# NCERT Solutions for Class 8 Maths Chapter 1- Rational Numbers

## NCERT Solutions for Class 8 Maths Chapter 1 Rational Numbers

It is a very important chapter for the students from the examination point of view and also to build basic knowledge about the numbers. Chapter solutions will give you a strong idea to categorize numbers into rational and Irrational. To help the students, the NCERT Solutions for Class 8 Maths Chapter 1 Rational Numbers are solved by the experts.

In this chapter, we learn about the rational numbers, real numbers, whole numbers, integers, and natural numbers and also study their properties like Closure, Commutativity, Associativity. For the students to understand in an easy way, properties of rational numbers explained in tabular form, and also compare with integers and whole numbers in  NCERT Grade 8 Maths Chapter  1 Rational Numbers. Rational Number is a very important number category under the topic number system. The following discussion is brief about the chapter-

## What are Rational Numbers-

Rational numbers are those numbers which we can represent in the form of a fraction or numerator(p) upon a denominator(q) where the denominator can be any value except 0 or we can say that rational numbers are those numbers which can be represented in p/q form where q ≠0 and p & q are integers. From this definition, we are saying that all the integers come under the category of a rational number.

To compare commutative property over addition, subtraction, multiplication and division of rational numbers with integers, whole numbers and natural numbers lets take a table of NCERT Class 8 Maths Chapter 1 Rational Numbers-

 Addition Subtraction Multiplication Division Rational numbers Yes No Yes No Integers Yes No Yes No Whole numbers Yes No Yes No Natural numbers Yes No Yes No

Let’s understand the concept by taking some examples:-

1. 6:- It is a rational number because it can be written as 6/1 and where the denominator is not zero.

2. 3/2:- It is also a rational number because it is already in the form of A/B where the denominator is also not equal to 0.

3. 0.333333333:- It is the third type of rational number where the decimal number is recurring and thus if we convert this recurring number into fraction then it will become 1/3 and we already know that any number which is infraction where the denominator is not zero is a rational number. -0.324569576:- In this decimal number the digits after the decimal point are not recurring and every next digit is different from the previous digit. Thus we can not represent in a fraction and anything which cannot be represented in the fraction is not a rational number or we can say that it is an irrational number.

4.  Most of the students are in doubt whether $\pi$ fall under the category of rational number or irrational number? This is the perfect value $\pi$ is 3.14159 26535 89793 23846 26433 83279…... So if we want to convert this decimal number into fraction then we cannot convert it because the numbers are not recurring and digits coming in the value are endless. Now you will think that 22/7 is the exact value of $\pi$but it is not then It is just an approximation of above-written value. Thus,$\pi$ it is not a rational number.

In this chapter, there are 2 exercises with 18 questions. Following are the important topics and sub-topics of NCERT Class 8 Maths Chapter 1 Rational Numbers.

1.1 Introduction

1.2 Properties of Rational Numbers

1.2.1 Closure

1.2.2 Commutativity

1.2.3 Associativity

1.2.4 The role of zero (0)

1.2.5 The role of 1

1.2.6 Negative of a number

1.2.7 Reciprocal

1.2.8 Distributivity of multiplication over addition for rational numbers

1.3 Representation of Rational Numbers on the Number Line

1.4 Rational Numbers between Two Rational Numbers

NCERT Solutions for Class 8 Maths Chapter 1- Rational numbers, Will give you a strong idea to categorize numbers into the Rational and Irrational. Along with you will find the detailed solutions of all the question of the topic.

Below mentioned are the questions and solutions for Class 8 Maths Rational Numbers:

NCERT Solutions for Class 8 Maths Rational Numbers In Text Question

NCERT Solutions for Class 8 Maths Rational Numbers Exercise 1.1

NCERT Solutions for Class 8 Maths Rational Numbers Exercise 1.2

## NCERT Solutions For Class 8 Maths- Chapter-wise

 Chapter -2 Linear Equations in One Variable Chapter-3 Understanding Quadrilaterals Chapter-4 Practical Geometry Chapter-5 Data Handling Chapter-6 Squares and Square Roots Chapter-7 Cubes and Cube Roots Chapter-8 Comparing Quantities Chapter-9 Algebraic Expressions and Identities Chapter-10 Visualizing Solid Shapes Chapter-11 Mensuration Chapter-12 Exponents and Powers Chapter-13 Direct and Inverse Proportions Chapter-14 Factorization Chapter-15 Introduction to Graphs Chapter-16 Playing with Numbers

## NCERT Solution to Rational Numbers: 1.1

### Question:

Q.1 Using appropriate properties find.

