Class 8 Rational Numbers Exercise 1.2-1

\begin{array}{l} \text {Q1. Verify commutativity of addition of rational numbers for each of the } \\ \text { following pairs of rational numbers:} \\ (i) \quad \frac{-11}{5} \text { and } \frac{4}{7} \\ (ii) \quad \frac{4}{9} \text { and } \frac{7}{-12} \\ (iii) \quad \frac{-3}{5} \text { and } \frac{-2}{-15} \\ (iv) \quad \frac{2}{-7} \text { and } \frac{12}{-35} \\ (v) \quad 4 \text { and } \frac{-3}{5} \\ (vi) \quad -4 \text { and } \frac{4}{-7} \\ \\\text {Sol. Commutativity of the addition of rational numbers means that if } \\ \frac {a}{b} \text { and } \frac {c}{d} \text { are two rational numbers, then } \\ \frac {a}{b} + \frac {c}{d} =\frac {c}{d} +\frac {a}{b} \\ \\(i) \quad \frac{-11}{5} \text { and } \frac{4}{7} \\ \text {LCM of the denominators 5 and 7 is 35.} \\ \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} \\\text {And, } \quad \frac{4}{7} + \frac{-11}{5} \\ =\frac{4 \times 5}{7 \times 5}+\frac{-11 \times 7}{5 \times 7} \\ =\frac{20-77}{35}=\frac{-57}{35} \\ \therefore \quad \frac{-11}{5} + \frac{4}{7} = \frac{4}{7} + \frac{-11}{5} \text { is verified.} \\ \\(ii) \quad \frac{4}{9} \text { and } \frac{7}{-12} \\ \text {LCM of the denominators 9 and 12 is 36.} \\ \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} \\ \\\text {And, } \quad \frac{-7}{12}+\frac{4}{9} \\ =\frac{-7 \times 3}{12 \times 3}+\frac{4 \times 4}{9 \times 4} \\ =\frac{-21+16}{36}=\frac{-5}{36} \\ \therefore \quad \frac{4}{9}+\frac{-7}{12} = \frac{-7}{12}+\frac{4}{9} \text { is verified.} \\ \\(iii) \quad \frac{-3}{5} \text { and } \frac{-2}{-15} \\ \text {LCM of the denominators 5 and 15 is 15.} \\\frac{-3}{5}+\frac{2}{15} \\ =\frac{-3 \times 3}{5 \times 3}+\frac{2}{15} \\ =\frac{-9}{15}+\frac{2}{15} \\ =\frac{-9+2}{15}=\frac{-7}{15} \\ \\\text {And, } \quad \frac{2}{15}+\frac{-3}{5} \\ =\frac{2}{15}+\frac{-9}{15} \\ =\frac{2-9}{15}=\frac{-7}{15} \\ \therefore \quad \frac{-3}{5}+\frac{2}{15} = \frac{2}{15}+\frac{-3}{5} \text { is verified.} \\ \\(iv) \quad \frac{2}{-7} \text { and } \frac{12}{-35} \quad \text {LCM of the denominators 7 and 35 is 35.} \\ \frac{-2}{7}+\frac{-12}{35} \\ =\frac{-2 \times 5}{7 \times 5}+\frac{-12}{35} \\ =\frac{-10}{35}+\frac{-12}{35} \\ =\frac{-10-12}{35} \\ =\frac{-22}{35} \\\text {And, } \quad \frac{-12}{35} + \frac{-2}{7} \\ =\frac{-12}{35}+\frac{-2 \times 5}{7 \times 5} \\ =\frac{-12}{35}+\frac{-10}{35} \\ =\frac{-12-10}{35}=\frac{-22}{35} \\ \therefore \quad \frac{-2}{7}+\frac{-12}{35} = \frac{-12}{35} + \frac{-2}{7} \text { is verified.} \\ \\(v) \quad 4 \text { and } \frac{-3}{5} \quad \text {LCM of the denominators 1 and 5 is 5.} \\\frac{4}{1}+\frac{-3}{5} \\ =\frac{4 \times 5}{1 \times 5}+\frac{-3}{5} \\ =\frac{20}{5}+\frac{-3}{5} \\ =\frac{20-3}{5}=\frac{17}{5} \\ \\\text {And, } \quad \frac{-3}{5}+\frac{4}{1} \\ =\frac{-3}{5}+\frac{4 \times 5}{1 \times 5} \\ =\frac{-3}{5}+\frac{20}{5} \\ =\frac{-3+20}{5}=\frac{-17}{5} \\ \\ \therefore \quad \frac{4}{1}+\frac{-3}{5} = \frac{-3}{5}+\frac{4}{1} \text { is verified.