Verify associativity of addition of rational numbers i.e., (x + y) + z = x + (y + z), when:

Solution:

We have to verify that $(x+y)+z=x+(y+z)$

(i) $x=\frac{1}{2}, y=\frac{2}{3}, z=\frac{-1}{5}$

$(\mathrm{x}+\mathrm{y})+\mathrm{z}=\left(\frac{1}{2}+\frac{2}{3}\right)+\frac{-1}{5}=\left(\frac{3}{6}+\frac{4}{6}\right)+\frac{-1}{5}=\frac{7}{6}+\frac{-1}{5}=\frac{35}{30}+\frac{-6}{30}=\frac{35-6}{30}=\frac{29}{30}$

$\mathrm{x}+(\mathrm{y}+\mathrm{z})=\frac{1}{2}+\left(\frac{2}{3}+\frac{-1}{5}\right)=\frac{1}{2}+\left(\frac{10}{15}+\frac{-3}{15}\right)=\frac{1}{2}+\frac{7}{15}=\frac{15}{30}+\frac{14}{30}=\frac{15+14}{30}=\frac{29}{30}$

$\therefore\left(\frac{1}{2}+\frac{2}{3}\right)+\frac{-1}{5}=\frac{1}{2}+\left(\frac{2}{3}+\frac{-1}{5}\right)$

Hence verified.

(ii) $x=\frac{-2}{5}, y=\frac{4}{3}, z=\frac{-7}{10}$

$(\mathrm{x}+\mathrm{y})+\mathrm{z}=\left(\frac{-2}{5}+\frac{4}{3}\right)+\frac{-7}{10}=\left(\frac{-6}{15}+\frac{20}{15}\right)+\frac{-7}{10}=\frac{14}{15}+\frac{-7}{10}=\frac{28}{30}+\frac{-21}{30}=\frac{28-21}{30}=\frac{7}{30}$

$x+(y+z)=\frac{-2}{5}+\left(\frac{4}{3}+\frac{-7}{10}\right)=\frac{-2}{5}+\left(\frac{40}{30}+\frac{-21}{30}\right)=\frac{-2}{5}+\frac{19}{30}=\frac{-12}{30}+\frac{19}{30}=\frac{-12+19}{30}=\frac{7}{30}$

$\left.\therefore \frac{-2}{5}+\frac{4}{3}\right)+\frac{-7}{10}=\frac{-2}{5}+\left(\frac{4}{3}+\frac{-7}{10}\right)$

Hence verified.

(iii) $x=\frac{-7}{11}, y=\frac{2}{-5}, z=\frac{-3}{22}$

$(x+y)+z=\left(\frac{-7}{11}+\frac{2}{-5}\right)+\frac{-3}{22}=\left(\frac{-35}{55}+\frac{-22}{55}\right)+\frac{-3}{22}=\frac{-57}{55}+\frac{-3}{22}=\frac{-114}{110}+\frac{-15}{110}=\frac{-114-15}{110}=\frac{-129}{110}$

$x+(y+z)=\frac{-7}{11}+\left(\frac{2}{-5}+\frac{-3}{22}\right)=\frac{-7}{11}+\left(\frac{-44}{110}+\frac{-15}{110}\right)=\frac{-7}{11}+\frac{-59}{110}=\frac{-70}{110}+\frac{-59}{110}=\frac{-70-59}{110}=\frac{-129}{110}$

$\therefore\left(\frac{-7}{11}+\frac{2}{-5}\right)+\frac{-3}{22}=\frac{-7}{11}+\left(\frac{2}{-5}+\frac{-3}{22}\right)$

Hence verified.

(iv) $x=-2, y=\frac{3}{5}, z=\frac{-4}{3}$

so, $(x+y)+z=\left(-2+\frac{3}{5}\right)+\frac{-4}{3}=\left(\frac{-10}{5}+\frac{3}{5}\right)+\frac{-4}{3}=\frac{-7}{5}+\frac{-4}{3}=\frac{-21}{15}+\frac{-20}{15}=\frac{-21-20}{15}=\frac{-41}{15}$

$x+(y+z)=-2+\left(\frac{3}{5}+\frac{-4}{3}\right)=\frac{-2}{1}+\left(\frac{9}{15}+\frac{-20}{15}\right)=\frac{-2}{1}+\frac{-11}{15}=\frac{-30}{15}+\frac{-11}{15}=\frac{-30-11}{15}=\frac{-41}{15}$

$\therefore\left(-2+\frac{3}{5}\right)+\frac{-4}{3}=-2+\left(\frac{3}{5}+\frac{-4}{3}\right)$

Hence verified.