Which of the two rational numbers is greater in the given pair?

Solution:

1. The two rational numbers are $\frac{-4}{3}$ and $\frac{-8}{7}$.

The LCM of the denominators 3 and 7 is 21.

Now,

$\frac{-4}{3}=\frac{-4 \times 7}{3 \times 7}=\frac{-28}{21}$

Also,

$\frac{-8}{7}=\frac{-8 \times 3}{7 \times 3}=\frac{-24}{21}$

Further,

$\frac{-28}{21}<\frac{-24}{21}$

$\therefore \frac{-4}{3}<\frac{-8}{7}$

2. The two rational numbers are $\frac{7}{-9}$ and $\frac{-5}{8}$.

The first fraction can be expressed as $\frac{7}{-9}=\frac{7 \times-1}{-9 \times-1}=\frac{-7}{9}$.

The LCM of the denominators 9 and 8 is 72.

Now, 

$\frac{-7}{9}=\frac{-7 \times 8}{9 \times 8}=\frac{-56}{72}$

Also,

$\frac{-5}{8}=\frac{-5 \times 9}{8 \times 9}=\frac{-45}{72}$

Further,

$\frac{-56}{72}<\frac{-45}{72}$

$\therefore \frac{7}{-9}<\frac{-5}{8}$

3. The two rational numbers are $\frac{-1}{3}$ and $\frac{4}{-5}$.

$\frac{4}{-5}=\frac{4 \times-1}{-5 \times-1}=\frac{-4}{5}$

The LCM of the denominators 3 and 5 is 15.

Now, 

$\frac{-1}{3}=\frac{-1 \times 5}{3 \times 5}=\frac{-5}{15}$

Also,

$\frac{-4}{5}=\frac{-4 \times 3}{5 \times 3}=\frac{-12}{15}$

Further,

$\frac{-12}{15}<\frac{-5}{15}$

$\therefore \frac{4}{-5}<\frac{-1}{3}$

4. The two rational numbers are $\frac{9}{-13}$ and $\frac{7}{-12}$.

Now, $\frac{9}{-13}=\frac{9 \times-1}{-13 \times-1}=\frac{-9}{13}$ and $\frac{7}{-12}=\frac{7 \times-1}{-12 \times-1}=\frac{-7}{12}$

The LCM of the denominators 13 and 12 is 156.

Now, 

$\frac{-9}{13}=\frac{-9 \times 12}{13 \times 12}=\frac{-108}{156}$

Also,

$\frac{-7}{12}=\frac{-7 \times 13}{12 \times 13}=\frac{-91}{156}$

Further,

$\frac{-108}{156}<\frac{-91}{156}$

 

$\therefore \frac{9}{-13}<\frac{7}{-12}$

$\therefore \frac{9}{-13}<\frac{7}{-12}$

5. The two rational numbers are $\frac{4}{-5}$ and $\frac{-7}{10}$.

$\therefore \frac{4}{-5}=\frac{4 \times-1}{-5 \times-1}=\frac{-4}{5}$

The LCM of the denominators 5 and 10 is 10.

Now,

$\frac{-4}{5}=\frac{-4 \times 2}{5 \times 2}=\frac{-8}{10}$

Also,

$\frac{-7}{10}=\frac{-7 \times 1}{10 \times 1}=\frac{-7}{10}$

Further,

$\frac{-8}{10}<\frac{-7}{10}$

$\therefore \frac{-4}{5}<\frac{-7}{10}$, or,$\frac{4}{-5}<\frac{-7}{10}$

6. The two rational numbers are $\frac{-12}{5}$ and $-3$.

$-3$ can be written as $\frac{-3}{1}$.

The LCM of the denominators is 5.

Now,

$\frac{-3}{1}=\frac{-3 \times 5}{1 \times 5}=\frac{-15}{5}$

Because $\frac{-15}{5}<\frac{-12}{5}$, we can conclude that $-3<\frac{-12}{5}$.