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KC Sinha Mathematics Solution Class 9 Chapter 1 Real numbers exercise 1.1

Page 1.24

Exercise 1.1


Type 1

Question 1 

State whether the following statements are true or false ?

Also give reasons for your answer.

(i) Every whole number is a natural number

Sol :  False, since 0 (zero) is a whole number but not a natural number.

(ii) Every integer is a rational number

Sol :  True, since every integer m may be written in the form \dfrac{m}{1} so it is a rational number.

(iii) Every rational number is an integer

Sol :  False, since \dfrac{2}{3} is a rational number but it is not an integer

(iv) If any rational number \dfrac{p}{q} is an integer , then q=±1

Sol : False, since \dfrac{4}{2}=2(integer) but q=2

 


Question 2

Write the following integers in the form of rational number \dfrac{p}{q}

(i) 9

Sol : 9=\dfrac{9}{1}

(ii) -13

Sol : -13=\dfrac{-13}{1}

(iii) 20

Sol : 20=\dfrac{20}{1}


Question 3

(i) Is \dfrac{p}{q} a rational number, if p=0 ?

Sol :

Yes , because \dfrac{0}{\text{some integer}}=0 by definition of rational number which can be expressed in term form of p/q when q is non zero integer  and p is a integer .

(ii) Is \dfrac{p}{q} a rational number , if q=0 ?

Sol :

No , According to definition of rational number which can be expressed in term form of p/q when q is non zero integer  and p is a integer .


Question 4

Fill up the blanks with the word terminating, non—terminating, repeating.

[Note:Let x = p/q be a rational number, such that the prime factorization of q is of the form 2n 5m, where n, m are non-negative integers. Then x has a decimal expansion which terminates.]

(i) On changing \dfrac{7}{16} to a decimal , it will be __ decimal

Sol:

Terminating

(ii) On changing \dfrac{3}{25} to a decimal, it will be __ decimal

Sol:

Terminating

(iii) On changing \dfrac{7}{12} to a decimal, it will be __ decimal

Sol:

Non-Terminating

(iv) If denominator of a rational number \dfrac{p}{q} has prime factors 2 and 5 only , the \dfrac{p}{q} can be written in __ decimal form

Sol:

Terminating

 


Question 5

Without writing following rational numbers in decimal forms , state which will have terminating decimal expansion ?

[Note:Let x = p/q be a rational number, such that the prime factorization of q is of the form 2n 5m, where n, m are non-negative integers. Then x has a decimal expansion which terminates.]

(i) \dfrac{2}{11}

Sol:

Here q=11 and it is not in the form of 2n5m, it is not a terminating decimal.

 

(ii) \dfrac{5}{9}

Sol:

Here q=9

Can be written as 32 and it is not in the form of 2n5m, it is not a terminating decimal.

 

(iii) \dfrac{9}{16}

Sol:

Here q=16

Can be written as 2and it is in the form of 2n5m, it is a terminating decimal.

 

(iv) \dfrac{11}{30}

Sol :

Here q=30

Can be written as 2×3×5

Since 3 is also there and it is not in the form of 2n5m, it is not a terminating decimal.


Question 6

Without writing following rational numbers in decimal forms , state which will have non-terminating decimal expansion ?

[Note:Let x = p/q be a rational number, such that the prime factorization of q is of the form 2n 5m, where n, m are non-negative integers. Then x has a decimal expansion which terminates.]

(i) \dfrac{5}{9}

Sol :

Here q=9

Can be written as 32 and it is not in the form of 2n5m, it is not a terminating decimal.

 

(ii) \dfrac{3}{8}

Sol :

Here q=8

Can be written as 23 and it is in the form of 2n5m, it is a terminating decimal.

 

(iii) \dfrac{7}{25}

Sol :

Here q=25

Can be written as 52 and it is in the form of 2n5m, it is a terminating decimal.

 

(iv) \dfrac{21}{20}

Sol :

Here q=20

Can be written as 22×5 and it is in the form of 2n5m, it is a terminating decimal.

 


Question 7

State which of the following rational numbers represent terminating decimal expansion ?

[Note:Let x = p/q be a rational number, such that the prime factorization of q is of the form 2n 5m, where n, m are non-negative integers. Then x has a decimal expansion which terminates.]

