ML Aggarwal Solution Class 9 Chapter 1 Rational and Irrational Numbers Exercise 1.3

 Exercise 1.3

Question 1

Sol :

PQ is a number line

We have two right angle triangle. They are 

ΔOAB and ΔOCD

In a right angled triangle

(hypotenuse) $^{2}=(\text { side } 1)^{2}+(\text { Side } 2)^{2}$

$\therefore O B^{2}=O A^{2}+A B^{2}$

$O B^{2}=3^{2}+1^{2}$ (∵OA=3, AB=1)

$O B^{2}=9+1$

$O B=\sqrt{10}$

Similarly, in ΔOCD

$O D^{2}=O C^{2}+C D^{2}$

$O D^{2}=4^{2}+1^{2}$ (∵OC=4 , CD=1)

$O D^{2}=16+1$

$O D=\sqrt{17}$

Question 2

Sol :

(i) $\frac{36}{100}$

Remainder becomes zero.

Decimal Expansion of $\frac{36}{100}(=0.36)$ is terminating.

(ii) $4 \frac{1}{8}$

$=\frac{4 \times 8+1}{8}$

$=\frac{33}{8}$

Remainder becomes zero

Decimal expansion of $4 \frac{1}{8}(=4.125)$ is terminating.

(iii) $\frac{2}{9}$

In the above decimal expansion, remainder is repeating. So, it is a non-terminating decimal . 

So, $\frac{2}{9}$=0.222.....$=0 . \dot{2}=0 . \overline{2}$

(iv) $\frac{2}{11}$

Decimal Expansion of $\frac{2}{11}$=0.181818...

Here, remainder is repeating. So, it a non-terminating repeating decimal

$\therefore \frac{2}{9}=0.\overline{18}$

(v) $\frac{3}{13}$

Decimal Expansion of $\frac{3}{13}$ is 0.230769....

Here, remainder is repeating. So, it a non-terminating repeating decimal.

$\therefore \frac{3}{13}=0 . \overline{230769}$

(vi) $\frac{329}{400}$

Decimal Expansion of $\frac{329}{400}$=0.8225

∴It is a terminating decimal Expansion.

Question 3

Sol :

(i) $\frac{13}{3125}$

Prime factorization of denominator 3125 .

$\begin{array}{l|l}5 & 3125 \\\hline 5 & 625 \\\hline 5 & 125 \\\hline 5 & 25 \\\hline 5 & 5 \\\hline & 1\end{array}$

3125=5×5×5×5×5×1

$=5^{5} \times 1$

$=1 \times 5^{5}$

$\therefore 3125=2^{\circ} \times 5^{5}$ $\left(\because 2^{\circ}=1\right)$

Since, denominator is in the form of $2 \times 5^{5}$ the decimal Expansion of $\frac{13}{3125}$ is terminating.

(ii) $\frac{17}{8}$

$\begin{array}{l|l}2 & 8 \\\hline 2 & 4 \\\hline 2 & 2 \\\hline &1\end{array}$

Prime factorization of denominator 8

8=2×2×2

$8=2^{3} \times 1$

$8=2^{3} \times 5^{\circ}$

Since, denominator is in the form of $2^{3} \times 5^{0}$ decimal Expansion of $\frac{17}{8}$ is terminating

(iii) $\frac{23}{75}$

Prime factorization of 75

$\begin{array}{l|l}3 & 75 \\\hline 5 &25 \\\hline 5 & 5 \\\hline &1\end{array}$

75=3×5×5×1

$75=3 \times 5^{2} \times 1$

$75=3 \times 2^{0} \times 5^{2}$ $\left(\because 2^{0}=1\right)$

Since, denominator contains prime factor 3 other than 2 or 5

Decimal Expansion of $\frac{23}{75}$ is non-terminating

(iv) $\frac{6}{15}$

Both numerator and denominator Contains common factor 3

$\frac{6}{15}=\frac{3 \times 2}{3 \times 5}=\frac{2}{5}$

$\therefore \frac{6}{15}=\frac{2}{5}$

Since, denominator is in the form $2^{0} \times 5^{1}$

Decimal Expansion of $\frac{6}{15}\left(=\frac{2}{5}\right)$ is terminating.

