\displaystyle \textbf{Question 1: }\text{Compare the following pairs of rational numbers:}
\displaystyle \text{(i) }\frac{15}{32}\text{ and }\frac{17}{24}\qquad\text{(ii) }\frac{10}{11}\text{ and }\frac{17}{18}
\displaystyle \text{(iii) }\frac{-5}{12}\text{ and }\frac{-3}{4}\qquad\text{(iv) }\frac{-7}{24}\text{ and }\frac{9}{-20}
\displaystyle \text{Answer:}

\displaystyle \text{(i) }\frac{15}{32}\text{ and }\frac{17}{24}
\displaystyle \text{LCM of }32\text{ and }24=96.
\displaystyle \frac{15}{32}=\frac{45}{96}\qquad\text{and}\qquad\frac{17}{24}=\frac{68}{96}.
\displaystyle \text{Since }\frac{45}{96}<\frac{68}{96},
\displaystyle \therefore \frac{15}{32}<\frac{17}{24}.

\displaystyle \text{(ii) }\frac{10}{11}\text{ and }\frac{17}{18}
\displaystyle \text{LCM of }11\text{ and }18=198.
\displaystyle \frac{10}{11}=\frac{180}{198}\qquad\text{and}\qquad\frac{17}{18}=\frac{187}{198}.
\displaystyle \text{Since }\frac{180}{198}<\frac{187}{198},
\displaystyle \therefore \frac{10}{11}<\frac{17}{18}.

\displaystyle \text{(iii) }\frac{-5}{12}\text{ and }\frac{-3}{4}
\displaystyle \text{LCM of }12\text{ and }4=12.
\displaystyle \frac{-5}{12}=\frac{-5}{12}\qquad\text{and}\qquad\frac{-3}{4}=\frac{-9}{12}.
\displaystyle \text{Since }\frac{-5}{12}>\frac{-9}{12},
\displaystyle \therefore \frac{-5}{12}>\frac{-3}{4}.

\displaystyle \text{(iv) }\frac{-7}{24}\text{ and }\frac{9}{-20}
\displaystyle \frac{9}{-20}=\frac{-9}{20}.
\displaystyle \text{LCM of }24\text{ and }20=120.
\displaystyle \frac{-7}{24}=\frac{-35}{120}\qquad\text{and}\qquad\frac{-9}{20}=\frac{-54}{120}.
\displaystyle \text{Since }\frac{-35}{120}>\frac{-54}{120},
\displaystyle \therefore \frac{-7}{24}>\frac{9}{-20}.
\displaystyle \\

\displaystyle \textbf{Question 2: }\text{Arrange the following rational numbers in ascending order:}
\displaystyle \text{(i) }\frac{5}{6},\ \frac{7}{9},\ \frac{11}{12},\ \frac{13}{18}
\displaystyle \text{(ii) }\frac{5}{-7},\ \frac{-9}{14},\ \frac{-5}{6},\ \frac{7}{-12}
\displaystyle \text{(iii) }2,\ \frac{1}{3},\ \frac{-13}{6},\ \frac{8}{-3}
\displaystyle \text{(iv) }\frac{13}{-28},\ \frac{-23}{42},\ \frac{-4}{7},\ \frac{-9}{14}
\displaystyle \text{Answer:}

\displaystyle \text{(i) }\frac{5}{6},\ \frac{7}{9},\ \frac{11}{12},\ \frac{13}{18}
\displaystyle \text{LCM of }6,\ 9,\ 12\text{ and }18=36.
\displaystyle \frac{5}{6}=\frac{30}{36},\qquad\frac{7}{9}=\frac{28}{36},\qquad\frac{11}{12}=\frac{33}{36},\qquad\frac{13}{18}=\frac{26}{36}.
\displaystyle \text{Since }\frac{26}{36}<\frac{28}{36}<\frac{30}{36}<\frac{33}{36},
\displaystyle \therefore \frac{13}{18}<\frac{7}{9}<\frac{5}{6}<\frac{11}{12}.

\displaystyle \text{(ii) }\frac{5}{-7},\ \frac{-9}{14},\ \frac{-5}{6},\ \frac{7}{-12}
\displaystyle \text{We know that }\frac{a}{-b}=\frac{-a}{b}.
\displaystyle \therefore \frac{5}{-7}=\frac{-5}{7}\qquad\text{and}\qquad\frac{7}{-12}=\frac{-7}{12}.
\displaystyle \text{LCM of }7,\ 14,\ 6\text{ and }12=84.
\displaystyle \frac{-5}{7}=\frac{-60}{84},\qquad\frac{-9}{14}=\frac{-54}{84},\qquad\frac{-5}{6}=\frac{-70}{84},
\displaystyle \frac{-7}{12}=\frac{-49}{84}.
\displaystyle \text{Since }\frac{-70}{84}<\frac{-60}{84}<\frac{-54}{84}<\frac{-49}{84},
\displaystyle \therefore \frac{-5}{6}<\frac{5}{-7}<\frac{-9}{14}<\frac{7}{-12}.

\displaystyle \text{(iii) }2,\ \frac{1}{3},\ \frac{-13}{6},\ \frac{8}{-3}
\displaystyle \frac{8}{-3}=\frac{-8}{3}.
\displaystyle \text{LCM of }1,\ 3\text{ and }6=6.
\displaystyle 2=\frac{12}{6},\qquad\frac{1}{3}=\frac{2}{6},\qquad\frac{-13}{6}=\frac{-13}{6},\qquad\frac{-8}{3}=\frac{-16}{6}.
\displaystyle \text{Since }\frac{-16}{6}<\frac{-13}{6}<\frac{2}{6}<\frac{12}{6},
\displaystyle \therefore \frac{-8}{3}<\frac{-13}{6}<\frac{1}{3}<2.

