Question
Find three rational numbers lying between $\frac{3}{5}$ and $\frac{7}{8}.$ How many rational numbers can be determined between these two numbers?
✓
Answer
$\text{x}=\frac{3}{5}$ and $\text{y}=\frac{7}{8}$
$\text{n}=3$
$\text{d}=\frac{(\text{y}-\text{x})}{\text{n}+1}=\frac{\frac{7}{8}-\frac{3}{5}}{3+1}=\frac{11}{40}\times\frac{1}{4}=\frac{11}{160}$ Rational numbers between $\text{x}=\frac{3}{5}$ and $\text{y}=\frac{7}{8}$ will be $(\text{x}+\text{d}),(\text{x}+2\text{d}),...,(\text{x}+\text{n}\text{d})$
$\Rightarrow\Big(\frac{3}{5}+\frac{11}{160}\Big),\Big(\frac{3}{5}+2\times\frac{11}{160}\Big),\Big(\frac{3}{5}+3\times\frac{11}{160}\Big)$
$\Rightarrow\Big(\frac{107}{160}\Big),\Big(\frac{118}{160}\Big),\Big(\frac{129}{160}\Big)$
$\Rightarrow\Big(\frac{107}{160}\Big),\Big(\frac{59}{80}\Big),\Big(\frac{129}{160}\Big)$
There are infinitely many rational numbers between two given rational numbers.
$\text{n}=3$
$\text{d}=\frac{(\text{y}-\text{x})}{\text{n}+1}=\frac{\frac{7}{8}-\frac{3}{5}}{3+1}=\frac{11}{40}\times\frac{1}{4}=\frac{11}{160}$ Rational numbers between $\text{x}=\frac{3}{5}$ and $\text{y}=\frac{7}{8}$ will be $(\text{x}+\text{d}),(\text{x}+2\text{d}),...,(\text{x}+\text{n}\text{d})$
$\Rightarrow\Big(\frac{3}{5}+\frac{11}{160}\Big),\Big(\frac{3}{5}+2\times\frac{11}{160}\Big),\Big(\frac{3}{5}+3\times\frac{11}{160}\Big)$
$\Rightarrow\Big(\frac{107}{160}\Big),\Big(\frac{118}{160}\Big),\Big(\frac{129}{160}\Big)$
$\Rightarrow\Big(\frac{107}{160}\Big),\Big(\frac{59}{80}\Big),\Big(\frac{129}{160}\Big)$
There are infinitely many rational numbers between two given rational numbers.
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