(v) $\frac{1258}{625}$

Prime factorization of denominator 625

$\begin{array}{l|l}5 & 625 \\\hline 5 & 125 \\\hline 5 & 25 \\\hline 5 & 5\\\hline &1\end{array}$

625=5×5×5×5×1

625=$5^{4} \times 2^{0}$

Since, denominator is in the form $2^{0} \times 5^{4}$, decimal Expansion of $\frac{1258}{625}$ is terminating

(vi) $\frac{77}{210}$

Both numerator and denominator contains common factor 7

$\frac{77}{210}=\frac{7 \times 11}{7 \times 30}=\frac{11}{30}$

$\therefore \frac{77}{210}=\frac{11}{30}$

Prime factorization of denominator 30.

$\begin{array}{l|l}2 & 30 \\\hline 3 & 15 \\\hline 5 & 5 \\\hline &1\end{array}$

30=2×3×5×1

30=3×2×5

Since, denominator contains prime factor 3 other 2 or 5 .

Decimal expansion of $\frac{77}{210}$ is non-terminating.

Question 4

Sol :

Expressing both numerator and denominator of fraction $\frac{987}{10500}$ as product of prime number by prime factorization method

$\begin{array}{l|l}3 & 987 \\\hline 7 & 329 \\\hline 47 & 47 \\\hline &1\end{array}$

$\therefore 987=3 \times 7 \times 47$

$\begin{array}{l|l}2 & 10500 \\ \hline2 & 5250 \\\hline 3 & 2625 \\\hline 5 & 875 \\\hline 5 & 175 \\\hline 5 & 35 \\\hline 7&7\\\hline &1\end{array}$

$\therefore 10500=2 \times 2 \times 3 \times 5 \times 5 \times 5 \times 7$

$\frac{987}{10500}=\frac{3 \times 7 \times 47}{2 \times 2 \times 3 \times 5 \times 5 \times 5 \times 7}$

$=\frac{47}{2^{2} \times 5^{3}}$

Since, denominator is in the form $2^{2} \times 5^{3}$, decimal expansion of $\frac{987}{10500}$ is terminating

Question 5

Sol :

(i) $\frac{17}{8}$

Prime factorization of denominator 8

$\begin{array}{l|l}2 & 8 \\ \hline2 & 4 \\\hline2 & 2 \\\hline &1\end{array}$

8=2×2×2×1 

$8=2^{3} \times 5^{0}$ $\left(\because a^{0}=1\right)$

$\frac{17}{8}=\frac{17}{2^{3}}$

$=\frac{17 \times 5^{3}}{2^{3} \times 5^{3}}$ (By multiplying both numerator and denominator with $5^{3}$)

$=\frac{17 \times 125}{(2 \times 5)^{3}}$

$=\frac{2125}{10^{3}}$

=2.125

$\begin{array}{r}125 \\\times 17 \\\hline 875 \\125 \times \\\hline 2125 \\\hline \end{array}$

(since, denominator is in the form $10^{3}$, decimal expansion is obtained by moving decimal print to three digits from)

$\therefore \frac{17}{8}=2.125$

(ii) $\frac{13}{3125}$

Prime factorization of 3125

$\begin{array}{l|l}5 & 3125 \\ \hline 5 & 625 \\\hline 5 & 125 \\\hline 5 & 25 \\\hline 5 & 5 \\\hline &1\end{array}$

3125=5×5×5×5×5$=5^5$ 

$\frac{13}{3125}=\frac{13}{5^{5}}$

$=\frac{13 \times 2^{5}}{5^{5} \times 2^{5}}$

$=\frac{13 \times 32}{(2 \times 5)^{5}}$

$=\frac{416}{10^{5}}$

=0.00416

$\begin{array}{r}32 \\\times 13 \\\hline 96 \\32 \times \\\hline 416 \\\hline\end{array}$

$\therefore \frac{13}{3125}=0.00416$

(iii) $\frac{7}{80}$

Prime factorization of 80

$\begin{array}{l|l}2 & 80 \\\hline 2 & 40 \\\hline 2 & 20 \\\hline 2 & 10 \\\hline  5 & 5 \\\hline &1\end{array}$

80=2×2×2×2×5

$80=2^{4} \times 5^{1}$

$\frac{7}{80}=\frac{7}{2^{4} \times 5^{1}}$

$=\frac{7 \times 5^{3}}{2^{4} \times 5^{1} \times 5^{3}}$ (Multiplying numerator and denominator by $5^{3}$)

$=\frac{7 \times 125}{2^{4} \times 5^{4}}$

$=\frac{7 \times 125}{(2 \times 5)^{4}}$

$=\frac{875}{10^{4}}$

=0.0875

(Since, denominator is $10^{4}$, decimal expansion can be obtained by moving decimal point 7 numerator to four digits from right.)

