NCERT Solutions For Class 9 Maths Chapter 1 Number System Exercise 1.3 (2025-26)

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.

Conclusion

The number system-related NCERT Solutions for Class 9, Chapter 1, Exercise 1.3, give an excellent foundation in fundamental mathematical topics, such as real and irrational numbers. Understanding and identifying between various numbers is the main objective of this particular activity, which is essential for comprehending higher-level math in later grades. As this is a foundational exercise, students should concentrate on developing their capacity to differentiate between and work with rational and irrational numbers because they are skills that are often expanded upon in other math areas. Gaining a thorough understanding of these ideas can also help you solve increasingly challenging mathematical and geometrical issues.

NCERT Solutions for Class 9 Maths Chapter 1 Other Exercises

CBSE Class 9 Maths Chapter 1 Other Study Materials

Chapter-Specific NCERT Solutions for Class 9 Maths

Given below are the chapter-wise NCERT Solutions for Class 9 Maths. Go through these chapter-wise solutions to be thoroughly familiar with the concepts.

Important Study Materials for CBSE Class 9 Maths