\displaystyle \text{(iv) }\frac{13}{-28},\ \frac{-23}{42},\ \frac{-4}{7},\ \frac{-9}{14}
\displaystyle \frac{13}{-28}=\frac{-13}{28}.
\displaystyle \text{LCM of }28,\ 42,\ 7\text{ and }14=84.
\displaystyle \frac{-13}{28}=\frac{-39}{84},\qquad\frac{-23}{42}=\frac{-46}{84},\qquad\frac{-4}{7}=\frac{-48}{84},
\displaystyle \frac{-9}{14}=\frac{-54}{84}.
\displaystyle \text{Since }\frac{-54}{84}<\frac{-48}{84}<\frac{-46}{84}<\frac{-39}{84},
\displaystyle \therefore \frac{-9}{14}<\frac{-4}{7}<\frac{-23}{42}<\frac{13}{-28}.
\displaystyle \\

Question 3: Represent each of these numbers on a Number Line:

(i) \frac{5}{6}     (ii) \frac{14}{3}     (iii) \frac{(-3)}{7}      (iv) \frac{(-17)}{5}     (v) \displaystyle -2  \frac{2}{7}

Answer:

(i) \frac{5}{6} :     Divide the unit length between 0 and 1 in 6 equal parts and then mark \frac{5}{6}

2020-05-30_9-18-46

(ii) \frac{14}{3} = 4 \frac{2}{3} :      Divide the unit length between 4 and 5 in 3 equal parts and then mark \frac{14}{3}

2020-05-30_9-22-14

(iii) \frac{(-3)}{7} :      Divide the unit length between 0 and -1 in 7 equal parts and then mark \frac{(-3)}{7}

2020-05-30_9-26-09

(iv) \frac{(-17)}{5} = -3 \frac{2}{5} :      Divide the unit length between -3 and -4 in 5 equal parts and then mark \frac{(-17)}{5}

2020-05-30_9-29-56

(v) \displaystyle -2  \frac{2}{7} :      Divide the unit length between -2 and -3 in 7 equal parts and then mark \displaystyle -2  \frac{2}{7}

2020-05-30_9-38-50

\\

\displaystyle \textbf{Question 4: }\text{Find the additive inverse of:}
\displaystyle \text{(i) }\frac{9}{13}\qquad\text{(ii) }\frac{14}{3}\qquad\text{(iii) }\frac{-3}{23}
\displaystyle \text{(iv) }\frac{8}{-11}\qquad\text{(v) }\frac{-22}{15}\qquad\text{(vi) }\frac{-11}{-9}
\displaystyle \text{Answer:}

\displaystyle \text{(i) }\frac{9}{13}
\displaystyle \frac{9}{13}+a=0
\displaystyle \Rightarrow a=-\frac{9}{13}.
\displaystyle \therefore \text{The additive inverse is }-\frac{9}{13}.

\displaystyle \text{(ii) }\frac{14}{3}
\displaystyle \frac{14}{3}+a=0
\displaystyle \Rightarrow a=-\frac{14}{3}.
\displaystyle \therefore \text{The additive inverse is }-\frac{14}{3}.

\displaystyle \text{(iii) }\frac{-3}{23}
\displaystyle \frac{-3}{23}+a=0
\displaystyle \Rightarrow a=\frac{3}{23}.
\displaystyle \therefore \text{The additive inverse is }\frac{3}{23}.

\displaystyle \text{(iv) }\frac{8}{-11}
\displaystyle \frac{8}{-11}+a=0
\displaystyle \Rightarrow a=\frac{8}{11}.
\displaystyle \therefore \text{The additive inverse is }\frac{8}{11}.

\displaystyle \text{(v) }\frac{-22}{15}
\displaystyle \frac{-22}{15}+a=0
\displaystyle \Rightarrow a=\frac{22}{15}.
\displaystyle \therefore \text{The additive inverse is }\frac{22}{15}.

\displaystyle \text{(vi) }\frac{-11}{-9}
\displaystyle \frac{-11}{-9}=\frac{11}{9}.
\displaystyle \frac{11}{9}+a=0
\displaystyle \Rightarrow a=-\frac{11}{9}.
\displaystyle \therefore \text{The additive inverse is }-\frac{11}{9}.
\displaystyle \\

\displaystyle \textbf{Question 5: }\text{Find the sum:}
\displaystyle \text{(i) }\frac{-7}{17}+\frac{6}{17}\qquad\text{(ii) }\frac{-5}{12}+\frac{7}{-12}
\displaystyle \text{(iii) }\frac{8}{15}+\frac{5}{12}\qquad\text{(iv) }\frac{-11}{18}+\frac{5}{-12}
\displaystyle \text{(v) }\frac{-11}{6}+\frac{-3}{4}+\frac{5}{8}+\frac{-7}{3}
\displaystyle \text{(vi) }\frac{4}{7}+\frac{2}{-3}+\frac{5}{21}+\frac{-8}{9}
\displaystyle \text{Answer:}

\displaystyle \text{(i) }\frac{-7}{17}+\frac{6}{17}=\frac{-7+6}{17}=-\frac{1}{17}

\displaystyle \text{(ii) }\frac{-5}{12}+\frac{7}{-12}=\frac{-5}{12}+\frac{-7}{12}
\displaystyle =\frac{-12}{12}=-1

\displaystyle \text{(iii) }\frac{8}{15}+\frac{5}{12}
\displaystyle \text{LCM of }15\text{ and }12=60.
\displaystyle =\frac{32}{60}+\frac{25}{60}=\frac{57}{60}=\frac{19}{20}

\displaystyle \text{(iv) }\frac{-11}{18}+\frac{5}{-12}
\displaystyle \text{LCM of }18\text{ and }12=36.
\displaystyle =\frac{-22}{36}+\frac{-15}{36}=\frac{-37}{36}

