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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.**So, between any two consecutive integers the number of integers is \(0\).**

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,

$\frac{-9}{6}<\frac{-8}{6}<\frac{-7}{6}<\frac{-6}{6}<\frac{-5}{6}<\frac{-4}{6}$

(or)

$\frac{-3}{2}<\frac{-8}{6}<\frac{-7}{6}<\frac{-6}{6}<\frac{-5}{6}<\frac{-2}{3}$

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.