 # Find three rational numbers between `
Question:

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 $d=(y-x) /(n+1)$ , $x Solution: (i) Let$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\becaused=\frac{y-x}{n+1}\therefored=\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 $\frac{9}{40}, \frac{19}{80}, \frac{17}{80}$