\displaystyle \text{(v) }\frac{-11}{6}+\frac{-3}{4}+\frac{5}{8}+\frac{-7}{3}
\displaystyle \text{LCM of }6,\ 4,\ 8\text{ and }3=24.
\displaystyle =\frac{-44}{24}+\frac{-18}{24}+\frac{15}{24}+\frac{-56}{24}
\displaystyle =\frac{-44-18+15-56}{24}=\frac{-103}{24}

\displaystyle \text{(vi) }\frac{4}{7}+\frac{2}{-3}+\frac{5}{21}+\frac{-8}{9}
\displaystyle \text{LCM of }7,\ 3,\ 21\text{ and }9=63.
\displaystyle =\frac{36}{63}+\frac{-42}{63}+\frac{15}{63}+\frac{-56}{63}
\displaystyle =\frac{36-42+15-56}{63}=\frac{-47}{63}
\displaystyle \\

\displaystyle \textbf{Question 6: }\text{Subtract:}
\displaystyle \text{(i) }\frac{2}{3}\text{ from }\frac{5}{6}\qquad\text{(ii) }\frac{-2}{5}\text{ from }\frac{-5}{7}
\displaystyle \text{(iii) }\frac{4}{9}\text{ from }\frac{-7}{8}\qquad\text{(iv) }\frac{-11}{6}\text{ from }\frac{8}{3}
\displaystyle \text{Answer:}

\displaystyle \text{(i) }\frac{5}{6}-\frac{2}{3}=\frac{5}{6}-\frac{4}{6}=\frac{1}{6}

\displaystyle \text{(ii) }\frac{-5}{7}-\frac{-2}{5}
\displaystyle =\frac{-25}{35}+\frac{14}{35}=\frac{-11}{35}

\displaystyle \text{(iii) }\frac{-7}{8}-\frac{4}{9}
\displaystyle =\frac{-63}{72}-\frac{32}{72}=\frac{-95}{72}

\displaystyle \text{(iv) }\frac{8}{3}-\frac{-11}{6}
\displaystyle =\frac{16}{6}+\frac{11}{6}=\frac{27}{6}=\frac{9}{2}
\displaystyle \\

\displaystyle \textbf{Question 7: }\text{The sum of two rational numbers is }\frac{-4}{9}.\text{ If one of them is }\frac{13}{6},
\displaystyle \text{find the other rational number.}
\displaystyle \text{Answer:}
\displaystyle \text{Let the other rational number be }a.
\displaystyle \frac{13}{6}+a=\frac{-4}{9}
\displaystyle \Rightarrow a=\frac{-4}{9}-\frac{13}{6}
\displaystyle =\frac{-8}{18}-\frac{39}{18}=\frac{-47}{18}
\displaystyle \therefore \text{The other rational number is }\frac{-47}{18}.
\displaystyle \\
\displaystyle \textbf{Question 8: }\text{What number should be added to }\frac{-2}{3}\text{ to get }\frac{-1}{7}\text{?}
\displaystyle \text{Answer:}
\displaystyle \text{Let the required number be }a.
\displaystyle \frac{-2}{3}+a=\frac{-1}{7}
\displaystyle \Rightarrow a=\frac{-1}{7}+\frac{2}{3}
\displaystyle =\frac{-3}{21}+\frac{14}{21}=\frac{11}{21}
\displaystyle \therefore \text{The required number is }\frac{11}{21}.
\displaystyle \\
\displaystyle \textbf{Question 9: }\text{What number should be subtracted from }-2\text{ to get }\frac{7}{11}\text{?}
\displaystyle \text{Answer:}
\displaystyle \text{Let the required number be }a.
\displaystyle -2-a=\frac{7}{11}
\displaystyle \Rightarrow -a=\frac{7}{11}+2
\displaystyle \Rightarrow -a=\frac{7}{11}+\frac{22}{11}=\frac{29}{11}
\displaystyle \Rightarrow a=\frac{-29}{11}
\displaystyle \therefore \text{The required number is }\frac{-29}{11}.
\displaystyle \\

\displaystyle \textbf{Question 10: }\text{Find the products:}
\displaystyle \text{(i) }\frac{4}{9}\times\frac{7}{12}\qquad\text{(ii) }-9\times\frac{7}{18}
\displaystyle \text{(iii) }\frac{-3}{16}\times\frac{8}{-15}\qquad\text{(iv) }\frac{6}{7}\times\frac{-21}{12}
\displaystyle \text{Answer:}

\displaystyle \text{(i) }\frac{4}{9}\times\frac{7}{12}=\frac{28}{108}=\frac{7}{27}

\displaystyle \text{(ii) }-9\times\frac{7}{18}=\frac{-63}{18}=\frac{-7}{2}

\displaystyle \text{(iii) }\frac{-3}{16}\times\frac{8}{-15}=\frac{24}{240}=\frac{1}{10}

\displaystyle \text{(iv) }\frac{6}{7}\times\frac{-21}{12}=\frac{-126}{84}=\frac{-3}{2}
\displaystyle \\

\displaystyle \textbf{Question 11: }\text{Find the quotient:}
\displaystyle \text{(i) }\frac{17}{8}\div\frac{51}{4}\qquad\text{(ii) }\frac{-16}{35}\div\frac{15}{14}
\displaystyle \text{(iii) }\frac{-12}{7}\div(-16)\qquad\text{(iv) }-9\div\frac{-5}{18}
\displaystyle \text{Answer:}

\displaystyle \text{(i) }\frac{17}{8}\div\frac{51}{4}=\frac{17}{8}\times\frac{4}{51}
\displaystyle =\frac{68}{408}=\frac{1}{6}

\displaystyle \text{(ii) }\frac{-16}{35}\div\frac{15}{14}=\frac{-16}{35}\times\frac{14}{15}
\displaystyle =\frac{-224}{525}=\frac{-32}{75}

