Question 1: Compare the following pairs of rational numbers:

$\displaystyle (i) \ \frac{15}{32} \text{ and } \frac{17}{24}$     $\displaystyle (ii) \ \frac{10}{11} \text{ and } \frac{17}{18}$      $\displaystyle (iii) \ \frac{ (-5)}{12} \text{ and } \frac{ (-3)}{4}$      $\displaystyle \ (iv) \frac{ (-7)}{24} \text{ and } \frac{9}{(-20)}$

Answer:

(i) $\frac{15}{32}$ and $\frac{17}{24}$

First take the LCM of $24$ and $32$. LCM $= 96$

Therefore: $\frac{15}{32}= \frac{45}{96}$ and $\frac{17}{24} = \frac{68}{96}$

Hence, we see that $\frac{45}{96}$ $<$ $\frac{68}{96}$ or we can say that $\frac{15}{32}$ $<$ $\frac{17}{24}$

(ii) $\frac{10}{11}$ and $\frac{17}{18}$

First take the LCM of $11$ and $18$. LCM $= 198$

Therefore: $\frac{10}{11}= \frac{180}{198}$ and $\frac{17}{18}= \frac{170}{180}$

Hence, we see that $\frac{180}{198}$ $>$ $\frac{170}{180}$ or we can say that $\frac{10}{11}$ $>$ $\frac{17}{18}$

(iii) $\frac{ (-5)}{12}$ and $\frac{ (-3)}{4}$

First take the LCM of $12$ and $4$. LCM $= 12$

Therefore: $\frac{ (-5)}{12}= \frac{ (-5)}{12}$ and $\frac{ (-3)}{4}= \frac{ (-9)}{12}$

Hence we see that $\frac{ (-5)}{12}$ $>$ $\frac{ (-9)}{12}$ or we can say that $\frac{ (-5)}{12}$ $>$ $\frac{ (-3)}{4}$

(iv) $\frac{ (-7)}{24}$ and $\frac{9}{(-20)}$

First take the LCM of $24$ and $20$. LCM $= 120$

Therefore: $\frac{ (-7)}{24}= \frac{ (-35)}{120}$ and $\frac{ (-9)}{20}= \frac{ (-54)}{120}$

Hence we see that $\frac{ (-35)}{120}$ $>$ $\frac{ (-54)}{120}$ or we can say that $\frac{ (-7)}{24}$ $>$ $\frac{9}{(-20)}$

$\\$

Question 2: Arrange in ascending order:

(i) $\frac{5 }{ 6}$ $,$ $\frac{7 }{ 9}$ $,$ $\frac{11 }{ 12}$ $,$ $\frac{13 }{ 18}$      (ii) $\frac{5 }{ (-7)}$ $,$ $\frac{ (-9) }{ 14}$ $,$ $\frac{ (-5) }{ 6}$ $,$ $\frac{7 }{ (-12)}$

(iii) $2,$ $\frac{1 }{ 3}$ $,$ $\frac{ (-13) }{ 6}$ $,$ $\frac{8 }{ (-3)}$       (iv) $\frac{13 }{ (-28)}$ $,$ $\frac{ (-23) }{ 42}$ $,$ $\frac{ (-4) }{ 7}$ $,$ $\frac{ (-9) }{ 14}$

Answer:

(i) $\frac{5 }{ 6}$ $,$ $\frac{7 }{ 9}$ $,$ $\frac{11 }{ 12}$ $,$ $\frac{13 }{ 18}$

LCM of $6, 9, 12, 18 = 36$

The fractions can be written as $\frac{30 }{ 36}$ $,$ $\frac{28 }{ 36}$ $,$ $\frac{33 }{ 36}$ $,$ $\frac{26 }{ 36}$

Therefore}, the order would be

$\frac{26 }{ 36}$ $<$ $\frac{28 }{ 36}$ $<$ $\frac{30 }{ 36}$ $<$ $\frac{33 }{ 36}$      or      $\frac{13 }{ 18}$ $<$ $\frac{7 }{ 9}$ $<$ $\frac{5 }{ 6}$ $<$ $\frac{11 }{ 12}$

(ii) $\frac{5 }{ (-7)}$ $,$ $\frac{ (-9) }{ 14}$ $,$ $\frac{ (-5) }{ 6}$ $,$ $\frac{7 }{ (-12)}$

Note: $\frac{a }{ (-b)}$ $=$ $\frac{(-a) }{ b}$

LCM of $7, 14, 6, 12 = 84$

The fractions can be written as $\frac{(-60) }{ 84}$ $,$ $\frac{(-54) }{ 84}$ $,$ $\frac{(-70) }{ 84}$ $,$ $\frac{(-49) }{ 84}$