$(i)\:\frac{-2}{3}\times \frac{3}{5}+\frac{5}{2}-\frac{3}{5}\times \frac{1}{6}$

By using commutativity property of numbers, we get,

$-\frac{2}{3}\times \frac{3}{5}+\frac{5}{2}-\frac{3}{5}\times\frac{1}{6} = -\frac{2}{3}\times \frac{3}{5}-\frac{3}{5}\times\frac{1}{6}+\frac{5}{2}$

(Now we will use distributivity of numbers)

= $= \frac{3}{5}\left ( \frac{-2}{3}+\frac{-1}{6} \right )+\frac{5}{2}\\\\\\=\frac{3}{5}\times \frac{-5}{6}+\frac{5}{2}\\\\\\=\frac{-1}{2}+\frac{5}{2}\\\\\\=\frac{-1+5}{2} = \frac{4}{2}\\\\=2$

## NCERT Solution to Rational Numbers: 1.1

### Question:

Q.1 Using appropriate properties find.

$(ii)\frac{2}{5}\times \frac{-3}{7}- \frac{1}{6}\times \frac{3}{2}+\frac{1}{14}\times \frac{2}{5}$

By using commutativity, we get

$\frac{2}{5}\times \left ( \frac{-3}{7} \right )-\frac{1}{6}\times \frac{3}{2}+\frac{1}{14}\times \frac{2}{5} = \frac{2}{5}\times \left ( \frac{-3}{7} \right ) +\frac{1}{14}\times \frac{2}{5}-\frac{1}{6}\times \frac{3}{2}$

Now by distributivity,

$= \frac{2}{5}\left ( \frac{-3}{7}+\frac{1}{14} \right )-\frac{1}{4}\\\\\\ =\frac{2}{5}\times \frac{-5}{14}-\frac{1}{4} = \frac{-1}{7}-\frac{1}{4} = \frac{-4-7}{28} = \frac{-11}{28}$

## NCERT Solution to Rational Numbers: 1.1

### Question:

Write the additive inverse of each of the following:

(i) $\frac{2}{8}$                           (ii)$\frac{-5}{9}$                    (iii) $\frac{-6}{-5}$               (iv) $\frac{2}{-9}$              (v)$\frac{19}{-6}$

(i) The additive inverse of  $\\ \\ \frac{2}{8}$ is $\frac{-2}{8}$  because  $\frac{2}{8}+\frac{-2}{8} = \frac{2-2}{8} = 0$

(ii) The additive inverse of $\frac{-5}{9}$ is $\frac{5}{9}$  because  $\frac{-5}{9}+\frac{5}{9} = \frac{-5+5}{9} = 0$

(iii) The additive inverse of  $\frac{-6}{-5}$  is $\frac{6}{-5}$  because $\frac{-6}{-5}+\frac{6}{-5} = \frac{-6+6}{-5} = 0$

(iv) The additive inverse of $\frac{2}{-9}$ is $\frac{-2}{-9}$  because $\frac{2}{-9}+\frac{-2}{-9} = \frac{2-2}{-9} = 0$

(v) The additive inverse of $\frac{19}{-6}$ is $\frac{-19}{-6}$  because $\frac{19}{-6}+\frac{-19}{-6} = \frac{19-19}{-6} =0$

Write the additive inverse of each of the following: (i) (ii) (iii) (iv) (v)

## NCERT Solution to Rational Numbers: 1.1

### Question:

3.  Verify that – (– x) = x for

(i) x = $\frac{11}{15}$          (ii) x = $\frac{-13}{17}$

(i)  We have  x = $\frac{11}{15}$

The additive inverse of x = $\frac{11}{15}$  is -x = $\frac{-11}{15}$

The same equality  $\frac{11}{15}+\left ( \frac{-11}{15} \right ) = 0$

which  implies  $-\left ( \frac{-11}{15} \right ) = \frac{11}{15}$      shows that  -(-x) = x