\displaystyle \text{(iii) }\frac{-12}{7}\div(-16)=\frac{-12}{7}\times\frac{-1}{16}
\displaystyle =\frac{12}{112}=\frac{3}{28}

\displaystyle \text{(iv) }-9\div\frac{-5}{18}=-9\times\frac{-18}{5}
\displaystyle =\frac{162}{5}
\displaystyle \\

\displaystyle \textbf{Question 12: }\text{The product of two rational numbers is }-7.\text{ If one of the numbers is }\frac{-8}{11},
\displaystyle \text{find the other rational number.}
\displaystyle \text{Answer:}
\displaystyle \text{Let the other rational number be }b.
\displaystyle \frac{-8}{11}\times b=-7
\displaystyle \Rightarrow b=-7\div\frac{-8}{11}
\displaystyle \Rightarrow b=-7\times\frac{-11}{8}=\frac{77}{8}
\displaystyle \therefore \text{The other rational number is }\frac{77}{8}.
\displaystyle \\

\displaystyle \textbf{Question 13: }\text{By what number must }\frac{1}{26}\text{ be divided to get }\frac{-8}{39}\text{?}
\displaystyle \text{Answer:}
\displaystyle \text{Let the required number be }a.
\displaystyle \frac{1}{26}\div a=\frac{-8}{39}
\displaystyle \Rightarrow \frac{1}{26a}=\frac{-8}{39}
\displaystyle \Rightarrow 39=-208a
\displaystyle \Rightarrow a=\frac{-39}{208}=\frac{-3}{16}
\displaystyle \therefore \text{The required number is }\frac{-3}{16}.
\displaystyle \\

\displaystyle \textbf{Question 14: }\text{Find a rational number between each of the following pairs of rational numbers.}
\displaystyle \text{(i) }\frac{7}{10}\text{ and }\frac{10}{17}\qquad\text{(ii) }1\frac{3}{8}\text{ and }2
\displaystyle \text{(iii) }\frac{-3}{5}\text{ and }\frac{-4}{7}\qquad\text{(iv) }-2\text{ and }\frac{-17}{21}
\displaystyle \text{Answer:}

\displaystyle \text{(i) }\frac{7}{10}\text{ and }\frac{10}{17}
\displaystyle \text{LCM of }10\text{ and }17=170.
\displaystyle \frac{7}{10}=\frac{119}{170}\qquad\text{and}\qquad\frac{10}{17}=\frac{100}{170}.
\displaystyle \text{Since }\frac{100}{170}<\frac{108}{170}<\frac{119}{170},
\displaystyle \therefore \frac{108}{170}\text{ is a rational number between }\frac{7}{10}\text{ and }\frac{10}{17}.

\displaystyle \text{(ii) }1\frac{3}{8}\text{ and }2
\displaystyle 1\frac{3}{8}=\frac{11}{8}\qquad\text{and}\qquad2=\frac{16}{8}.
\displaystyle \text{Since }\frac{11}{8}<\frac{12}{8}<\frac{16}{8},
\displaystyle \therefore \frac{12}{8}=\frac{3}{2}\text{ is a rational number between }1\frac{3}{8}\text{ and }2.

\displaystyle \text{(iii) }\frac{-3}{5}\text{ and }\frac{-4}{7}
\displaystyle \frac{1}{2}\left(\frac{-3}{5}+\frac{-4}{7}\right)=\frac{1}{2}\left(\frac{-21-20}{35}\right)
\displaystyle =\frac{1}{2}\times\frac{-41}{35}=\frac{-41}{70}.
\displaystyle \therefore \frac{-41}{70}\text{ is a rational number between }\frac{-3}{5}\text{ and }\frac{-4}{7}.

\displaystyle \text{(iv) }-2\text{ and }\frac{-17}{21}
\displaystyle -2=\frac{-42}{21}.
\displaystyle \text{Since }\frac{-42}{21}<\frac{-18}{21}<\frac{-17}{21},
\displaystyle \therefore \frac{-18}{21}=\frac{-6}{7}\text{ is a rational number between }-2\text{ and }\frac{-17}{21}.
\displaystyle \\

\displaystyle \textbf{Question 15: }\text{Find three rational numbers between:}
\displaystyle \text{(i) }4\text{ and }4\frac{2}{3}\qquad\text{(ii) }\frac{-1}{2}\text{ and }\frac{-1}{4}
\displaystyle \text{Answer:}

\displaystyle \text{(i) }4\text{ and }4\frac{2}{3}
\displaystyle 4=\frac{12}{3}\qquad\text{and}\qquad4\frac{2}{3}=\frac{14}{3}.
\displaystyle \frac{12}{3}=\frac{120}{30}\qquad\text{and}\qquad\frac{14}{3}=\frac{140}{30}.
\displaystyle \text{Since }\frac{120}{30}<\frac{125}{30}<\frac{126}{30}<\frac{127}{30}<\frac{140}{30},
\displaystyle \therefore \frac{125}{30},\ \frac{126}{30},\ \frac{127}{30}\text{ are three rational numbers between them.}
\displaystyle \text{In simplest form, they are }\frac{25}{6},\ \frac{21}{5},\ \frac{127}{30}.