(i) \dfrac{3}{8}

Sol :

Denominator = 8 = 2×2×2

Can be written as terminating expansion [is in form 2n5m]

 

(ii) \dfrac{23}{7}

Sol :

Denominator = 7

Can not be written as terminating expansion [not in form 2n5m]

 

(iii) \dfrac{27}{40}

Sol :

Denominator = 40 = 2×2×2×5

Can be written as terminating expansion [is in form 2n5m]

 

(iv) \dfrac{27}{130}

Sol :

Denominator = 130 = 2×5×13

Can not be written as terminating expansion because it is not in the form of 2n5m [as 13 is present in prime factorization]

 

(v) \dfrac{38}{35}

Sol :

Denominator = 35 = 7×5

Can not be written as terminating expansion because it is not in the form of 2n5m [as 7 is present in prime factorization]

 

(vi) \dfrac{25}{128}

Sol :

Denominator = 128 = 2×2×2×2×2×2×2

Can be written as terminating expansion [ is in the form of 2n5m]

(vii) \dfrac{17}{138}

Sol :

Denominator = 138 = 2×3×23

Can not be written as terminating expansion because it is not in the form of 2n5m [as 3 and 23 is present in prime factorization]

 

(viii) \dfrac{29}{50}

Sol :

Denominator = 50 = 2×5×5

Can be written as terminating expansion because it is in the form of 2n5m

 


Question 8

State which of the following rational numbers represent non-terminating decimal expansion ?

[Note:Let x = p/q be a rational number, such that the prime factorization of q is of the form 2n 5m, where n, m are non-negative integers. Then x has a decimal expansion which terminates.]

(i) \dfrac{11}{7}

Sol :

Denominator = 7

Can not be written as terminating expansion [not in form 2n5m]

 

(ii) \dfrac{3}{10}

Sol :

Denominator = 10 = 2×5

Can be written as terminating expansion [is in form 2n5m]

 

(iii) \dfrac{7}{18}

Sol :

Denominator = 18 = 2×3×3

Can not be written as terminating expansion because it is not in the form of 2n5m [as 3 is present in prime factorization]

 

(iv) \dfrac{23}{250}

Sol :

Denominator = 250 =2×5×5×5

Can be written as terminating expansion [is in form 2n5m]

 

(v) \dfrac{17}{21}

Sol :

Denominator = 21 = 3×7

Can not be written as terminating expansion because it is not in the form of 2n5m [as 3 and 7 is present in prime factorization]

 

(vi) \dfrac{29}{30}

Sol :

Denominator = 30 = 2×3×5

Can not be written as terminating expansion because it is not in the form of 2n5m [as 3 is present in prime factorization]

 

(vii) \dfrac{29}{121}

Sol :

Denominator = 121 = 11×11

Can not be written as terminating expansion because it is not in the form of 2n5m [as 11 is present in prime factorization]

 

(viii) \dfrac{31}{60}

Sol :

Denominator = 60 = 2×2×3×5

Can not be written as terminating expansion because it is not in the form of 2n5m [as 3 is present in prime factorization]

 


Question 9

Write the following in decimal form and state, what kind of decimal expansion each has ?

(i) \dfrac{36}{100}

Sol :

Its denominator 100 = 22×52=(2×5)2

Since denominator has prime factors 2 and 5 only , so its decimal expansion is terminating

Now , \dfrac{36}{100}=0.36

 


(ii) \dfrac{10}{3}

Sol :

Its denominator 3

Since its denomination is not in the form of 2m×5n , therefore its decimal expansion is non-terminating recuring

\\3\overline{)10(}3.3\dots\\\phantom{3)}\underline{\phantom{0}9\phantom{0}}\\\phantom{3)9}10\\\phantom{3)}\underline{\phantom{11}9\phantom{0}}\\\phantom{3)11}10\\\phantom{3)110} \vdots

=3.\overline{3}

 

(iii) \dfrac{1}{11}

Its denominator 11

Since its denomination is not in the form of 2m×5n , therefore its decimal expansion is non-terminating recuring

\\11\overline{)100(}0.09\dots\\\phantom{11)}\underline{\phantom{0}99\phantom{0}}\\\phantom{11)1}100\\\phantom{11)}\underline{\phantom{11}99\phantom{0}}\\\phantom{11)100}1\\\phantom{11)111} \vdots