Therefore}, the order would be

$\frac{(-70) }{ 84}$ $<$ $\frac{(-60) }{ 84}$ $<$ $\frac{(-54) }{ 84}$ $<$ $\frac{(-49) }{ 84}$      or      $\frac{(-5) }{ 6}$ $<$ $\frac{5 }{ (-7)}$ $<$ $\frac{(-9) }{ 14}$ $<$ $\frac{7 }{ (-12)}$

(iii) $2,$ $\frac{1 }{ 3}$ $,$ $\frac{ (-13) }{ 6}$ $,$ $\frac{8 }{ (-3)}$

Note: $\frac{a }{ (-b)}$ $=$ $\frac{(-a) }{ b}$

LCM of $3, 6 = 6$

The fractions can be written as $\frac{(-12) }{ 6}$ $,$ $\frac{2 }{ 6}$ $,$ $\frac{(-13) }{ 6}$ $,$ $\frac{(-16) }{ 6}$

Therefore}, the order would be

$\frac{(-16) }{ 6}$ $<$ $\frac{(-13) }{ 6}$ $<$ $\frac{(-12) }{ 6}$ $<$ $\frac{2 }{ 6}$      or      $\frac{(-8) }{ 3}$ $<$ $\frac{(-13) }{ 6}$ $< -2 <$ $\frac{1 }{ 3}$

(iv) $\frac{5 }{ 6}$ $,$ $\frac{7 }{ 9}$ $,$ $\frac{11 }{ 12}$ $,$ $\frac{13 }{ 18}$

Note: $\frac{a }{ (-b)}$ $=$ $\frac{(-a) }{ b}$

LCM of $3, 6 = 6$

The fractions can be written as $\frac{(-12) }{ 6}$ $,$ $\frac{2 }{ 6}$ $,$ $\frac{(-13) }{ 6}$ $,$ $\frac{(-16) }{ 6}$

Therefore}, the order would be

$\frac{(-16) }{ 6}$ $<$ $\frac{(-13) }{ 6}$ $<$ $\frac{(-12) }{ 6}$ $<$ $\frac{2 }{ 6}$      or      $\frac{(-8) }{ 3}$ $<$ $\frac{(-13) }{ 6}$ $< -2 <$ $\frac{1 }{ 3}$

$\\$

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 $\frac{5}{6}$

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

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

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

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

$\\$

Question 4: Find the additive inverse of:

$\displaystyle i) \frac{9}{13}$           $\displaystyle (ii) \frac{14}{3}$            $\displaystyle iii) \frac{(-3)}{23}$         $\displaystyle iv) \frac{8}{(-11)}$            $v) \displaystyle \frac{(-22)}{15}$           $\displaystyle vi) \frac{(-11)}{(-9)}$

Answer:

(i) $\displaystyle \frac{9}{13}$

$\displaystyle \frac{9}{13} +a=0 \text{ Therefore the additive inverse is } \frac{(-9)}{13}$

(ii) $\displaystyle \frac{14}{3}$

$\displaystyle \frac{14}{3} +a=0 \text{ Therefore the additive inverse is } \frac{(-14)}{3}$

(iii) $\displaystyle \frac{(-3)}{23}$

$\displaystyle \frac{(-3)}{23} +a=0 \text{ Therefore the additive inverse is } \frac{(3)}{23}$

(iv) $\displaystyle \frac{8}{(-11)}$

$\displaystyle \frac{8}{(-11)} +a=0 \text{ Therefore the additive inverse is } \frac{8}{11}$

(v) $\displaystyle \frac{(-22)}{15}$

$\displaystyle \frac{(-22)}{15} +a=0 \text{ Therefore the additive inverse is } \frac{22}{15}$

(vi) $\displaystyle \frac{(-11)}{(-9)}$

$\displaystyle \frac{(-11)}{(-9)} +a=0 \text{ Therefore the additive inverse is } \frac{(-11)}{9}$

$\displaystyle \\$

Question 5: Find the sum:

$\displaystyle (i) \ \ \frac{(-7)}{17} + \frac{6}{17}$          $\displaystyle (ii) \ \ \frac{(-5)}{12} + \frac{7}{(-12)}$     $\displaystyle (iii) \ \ \frac{8}{15} + \frac{5}{12}$           $\displaystyle (iv) \ \ \frac{ (-11)}{18} + \frac{5}{(-12)}$     $\displaystyle (v) \ \ \frac{(-11)}{6} + \frac{(-3)}{4} + \displaystyle \frac{5}{8} + \frac{(-7)}{3}$           $\displaystyle (vi) \ \ \frac{4}{7} + \frac{2}{(-3)} + \displaystyle \frac{5}{21} + \frac{(-8)}{9}$

Answer:

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

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

$\displaystyle (iii) \ \ \frac{8}{15} + \frac{5}{12} = \frac{(32+25)}{60} = \frac{19}{20} \text{ (Note: LCM of 15 and 12 is 60 ) }$

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

$\displaystyle (v) \ \ \frac{(-11)}{6} + \frac{(-3)}{4} + \displaystyle \frac{5}{8} + \frac{(-7)}{3} = \frac{(-103)}{24} \text{ (Note: LCM of 6, 4, 8 and 3 is 24) }$

$\displaystyle (vi) \ \ \frac{4}{7} + \frac{2}{(-3)} + \displaystyle \frac{5}{21} + \frac{(-8)}{9} = \frac{(-47)}{63} \text{ Note: LCM of 7, 3, 21 , and 9 is 63}$

$\displaystyle \\$

Question 6: Subtract:

$\displaystyle (i) \ \ \frac{2}{3} \text{ from } \frac{5}{6}$          $\displaystyle (ii) \ \ \frac{(-2)}{5} \text{ from } \frac{(-5)}{7}$           $\displaystyle (iii) \ \ \frac{4}{9} \text{ from } \frac{(-7)}{8}$            $\displaystyle (iv) \ \ \frac{ (-11)}{6} \text{ from } \frac{8}{3}$

Answer:

$\displaystyle (i) \ \ \frac{2}{3} \text{ from } \frac{5}{6}$

$\displaystyle \frac{5}{6} - \frac{2}{3} = \frac{5}{6}- \frac{4}{6}= \frac{1}{6}$

$\displaystyle (ii) \ \ \frac{(-2)}{5} \text{ from } \frac{(-5)}{7}$

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

$\displaystyle (iii) \ \ \frac{4}{9} \text{ from } \frac{(-7)}{8}$

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

$\displaystyle (iv) \ \ \frac{ (-11)}{6} \text{ from } \frac{8}{3}$

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

$\displaystyle \\$

Question 7: The sum of two rational numbers is $\displaystyle \frac{-4}{9}$ . If one of them is $\displaystyle \frac{13}{6}$ then find the other.

Answer:

$\displaystyle \frac{13}{6} +a= \frac{-4}{9}$

$\displaystyle \Rightarrow a= \frac{(-4)}{9} - \frac{13}{6} = \frac{(-47)}{18}$

$\displaystyle \\$

Question 8: What number should be added to $\displaystyle \frac{-2}{3}$ to get $\displaystyle \frac{-1}{7}$

Answer:

$\displaystyle \frac{-2}{3} +a= \frac{-1}{7}$

$\displaystyle \Rightarrow a= \frac{-1}{7} + \frac{2}{3} = \frac{11}{21}$

$\displaystyle \\$

Question 9: What number should be subtracted from $\displaystyle -2$ to get $\displaystyle \frac{7}{11}$

Answer:

$\displaystyle -2-a= \frac{7}{11}$

$\displaystyle \Rightarrow a= -2- \frac{7}{11} = \frac{-29}{11}$

$\displaystyle \\$

Question 10: Find the products:

$\displaystyle (i) \ \ \frac{4}{9} \times \frac{7}{12}$          $\displaystyle (ii) \ \ -9 \times \frac{7}{18}$          $\displaystyle (iii) \ \ \frac{-3}{16} \times \frac{8}{-15}$          $\displaystyle (iv) \ \ \frac{6}{7} \times \frac{-21}{12}$

Answer:

$\displaystyle (i) \ \ \frac{4}{9} \times \frac{7}{12} = \frac{7}{27}$

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

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

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

$\displaystyle \\$

Question 11: Find the quotient:

$\displaystyle (i) \ \ \frac{ 17}{8} \div \frac{51}{4}$     $\displaystyle (ii) \ \ \frac{ (-16)}{35} \div \frac{15}{14}$      $\displaystyle (iii) \ \ \frac{ (-12)}{7} \div (-16)$      $\displaystyle (iv) \ \ -9 \div \frac{(-5)}{18}$

Answer:

$\displaystyle (i) \ \ \frac{ 17}{8} \div \frac{51}{4} = \frac{17}{8} \times \frac{4}{51} = \frac{4}{24} = \frac{1}{6}$

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

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

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

$\displaystyle \\$

Question 12: The product of two rational numbers is $\displaystyle -7$. If one of the number is $\displaystyle \frac{-8}{11}$ , then find the other.

Answer:

$\displaystyle a \times b= -7$

$\displaystyle \Rightarrow \frac{-8}{11} \times b= -7$

$\displaystyle \Rightarrow b= \frac{77}{8}$

$\displaystyle \\$

Question 13: By what number must $\displaystyle \frac{1}{26}$ be divided to get $\displaystyle \frac{-8}{39}$ ?

