# 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. The number system is the foundation of mathematics as mathematics starts with numbers. Solutions of NCERT for class 8 maths chapter 1 Rational Numbers is an important tool to enhance your preparation of the number system. Rational Numbers is the subpart of the unit numbers system. Rational numbers are those numbers which you 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 can conclude that all the integers come under the category of a rational number. CBSE NCERT solutions for class 8 maths chapter 1 Rational Numbers are designed to provide you proper step by step solutions to a particular question. In this chapter, you will 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 are explained in tabular form, and also the comparison is given with integers and whole numbers. In this particular chapter, there are a total of 24 questions in 2 exercises. NCERT solutions for class 8 maths chapter 1 Rational Numbers are covering detailed answers to all the 24 questions. NCERT solutions are downloadable for free through the link.

## Topic: Properties of Rational Numbers

 Addition Subtraction Multiplication Division 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 numbers. Any number divided by zero is not defined.

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

Commutative for

 Addition Subtraction Multiplication Division 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

 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

Associative for

 Addition Subtraction Multiplication Division Rational Numbers ... ... ... No Integers ... ... Yes ... Whole Numbers Yes ... ... ... Natural Numbers ... No ... ...

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

 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

(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,

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

(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 $\frac{-11}{6}$

## Q1 (i) 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)

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

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,

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

(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 $\frac{19}{-6}+\frac{-19}{-6} = \frac{19-19}{-6} =0$

(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

(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 $-1\times -1 = 1$

(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.,  $a\times\frac{1}{-a} = 1$

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

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

By the Associativity property for multiplication, we know that

a × (b × c) = (a × b) × c

Thus property used here is the associativity.

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 $\frac{-9}{8}$

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 $\frac{10}{3}$.

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)

(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 solutions for class 8 maths chapter 1 Rational Numbers Excercise: 1.2

(i) To represent  $\frac{7}{4}$  on a 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 the 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.

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 drawn number line.

The 5 rational numbers smaller than 2 can be any number in the 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

Rational numbers between any 2 numbers can easily 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 the mean between $\frac{-2}{5}$ and $\frac{1}{20}$ .

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

Similarly, we will find a 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  $\frac{1}{2}$.

(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 denominator 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 $\frac{9}{32}, \frac{10}{32}, \frac{11}{32}, \frac{12}{32}, \frac{13}{32}$

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.

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 denominators 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 a difference of at least 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.

## NCERT solutions for class 8 maths: Chapter-wise

 Chapter -1 NCERT solutions for class 8 maths chapter 1 Rational Numbers Chapter -2 Solutions of NCERT for class 8 maths chapter 2 Linear Equations in One Variable Chapter-3 CBSE NCERT solutions for class 8 maths chapter 3 Understanding Quadrilaterals Chapter-4 NCERT solutions for class 8 maths chapter 4 Practical Geometry Chapter-5 Solutions of NCERT for class 8 maths chapter 5 Data Handling Chapter-6 CBSE NCERT solutions for class 8 maths chapter 6 Squares and Square Roots Chapter-7 NCERT solutions for class 8 maths chapter 7 Cubes and Cube Roots Chapter-8 Solutions of NCERT for class 8 maths chapter 8 Comparing Quantities Chapter-9 NCERT solutions for class 8 maths chapter 9 Algebraic Expressions and Identities Chapter-10 CBSE NCERT solutions for class 8 maths chapter 10 Visualizing Solid Shapes Chapter-11 NCERT solutions for class 8 maths chapter 11 Mensuration Chapter-12 NCERT Solutions for class 8 maths chapter 12 Exponents and Powers Chapter-13 CBSE NCERT solutions for class 8 maths chapter 13 Direct and Inverse Proportions Chapter-14 NCERT solutions for class 8 maths chapter 14 Factorization Chapter-15 Solutions of NCERT for class 8 maths chapter 15 Introduction to Graphs Chapter-16 CBSE NCERT solutions for class 8 maths chapter 16 Playing with Numbers

## How to use NCERT solutions for class 8 maths chapter 1 Rational Numbers?

• Clear your concepts about various types of numbers using the text given in the textbook.
• Learn the proper usage of those concepts while solving a particular problem.
• During the practice, keep using NCERT solutions for class 8 maths chapter 1 Rational Numbers to boost your preparation.

Keep working hard and happy learning!