KC Sinha: Exercise 1.1 - Mathematics Solution Class 9 Chapter 1 Real numbers

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Exercise 1.1

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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

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

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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 .

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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 5^m, 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

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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 5^m, 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 2ⁿ5^m, 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 2ⁿ5^m, it is not a terminating decimal.

(iii) $\dfrac{9}{16}$
Sol:
Here q=16
Can be written as 2⁴ and it is in the form of 2ⁿ5^m, 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 2ⁿ5^m, it is not a terminating decimal.

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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 5^m, 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 2ⁿ5^m, 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 2ⁿ5^m, 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 2ⁿ5^m, 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 2ⁿ5^m, it is a terminating decimal.

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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 5^m, 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 2ⁿ5^m]

(ii) $\dfrac{23}{7}$
Sol :
Denominator = 7
Can not be written as terminating expansion [not in form 2ⁿ5^m]

(iii) $\dfrac{27}{40}$
Sol :
Denominator = 40 = 2×2×2×5
Can be written as terminating expansion [is in form 2ⁿ5^m]

(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 2ⁿ5^m [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 2ⁿ5^m [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 2ⁿ5^m]
(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 2ⁿ5^m [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 2ⁿ5^m

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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 5^m, 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 2ⁿ5^m]

(ii) $\dfrac{3}{10}$
Sol :
Denominator = 10 = 2×5
Can be written as terminating expansion [is in form 2ⁿ5^m]

(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 2ⁿ5^m [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 2ⁿ5^m]

(v) $\dfrac{17}{21}$
Sol :
Denominator = 21 = 3×7
Can not be written as terminating expansion because it is not in the form of 2ⁿ5^m [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 2ⁿ5^m [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 2ⁿ5^m [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 2ⁿ5^m [as 3 is present in prime factorization]

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

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(ii) $\dfrac{10}{3}$
Sol :
Its denominator 3
Since its denomination is not in the form of 2^m×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 2^m×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 2^m×5ⁿ , therefore its decimal expansion is non-terminating recuring
$\\3\overline{)10(}0.3\dots\\\phantom{3)}\underline{\phantom{0}9\phantom{0}}\\\phantom{3)9}10\\\phantom{3)}\underline{\phantom{11}9\phantom{0}}\\\phantom{3)11}1\\\phantom{3)11} \vdots$
$=0.\overline{3}$

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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

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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 10² or 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 10¹ or 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 10¹ or 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 10⁶ or 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 10⁶ or 1000000
1000000x=1000000×(3.142678..)
1000000x=3142678.142678.. (ii)
Subtracting (i) from (ii) , we get
1000000x-x=3142678.142678.. - 3.142678..
99999x=3142675
$x=\dfrac{3142675}{99999}$

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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 10² or 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 10^m=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 10^m=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 10² or 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 10² or 100
100x=100×(18.48..)
100x=1848.48.. (ii)
Subtracting (i) from (ii) , we get
100x-x=1848.48.. - 18.48..
99x=1830
$x=\dfrac{1830}{99}=\dfrac{610}{33}$

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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
⇒$a+\dfrac{b-a}{n+1}$ , $a+\dfrac{2(b-a)}{n+1}$ , $a+\dfrac{3(b-a)}{n+1}$ , $a+\dfrac{4(b-a)}{n+1}$ , $a+\dfrac{5(b-a)}{n+1}$
⇒$1+\dfrac{1}{6}$ , $1+\dfrac{2}{6}$  ,  $1+\dfrac{3}{6}$ , $1+\dfrac{4}{6}$ , $1+\dfrac{5}{6}$
⇒$\dfrac{6+1}{6}$ , $\dfrac{6+2}{6}$  ,  $\dfrac{6+3}{6}$ , $\dfrac{6+4}{6}$ , $\dfrac{6+5}{6}$
⇒$\dfrac{7}{6}$ , $\dfrac{8}{6}$  ,  $\dfrac{9}{6}$ , $\dfrac{10}{6}$ , $\dfrac{11}{6}$
⇒$\dfrac{7}{6}$ , $\dfrac{4}{3}$  ,  $\dfrac{3}{2}$ , $\dfrac{5}{3}$ , $\dfrac{11}{6}$

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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 $=\dfrac{31}{50},\dfrac{32}{50},\dfrac{33}{50},\dfrac{34}{50} ,\dfrac{35}{50}$

