Question

Fill in the blanks with the correct symbol out of $>,=$ and $<:$

(i) $-\frac{3}{7}......\frac{6}{-13}$

(ii) $\frac{5}{-13}......-\frac{35}{91}$

(iii) $-\mathrm{2......}-\frac{13}{5}$

(iv) $-\frac{2}{3}......\frac{-5}{8}$

(v) $\mathrm{0......}\frac{-3}{-5}$

(vi) $-\frac{8}{9}......-\frac{9}{10}$

Answer 1

(i) We will write each of the given numbers with positive denominators.
One number $=-\frac{3}{7}$
Other number $=\frac{6}{-13}$
$=\frac{6\times (-1)}{-13\times (-1)}=\frac{-6}{13}$
LCM of $7$ and $13=91$
$\therefore -\frac{3}{7}=-\frac{3\times 13}{7\times 13}=-\frac{39}{91}$
And,
$-\frac{6}{13}=-\frac{6\times 7}{13\times 7}=-\frac{42}{91}$
Clearly, $-39>-41$
$\therefore -\frac{39}{91}>-\frac{42}{91}$
Thus, $-\frac{3}{7}>\frac{6}{-13}$
(ii) We will write each of the given numbers with positive denominators.
One number $=\frac{5}{-13}=\frac{5\times (-1)}{-13\times (-1)}=-\frac{5}{13}$
Other number $=-\frac{35}{91}$
LCM of $13$ and $91=91$

$\therefore -\frac{5}{13}=-\frac{5\times 7}{13\times 7}=-\frac{35}{91}$ and $-\frac{35}{91}$
Clearly,$-35=-35$
$\therefore -\frac{35}{91}=-\frac{35}{91}$
Thus,$-\frac{5}{13}=-\frac{35}{91}$

(iii) We will write each of the given numbers with positive denominators.
One number $=-2$
We can write $-2$ as$-\frac{2}{1}.$
Other number $=-\frac{13}{5}$
LCM of $1$ and $5=5$
​$\therefore -\frac{2}{1}=-\frac{2\times 5}{1\times 5}=-\frac{10}{5}$ and $\frac{-13}{5}=-\frac{13\times 1}{5\times 1}=-\frac{13}{5}$
Clearly, $-10>-13$
$\therefore -\frac{10}{5}>-\frac{13}{5}$
Thus,$-\frac{2}{1}>-\frac{13}{5}$
$-2>-\frac{13}{5}$
(iv) We will write each of the given numbers with positive denominators.
One number $=-\frac{2}{3}$
Other number $=\frac{5}{-8}=\frac{5\times (-1)}{-8\times (-1)}=-\frac{5}{8}$
LCM of $3$ and $8=24$

$\therefore -\frac{2}{3}=-\frac{2\times 8}{3\times 8}=-\frac{16}{24}$ and $-\frac{5}{8}=\frac{-5\times 3}{8\times 3}=-\frac{15}{24}$
Clearly, $-16<-15$
$-\frac{16}{24}<-\frac{15}{24}$
Thus,$-\frac{2}{3}<-\frac{5}{8}$
$-\frac{2}{3}<\frac{5}{-8}$
(v) $\frac{-3}{-5}=\frac{-3\times -1}{-5\times -1}=\frac{3}{5}$
$\frac{3}{5}$ is a positive number.
Because every positive rational number is greater than $0$
$\Rightarrow \frac{3}{5}>0$
$\therefore 0<\frac{3}{5}$.
(vi) We will write each of the given numbers with positive denominators.
One number $=-\frac{8}{9}$
Other number $=-\frac{9}{10}$
LCM of $9$ and $10$ = $90$

$\therefore -\frac{8}{9}=-\frac{8\times 10}{9\times 10}=-\frac{80}{90}$ and $\frac{-9}{10}=-\frac{9\times 9}{10\times 9}=-\frac{81}{90}$
Clearly,$-81<-80$
$\therefore -\frac{81}{90}<-\frac{80}{90}$
Thus,$-\frac{9}{10}<-\frac{8}{9}$