Question

Find ten rational numbers between $-\frac{3}{4}$ and $\frac{5}{6}$

Answer 1

Find the rational number between two rational numbers

Rational number is a type of real numbers, which is in the form of $\raisebox{1ex}{$p$}\!\left/ \!\raisebox{-1ex}{$q$}\right.$ where $q$ not equal to $0$.

Any fraction with non-zero denominators is a rational number.

Let $x$ be a rational number such that $x$ is less than ${\frac{1}{10}}^{th}$of the difference between the two number.

$$\begin{gathered} \therefore x<\frac{1}{10}·\left(\frac{5}{6}+\frac{3}{4}\right) \\ \Rightarrow x<\frac{1}{10}·\frac{19}{12} \\ \Rightarrow x<\frac{19}{120} \end{gathered}$$

Say $x=\frac{1}{12}$.

From the property of rational number we know that sum of two rational number is a rational number.

So we can say the rational in between $-\frac{3}{4}$ and $\frac{5}{6}$ are of form $-\frac{3}{4}+n\frac{1}{12}$ where $n=1,2,...,10$

$\therefore$the rational numbers are

$-\frac{3}{4}+\frac{1}{12},-\frac{3}{4}+\frac{2}{12},-\frac{3}{4}+\frac{3}{12},-\frac{3}{4}+\frac{4}{12},-\frac{3}{4}+\frac{5}{12},-\frac{3}{4}+\frac{6}{12},-\frac{3}{4}+\frac{7}{12},-\frac{3}{4}+\frac{8}{12},-\frac{3}{4}+\frac{9}{12},-\frac{3}{4}+\frac{10}{12},$

$=-\frac{2}{3},-\frac{7}{12},-\frac{1}{2},-\frac{5}{12},-\frac{1}{3},-\frac{1}{4},-\frac{1}{6},-\frac{1}{12},0,\frac{1}{12}$

Hence, the ten rational numbers between $-\frac{3}{4}$ and $\frac{5}{6}$ are $-\frac{2}{3},-\frac{7}{12},-\frac{1}{2},-\frac{5}{12},-\frac{1}{3},-\frac{1}{4},-\frac{1}{6},-\frac{1}{12},0,\frac{1}{12}$.