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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
⇒$a+\dfrac{b-a}{n+1}$
⇒$\dfrac{4}{5}+\dfrac{\dfrac{7}{13}-\dfrac{4}{5}}{1+1}$
⇒$\dfrac{4}{5}+\dfrac{\dfrac{7\times5-4\times 13}{65}}{1+1}$
⇒$\dfrac{4}{5}+\dfrac{\dfrac{35-52}{65}}{2}$
⇒$\dfrac{4}{5}+\dfrac{\dfrac{-17}{65}}{2}$
⇒$\dfrac{4}{5}+\left(\dfrac{-17}{65} \times \dfrac{1}{2}\right)$
⇒$\dfrac{4}{5}+\left(\dfrac{-17}{130}\right)$
⇒$\dfrac{4}{5}-\dfrac{17}{130}$
⇒$\dfrac{4\times 26 -17 \times 1}{130}$
⇒$\dfrac{104-17}{130}$
⇒$\dfrac{87}{130}$

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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
⇒$a+\dfrac{b-a}{n+1}$ , $a+\dfrac{2(b-a)}{n+1}$ , $a+\dfrac{3(b-a)}{n+1}$
⇒$\dfrac{-2}{5}+\dfrac{1}{20}$ , $\dfrac{-2}{5}+2\times \dfrac{1}{20}$ , $\dfrac{-2}{5}+3\times \dfrac{1}{20}$
⇒$\dfrac{-2\times 4 +1\times 1}{20}$ , $\dfrac{-2}{5} + \dfrac{2}{20}$  , $\dfrac{-2}{5}+\dfrac{3}{20}$
⇒$\dfrac{-8+1}{20}$ , $\dfrac{-2\times 4 +2\times 1}{20}$  , $\dfrac{-2\times 4+3\times 1}{20}$
⇒$\dfrac{-7}{20}$ , $\dfrac{-8+2}{20}$  , $\dfrac{-8+3}{20}$
⇒$\dfrac{-7}{20}$ , $\dfrac{-6}{20}$  , $\dfrac{-5}{20}$
⇒$\dfrac{-7}{20}$ , $\dfrac{-3}{10}$  , $\dfrac{-1}{4}$

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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

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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
⇒$a+\dfrac{b-a}{n+1}$ , $a+\dfrac{2(b-a)}{n+1}$
⇒$\dfrac{5}{6}+\dfrac{1}{126}$ , $\dfrac{5}{6}+2\times \dfrac{1}{126}$
⇒$\dfrac{5\times 21+1\times 1}{126}$ , $\dfrac{5}{6}+\dfrac{2}{126}$
⇒$\dfrac{105+1}{126}$ , $\dfrac{5\times 21 + 2\times 1}{126}$
⇒$\dfrac{106}{126}$ , $\dfrac{105+2}{126}$
⇒$\dfrac{53}{63}$ , $\dfrac{107}{126}$

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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 $=\dfrac{\dfrac{3}{4}+\dfrac{4}{3}}{2}$
$=\dfrac{\dfrac{3\times 3+4\times 4}{12}}{2}$
$=\dfrac{\dfrac{9+16}{12}}{2}$
$=\dfrac{\dfrac{25}{12}}{2}$
$=\dfrac{25}{12}\times \dfrac{1}{2}$
⇒$\dfrac{25}{24}$

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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
⇒$a+\dfrac{b-a}{n+1}$ , $a+\dfrac{2(b-a)}{n+1}$
⇒$\dfrac{1}{2}+\dfrac{1}{12}$ , $\dfrac{1}{2}+2\times \dfrac{1}{12}$
⇒$\dfrac{1\times 6+1}{12}$ , $\dfrac{1}{2}+\dfrac{2}{12}$
⇒$\dfrac{6+1}{12}$ , $\dfrac{1\times 6+2\times 1}{12}$
⇒$\dfrac{7}{12}$ , $\dfrac{6+2}{12}$
⇒$\dfrac{7}{12}$ , $\dfrac{8}{12}$
⇒$\dfrac{7}{12}$ , $\dfrac{4}{6}$
⇒$\dfrac{7}{12}$ $\dfrac{2}{3}$

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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
⇒$a+\dfrac{b-a}{n+1}$ , $a+\dfrac{2(b-a)}{n+1}$
⇒$\dfrac{1}{3}+\dfrac{1}{18}$ , $\dfrac{1}{3}+2\times \dfrac{1}{18}$
⇒$\dfrac{1\times 6+1\times 1}{18}$ , $\dfrac{1}{3}+ \dfrac{2}{18}$
⇒$\dfrac{6+1}{18}$ , $\dfrac{1\times 6+2\times 1}{18}$
⇒$\dfrac{7}{18}$ , $\dfrac{1\times 6+2\times 1}{18}$
⇒$\dfrac{7}{18}$ , $\dfrac{6+2}{18}$
⇒$\dfrac{7}{18}$ , $\dfrac{8}{18}$
⇒$\dfrac{7}{18}$ , $\dfrac{4}{9}$

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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

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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 .

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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-y² 
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)

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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 = a^n-1×a is a rational number.