RD Sharma solution class 8 chapter 1 Rational Numbers Exercise 1.6

Exercise 1.6

Page-1.31

Question 1:

Verify the property: x × y = y × x by taking:
(i) $x=-\frac{1}{3}, y=\frac{2}{7}$
(ii) $x=\frac{-3}{5}, y=\frac{-11}{13}$
(iii) $x=2, y=\frac{7}{-8}$
(iv) $x=0, y=\frac{-15}{8}$

Answer 1:

$$\begin{gathered} \mathrm{We} \mathrm{have} \mathrm{to} \mathrm{verify} \mathrm{that} x\times y=y\times x. \\ \left(\mathrm{i}\right) x=\frac{-1}{3}, y=\frac{2}{7} \\ x\times y=\frac{-1}{3}\times \frac{2}{7}=\frac{-2}{21} \\ y\times x=\frac{2}{7}\times \frac{-1}{3}=\frac{-2}{21} \\ \therefore \frac{-1}{3}\times \frac{2}{7}=\frac{2}{7}\times \frac{-1}{3} \\ \mathrm{Hence} \mathrm{verified}. \\ \left(\mathrm{ii}\right) x=\frac{-3}{5}, y=\frac{-11}{13} \\ \mathrm{x}\times \mathrm{y}=\frac{-3}{5}\times \frac{-11}{13}=\frac{33}{65} \\ y\times x=\frac{-11}{13}\times \frac{-3}{5}=\frac{33}{65} \\ \therefore \frac{-3}{5}\times \frac{-11}{13}=\frac{-11}{13}\times \frac{-3}{5} \\ \mathrm{Hence} \mathrm{verified}. \\ \left(\mathrm{iii}\right) x=2, y=\frac{7}{-8} \\ x\times y=2\times \frac{7}{-8}= \frac{7}{-4} \\ y\times x=\frac{7}{-8}\times 2= \frac{7}{-4} \\ \therefore 2\times \frac{7}{-8}=\frac{7}{-8}\times 2 \\ \mathrm{Hence} \mathrm{verified}. \\ \left(\mathrm{iv}\right) x=0, y=\frac{-15}{8} \\ x\times y=0\times \frac{-15}{8}=0 \\ y\times x=\frac{-15}{8}\times 0=0 \\ \therefore 0\times \frac{-15}{8}=\frac{-15}{8}\times 0 \\ \mathrm{Hence} \mathrm{verified}. \end{gathered}$$

Question 2:

Verify the property: x × (y × z) = (x × y) × z by taking:

(i) $x=\frac{-7}{3}, y=\frac{12}{5}, z=\frac{4}{9}$

(ii) $x=0, y=\frac{-3}{5}, z=\frac{-9}{4}$

(iii) $x=\frac{1}{2}, y=\frac{5}{-4}, z=\frac{-7}{5}$

(iv) $x=\frac{5}{7}, y=\frac{-12}{13}, z=\frac{-7}{18}$

Answer 2:

