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**Hint:**Here we will find three rational numbers between $ - 2$ and $5$. Firstly we will find the average of the two numbers then we will find the average of the number we got and any one of the numbers given. Finally we will again find the average of the two numbers we got and get our desired answer.

**Complete step-by-step answer:**

The two numbers given are$ - 2$ and $5$.

Firstly we will find average of the given numbers as,

Average = Sum of the two numbers/ Number of numbers

Average $ = \dfrac{{ - 2 + 5}}{2} = \dfrac{3}{2}$….$\left( 1 \right)$

Next, we will find the average of 5 and the value in equation $\left( 1 \right)$ as,

Average $ = \dfrac{{ - 2 + \dfrac{3}{2}}}{2} = \dfrac{{\dfrac{{ - 4 + 3}}{2}}}{2}$

Average $ = \dfrac{{\dfrac{{ - 1}}{2}}}{2} = - \dfrac{1}{4}$….$\left( 2 \right)$

Finally we will take average of the values in equation $\left( 1 \right)$ and $\left( 2 \right)$ as,

Average $ = \dfrac{{\dfrac{3}{2} + \left( { - \dfrac{1}{4}} \right)}}{2} = \dfrac{{\dfrac{{6 - 1}}{4}}}{2}$

Average $ = \dfrac{{\dfrac{5}{4}}}{2} = \dfrac{5}{8}$…..$\left( 3 \right)$

**So, from equation $\left( 1 \right)$ $\left( 2 \right)$ and $\left( 3 \right)$ we get the three rational numbers between $ - 2$ and $5$ as,**

$\dfrac{3}{2}, - \dfrac{1}{4},\dfrac{5}{8}$

$\dfrac{3}{2}, - \dfrac{1}{4},\dfrac{5}{8}$

**Note:**

Rational numbers are those numbers that can be expressed in the form of $\dfrac{p}{q}$ where$q \ne 0$.

1. If we find the decimal expansion of a rational number it either terminates after a finite number or the digit starts to repeat themselves over and over again.

2. If it is not a rational number that means it is an irrational number.

3. The set of all rational numbers together with addition and multiplication operations forms a field.

4. The set of a rational number is countable but the set of irrational numbers is uncountable and as real number is a union of rational and irrational numbers so it is also uncountable.

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