=0.\overline{09}

 

(iv) \dfrac{7}{8}

Its denominator 8 = 23

Since denominator has prime factors 2 only , so its decimal expansion is terminating

\\8\overline{)70(}0.875\\\phantom{8)}\underline{64\phantom{0}}\\\phantom{8)7}60\\\phantom{3)}\underline{\phantom{7}56\phantom{0}}\\\phantom{8)70}40\\\phantom{8)70}\underline{40}\\\phantom{8)700}0

=0.875

 

(v) \dfrac{3}{10}

Its denominator 10 = 2×5

Since denominator has prime factors 2 and 5 only , so its decimal expansion is terminating

\\10\overline{)30(}0.3\\\phantom{10)}\underline{30}\\\phantom{100)}0

=0.3

 

(vi) \dfrac{1}{3}

Its denominator 3

Since its denomination is not in the form of 2m×5n , therefore its decimal expansion is non-terminating recuring

\\3\overline{)10(}0.3\dots\\\phantom{3)}\underline{\phantom{0}9\phantom{0}}\\\phantom{3)9}10\\\phantom{3)}\underline{\phantom{11}9\phantom{0}}\\\phantom{3)11}1\\\phantom{3)11} \vdots

=0.\overline{3}

 


Question 10

Write the following rational numbers in the decimal form:

(i) \dfrac{5}{16}

Sol : 0.3125

(ii) \dfrac{10}{3}

Sol : 3.3

(iii) \dfrac{3}{11}

Sol : 0.27

(iv) \dfrac{6}{7}

Sol : 0.\overline{857142}

(v) \dfrac{5}{21}

Sol : 0.\overline{238095}

(vi) \dfrac{327}{500}

Sol : 0.654

(vii) \dfrac{5}{6}

Sol : 0.83

(viii) \dfrac{33}{26}

Sol : 1.2692307

 


Type 2

Question 11

Show that the following numbers can be represented in the form \dfrac{p}{q} where p and q are integers and q≠0:

(i) 1.\overline{27}

Sol :

Let x=1.\overline{27}

Then , x=1.2727..   (i)

In x , number of digits after decimal which are not in the repeating block of digits is zero as there is no such digit. Here, repeating block is 27 , in which number of digits is 2 .

Multiply both sides of (i) by 10or 100

100x=100×(1.2727..)

100x=127.2727.. (ii)

Subtracting (i) from (ii) , we get

100x-x=127.2727.. – 1.2727..

99x=126

x=\dfrac{126}{99}=\dfrac{42}{33}=\dfrac{14}{11}

 

(ii) 0.3333…

Sol :

Let x=0.3.. (i)

In x , number of digits after decimal which are not in the repeating block of digits is zero as there is no such digit. Here, repeating block is 3 , in which number of digits is 1 .

Multiply both sides of (i) by 10or 10

10x=10×(0.3..)

10x=3.3.. (ii)

Subtracting (i) from (ii) , we get

10x-x=3.3.. – 0.3..

9x=3

x=\dfrac{3}{9}=\dfrac{1}{3}

 

 

(iii) 0.\overline{6}

Sol :

Let x=0.6.. (i)

In x , number of digits after decimal which are not in the repeating block of digits is zero as there is no such digit. Here, repeating block is 6 , in which number of digits is 1 .

Multiply both sides of (i) by 10or 10

10x=10×(0.6..)

10x=6.6.. (ii)

Subtracting (i) from (ii) , we get

10x-x=6.6.. – 0.6..

9x=6

x=\dfrac{6}{9}=\dfrac{2}{3}

 

(iv) 0.2353532…

Sol :

Let x=0.235353.. (i)

In x , number of digits after decimal which are not in the repeating block of digits is zero as there is no such digit. Here, repeating block is 235353 , in which number of digits is 6 .

Multiply both sides of (i) by 10or 1000000

1000000x=1000000×(0.235353..)

1000000x=235353.235353.. (ii)

Subtracting (i) from (ii) , we get

1000000x-x=235353.235353.. – 0.235353..

99999x=235353

x=\dfrac{235353}{99999}

 

(v) 3.\overline{142678}

Sol :

Let x=3.142678.. (i)

In x , number of digits after decimal which are not in the repeating block of digits is zero as there is no such digit. Here, repeating block is 142678 , in which number of digits is 6 .

