Question

Which of the two rational numbers in each of the following pairs of rational numbers is greater?
(i) $\frac{-3}{8},0$
(ii) $\frac{5}{2},0$
(iii) $\frac{-4}{11},\frac{3}{11}$
(iv) $\frac{-7}{12},\frac{5}{-8}$
(v) $\frac{4}{-9},\frac{-3}{-7}$
(vi) $\frac{-5}{8},\frac{3}{-4}$
(vii) $\frac{5}{9},\frac{-3}{-8}$
(viii) $\frac{5}{-8},\frac{-7}{12}$

Answer 1

(i) We know that every positive rational number is greater than zero and every negative rational number is smaller than zero. Thus,

$\frac{-3}{8}>0$
(ii) $\frac{5}{2}>0.$ Because every positive rational number is greater than zero and every negative rational number is smaller than zero.
(iii) $\frac{-4}{8}<\frac{3}{11}.$ Because every positive rational number is greater than zero and every negative rational number is smaller than zero.
(iv)

$$\begin{gathered} \frac{-7}{12}=\frac{-7\times 2}{12\times 2}=\frac{-14}{24}\mathrm{and}\frac{5}{-8}=\frac{-5\times 3}{8\times 3}=\frac{-15}{24} \\ \mathrm{Therefore},\frac{-7}{12}>\frac{5}{-8} \end{gathered}$$

(v)

$$\begin{gathered} \frac{4}{-9}=\frac{-4\times 7}{9\times 7}=\frac{-28}{63}\mathrm{and}\frac{-3}{-7}=\frac{3\times 7}{7\times 9}=\frac{21}{63} \\ \mathrm{Therefore},\frac{4}{-9}<\frac{-3}{-7} \end{gathered}$$
(vi)

$$\begin{gathered} \frac{-5}{8}\mathrm{and}\frac{3}{-4}=\frac{-3\times 2}{4\times 2}=\frac{-6}{8} \\ \mathrm{Therefore},\frac{-5}{8}>\frac{3}{-4} \end{gathered}$$

(vii)

$$\begin{gathered} \frac{5}{9}=\frac{5\times 8}{9\times 8}=\frac{40}{72}\mathrm{and}\frac{-3}{-8}=\frac{3\times 9}{8\times 9}=\frac{27}{72} \\ \mathrm{Therefore},\frac{5}{9}>\frac{-3}{-8} \end{gathered}$$

(viii)
$$\begin{gathered} \frac{-7}{12}=\frac{-7\times 2}{12\times 2}=\frac{-14}{24}\mathrm{and}\frac{5}{-8}=\frac{-5\times 3}{8\times 3}=\frac{-15}{24} \\ \mathrm{Therefore},\frac{-7}{12}>\frac{5}{-8} \end{gathered}$$