# NCERT Solutions for Class 9 Maths Chapter 1 Number Systems

NCERT solutions for class 9 maths chapter 1 Number Systems- Are you looking for in-depth knowledge of numbers and its applications? Then you must read this chapter. You will have a great conceptual clarity related to number system after going through it. Solutions of NCERT class 9 maths chapter 1 Number Systems are also there to provide assistance whenever you feel trouble while facing any problem related to this particular chapter. In this particular chapter, there is a total of 6 exercises which consist of 27 questions. It has a weightage of 8 marks in the CBSE class 9 final examination. CBSE NCERT solutions for class 9 maths chapter 1 Number Systems are covering the solutions to each and every question present in the practice exercises.

Apart from school exams, the solutions are also beneficial if you are aiming for exams like- National Talent Search Examination (NTSE), Indian National Olympiad (INO), SSC, CAT, etc. In NCERT class 9 maths chapter 1 Number Systems, you will learn the extended form of the number line and learn how to represent various types of the number on it. The Chapter starts with an introduction which includes rational numbers, whole numbers, and integers followed by important topics like irrational numbers, real numbers, how to represent real numbers on the number line, operations on real numbers and many more. With the understanding of rational numbers, we will also study how to represent square roots of 2, 3, 5 and other non-rational numbers. NCERT solutions for class 9 maths chapter 1 Number Systems are written keeping important basics in mind so that a student can get 100% out of it. NCERT solutions are also available for different subjects and classes which you can get by clicking on the link.

## Q1 Is zero a rational number? Can you write it in the form $\frac{p}{q}$ , where p and q are integers and q ≠ 0?

Any number that can represent  in the form of $\frac{p}{q}$  $(where \ q \neq 0)$ is a rational number

Now, we can write 0 in the form of  $\frac{p}{q}$  for eg. $\frac{0}{1},\frac{0}{2},\frac{0}{-1}$  etc.

Therefore, 0 is a  rational number.

There are an infinite number of rational numbers between 3 and 4. one way to take them is
$\Rightarrow 3 = \frac{21}{7} \ and \ 4 = \frac{28}{7}$
Therefore, six rational numbers between 3 and 4 are $\frac{22}{7}, \frac{23}{7},\frac{24}{7},\frac{25}{7},\frac{26}{7},\frac{27}{7}$

We can write
$\Rightarrow \frac{3}{5}= \frac{3\times 6}{5\times 6} = \frac{18}{30}$
And
$\Rightarrow \frac{4}{5}= \frac{4\times 6}{5\times 6} = \frac{24}{30}$
Therefore, five rational numbers between  $\dpi{100} \frac{3}{5}$ and $\dpi{100} \frac{4}{5}$ . are $\dpi{100} \frac{19}{30},\frac{20}{30},\frac{21}{30},\frac{22}{30},\frac{23}{30},$

(i) TRUE
The number that is starting from 1, i.e 1, 2, 3, 4, 5, 6, .................. are natural numbers
The number that is starting from 0. i.e, 0, 1, 2, 3, 4, 5.............are whole numbers
Therefore, we can clearly see that the collection of whole numbers contains all natural numbers.

(ii) FALSE
Because integers may be negative or positive but whole numbers are always positive. for eg. -1 is an integer but not a whole number.

## Q1 (i) State whether the following statements are true or false. Justify your answers. (i) Every irrational number is a real number.

(i) TRUE
Since the real numbers are the collection of all rational and irrational numbers.

(ii) FALSE
Because negative numbers are also present on the number line and no negative number can be the square root of any natural number

(iii) FALSE
As real numbers include both rational and irrational numbers. Therefore, every real number cannot be an irrational number.

For eg. 4 is a real number but not an irrational number

NO, Square root of all positive integers is not irrational. For the eg  square of 4 is 2 which is a rational number.

We know that
$\sqrt{5} = \sqrt{(2)^2+(1)^2}$
Now,

Let OA be a line of length 2 unit on the number line. Now, construct AB of unit length perpendicular to OA. and join OB.
Now, in right angle triangle OAB, by Pythagoras theorem
$OB = \sqrt{(2)^2+(1)^2}=\sqrt{5}$
Now,  take O as centre and OB as radius, draw an arc intersecting number line at C.  Point C represent $\sqrt{5}$ on a number line.

## Q1 (i) Write the following in decimal form and say what kind of decimal expansion each has : (i) $\frac{36}{100}$

We can write  $\frac{36}{100}$ as
$\Rightarrow \frac{36}{100}= 0.36$
Since the decimal expansion ends after a finite number of steps. Hence, it is terminating

We can rewrite  $\frac{1}{11}$  as

$\Rightarrow \frac{1}{11}= 0.09090909..... = 0.\overline{09}$
Since decimal expansion repeats itself so it is a non-terminating recurring decimal expansion.