$$\begin{gathered} \mathrm{We} \mathrm{have} \mathrm{to} \mathrm{verify} \mathrm{that} x\times (y\times z)=(x\times y)\times z. \\ \left(\mathrm{i}\right) x=\frac{-7}{3}, y=\frac{12}{5}, z=\frac{4}{9} \\ \mathrm{x}\times (\mathrm{y}\times \mathrm{z})=\frac{-7}{3}\times (\frac{12}{5}\times \frac{4}{9})=\frac{-7}{3}\times \frac{16}{15}=\frac{-112}{45} \\ (x\times y)\times z=(\frac{-7}{3}\times \frac{12}{5})\times \frac{4}{9}=\frac{-28}{5}\times \frac{4}{9}=\frac{-112}{45} \\ \therefore \frac{-7}{8}\times (\frac{15}{5}\times \frac{4}{9})=(\frac{-7}{8}\times \frac{15}{5})\times \frac{4}{9} \\ \mathrm{Hence} \mathrm{verified}. \\ \left(\mathrm{ii}\right) x=0, y=\frac{-3}{5}, z=\frac{-9}{4} \\ x\times (y\times z)= 0\times (\frac{-3}{5}\times \frac{-9}{4})= 0\times \frac{27}{20}=0 \\ (x\times y)\times z=(0\times \frac{-3}{5})\times \frac{-9}{4}= 0\times \frac{-9}{4}=0 \\ \therefore 0\times (\frac{-3}{5}\times \frac{-9}{4})=(0\times \frac{-3}{5})\times \frac{-9}{4} \\ \mathrm{Hence} \mathrm{verified}. \\ \left(\mathrm{iii}\right) x=\frac{1}{2}, y=\frac{5}{-4}, z=\frac{-7}{4} \\ x\times (y\times z)= \frac{1}{2}\times (\frac{5}{-4}\times \frac{-7}{4})=\frac{1}{2}\times \frac{35}{16}=\frac{35}{32} \\ (x\times y)\times z=(\frac{1}{2}\times \frac{5}{-4})\times \frac{-7}{4}=\frac{5}{-8}\times \frac{-7}{4}=\frac{35}{32} \\ \therefore \frac{1}{2}\times (\frac{5}{-4}\times \frac{-7}{4})=(\frac{1}{2}\times \frac{5}{-4})\times \frac{-7}{4} \\ \mathrm{Hence} \mathrm{verified}. \\ \left(\mathrm{iv}\right) x=\frac{5}{7}, y=\frac{-12}{13}, z=\frac{-7}{18} \\ x\times (y\times z)=\frac{5}{7}\times (\frac{-12}{13}\times \frac{-7}{18})=\frac{5}{7}\times \frac{14}{39}=\frac{10}{39} \\ (x\times y)\times z=\frac{5}{7}\times \frac{-12}{13})\times \frac{-7}{18}=\frac{-60}{91}\times \frac{-7}{18}=\frac{10}{39} \\ \therefore \frac{5}{7}\times (\frac{-12}{13}\times \frac{-7}{18})=(\frac{5}{7}\times \frac{-12}{13})\times \frac{-7}{18} \\ \mathrm{Hence} \mathrm{verified}. \end{gathered}$$

Page-1.32

Question 3:

Verify the property: x × (y + z) = x × y + x × z by taking:

(i) $x=\frac{-3}{7}, y=\frac{12}{13}, z=\frac{-5}{6}$

(ii) $x=\frac{-12}{5}, y=\frac{-15}{4}, z=\frac{8}{3}$

(iii) $x=\frac{-8}{3}, y=\frac{5}{6}, z=\frac{-13}{12}$

(iv) $x=\frac{-3}{4}, y=\frac{-5}{2}, z=\frac{7}{6}$

Answer 3:

