Fill in the boxes with the correct symbol out of >, <, and =
(i) $\frac{- 5}{7} □ \frac{2}{3}$ (ii) $\frac{- 4}{5} □ \frac{- 5}{7}$ (iii) $\frac{- 7}{8} □ \frac{14}{- 16}$
(iv) $\frac{- 8}{5} □ \frac{- 7}{4}$ (v) $\frac{1}{- 3} □ \frac{- 1}{4}$ (vi) $\frac{5}{- 11} □ \frac{- 5}{11}$
(vii) $0 □ \frac{- 7}{6}$

Solution

(i)

$\frac{- 5}{7} = \frac{- 5 \times 3}{7 \times 3} = \frac{- 15}{21} \frac{2}{3} = \frac{2 \times 7}{3 \times 7} = \frac{14}{21}$

As −15 < 14,

Therefore, $\frac{- 5}{7} [<] \frac{2}{3}$

(ii)

$\frac{- 4}{5} = \frac{- 4 \times 7}{5 \times 7} = \frac{- 28}{35} \frac{- 5}{7} = \frac{- 5 \times 5}{7 \times 5} = \frac{- 25}{35}$

As −28 < −25

Therefore, $\frac{- 4}{5} [<] \frac{- 5}{7}$

(iii) Here, $\frac{14}{- 16} = \frac{7 \times 2}{- 8 \times 2} = \frac{7}{- 8} = \frac{- 7}{8}$

Therefore, $\frac{- 7}{8} [=] \frac{14}{- 16}$

(iv)

$\frac{- 8}{5} = \frac{- 8 \times 4}{5 \times 4} = \frac{- 32}{20} \frac{- 7}{4} = \frac{- 7 \times 5}{4 \times 5} = \frac{- 35}{20}$

As −32 > −35,

Therefore, $\frac{- 8}{5} [>] \frac{- 7}{4}$

(v)

$\frac{- 1}{3} = \frac{- 1 \times 4}{3 \times 4} = \frac{- 4}{12} \frac{- 1}{4} = \frac{- 1 \times 3}{4 \times 3} = \frac{- 3}{12}$

As −4 < −3,

Therefore, $\frac{- 1}{3} [<] \frac{- 1}{4}$

(vi) $\frac{5}{- 11} [=] \frac{- 5}{11}$

(vii) $0 [>] \frac{- 7}{6}$

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