$\therefore \frac{7}{80}=0.0875$

(iv) $\frac{6}{15}$

Prime factorization of 6 and 15

$\begin{array}{l|l}2 & 6 \\\hline 3 & 3 \\\hline & 1\end{array}$

$\therefore 6=2 \times 3$

$\begin{array}{l|l}3 &15 \\\hline5 & 15 \\\hline & 1\end{array}$

$\therefore 15=3 \times 5$

$\frac{6}{15}=\frac{2 \times 3}{3 \times 5}=\frac{2}{5}$ (By multiplying both numerator and denominator by 2)

$=\frac{2 \times 2}{5 \times 2}$

$=\frac{4}{10}$

=0.4

(v) $\frac{2^{2} \times 7}{5^{4}}$

$=\frac{2^{2} \times 7 \times 2^{4}}{5^{4} \times 2^{4}}$ (By multiplying both numerator and denominator by $2^{4}$ )

$=\frac{4 \times 7 \times 16}{(2 \times 5)^{4}}$

$=\frac{28 \times 16}{10^{4}}$

$=\frac{448}{10^{4}}$

=0.0448

$\therefore \frac{2^{2} \times 7}{5^{4}}=0.0448$

(vi) $\frac{237}{1500}$

Prime factorization of 237 and 1500

$\begin{array}{l|l}3 &237 \\\hline 79 & 79 \\\hline &1\end{array}$

$\therefore \quad 237=3 \times 79$

$\begin{array}{l|l}2 & 1500 \\\hline 2 & 750 \\\hline 3 & 375 \\\hline 5 & 125 \\\hline 5 & 25 \\\hline 5 & 5 \\\hline & 1\end{array}$

$\therefore 1500=2 \times 2 \times 3 \times 5 \times 5 \times 5$

$\frac{237}{1500}=\frac{3 \times 79}{2^{2} \times 3 \times 5^{3}}$

$=\frac{79}{2^{2} \times 5^{3}}$ (Multiplying both numerator and denominator by 2)

$=\frac{79 \times 2}{2^{2} \times 5^{3} \times 2}$

$=\frac{158}{2^{3} \times 5^{3}}$

$=\frac{158}{(10)^{3}}$

=0.158

$\therefore \frac{237}{1500}=0.158$

Question 6

Sol :

Given rational number $\frac{257}{5000}$

Prime factorization of denominator 5000

$\begin{array}{l|l}2 & 5000 \\\hline 2 & 2500 \\\hline 2 & 1250 \\\hline 5 & 625 \\\hline 5 & 125 \\\hline 5 & 25\\\hline 5&5 \\\hline & 1\end{array}$

∴5000=2×2×2×5×5×5×5

$=2^{3} \times 5^{4}$

Thus, denominator of rational number is in the form $2^{m} \times 5^{n}$ where m=3 and n=4

$\therefore \frac{257}{5000}=\frac{257}{2^{3} \times 5^{4}}$

$=\frac{257 \times 2}{2^{3} \times 5^{4} \times 2}$ (By multiplying numerator and denominator by 2)

$=\frac{514}{2^{4} \times 5^{4}}$

$=\frac{514}{(2 \times 5)^{4}}$

$=\frac{514}{10^{4}}$

=0.0514

∴Decimal expansion of $\frac{257}{5000}$ is 0.0514

Question 7

Sol :

Decimal expansion of $\frac{1}{7}$

$\therefore$ Decimal Expansion of $\frac{1}{7}$ is non-terminating repeating

$\frac{1}{7}=0.\overline{142857}$

$\frac{2}{7}$ can be written as $2 \times \frac{1}{7}$

Decimal Expansion of $\frac{2}{7}=2 \times \frac{1}{7}$

$=2 \times 0.\overline{142857}$

$=0 . \overline{285714}$

Similarly,

$\frac{3}{7}=3 \times \frac{1}{7}$ $=3 \times 0.\overline{142857}=0.\overline{428571}$

$\frac{4}{7}=4 \times \frac{1}{7}$ $=4 \times 0.\overline{142857}=0.\overline{571428}$

$\frac{5}{7}=5 \times \frac{1}{7}$ $=5 \times 0.\overline{142857}=0.\overline{714285}$

$\frac{6}{7}=6 \times \frac{1}{7}$ $=6 \times 0.\overline{142857}=0.\overline{857142}$

Question 8

Sol :

(i) Let $x=0 . \overline{3}$=0.3333...(i)

As there is one repeating digit after the decimal point, so multiplying both sides of eq-(i) by 10

10x=3.333...(2)

Subtracting (1) from (2), we get

10x-x=3.333...-0.333...