Answer:

$\displaystyle \frac{1}{26} \div a= \frac{-8}{39}$

$\displaystyle \Rightarrow \frac{1}{26} \times \frac{1}{a} = \frac{-8}{39}$

$\displaystyle \Rightarrow a= \frac{-1}{26} \times \frac{39}{8} = \frac{-3}{16}$

$\displaystyle \\$

Question 14: Find a rational number between each of the following pairs of rational numbers.

$\displaystyle (i) \ \ \frac{7}{10} \text{ and } \frac{10}{17}$    $\displaystyle (ii) \ \ 1 \frac{3}{8} \text{ and } 2$    $\displaystyle (iii) \ \ \frac{-3}{5} \text{ and } \frac{-4}{7}$    $\displaystyle (iv) \ \ - 2 \text{ and } \frac{-17}{21}$

Answer:

$\displaystyle (i) \ \ \frac{7}{10} \text{ and } \frac{10}{17}$

First take the LCM of $\displaystyle 10 \text{ and } 17$ which is $\displaystyle 170$.

Convert the numbers with $\displaystyle 170$ as the denominator.

Hence we get $\displaystyle \frac{109}{170} \text{ and } \frac{100}{170}$

$\displaystyle \text{ Therefore the rational numbers between } \frac{7}{10} \text{ and } \frac{10}{17} \text{ are } \frac{108}{170} \text{ or } \frac{107}{170}$

$\displaystyle (ii) \ \ 1 \frac{3}{8} \text{ and } 2 \text{ or } \frac{11}{8 } \text{ and } \frac{16}{8}$

$\displaystyle \text{ Therefore the rational numbers between } 1 \frac{3}{8} \text{ and 2 are} \frac{12}{8} \text{ or } \frac{13}{8}$

$\displaystyle (iii) \ \ \frac{-3}{5} \text{ and } \frac{-4}{7}$

First take the LCM of $\displaystyle 5 \text{ and } 7$ which is $\displaystyle 35$.

Convert the numbers with $\displaystyle 35$ as the denominator.

$\displaystyle \text{ Hence we get } \frac{-21}{35} \text{ and } \frac{-20}{35}$

$\displaystyle \text{ Therefore the rational numbers between } \frac{-3}{5} \text{ and } \frac{-4}{7} \text{ are} \frac{-20.5}{35} \text{ or } \frac{-41}{70}$

$\displaystyle (iv) \ \ - 2 \text{ and } \frac{-17}{21} \text{ or } \frac{-42}{21} \text{ and } \frac{-17}{21}$

$\displaystyle \text{ Therefore the rational number between } -2 \text{ and } \frac{-17}{21} \text{ are } \frac{-18}{21} \text{ or } \frac{-6}{7}$

$\displaystyle \\$

Question 15: Find three rational numbers between:

$\displaystyle (i) \ \ 4 \text{ and } 4\frac{2}{3}$       $\displaystyle (ii) \ \ \frac{-1 }{2} \text{ and } \frac{-1 }{4}$

Answer:

$\displaystyle (i) \ \ 4 \text{ and } 4\frac{2}{3} \text{ or } \frac{12}{3} \text{ and } \frac{14}{3}$

$\displaystyle \text{ Therefore the rational numbers between } 4 \text{ and } 4\frac{2}{3}$

$\displaystyle \frac{125}{30} \text{ and } \frac{126}{30} \text{ and } \frac{127}{30} \text{ or } \frac{25}{6} \text{ and } \frac{43}{10} \text{ and } \frac{127}{30}$

$\displaystyle (ii) \ \ \frac{-1 }{2} \text{ and } \frac{-1 }{4} \text{ or } \frac{-2 }{4} \text{ and } \frac{-1 }{4}$

$\displaystyle \text{ Therefore the rational numbers between } \frac{-1 }{2} \text{ and } \frac{-1 }{4} \text{ are }$

$\displaystyle \frac{ -11 }{40} \text{ and } \frac{-12 }{40} \text{ and } \frac{-13 }{40} \text{ or } \frac{-11 }{40} \text{ and } \frac{-3 }{10} \text{ and } \frac{-13 }{40}$

$\displaystyle \\$

Question 16: Find 5 rational numbers between:

$\displaystyle (i) \ \ \frac{3}{5} \text{ and } \frac{2}{3}$    $\displaystyle (ii) \ \ -2 \text{ and } -1 \frac{1}{2}$

Answer:

$\displaystyle (i) \ \ \frac{3}{5} \text{ and } \frac{2}{3} \text{ or } \frac{9}{15} \text{ and } \frac{10}{15}$