\displaystyle \text{(ii) }\frac{-1}{2}\text{ and }\frac{-1}{4}
\displaystyle \frac{-1}{2}=\frac{-20}{40}\qquad\text{and}\qquad\frac{-1}{4}=\frac{-10}{40}.
\displaystyle \text{Since }\frac{-20}{40}<\frac{-13}{40}<\frac{-12}{40}<\frac{-11}{40}<\frac{-10}{40},
\displaystyle \therefore \frac{-13}{40},\ \frac{-12}{40},\ \frac{-11}{40}\text{ are three rational numbers between them.}
\displaystyle \text{In simplest form, they are }\frac{-13}{40},\ \frac{-3}{10},\ \frac{-11}{40}.
\displaystyle \\

\displaystyle \textbf{Question 16: }\text{Find 5 rational numbers between:}
\displaystyle \text{(i) }\frac{3}{5}\text{ and }\frac{2}{3}\qquad\text{(ii) }-2\text{ and }-1\frac{1}{2}
\displaystyle \text{Answer:}

\displaystyle \text{(i) }\frac{3}{5}\text{ and }\frac{2}{3}
\displaystyle \frac{3}{5}=\frac{9}{15}\qquad\text{and}\qquad\frac{2}{3}=\frac{10}{15}.
\displaystyle \text{Multiplying the numerator and denominator of both fractions by }10,\text{ we get}
\displaystyle \frac{9}{15}=\frac{90}{150}\qquad\text{and}\qquad\frac{10}{15}=\frac{100}{150}.
\displaystyle \text{Therefore, five rational numbers between them are}
\displaystyle \frac{91}{150},\ \frac{92}{150},\ \frac{93}{150},\ \frac{94}{150},\ \frac{95}{150}.
\displaystyle \text{In simplest form, they are }\frac{91}{150},\ \frac{46}{75},\ \frac{31}{50},\ \frac{47}{75},\ \frac{19}{30}.

\displaystyle \text{(ii) }-2\text{ and }-1\frac{1}{2}
\displaystyle -2=\frac{-4}{2}\qquad\text{and}\qquad-1\frac{1}{2}=\frac{-3}{2}.
\displaystyle \text{Multiplying the numerator and denominator of both fractions by }10,\text{ we get}
\displaystyle \frac{-4}{2}=\frac{-40}{20}\qquad\text{and}\qquad\frac{-3}{2}=\frac{-30}{20}.
\displaystyle \text{Therefore, five rational numbers between them are}
\displaystyle \frac{-35}{20},\ \frac{-34}{20},\ \frac{-33}{20},\ \frac{-32}{20},\ \frac{-31}{20}.
\displaystyle \text{In simplest form, they are }\frac{-7}{4},\ \frac{-17}{10},\ \frac{-33}{20},\ \frac{-8}{5},\ \frac{-31}{20}.
\displaystyle \\

\displaystyle \textbf{Question 17: }\text{Determine whether the following numbers are rational or irrational:}
\displaystyle \text{(i) }\frac{2}{-17}\qquad\text{(ii) }0.6\qquad\text{(iii) }\sqrt{\frac{121}{169}}\qquad\text{(iv) }3\sqrt{12}-6\sqrt{3}
\displaystyle \text{(v) }0.142857\ldots\qquad\text{(vi) }\frac{4\sqrt{18}}{3\sqrt{2}}\qquad\text{(vii) }9\sqrt{2}-\sqrt{32}
\displaystyle \text{(viii) }2\sqrt{3}-3\sqrt{2}\qquad\text{(ix) }\sqrt{\frac{12}{75}}\qquad\text{(x) }\pi
\displaystyle \text{(xi) }-\sqrt{144}\qquad\text{(xii) }1.411443143\ldots\qquad\text{(xiii) }\sqrt{0.9}
\displaystyle \text{(xiv) }\sqrt{0.09}\qquad\text{(xv) }\frac{1}{\sqrt{2}}
\displaystyle \text{Answer:}

\displaystyle \text{(i) }\frac{2}{-17}=\frac{-2}{17}\text{ : Rational}
\displaystyle \text{It is of the form }\frac{p}{q},\text{ where }p,q\in\mathbb{Z}\text{ and }q\ne0.

\displaystyle \text{(ii) }0.6\text{ : Rational}
\displaystyle 0.6=\frac{6}{10}=\frac{3}{5}.
\displaystyle \therefore 0.6\text{ is rational.}

\displaystyle \text{(iii) }\sqrt{\frac{121}{169}}=\frac{\sqrt{121}}{\sqrt{169}}=\frac{11}{13}\text{ : Rational}
\displaystyle \text{Since }\frac{11}{13}\text{ is of the form }\frac{p}{q},\text{ it is rational.}

\displaystyle \text{(iv) }3\sqrt{12}-6\sqrt{3}\text{ : Rational}
\displaystyle 3\sqrt{12}-6\sqrt{3}=3(2\sqrt{3})-6\sqrt{3}=6\sqrt{3}-6\sqrt{3}=0.
\displaystyle \text{Since }0=\frac{0}{1},\text{ it is rational.}

\displaystyle \text{(v) }0.142857142857\ldots\text{ : Rational}
\displaystyle \text{The decimal is non-terminating but recurring, so it is rational.}
\displaystyle \text{In fact, }0.142857142857\ldots=\frac{1}{7}.

\displaystyle \text{(vi) }\frac{4\sqrt{18}}{3\sqrt{2}}\text{ : Rational}
\displaystyle \frac{4\sqrt{18}}{3\sqrt{2}}=\frac{4(3\sqrt{2})}{3\sqrt{2}}=4.
\displaystyle \text{Since }4=\frac{4}{1},\text{ it is rational.}

\displaystyle \text{(vii) }9\sqrt{2}-\sqrt{32}\text{ : Irrational}
\displaystyle 9\sqrt{2}-\sqrt{32}=9\sqrt{2}-4\sqrt{2}=5\sqrt{2}.
\displaystyle \text{Since }\sqrt{2}\text{ is irrational, }5\sqrt{2}\text{ is irrational.}