(ii) Additive inverse of x = $\frac{-13}{17}$ is -x = $\frac{13}{17}$                        (since $\frac{-13}{17}+\frac{13}{17} = 0$)

The same quality shows that the additive inverse of $\frac{13}{17}$ is $\frac{-13}{17}$

i.e., -(-x) = x

3. Verify that &ndash; (&ndash; x) = x for (i) x = (ii) x =

## NCERT Solution to Rational Numbers: 1.1

### Question:

4. Find the multiplicative inverse of the following.

(i)  - 13               (ii) $\frac{-13}{19}$           (iii) $\frac{1}{5}$            (iv) $\frac{-5}{8}\times \frac{-3}{7}$          (v)  $-1\times \frac{-2}{5}$         (vi)  - 1

(i) The multiplicative inverse of -13 is $\frac{-1}{13}$ because  $-13\times \frac{-1}{13} = 1$

(ii) The multiplicative inverse of $\frac{-13}{19}$ is $\frac{-19}{13}$ because of i

(iii) The multiplicative inverse of $\frac{1}{5}$  is 5 because   $\frac{1}{5}\times 5 = 1$

(iv) The multiplicative inverse of $\frac{-5}{8}\times \frac{-3}{7}$ is  $\frac{56}{15}$  because  $\frac{15}{56}\times \frac{56}{15} = 1$

(v) The multiplicative inverse of $-1\times \frac{-2}{5}$ is  $\frac{5}{2}$ because   $\frac{2}{5}\times \frac{5}{2} = 1$

(vi) The multiplicative inverse of -1 is -1 because $-1\times -1 = 1$

4. Find the multiplicative inverse of the following. (i) -13 (ii) (iii) (iv) (v) (vi) - 1

## NCERT Solution to Rational Numbers: 1.1

### Question:

Q5. Name the property under multiplication used in each of the following.

(i) $\frac{-4}{5}\times 1= 1\times \frac{-4}{5} = \frac{-4}{5}$            (ii) $-\frac{13}{17}\times \frac{-2}{7} = \frac{-2}{7}\times \frac{-13}{17}$       (iii) $\frac{-19}{29}\times \frac{29}{-19} = 1$

(i) Multiplying any number with 1 we get the same number back.

i.e., a$\times$1 = 1$\times$a = a

Hence 1 is the multiplicative identity for rational numbers.

(ii) Commutavity property states that  a$\times$b = b$\times$a

(iii) It is the multipicative inverse identity, i.e.,  $a\times\frac{1}{-a} = 1$

Q5. Name the property under multiplication used in each of the following. (i) (ii) (iii)

## NCERT Solution to Rational Numbers: 1.1

### Question:

Q6. Multiply  $\frac{6}{13}$  by the reciprocal of  $\frac{-7}{16}$ .

We know that the reciprocal of  $\frac{-7}{16}$ is $\frac{16}{-7}$.

Now,    $\frac{6}{13}\times \frac{16}{-7} = \frac{-96}{91}$

Q6. Multiplyby the reciprocal of .

## NCERT Solution to Rational Numbers: 1.1

### Question:

Q7. Tell what property allows you to compute  $\frac{1}{3}\times \left ( 6\times \frac{4}{3} \right )$  as $\left ( \frac{1}{3}\times 6 \right )\times \frac{4}{3}$.

By the Associativity property for multiplication, we know that

a × (b × c) = (a × b) × c

Thus property used here is the associativity.

Q7. Tell what property allows you to compute as.

## NCERT Solution to Rational Numbers: 1.1

### Question:

8. Is  $\frac{8}{9}$  the multiplicative inverse of $-1\frac{1}{8}$ ? Why or why not?

We know that A is the multiplicative inverse of B if   B$\times$A = 1

Applying this in given question, we get,

$\frac{-9}{8}\times \frac{8}{9} = -1$    Thus $\frac{8}{9}$ is not the multiplicative inverse of $\frac{-9}{8}$

8. Is the multiplicative inverse of ? Why or why not?

## NCERT Solution to Rational Numbers: 1.1

### Question:

9. Is 0.3 the multiplicative inverse of  $3\frac{1}{3}$ ? Why or why not?

We know that if A is multiplicative inverse of B then B$\times$A = 1

In this question, $\frac{10}{3}\times\frac{3}{10} = 1$                                               (Since 0.3  =  $\frac{3}{10}$ )

This 0.3 is the multiplicative inverse of $\frac{10}{3}$.