Multiply both sides of (i) by 10or 1000000

1000000x=1000000×(3.142678..)

1000000x=3142678.142678.. (ii)

Subtracting (i) from (ii) , we get

1000000x-x=3142678.142678.. – 3.142678..

99999x=3142675

x=\dfrac{3142675}{99999}

 

 


Type 3

Question 12

Write the following in the form of \dfrac{p}{q} , where p and q are integers and q≠0 :

(i) 0.25

Sol :

=\dfrac{25}{100}=\dfrac{5}{20}=\dfrac{1}{4}

 

(ii) 0.54

Sol :

=\dfrac{54}{100}=\dfrac{27}{50}

 

(iii) 6.\overline{46}

Sol :

Let x=6.4646..   (i)

In x , number of digits after decimal which are not in the repeating block of digits is zero as there is no such digit. Here, repeating block is 46 , in which number of digits is 2 .

Multiply both sides of (i) by 10or 100

100x=100×(6.4646..)

100x=646.4646.. (ii)

Subtracting (i) from (ii) , we get

100x-x=646.4646.. – 6.4646..

99x=145

x=\dfrac{640}{99}

 

(iv) 0.0\overline{3}

Sol :

Let x=0.033..  (i)

Here, number of digits af‌ter decimal which are not in the repeating block, m=1 .

Now multiply both sides of (i) by 10m=101=10 we get

10x=10×0.03..

10x=0.3.. (ii)

Again, number of digits in repeating block, n=1

Multiplying both sides of (ii) by 10n=101=10 we get

100x=10×0.3..

100x=3.3.. (iii)

Subtracting (ii) from (iii) , we get

100x-10x=3.3.. – 0.3..

90x=3

x=\dfrac{3}{90}=\dfrac{1}{30}

 

(v) 4.6\overline{732}

Sol :

Let x=4.6732732..  (i)

Here, number of digits af‌ter decimal which are not in the repeating block, m=1 .

Now multiply both sides of (i) by 10m=101=10 we get

10x=10×4.6732732..

10x=46.732732.. (ii)

Again, number of digits in repeating block, n=3

Multiplying both sides of (ii) by 103=103=1000 we get

10000x=1000×46.732732

10000x=46732.732.. (iii)

Subtracting (ii) from (iii) , we get

10000x-10x=46732.732.. – 46.732732..

9990x=46686

x=\dfrac{46686}{9990}=\dfrac{23343}{4995}

\dfrac{7781}{1665}

 

(vi) 4.\overline{27}

Sol :

Let x=4.2727..  (i)

In x , number of digits after decimal which are not in the repeating block of digits is zero as there is no such digit. Here, repeating block is 27 , in which number of digits is 2 .

Multiply both sides of (i) by 10or 100

100x=100×(4.2727..)

100x=427.27.. (ii)

Subtracting (i) from (ii) , we get

100x-x=427.27.. – 4.27..

99x=423

x=\dfrac{423}{99}=\dfrac{141}{33}

=\dfrac{47}{11}

 

(vii) 3.\overline{7}

Sol :

Let x=3.7..  (i)

In x , number of digits after decimal which are not in the repeating block of digits is zero as there is no such digit. Here, repeating block is 7 , in which number of digits is 1 .

Multiply both sides of (i) by 101 or 10

10x=10×(3.7..)

10x=37.7.. (ii)

Subtracting (i) from (ii) , we get

10x-x=37.7.. – 3.7..

9x=34

=\dfrac{34}{9}

 

(viii) 18.\overline{48}

Sol :

Let x=18.48..  (i)

In x , number of digits after decimal which are not in the repeating block of digits is zero as there is no such digit. Here, repeating block is 48 , in which number of digits is 2 .

Multiply both sides of (i) by 10or 100

100x=100×(18.48..)

100x=1848.48.. (ii)

Subtracting (i) from (ii) , we get

100x-x=1848.48.. – 18.48..

99x=1830

x=\dfrac{1830}{99}=\dfrac{610}{33}

 


Type 4

Question 13

Find f‌ive rational numbers between 1 and 2.

