Question

Verify associativity of addition of rational numbers i.e., (x + y) + z = x + (y + z), when:
(i) $x=\frac{1}{2},y=\frac{2}{3},z=-\frac{1}{5}$
(ii) $x=\frac{-2}{5},y=\frac{4}{3},z=\frac{-7}{10}$
(iii) $x=\frac{-7}{11},y=\frac{2}{-5},z=\frac{-3}{22}$
(iv) $x=-2,y=\frac{3}{5},z=\frac{-4}{3}$

Answer 1

$$\begin{gathered} \mathrm{We}\mathrm{have}\mathrm{to}\mathrm{verify}\mathrm{that}(\mathrm{x}+\mathrm{y})+\mathrm{z}=\mathrm{x}+(\mathrm{y}+\mathrm{z}). \\ \left(\mathrm{i}\right) \\ x=\frac{1}{2},y=\frac{2}{3},z=\frac{-1}{5} \\ (\mathrm{x}+\mathrm{y})+\mathrm{z}=(\frac{1}{2}+\frac{2}{3})+\frac{-1}{5}=(\frac{3}{6}+\frac{4}{6})+\frac{-1}{5}=\frac{7}{6}+\frac{-1}{5}=\frac{35}{30}+\frac{-6}{30}=\frac{35-6}{30}=\frac{29}{30} \\ \mathrm{x}+(\mathrm{y}+\mathrm{z})=\frac{1}{2}+(\frac{2}{3}+\frac{-1}{5})=\frac{1}{2}+(\frac{10}{15}+\frac{-3}{15})=\frac{1}{2}+\frac{7}{15}=\frac{15}{30}+\frac{14}{30}=\frac{15+14}{30}=\frac{29}{30} \\ \therefore (\frac{1}{2}+\frac{2}{3})+\frac{-1}{5}=\frac{1}{2}+(\frac{2}{3}+\frac{-1}{5}) \\ \mathrm{Hence}\mathrm{verified}. \\ \left(\mathrm{ii}\right) \\ x=\frac{-2}{5},y=\frac{4}{3},z=\frac{-7}{10} \\ (\mathrm{x}+\mathrm{y})+\mathrm{z}=(\frac{-2}{5}+\frac{4}{3})+\frac{-7}{10}=(\frac{-6}{15}+\frac{20}{15})+\frac{-7}{10}=\frac{14}{15}+\frac{-7}{10}=\frac{28}{30}+\frac{-21}{30}=\frac{28-21}{30}=\frac{7}{30} \\ x+(y+z)=\frac{-2}{5}+(\frac{4}{3}+\frac{-7}{10})=\frac{-2}{5}+(\frac{40}{30}+\frac{-21}{30})=\frac{-2}{5}+\frac{19}{30}=\frac{-12}{30}+\frac{19}{30}=\frac{-12+19}{30}=\frac{7}{30} \\ \therefore \frac{-2}{5}+\frac{4}{3})+\frac{-7}{10}=\frac{-2}{5}+(\frac{4}{3}+\frac{-7}{10}) \\ \mathrm{Hence}\mathrm{verified}. \\ (\mathrm{iii}) \\ x=\frac{-7}{11},y=\frac{2}{-5},z=\frac{-3}{22} \\ (x+y)+z=(\frac{-7}{11}+\frac{2}{-5})+\frac{-3}{22}=(\frac{-35}{55}+\frac{-22}{55})+\frac{-3}{22}=\frac{-57}{55}+\frac{-3}{22}=\frac{-114}{110}+\frac{-15}{110}=\frac{-114-15}{110}=\frac{-129}{110} \\ x+(y+z)=\frac{-7}{11}+(\frac{2}{-5}+\frac{-3}{22})=\frac{-7}{11}+(\frac{-44}{110}+\frac{-15}{110})=\frac{-7}{11}+\frac{-59}{110}=\frac{-70}{110}+\frac{-59}{110}=\frac{-70-59}{110}=\frac{-129}{110} \\ \therefore (\frac{-7}{11}+\frac{2}{-5})+\frac{-3}{22}=\frac{-7}{11}+(\frac{2}{-5}+\frac{-3}{22}) \\ \mathrm{Hence}\mathrm{verified}. \\ (\mathrm{iv}) \\ x=-2,y=\frac{3}{5},z=\frac{-4}{3} \\ \mathrm{so},(x+y)+z=(-2+\frac{3}{5})+\frac{-4}{3}=(\frac{-10}{5}+\frac{3}{5})+\frac{-4}{3}=\frac{-7}{5}+\frac{-4}{3}=\frac{-21}{15}+\frac{-20}{15}=\frac{-21-20}{15}=\frac{-41}{15} \\ x+(y+z)=-2+(\frac{3}{5}+\frac{-4}{3})=\frac{-2}{1}+(\frac{9}{15}+\frac{-20}{15})=\frac{-2}{1}+\frac{-11}{15}=\frac{-30}{15}+\frac{-11}{15}=\frac{-30-11}{15}=\frac{-41}{15} \\ \therefore (-2+\frac{3}{5})+\frac{-4}{3}=-2+(\frac{3}{5}+\frac{-4}{3}) \\ \mathrm{Hence}\mathrm{verified}. \end{gathered}$$