Question

Question 2
Find three rational numbers between
(i) -1 and -2
(ii) 0.1 and 0.11
(iii) $\frac{5}{7}$ and $\frac{6}{7}$
(iv) $\frac{1}{4}$ and $\frac{1}{5}$
Thinking process
Use the concept that three numbers between x and y are x+d, x+2d and x+3d , where
$\text{d}=\frac{y-x}{n+1},x<y$ and n =3

Answer 1

(i) Let x = - 2 and y = -1
Here, x < y and we have to find three rational numbers, so n=3.
$\because d=\frac{y-x}{n+1}=\frac{-1+2}{3+1}=\frac{1}{4}$
Since, the three rational numbers between x and y are x+d, x+2d and x+3d.
Now,
$x+d=-2+\frac{1}{4}=\frac{-8+1}{4}=-\frac{7}{4}x+2d=-2+\frac{2}{4}=\frac{-8+2}{4}=\frac{-6}{4}=\frac{-3}{2}$
$\text{and}x+3d=-2+\frac{3}{4}=\frac{-8+3}{4}=\frac{-5}{4}$
Hence, three rational numbers between -1 and -2 are
$\frac{-7}{4},\frac{-3}{2}$ and $\frac{-5}{4}$.

(ii) Let x = 0.1 and y = 0.11
Here x < y and we have to find three rational numbers, so consider, n = 3.
$\therefore d=\frac{y-x}{n+1}=\frac{0.11-0.1}{3+1}=\frac{0.01}{4}$
Since, the three rational numbers between x and y are (x+d),(x+2d) and (x+3d)
Now,
$x+d=0.1+\frac{0.01}{4}=\frac{0.4+0.01}{4}=0.1025x+2d=0.1+\frac{0.02}{4}=\frac{0.4+0.02}{4}=\frac{0.42}{4}=0.105\text{and}x+3d=0.1+\frac{0.03}{4}=\frac{0.4+0.03}{4}=\frac{0.43}{4}=0.01075$
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.

$\text{(iii) Let}x=\frac{5}{7}$ and $y=\frac{6}{7}$
Here x < y
Here, we have to find three rational numbers.
Consider, n=3
$\because d=\frac{y-x}{n+1}\because d=\frac{\frac{6}{7}-\frac{5}{7}}{4}=\frac{\frac{1}{7}}{4}=\frac{1}{28}$
Since, the three rational numbers between x and y are (x+d), (x+2d) and (x+3d)
Now,
$x+d=\frac{5}{7}+\frac{1}{28}=\frac{20+1}{28}=\frac{21}{28}$
$x+2d=\frac{5}{7}+\frac{2}{28}=\frac{20+2}{28}=\frac{22}{28}x+3d=\frac{5}{7}+\frac{3}{28}=\frac{20+3}{28}=\frac{23}{28}$
$\text{Hence, three rational numbers between}\frac{5}{7}\text{and}\frac{6}{7}\text{are}\frac{21}{28},\frac{22}{28},\frac{23}{28}$
Also, without using above formula, the three rational numbers
between $\frac{5}{7}$ and $\frac{6}{7}$ are $\frac{51}{70},\frac{52}{70},\frac{53}{70}$.

$\text{(iv) Let}x=\frac{1}{5}\text{and}y=\frac{1}{4}$
Here x<y
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 +2d and x +3d.
$\text{Now,}x+d=\frac{1}{5}+\frac{1}{80}=\frac{16+1}{80}=\frac{17}{80}$
$x+2d=\frac{1}{5}+\frac{2}{80}=\frac{16+2}{80}=\frac{18}{80}=\frac{9}{40}\text{and}x+3d=\frac{1}{5}+\frac{3}{80}=\frac{16+3}{80}=\frac{19}{80}$
$\text{Hence, three rational numbers between}\frac{1}{4}and\frac{1}{5}are\frac{17}{80},\frac{9}{40},\frac{19}{80}$.