Question

List 5 rational numbers between -2 and -1.

Answer 1

Hints: A rational number is of the form \[\dfrac{p}{q}\] where p and q must be integers only, like -1.5 is a rational number because it can be converted into \[\dfrac{{ - 3}}{2}\] where p and q are integers as -3 and 2 respectively.

Complete step by step solution:
Given that we have to find 5 rational numbers which must be greater than -2 and less than -1
For better understanding let's try to make a generalised formula in which whatever number we will find will always be between -1 and -2. As -1 and -2 are also rational numbers we will try to find the numbers excluding them.
Let us assume any number in the form of \[\dfrac{1}{n}\] where n belongs to the natural number.
The property of \[\dfrac{1}{n}\] is that no matter what we put in place of n we will always get a number greater than 0 and less than 1.
Now if we add a 1 in \[\dfrac{1}{n}\] the range will also be increased by 1 i.e., now the numbers we will be getting will be greater than 1 and less than 2 and it will be of the form \[\left( {1 + \dfrac{1}{n}} \right)\] where again n belongs to natural number now all we need to do is to place a minus sign in front of it and we will get the numbers in the range, less than -1 and greater than -2 and the number will be in the form of \[ - \left( {1 + \dfrac{1}{n}} \right)\] now whatever we put the value of n we will always get rational numbers between -1 and -2.
For instance let's try to put \[n = 2,3,4,5,6\] Now let's try to put one by one.
\[\begin{array}{l}
n = 2\\
 = - \left( {1 + \dfrac{1}{n}} \right)\\
 = - \left( {1 + \dfrac{1}{2}} \right)\\
 = - \dfrac{3}{2}\\
n = 3\\
 = - \left( {1 + \dfrac{1}{n}} \right)\\
 = - \left( {1 + \dfrac{1}{3}} \right)\\
 = - \dfrac{4}{3}\\
n = 4\\
 = - \left( {1 + \dfrac{1}{n}} \right)\\
 = - \left( {1 + \dfrac{1}{4}} \right)\\
 = - \dfrac{5}{4}\\
n = 5\\
 = - \left( {1 + \dfrac{1}{n}} \right)\\
 = - \left( {1 + \dfrac{1}{5}} \right)\\
 = - \dfrac{6}{5}\\
n = 6\\
 = - \left( {1 + \dfrac{1}{6}} \right)\\
 = - \left( {1 + \dfrac{1}{6}} \right)\\
 = - \dfrac{7}{6}
\end{array}\]
There could be infinitely many rational numbers in between -2 and -1.
For now let us have \[ - \dfrac{3}{2}, - \dfrac{4}{3}, - \dfrac{5}{4}, - \dfrac{6}{5}, - \dfrac{7}{6}\] as the 5 rational number we were looking for.

Note: There cannot be a zero in denominator for a rational number because the ans is not defined and also square root, cube root or any power root is not to be present in a denominator of a rational number it can be in the numerator but not denominator. In case you came across one just rationalise it by multiplying the same number in both numerator and denominator.