9x=3

$x=\frac{3}{9}$

$x=\frac{1}{3}$

$\therefore x=0 . \overline{3}=\frac{1}{3}$, which is in $\frac{p}{q}$ form

(ii) Let $x=5 \cdot \overline{2}$

=5.222...(i)

As there is one repeating digit after the decimal print, so multiplying both sides of eq-(1) by 10 .

10x=52.222...(2)

Subtracting (1) from (2) , we get

$\begin{aligned} 10 x &=52.222 \\ (-)x &=\phantom{5}5.222 \\ \hline 9 x &=47.000 \end{aligned}$

$x=\frac{47}{9}$

$\therefore x=5 . \overline{2}=\frac{47}{9}$, which is in $\frac{p}{q}$ form.

(iii) Let x=0.404040....(1)

As there is two repeating digit after the decimal point , so multiplying both sides of equation-(i) by 100

100x=40.404040..(2)

Subtracting (1) from (2), we get

$\begin{aligned} 100 x &=40.404040 \ldots . \\ x &=\phantom{5}0.404040 \ldots . \\ \hline 99 x &=40.000 \end{aligned}$

99x=40

$x=\frac{40}{99}$

∴x=0.404040....$=\frac{40}{99}$, which is in $\frac{p}{q}$ form

(iv) Let $x=0.4 \overline{7}$=0.4777...

x=0.4777...(1)

There is one non-repeating digit after the decimal point , multiplying both sides of eq-(1) by 10

10x=4.777...(2)

As there is one repeating digit after the decimal point, multiplying both sides of eq-(2) by 10.

100x=47.777....(3)

Subtracting (2) from (3) , we get

$\begin{aligned} 100 x &=47.777 \\ (-)10 x &=\phantom{4}4.777 \\\hline 90 x&=43.000 \end{aligned}$

90x=43

$x=\frac{43}{90}$

∴x=0.4777...$=\frac{43}{90}$, which is in $\frac{p}{q}$ form

(v) $0.\overline{134}$

Let x=0.1343434...(1)

There is one non-repeating digit after the decimal point , multiplying both sides of (1) by 10

10x=1.343434..(2)

As there are two repeating digits after the decimal point . So, multiplying both sides of (2) by 100

1000x=134.343434...(3)

Subtracting (2) from (3), we get

$\begin{aligned}1000 x&=134.343434 \ldots \\10 x&=\phantom{00}1.343434 \ldots . \\ \hline 990 x&=133.00000\end{aligned}$

990x=133

$x=\frac{133}{990}$

$\therefore x=\frac{133}{990}$, which is in $\frac{p}{q}$ form

(vi) Let $x=0 . \overline{001}$

x=0.001001...(1)

As there are three repeating digits after the decimal point , so multiplying both sides eq-(1) by 1000

1000x=1.001001....(2)

Subtracting (1) from (2), we get

$\begin{aligned}1000x=&1.001001...\\  (-)x=&0.001001...\\ \hline 999x=& 1\end{aligned}$

$x=\frac{1}{999}$

$\therefore x=0 . \overline{001}=\frac{1}{999}$, which is in $\frac{p}{q}$ form

Question 9

Sol :

(i) √23

Square root of 23 by long division method 

∴√23=4.79583, Which has non-terminating and non-repeating decimal Expansion

So, it is an irrational number

(ii) √225

Prime factorization of 225

$\begin{array}{l|l}3 & 225 \\ \hline 3 & 75 \\\hline 5 & 25 \\\hline 5 & 5 \\\hline &1\end{array}$

225=3×3×5×5

225=$(3 \times 5)^{2}$

$\sqrt{225}=\sqrt{(3 \times 5)^{2}}=\left((3 \times 5)^{2}\right)^{\frac{1}{2}}$

$\therefore \sqrt{225}=3 \times 5=15$

$\sqrt{295}=15$ which is a rational number.