9. Is 0.3 the multiplicative inverse of? Why or why not?

## NCERT Solution to Rational Numbers: 1.1

### Question:

Q10.Write.

(i) The rational number that does not have a reciprocal.

(ii) The rational numbers that are equal to their reciprocals.

(iii) The rational number that is equal to its negative.

(i) Zero(0).      We know that reciprocal of A is $\frac{1}{A}$. So for 0, its reciprocal is not defined.

(ii) 1 and -1 .                                                    (Since  $\frac{1}{1} = 1$  and   $\frac{1}{-1} = -1$  )

(iii) Zero,0.                                                             (as -0=0)

Q10.Write. (i) The rational number that does not have a reciprocal. (ii) The rational numbers that are equal to their reciprocals. (iii) The rational number that is equal to its negative.

## NCERT Solution to Rational Numbers: 1.1

### Question:

Q11. Fill in the blanks.

(i) Zero has ________ reciprocal.

(ii) The numbers ________ and ________ are their own reciprocals.

(iii) The reciprocal of – 5 is ________.

(iv) Reciprocal of  $\frac{1}{x}$ , where x ≠ 0 is ________.

(v) The product of two rational numbers is always a _______.

(vi) The reciprocal of a positive rational number is ________.

(i) Zero has no reciprocal as it's reciprocal is not defined.

(ii) The numbers 1 and -1 are their own reciprocals as  $\frac{1}{1} = 1$ and  $\frac{1}{-1} = -1$

(iii) $\frac{1}{-5}$.   We know that reciprocal of A is  $\frac{1}{A}$.

(iv)  Since    $\frac{1}{\frac{1}{x}} = x$.

(v) rational number. We know that if p and q are 2 rational numbers then pq is also a rational number.

(vi) Positive. Since reciprocal of A is  $\frac{1}{A}$ , now if A is positive then reciprocal is also positive.

## NCERT Solution to Rational Numbers: 1.2

### Question:

1. Represent these numbers on the number line.   (i) $\frac{7}{4}$         (ii) $\frac{-5}{6}$

(i) To represent  $\frac{7}{4}$  on number line, firstly we will divide 1 in 4 parts and draw it on a line such       as  1/4, 2/4, 3/4, ........, 9/4. Then  will mark the required number.

(ii) To represent $\frac{-5}{6}$ on number line, firstly we will divide 1 in 6 parts and draw it on the left side of zero on number line such as -1/6, -2/6, .......,-9/4. Then mark the required         number on the number line.

1. Represent these numbers on the number line. (i) (ii)

## NCERT Solution to Rational Numbers: 1.2

### Question:

Q2. Represent  $\frac{-2}{11}$,$\frac{-5}{11}$,$\frac{-9}{11}$  on the number line.

We will divide 1 into 11 parts, then start marking numbers on left side of zero such as -1/11, -2/11, -3/11,.........,-12/11. Mark the required numbers on the drwan number line.

Q2. Represent,,on the number line.

## NCERT Solution to Rational Numbers: 1.2

### Question:

Q3. Write five rational numbers which are smaller than 2.

The 5 rational numbers smaller than 2 can be any number any number in form of p/q where

$\not\equiv$ 0.  Hence infinite numbers are possible.

Examples of 5 such numbers are 1, 1/3, 0, -1, -2

Q3. Write five rational numbers which are smaller than 2.

## NCERT Solution to Rational Numbers: 1.2

### Question:

4. Find ten rational numbers between  $\frac{-2}{5}$ and $\frac{1}{2}$ .

Rational numbers between any 2 numbers can easily be find out by taking  their means.

i.e., For $\\\frac{-2}{5}$ and $\\\frac{1}{2}$

Their mean is $\left ( \frac{-2}{5}+\frac{1}{2} \right )\div 2 = \frac{1}{20}$ .     Hence 1 rational number between $\frac{-2}{5}$ and $\frac{1}{2}$ is $\frac{1}{20}$.

Now we will find mean between $\frac{-2}{5}$ and $\frac{1}{20}$ .

