# NCERT Solutions Class 9 Maths Chapter 1 Exercise 1.3

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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

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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Q 1: Give an overview of the chapter Number system.

Ans: In this Chapter, you will get to understand about the different types of numbers along with their varied characteristics. The chapter primarily deals with irrational numbers, real numbers and their decimal expansion. It is followed by representing real numbers on the numbers, operating on real numbers, and lastly laws of exponents for real numbers.

You will also be able to learn about rational numbers and irrational numbers with their properties. The chapter also introduces you to:

• Classification of expressions into rational or irrational numbers

• Simplification of expressions

• Number Line representation

• Rationalization

Q 2: Give a brief on decimal and decimal classification.

Ans: Decimal is a very interesting and fun part of the Number System. Decimal fractions were first introduced and used by Chinese at the end of the 4th century and then spread to the Middle East and from there to Europe. Decimals are used in our daily life without consciousness. For instance – counting of money, filling fuel into our vehicle or while measuring our weight. Decimals can never be whole numbers. Few examples: 1.8, 000.23 etc.

Decimals classifications: Terminating decimal fractions are 17/4= 4.25, 21/5 = 4.2 and so on.

Non - terminating decimal fractions are 16/3 = 5.33333 , 15.35353535 etc.

Integer: All the numbers that does not have decimals in them are known as integers. Thus, -9, 4, 1476 etc. Do you know? All integers are whole numbers including the negative numbers.

Q 3: What does this chapter mainly focuses on?

Ans: The chapter mainly focuses on different types of numbers. They are:

• Complex Number

• Imaginary Number

• Real Number

• Rational Number

• Irrational Numbers

• Integers

• Whole Numbers

• Natural Numbers

• Natural Numbers - any of the given numbers that are used for counting purpose, starting from one, is considered a natural number.

• Whole Numbers - the total union set of all the Natural numbers which includes zero are the set of whole numbers.

• Integers - the set of all the whole numbers including their negative terms is called the set of integers.

• Rational Numbers - any number which can be written as a ratio of two natural numbers is known as a rational number.

• Irrational Numbers - any number which cannot be written in the form of a ratio of two natural numbers is known as an irrational number.

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Ans: Solutions to the questions are prepared by our science experts as per the latest guidelines provided by the CBSE and NCERT. We have drafted the solutions in a step by step way to make your preparation simple and easy to understand. We have given all the relevant exams and diagrams to the questions with real-life examples.

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