RD Sharma solution class 8 chapter 1 Rational Numbers Exercise 1.2

Exercise 1.2

Page-1.14

Question 1:

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}$$

Question 2:

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 2:

$$\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}$$

Question 3:

Write the additive inverse of each of the following rational numbers:
(i) $\frac{-2}{17}$
(ii) $\frac{3}{-11}$
(iii) $\frac{-17}{5}$
(iv) $\frac{-11}{-25}$

Answer 3:

$$\begin{gathered} \left(\mathrm{i}\right) \\ \mathrm{Additive} \mathrm{inverse} \mathrm{is} \mathrm{the} \mathrm{negative} \mathrm{of} \mathrm{the} \mathrm{given} \mathrm{number}. \\ \mathrm{So}, \mathrm{additive} \mathrm{inverse} \mathrm{of} \frac{-2}{17} \mathrm{is} \frac{2}{17}. \\ \left(\mathrm{ii}\right) \\ \mathrm{Additive} \mathrm{inverse} \mathrm{is} \mathrm{the} \mathrm{negative} \mathrm{of} \mathrm{the} \mathrm{given} \mathrm{number}. \\ \mathrm{So}, \mathrm{additive} \mathrm{inverse} \mathrm{of} \frac{3}{-11} \mathrm{is} \frac{3}{11}. \\ \left(\mathrm{iii}\right) \\ \mathrm{Additive} \mathrm{inverse} \mathrm{is} \mathrm{the} \mathrm{negative} \mathrm{of} \mathrm{the} \mathrm{given} \mathrm{number}. \\ \mathrm{So}, \mathrm{additive} \mathrm{inverse} \mathrm{of} \frac{-17}{5} \mathrm{is} \frac{17}{5}. \\ \left(\mathrm{iv}\right) \\ \mathrm{Additive} \mathrm{inverse} \mathrm{is} \mathrm{the} \mathrm{negative} \mathrm{of} \mathrm{the} \mathrm{given} \mathrm{number}. \\ \mathrm{So}, \mathrm{additive} \mathrm{inverse} \mathrm{of} \frac{-11}{-25} \mathrm{is} \frac{-11}{25}. \end{gathered}$$

Question 4:

Write the negative (additive inverse) of each of the following:
(i) $\frac{-2}{5}$
(ii) $\frac{7}{-9}$
(iii) $\frac{-16}{13}$
(iv) $\frac{-5}{1}$
(v) 0
(vi) 1
(vii) −1

Answer 4:

$$\begin{gathered} \left(\mathrm{i}\right) \\ \mathrm{Negative} \mathrm{of} \frac{-2}{5} \mathrm{is} \frac{2}{5}. \\ \left(\mathrm{ii}\right) \\ \mathrm{Negative} \mathrm{of} \frac{7}{-9} \mathrm{is} \frac{7}{9}. \\ \left(\mathrm{iii}\right) \\ \mathrm{Negative} \mathrm{of} \frac{-16}{13} \mathrm{is} \frac{16}{13}. \\ \left(\mathrm{iv}\right) \\ \mathrm{Negative} \mathrm{of} \frac{-5}{1} \mathrm{is} \frac{5}{1}. \\ \left(\mathrm{v}\right) \\ \mathrm{Negative} \mathrm{value} \mathrm{of} 0 \mathrm{is} 0. \\ \left(\mathrm{vi}\right) \\ \mathrm{Negative} \mathrm{of} 1 \mathrm{is} -1. \\ \left(\mathrm{vi}\right) \\ \mathrm{Negative} \mathrm{of} -1 \mathrm{is} 1. \end{gathered}$$

Question 5:

Using commutativity and associativity of addition of rational numbers, express each of the following as a rational number:
(i) $\frac{2}{5}+\frac{7}{3}+\frac{-4}{5}+\frac{-1}{3}$
(ii) $\frac{3}{7}+\frac{-4}{9}+\frac{-11}{7}+\frac{7}{9}$
(iii) $\frac{2}{5}+\frac{8}{3}+\frac{-11}{15}+\frac{4}{5}+\frac{-2}{3}$
(iv) $\frac{4}{7}+0+\frac{-8}{9}+\frac{-13}{7}+\frac{17}{21}$

Answer 5:

$$\begin{gathered} \left(\mathrm{i}\right) \\ \text{We have:} \\ \frac{2}{5}+\frac{7}{3}+\frac{-4}{5}+\frac{-1}{3} \\ =(\frac{2}{5}+\frac{-4}{5})+(\frac{7}{3}+\frac{-1}{3}) \\ =\left(\frac{2-4}{5}\right)+\left(\frac{7-1}{3}\right) \\ =\frac{-2}{5}+\frac{6}{3} \\ =\frac{-6+30}{15} \\ =\frac{24}{15} \\ =\frac{8}{5} \\ \left(\mathrm{ii}\right) \\ \text{We have:} \\ \frac{3}{7}+\frac{-4}{9}+\frac{-11}{7}+\frac{7}{9} \\ =(\frac{3}{7}+\frac{-11}{7})+(\frac{-4}{9}+\frac{7}{9}) \\ =\left(\frac{3-11}{7}\right)+\left(\frac{-4+7}{9}\right) \\ =\frac{-8}{7}+\frac{3}{9} \\ =\frac{-72+21}{63} \\ =\frac{-51}{63} \\ =\frac{-17}{21} \\ \left(\mathrm{iii}\right) \\ \text{We have:} \\ \frac{2}{5}+\frac{8}{3}+\frac{-11}{15}+\frac{4}{5}+\frac{-2}{3} \\ =(\frac{2}{5}+\frac{4}{5})+(+\frac{8}{3}+\frac{-2}{3})+\frac{-11}{15} \\ =\left(\frac{2+4}{5}\right)+\left(\frac{8-2}{3}\right)+\frac{-11}{15} \\ =\frac{6}{5}+\frac{6}{3}+\frac{-11}{15} \\ =\frac{18+30-11}{15} \\ =\frac{37}{15} \\ \left(\mathrm{iv}\right) \\ \text{We have:} \\ \frac{4}{7}+0+\frac{-8}{9}+\frac{-13}{7}+\frac{17}{21} \\ =(\frac{4}{7}+\frac{-13}{7})+\left(\frac{-8}{9}\right)+\frac{17}{21} \\ =\left(\frac{4-13}{7}\right)+\left(\frac{-8}{9}\right)+\frac{17}{21} \\ =\frac{-9}{7}+\frac{-8}{9}+\frac{17}{21} \\ =\frac{-81-56+51}{63} \\ =\frac{-86}{63} \end{gathered}$$

Question 6:

Re-arrange suitably and find the sum in each of the following:
(i) $\frac{11}{12}+\frac{-17}{3}+\frac{11}{2}+\frac{-25}{2}$
(ii) $\frac{-6}{7}+\frac{-5}{6}+\frac{-4}{9}+\frac{-15}{7}$
(iii) $\frac{3}{5}+\frac{7}{3}+\frac{9}{5}+\frac{-13}{15}+\frac{-7}{3}$
(iv) $\frac{4}{13}+\frac{-5}{8}+\frac{-8}{13}+\frac{9}{13}$
(v) $\frac{2}{3}+\frac{-4}{5}+\frac{1}{3}+\frac{2}{5}$
(vi) $\frac{1}{8}+\frac{5}{12}+\frac{2}{7}+\frac{7}{12}+\frac{9}{7}+\frac{-5}{16}$

Answer 6:

$$\begin{gathered} \left(\mathrm{i}\right) \\ (\frac{11}{2}+\frac{-25}{2})+\frac{-17}{3}+\frac{11}{12} \\ =\left(\frac{11-25}{2}\right)+\frac{-17}{3}+\frac{11}{12} \\ =\frac{-14}{2}+\frac{-17}{3}+\frac{11}{12} \\ =-7+\frac{-17}{3}+\frac{11}{12} \\ =\frac{-84-68+11}{12} \\ =\frac{-141}{12} \\ =\frac{-47}{4} \\ \left(\mathrm{ii}\right) \\ (\frac{-6}{7}+\frac{-15}{7})+\frac{-5}{6}+\frac{-4}{9} \\ =\left(\frac{-6-15}{7}\right)+\frac{-5}{6}+\frac{-4}{9} \\ =\frac{-21}{7}+\frac{-5}{6}+\frac{-4}{9} \\ =\frac{-378-105-56}{126} \\ =\frac{-539}{126} \\ =\frac{-77}{18} \\ \left(\mathrm{iii}\right) \\ (\frac{3}{5}+\frac{9}{5})+(\frac{-7}{3}+\frac{7}{3})+\frac{-13}{15} \\ =\left(\frac{3+9}{5}\right)+\left(\frac{-7+7}{3}\right)+\frac{-13}{15} \\ =\frac{12}{5}+\frac{-13}{15} \\ =\frac{36-13}{15} \\ =\frac{23}{15} \\ \left(\mathrm{iv}\right) \\ (\frac{4}{13}-\frac{8}{13}+\frac{9}{13})+\frac{-5}{8} \\ =\left(\frac{4-8+9}{13}\right)+\frac{-5}{8} \\ =\frac{5}{13}+\frac{-5}{8} \\ =\frac{40-65}{104} \\ =\frac{-25}{104} \\ \left(\mathrm{v}\right) \\ (\frac{2}{3}+\frac{1}{3})+(\frac{-4}{5}+\frac{2}{5}) \\ =\frac{3}{3}+\frac{-2}{5} \\ =\frac{15-6}{15} \\ =\frac{9}{15} \\ =\frac{3}{5} \\ \left(\mathrm{vi}\right) \\ (\frac{1}{8}+\frac{-5}{16})+(\frac{5}{12}+\frac{7}{12})+(\frac{2}{7}+\frac{9}{7}) \\ =\left(\frac{2-5}{16}\right)+\left(\frac{5+7}{12}\right)+\left(\frac{2+9}{7}\right) \\ =\frac{-3}{16}+\frac{12}{12}+\frac{11}{7} \\ =\frac{-63+336+528}{336} \\ =\frac{801}{336} \\ =\frac{267}{112} \end{gathered}$$