\displaystyle \text{(viii) }2\sqrt{3}-3\sqrt{2}\text{ : Irrational}
\displaystyle \text{Let }x=2\sqrt{3}-3\sqrt{2}\text{ and suppose that }x\text{ is rational.}
\displaystyle x^2=(2\sqrt{3}-3\sqrt{2})^2=12+18-12\sqrt{6}=30-12\sqrt{6}.
\displaystyle \Rightarrow \sqrt{6}=\frac{30-x^2}{12}.
\displaystyle \text{If }x\text{ is rational, then }\frac{30-x^2}{12}\text{ is rational.}
\displaystyle \text{But }\sqrt{6}\text{ is irrational. This is a contradiction.}
\displaystyle \therefore 2\sqrt{3}-3\sqrt{2}\text{ is irrational.}

\displaystyle \text{(ix) }\sqrt{\frac{12}{75}}\text{ : Rational}
\displaystyle \sqrt{\frac{12}{75}}=\sqrt{\frac{4}{25}}=\frac{2}{5}.
\displaystyle \therefore \sqrt{\frac{12}{75}}\text{ is rational.}

\displaystyle \text{(x) }\pi\text{ : Irrational}
\displaystyle \pi\text{ cannot be expressed in the form }\frac{p}{q},\text{ where }p,q\in\mathbb{Z}\text{ and }q\ne0.
\displaystyle \therefore \pi\text{ is irrational. The fraction }\frac{22}{7}\text{ is only an approximation of }\pi.

\displaystyle \text{(xi) }-\sqrt{144}=-12\text{ : Rational}
\displaystyle -12=\frac{-12}{1}.
\displaystyle \therefore -\sqrt{144}\text{ is rational.}

\displaystyle \text{(xii) }1.411443143\ldots\text{ : Irrational}
\displaystyle \text{The decimal is non-terminating and non-repeating, so it is irrational.}

\displaystyle \text{(xiii) }\sqrt{0.9}\text{ : Irrational}
\displaystyle \sqrt{0.9}=\sqrt{\frac{9}{10}}=\frac{3}{\sqrt{10}}=\frac{3\sqrt{10}}{10}.
\displaystyle \text{Since }\sqrt{10}\text{ is irrational, }\sqrt{0.9}\text{ is irrational.}

\displaystyle \text{(xiv) }\sqrt{0.09}\text{ : Rational}
\displaystyle \sqrt{0.09}=\sqrt{\frac{9}{100}}=\frac{3}{10}.
\displaystyle \therefore \sqrt{0.09}\text{ is rational.}

\displaystyle \text{(xv) }\frac{1}{\sqrt{2}}\text{ : Irrational}
\displaystyle \frac{1}{\sqrt{2}}=\frac{\sqrt{2}}{2}.
\displaystyle \text{Since }\sqrt{2}\text{ is irrational, }\frac{\sqrt{2}}{2}\text{ is irrational.}
\displaystyle \\

\displaystyle \textbf{Question 19: }\text{State whether True or False:}

\displaystyle \text{(i) Every real number is either rational or irrational.}
\displaystyle \text{Answer: True}
\displaystyle \text{Every real number belongs to either the set of rational numbers or the set of irrational numbers.}

\displaystyle \text{(ii) Every real number can be represented on a number line.}
\displaystyle \text{Answer: True}
\displaystyle \text{Every real number corresponds to a unique point on the number line.}

\displaystyle \text{(iii) There exists an integer which is not a rational number.}
\displaystyle \text{Answer: False}
\displaystyle \text{Every integer }a\text{ can be written as }\frac{a}{1},\text{ and hence every integer is rational.}

\displaystyle \text{(iv) There exists a point on the number line which does not represent any real number.}
\displaystyle \text{Answer: False}
\displaystyle \text{Every point on the number line represents exactly one real number.}

\displaystyle \text{(v) An infinite number of rational numbers can be inserted between any two rational numbers.}
\displaystyle \text{Answer: True}
\displaystyle \text{There are infinitely many rational numbers between any two distinct rational numbers.}

\displaystyle \text{(vi) The multiplicative inverse of any rational number }a\text{ is }\frac{1}{a}.
\displaystyle \text{Answer: False}
\displaystyle \text{The multiplicative inverse of }a\text{ is }\frac{1}{a}\text{ only when }a\ne0.
\displaystyle \text{The rational number }0\text{ has no multiplicative inverse.}
\displaystyle \\

\displaystyle \textbf{Question 20: }\text{Fill in the blanks:}

\displaystyle \text{(i) }\underline{\qquad}\text{ is a rational number that is its own additive inverse.}
\displaystyle \text{Answer: }0

\displaystyle \text{(ii) }\underline{\qquad}\text{ is a rational number that does not have a multiplicative inverse.}
\displaystyle \text{Answer: }0

\displaystyle \text{(iii) }\underline{\qquad}\text{ and }\underline{\qquad}\text{ are two rational numbers that are equal to their own reciprocals.}
\displaystyle \text{Answer: }1\text{ and }-1

\displaystyle \text{(iv) The product of a non-zero rational number with its reciprocal is }\underline{\qquad}.
\displaystyle \text{Answer: }1

\displaystyle \text{(v) The reciprocal of a negative rational number is }\underline{\qquad}.
\displaystyle \text{Answer: Negative}

\displaystyle \text{(vi) The multiplicative inverse of the multiplicative inverse }\frac{1}{a},\ a\ne0,\text{ is }\underline{\qquad}.
\displaystyle \text{Answer: }a

\displaystyle \text{(vii) The number of irrational numbers between any two distinct rational numbers is }\underline{\qquad}.
\displaystyle \text{Answer: Infinite}
\displaystyle \\

\displaystyle \textbf{Question 21: }\text{Arrange in ascending order:}
\displaystyle \text{(i) }8\sqrt{3},\ 2\sqrt{15},\ 11,\ 2\sqrt{6},\ 3\sqrt{7}
\displaystyle \text{(ii) }4\sqrt{5},\ \sqrt{122},\ 7\sqrt{3},\ 13,\ 8\sqrt{2}
\displaystyle \text{Answer:}

