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# List five rational numbers between $- \dfrac{1}{2}$ and $\dfrac{2}{3}$.

Last updated date: 20th Jun 2024
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Hint: We can identify the given numbers as rational numbers. If $a$ and $b$ are rational numbers, then $\dfrac{{a + b}}{2}$ is also a rational number. So using this idea we can find any number of rational numbers between them.

Complete step-by-step solution:
We are given two numbers $- \dfrac{1}{2}$ and $\dfrac{2}{3}$.
We are asked to find five rational numbers in between them.
A rational number is a number which can be written in the form $\dfrac{p}{q}$, where $p$ and $q$ are integers with no common factors and $q \ne 0$.
Now we can see that the given numbers $- \dfrac{1}{2}$ and $\dfrac{2}{3}$ satisfy these conditions and so they are rational numbers.
Since one of the numbers is positive and other is negative, $0$ lies in between them. Clearly $0$ is a rational number and can be written as $\dfrac{0}{1}$.
So we can write one of the rational numbers between the given numbers as $0$.
We know if $a$ and $b$ are rational numbers, then $\dfrac{{a + b}}{2}$ is also a rational number. And also $\dfrac{{a + b}}{2}$ lies in between the numbers $a$ and $b$.
Using this fact, we can find the rational numbers between the given numbers.
Consider, $- \dfrac{1}{2},0$ and $0,\dfrac{2}{3}$.
$- \dfrac{1}{2}$ and $0$ are rational implies, $\dfrac{{ - \dfrac{1}{2} + 0}}{2} = - \dfrac{1}{4}$ is a rational number.
$\dfrac{2}{3}$ and $0$ are rational implies, $\dfrac{{\dfrac{2}{3} + 0}}{2} = \dfrac{2}{6} = \dfrac{1}{3}$ is a rational number.
So we got two rational numbers $- \dfrac{1}{4}$ and $\dfrac{1}{3}$ between the given numbers.
Now consider $- \dfrac{1}{4},0$ and $0,\dfrac{1}{3}$.
Again using the same logic we can see $\dfrac{{0 - \dfrac{1}{4}}}{2} = - \dfrac{1}{8}$ and $\dfrac{{0 + \dfrac{1}{3}}}{2} = \dfrac{1}{6}$ and this gives $- \dfrac{1}{8}$ and $\dfrac{1}{6}$ as rational numbers between the given numbers.

Thus we get five rational numbers between $- \dfrac{1}{2}$ and $\dfrac{2}{3}$. They are $- \dfrac{1}{8}, - \dfrac{1}{4},0,\dfrac{1}{6},\dfrac{1}{3}$.

Note: Here we found the first number as $0$. We can also start from the given numbers and take the mean.
$- \dfrac{1}{2}$ and $\dfrac{2}{3}$ are rational numbers implies $\dfrac{{ - \dfrac{1}{2} + \dfrac{2}{3}}}{2}$ is also a rational number.
And $\dfrac{{ - \dfrac{1}{2} + \dfrac{2}{3}}}{2} = \dfrac{{\dfrac{{4 - 3}}{6}}}{2} = \dfrac{1}{{12}}$
This gives $\dfrac{1}{{12}}$ is a rational number between $- \dfrac{1}{2}$ and $\dfrac{2}{3}$.
And we can continue the process. But we chose $0$ to make the calculation easier.