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

Given:
$(i)$. $-\frac{5}{7}\square\ \frac{2}{3}$

$(ii)$. $-\frac{4}{5}\square\ -\frac{5}{7}$

$(iii)$. $-\frac{7}{8}\square\ \frac{14}{-16}$

$(iv)$. $-\frac{8}{5}\square\ -\frac{7}{4}$

$(v)$. $\frac{1}{-3}\square\ \frac{-1}{4}$

$(vi)$. $\frac{5}{-11}\ \square\ \frac{-5}{11}$

$(vii)$. $0\ \square\ \frac{-7}{6}$

To do: To fill in the boxes with the correct symbol out of >, <, and =.

Solution:

$(i)$. $-\frac{5}{7}\square \frac{2}{3}$

Here LCM of $7$ and $3$ is $21$.

So, $-\frac{5}{7}=-\frac{5\times 3}{7\times 3}$

$=-\frac{15}{21}$

And $\frac{2}{3}=\frac{(2\times 7)}{(3\times 7)}$

$=\frac{14}{21}$

On comparing both fractions:

$-\frac{15}{21}$ < $\frac{14}{21}$

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

$(ii)$. $\frac{-4}{5}\square -\frac{5}{7}$

Here LCM of $5$ and $7$ is $35$

$-\frac{4}{5}=\frac{(-4\times 7)}{(5\times 7)}$

$=-\frac{28}{35}$

And $-\frac{5}{7}=\frac{(-5\times 5)}{(7\times 5)}$

$=-\frac{25}{35}$

Here, $-\frac{28}{35}$ <$-\frac{25}{35}$

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

$(iii)$. $\frac{-7}{8}\square \frac{14}{-16}$

Here LCM of $8$ and $16$ is $16$.

So, $-\frac{7}{8}=\frac{(-7\times 2)}{(8\times 2)}$

$=-\frac{14}{16}$

And $\frac{14}{-16}=-\frac{14\times 1}{16\times 1}=-\frac{14}{16}$

On comparing,  $-\frac{14}{16}$=-$\frac{14}{16}$

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

$(iv).$ $-\frac{8}{5}\square -\frac{7}{4}$

Here LCM of $5$ and $4$ is $20$.

So, $-\frac{8}{5}=\frac{(-8\times 4)}{(5\times 4)}$

$=-\frac{32}{20}$

And $-\frac{7}{4}=\frac{(-7\times 5)}{(4\times 5)}$

$=-\frac{35}{20}$

Or $-\frac{32}{20}$ > $-\frac{35}{20}$

On comparing both rational numbers we have : $-\frac{8}{5}$ > $-\frac{7}{4}$

$(v)$. $\frac{1}{-3}\square -\frac{1}{4}$

Here LCM of $3$ and $4$ is $12$.

So, $\frac{1}{-3}=\frac{(1\times 4)}{(-3\times 4)}$

$=-\frac{4}{12}$

And $-\frac{1}{4}=\frac{(-1\times 3)}{(4\times 3)}$

$=-\frac{3}{12}$

$-\frac{4}{12}$   <   $-\frac{3}{12}$

On comparing both rational numbers we have: $\frac{1}{-3}$ < $-\frac{1}{4}$

$(vi)$. $\frac{5}{-11}\square -\frac{5}{11}$

On comparing both given rational numbers we have:

$-\frac{5}{11}$= $-\frac{5}{11}$

$(vii)$. $0\square \frac{-7}{6}$

Here, since all the negative numbers are less than zero, hence,

$0$ > $\frac{-7}{6}$

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Updated on: 10-Oct-2022

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