We can rewrite $4\frac{1}{8}$  as

$\Rightarrow 4\frac{1}{8} = \frac{33}{8}= 4.125$
Since the decimal expansion ends after a finite number. Therefore, it is terminating

We can rewrite $\frac{3}{13}$  as

$\Rightarrow \frac{3}{13} = 0.230769230769 = 0.\overline{230769}$
Since decimal expansion repeats itself so it is a non-terminating recurring decimal expansion.

We can rewrite $\frac{2}{11}$ as

$\Rightarrow \frac{2}{11} = 0.181818......= 0.\overline{18}$
Since decimal expansion repeats itself so it is a non-terminating recurring decimal expansion.

We can rewrite $\frac{329}{400}$ as

$\Rightarrow \frac{329}{400}= 0.8225$
Since decimal expansion ends after finite no. of figures. Hence, it is terminating.

It is given  that  $\frac{1}{7}=0.\overline{142857}$

Therefore,

$\Rightarrow \frac{2}{7} = 2\times \frac{1}{7} = 2 \times 0.\overline{142857}= 0.\overline{285714}$

Similarly,

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

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

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

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

Let $x = 0.\overline6= 0.6666....$             -(i)

Now, multiply by 10 on both sides

$10x= 6.6666...$

$\Rightarrow 10x = 6 + x \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \(using \ (i))$

$\Rightarrow 9x = 6$

$\Rightarrow x = \frac{6}{9} = \frac{2}{3}$

Therefore,  $\frac{p}{q}$  form of  $0.\bar{6}$  is  $\frac{2}{3}$

We can write $0.4\overline7$ as

$\Rightarrow 0.4\overline7 = \frac{4}{10}+ \frac{0.777..}{10}$             -(i)

Now,

Let $x = 0.\overline7= 0.7777....$             -(ii)

Now, multiply by 10 on both sides

$10x= 7.7777...$

$\Rightarrow 10x = 7 + x \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \(using \ (ii))$

$\Rightarrow 9x = 7$

$\Rightarrow x = \frac{7}{9}$
Now, put the value of x in equation (i). we will get

$\Rightarrow 0.4\overline7 = \frac{4}{10}+ \frac{7}{10\times 9}= \frac{4}{10}+ \frac{7}{90} = \frac{36+7}{90} = \frac{43}{90}$

Therefore,  $\frac{p}{q}$  form of  $0.4\overline7$  is  $\frac{43}{90}$

Let $x = 0.\overline{001}= 0.001001....$             -(i)

Now, multiply by 1000 on both sides

$1000x= 1.001001...$

$\Rightarrow 1000x = 1 + x \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \(using \ (i))$

$\Rightarrow 999x = 1$

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

Therefore,  $\frac{p}{q}$  form of  $0.\overline{001}$  is  $\frac{1}{999}$

Let $x = 0.\overline{9}= 0.9999....$             -(i)

Now, multiply by 10 on both sides

$10x= 9.999....$

$\Rightarrow 10x = 9 + x \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \(using \ (i))$

$\Rightarrow 9x = 9$

$\Rightarrow x = \frac{9}{9} = 1$

Therefore,  $\frac{p}{q}$  form of  $0.999....$  is  1

The difference between 1 and 0.999999 is o.000001 which is almost negligible.

Therefore, 0.999 is too much closer to 1. Hence, we can write 0.999999.... as 1

We can rewrite  $\frac{1}{17}$  as

$\Rightarrow \frac{1}{17} = 0.05882352941176470588235294117647= 0.\overline{0588235294117647}$
Therefore, there are total  16  number of digits be in the repeating block of digits in the decimal expansion of $\frac{1}{17}$

We can observe that when q is 2, 4, 5, 8, 10… then the decimal expansion is terminating. For example:

$\frac{3}{2}= 1.5$, denominator $q = 2^1$

$\frac{8}{5}= 1.6$, denominator $q = 5^1$

$\frac{15}{10} = 1.5$ , denominator $q =10=2\times 5= 2^1 , 5^1$

Therefore,

It can be observed that the terminating decimal can be obtained in a condition where prime factorization of the denominator of the given fractions has the power of 2 only or 5 only or both.

Three numbers whose decimal expansions are non-terminating non-recurring are
1) 0.02002000200002......
2) 0.15115111511115.......
3) 0.27227222722227.......

