Arrange the following rational numbers in ascending order:

Solution:

(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

$\frac{-9}{14}<\frac{-4}{7}<\frac{-23}{42}<\frac{13}{-28}$