Question

# Find six rational numbers between 3 and 4.

Hint: Here, we have to consider the average of 3 and 4 which lies in between 3 and 4 and is a rational number. i.e consider the formula $x<$\dfrac{x+y}{2}<$y$ With the help of this formula find 6 rational numbers.

Here, two numbers are given as 3 and 4. We have to find 6 rational numbers between 3 and 4.
First, we should know what a rational number is? We know that a rational number is a number of the form $\dfrac{p}{q}$, where $p$ and $q$ are integers and $q\ne 0$.
We also know that the average of two numbers lies between them. i.e. If $x$ and $y$ are two numbers then $\dfrac{x+y}{2}$ lies between them.
$x<$\dfrac{x+y}{2}<$y$
Here, $x=3$ and $y=4$ , so we can write:
\begin{align} & 3<\dfrac{3+4}{2}<4 \\ & 3<\dfrac{7}{2}<4 \\ \end{align}
Hence, we can say that $\dfrac{7}{2}$ is a rational number that lies between $3$ and $4$.
Now to find the other rational numbers we have to follow the same procedure.
Take $x=3$ and $y=\dfrac{7}{2}$, then we can write:
$\dfrac{x+y}{2}=\dfrac{1}{2}\left( 3+\dfrac{7}{2} \right)$
Next by taking the LCM and cross multiplying we get:
\begin{align} & \dfrac{x+y}{2}=\dfrac{1}{2}\left( 3+\dfrac{7}{2} \right) \\ & \dfrac{x+y}{2}=\dfrac{1}{2}\left( \dfrac{3\times 2+7}{2} \right) \\ & \dfrac{x+y}{2}=\dfrac{1}{2}\left( \dfrac{6+7}{2} \right) \\ & \dfrac{x+y}{2}=\dfrac{1}{2}\left( \dfrac{13}{2} \right) \\ & \dfrac{x+y}{2}=\dfrac{13}{4} \\ \end{align}
i.e. we can write $3<\dfrac{13}{4}<\dfrac{7}{2}$
Hence the, the rational number is $\dfrac{13}{4}$.
Now, consider $x=\dfrac{13}{4}$ and $y=\dfrac{7}{2}$.
$\dfrac{x+y}{2}=\dfrac{1}{2}\left( \dfrac{13}{4}+\dfrac{7}{2} \right)$
Next, by taking LCM and doing cross multiplication we obtain:
\begin{align} & \dfrac{x+y}{2}=\dfrac{1}{2}\left( \dfrac{13}{4}+\dfrac{7}{2} \right) \\ & \dfrac{x+y}{2}=\dfrac{1}{2}\left( \dfrac{13+7\times 2}{4} \right) \\ & \dfrac{x+y}{2}=\dfrac{1}{2}\left( \dfrac{13+14}{2} \right) \\ & \dfrac{x+y}{2}=\dfrac{1}{2}\times \dfrac{27}{4} \\ & \dfrac{x+y}{2}=\dfrac{27}{8} \\ \end{align}
Thefore, we can write $\dfrac{13}{4}<\dfrac{27}{8}<\dfrac{7}{2}$
i.e. the rational number is $\dfrac{27}{8}$.
Now, we have to consider a rational number between $\dfrac{7}{2}$ and $4$.
Let $x=\dfrac{7}{2}$ and $y=4$.
$\dfrac{x+y}{2}=\dfrac{1}{2}\left( \dfrac{7}{2}+4 \right)$
Next, by taking the LCM and by cross multiplying we get:
\begin{align} & \dfrac{x+y}{2}=\dfrac{1}{2}\left( \dfrac{7+8}{2} \right) \\ & \dfrac{x+y}{2}=\dfrac{1}{2}\times \dfrac{15}{2} \\ & \dfrac{x+y}{2}=\dfrac{15}{4} \\ \end{align}
Therefore, we can write:
$\dfrac{7}{2}<\dfrac{15}{4}<4$
Hence, the rational number is $\dfrac{15}{4}$.
Now, we have to consider a rational number between $\dfrac{7}{2}$ and $\dfrac{15}{4}$.
Let $x=\dfrac{7}{2}$ and $y=\dfrac{15}{4}$.
$\dfrac{x+y}{2}=\dfrac{1}{2}\left( \dfrac{7}{2}+\dfrac{15}{4} \right)$
Next, by taking the LCM and by cross multiplying we get:
\begin{align} & \dfrac{x+y}{2}=\dfrac{1}{2}\left( \dfrac{7\times 2+15}{4} \right) \\ & \dfrac{x+y}{2}=\dfrac{1}{2}\left( \dfrac{14+15}{4} \right) \\ & \dfrac{x+y}{2}=\dfrac{1}{2}\times \dfrac{29}{4} \\ & \dfrac{x+y}{2}=\dfrac{29}{8} \\ \end{align}
Therefore, we can write:
$\dfrac{7}{2}<\dfrac{29}{8}<\dfrac{15}{4}$
Hence, the rational number is $\dfrac{29}{8}$.
Next, consider a rational number between $\dfrac{15}{4}$ and $4$.
Let $x=\dfrac{15}{4}$ and $y=4$.
$\dfrac{x+y}{2}=\dfrac{1}{2}\left( \dfrac{15}{4}+4 \right)$
Next, by taking the LCM and by cross multiplying we get:
\begin{align} & \dfrac{x+y}{2}=\dfrac{1}{2}\left( \dfrac{15+4\times 4}{4} \right) \\ & \dfrac{x+y}{2}=\dfrac{1}{2}\left( \dfrac{15+16}{4} \right) \\ & \dfrac{x+y}{2}=\dfrac{1}{2}\times \dfrac{31}{4} \\ & \dfrac{x+y}{2}=\dfrac{31}{8} \\ \end{align}
Therefore, we can write:
$\dfrac{15}{4}<\dfrac{31}{8}<4$
Hence, the rational number is $\dfrac{31}{8}$.
Therefore the six rational numbers between $3$ and $4$ are $\dfrac{7}{2},\dfrac{13}{4},\dfrac{27}{8},\dfrac{15}{4},\dfrac{29}{8}$ and $\dfrac{31}{8}$.

Note: An alternate method to find the rational numbers is by multiplying 7 in the numerator and denominator i.e. we get $\dfrac{21}{7}$ and $\dfrac{28}{7}$. Then find the rational numbers between them.