We can write  $\frac{5}{7}$  as

$\Rightarrow \frac{5}{7} = 0.714285714285.... = 0.\overline{714285}$

And  $\frac{9}{11}$ as

$\Rightarrow \frac{9}{11} = 0.818181.... = 0.\overline{81}$
Therefore,  three different irrational numbers between the rational numbers $\dpi{80} \frac{5}{7}$ and $\dpi{80} \frac{9}{11}$  are

1) 0.72737475....
2) 0.750760770780...
3) 0.790780770760....

We can rewrite $\sqrt{23}$ in decimal form as

$\Rightarrow \sqrt{23} = 4.7958152....$

Now, as the decimal expansion of this number is non-terminating non-recurring.

Therefore, it is an irrational number.

We can rewrite  $\sqrt{225}$  as

$\Rightarrow \sqrt{225} = 15$
We can clearly see that it is a rational number because we can represent it in  $\frac{p}{q}$  form

We can rewrite 0.3796 as

$\Rightarrow 0.3796 = \frac{3796}{10000}$
Now, we can clearly see that it is a rational number as the decimal expansion of this number is terminating and we can also write it in $\frac{p}{q}$  form.

We can rewrite  7.478478.... as

$\Rightarrow 7.478478.... = 7.\overline{478}$
Now,  as the decimal expansion of this number is non-terminating recurring. Therefore, it is a rational number.

In the case of number  1.101001000100001...
As the decimal expansion of this number is non-terminating non-repeating. Therefore, it is an irrational number.

NCERT solutions for class 10 maths chapter 1 Number Systems Excercise: 1.4

3.765 can be visualised as in the following steps.
First, we draw a number line and mark points on it after that we will divide the number line between points 3 and 4. And then we will divide the points between 3.7 and 3.8 as the number is between both of them.

We can rewrite $4.\overline{26}$  as
$\Rightarrow 4.\overline{26}= 4.262626...$
Now,  4.2626 can be visualised as in the following steps.

NCERT solutions for class 9 maths chapter 1 Number Systems Excercise: 1.5

Value of  $\sqrt{5}$  is  2.23606798....
Now,
$\Rightarrow 2 - \sqrt{5} = 2 - 2.23606798... = -0.23606798...$
Now,
Since the number is in non-terminating non-recurring. Therefore, it is an irrational number.

Given number is  $\left ( 3+\sqrt{23} \right )-\sqrt{23}$

$\Rightarrow \left ( 3+\sqrt{23} \right )-\sqrt{23} = 3+\sqrt{23}-\sqrt{23} = 3$
Now, it is clearly a rational number because we can represent it in the form of  $\frac{p}{q}$

Given number is   $\frac{2\sqrt{7}}{7\sqrt{7}}$

$\Rightarrow \frac{2\sqrt{7}}{7\sqrt{7}} = \frac{2}{7}$
As we can clearly see that it can be represented in  $\dpi{80} \frac{p}{q}$  form. Therefore, it is a rational number.

Given number is $\frac{1}{\sqrt{2}}$

$\Rightarrow \frac{1}{\sqrt{2}}= \frac{\sqrt{2}}{2}=\frac{1.41421356..}{2} =0.7071068...$
Now,
Clearly, as the decimal expansion of this expression is non-terminating and non-recurring. Therefore, it is an irrational number.

Given number is $2\pi$
We know that the value of $\pi = 3.14159265...$
Now,
$\Rightarrow 2\pi = 2\times 3.14159265... = 6.2831853...$
Now,
Clearly, as the decimal expansion of this expression is non-terminating and non-recurring. Therefore, it is an irrational number.

Given number is  $(3+\sqrt{3})(2+\sqrt{2})$
Now, we will reduce it into
$\Rightarrow (3+\sqrt{3})(2+\sqrt{2})= 3.2+3.\sqrt{2}+\sqrt{3}.2+\sqrt{3}.\sqrt{2}$
$= 6+3\sqrt{2}+2\sqrt{3}+\sqrt{6}$

Therefore, answer is  $6+3\sqrt{2}+2\sqrt{3}+\sqrt{6}$

Given number is

$\left (3+\sqrt{3} \right )\left ( 3-\sqrt{3} \right )$
Now, we will reduce it into
$\Rightarrow (3+\sqrt{3})(3-\sqrt{3})= \left ( (3)^2-(\sqrt{3})^2 \right )$                  $\left ( using \ (a+b)(a-b)=a^2-b^2 \right )$
$=9 - 3 = 6$

Given number is  $\left (\sqrt{5}+\sqrt{2} \right )^{2}$

Now, we will reduce it into
$\Rightarrow \left (\sqrt{5}+\sqrt{2} \right )^{2} = \left ( (\sqrt{5})^2+(\sqrt{2})^2+2.\sqrt{5}.\sqrt{2} \right )$            $\left ( using \ (a+b)^2=a^2+b^2 +2ab\right )$
$=5+2+2\sqrt{10}$
$=7+2\sqrt{10}$

