Page 1.24
Type 1
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
(i) 9
Sol : 9=\dfrac{9}{1}
(ii) -13
Sol : -13=\dfrac{-13}{1}
(iii) 20
Sol : 20=\dfrac{20}{1}
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 .
[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
[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 24 and 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.
[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.
[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
[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]
(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}
(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
(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}
Type 3
(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 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 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}
Type 4
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}
[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}
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}
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}
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
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
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}
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}
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}
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}
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}
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
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
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)
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.
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 24 and 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 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}
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 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 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}
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 0.2Sol :
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=3Now, \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=2Now, \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 0.1Sol :
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 numbersSol :
⇒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-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)
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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