\displaystyle \text{(i) }8\sqrt{3},\ 2\sqrt{15},\ 11,\ 2\sqrt{6},\ 3\sqrt{7}
\displaystyle \text{Expressing each number in the form }\sqrt{a}\text{, we get}
\displaystyle 8\sqrt{3}=\sqrt{192},\quad2\sqrt{15}=\sqrt{60},\quad11=\sqrt{121},
\displaystyle 2\sqrt{6}=\sqrt{24},\quad3\sqrt{7}=\sqrt{63}.
\displaystyle \text{Since }24<60<63<121<192,
\displaystyle \sqrt{24}<\sqrt{60}<\sqrt{63}<\sqrt{121}<\sqrt{192}.
\displaystyle \therefore 2\sqrt{6}<2\sqrt{15}<3\sqrt{7}<11<8\sqrt{3}.

\displaystyle \text{(ii) }4\sqrt{5},\ \sqrt{122},\ 7\sqrt{3},\ 13,\ 8\sqrt{2}
\displaystyle \text{Expressing each number in the form }\sqrt{a}\text{, we get}
\displaystyle 4\sqrt{5}=\sqrt{80},\quad\sqrt{122}=\sqrt{122},\quad7\sqrt{3}=\sqrt{147},
\displaystyle 13=\sqrt{169},\quad8\sqrt{2}=\sqrt{128}.
\displaystyle \text{Since }80<122<128<147<169,
\displaystyle \sqrt{80}<\sqrt{122}<\sqrt{128}<\sqrt{147}<\sqrt{169}.
\displaystyle \therefore 4\sqrt{5}<\sqrt{122}<8\sqrt{2}<7\sqrt{3}<13.
\displaystyle \\

\displaystyle \textbf{Question 22: }\text{Write the rationalizing factors of the following:}
\displaystyle \text{(i) }\sqrt{3}\qquad\text{(ii) }6\sqrt{3}\qquad\text{(iii) }3+\sqrt{2}
\displaystyle \text{(iv) }\sqrt{7}+\sqrt{3}\qquad\text{(v) }5-\sqrt{11}\qquad\text{(vi) }3\sqrt{5}-3
\displaystyle \text{Answer:}

\displaystyle \text{(i) }\sqrt{3}
\displaystyle \sqrt{3}\times\sqrt{3}=3.
\displaystyle \therefore \text{The rationalizing factor is }\sqrt{3}.

\displaystyle \text{(ii) }6\sqrt{3}
\displaystyle 6\sqrt{3}\times\sqrt{3}=18.
\displaystyle \therefore \text{The rationalizing factor is }\sqrt{3}.

\displaystyle \text{(iii) }3+\sqrt{2}
\displaystyle (3+\sqrt{2})(3-\sqrt{2})=9-2=7.
\displaystyle \therefore \text{The rationalizing factor is }3-\sqrt{2}.

\displaystyle \text{(iv) }\sqrt{7}+\sqrt{3}
\displaystyle (\sqrt{7}+\sqrt{3})(\sqrt{7}-\sqrt{3})=7-3=4.
\displaystyle \therefore \text{The rationalizing factor is }\sqrt{7}-\sqrt{3}.

\displaystyle \text{(v) }5-\sqrt{11}
\displaystyle (5-\sqrt{11})(5+\sqrt{11})=25-11=14.
\displaystyle \therefore \text{The rationalizing factor is }5+\sqrt{11}.

\displaystyle \text{(vi) }3\sqrt{5}-3
\displaystyle (3\sqrt{5}-3)(3\sqrt{5}+3)=45-9=36.
\displaystyle \therefore \text{The rationalizing factor is }3\sqrt{5}+3.
\displaystyle \\

\displaystyle \textbf{Question 23: }\text{Rationalize the denominator of each of the following:}
\displaystyle \text{(i) }\frac{6}{\sqrt{3}}\qquad\text{(ii) }\frac{5}{3\sqrt{2}}\qquad\text{(iii) }\frac{4}{\sqrt{5}-\sqrt{3}}
\displaystyle \text{(iv) }\frac{11}{6-2\sqrt{2}}\qquad\text{(v) }\frac{7}{3\sqrt{3}-2\sqrt{2}}
\displaystyle \text{(vi) }\frac{7+\sqrt{10}}{7-\sqrt{10}}\qquad\text{(vii) }\frac{2\sqrt{5}-4}{2\sqrt{5}+4}
\displaystyle \text{(viii) }\frac{\sqrt{8}-\sqrt{6}}{\sqrt{8}+\sqrt{6}}
\displaystyle \text{Answer:}

\displaystyle \text{(i) }\frac{6}{\sqrt{3}}\times\frac{\sqrt{3}}{\sqrt{3}}=\frac{6\sqrt{3}}{3}=2\sqrt{3}

\displaystyle \text{(ii) }\frac{5}{3\sqrt{2}}\times\frac{\sqrt{2}}{\sqrt{2}}=\frac{5\sqrt{2}}{6}

\displaystyle \text{(iii) }\frac{4}{\sqrt{5}-\sqrt{3}}\times\frac{\sqrt{5}+\sqrt{3}}{\sqrt{5}+\sqrt{3}}
\displaystyle =\frac{4(\sqrt{5}+\sqrt{3})}{5-3}=2(\sqrt{5}+\sqrt{3})

\displaystyle \text{(iv) }\frac{11}{6-2\sqrt{2}}\times\frac{6+2\sqrt{2}}{6+2\sqrt{2}}
\displaystyle =\frac{11(6+2\sqrt{2})}{36-8}=\frac{11(6+2\sqrt{2})}{28}
\displaystyle =\frac{11(3+\sqrt{2})}{14}

