Find three rational numbers between
(i) $-1$ and $-2$
(ii) $0.1$ and $0.11$
(iii) $5 / 7$ and $6 / 7$
(iv) $1 / 4$ and $1 / 5$
Thinking Process
Use the concept that three rational numbers between $x$ and $y$ are $x+d, x+2 d$ and $x+3 d$, where $d=(y-x) /(n+1)$ , $x
(i) Let $y=-1$ and $x=-2$
Here $x<$ y and we have to find three rational numbers, so $n=3$.
$\because \quad d=\frac{y-x}{n+1}=\frac{-1+2}{3+1}=\frac{1}{4}$
Since, the three rational numbers between $x$ and $y$ are $x+d, x+2 d$ and $x+3 d$.
Now, $x+d=-2+\frac{1}{4}=\frac{-8+1}{4}=-\frac{7}{4}$
$x+2 d=-2+\frac{2}{4}=\frac{-8+2}{4}=\frac{-6}{4}=\frac{-3}{2}$
and $x+3 d=-2+\frac{3}{4}=\frac{-8+3}{4}=\frac{-5}{4}$
Hence, three rational numbers between $-1$ and $-2$ are $\frac{-7}{4}, \frac{-3}{2}$ and $\frac{-5}{4}$.
Alternate Method
Let $x=-1$ and $y=-2$
We kno'w, a rational number between $x$ and $y=\frac{x+y}{2}$
$\therefore$ A rational number between $-1$ and $-2=\frac{-1-2}{2}=-\frac{3}{2}$
and a rational number between $-1$ and $-\frac{3}{2}=\frac{-1-\frac{3}{2}}{2}=\frac{-2-3}{4}=-\frac{5}{4}$
Similarly, $-\frac{7}{4}$ is a rational number between $-1$ and $-2$.
Hence, required solution $=-\frac{3}{2},-\frac{5}{4},-\frac{7}{4}$
(ii) Let $x=0.1$ and $y=0.11$
Here, $x $\therefore$ $d=\frac{y-x}{n+1}=\frac{0.11-0.1}{3+1}=\frac{0.01}{4}$ Since, the three rational numbers between $x$ and $y \operatorname{are}(x+d),(x+2 d)$ and $(x+3 d)$. Now, $\quad x+d=0.1+\frac{0.01}{4}=\frac{0.4+0.01}{4}=\frac{0.41}{4}=0.1025$ $x+2 d=0.1+\frac{0.02}{4}=\frac{0.4+0.02}{4}=\frac{0.42}{4}=0.105$ and $x+3 d=0.1+\frac{0.03}{4}=\frac{0.4+0.03}{4}=\frac{0.43}{4}=0.1075$ Hence, three rational numbers between $0.1$ and $0.11$ are $0.1025,0.105,0.1075$. Also, without using above formula the three rational numbers between $0.1$ and $0.11$ are $0.101,0.102,0.103 .$ (iii) Let $\quad x=\frac{5}{7}$ and $y=\frac{6}{7}$ Here, $x Here, we have to find three rational numbers. Consider, $n=3$ $\because$ $d=\frac{y-x}{n+1}$ $\therefore$ $d=\frac{\frac{6}{7}-\frac{5}{7}}{4}=\frac{\frac{1}{7}}{4}=\frac{1}{28}$ Since, the three rational numbers between $x$ and $y$ are $(x+d)_{1}(x+2 d)$ and $(x+3 d)$. Now, $x+d=\frac{5}{7}+\frac{1}{28}=\frac{20+1}{28}=\frac{21}{28}$ $x+2 d=\frac{5}{7}+\frac{2}{28}=\frac{20+2}{28}=\frac{22}{28}$ and $x+3 d=\frac{5}{7}+\frac{3}{28}=\frac{20+3}{28}=\frac{23}{28}$ Hence, three rational numbers between $\frac{5}{7}$ and $\frac{6}{7}$ are $\frac{21}{28}, \frac{22}{28}, \frac{23}{28}$ Also, without using above formula, the three rational numbers between $\frac{5}{7}$ and $\frac{6}{7}$ are $\frac{51}{70}, \frac{52}{70}, \frac{53}{70}$ (iv) Let $x=\frac{1}{5}$ and $y=\frac{1}{4}$ Here, $x Here, we have to find three rational numbers. Consider, $n=3$ $\because$ $d=\frac{y-x}{n+1}=\frac{\frac{1}{4}-\frac{1}{5}}{3+1}=\frac{\frac{5-4}{20}}{4}=\frac{1}{80}$ Since, the three rational numbers between $x$ and $y$ are $x+d, x+2 d$ and $x+3 d$. Now, $x+d=\frac{1}{5}+\frac{1}{80}=\frac{16+1}{80}=\frac{17}{80}$ $x+2 d=\frac{1}{5}+\frac{2}{80}=\frac{16+2}{80}=\frac{18}{80}=\frac{9}{40}$ and $x+3 d=\frac{1}{5}+\frac{3}{80}=\frac{16+3}{80}=\frac{19}{80}$ Hence, three rational numbers between $\frac{1}{4}$ and $\frac{1}{5}$ are $\frac{17}{80}, \frac{9}{40}, \frac{19}{80}$. Alternate Method Let $x=\frac{1}{4}$ and $y=\frac{1}{5}$ So, a rational number between $x$ and $y=\frac{x+y}{2}$ $\therefore$ A rational number between $\frac{1}{4}$ and $\frac{1}{5}=\frac{\frac{1}{4}+\frac{1}{5}}{2}=\frac{\frac{5+4}{20}}{2}$ $=\frac{9}{2 \times 20}=\frac{9}{40}$ Again, a rational number between $\frac{1}{4}$ and $\frac{9}{40}=\frac{\frac{1}{4}+\frac{9}{40}}{2}=\frac{\frac{10+9}{40}}{2}=\frac{19}{2 \times 40}=\frac{19}{80}$ Again, a rational number between $\frac{1}{5}$ and $\frac{9}{40}=\frac{\frac{1}{5}+\frac{9}{40}}{2}=\frac{\frac{8+9}{40}}{2}=\frac{\frac{17}{40}}{2}=\frac{17}{40 \times 2}=\frac{17}{80}$ Hence, three rational numbers between $\frac{1}{4}$ and $\frac{1}{5}$ are $\frac{9}{40}, \frac{19}{80}, \frac{17}{80}$