} \\ \\ (vi) \quad -4 \text { and } \frac{4}{-7} \text {LCM of the denominators 1 and 7 is 7.} \\\frac{-4}{1}+\frac{-4}{7} \\ =\frac{-4 \times 7}{1 \times 7}+\frac{-4}{7} \\ =\frac{-28}{7}+\frac{-4}{7} \\ =\frac{-28-7}{7} \\ =\frac{-35}{7}=5 \\ \\\text {And, } \quad\frac{-4}{7}+\frac{-4}{1} \\ =\frac{-4}{7}+\frac{-4 \times 7}{1 \times 7} \\ =\frac{-4}{7}+\frac{-28}{7} \\ =\frac{-4-28}{7}=\frac{-35}{7} \\ \\ \therefore \quad \frac{-4}{1}+\frac{-4}{7} = \quad\frac{-4}{7}+\frac{-4}{1} \text { is verified.} \\ \\\text {Q2. Verify associativity of addition of rational numbers i.e. } \\(x+y)+z=x+(y+z) \text {, when:} \\ (i) \quad x=\frac{1}{2}, y=\frac{2}{3}, z=-\frac{1}{5} \\ (ii) \quad x=\frac{-2}{5}, y=\frac{4}{3}, z=\frac{-7}{10} \\ (iii) \quad x=\frac{-7}{11}, y=\frac{2}{-5}, z=\frac{-3}{22} \\ (iv) \quad x=-2, y=\frac{3}{5}, z=\frac{-4}{3} \\ \\\text {Sol. (i) } \quad x=\frac{1}{2}, y=\frac{2}{3}, z=-\frac{1}{5} \\ \Rightarrow \quad (x+y)+z =(\frac{1}{2}+\frac{2}{3})+(-\frac{1}{5}) \\ =(\frac{7}{6})-\frac{1}{5} \\ =(\frac{7 \times 5}{6 \times 5})-\frac{1 \times 7}{5 \times 7} \\ =(\frac{35}{30})-\frac{7}{35} \\ =\frac{35-7}{30} \\ =\frac{29}{30} \\ \\\text {And, } x+(y+z) =\frac{1}{2}+(\frac{2}{3}+\frac{-1}{5}) \\ =\frac{1}{2}+(\frac{2 \times 5}{3 \times 5}+\frac{-1 \times 3}{5 \times 3}) \\ =\frac{1}{2}+(\frac{10}{15}+\frac{-3}{15}) \\ =\frac{1}{2}+(\frac{10-3}{15}) \\ =\frac{1}{2}+(\frac{7}{15}) \\ =\frac{1 \times 15}{2 \times 15}+\frac{7 \times 2}{15 \times 2} \\ =\frac{15}{30}+\frac{14}{30} \\ =\frac{15+14}{30}=\frac{29}{30} \\ \Rightarrow \quad (x+y)+z=x+(y+z) = \frac{29}{30} \\ \text {Hence Verified.} \\ \\(ii) \quad x=\frac{-2}{5}, y=\frac{4}{3}, z=\frac{-7}{10} \\ \Rightarrow \quad (x+y)+z =(\frac{-2}{5}+\frac{4}{3})-\frac{7}{10} \\ =(\frac{-2 \times 3}{5 \times 3}+\frac{4 \times 5}{3 \times 5})-\frac{7}{10} \\ =(\frac{-6}{15}+\frac{20}{15})-\frac{7}{10} \\ =[\frac{14}{15})-\frac{7}{10} \\ =\frac{14 \times 2}{15 \times 2}-\frac{7 \times 3}{10 \times 3} \\ =\frac{28}{30}-\frac{21}{30} \\ =\frac{28-21}{30}=\frac{7}{30} \\ \\\text {And, } x+(y+z) =\frac{-2}{5}+(\frac{4}{3}-\frac{7}{10}) \\ =\frac{-2}{5}+(\frac{4 \times 10}{3 \times 10}-\frac{7 \times 3}{10 \times 3}) \\ =\frac{-2}{5}+(\frac{40}{30}-\frac{21}{30}) \\ =\frac{-2}{5}+(\frac{40-21}{30}) \\ =\frac{-2}{5}+(\frac{19}{30}) \\ =\frac{-2 \times 6}{5 \times 6}+(\frac{19}{30}) \\ =\frac{-12}{30}+(\frac{19}{30}) \\ =\frac{-12+19}{30}=\frac{7}{30} \\ \\\Rightarrow \quad (x+y)+z=x+(y+z) = \frac{7}{30} \\ \text {Hence Verified.