This implies new required rational number is $\left ( \frac{-2}{5}+\frac{1}{20} \right )/2 = \frac{-7}{40}$.

Similarly, we will find mean between  $\frac{1}{20}$  and  $\frac{1}{2}$

New required rational number is $\left ( \frac{1}{20}+\frac{1}{2} \right ) /2 = \frac{11}{40}$

Similarly, we will take means of new numbers generated between $\frac{-2}{5}$ and  $\frac{1}{2}$.

4. Find ten rational numbers betweenand .

## NCERT Solution to Rational Numbers: 1.2

### Question:

5. Find five rational numbers between.

(i)  $\frac{2}{3}$  and  $\frac{4}{5}$              (ii)  $\frac{-3}{2}$ and  $\frac{5}{3}$             (iii)  $\frac{1}{4}$ and  $\frac{1}{2}$

(i)     For finding rational numbers between 2 numbers one method is to find means between the numbers repeatedly.

Another method is:- For    $\frac{2}{3}$  and  $\frac{4}{5}$

$\frac{2}{3}$  can be written as  $\frac{10}{15}$                            $\left ( \frac{2}{3}\times \frac{5}{5} = \frac{10}{15}\right )$

and  $\frac{4}{5}$ can be written as  $\frac{12}{15}$                              $\left ( \frac{4}{5}\times \frac{3}{3} = \frac{12}{15}\right )$

Thus numbers between $\frac{10}{15}$ and $\frac{12}{15}$ are the required numbers.

Now since we require 5 numbers in between, thus we multiply numerator and denomenator both by 4.

It becomes numbers between $\frac{40}{60}$ and $\frac{48}{60}$.

Thus numbers are $\frac{41}{60}, \frac{42}{60}, \frac{43}{60}, \frac{44}{60}, \frac{45}{60}$.

(ii)      Similarly for $\frac{-3}{2}$and $\frac{5}{3}$

Required numbers fall between $\frac{-9}{6}$ and $\frac{10}{6}$                 $\left \{ \left ( \frac{-3}{2}\times \frac{3}{3} \right )= \frac{-9}{6} \right \}$

Thus numbers are $\frac{-8}{6}, \frac{-7}{6}, \frac{-6}{6}, \frac{-5}{6}, \frac{-4}{6}$

(iii)      For   $\frac{1}{4}$  and  $\frac{1}{2}$

Required numbers lie between $\frac{1}{4}$ and $\frac{2}{4}$   or we can say between $\frac{8}{32}$ and $\frac{16}{32}$

Thus numbers are $\frac{9}{32}, \frac{10}{32}, \frac{11}{32}, \frac{12}{32}, \frac{13}{32}$

5. Find five rational numbers between. (i) and (ii)and (iii)and

## NCERT Solution to Rational Numbers: 1.2

### Question:

6. Write five rational numbers greater than –2.

There exist infinitely many rational numbers (can be expressed in the form of p/q where q $\not\equiv$0) greater than -2 .

Few such examples are -1, -1/2, 0, 1, 1/3 etc.

6. Write five rational numbers greater than &ndash;2.

## NCERT Solution to Rational Numbers: 1.2

### Question:

7. Find ten rational numbers between  $\frac{3}{5}$ and $\frac{3}{4}$.

Finding rational numbers between $\frac{3}{5}$ and $\frac{3}{4}$ is equivalent to find rational numbers between rational numbers between  $\frac{12}{20}$ and  $\frac{15}{20}$ ,since these numbers are obtained by just making their denomenators equal.

Further it is equivalent to find rational number between $\frac{96}{160}$ and $\frac{120}{160}$

(We obtained above numbers by multiplying and dividing numbers by 8 to create differnce of atleast 10 numbers).

Thus required numbers are $\frac{97}{160},\frac{98}{160},\frac{99}{160},........,\frac{106}{160}$

Alternate:- Rational numbers can also be found by taking mean of the given numbers and the newly obtained number.

7. Find ten rational numbers betweenand.

## NCERT Solution to Rational Numbers: In Text Question

### Question:

Fill in the blanks in the following table.

Rational Numbers Yes Yes ... No
Integers ... Yes ... No
Whole Numbers ... ... Yes ...
Natural Numbers ... No ... ...

