Question

Verify the following:
(i) $\left(\frac{3}{4}+\frac{-2}{5}\right)+\frac{-7}{10}=\frac{3}{4}+\left(\frac{-2}{5}+\frac{-7}{10}\right)$
(ii) $\left(\frac{-7}{11}+\frac{2}{-5}\right)+\frac{-13}{22}=\frac{-7}{11}+\left(\frac{2}{-5}+\frac{-13}{22}\right)$
(iii) $-1+\left(\frac{-2}{3}+\frac{-3}{4}\right)=\left(-1+\frac{-2}{3}\right)+\frac{-3}{4}$

Answer 1

1.
LHS = $\left\{\left(\frac{3}{4}+\frac{-2}{5}\right)+\frac{-7}{10}\right\}$

$\left\{\left(\frac{15-8}{20}\right)+\frac{-7}{10}\right\}=\left(\frac{7}{20}+\frac{-7}{10}\right)=\left(\frac{7}{20}+\frac{-14}{20}\right)=\left(\frac{7+(-14)}{20}\right)=\frac{-7}{20}$

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

$\left\{\frac{3}{4}+\left(\frac{-4}{10}+{\displaystyle \frac{-7}{10}}\right)\right\}=\left\{{\displaystyle \frac{3}{4}}+\left({\displaystyle \frac{-4-7}{10}}\right)\right\}=\left\{{\displaystyle \frac{3}{4}}+\left({\displaystyle \frac{-11}{10}}\right)\right\}=\left({\displaystyle \frac{3}{4}}+{\displaystyle \frac{-11}{10}}\right)=\left({\displaystyle \frac{15}{20}}+{\displaystyle \frac{-22}{20}}\right)=\left({\displaystyle \frac{15-22}{20}}\right)=\frac{-7}{20}$

∴​ $\left(\frac{3}{4}+\frac{-2}{5}\right)+\frac{-7}{10}=\frac{3}{4}+\left({\displaystyle \frac{-2}{5}}+{\displaystyle \frac{-7}{10}}\right)$


2.
LHS = $\left\{\left(\frac{-7}{11}+\frac{2}{-5}\right)+\frac{-13}{22}\right\}$

We will first make the denominator positive.

$\left\{\left(\frac{-7}{11}+\frac{2\times (-1)}{-5\times (-1)}\right)+\frac{-13}{22}\right\}=\left\{\left(\frac{-7}{11}+\frac{-2}{5}\right)+\frac{-13}{22}\right\}$

$\left\{\left(\frac{-7}{11}+\frac{-2}{5}\right)+\frac{-13}{22}\right\}=\left\{\left({\displaystyle \frac{-35}{55}}+{\displaystyle \frac{-22}{55}}\right)+{\displaystyle \frac{-13}{22}}\right\}=\left\{\left({\displaystyle \frac{-35-22}{55}}\right)+{\displaystyle \frac{-13}{22}}\right\}=\left({\displaystyle \frac{-57}{55}}+{\displaystyle \frac{-13}{22}}\right)=\frac{-114}{110}+\frac{-65}{110}=\frac{-114-65}{110}=\frac{-179}{110}$

RHS = $\left\{\frac{-7}{11}+\left(\frac{2}{-5}+{\displaystyle \frac{-13}{22}}\right)\right\}$

We will first make the denominator positive.

$\left\{\frac{-7}{11}+\left({\displaystyle \frac{2\times (-1)}{-5\times (-1)}}+{\displaystyle \frac{-13}{22}}\right)\right\}=\left\{{\displaystyle \frac{-7}{11}}+\left({\displaystyle \frac{-2}{5}}+{\displaystyle \frac{-13}{22}}\right)\right\}$

$\left\{{\displaystyle \frac{-7}{11}}+\left({\displaystyle \frac{-2}{5}}+{\displaystyle \frac{-13}{22}}\right)\right\}=\left\{{\displaystyle \frac{-7}{11}}+\left({\displaystyle \frac{-44}{110}}+{\displaystyle \frac{-65}{110}}\right)\right\}=\left\{{\displaystyle \frac{-7}{11}}+\left({\displaystyle \frac{-44+(-65)}{110}}\right)\right\}=\frac{-7}{11}+\frac{-109}{110}=\frac{-70}{110}+\frac{-109}{110}=\frac{-70-109}{110}=\frac{-179}{110}$

∴​ $\left(\frac{-7}{11}+\frac{2}{-5}\right)+\frac{-13}{22}=\frac{-7}{11}+\left({\displaystyle \frac{2}{-5}}+{\displaystyle \frac{-13}{22}}\right)$


3.
LHS = $-1+\left({\displaystyle \frac{-2}{3}}+{\displaystyle \frac{-3}{4}}\right)$

$\left\{\frac{-1}{1}+\left({\displaystyle \frac{-2}{3}}+{\displaystyle \frac{-3}{4}}\right)\right\}=\left\{{\displaystyle \frac{-1}{1}}+\left({\displaystyle \frac{-8}{12}}+{\displaystyle \frac{-9}{12}}\right)\right\}=\left\{{\displaystyle \frac{-1}{1}}+\left({\displaystyle \frac{-8-9}{12}}\right)\right\}=\left\{{\displaystyle \frac{-1}{1}}+\left({\displaystyle \frac{-17}{12}}\right)\right\}=\left({\displaystyle \frac{-1}{1}}+{\displaystyle \frac{-17}{12}}\right)=\left({\displaystyle \frac{-1\times 12}{1\times 12}}+{\displaystyle \frac{-17\times 1}{12\times 1}}\right)=\left(\frac{-12+(-17)}{12}\right)=\left(\frac{-12-17}{12}\right)=\frac{-29}{12}$

RHS = $\left\{\left(-1+{\displaystyle \frac{-2}{3}}\right)+\frac{-3}{4}\right\}$

$\left\{\left({\displaystyle \frac{-1}{1}}+{\displaystyle \frac{-2}{3}}\right)+\frac{-3}{4}\right\}=\left\{\left({\displaystyle \frac{-3}{3}}+{\displaystyle \frac{-2}{3}}\right)+{\displaystyle \frac{-3}{4}}\right\}=\left\{\left({\displaystyle \frac{-3-2}{3}}\right)+{\displaystyle \frac{-3}{4}}\right\}=\left\{\left({\displaystyle \frac{-5}{3}}\right)+{\displaystyle \frac{-3}{4}}\right\}=\left({\displaystyle \frac{-5}{3}}+{\displaystyle \frac{-3}{4}}\right)=\left({\displaystyle \frac{-20}{12}}+{\displaystyle \frac{-9}{12}}\right)=\left(\frac{-20-9}{12}\right)=\frac{-29}{12}$

∴ $-1+\left({\displaystyle \frac{-2}{3}}+{\displaystyle \frac{-3}{4}}\right)=\left(-1+{\displaystyle \frac{-2}{3}}\right)+\frac{-3}{4}$