Question

Find 10 rational numbers between $\dfrac{-3}{4}$ and $\dfrac{5}{6}$.

Answer 1

Hint: We first explain the general conditions to find the in between rational numbers. We convert them in fractions and find the equal forms in fraction. We find the in between numbers of the numerators keeping the denominator fixed.

Complete step by step answer:
To find in between rational numbers we first transform the given fractions to their equivalent fraction form such that the denominator of the fractions remains same. Then we add 1 to the numbers of in between rational numbers we needed. Then we multiply the number with both denominator and numerator of the fractions.
The equivalent form of $\dfrac{-3}{4}$ and $\dfrac{5}{6}$ is $\dfrac{\left( -3 \right)\times 3}{4\times 3}=\dfrac{-9}{12}$ and $\dfrac{5\times 2}{6\times 2}=\dfrac{10}{12}$ respectively as the LCM of 4 and 6 is 12.

To find ten rational number between $\dfrac{-9}{12}$ and $\dfrac{10}{12}$, we add 1 to 10 and get $1+10=11$. Now we have to multiply 11 to both denominator and numerator of the fractions of $\dfrac{-9}{12}$ and $\dfrac{10}{12}$.
Multiplying 11 we get
$\dfrac{\left( -9 \right)\times 11}{12\times 11}=\dfrac{-99}{132}$ and $\dfrac{10\times 11}{12\times 11}=\dfrac{110}{132}$.
We can see that $\dfrac{-99}{132}$ and $\dfrac{110}{132}$ are basically the extended form of $\dfrac{-3}{4}$ and $\dfrac{5}{6}$ respectively.
We now take the in between numbers of the numerators keeping the denominator fixed.

So, the in between rational numbers are $\dfrac{-90}{132},\dfrac{-47}{132},\dfrac{-31}{132},\dfrac{-7}{132},0,\dfrac{11}{132},\dfrac{26}{132},\dfrac{49}{132},\dfrac{76}{132},\dfrac{100}{132}$.

Note: We need to remember that the requirement of converting into fraction is to keep the denominator same for all the fractions. So, in case of fractions we just need to care about their denominators. To simplify the particular given problem, we could have also taken the in between fractions of $\dfrac{-9}{12}$ and $\dfrac{10}{12}$ without multiplying 11.