Therefore, the answer is  $7+2\sqrt{10}$

Given number is  $\left (\sqrt{5}-\sqrt{2} \right )\left ( \sqrt{5}+\sqrt{2} \right )$

Now, we will reduce it into
$\Rightarrow \left (\sqrt{5}-\sqrt{2} \right )\left ( \sqrt{5}+\sqrt{2} \right )= \left ( (\sqrt{5})^2-(\sqrt{2})^2 \right )$           $\left ( using \ (a+b)(a-b)=a^2-b^2\right )$
$=5-2$
$=3$

When we measure a length with scale or any other instrument, we only obtain an approximate rational value. We never obtain an exact value.
For this reason, we cannot say that either c or d is irrational.
Therefore, the fraction  $\frac{c}{d}$  is irrational. Hence, the value of $\pi$ is approximately equal to $\frac{22}{7} = 3.142857....$

Therefore, $\pi$ is irrational.

Draw a line segment OB of 9.3 unit. Then, extend it to C so that BC = 1 unit. Find the mid-point D of OC and draw a semi-circle on OC while taking D as its centre and OD as the radius. Now, Draw a perpendicular to line OC passing through point B and intersecting the semi-circle at E. Now, Take B as the centre and BE as radius, draw an arc intersecting the number line at F.  the length BF is $9\sqrt{3}$ units.

Given number is  $\frac{1}{\sqrt{7}}$

Now, on rationalisation, we will get

$\Rightarrow \frac{1}{\sqrt 7} = \frac{1}{\sqrt 7}\times \frac{\sqrt 7}{\sqrt 7 } = \frac{\sqrt7}{7}$

Therefore, the answer is  $\frac{\sqrt7}{7}$

Given number is  $\frac{1}{\sqrt{7}-\sqrt{6}}$

Now, on rationalisation, we will get

$\Rightarrow \frac{1}{\sqrt 7-\sqrt6} = \frac{1}{\sqrt 7-\sqrt 6}\times \frac{\sqrt 7+\sqrt 6 }{\sqrt 7+\sqrt 6 } = \frac{\sqrt7+\sqrt 6}{(\sqrt7)^2-(\sqrt6)^2} = \frac{\sqrt7+\sqrt6}{7-6}$ $= \sqrt7+\sqrt6$

Therefore, the answer is  $\sqrt7+\sqrt6$

Given number is  $\frac{1}{\sqrt{5}+\sqrt{2}}$

Now, on rationalisation, we will get

$\\= \frac{1}{\sqrt 5+\sqrt2} \\\\= \frac{1}{\sqrt 5+\sqrt 2}\times \frac{\sqrt 5-\sqrt 2 }{\sqrt 5-\sqrt 2 }\\\\ = \frac{\sqrt5-\sqrt 2}{(\sqrt5)^2-(\sqrt2)^2} \\\\= \frac{\sqrt5-\sqrt2}{5-2}$
$= \frac{\sqrt5-\sqrt2}{3}$

Therefore, the answer is  $\frac{\sqrt5-\sqrt2}{3}$

Given number is  $\frac{1}{\sqrt{7}-2}$

Now, on rationalisation, we will get

$\Rightarrow \frac{1}{\sqrt 7-2} = \frac{1}{\sqrt 7- 2}\times \frac{\sqrt 7+ 2 }{\sqrt 7+ 2 } = \frac{\sqrt7+ 2}{(\sqrt7)^2-(2)^2} = \frac{\sqrt7+2}{7-4}$
$= \frac{\sqrt7+2}{3}$

