Question
Find five rational numbers between $\frac{3}{5}$ and $\frac{2}{3}.$
✓
Answer
n = 5 n + 1 = 6$\text{x}=\frac{3}{5},\text{y}=\frac{2}{3}$
$\text{d}=\frac{\text{y}-\text{x}}{\text{n}+1}=\frac{\frac{2}{3}-\frac{3}{5}}{6}=\frac{10-9}{90}=\frac{1}{90}$
Thus, rational numbers between $\frac{3}{5}$ and $\frac{2}{3}$ will be$(\text{x}+\text{d}),(\text{x}+2\text{d}),(\text{x}+3\text{d}),(\text{x}+4\text{d}),(\text{x}+5\text{d})$
$=\Big(\frac{3}{5}+\frac{1}{90}\Big),\Big(\frac{3}{5}+\frac{2}{90}\Big),\Big(\frac{3}{5}+\frac{3}{90}\Big),\\\Big(\frac{3}{5}+\frac{4}{90}\Big),\Big(\frac{3}{5}+\frac{5}{90}\Big)$
$=\Big(\frac{55}{90}\Big),\Big(\frac{56}{90}\Big),\Big(\frac{57}{90}\Big),\Big(\frac{58}{90}\Big),\Big(\frac{59}{90}\Big)$
$=\Big(\frac{11}{18}\Big),\Big(\frac{28}{45}\Big),\Big(\frac{19}{30}\Big),\Big(\frac{29}{46}\Big),\Big(\frac{59}{90}\Big)$
$\text{d}=\frac{\text{y}-\text{x}}{\text{n}+1}=\frac{\frac{2}{3}-\frac{3}{5}}{6}=\frac{10-9}{90}=\frac{1}{90}$
Thus, rational numbers between $\frac{3}{5}$ and $\frac{2}{3}$ will be$(\text{x}+\text{d}),(\text{x}+2\text{d}),(\text{x}+3\text{d}),(\text{x}+4\text{d}),(\text{x}+5\text{d})$
$=\Big(\frac{3}{5}+\frac{1}{90}\Big),\Big(\frac{3}{5}+\frac{2}{90}\Big),\Big(\frac{3}{5}+\frac{3}{90}\Big),\\\Big(\frac{3}{5}+\frac{4}{90}\Big),\Big(\frac{3}{5}+\frac{5}{90}\Big)$
$=\Big(\frac{55}{90}\Big),\Big(\frac{56}{90}\Big),\Big(\frac{57}{90}\Big),\Big(\frac{58}{90}\Big),\Big(\frac{59}{90}\Big)$
$=\Big(\frac{11}{18}\Big),\Big(\frac{28}{45}\Big),\Big(\frac{19}{30}\Big),\Big(\frac{29}{46}\Big),\Big(\frac{59}{90}\Big)$
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