(iii) 0.3796

Decimal expansion of 0.3796 is terminating 

So, $0.3796=\frac{3796}{10000}$ which is in $\frac{p}{q}$ form

∴0.3796 is a rational number

(iv) x=7.478478..(1)

As there are three repeating digits after the decimal point. So, multiplying both sides of eq-(1) by 1000

1000x=7478.478478...(2)

Subtracting (1) from (2) , we get

$\begin{aligned}1000 x=&7478.478478...\\x=&\phantom{000}7.478478. . .\\ \hline999 x=&7471.0\end{aligned}$

$x=\frac{7471}{999}$

∴x=7.478478...$=\frac{7471}{999}$ , which is in '$\frac{p}{q}$' form

So, 7.478478....is a rational number

(v) 1.101001000100001...

From the above decimal Expansion, we observed that after decimal point, number of zeros between two consecutive tones are increasing . So, it a non-terminating and non-repeating decimal Expansion.

∴1.101001000100001.... is an irrational number

(vi) $345.0\overline{456}$

Let x=345.0456456....(1)

Multiplying by 10 on both sides of eq-(1)

10x=3450.456456...(2)

As there are three repeating digits after the decimal point. So multiplying both sides of (2) by 1000

10000x=3450456.456456...(3)

eq-(3)-eq-(2)

$\begin{aligned}10000x=&3450456.456456...\\ 10x=&\phantom{0000}345.456456...\\ \hline 9990x=&3450111.0\end{aligned}$

∴9990x=3450111

x=\frac{3450111}{9990}

which is in the form $\frac{p}{q}$

So, $345.0\overline{456}$ is a rational number 

Question 10

Sol :

(i) Decimal expansion of $\frac{1}{3}$ and $\frac{1}{2}$

$\therefore \frac{1}{3}=0.333 \ldots$

$=0 . \overline{3}$

$\begin{array}{l|l}& 0.5 \\\hline 2 & 10 \\& 10 \\\hline &0\end{array}$

$\therefore \frac{1}{2}=0.5$

There are infinite rational numbers between

$\frac{1}{3}(=0 . \overline{3})$ and $\frac{1}{2}(=0.5) .$

One among them is 0.4040040004...

(ii) $\frac{-2}{5}$ and $\frac{1}{2}$

Decimal Expansion of $\frac{-2}{5}$ and $\frac{1}{2}$

$\begin{array}{l|l}&0.4 \\\hline 5 & 20 \\& 20 \\\hline &0\end{array}$

$\therefore-\frac{2}{5}=-0.4$

$\begin{array}{l|l}& 0.5 \\\hline 2 & 10 \\& 10 \\\hline &0\end{array}$

$\therefore \frac{1}{2}=0.5$

There are many irrational numbers between $\frac{-2}{5}$ and $\frac{1}{2}$ One among them is 0.1010010001..

(iii) 0 and 0.1

There are infinite irrational number between 0 and 0.1 . One among them is

0.06006000600006...

Question 11

Sol :

There are infinite irrational number between 2 and 3 . Two among them are

2.0101001000100001...

2.919119111911119....

Question 12

Sol :

Decimal Expansion of $\frac{4}{9}$ and $\frac{7}{11}$.

$\frac{4}{9}=0.44 \ldots .$

$=0 . \overline{4}$

$\frac{7}{11}=0.6363 \ldots .$

$=0 . \overline{63}$

there are infinite rational number between 

$\frac{4}{9}(=0.4)$ and $\frac{7}{11}(=0.\overline{63})$

Two among them are 0.404004000400004....

0.51511511151111115....

Question 13

Sol :

Value of √2=1.414...

Value of √3=1.732...

There are many rational numbers between √2 and √3. One among them 1.6

Finding value of √2 and √3 by long division method.

∴√2=1.414...

∴√3=1.732...

Question 14

Sol :

$2 \sqrt{3}=\sqrt{2^{2}} \times \sqrt{3}$

$=\sqrt{4} \times \sqrt{3}=\sqrt{4 \times 3}$

∴2√3=√12

We have ,

12<12.25<12.96<15

√12<√12.25<√12.96<√15

$\sqrt{12}<\sqrt{(3.5)^{2}}<\sqrt{(3.6)^{2}}<\sqrt{15}$

$\sqrt{12}<3.5<3.6<\sqrt{15}$

∴3.5 and 3.6 are two rational numbers between √12 and √15

Question 15

Sol :

We have 5<6<7

√5<√6<√7

∴√6 is an irrational number between √5 and √7 

Question 16

Sol :

We have 3<5<6<7

√3<√5<√6<√7

∴√5 and √6 are two irrational numbers between √3 and √7