Therefore, answer is  $\frac{\sqrt7+2}{3}$

NCERT solutions for class 9 maths chapter 1 Number Systems Excercise: 1.6

Given number is $64^{\frac{1}{2}}$

Now, on simplifying it we will get

$\Rightarrow 64^{\frac{1}{2}} = (8^2)^\frac{1}{2} = 8$

Given number is $32^{\frac{1}{5}}$

Now, on simplifying it we will get

$\Rightarrow 32^{\frac{1}{5}} = (2^5)^\frac{1}{5} = 2$

Given number is $125^{\frac{1}{3}}$

Now, on simplifying it we will get

$\Rightarrow 125^{\frac{1}{3}} = (5^3)^\frac{1}{3} = 5$

Given number is $9^{\frac{3}{2}}$

Now, on simplifying it we will get

$\Rightarrow 9^{\frac{3}{2}} = (3^2)^\frac{3}{2} = 3^3 = 27$

Given number is $32^{\frac{2}{5}}$

Now, on simplifying it we will get

$\Rightarrow 32^{\frac{2}{5}} = (2^5)^\frac{2}{5} = 2^2 = 4$

Q2 (iii) Find : $16^{\frac{3}{4}}$

Given number is $16^{\frac{3}{4}}$

Now, on simplifying it we will get

$\Rightarrow 16^{\frac{3}{4}} = (2^4)^\frac{3}{4} = 2^3 = 8$

Given number is $125^{\frac{-1}{3}}$

Now, on simplifying it we will get

$\Rightarrow 125^{\frac{-1}{3}} = (5^3)^\frac{-1}{3} = 5^{-1} = \frac{1}{5}$
Therefore, the answer is $\frac{1}{5}$

Given number is $2^{\frac{2}{3}}.2^{\frac{1}{5}}$

Now, on simplifying it we will get

$\Rightarrow 2^{\frac{2}{3}}.2^{\frac{1}{5}} = 2^{\frac{2}{3}+\frac{1}{5}} = 2^{\frac{10+3}{15}} = 2^\frac{13}{15}$            $\left ( \because a^n.a^m = a^{n+m} \right )$

Therefore, the answer is $2^{\frac{13}{15}}$

Given number is $\left (\frac{1}{3^{3}} \right )^{7}$

Now, on simplifying it we will get

$\Rightarrow \left ( \frac{1}{3^3} \right )^7= \frac{1^7}{3^{3\times7}} = \frac{1}{3^{21}} = 3^{-21}$                   $\left ( \because (a^n)^m = a^{n.m} \ and \ \frac{1}{a^m}= a^{-m}\right )$

Therefore, the answer is $3^{-21}$

Given number is  $\frac{11^{\frac{1}{2}}}{11^{\frac{1}{4}}}$

Now, on simplifying it we will get

$\Rightarrow \frac{11^{\frac{1}{2}}}{11^{\frac{1}{4}}} = 11^{\frac{1}{2}-\frac{1}{4}}= 11^{\frac{2-1}{4}} = 11^\frac{1}{4}$                $\left ( \because \frac{a^n}{a^m}= a^{n-m}\right )$
Therefore, the answer is  $11^{\frac{1}{4}}$

Given number is  $7^{\frac{1}{2}}.8^{\frac{1}{2}}$

Now, on simplifying it we will get

$\Rightarrow 7^{\frac{1}{2}}.8^{\frac{1}{2}}= (7\times8)^{\frac{1}{2}} = 56^{\frac{1}{2}}$                                                          $\left ( \because a^n.b^n=(a.b)^n\right )$
Therefore, the answer is  $56^{\frac{1}{2}}$

## NCERT solutions for class 9 maths chapter wise

 Chapter No. Chapter Name Chapter 1 NCERT solutions for class 9 maths chapter 1 Number Systems Chapter 2 CBSE NCERT solutions for class 9 maths chapter 2 Polynomials Chapter 3 Solutions of NCERT class 9 maths chapter 3 Coordinate Geometry Chapter 4 NCERT solutions for class 9 maths chapter 4 Linear Equations In Two Variables Chapter 5 CBSE NCERT solutions for class 9 maths chapter 5 Introduction to Euclid's Geometry Chapter 6 Solutions of NCERT class 9 maths chapter 6 Lines And Angles Chapter 7 NCERT solutions for class 9 maths chapter 7 Triangles Chapter 8 CBSE NCERT solutions for class 9 maths chapter 8 Quadrilaterals Chapter 9 Solutions of NCERT class 9 maths chapter 9 Areas of Parallelograms and Triangles Chapter 10 NCERT solutions for class 9 maths chapter 10 Circles Chapter 11 CBSE NCERT solutions for class 9 maths chapter 11 Constructions Chapter 12 Solutions of NCERT class 9 maths chapter 12 Heron’s Formula Chapter 13 NCERT solutions for class 9 maths chapter 13 Surface Area and Volumes Chapter 14 CBSE NCERT solutions for class 9 maths chapter 14 Statistics Chapter 15 Solutions of NCERT class 9 maths chapter 15 Probability

## How to use NCERT Solutions for class 9 maths chapter 1 Number Systems?

• First of all, go through the conceptual text given in the book before the exercises.

• After covering the conceptual theory, you must go through some examples to understand the application of that particular concept.

• After observing the application, come to the practice exercises available in the textbook

• While solving the practice exercises, you can take the help of NCERT solutions for class 9 maths chapter 1 Number Systems to boost your preparation.

Keep working hard & happy learning!