NCERT Solutions for Class 9 Maths Chapter 1: Number System - Exercise 1.3

Exercise (1.3)

1. Write the following in decimal form and say what kind of decimal expansion each has:

(i) $\dfrac{ {36}}{ {100}}$

Ans: Divide $36$ by $100$. 

$\begin{matrix} &{0.36}\\ 100&{\overline{)\;36\quad}}\\ &\underline{-0\quad}\\ &360\\ &\underline{-300\quad}\\ &\;\;600\\ &\underline{-600}\\ &\underline{\quad 0 \;\;} \end{matrix}$

So, $\dfrac{36}{100}=0.36$ and it is a terminating decimal number.

(ii) $\dfrac{ {1}}{ {11}}$

Ans: Divide $1$ by $11$.

$ \begin{matrix} &{0.0909..}\\ 11&{\overline{)\;1\quad}}\\ &\underline{-0\quad}\\ &10\\ &\underline{\;\;-0\quad}\\ &\;\;100\\ &\underline{\;-99}\\ &\quad10\\ &\quad\underline{\;\;-0\quad}\\ &\;\;100\\ &\underline{\;-99}\\ &\quad1\quad\\ \end{matrix} $

It is noticed that while dividing $1$ by $11$, in the quotient $09$ is repeated.

So, $\dfrac{1}{11}=0.0909.....$ or 

       $\dfrac{1}{11}=0.\overline{09}$ 

and it is a non-terminating and recurring decimal number.

(iii)  $ {4}\dfrac{ {1}}{ {8}}$

Ans: $4\dfrac{1}{8}=4+\dfrac{1}{8}=\dfrac{32+1}{8}=\dfrac{33}{8}$

Divide $33$ by $8$.

$\begin{matrix} &{4.125}\\ 8&{\overline{)\;33\quad}}\\ &\underline{-32\quad}\\ &10\\ &\underline{\;\;-8\quad}\\ &\;\;20\\ &\underline{-16}\\ &\;\quad\;40\\ &\quad\underline{\quad-40\quad}\\ &\quad\underline{\quad 0\quad}\\ \end{matrix}$

Notice that, after dividing $33$ by $8$, the remainder is found as $0$.

So, $4\dfrac{1}{8}=4.125$ and it is a terminating decimal number.

(iv)  $\dfrac{ {3}}{ {13}}$

Ans: Divide $3$ by $13$.

$\begin{matrix} &{0.230769}\\ 13&{\overline{)\;3\quad}}\\ &\underline{-0\quad}\\ &30\\ &\underline{\;-26\quad}\\ &\;40\\ &\underline{-39\quad}\\ &\;\quad\;10\\ &\quad\underline{\quad-0\quad}\\ &\quad{\quad100}\\ &\quad\quad\underline{-91\quad}\\ &\quad\quad90\\ &\quad\quad\underline{-78\quad}\\ &\quad\quad 120\\ &\quad \quad\underline{-117\quad}\\ &\quad\quad\underline{\quad 3\quad} \end{matrix}$

It is observed that while dividing $3$ by $13$, the remainder is found as $3$ and that is repeated after each $6$ continuous divisions.

So, $\dfrac{3}{13}=0.230769.......$ or

       $\dfrac{3}{13}=0.\overline{230769}$ 

and it is a non-terminating and recurring decimal number.

(v)   $\dfrac{ {2}}{ {11}}$

Ans: Divide $2$ by $11$.

$\begin{matrix} &{0.1818}\\ 11&{\overline{)\;2\quad}}\\ &\underline{-0\quad}\\ &20\\ &\underline{\;-11\quad}\\ &\;90\\ &\underline{-88\;}\\ &\;\quad\;20\\ &\quad\underline{\quad-11\quad}\\ &\quad{\quad90}\\ &\quad\underline{-88}\\ &\quad2\\ \end{matrix}$

It can be noticed that while dividing $2$ by $11$, the remainder is obtained as $2$ and then $9$, and these two numbers are repeated infinitely as remainders.

So, $\dfrac{2}{11}=0.1818.....$ or 

       $\dfrac{2}{11}=0.\overline{18}$ 

and it is a non-terminating and recurring decimal number.

(vi) $\dfrac{ {329}}{ {400}}$

Ans: Divide $329$ by $400$.

$\begin{matrix} &{0.8225}\\ 400&{\overline{)\;329\quad}}\\ &\underline{-0\quad}\\ &3290\\ &\underline{\;-3200\quad}\\ &\;900\\ &\underline{-800\;}\\ &\quad\;1000\\ &\quad\underline{\quad-800\quad}\\ &\quad{\quad2000}\\ &\quad\underline{\quad-2000\quad}\\ &\quad\underline{\quad 0 \quad}\\ \end{matrix}$

It can be seen that while dividing $329$ by $400$, the remainder is obtained as $0$.

So, $\dfrac{329}{400}=0.8225$ and is a terminating decimal number.

2. You know that $\dfrac{ {1}}{ {7}} {=0} {.142857}...$. Can you predict what the decimal expansions of $\dfrac{ {2}}{ {7}} {,}\dfrac{ {3}}{ {7}} {,}\dfrac{ {4}}{ {7}} {,}\dfrac{ {5}}{ {7}} {,}\dfrac{ {6}}{ {7}}$  are, without actually doing the long division? If so, how?

(Hint: Study the remainders while finding the value of $\frac{ {1}}{ {7}}$ carefully.)

Ans: Note that,  $\dfrac{2}{7},\dfrac{3}{7},\dfrac{4}{7},\dfrac{5}{7}$ and $\dfrac{6}{7}$ can be rewritten as $2\times \dfrac{1}{7},\text{ 3}\times \dfrac{1}{7},\text{ 4}\times \dfrac{1}{7},\text{ 5}\times \dfrac{1}{7},$ and $6\times \dfrac{1}{7}$

Substituting the value of $\dfrac{1}{7}=0.142857$ , gives 

$2\times \dfrac{1}{7}=2\times 0.142857...$$=0.285714...$

 $3\times \dfrac{1}{7}=3\times 0.428571...$\[\text{=}\,\text{0}\text{.428571}...\]

\[4\times \dfrac{1}{7}=4\times 0.142857...\]\[\text{=}\,\text{0}\text{.571428}...\]

$5\times \dfrac{1}{7}=5\times 0.71425...$  \[\text{=}\,\text{0}\text{.714285}...\]

$6\times \dfrac{1}{7}=6\times 0.142857...$\[\text{=}\,\text{0}\text{.857142}...\]

So, the values of $\dfrac{2}{7},\text{ }\dfrac{3}{7},\text{ }\dfrac{4}{7},\text{ }\dfrac{5}{7}$ and $\dfrac{6}{7}$ obtained without performing long division are

\[\dfrac{2}{7}=0.\overline{285714}\]

$\dfrac{3}{7}=0.\overline{428571}$

$\dfrac{4}{7}=0.\overline{571428}$

\[\dfrac{5}{7}=0.\overline{714285}\]

$\dfrac{6}{7}=0.\overline{857142}$

3. Express the following in the form \[\dfrac{ {p}}{ {q}}\], where $ {p}$ and $ {q}$ are integers and $ {q}\ne  {0}$.

(i) $ {0} {.}\overline{ {6}}$

Ans: Let $x=0.\overline{6}$  

 $\Rightarrow x=0.6666$                                                   ….… (1)

 Multiplying both sides of the equation (1) by $10$, gives

$10x=0.6666\times 10$

$10x=6.6666$…..                 …… (2)

Subtracting the equation $\left( 1 \right)$ from $\left( 2 \right)$, gives

\[\begin{align} & 10x=6.6666..... \\ & \underline{-x=0.6666.....} \\ & 9x=6 \\ & 9x=6 \\ & x=\dfrac{6}{9}=\dfrac{2}{3} \\ \end{align}\]

So, the decimal number becomes

$0.\overline{6}=\dfrac{2}{3}$  and it is in the required  $\dfrac{p}{q}$ form.

(ii) $ {0} {.}\overline{ {47}}$

Ans: Let  $x=0.\overline{47}$

$\text{   }\Rightarrow x=0.47777.....$                                             ……(a)

Multiplying both sides of the equation (a) by $10$, gives

$10x=4.7777.....$         ……(b)

Subtracting the equation $\left( a \right)$ from $\left( b \right)$, gives

$\begin{align} & 10x=4.7777..... \\ & \underline{-x=0.4777.....} \\ & 9x=4.3 \\ \end{align}$

Therefore,

$\begin{align} & \,\,\,\,\,\,x=\dfrac{4.3}{9}\times \dfrac{10}{10} \\ & \Rightarrow x=\dfrac{43}{90} \\ \end{align}$

So, the decimal number becomes 

$0.\overline{47}=\dfrac{43}{90}$  and it is in the required $\dfrac{p}{q}$ form.

(iii) $ {0} {.}\overline{ {001}}$

Ans: Let $x=0.\overline{001}\Rightarrow $           …… (1)

Since the number of recurring decimal number is $3$, so multiplying both sides of the equation (1) by $1000$, gives

$1000\times x=1000\times 0.001001.....$ …… (2)

Subtracting the equation (1) from (2) gives

$\begin{align} & 1000x=1.001001..... \\ & \underline{\text{ }-x=0.001001.....} \\ & 999x=1 \\ \end{align}$

 $\Rightarrow x=\dfrac{1}{999}$

Hence, the decimal number becomes 

$0.\overline{001}=\dfrac{1}{999}$ and it is in the $\dfrac{p}{q}$ form.

4. Express $ {0} {.99999}.....$ in the form of $\dfrac{ {p}}{ {q}}$ . Are you surprised by your answer? With your teacher and classmates, discuss why the answer makes sense.

Ans:

Let $x=0.99999.....$                                                             ....... (a)

Multiplying by $10$ both sides of the equation (a), gives

$10x=9.9999.....$                                                             …… (b)

Now, subtracting the equation (a) from (b), gives

$\begin{align} & 10x=9.99999..... \\ & \underline{\,-x=0.99999.....} \\ & \,\,9x=9 \\ \end{align}$

$\Rightarrow x=\dfrac{9}{9}$

 $\Rightarrow x=1$.

So, the decimal number becomes

$0.99999...=\dfrac{1}{1}$ which is in the $\dfrac{p}{q}$ form.

Yes, for a moment we are amazed by our answer, but when we observe that $0.9999.........$ is extending infinitely, then the answer makes sense.

Therefore, there is no difference between $1$ and $0.9999.........$ and hence these two numbers are equal.

5. What can the maximum number of digits be in the repeating block of digits in the decimal expansion of $\dfrac{ {1}}{ {17}}$ ? Perform the division to check your answer.

Ans: Here the number of digits in the recurring block of $\dfrac{1}{17}$ is to be determined. So, let us calculate the long division to obtain the recurring block of $\dfrac{1}{17}$. Dividing $1$ by $17$ gives

\begin{matrix} &{0.0588235294117646}\\ 17&{\overline{)\quad1\quad\quad\quad\quad\quad\quad\quad\quad}}\\ &\underline{-0\quad}\qquad\qquad\qquad\\ &10\qquad\qquad\quad\quad\\ &\underline{\;-0\quad}\qquad\qquad\quad\\ &\;100\qquad\qquad\qquad\\ &\underline{-85\;}\qquad\qquad\quad\\ &\quad\;150\qquad\qquad\quad\\ &\quad\underline{-136\;}\qquad\qquad\quad\\ &\quad{\quad140}\qquad\qquad\;\;\\ &\quad\underline{-136\quad}\qquad\quad\\ &{\quad 40 \quad}\quad\\ &\underline{-34\;\;}\quad\\ &\;60\\ &\underline{-51}\\ &\quad\quad 90\\ &\quad\;\;\underline{-85}\\ &\qquad\quad\; 50\\ &\quad\quad\;\;\underline{-34}\\ &\quad\quad\qquad 160\\ &\qquad\quad\;\underline{-153}\\ &\qquad\qquad\quad\;70\\ &\qquad\quad\quad\;\;\underline{-68}\\ &\qquad\qquad\qquad\; 20\\ &\qquad\qquad\quad\;\underline{-17}\\ &\qquad\qquad\quad\quad\; 130\\ &\qquad\qquad\quad\;\;\underline{-119}\\ &\qquad\qquad\qquad\quad 110\\ &\qquad\qquad\qquad\;\;\underline{-102}\\ &\qquad\qquad\qquad\quad\quad\quad 80\\ &\qquad\qquad\qquad\qquad\;\underline{-68}\\ &\qquad\qquad\qquad\quad\quad\quad\; 120\\ &\qquad\qquad\qquad\qquad\;\;\underline{-119}\\ &\qquad\qquad\qquad\quad\quad\quad\; 1\\ \end{matrix}

Thus, it is noticed that while dividing $1$ by $17$, we found $16$ number of digits in the

repeating block of decimal expansion that will continue to be $1$ after going through $16$ continuous divisions.

Hence, it is concluded that $\dfrac{1}{17}=0.0588235294117647.....$ or 

 $\dfrac{1}{17}=0.\overline{0588235294117647}$ and it is a recurring and non-terminating decimal number.

6. Look at several examples of rational numbers in the form $\dfrac{ {p}}{ {q}}\left(  {q}\ne  {0} \right)$, where $ {p}$ and $ {q}$ are integers with no common factors other than $ {1}$ and having terminating decimal representations (expansions). Can you guess what property $ {q}$ must satisfy?

Ans: Let us consider the examples of such rational numbers $\dfrac{5}{2},\dfrac{5}{4},\dfrac{2}{5},\dfrac{2}{10},\dfrac{5}{16}$ of the form $\dfrac{p}{q}$ which have terminating decimal representations.

$\begin{align} & \dfrac{5}{2}=2.5 \\ & \dfrac{5}{4}=1.25 \\ & \dfrac{2}{5}=0.4 \\ & \dfrac{2}{10}=0.2 \\ & \dfrac{5}{16}=0.3125 \\ \end{align}$

In each of the above examples, it can be noticed that the denominators of the rational numbers have powers of $2,5$ or both. 

So, $q$ must satisfy the form either ${{2}^{m}}$, or ${{5}^{n}}$, or  both ${{2}^{m}}\times {{5}^{n}}$ (where $m=0,1,2,3.....$ and $n=0,1,2,3.....$) in the form of $\dfrac{p}{q}$.

7. Write three numbers whose decimal expansions are non-terminating non-recurring.

Ans: All the irrational numbers are non-terminating and non-recurring, because irrational numbers do not have any representations of the form of $\dfrac{p}{q}$ $\left( q\ne 0 \right)$, where $p$ and $q$are integers. For example: 

$\sqrt{2}=1.41421.....$,

$\sqrt{3}=1.73205...$

$\sqrt{7}=2.645751....$

are the numbers whose decimal representations are non-terminating and non-recurring.

8. Find any three irrational numbers between the rational numbers $\dfrac{ {5}}{ {7}}$ and $\dfrac{ {9}}{ {11}}$.

Ans: Converting  $\dfrac{5}{7}$and $\dfrac{9}{11}$ into the decimal form gives

$\dfrac{5}{7}=0.714285.....$ and 

$\dfrac{9}{11}=0.818181.....$

Therefore, $3$ irrational numbers that are contained between $0.714285......$ and $0.818181.....$

are:

$\begin{align} & 0.73073007300073...... \\ & 0.74074007400074...... \\ & 0.76076007600076...... \\ \end{align}$

Hence, three irrational numbers between the rational numbers $\dfrac{5}{7}$ and $\dfrac{9}{11}$ are

$\begin{align} & 0.73073007300073...... \\ & 0.74074007400074...... \\ & 0.76076007600076...... \\ \end{align}$

9. Classify the following numbers as rational or irrational:

(i) $\sqrt{ {23}}$

Ans: The following diagram reminds us of the distinctions among the types of rational and irrational numbers.

After evaluating the square root gives

$\sqrt{23}=4.795831.....$ , which is an irrational number.

(ii) $\sqrt{ {225}}$

Ans: After evaluating the square root gives

$\sqrt{225}=15$, which is a rational number.

That is, $\sqrt{225}$ is a rational number.

(iii) $ {0} {.3796}$

Ans: The given number is $0.3796$. It is terminating decimal. 

So, $0.3796$ is a rational number.

(iv) $ {7} {.478478}$

Ans: The given number is \[7.478478\ldots .\] 

It is a non-terminating and recurring decimal that can be written in the $\dfrac{p}{q}$ form.

Let      $x=7.478478\ldots .$                                   ……(a)

Multiplying the equation (a) both sides by $100$ gives

$\Rightarrow 1000x=7478.478478.....$                                               ……(b)

Subtracting the equation (a) from (b), gives

$\begin{align} & 1000x=7478.478478.... \\ & \underline{\text{ }-x=\text{ }7.478478\ldots .} \\ & 999x=7471 \\ & 999x=7471 \\ & \text{ }x=\dfrac{7471}{999} \\ \end{align}$

Therefore, $7.478478.....=\dfrac{7471}{999}$, which is in the form of $\dfrac{p}{q}$

So, $7.478478...$ is a rational number.

(v) $ {1} {.101001000100001}.....$

Ans: The given number is \[1.101001000100001....\]

It can be clearly seen that the number \[1.101001000100001....\] is a non-terminating and non-recurring decimal and it is known that non-terminating non-recurring decimals cannot be written in the form of $\dfrac{p}{q}$.

Hence, the number \[1.101001000100001....\] is an irrational number.

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