Question

Which of the two rational numbers is greater in each of the following pairs?

(i) $\frac{5}{9}\mathrm{or}\frac{-3}{-8}$
(ii) $\frac{4}{-3}\mathrm{or}\frac{-8}{7}$
(iii) $\frac{-12}{5}\mathrm{or}-3$
(iv) $\frac{7}{-9}\mathrm{or}\frac{-5}{8}$
(v) $\frac{4}{-5}\mathrm{or}\frac{-7}{8}$
(vi) $\frac{9}{-13}\mathrm{or}\frac{7}{-12}$

Answer 1

$\begin{array}{l}\left(\text{i}\right)\frac{5}{9},\frac{-3}{-8}\\ \frac{(-3)\times (-1)}{(-8)\times (-1)}=\frac{3}{8}\\ \text{L}.\text{C}.\text{M. of 9 and 8}is\text{72.}\\ \frac{5\times 8}{9\times 8}=\frac{40}{72}\\ \frac{3\times 9}{8\times 9}=\frac{27}{72}\\ 27<40\\ \frac{-3}{-8}<\frac{5}{9}\\ \text{So,}\frac{5}{9}\text{is greater}\text{.}\end{array}$
$\begin{array}{l}\left(\text{ii}\right)\\ \frac{4}{-3},\frac{-8}{7}\\ \mathrm{We}\mathrm{will}\mathrm{convert}\mathrm{each}\mathrm{negative}\mathrm{denominator}\mathrm{into}\mathrm{positive}.\\ \frac{4\times -1}{-3\times -1}=\frac{-4}{3}\\ \\ \begin{array}{l}\text{L}.\text{C}.\text{M. of 3 and 7 is 21.}\\ \\ \frac{-4\times \left(7\right)}{\left(3\right)\times \left(7\right)}=\frac{-28}{21}\\ \frac{(-8)\times 3}{7\times 3}=\frac{-24}{21}\\ (-\text{24})>(-\text{28})\\ \frac{-8}{7}>\frac{4}{(-3)}\\ \text{So,}\frac{-8}{7}\text{is greater}\text{.}\end{array}\end{array}$


$\begin{array}{l}\\ \left(\text{iii}\right)\\ \frac{-12}{5},-3\\ \begin{array}{l}\text{L}.\text{C}.\text{M. of 5 and 1 is 5.}\\ \\ \frac{-12\times 1}{5\times 1}=\frac{-12}{5}\\ \frac{-3\times 5}{1\times 5}=\frac{-15}{5}\\ -\text{12}>-15\\ \frac{-12}{5}>-3\\ \frac{-12}{5}\text{}\text{is greater.}\end{array}\\ \end{array}$
$\begin{array}{l}\left(\text{iv}\right)\\ \frac{-7}{9},\frac{-5}{8}\\ \text{L}.\text{C}.\text{M. of 9 and 8 is 72.}\\ \\ \frac{-7\times 8}{9\times 8}=\frac{-56}{72}\\ \frac{-5\times 9}{8\times 9}=\frac{-45}{72}\\ -\text{56}<-\text{45}\\ \frac{-7}{9}<\frac{-5}{8}\end{array}$

$\begin{array}{l}\left(\text{v}\right)\frac{4}{-5},\frac{-7}{8}\\ \mathrm{We}\mathrm{will}\mathrm{convert}\mathrm{each}\mathrm{negative}\mathrm{denominator}\mathrm{into}\mathrm{positive}.\\ \begin{array}{l}\frac{4\times -1}{-5\times -1}=\frac{-4}{5}\\ \begin{array}{l}\text{L}.\text{C}.\text{M. of 5 and 8 is 4}0.\\ \\ \frac{-4\times 8}{5\times 8}=\frac{-32}{40}\\ \frac{-7\times 5}{8\times 5}=\frac{-35}{40}\\ -\text{32}>-\text{35}\end{array}\\ \frac{-4}{5}>\frac{-7}{8}\\ \end{array}\end{array}$


$\begin{array}{l}\left(\text{vi}\right)\frac{9}{-13},\frac{7}{-12}\\ \mathrm{We}\mathrm{will}\mathrm{convert}\mathrm{each}\mathrm{negative}\mathrm{denominator}\mathrm{into}\mathrm{positive}.\\ \frac{9\times -1}{-13\times -1}=\frac{-9}{13}\\ \frac{7\times -1}{-12\times -1}=\frac{-7}{12}\\ \text{L}.\text{C}.\text{M. of 13 and 12 is}156.\\ \\ \frac{-9\times 12}{13\times (-12)}=\frac{-108}{156}\\ \frac{-7\times 13}{(-12)\times 13}=\frac{-91}{156}\\ -\text{1}0\text{8}<-\text{91}\\ \frac{9}{-13}<\frac{7}{-12}\end{array}$