Sol :

Here a=1 and b=2 and n=5

Now, \dfrac{b-a}{n+1}=\dfrac{2-1}{5+1}=\dfrac{1}{6}

The required rational numbers will be

a+\dfrac{b-a}{n+1} , a+\dfrac{2(b-a)}{n+1} , a+\dfrac{3(b-a)}{n+1} , a+\dfrac{4(b-a)}{n+1} , a+\dfrac{5(b-a)}{n+1}

1+\dfrac{1}{6} , 1+\dfrac{2}{6}  ,  1+\dfrac{3}{6} , 1+\dfrac{4}{6} , 1+\dfrac{5}{6}

\dfrac{6+1}{6} , \dfrac{6+2}{6}  ,  \dfrac{6+3}{6} , \dfrac{6+4}{6} , \dfrac{6+5}{6}

\dfrac{7}{6} , \dfrac{8}{6}  ,  \dfrac{9}{6} , \dfrac{10}{6} , \dfrac{11}{6}

\dfrac{7}{6} , \dfrac{4}{3}  ,  \dfrac{3}{2} , \dfrac{5}{3} , \dfrac{11}{6}

 


Question 14

Find five rational numbers between \dfrac{3}{5} and \dfrac{4}{5}

[Hint: \dfrac{3}{5}=\dfrac{30}{50};\dfrac{4}{5}=\dfrac{40}{50}] Now write five rational numbers between \dfrac{30}{50}\text{ and }\dfrac{40}{50}]

Sol :

Let a=\dfrac{3}{5}=\dfrac{30}{50} and b=\dfrac{4}{5}=\dfrac{40}{50}

Five rational numbers are =\dfrac{31}{50},\dfrac{32}{50},\dfrac{33}{50},\dfrac{34}{50} ,\dfrac{35}{50}

 


Question 15

Write one rational number between \dfrac{4}{5}\text{ and }\dfrac{7}{13}

Sol :

Here a=\dfrac{4}{5} , b=\dfrac{7}{13} and n=1

The required rational numbers will be

a+\dfrac{b-a}{n+1}

\dfrac{4}{5}+\dfrac{\dfrac{7}{13}-\dfrac{4}{5}}{1+1}

\dfrac{4}{5}+\dfrac{\dfrac{7\times5-4\times 13}{65}}{1+1}

\dfrac{4}{5}+\dfrac{\dfrac{35-52}{65}}{2}

\dfrac{4}{5}+\dfrac{\dfrac{-17}{65}}{2}

\dfrac{4}{5}+\left(\dfrac{-17}{65} \times \dfrac{1}{2}\right)

\dfrac{4}{5}+\left(\dfrac{-17}{130}\right)

\dfrac{4}{5}-\dfrac{17}{130}

\dfrac{4\times 26 -17 \times 1}{130}

\dfrac{104-17}{130}

\dfrac{87}{130}

 


Question 16

Find three rational numbers between -\dfrac{2}{5}\text{ and }-\dfrac{1}{5}

Sol :

Here a=-\dfrac{2}{5}  , b=-\dfrac{1}{5} and n=3

Now, \dfrac{b-a}{n+1}=\dfrac{\dfrac{-1}{5}-\dfrac{-2}{5}}{3+1} =\dfrac{\dfrac{-1+2}{5}}{4} =\dfrac{1}{5}\times \dfrac{1}{4} =\dfrac{1}{20}

The required rational numbers will be

a+\dfrac{b-a}{n+1} , a+\dfrac{2(b-a)}{n+1} , a+\dfrac{3(b-a)}{n+1}

\dfrac{-2}{5}+\dfrac{1}{20} , \dfrac{-2}{5}+2\times \dfrac{1}{20} , \dfrac{-2}{5}+3\times \dfrac{1}{20}

\dfrac{-2\times 4 +1\times 1}{20} , \dfrac{-2}{5} + \dfrac{2}{20}  , \dfrac{-2}{5}+\dfrac{3}{20}

\dfrac{-8+1}{20} , \dfrac{-2\times 4 +2\times 1}{20}  , \dfrac{-2\times 4+3\times 1}{20}

\dfrac{-7}{20} , \dfrac{-8+2}{20}  , \dfrac{-8+3}{20}

\dfrac{-7}{20} , \dfrac{-6}{20}  , \dfrac{-5}{20}

\dfrac{-7}{20} , \dfrac{-3}{10}  , \dfrac{-1}{4}

 


