Question

Which of the two rational numbers is greater in the given pair?
(i) $\frac{3}{8}\mathrm{or}0$
(ii) $\frac{-2}{9}\mathrm{or}0$
(iii) $\frac{-3}{4}\mathrm{or}\frac{1}{4}$
(iv) $\frac{-5}{7}\mathrm{or}\frac{-4}{7}$
(v) $\frac{2}{3}\mathrm{or}\frac{3}{4}$
(vi) $\frac{-1}{2}\mathrm{or}-1$

Answer 1

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.
∴ $\frac{3}{8}>0$

(ii)$\frac{-2}{9}$ is a negative rational number.
∴ $\frac{-2}{9}<0$

(iii) $\frac{-3}{4}$ is a negative rational number.
∴ $\frac{-3}{4}<0$
Also,
$\frac{1}{4}$ is a positive rational number.
∴ $\frac{1}{4}>0$
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.

∴ $\frac{-5}{7}<\frac{-4}{7}$

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

∴$\frac{2}{3}<\frac{3}{4}$

(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,
$\frac{-1}{1}=\frac{-1\times 2}{1\times 2}=\frac{-2}{2}$
∵ $\frac{-2}{1}<\frac{-1}{1}$
∴ $-1<\frac{-1}{2}$