Question

Verify commutativty of addition of rational numbers for each of the following pairs of rotional numbers:
(i) $\frac{-11}{5}\mathrm{and}\frac{4}{7}$

(ii) $\frac{4}{9}\mathrm{and}\frac{7}{-12}$

(iii) $\frac{-3}{5}\mathrm{and}\frac{-2}{-15}$

(iv) $\frac{2}{-7}\mathrm{and}\frac{12}{-35}$

(v) $4\mathrm{and}\frac{-3}{5}$

(vi) $-4\mathrm{and}\frac{4}{-7}$

Answer 1

$$\begin{gathered} C\mathrm{ommutativity}\mathrm{of}the\mathrm{addition}\mathrm{of}\mathrm{rational}\mathrm{numbers}\mathrm{means}that\mathrm{if}\frac{\mathrm{a}}{\mathrm{b}}and\frac{\mathrm{c}}{\mathrm{d}}\mathrm{are}\mathrm{two}\mathrm{rational}\mathrm{numbers},\mathrm{then}\frac{\mathrm{a}}{\mathrm{b}}+\frac{\mathrm{c}}{\mathrm{d}}=\frac{\mathrm{c}}{\mathrm{d}}+\frac{\mathrm{a}}{\mathrm{b}}. \\ \left(\mathrm{i}\right) \\ \text{We have}\frac{-11}{5}\mathrm{and}\frac{4}{7}. \\ \therefore \frac{-11}{5}+\frac{4}{7}=\frac{-11\times 7}{5\times 7}+\frac{4\times 5}{7\times 5}=\frac{-77}{35}+\frac{20}{35}=\frac{-77+20}{35}=\frac{-57}{35} \\ \frac{4}{7}+\frac{-11}{5}=\frac{4\times 5}{7\times 5}+\frac{-11\times 7}{5\times 7}=\frac{20}{35}+\frac{-77}{35}=\frac{20-77}{35}=\frac{-57}{35} \\ \therefore \frac{-11}{5}+\frac{4}{7}=\frac{4}{7}+\frac{-11}{5} \\ \mathrm{Hence}\mathrm{verified}. \\ \left(\mathrm{ii}\right) \\ \text{We have}\frac{4}{9}\mathrm{and}\frac{7}{-12}. \\ \therefore \frac{4}{9}+\frac{-7}{12}=\frac{4\times 4}{9\times 4}+\frac{-7\times 3}{12\times 3}=\frac{16}{36}+\frac{-21}{36}=\frac{16-21}{36}=\frac{-5}{36} \\ \frac{-7}{12}+\frac{4}{9}=\frac{-7\times 3}{12\times 3}+\frac{4\times 4}{9\times 4}=\frac{-21}{36}+\frac{16}{36}=\frac{-21+16}{36}=\frac{-5}{36} \\ \therefore \frac{4}{9}+\frac{-7}{12}=\frac{-7}{12}+\frac{4}{9} \\ \mathrm{Hence}\mathrm{verified}. \\ \left(\mathrm{iii}\right) \\ \text{We have}\frac{-3}{5}\mathrm{and}\frac{-2}{-15}\text{or}\frac{-3}{5}\mathrm{and}\frac{2}{15}. \\ \therefore \frac{-3}{5}+\frac{2}{15}=\frac{-9}{15}+\frac{2}{15}=\frac{-9+2}{15}=\frac{-7}{15} \\ \frac{2}{15}+\frac{-3}{5}=\frac{2}{15}+\frac{-9}{15}=\frac{2-9}{15}=\frac{-7}{15} \\ \therefore \frac{-3}{5}+\frac{-2}{-15}=\frac{-2}{-15}+\frac{-3}{5} \\ \mathrm{Hence}\mathrm{verified}. \\ \left(\mathrm{iv}\right) \\ \text{We have}\frac{2}{-7}\mathrm{and}\frac{12}{-35}. \\ \therefore \frac{-2}{7}+\frac{-12}{35}=\frac{-2\times 5}{7\times 5}+\frac{-12}{35}=\frac{-10-12}{35}=\frac{-22}{35} \\ \frac{12}{-35}+\frac{2}{-7}=\frac{-12}{35}+\frac{-2\times 5}{7\times 5}=\frac{-12-10}{35}=\frac{-22}{35} \\ \therefore \frac{2}{-7}+\frac{12}{-35}=\frac{12}{-35}+\frac{2}{-7} \\ \mathrm{Hence}\mathrm{verified}. \\ \left(\mathrm{v}\right) \\ \text{We have}4\mathrm{and}\frac{-3}{5}. \\ \therefore 4+\frac{-3}{5}=\frac{4\times 5}{1\times 5}+\frac{-3}{5}=\frac{20-3}{5}=\frac{17}{5} \\ \frac{-3}{5}+4=\frac{-3}{5}+\frac{4\times 5}{1\times 5}=\frac{-3+20}{5}=\frac{17}{5} \\ \therefore 4+\frac{-3}{5}=\frac{-3}{5}+4 \\ \mathrm{Hence}\mathrm{verified}. \\ \left(\mathrm{vi}\right) \\ \text{We have}-4\mathrm{and}\frac{4}{-7}. \\ \therefore \frac{-4}{1}+\frac{-4}{7}=\frac{-4\times 7}{1\times 7}+\frac{-4}{7}=\frac{-28-4}{7}=\frac{-32}{7} \\ \frac{-4}{7}+\frac{-4}{1}=\frac{-4}{7}+\frac{-4\times 7}{1\times 7}=\frac{-4-28}{7}=\frac{-32}{7} \\ \therefore -4+\frac{4}{-7}=\frac{4}{-7}-4 \\ \mathrm{Hence}\mathrm{verified}. \end{gathered}$$