\displaystyle \text{(v) }\frac{7}{3\sqrt{3}-2\sqrt{2}}\times\frac{3\sqrt{3}+2\sqrt{2}}{3\sqrt{3}+2\sqrt{2}}
\displaystyle =\frac{7(3\sqrt{3}+2\sqrt{2})}{27-8}=\frac{7(3\sqrt{3}+2\sqrt{2})}{19}

\displaystyle \text{(vi) }\frac{7+\sqrt{10}}{7-\sqrt{10}}\times\frac{7+\sqrt{10}}{7+\sqrt{10}}
\displaystyle =\frac{(7+\sqrt{10})^2}{49-10}=\frac{59+14\sqrt{10}}{39}

\displaystyle \text{(vii) }\frac{2\sqrt{5}-4}{2\sqrt{5}+4}\times\frac{2\sqrt{5}-4}{2\sqrt{5}-4}
\displaystyle =\frac{(2\sqrt{5}-4)^2}{20-16}=\frac{36-16\sqrt{5}}{4}=9-4\sqrt{5}

\displaystyle \text{(viii) }\frac{\sqrt{8}-\sqrt{6}}{\sqrt{8}+\sqrt{6}}\times\frac{\sqrt{8}-\sqrt{6}}{\sqrt{8}-\sqrt{6}}
\displaystyle =\frac{(\sqrt{8}-\sqrt{6})^2}{8-6}=\frac{14-8\sqrt{3}}{2}=7-4\sqrt{3}
\displaystyle \\

\displaystyle \textbf{Question 24: }\text{Insert 5 rational numbers between:}
\displaystyle \text{(i) }\sqrt{5}\text{ and }\sqrt{13}\qquad\text{(ii) }\sqrt{7}\text{ and }3\sqrt{3}\qquad\text{(iii) }2\text{ and }2.5
\displaystyle \text{Answer:}

\displaystyle \text{(i) }\sqrt{5}\text{ and }\sqrt{13}
\displaystyle \sqrt{5}\approx2.236\qquad\text{and}\qquad\sqrt{13}\approx3.606.
\displaystyle \text{Therefore, five rational numbers between them are}
\displaystyle \frac{5}{2},\ \frac{11}{4},\ 3,\ \frac{13}{4},\ \frac{7}{2}.

\displaystyle \text{(ii) }\sqrt{7}\text{ and }3\sqrt{3}
\displaystyle \sqrt{7}\approx2.646\qquad\text{and}\qquad3\sqrt{3}\approx5.196.
\displaystyle \text{Therefore, five rational numbers between them are}
\displaystyle 3,\ \frac{7}{2},\ 4,\ \frac{9}{2},\ 5.

\displaystyle \text{(iii) }2\text{ and }2.5
\displaystyle 2=\frac{20}{10}\qquad\text{and}\qquad2.5=\frac{25}{10}.
\displaystyle \text{Writing them with denominator }100,\text{ we get }\frac{200}{100}\text{ and }\frac{250}{100}.
\displaystyle \text{Therefore, five rational numbers between them are}
\displaystyle \frac{21}{10},\ \frac{11}{5},\ \frac{23}{10},\ \frac{12}{5},\ \frac{49}{20}.
\displaystyle \\

\displaystyle \textbf{Question 25: }\text{State whether True or False:}

\displaystyle \text{(i) }\sqrt{3}+\sqrt{7}=\sqrt{10}.
\displaystyle \text{Answer: False}
\displaystyle \sqrt{3}+\sqrt{7}\ne\sqrt{3+7}=\sqrt{10}.

\displaystyle \text{(ii) }5\sqrt{2}+3\sqrt{2}=8\sqrt{2}.
\displaystyle \text{Answer: True}
\displaystyle 5\sqrt{2}+3\sqrt{2}=(5+3)\sqrt{2}=8\sqrt{2}.

\displaystyle \text{(iii) }8\sqrt{3}-3\sqrt{3}=5\sqrt{3}.
\displaystyle \text{Answer: True}
\displaystyle 8\sqrt{3}-3\sqrt{3}=(8-3)\sqrt{3}=5\sqrt{3}.

\displaystyle \text{(iv) }(9+\sqrt{3})+(3-\sqrt{3})\text{ is a rational number.}
\displaystyle \text{Answer: True}
\displaystyle (9+\sqrt{3})+(3-\sqrt{3})=12.
\displaystyle \text{Since }12\text{ is rational, the statement is true.}

\displaystyle \text{(v) }(7\sqrt{3})(2\sqrt{48})\text{ is an irrational number.}
\displaystyle \text{Answer: False}
\displaystyle (7\sqrt{3})(2\sqrt{48})=14\sqrt{144}=14\times12=168.
\displaystyle \text{Since }168\text{ is rational, the statement is false.}

\displaystyle \text{(vi) }(3\sqrt{2})(5\sqrt{8}-7\sqrt{18}+3\sqrt{32})\text{ is a rational number.}
\displaystyle \text{Answer: True}
\displaystyle (3\sqrt{2})(10\sqrt{2}-21\sqrt{2}+12\sqrt{2})
\displaystyle =(3\sqrt{2})(\sqrt{2})=6.
\displaystyle \text{Since }6\text{ is rational, the statement is true.}

\displaystyle \text{(vii) }\frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}}+\frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}+\sqrt{2}}\text{ is a rational number.}
\displaystyle \text{Answer: True}
\displaystyle \frac{(\sqrt{3}+\sqrt{2})^2+(\sqrt{3}-\sqrt{2})^2}{(\sqrt{3}-\sqrt{2})(\sqrt{3}+\sqrt{2})}
\displaystyle =\frac{(5+2\sqrt{6})+(5-2\sqrt{6})}{3-2}=\frac{10}{1}=10.
\displaystyle \text{Since }10\text{ is rational, the statement is true.}
\displaystyle \\

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