The rational numbers are

$\displaystyle \frac{91}{150} \text{ , } \frac{92}{150} \text{ , } \frac{93}{150} \text{ , } \frac{94}{150} \text{ , } \frac{95}{150} \text{ or } \frac{91}{150} \text{ , } \frac{46}{75} \text{ , } \frac{31}{50} \text{ , } \frac{47}{75} \text{ , } \frac{19}{30}$

$\displaystyle (ii) \ \ -2 \text{ and } -1 \frac{1}{2} \text{ or } \frac{-4 }{2} \text{ and } \frac{-3 }{2}$

The rational numbers are

$\displaystyle \frac{-31 }{20} \text{ , } \frac{-32 }{20} \text{ , } \frac{-33 }{20} \text{ , } \frac{-34 }{20} \text{ , } \frac{-35 }{20} \text{ or } \frac{-31 }{20} \text{ , } \frac{-8 }{5} \text{ , } \frac{-33 }{20} \text{ , } \frac{-17 }{10} \text{ , } \frac{-7 }{4}$

$\displaystyle \\$

Question 17: Determine whether the numbers are rational or irrational:

(i) $\displaystyle \frac{2}{-17}$     (ii) $\displaystyle 0.6$     (iii) $\displaystyle \sqrt{\frac{121}{169}}$      (iv) $\displaystyle 3 \sqrt{12} - 6 \sqrt{3}$      (v) $\displaystyle 0.142857 \ldots$

(vi) $\displaystyle \frac{4 \sqrt{18}}{3 \sqrt{2}}$     (vii) $\displaystyle 9 \sqrt{2}- \sqrt{32}$     (viii) $\displaystyle 2 \sqrt{3} - 3 \sqrt{2}$      (ix) $\displaystyle \sqrt{\frac{12}{75}}$     (x) $\displaystyle \pi$

(xi) $\displaystyle - \sqrt{144}$      (xii) $\displaystyle 1.411443143 \ldots$      (xiii) $\displaystyle \sqrt{0.9}$     (xiv)  $\displaystyle \sqrt{0.09}$      $\displaystyle (xv) \ \ \frac{1}{\sqrt{2}}$

Answer:

$\displaystyle (i) \ \ \frac{2}{-17} \text{ : Rational }$

Rational numbers are those numbers which can be expressed in the form of $\displaystyle \frac{p}{q}$ where $\displaystyle p \text{ and } q$ are integers. So, 1 and 17 are $\displaystyle p \text{ and } q$ respectively and they are integers too.! … Rational numbers are in the form of $\displaystyle \frac{p}{q} \text{ and } q$ is not equal to zero.

(ii) $\displaystyle 0.6$ : Rational

The decimal $\displaystyle 0.6$ is a rational number. It is the decimal form of the fraction $\displaystyle \frac{6}{10} = \frac{5}{6}$. Rational numbers are in the form of $\displaystyle \frac{p}{q} \text{ and } q$ is not equal to zero

$\displaystyle (iii) \ \ \sqrt{\frac{121}{169}} = \frac{11}{13} \text{ : Rational }$

Rational numbers are those numbers which can be expressed in the form of $\displaystyle \frac{p}{q}$ where $\displaystyle p \text{ and } q$ are integers. So, 11 and 13 are $\displaystyle p \text{ and } q$ respectively and they are integers too.! … Rational numbers are in the form of $\displaystyle \frac{p}{q} \text{ and } q$ is not equal to zero.

(iv) $\displaystyle 3 \sqrt{12} - 6 \sqrt{3} = 6 \sqrt{3} - 6 \sqrt{3} = 0 \text{ : Rational }$

$\displaystyle 0$ can be written as, $\displaystyle \frac{p}{q}$, where $\displaystyle p = 0 \text{ and } b$ is any non-zero integer. Hence, 0 is a rational number.

(v) $\displaystyle 0.142857 \ldots$ : Irrational

$\displaystyle 0.142857142857 \ldots$ is non terminating recurring, so it is a rational number.

$\displaystyle (vi) \ \ \frac{4 \sqrt{18}}{3 \sqrt{2}} = \frac{4 \times 3 \sqrt{2} }{3\sqrt{2}} = 4 \text{ : Rational }$

Every whole number is a rational number, because any whole number can be written as a fraction. For example, $\displaystyle =4$ can be written as $\displaystyle \frac{4}{1}$.

(vii) $\displaystyle 9 \sqrt{2}- \sqrt{32}=9 \sqrt{2}-4 \sqrt{2} = 5 \sqrt{2}$ : Irrational

$\displaystyle \sqrt{2}$ is an irrational number. Hence $\displaystyle 5 \sqrt{2}$ is irrational.

Proof: Lets first assume $\displaystyle \sqrt{2}$ to be rational.

