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)}

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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}

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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

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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) }