frac{-8}{12}=\frac{-2\times 10}{3\times 10}=\frac{-20}{30}$

 

Now, between $\frac{-30}{20}$ and $\frac{-20}{30}$, there are \(9\) rational numbers present.