Question 1: Compare the following pairs of rational numbers:

(i) \frac{15}{32} and \frac{17}{24}      (ii) \frac{10}{11} and \frac{17}{18}      (iii) \frac{ (-5)}{12} and \frac{ (-3)}{4}      (iv) \frac{ (-7)}{24} 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) -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) -2 \frac{2}{7} :      Divide the unit length between -2 and -3 in 7 equal parts and then mark -2 \frac{2}{7}

2020-05-30_9-38-50

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Question 4: Find the additive inverse of:

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

Answer:

(i) \frac{9}{13}

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

(ii) \frac{14}{3}

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

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

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

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

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

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

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

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

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

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Question 5: Find the sum:

(i)  \frac{(-7)}{17} + \frac{6}{17}      (ii) \frac{(-5)}{12} + \frac{7}{(-12)}      (iii)  \frac{8}{15} + \frac{5}{12}      (iv)  \frac{ (-11)}{18} + \frac{5}{(-12)}

(v)  \frac{(-11)}{6} + \frac{(-3)}{4} +\frac{5}{8} + \frac{(-7)}{3}      (vi) \frac{4}{7} + \frac{2}{(-3)} +\frac{5}{21} + \frac{(-8)}{9}

Answer:

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

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

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

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

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

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

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Question 6: Subtract:

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

Answer:

(i) \frac{2}{3}   from  \frac{5}{6}

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

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

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

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

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

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

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

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

Answer:

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

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

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

Answer:

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

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

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

Answer:

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

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

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Question 10: Find the products:

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

Answer:

(i) \frac{4}{9} \times \frac{7}{12} = \frac{7}{27}     (ii)  -9 \times \frac{7}{18} = \frac{(-7)}{2 }    (iii) \frac{-3}{16} \times \frac{8}{-15} = \frac{1}{10 }     (iv) \frac{6}{7} \times \frac{-21}{12} = \frac{-3}{2}

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Question 11: Find the quotient:

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

Answer:

(i) \frac{ 17}{8} \div \frac{51}{4} = \frac{17}{8} \times \frac{4}{51} = \frac{4}{24} = \frac{1}{6}      (ii) \frac{ (-16)}{35} \div \frac{15}{14} = \frac{(-16)}{35} \times \frac{14}{15} = \frac{(-32)}{75}

(iii) \frac{ (-12)}{7} \div (-16) = \frac{ (-12)}{7} \times \frac{(-1)}{16} = \frac{3}{28}      (iv)  -9 \div \frac{(-5)}{18} = -9 \times \frac{18}{-5} = \frac{162}{5}

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

Answer:

a \times b= -7

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

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

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

Answer:

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

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

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

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

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

Answer:

(i) \frac{7}{10} and \frac{10}{17}

First take the LCM of 10 and 17 which is 170 .

Convert the numbers with 170 as the denominator.

Hence we get \frac{109}{170} and \frac{100}{170}

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

(ii) 1 \frac{3}{8} and 2 Or \frac{11}{8 } and \frac{16}{8}

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

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

First take the LCM of 5 and 7 which is 35 .

Convert the numbers with 35 as the denominator.

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

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

(iv) - 2 and \frac{-17}{21} Or \frac{-42}{21} and \frac{-17}{21}

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

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Question 15: Find three rational numbers between:

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

Answer:

(i) 4 and 4   \frac{2}{3}    Or  \frac{12}{3} and  \frac{14}{3}

Therefore the rational numbers between 4 and 4  \frac{2}{3} are \frac{125}{30} and  \frac{126}{30} and  \frac{127}{30} Or  \frac{25}{6} and  \frac{43}{10} and  \frac{127}{30}

(ii) \frac{-1 }{2} and  \frac{-1 }{4} Or  \frac{-2 }{4} and  \frac{-1 }{4}

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

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

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Question 16: Find 5 rational numbers between:

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

Answer:

(i) \frac{3}{5} and  \frac{2}{3} Or  \frac{9}{15} and  \frac{10}{15}

The rational numbers are

 \frac{91}{150} ,  \frac{92}{150} ,  \frac{93}{150} ,  \frac{94}{150} ,  \frac{95}{150} Or  \frac{91}{150} ,  \frac{46}{75} ,  \frac{31}{50} ,  \frac{47}{75} ,  \frac{19}{30}