Question 17

Find three rational numbers between 0 and 0.2

Sol :

Average of 0 and 0.2 =\dfrac{0+0.2}{2}=0.1

Average of 0.1 and 0 =\dfrac{0.1+0}{2}=0.05

Average of 0.05 and 0.2 =\dfrac{0.05+0.2}{2}=0.12

The required rational numbers are 0.1 , 0.05 , 0.12

 

Alternate method

Here a=0 , b=0.2 and n=3

Now, \dfrac{b-a}{n+1}=\dfrac{0.2-0}{3+1}=\dfrac{0.2}{4}

The required rational numbers will be

a+\dfrac{b-a}{n+1} , a+\dfrac{2(b-a)}{n+1} , a+\dfrac{3(b-a)}{n+1}

0+\dfrac{0.2}{4} , 0+2\times \dfrac{0.2}{4} , 0+3\dfrac{0.2}{4}

\dfrac{0.2}{4} , \dfrac{0.4}{4} , \dfrac{0.6}{4}

⇒0.05 , 0.1 , 0.15

 


Question 18

Find two rational numbers between \dfrac{5}{6}\text{ and }\dfrac{6}{7}

Sol :

Here a=\dfrac{5}{6} , b=\dfrac{6}{7} and n=2

Now, \dfrac{b-a}{n+1}=\dfrac{\dfrac{6}{7}-\dfrac{5}{6}}{2+1} =\dfrac{\dfrac{6\times 6 - 5\times 7}{42}}{3} =\dfrac{\dfrac{36-35}{42}}{3}=\dfrac{1}{42}\times \dfrac{1}{3} =\dfrac{1}{126}

The required rational numbers will be

a+\dfrac{b-a}{n+1} , a+\dfrac{2(b-a)}{n+1}

\dfrac{5}{6}+\dfrac{1}{126} , \dfrac{5}{6}+2\times \dfrac{1}{126}

\dfrac{5\times 21+1\times 1}{126} , \dfrac{5}{6}+\dfrac{2}{126}

\dfrac{105+1}{126} , \dfrac{5\times 21 + 2\times 1}{126}

\dfrac{106}{126} , \dfrac{105+2}{126}

\dfrac{53}{63} , \dfrac{107}{126}

 


Question 19

Find one rational numbers between \dfrac{3}{4}\text{ and }\dfrac{4}{3}

Sol :

Here a=\dfrac{3}{4} , b=\dfrac{4}{3}

The required rational numbers will be

Average =\dfrac{\dfrac{3}{4}+\dfrac{4}{3}}{2}

=\dfrac{\dfrac{3\times 3+4\times 4}{12}}{2}

=\dfrac{\dfrac{9+16}{12}}{2}

=\dfrac{\dfrac{25}{12}}{2}

=\dfrac{25}{12}\times \dfrac{1}{2}

\dfrac{25}{24}

 


Question 20

Find two rational numbers between \dfrac{1}{2}\text{ and }\dfrac{3}{4}

Sol :

Here a=\dfrac{1}{2} , b=\dfrac{3}{4} and n=2

Now, \dfrac{b-a}{n+1}=\dfrac{\dfrac{3}{4}-\dfrac{1}{2}}{2+1} =\dfrac{\dfrac{3\times 1-1\times 2}{4}}{3}=\dfrac{\dfrac{3-2}{4}}{3}=\dfrac{\dfrac{1}{4}}{3}  =\dfrac{1}{4}\times \dfrac{1}{3}=\dfrac{1}{12}

The required rational numbers will be

a+\dfrac{b-a}{n+1} , a+\dfrac{2(b-a)}{n+1}

\dfrac{1}{2}+\dfrac{1}{12} , \dfrac{1}{2}+2\times \dfrac{1}{12}

\dfrac{1\times 6+1}{12} , \dfrac{1}{2}+\dfrac{2}{12}

\dfrac{6+1}{12} , \dfrac{1\times 6+2\times 1}{12}

\dfrac{7}{12} , \dfrac{6+2}{12}

\dfrac{7}{12} , \dfrac{8}{12}

\dfrac{7}{12} , \dfrac{4}{6}

\dfrac{7}{12} \dfrac{2}{3}

 


