Question

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

Answer 1

(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 $\frac{-4}{9},\frac{-5}{12},\frac{-7}{18}\mathrm{and}\frac{-2}{3}.$

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,

$\frac{-24}{36}<\frac{-16}{36}<\frac{-15}{36}<\frac{-14}{36}$

∴ ​$\frac{-2}{3}<\frac{-4}{9}<\frac{-5}{12}<\frac{-7}{18}$

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 $\frac{-3}{4},\frac{-5}{12},\frac{-7}{16}\mathrm{and}\frac{-9}{24}.$

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,

$\frac{-36}{48}<\frac{-21}{48}<\frac{-20}{48}<\frac{-18}{48}$

∴​ $\frac{-3}{4}<\frac{-7}{16}<\frac{-5}{12}<\frac{-9}{24}$

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 $\frac{-3}{5},\frac{-7}{10},\frac{-11}{15}\mathrm{and}\frac{-13}{20}.$

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,

$\frac{-44}{60}<\frac{-42}{60}<\frac{-39}{60}<\frac{-36}{60}$

∴​ $\frac{-11}{15}<\frac{-7}{10}<\frac{-13}{20}<\frac{-3}{5}$.

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 $\frac{-4}{7},\frac{-9}{14},\frac{-13}{28}\mathrm{and}\frac{-23}{42}.$

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,

$\frac{-54}{84}<\frac{-48}{84}<\frac{-46}{84}<\frac{-39}{84}$

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

That is

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