It can be seen that rational numbers, integers, whole numbers, natural numbers are not closed under division because of Zero is included in these number. Any number divided byzero is not defined.

Rational Numbers Yes Yes Yes No
Integers Yes Yes Yes No
Whole Numbers Yes No Yes No
Natural Numbers Yes No Yes Yes

Fill in the blanks in the following table. Addition Substraction Multiplication Division Rational Numbers Yes Yes ... No Integers ... Yes ... No Whole Numbers ... ... Yes ... Natural Numbers ... No ... ...

## NCERT Solution to Rational Numbers: In Text Question

### Question:

Complete the following table:

Commutative for
Rational Numbers Yes ... ... ...
Integers ... No ... ...
Whole numbers ... ... Yes ...
Natural numbers ... ... ... No

In rational numbers , a$\div$$\neq$ b$\div$a

also a-b $\neq$ b-a

Rational Numbers Yes No Yes No
Integers Yes No Yes No
Whole numbers Yes No Yes No
Natural numbers Yes No Yes No

Complete the following table: Commutative for Addition Subtraction multiplication division Rational Numbers Yes ... ... ... Integers ... No ... ... Whole numbers ... ... Yes ... Natural numbers ... ... ... No

## NCERT Solution to Rational Numbers: In Text Question

### Question:

Complete the following table:

Associative for
Rational Numbers ... ... ... No
Integers ... ... Yes ...
Whole Numbers Yes ... ... ...
Natural Numbers ... Yes ... ...

For associative in multiplication :- a$\times$(b$\times$c) = (a$\times$b)$\times$c

Associative for
Rational Numbers Yes No     Yes No
Integers Yes No Yes No
Whole Numbers Yes No Yes No
Natural Numbers Yes No     Yes No

Complete the following table: Associative for Addition Subtraction Multiplication Division Rational Numbers ... ... ... No Integers ... ... Yes ... Whole Numbers Yes ... ... ... Natural Numbers ... Yes ... ...

## NCERT Solution to Rational Numbers: In Text Question

### Question:

Find using distributivity:

(i) $\left \{ \frac{7}{5}\times \left ( \frac{-3}{12} \right ) \right \}+\left \{ \frac{7}{5}\times \frac{5}{12} \right \}$                  (ii) $\left \{ \frac{9}{16}\times \frac{4}{12} \right \}+\left \{ \frac{9}{16}\times \frac{-3}{9} \right \}$

(i)  Using distributivity,  a(b+c) = ab + ac

$\left \{ \frac{7}{5}\times \left ( \frac{-3}{12} \right ) \right \}+\left \{ \frac{7}{5}\times \frac{5}{12} \right \} = \frac{7}{5}\left ( \frac{-3}{12}+\frac{5}{12} \right )\\\\\\=\frac{7}{5}\times\frac{1}{6} = \frac{7}{30}$

(ii) Using distributivity of multiplication over addition and subtraction,

$\left \{ \frac{9}{16}\times \frac{4}{12} \right \}+\left \{ \frac{9}{16}\times \frac{-3}{9} \right \} = \frac{9}{16}\left ( \frac{4}{12}-\frac{3}{9} \right )\\\\\\=\frac{9}{16}\times0 = 0$



Find using distributivity: (i) (ii)

## NCERT Solution to Rational Numbers: In Text Question

### Question:

Write the rational number for each point labelled with a letter:

(i) In this, we can see that 1 is divided into 5 parts each, so when we are moving from zero to the right-hand side, it is easy to observe that

All the numbers should contain 5 in their denominator. Thus,   A is equal to $\frac{1}{5}$  ,  B is equal to $\frac{4}{5}$,  C is equal to $\frac{5}{5} = 1$   ,   D is equal to $\frac{8}{5}$ ,  E is equal to $\frac{9}{5}$

(ii) Here we see that 1 is divided in 6 parts each. So when we move from zero towards left we observe that

All the numbers should contain 6 in their denominator. Thus, F is equal to $\frac{-2}{6}$,  G is equal to $\frac{-5}{6}$,  H is equal to $\frac{-7}{6}$, I is equal to $\frac{-8}{6}$, J is equal to $\frac{-11}{6}$

Write the rational number for each point labelled with a letter: