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

Solution:

We know:

(i) Every positive rational number is greater than 0.

(ii) Every negative rational number is less than 0.

Thus, we have:

(i) $\frac{3}{8}$ is a positive rational number.

$\therefore \frac{3}{8}>0$

(ii) $\frac{-2}{9}$ is a negative rational number.

$\therefore \frac{-2}{9}<0$

(iii) $\frac{-3}{4}$ is a negative rational number.

$\therefore \frac{-3}{4}<0$

Also

$\frac{1}{4}$ is a positive rational number.

$\therefore \frac{1}{4}>0$

 Combining the two inequalities, we get:

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

(iv) Both $\frac{-5}{7}$ and $\frac{-4}{7}$ have the same denominator, that is, 7 .

So, we can directly compare the numerators.

$\therefore \frac{-5}{7}<\frac{-4}{7}$

(v) The two rational numbers are $\frac{2}{3}$ and $\frac{3}{4}$.

 The LCM of the denominators 3 and 4 is 12.

 Now

$\frac{2}{3}=\frac{2 \times 4}{3 \times 4}=\frac{8}{12}$

Also,

$\frac{3}{4}=\frac{3 \times 3}{4 \times 3}=\frac{9}{12}$

Further

$\frac{8}{12}<\frac{9}{12}$

$\therefore \frac{2}{3}<\frac{3}{4}$

(vi)The two rational numbers are $\frac{-1}{2}$ and $-1$.

We can write $-1=\frac{-1}{1}$.

The LCM of the denominators 2 and 1 is 2.

Now,

$\frac{-1}{2}=\frac{-1 \times 1}{2 \times 1}=\frac{-1}{2}$

Also,

$\frac{-1}{1}=\frac{-1 \times 2}{1 \times 2}=\frac{-2}{2}$

$\because \frac{-2}{1}<\frac{-1}{1}$

$\therefore-1<\frac{-1}{2}$