$$\begin{gathered} \mathrm{We} \mathrm{have} \mathrm{to} \mathrm{verify} \mathrm{that} x\times (y+z)=x\times y+x\times z. \\ \left(\mathrm{i}\right) x=\frac{-3}{7}, y=\frac{12}{13}, z=\frac{-5}{6} \\ \mathrm{x}\times (\mathrm{y}+\mathrm{z})=\frac{-3}{7}\times (\frac{12}{13}+\frac{-5}{6})=\frac{-3}{7}\times \frac{72-65}{78}=\frac{-3}{7}\times \frac{7}{78}= \frac{-1}{26} \\ x\times y+x\times z=\frac{-3}{7}\times \frac{12}{13}+\frac{-3}{7}\times \frac{-5}{6} \\ =\frac{-36}{91}+\frac{5}{14} \\ =\frac{-36\times 2+5\times 13}{182}=\frac{-72+65}{182} \\ =\frac{-1}{26} \\ \therefore \frac{-3}{7}\times (\frac{12}{13}+\frac{-5}{6})=\frac{-3}{7}\times \frac{12}{13}+\frac{-3}{7}\times \frac{-5}{6} \\ \mathrm{Hence} \mathrm{verified}. \\ \left(\mathrm{ii}\right) x=\frac{-12}{5}, y=\frac{-15}{4}, z=\frac{8}{3} \\ x\times (y+z)=\frac{-12}{5}\times (\frac{-15}{4}+\frac{8}{3})=\frac{-12}{5}\times \frac{-45+32}{12}=\frac{-12}{5}\times \frac{-13}{12}=\frac{13}{5} \\ x\times y+x\times z=\frac{-12}{5}\times \frac{-15}{4}+\frac{-12}{5}\times \frac{8}{3} \\ =\frac{9}{1}+\frac{-32}{5} \\ =\frac{45-32}{5} \\ =\frac{13}{5} \\ \therefore \frac{-12}{5}\times (\frac{-15}{4}+\frac{8}{3})=\frac{-12}{5}\times \frac{-15}{4}+\frac{-12}{5}\times \frac{8}{3} \\ \mathrm{Hence} \mathrm{verified}. \\ \left(\mathrm{iii}\right) x=\frac{-8}{3}, y=\frac{5}{6}, \mathrm{z}=\frac{-13}{12} \\ x\times (y+z)=\frac{-8}{3}\times (\frac{5}{6}+\frac{-13}{12})=\frac{-8}{3}\times \frac{10-13}{12}=\frac{-8}{3}\times \frac{-3}{12}=\frac{2}{3} \\ x\times y+x\times z=\frac{-8}{3}\times \frac{5}{6}+\frac{-8}{3}\times \frac{-13}{12} \\ =\frac{-20}{9}+\frac{26}{9} \\ =\frac{-20+26}{9} \\ =\frac{2}{3} \\ \therefore \frac{-8}{3}\times (\frac{5}{6}+\frac{-13}{12})=\frac{-8}{3}\times \frac{5}{6}+\frac{-8}{3}\times \frac{-13}{12} \\ \mathrm{Hence} \mathrm{verified} . \\ \left(\mathrm{iv}\right) x=\frac{-3}{4}, y=\frac{-5}{2}, \mathrm{z}=\frac{7}{6} \\ x\times (y+z)=\frac{-3}{4}\times (\frac{-5}{2}+\frac{7}{6})=\frac{-3}{4}\times \frac{-15+7}{6}=\frac{-3}{4}\times \frac{-8}{6}=1 \\ x\times y+x\times z=\frac{-3}{4}\times \frac{-5}{2}+\frac{-3}{4}\times \frac{7}{6} \\ =\frac{15}{8}+\frac{-7}{8} \\ =\frac{15-7}{8} \\ =1 \\ \therefore \frac{-3}{4}\times (\frac{-5}{2}+\frac{7}{6})=\frac{-3}{4}\times \frac{-5}{2}+\frac{-3}{4}\times \frac{7}{6} \\ \mathrm{Hence} \mathrm{verified}. \end{gathered}$$

Question 4:

Use the distributivity of multiplication of rational numbers over their addition to simplify:
(i) $\frac{3}{5}\times \left(\frac{35}{24}+\frac{10}{1}\right)$

(ii) $\frac{-5}{4}\times \left(\frac{8}{5}+\frac{16}{5}\right)$

(iii) $\frac{2}{7}\times \left(\frac{7}{16}-\frac{21}{4}\right)$

(iv) $\frac{3}{4}\times \left(\frac{8}{9}-40\right)$

Answer 4:

$$\begin{gathered} \left(\mathrm{i}\right) \frac{3}{5}\times (\frac{35}{24}+\frac{10}{1})=\frac{3}{5}\times \frac{35}{24}+\frac{3}{5}\times \frac{10}{1}=\frac{7}{8}+\frac{6}{1}=\frac{7+48}{8}=\frac{55}{8} \\ \left(\mathrm{ii}\right) \frac{-5}{4}\times (\frac{8}{5}+\frac{16}{5})=\frac{-5}{4}\times \frac{8}{5}+\frac{-5}{4}\times \frac{16}{5}=\frac{-2}{1}+\frac{-4}{1}=-6 \\ \left(\mathrm{iii}\right) \frac{2}{7}\times (\frac{7}{16}-\frac{21}{4})=\frac{2}{7}\times \frac{7}{16}-\frac{2}{7}\times \frac{21}{4}=\frac{1}{8}-\frac{3}{2}=\frac{1-12}{8}=\frac{-11}{8} \\ \left(\mathrm{iv}\right) \frac{3}{4}\times (\frac{8}{9}-40)=\frac{3}{4}\times \frac{8}{9}-\frac{3}{4}\times 40=\frac{2}{3}-30=\frac{2-90}{3}=\frac{-88}{3} \end{gathered}$$

