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Find a rational number between the following rational numbers:
(i) $\dfrac{3}{4}$ and $\dfrac{4}{3}$
(ii) 5 and 6
(iii) $\dfrac{-3}{4}$ and $\dfrac{1}{3}$

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Last updated date: 26th Feb 2024
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IVSAT 2024
Answer
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Hint: We start solving the problem by recalling the fact that the average of the two numbers a and b lies between them and its value is $\dfrac{a+b}{2}$. We then find the average of the given two rational numbers by using this formula. We then make the necessary calculations to get the value of the required rational number. We then follow the similar steps to find the rational number of the remaining two problems.

Complete step by step answer:
According to the problem, we need to find a rational number between the following rational numbers (i) $\dfrac{3}{4}$ and $\dfrac{4}{3}$
(ii) 5 and 6
(iii) $\dfrac{-3}{4}$ and $\dfrac{1}{3}$
(i) Now, let us find the rational that lies between the rational numbers $\dfrac{3}{4}$ and $\dfrac{4}{3}$. We know that the average of the two numbers a and b lies between them and its value is $\dfrac{a+b}{2}$.
So, let us find the average of the rational numbers $\dfrac{3}{4}$ and $\dfrac{4}{3}$. Let us assume that average as ${{r}_{1}}$.
$\Rightarrow {{r}_{1}}=\dfrac{\dfrac{3}{4}+\dfrac{4}{3}}{2}$.
$\Rightarrow {{r}_{1}}=\dfrac{\dfrac{9+16}{12}}{2}$.
$\Rightarrow {{r}_{1}}=\dfrac{\dfrac{25}{12}}{2}$.
$\Rightarrow {{r}_{1}}=\dfrac{25}{24}$.
∴ The rational numbers that lie between the rational numbers $\dfrac{3}{4}$ and $\dfrac{4}{3}$ is $\dfrac{25}{24}$.
(ii) Now, let us find the rational that lies between the rational numbers 5 and 6. We know that the average of the two numbers a and b lies between them and its value is $\dfrac{a+b}{2}$.
So, let us find the average of the rational numbers 5 and 6. Let us assume that average as ${{r}_{2}}$.
$\Rightarrow {{r}_{2}}=\dfrac{5+6}{2}$.
$\Rightarrow {{r}_{2}}=\dfrac{11}{2}$.
∴ The rational numbers that lie between the rational numbers 5 and 6 is $\dfrac{11}{2}$.
(iii) Now, let us find the rational that lies between the rational numbers $\dfrac{-3}{4}$ and $\dfrac{1}{3}$. We know that the average of the two numbers a and b lies between them and its value is $\dfrac{a+b}{2}$.
So, let us find the average of the rational numbers $\dfrac{-3}{4}$ and $\dfrac{1}{3}$. Let us assume that average as ${{r}_{3}}$.
$\Rightarrow {{r}_{3}}=\dfrac{\dfrac{-3}{4}+\dfrac{1}{3}}{2}$.
$\Rightarrow {{r}_{3}}=\dfrac{\dfrac{-9+4}{12}}{2}$.
$\Rightarrow {{r}_{3}}=\dfrac{\dfrac{-5}{12}}{2}$.
$\Rightarrow {{r}_{3}}=\dfrac{-5}{24}$.
∴ The rational numbers that lies between the rational numbers $\dfrac{-3}{4}$ and $\dfrac{1}{3}$ is $\dfrac{-5}{24}$.

Note: We can find any rational number that lies between the given rational numbers irrespective of their average. Here we found the average of them because it is easier to calculate than to find the other rational numbers. We should know that integers are also part of rational numbers. We can also solve this problem by using $a+r\left( b-a \right)$, where a, b are the two given rational numbers and r is any value in the interval in $\left( 0,1 \right)$.