Find three rational numbers between

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, $n=3$

$\because$                 $d=\frac{y-x}{n+1}$

$\therefore$               $d=\frac{\frac{6}{7}-\frac{5}{7}}{4}=\frac{\frac{1}{7}}{4}=\frac{1}{28}$

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 $\frac{9}{40}, \frac{19}{80}, \frac{17}{80}$