Question 5:

Find the multiplicative inverse (reciprocal) of each of the following rational numbers:
(i) 9
(ii) −7
(iii) $\frac{12}{5}$
(iv) $\frac{-7}{9}$
(v) $\frac{-3}{-5}$
(vi) $\frac{2}{3}\times \frac{9}{4}$
(vii) $\frac{-5}{8}\times \frac{16}{15}$
(viii) $-2\times \frac{-3}{5}$
(ix) −1
(x) $\frac{0}{3}$
(xi) 1

Answer 5:

$$\begin{gathered} \left(\mathrm{i}\right) \mathrm{Multiplicative} \mathrm{inverse} (\mathrm{reciprocal)} \mathrm{of} 9 = \frac{1}{9} \\ (\mathrm{ii}) \mathrm{Multiplicative} \mathrm{inverse} (\mathrm{reciprocal)} \mathrm{of} -7 = \frac{-1}{7} \\ \left(\mathrm{iii}\right) \mathrm{Multiplicative} \mathrm{inverse} (\mathrm{reciprocal)} \mathrm{of} \frac{12}{5} = \frac{5}{12} \\ (\mathrm{iv}) \mathrm{Multiplicative} \mathrm{inverse} (\mathrm{reciprocal)} \mathrm{of} \frac{-7}{9} = \frac{-9}{7} \\ \left(\mathrm{v}\right) \mathrm{Multiplicative} \mathrm{inverse} (\mathrm{reciprocal)} \mathrm{of} \frac{-3}{-5} = \frac{-5}{-3} or \frac{5}{3} \\ (\mathrm{vi}) \mathrm{Multiplicative} \mathrm{inverse} (\mathrm{reciprocal)} \mathrm{of} \frac{2}{3}\times \frac{9}{4} = \frac{3}{2}\times \frac{4}{9}=\frac{2}{3} \\ \left(\mathrm{vii}\right) \mathrm{Multiplicative} \mathrm{inverse} (\mathrm{reciprocal)} \mathrm{of} \frac{-5}{8}\times \frac{16}{15} = \frac{8}{-5}\times \frac{15}{16}=\frac{-3}{2} \\ (\mathrm{viii}) \mathrm{Multiplicative} \mathrm{inverse} (\mathrm{reciprocal)} \mathrm{of} -2\times \frac{-3}{5} = \frac{1}{-2}\times \frac{5}{-3}=\frac{5}{6} \\ \left(\mathrm{ix}\right) \mathrm{Multiplicative} \mathrm{inverse} (\mathrm{reciprocal)} \mathrm{of} -1 = \frac{1}{-1}=-1 \\ (\mathrm{x}) \mathrm{Multiplicative} \mathrm{inverse} (\mathrm{reciprocal)} \mathrm{of} \frac{0}{3} = \frac{3}{0}=\mathrm{undefined} \\ \left(\mathrm{ix}\right) \mathrm{Multiplicative} \mathrm{inverse} (\mathrm{reciprocal)} \mathrm{of} 1 = \frac{1}{1}=1 \end{gathered}$$

Question 6:

Name the property of multiplication of rational numbers illustrated by the following statements:
(i) $\frac{-5}{16}\times \frac{8}{15}=\frac{8}{15}\times \frac{-5}{16}$

