Arrange the following rational numbers in ascending order:
(i) $\frac{4}{-9}, \frac{-5}{12}, \frac{7}{-18}, \frac{-2}{3}$
(ii) $\frac{-3}{4}, \frac{5}{-12}, \frac{-7}{16}, \frac{9}{-24}$
(iii) $\frac{3}{-5}, \frac{-7}{10}, \frac{-11}{15}, \frac{-13}{20}$
(iv) $\frac{-4}{7}, \frac{-9}{14}, \frac{13}{-28}, \frac{-23}{42}$
(i) We will write each of the given numbers with positive denominators.
We have:
$\frac{4}{-9}=\frac{4 \times(-1)}{-9 \times(-1)}=\frac{-4}{9}$ and $\frac{7}{-18}=\frac{7 \times(-1)}{-18 \times(-1)}=\frac{-7}{18}$
Thus, the given numbers are $\frac{-4}{9}, \frac{-5}{12}, \frac{-7}{18}$ and $\frac{-2}{3}$.
LCM of 9, 12, 18 and 3 is 36.
Now,
$\frac{-4}{9}=\frac{-4 \times 4}{9 \times 4}=\frac{-16}{36}$
$\frac{-5}{12}=\frac{-5 \times 3}{12 \times 3}=\frac{-15}{36}$
$\frac{-7}{18}=\frac{-7 \times 2}{18 \times 2}=\frac{-14}{36}$
$\frac{-2}{3}=\frac{-2 \times 12}{3 \times 12}=\frac{-24}{36}$
Clearly,
$\frac{-24}{36}<\frac{-16}{36}<\frac{-15}{36}<\frac{-14}{36}$
$\therefore \frac{-2}{3}<\frac{-4}{9}<\frac{-5}{12}<\frac{-7}{18}$
That is
$\frac{-2}{3}<\frac{4}{-9}<\frac{-5}{12}<\frac{7}{-18}$
(ii) We will write each of the given numbers with positive denominators.
We have:
$\frac{5}{-12}=\frac{5 \times(-1)}{-12 \times(-1)}=\frac{-5}{12}$ and $\frac{9}{-24}=\frac{9 \times(-1)}{-24 \times(-1)}=\frac{-9}{24}$
Thus, the given numbers are $\frac{-3}{4}, \frac{-5}{12}, \frac{-7}{16}$ and $\frac{-9}{24}$.
LCM of 4, 12, 16 and 24 is 48.
Now,
$\frac{-3}{4}=\frac{-3 \times 12}{4 \times 12}=\frac{-36}{48}$
$\frac{-5}{12}=\frac{-5 \times 4}{12 \times 4}=\frac{-20}{48}$
$\frac{-7}{16}=\frac{-7 \times 3}{16 \times 3}=\frac{-21}{48}$
$\frac{-9}{24}=\frac{-9 \times 2}{24 \times 2}=\frac{-18}{48}$
Clearly,
$\frac{-36}{48}<\frac{-21}{48}<\frac{-20}{48}<\frac{-18}{48}$
$\therefore \frac{-3}{4}<\frac{-7}{16}<\frac{-5}{12}<\frac{-9}{24}$
That is
$\frac{-3}{4}<\frac{-7}{16}<\frac{5}{-12}<\frac{9}{-24}$
(iii) We will write each of the given numbers with positive denominators.
We have:
$\frac{3}{-5}=\frac{3 \times(-1)}{-5 \times(-1)}=\frac{-3}{5}$
Thus, the given numbers are $\frac{-3}{5}, \frac{-7}{10}, \frac{-11}{15}$ and $\frac{-13}{20}$.
LCM of 5, 10, 15 and 20 is 60.
Now,
$\frac{-3}{5}=\frac{-3 \times 12}{5 \times 12}=\frac{-36}{60}$
$\frac{-7}{10}=\frac{-7 \times 6}{10 \times 6}=\frac{-42}{60}$
$\frac{-11}{15}=\frac{-11 \times 4}{15 \times 4}=\frac{-44}{60}$
$\frac{-13}{20}=\frac{-13 \times 3}{20 \times 3}=\frac{-39}{60}$
Clearly,
$\frac{-44}{60}<\frac{-42}{60}<\frac{-39}{60}<\frac{-36}{60}$
$\therefore \frac{-11}{15}<\frac{-7}{10}<\frac{-13}{20}<\frac{-3}{5} .$
That is
$\frac{-11}{15}<\frac{-7}{10}<\frac{-13}{20}<\frac{3}{-5}$
(iv) We will write each of the given numbers with positive denominators.
We have:
$\frac{13}{-28}=\frac{13 \times(-1)}{-28 \times(-1)}=\frac{-13}{28}$
Thus, the given numbers are $\frac{-4}{7}, \frac{-9}{14}, \frac{-13}{28}$ and $\frac{-23}{42}$.
LCM of 7, 14, 28 and 42 is 84.
Now,
$\frac{-4}{7}=\frac{-4 \times 12}{7 \times 12}=\frac{-48}{84}$
$\frac{-9}{14}=\frac{-9 \times 6}{14 \times 6}=\frac{-54}{84}$
$\frac{-13}{28}=\frac{-13 \times 3}{28 \times 3}=\frac{-39}{84}$
$\frac{-23}{42}=\frac{-23 \times 2}{42 \times 2}=\frac{-46}{84}$
Clearly,
$\frac{-54}{84}<\frac{-48}{84}<\frac{-46}{84}<\frac{-39}{84}$
$\therefore \frac{-9}{14}<\frac{-4}{7}<\frac{-23}{42}<\frac{-13}{28}$
That is
$\frac{-9}{14}<\frac{-4}{7}<\frac{-23}{42}<\frac{13}{-28}$