$\displaystyle \sqrt{2} = \frac{a}{b}$, where $\displaystyle b$ is not equal to $\displaystyle 0$.

Here, $\displaystyle a \text{ and } b$ are co-primes whose HCF is $\displaystyle 1$.

$\displaystyle \sqrt{2} = \frac{a}{b}$ ( squaring both sides )…

$\displaystyle 2 = \frac{a^2}{b^2}$

$\displaystyle 2b^2 = a^2$     … … … … … (i)

Here, $\displaystyle 2$ divides $\displaystyle a^2$ also a ( because, if a prime number divides the square of a positive integer, then it divides the integer itself )

Now, let $\displaystyle a = 2c$ ( squaring both sides )…

$\displaystyle a^2 = 4c^2$     … … … … … (ii)

Substituting Equation (i) in Equation (ii),

$\displaystyle 2 b^2 = 4c^2$   $\displaystyle \Rightarrow b^2 = 2c^2$   $\displaystyle \Rightarrow 2c^2 = b^2$

$\displaystyle 2$ divides $\displaystyle b^2$ as well as $\displaystyle b$.

Conclusion:

Here, $\displaystyle a \text{ and } b$ are both divisible by $\displaystyle 2$. But our assumption that their HCF is $\displaystyle 1$ is being contradicted.

Therefore, our assumption that $\displaystyle \sqrt{2}$ is rational is wrong. Thus, it is irrational.

(viii) $\displaystyle 2 \sqrt{3} - 3 \sqrt{2}$ : Irrational

Proof: Let $\displaystyle x = 2 \sqrt{3} - 3 \sqrt{2}$ is a rational number.

Squaring both sides, $\displaystyle x^2 = (2 \sqrt{3} - 3 \sqrt{2})^2 = 30 - 12 \sqrt{6}$

$\displaystyle \Rightarrow 12\sqrt{6} = 30-x^2$

$\displaystyle \Rightarrow \frac{30-x^2}{12} = \sqrt{6} =$ Irrational number.

$\displaystyle \text{ Now if } x \text{ is rational, then} x^2 \text{ is also rational and hence} \frac{30-x^2}{12} \\ \text{ which contradicts our initial assumption. Hence} 2 \sqrt{3} - 3 \sqrt{2}$.

(ix) $\displaystyle \sqrt{\frac{12}{75}} = \frac{2}{5} \text{ : Rational }$

Rational numbers are those numbers which can be expressed in the $\displaystyle \text{ form of } \frac{p}{q} \text{ where } p \text{ and } q \text{ are integers. So, 2 and 5 are } p \text{ and } q \text{ respectively and }$ $\displaystyle \text{ they are integers too.! ... Rational numbers are in the form of } \frac{p}{q} \text{ and } q \\ \text{ is not equal to zero.}$

(x) $\displaystyle \pi$ : Irrational

The basic answer to this question is that $\displaystyle pi$ is irrational because it represents the ratio of the circumference of a circle to its diameter and that ratio is irrational.

$\displaystyle \pi$ is irrational. It cannot be represented as a ratio of $\displaystyle 2$ integers (with non zero denominator). $\displaystyle \frac{22}{7}$ is just an approximation of $\displaystyle \pi$. It is used to simplify the problems and achieve a result as less deviated as possible.

(xi) $\displaystyle - \sqrt{144} = -12$ Rational

$\displaystyle -12 \text{ can be written as, } \frac{p}{q} \text{ , where} p \text{ and } q \text{ are non-zero integer.}$

$\displaystyle \text{ For example } - 12 = \frac{-24}{2} = \frac{-12}{1} \text{ Hence, -12 is a rational number.}$

(xii) $\displaystyle 1.411443143 \ldots$ : Rational

It is in a repeating decimal and hence can be represented in the form of $\displaystyle \frac{p}{q} \text{ and } q \text{ is not equal to zero.}$

(xiii) $\displaystyle \sqrt{0.9}$ : Irrational

$\displaystyle \sqrt{10} \text{ is irrational, so it's inverse } \frac{1}{\sqrt{10}} \text{ (which is the same as } \sqrt{0.1}\text{ is irrational}$ .

All unresolved roots are known as an irrational number since $\displaystyle \sqrt{0.9}$ is an unresolved root which is also known as surds, so it is an irrational number.