(ii) $\frac{-17}{5}\times 9=9\times \frac{-17}{5}$

(iii) $\frac{7}{4}\times \left(\frac{-8}{3}+\frac{-13}{12}\right)=\frac{7}{4}\times \frac{-8}{3}+\frac{7}{4}\times \frac{-13}{12}$

(iv) $\frac{-5}{9}\times \left(\frac{4}{15}\times \frac{-9}{8}\right)=\left(\frac{-5}{9}\times \frac{4}{15}\right)\times \frac{-9}{8}$

(v) $\frac{13}{-17}\times 1=\frac{13}{-17}=1\times \frac{13}{-17}$

(vi) $\frac{-11}{16}\times \frac{16}{-11}=1$

(vii) $\frac{2}{13}\times 0=0=0\times \frac{2}{13}$

(viii) $\frac{-3}{2}\times \frac{5}{4}+\frac{-3}{2}\times \frac{-7}{6}=\frac{-3}{2}\times \left(\frac{5}{4}+\frac{-7}{6}\right)$

Answer 6:

(i) Commutative property
(ii) Commutative property
(iii) Distributivity of multiplication over addition
(iv) Associativity of multiplication
(v) Existence of identity for multiplication
(vi) Existence of multiplicative inverse
(vii) Multiplication by 0
(viii) Distributive property

Question 7:

Fill in the blanks:
(i) The product of two positive rational numbers is always .....
(ii) The product of a positive rational number and a negative rational number is always .....
(iii) The product of two negative rational numbers is always .....
(iv) The reciprocal of a positive rational number is .....
(v) The reciprocal of a negative rational number is .....
(vi) Zero has ..... reciprocal.
(vii) The product of a rational number and its reciprocal is .....
(viii) The numbers ..... and ..... are their own reciprocals.
(ix) If a is reciprocal of b, then the reciprocal of b is .....
(x) The number 0 is ..... the reciprocal of any number.
(xi) Reciprocal of $\frac{1}{a}, a\ne 0$ is .....
(xii) (17 × 12)⁻¹ = 17⁻¹ × .....

Answer 7:

(i) Positive
(ii) Negative
(iii) Positive
(iv) Positive
(v) Negative
(vi) No
(vii) 1
(viii) -1 and 1
(ix) a
(x) not
(xi) a
(xii) ${12}^{-1}$

Page-1.33

Question 8:

Fill in the blanks:
(i) $-4\times \frac{7}{9}=\frac{7}{9}\times ......$
(ii) $\frac{5}{11}\times \frac{-3}{8}=\frac{-3}{8}\times ......$
(iii) $\frac{1}{2}\times \left(\frac{3}{4}+\frac{-5}{12}\right)=\frac{1}{2}\times ......+......\times \frac{-5}{12}$
(iv) $\frac{-4}{5}\times \left(\frac{5}{7}+\frac{-8}{9}\right)=\left(\frac{-4}{5}\times .....\right)\times \frac{-8}{9}$

Answer 8:

$$\begin{gathered} \left(\mathrm{i}\right) -4 \\ \mathrm{x}\times \mathrm{y}=\mathrm{y}\times \mathrm{x} \left(\mathrm{commutativity}\right) \\ \left(\mathrm{ii}\right) \frac{5}{11} \\ \mathrm{x}\times \mathrm{y}=\mathrm{y}\times \mathrm{x} \left(\mathrm{commutativity}\right) \\ \left(\mathrm{iii}\right) \frac{3}{4};\frac{1}{2} \\ \mathrm{x}\times (\mathrm{y}+\mathrm{z})=\mathrm{x}\times \mathrm{y}+\mathrm{x}\times \mathrm{z} (d\mathrm{istributivity} \mathrm{of} \mathrm{multiplication} \mathrm{over} \mathrm{addition}) \\ \left(\mathrm{iv}\right) \frac{5}{7} \\ \mathrm{x}\times (\mathrm{y}\times \mathrm{z})=(\mathrm{x}\times \mathrm{y})\times \mathrm{z} (a\mathrm{ssociativity} \mathrm{of} \mathrm{multiplication} ) \end{gathered}$$