Question 21 

Find two rational numbers between \dfrac{1}{3}\text{ and }\dfrac{1}{2}

Sol :

Here a=\dfrac{1}{3} , b=\dfrac{1}{2}

Average =\dfrac{\dfrac{1}{3}+\dfrac{1}{2}}{2}=\dfrac{\dfrac{2+3}{6}}{2} =\dfrac{\dfrac{5}{6}}{2}=\dfrac{5}{6}\times \dfrac{1}{2} =\dfrac{5}{12}

and another is

Average =\dfrac{\dfrac{1}{3}+\dfrac{5}{12}}{2}=\dfrac{\dfrac{4+5}{12}}{2} =\dfrac{\dfrac{9}{12}}{2}=\dfrac{9}{12}\times \dfrac{1}{2} =\dfrac{3}{4}\times \dfrac{1}{2}=\dfrac{3}{8}

The required rational numbers will be

\dfrac{3}{8} , \dfrac{5}{12}

Alternate Method

Here a=\dfrac{1}{3} , b=\dfrac{1}{2} and n=2

Now, \dfrac{b-a}{n+1}=\dfrac{\dfrac{1}{2}-\dfrac{1}{3}}{2+1} =\dfrac{\dfrac{1\times 3-1\times 2}{6}}{3} =\dfrac{\dfrac{3-2}{6}}{3}=\dfrac{\dfrac{1}{6}}{3}  =\dfrac{1}{6}\times \dfrac{1}{3} =\dfrac{1}{18}

The required rational numbers will be

a+\dfrac{b-a}{n+1} , a+\dfrac{2(b-a)}{n+1}

\dfrac{1}{3}+\dfrac{1}{18} , \dfrac{1}{3}+2\times \dfrac{1}{18}

\dfrac{1\times 6+1\times 1}{18} , \dfrac{1}{3}+ \dfrac{2}{18}

\dfrac{6+1}{18} , \dfrac{1\times 6+2\times 1}{18}

\dfrac{7}{18} , \dfrac{1\times 6+2\times 1}{18}

\dfrac{7}{18} , \dfrac{6+2}{18}

\dfrac{7}{18} , \dfrac{8}{18}

\dfrac{7}{18} , \dfrac{4}{9}

 


Question 22 

Find two rational numbers between 0 and 0.1

Sol :

Here a=0 and b=0.1

The required rational numbers will be

Average of 0 and 0.1 =\dfrac{0+0.1}{2}=0.05

Average of 0.05 and 0.1 =\dfrac{0.05+0.1}{2}=0.07

Two rational numbers are

⇒0.05 , 0.07

 


Question 23

How many rational numbers can be written between 2.5 and 2.6 ? Write ten of these numbers

Sol :

⇒Infinitely many rational numbers can be written between 2.5 and 2.6

 

⇒ Ten rational numbers between 2.5 and 2.6 are 2.51 , 2.52 , 2.53 , 2.54 , 2.55 , 2.56 , 2.57  , 2.58 , 2.59 , 2.511 .

 


Type 5

Question 24

If x and y are rational numbers then, show that the following are also rational numbers :

 

Properties of rational numbers

⇒Sum of two rational numbers is rational number ..(i)

⇒Difference of two rational numbers is rational number ..(ii)

⇒Product of two rational numbers is rational number..(iii)

⇒Division of two rational numbers is rational number..(iv)

 

(i) x2-y2

Sol :

x2-y

can be written as (x+b)(x-b)

x+b is a rational number by (i)

x-b is a rational number by (ii)

So , x2-y2 is also a rational number

 

(ii) x-y

Sol :

x-y is a rational number by (ii)

 

(iii) \dfrac{x}{y} , where y≠0

Sol :

\dfrac{x}{y} , where y≠0 is a rational number by (iv)

 

(iv) x+y

Sol :

x+y is rational number by (i)

 


Question 25

If a is a rational number , then prove that an will be a rational number , where n is a rational number greater than 1 .

Sol :

Rational Number is a number that can be expressed in the form p/q, where q not equal to zero.

We know that product of two rational number is always a rational number.
Hence if a is a rational number then

a2 = a × a is a rational number.

a3 = a2× a is a rational number,

a4 = a3×a is a rational number,

……

……

 an = an-1×a is a rational number.

 


 

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