(xiv)  $\displaystyle \sqrt{0.09} = \frac{3}{10} \text{ : Rational }$

$\displaystyle \sqrt{0.09} = \text{ can be represented in the form of } \frac{p}{q} \text{ and } q \text{ is not equal to zero.}$

(xv) $\displaystyle \frac{1}{\sqrt{2}} \text{ : Irrational }$

$\displaystyle \sqrt{2} = \text{ is an irrational number as proved above. }$

$\displaystyle \text{ Hence it's inverse } \frac{1}{\sqrt{2}} \text{ is irrational number.}$

$\displaystyle \\$

Question / Answer 19: State whether True or False:

(i) Every real number is either rational or irrational: True

(ii) Every real number can be represented on a number line: True

(iii) There exists and integer which is not a rational number: False

(iv) There exist a point on a number line which do not represent any real number: False

(v) An infinite number of rational numbers can be inserted between any two rational numbers: True

(vi) The multiplicative inverse of any rational number a is 1/a : False

$\displaystyle \\$

Question / Answer 20: Fill in the blanks

(i)        0      is a rational number that is its own additive inverse.

(ii)           is a rational number that does not have a multiplicative inverse.

(iii)           and      -1      are two rational numbers that are equal to their own reciprocal.

(iv)  The product of a rational number with its reciprocal is      1.

(v)  The reciprocal of a negative number is negative.

$\displaystyle (vi) \ \ \text{ The multiplicative inverse of a rational number is } \frac{1}{a} , a \neq 0 \text{ is a .}$

(vii)  Number of irrational numbers between any two rational numbers is infinite.

$\displaystyle \\$

Question 21: Arrange in ascending order

(i) $\displaystyle 8 \sqrt{ 3} ,2 \sqrt{15},11,2 \sqrt{6},3 \sqrt{7}$    (ii)  $\displaystyle 4 \sqrt{ 5} , \sqrt{122},7 \sqrt{3},13,8 \sqrt{2}$

Answer:

(i) $\displaystyle 8 \sqrt{ 3} ,2 \sqrt{15},11,2 \sqrt{6},3 \sqrt{7}$

First take everything within under root sign. That way we can compare the numbers easily.

The numbers would then be $\displaystyle \sqrt{ 192} , \sqrt{60}, \sqrt{121}, \sqrt{24}, \sqrt{63}$

Now arrange in ascending order $\displaystyle \sqrt{24}, \sqrt{60}, \sqrt{63} , \sqrt{121}, \sqrt{ 192}$

Or $\displaystyle 2 \sqrt{6},2 \sqrt{15},3 \sqrt{7} ,11 ,8 \sqrt{ 3}$

(ii)  $\displaystyle 4 \sqrt{ 5} , \sqrt{122},7 \sqrt{3},13,8 \sqrt{2}$

First take everything within under root sign. That way we can compare the numbers easily.

The numbers would then be $\displaystyle \sqrt{ 80} , \sqrt{122}, \sqrt{147}, \sqrt{169}, \sqrt{128}$

Now arrange in ascending order $\displaystyle \sqrt{ 80} , \sqrt{122}, \sqrt{128}, \sqrt{147}, \sqrt{169}$

Or $\displaystyle 4 \sqrt{ 5} , \sqrt{122},8 \sqrt{2},7 \sqrt{3},13$

$\displaystyle \\$

Question 22: Write the rationalizing factors of the following:

(i)  $\displaystyle \sqrt{3}$      (ii)  $\displaystyle 6 \sqrt{3}$     (iii)  $\displaystyle 3+ \sqrt{2}$

(iv)  $\displaystyle \sqrt{7}+ \sqrt{3}$     (v)  $\displaystyle 5- \sqrt{11}$     (vi)  $\displaystyle (3 \sqrt{5}-3)$

Answer:

(i)  $\displaystyle \sqrt{3}$

$\displaystyle \sqrt{3} \times \sqrt{3}=3 \text{ Therefore rationalizing factor is } \sqrt{3}$

(ii)  $\displaystyle 6 \sqrt{3}$

$\displaystyle 6 \sqrt{3} \times \sqrt{3}=18 \text{ Therefore rationalizing factor is } \sqrt{3}$

(iii)  $\displaystyle 3+ \sqrt{2}$

$\displaystyle (3+ \sqrt{ 2})( 3- \sqrt{ 2}) =7 \text{ Therefore rationalizing factor is } 3- \sqrt{ 2}$

(iv)  $\displaystyle \sqrt{7}+ \sqrt{3}$

$\displaystyle (\sqrt{7}+ \sqrt{ 3})( \sqrt{7}- \sqrt{ 3}) =4 \text{ Therefore rationalizing factor is } (\sqrt{7}+ \sqrt{ 3})$

(v)  $\displaystyle 5- \sqrt{11}$

$\displaystyle (5- \sqrt{ 11})( 5+ \sqrt{ 11}) =14 \text{ Therefore rationalizing factor is } (5+ \sqrt{ 11})$

(vi)  $\displaystyle (3 \sqrt{5}-3)$

$\displaystyle (3 \sqrt{5}-3)( 3 \sqrt{5}+3) =14 \text{ Therefore rationalizing factor is } (3 \sqrt{5}+3)$