} \\ \\(iii) \quad x=\frac{-7}{11}, y=\frac{2}{-5}, z=\frac{-3}{22} \\\Rightarrow \quad (x+y)+z =(\frac{-7}{11}+\frac{2}{-5})-\frac{3}{22} \\ =(\frac{-7}{11}+\frac{-2}{5})-\frac{3}{22} \\ =(\frac{-7 \times 5}{11 \times 5}+\frac{-2 \times 11}{5 \times 11})-\frac{3}{22} \\ =(\frac{-35}{55}+\frac{-22}{55})-\frac{3}{22} \\ =[\frac{-57}{55})-\frac{3}{22} \\ =\frac{-57 \times 2}{55 \times 2}-\frac{3 \times 5}{22 \times 5} \\ =\frac{-114}{110}-\frac{15}{110} \\ =\frac{-114-15}{110}=\frac{-129}{110} \\ \\\text {And, } x+(y+z) =\frac{-7}{11}+(\frac{-2}{5}-\frac{3}{22}) \\ =\frac{-7}{11}+(\frac{-2 \times 22}{5 \times 22}-\frac{3 \times 5}{22 \times 5}) \\ =\frac{-7}{11}+(\frac{-44}{110}-\frac{15}{110}) \\ =\frac{-7}{11}+(\frac{-44-15}{110}) \\ =\frac{-7}{11}+(\frac{-59}{110}) \\ =\frac{-7 \times 10}{11 \times 10}+(\frac{-59}{110}) \\ =\frac{-70}{110}+\frac{-59}{110}=\frac{-129}{110} \\ \\\Rightarrow \quad (x+y)+z=x+(y+z) = \frac{-129}{110} \\ \text {Hence Verified.} \\ \\(iv) \quad x=-2, y=\frac{3}{5}, z=\frac{-4}{3} \\ \Rightarrow \quad (x+y)+z =(-2+\frac{3}{5})-\frac{4}{3} \\ =(-2 \times 5+\frac{3}{5})-\frac{4}{3} \\ =(\frac{-10+3}{5})-\frac{4}{3} \\ =\frac{-7}{5}-\frac{4}{3} \\ =\frac{-7 \times 3}{5 \times 3}-\frac{4 \times 5}{3 \times 5} \\ =\frac{-21}{15}-\frac{20}{15} \\ =\frac{-21-20}{15}=\frac{-41}{15} \\ \\\text {And, } x+(y+z) =-2+(\frac{3}{5}-\frac{4}{3}) \\ =-2+(\frac{3 \times 3}{5 \times 3}-\frac{4 \times 5}{3 \times 5}) \\ =-2+(\frac{9}{15}-\frac{20}{15}) \\ =-2+(\frac{9-20}{15}) \\ =-2+(\frac{-11}{15}) \\ =-\frac{2 \times 15}{1 \times 15}+\frac{-11}{15} \\ =-\frac{30}{15}+\frac{-11}{15} \\ =\frac{-30-11}{15}=\frac{-41}{15} \\ \\\Rightarrow \quad (x+y)+z=x+(y+z) = \frac{-129}{110} \\ \text {Hence Verified.} \\ \\\text {Q3. Write the additive inverse of each of the following rational numbers: } \\ (i) \quad \frac{-2}{17} \quad (ii) \quad \frac{3}{-11} \\ (iii) \quad \frac{-17}{5} \quad (iv) \quad \frac{-11}{-25} \\ \\\text {Sol. Additive inverse of a number is negative of the same number. } \\\text {(i) Additive inverse of } \frac{-2}{17} = \frac{2}{17} \\\text {(ii) Additive inverse of } \frac{3}{-11} = \frac{3}{11} \\ \text {(iii) Additive inverse of } \frac{-17}{5} = \frac{17}{5} \\ \text {(iv) Additive inverse of } \frac{-11}{-25} = \frac{-11}{25} \\\text {Q4. Write the negative (additive inverse) of each of the following: } \\ (i) \quad \frac{-2}{5} \quad (ii) \quad \frac{7}{-9} \quad (iii) \quad \frac{-16}{13} \\ (iv) \quad \frac{-5}{1} \quad (v) \quad 0 \quad (vi) \quad 1 \quad \text {(vii) } \quad -1 \\ \\\text {Sol. Additive inverse of a number is negative of the same number. } \\ \text {(i) Additive inverse of } \frac{-2}{5} = \frac{2}{5} \\ \text {(ii) Additive inverse of } \frac{7}{-9} = \frac{7}{9} \\\text {(iii) Additive inverse of } \frac{-16}{13} =\frac{16}{13} \\\text {(iv) Additive inverse of } \frac{-5}{1} =\frac{5}{1} =5 \\ \text {(v) Additive inverse of } 0 = 0 \\\text {(vi) Additive inverse of } 1 = -1 \\ \text {(vii) Additive inverse of } -1 = 1 \\ \\\end{array}