(ii) -2 and -1 \frac{1}{2} Or  \frac{-4 }{2} and  \frac{-3 }{2}

The rational numbers are \frac{-31 }{20} ,  \frac{-32 }{20} ,  \frac{-33 }{20} ,  \frac{-34 }{20} ,  \frac{-35 }{20} Or  \frac{-31 }{20} ,  \frac{-8 }{5} ,  \frac{-33 }{20} ,  \frac{-17 }{10} ,  \frac{-7 }{4}

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

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

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

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

Answer:

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

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

(ii) 0.6 : Rational

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

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

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

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

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

(v) 0.142857 \ldots : Irrational

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

(vi) \frac{4 \sqrt{18}}{3 \sqrt{2}} = $latex \frac{4 \times 3 \sqrt{2}}{3

\sqrt{2}} &s=2$ =4 : Rational

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

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

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

Proof: Lets first assume \sqrt{2} to be rational.
\sqrt{2} = \frac{a}{b} , where b is not equal to 0 .
Here, a and b are co-primes whose HCF is 1 .
\sqrt{2} = \frac{a}{b} ( squaring both sides )…
2 = \frac{a^2}{b^2}
2b^2 = a^2      … … … … … (i)  
Here, 2 divides a^2 also a ( because, if a prime number divides the square of a positive integer, then it divides the integer itself )
Now, let a = 2c ( squaring both sides )…
a^2 = 4c^2      … … … … … (ii)  
Substituting Equation (i) in Equation (ii),
2 b^2 = 4c^2    \Rightarrow b^2 = 2c^2    \Rightarrow 2c^2 = b^2
2 divides b^2 as well as b .
Conclusion:
Here, a and b are both divisible by 2 . But our assumption that their HCF is 1 is being contradicted.
Therefore, our assumption that \sqrt{2} is rational is wrong. Thus, it is irrational.

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

Proof: Let x = 2 \sqrt{3} - 3 \sqrt{2} is a rational number.
Squaring both sides, x^2 = (2 \sqrt{3} - 3 \sqrt{2})^2 = 30 - 12 \sqrt{6}
\Rightarrow 12\sqrt{6} = 30-x^2
\Rightarrow \frac{30-x^2}{12} = \sqrt{6} = Irrational number.
Now if x is rational, then x^2 is also rational and hence \frac{30-x^2}{12} which contradicts our initial assumption. Hence $latex 2 \sqrt{3} - 3 \sqrt{2}

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

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

(x) \pi : Irrational

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

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

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

-12 can be written as, \frac{p}{q} , where p and q are non-zero integer. For example - 12 = \frac{-24}{2} = \frac{-12}{1} Hence, -12 is a rational number.

(xii) 1.411443143 \ldots : Rational

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

(xiii)  \sqrt{0.9} : Irrational

\sqrt{10} is irrational,so it’s inverse \frac{1}{\sqrt{10}} (which is the same as \sqrt{0.1} ) is irrational. All unresolved roots are known as an irrational number since  \sqrt{0.9}  is an unresolved root which is also known as surds, so it is an irrational number.

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

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

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

\sqrt{2} = is an irrational number as proved above. Hence it’s inverse \frac{1}{\sqrt{2}} is irrational number.

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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

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Question / Answer 20: Fill in the blanks

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

(ii)       0      is a rational number that does not have a multiplicative inverse.

(iii)       1      and      -1      are two rational numbers which are equal it their own reciprocal.

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

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

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

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

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Question 21: Arrange in ascending order

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

Answer:

(i) 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 \sqrt{ 192} , \sqrt{60}, \sqrt{121}, \sqrt{24}, \sqrt{63}

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

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

(ii)  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 \sqrt{ 80} , \sqrt{122}, \sqrt{147}, \sqrt{169}, \sqrt{128}

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

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

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Question 22: Write the rationalizing factors of the following:

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

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

Answer:

(i)  \sqrt{3}

\sqrt{3} \times \sqrt{3}=3 Therefore rationalizing factor is \sqrt{3}

(ii)  6 \sqrt{3}

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

(iii)  3+ \sqrt{2}

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

(iv)  \sqrt{7}+ \sqrt{3}

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

(v)  5- \sqrt{11}

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

(vi)  (3 \sqrt{5}-3)

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

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

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

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

Answer:

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

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

(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})

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

(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}

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

(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}

(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}

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Question 24: Insert 5 rational numbers between:

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

Answer:

(i) \sqrt{5}   and \sqrt{13}

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

(ii) \sqrt{7}   and 3 \sqrt{3}

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

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

(iii) 2 and 2.5

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

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

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Question & Answer 25: State True or False:

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

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

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

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

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

(vi)  (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)

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