$\displaystyle \\$

Question 23: Rationalize the denominator of each of the following:

$\displaystyle (i) \ \ \frac{6}{\sqrt{3}}$     $\displaystyle (ii) \ \ \frac{5}{3 \sqrt{2}}$     $\displaystyle (iii) \ \ \frac{4}{\sqrt{5}- \sqrt{3}}$     $\displaystyle (iv) \ \ \frac{11}{6-2 \sqrt{2}}$

$\displaystyle (v) \ \ \frac{7}{3 \sqrt{3} -2 \sqrt{2}}$      $\displaystyle (vi) \ \ \frac{7+ \sqrt{10}}{7- \sqrt{10}}$     $\displaystyle (vii) \ \ \frac{2 \sqrt{5}-4}{2 \sqrt{5}+4}$     $\displaystyle (viii) \ \ \frac{ \sqrt{8}- \sqrt{6}}{ \sqrt{8}+ \sqrt{6}}$

Answer:

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

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

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

$\displaystyle (iv) \ \ \frac{11}{6-2 \sqrt{2}} \times \frac{6+2 \sqrt{2}}{6+2 \sqrt{2}} = \frac{11(6-2 \sqrt{2})}{28}$

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

$\displaystyle (vi) \ \ \frac{7+ \sqrt{10}}{7- \sqrt{10}} \times \frac{7+ \sqrt{10} }{7+ \sqrt{10}} = \frac{(7+ \sqrt{10})^2}{39}$

$\displaystyle (vii) \ \ \frac{2 \sqrt{5}-4}{2 \sqrt{5}+4} \times \frac{2 \sqrt{5}-4}{2 \sqrt{5} - 4} = \frac{(2 \sqrt{5} - 4 )^2}{4}$

$\displaystyle (viii) \ \ \frac{ \sqrt{8}- \sqrt{6}}{ \sqrt{8}+ \sqrt{6}} \times \frac{ \sqrt{8}- \sqrt{6}}{ \sqrt{8} - \sqrt{6}} = \frac{( \sqrt{8} - \sqrt{6} )^2}{2}$

$\displaystyle \\$

Question 24: Insert 5 rational numbers between:

(i) $\displaystyle \sqrt{5}$  and $\displaystyle \sqrt{13}$    (ii) $\displaystyle \sqrt{7}$  and $\displaystyle 3 \sqrt{3}$    (iii) $\displaystyle 2 \text{ and } 2.5$

Answer:

(i) $\displaystyle \sqrt{5}$  and $\displaystyle \sqrt{13}$

$\displaystyle \sqrt{6}, \sqrt{7} , \sqrt{8} , \sqrt{9} , \sqrt{10}$

(ii) $\displaystyle \sqrt{7}$  and $\displaystyle 3 \sqrt{3}$

First take everything within the root sign. So we need to find rational numbers between $\displaystyle \sqrt{7}$  and $\displaystyle \sqrt{27 }$

Hence the numbers are $\displaystyle \sqrt{8}, \sqrt{9} , \sqrt{10} , \sqrt{11} , \sqrt{12}$

(iii) $\displaystyle 2 \text{ and } 2.5$

We can make the numbers as square root. So we need to find rational numbers between $\displaystyle \sqrt{4}$  and $\displaystyle \sqrt{6.25}$

Hence the numbers are $\displaystyle \sqrt{4.1}, \sqrt{4.2} , \sqrt{4.3} , \sqrt{4.4} , \sqrt{4.5}$

$\displaystyle \\$

Question & Answer 25: State True or False:

(i)  $\displaystyle \sqrt{3} + \sqrt{7} = \sqrt{10}$  : False

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

(iii)  $\displaystyle 8 \sqrt{3} - 3 \sqrt{3} = 5 \sqrt{3}$   : True

(iv)  $\displaystyle (9+ \sqrt{3} )+(3- \sqrt{3})$   is a rational number: True (the value is $\displaystyle 12$ which is a rational number)

(v)  $\displaystyle (7 \sqrt{3} )(2 \sqrt{48} )$ is irrational number: False (the value is $\displaystyle 168$ which is a rational number)

(vi)  $\displaystyle (3 \sqrt{2} )(5 \sqrt{8} -7 \sqrt{18} +3 \sqrt{32} )$ is a rational number: True (the value is 6 which is a rational number)

$\displaystyle (vii) \ \ \frac{( \sqrt{3} + \sqrt{2} )}{( \sqrt{3} - \sqrt{2} )} + \frac{( \sqrt{3} - \sqrt{2})}{( \sqrt{3}+ \sqrt{2})} \\ \text{ Is a rational number. True (the value is 10 which is a rational number) }$