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

Question:

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

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

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

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

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

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

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

Solution:

(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,

$\frac{-6}{13}=\frac{-6 \times 7}{13 \times 7}=\frac{-42}{91} \frac{-6}{13}=\frac{-6 \times 7}{13 \times 7}=\frac{-42}{91} \frac{-6}{13}=\frac{-6 \times 7}{13 \times 7}=\frac{-42}{91}$

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 $=\frac{5}{-8}=\frac{5 \times(-1)}{-8 \times(-1)}=\frac{-5}{8}$

$\mathrm{LCM}$ 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,

$-16<-15$

$\therefore \frac{-16}{24}<\frac{-15}{24}$

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 $0, \frac{3}{5}>0 \Rightarrow 0<\frac{3}{5}$.

(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,

$\frac{-9}{10}<\frac{-8}{9}$