# Arrange the following rational numbers in descending order:

Question:

Arrange the following rational numbers in descending order:

(i) $-2, \frac{-13}{6}, \frac{8}{-3}, \frac{1}{3}$

(ii) $\frac{-3}{10}, \frac{7}{-15}, \frac{-11}{20}, \frac{14}{-30}$

(iii) $\frac{-5}{6}, \frac{-7}{12}, \frac{-13}{18}, \frac{23}{-24}$

(iv) $\frac{-10}{11}, \frac{-19}{22}, \frac{-23}{33}, \frac{-39}{44}$

Solution:

(i) We will first write each of the given numbers with positive denominators. We have:

$\frac{8}{-3}=\frac{8 \times(-1)}{-3 \times(-1)}=\frac{-8}{3}$

Thus, the given numbers are $-2, \frac{-13}{6}, \frac{-8}{3}$ and $\frac{1}{3}$

LCM of 1, 6, 3 and 3 is 6

Now,

$\frac{-2}{1}=\frac{-2 \times 6}{1 \times 6}=\frac{-12}{6}$

$\frac{-13}{6}=\frac{-13 \times 1}{6 \times 1}=\frac{-13}{6}$

$\frac{-8}{3}=\frac{-8 \times 2}{3 \times 2}=\frac{-16}{6}$

and

$\frac{1}{3}=\frac{1 \times 2}{3 \times 2}=\frac{2}{6}$

Clearly,Thus,

$\frac{2}{6}>\frac{-12}{6}>\frac{-13}{6}>\frac{-16}{6}$

$\therefore \frac{1}{3}>-2>\frac{-13}{6}>\frac{-8}{3} .$ i.e $\frac{1}{3}>-2>\frac{-13}{6}>\frac{8}{-3}$

(ii) We will first write each of the given numbers with positive denominators. We have:

$\frac{7}{-15}=\frac{7 \times(-1)}{-15 \times(-1)}=\frac{-7}{15}$ and $\frac{17}{-30}=\frac{17 \times(-1)}{-30 \times(-1)}=\frac{-17}{30}$

Thus, the given numbers are $\frac{-3}{10}, \frac{-7}{15}, \frac{-11}{20}$ and $\frac{-17}{30}$

LCM of 10, 15, 20 and 30 is 60

Now,

$\frac{-3}{10}=\frac{-3 \times 6}{10 \times 6}=\frac{-18}{60}$

$\frac{-7}{15}=\frac{-7 \times 4}{15 \times 4}=\frac{-28}{60}$

$\frac{-11}{20}=\frac{-11 \times 3}{20 \times 3}=\frac{-33}{60}$

and

$\frac{-17}{30}=\frac{-17 \times 2}{30 \times 2}=\frac{-34}{60}$

Clearly,

$\frac{-18}{60}>\frac{-28}{60}>\frac{-33}{60}>\frac{-34}{60}$

$\therefore \frac{-3}{10}>\frac{-7}{15}>\frac{-11}{20}>\frac{-17}{30} .$ i.e $\frac{-3}{10}>\frac{7}{-15}>\frac{-11}{20}>\frac{17}{-30}$

(iii) We will first write each of the given numbers with positive denominators. We have:

$\frac{23}{-24}=\frac{23 \times(-1)}{-24 \times(-1)}=\frac{-23}{24}$

Thus, the given numbers are $\frac{-5}{6}, \frac{-7}{12}, \frac{-13}{18}$ and $\frac{-23}{24}$

LCM of 6, 12, 18 and 24 is 72

Now,

$\frac{-5}{6}=\frac{-5 \times 12}{6 \times 12}=\frac{-60}{72}$

$\frac{-7}{12}=\frac{-7 \times 6}{12 \times 6}=\frac{-42}{72}$

$\frac{-13}{18}=\frac{-13 \times 4}{18 \times 4}=\frac{-52}{72}$

and

$\frac{-23}{24}=\frac{-23 \times 3}{24 \times 3}=\frac{-69}{72}$

Clearly,

$\frac{-42}{72}>\frac{-52}{72}>\frac{-60}{72}>\frac{-69}{72}$

$\therefore \frac{-7}{12}>\frac{-13}{18}>\frac{-5}{6}>\frac{-23}{24}$. i.e $\frac{-7}{12}>\frac{-13}{18}>\frac{-5}{6}>\frac{23}{-24}$

(iv) The given numbers are $\frac{-10}{11}, \frac{-19}{22}, \frac{-23}{33}$ and $\frac{-39}{44}$

LCM of 11, 22, 33 and 44 is 132

Now,

$\frac{-10}{11}=\frac{-10 \times 12}{11 \times 12}=\frac{-120}{132}$

$\frac{-19}{22}=\frac{-19 \times 6}{22 \times 6}=\frac{-114}{132}$

$\frac{-23}{33}=\frac{-23 \times 4}{33 \times 4}=\frac{-92}{132}$

and

$\frac{-39}{44}=\frac{-39 \times 3}{44 \times 3}=\frac{-117}{132}$

Clearly,

$\frac{-92}{132}>\frac{-114}{132}>\frac{-117}{132}>\frac{-120}{132}$

$\therefore \frac{-23}{33}>\frac{-19}{22}>\frac{-39}{44}>\frac{-10}{11}$