How can rational numbers be represented on a number line?

Rational numbers can be represented on a number line by marking points that correspond to their values. Positive rational numbers are placed to the right of zero, while negative rational numbers are placed to the left.

Why is it important to convert a rational number into its standard form?

Converting a rational number into its standard form simplifies comparisons and calculations, making it easier to work with and understand.

How do the NCERT Solutions help in understanding rational numbers?

The NCERT Solutions provide clear, step-by-step explanations that help students understand how to apply the properties of rational numbers in solving problems. They also enhance problem-solving skills and exam preparation.

Exploring Rational Numbers: NCERT Class 8 Insights

Using appropriate properties find (i)

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

(ii)

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

Sol. (i)

- \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}

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

(ii)

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

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

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

= \frac{2}{5} \times [(- \frac{3}{7}) + \frac{1}{14}] - \frac{1}{6} \times \frac{3}{2}

= \frac{2}{5} \times (\frac{- 6 + 1}{14}) - \frac{3}{12}

= \frac{2}{5} \times \frac{- 5}{14} - \frac{1}{4} = \frac{- 1}{7} - \frac{1}{4} = \frac{- 4 - 7}{28} = \frac{- 11}{28}

.

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}

Sol. (i) The additive inverse of

\frac{2}{8}

is

(\frac{- 2}{8}) = \frac{- 2}{8}

. (ii) The additive inverse of

\frac{- 5}{9}

is

- (\frac{- 5}{9}) = \frac{5}{9}

. (iii) The additive inverse of

\frac{- 6}{- 5}

is

\frac{- 6}{5}

. (iv) The additive inverse of

\frac{2}{- 9}

is

\frac{2}{9}

. (v) The additive inverse of

\frac{19}{- 6}

is

\frac{19}{6}

.

Verify that

- (- x) = x

for (i)

x = \frac{11}{15}

(ii)

x = - \frac{13}{17}

Sol. (i)

x = \frac{11}{15}

∴ - (- x) = - (- \frac{11}{15}) = \frac{11}{15} = x

(ii)

x = \frac{- 13}{17}

∴ - (- x) = - [- (\frac{- 13}{17})] = - [\frac{13}{17}] = - \frac{13}{17}

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}{7}

(vi) -1 Sol. (i) The multiplicative inverse of -13 is

(- 13)^{- 1} = \frac{1}{- 13}

(ii) The multiplicative inverse of

\frac{- 13}{19} is (\frac{- 13}{19})^{- 1} = \frac{19}{- 13} = \frac{- 19}{13} .

(iii) The multiplicative inverse of

\frac{1}{5}

is 5 . (iv) We have,

\frac{- 5}{8} \times \frac{- 3}{7} = \frac{- 5 \times- 3}{8 \times 7} = \frac{15}{56}

The multiplicative inverse of

\frac{15}{56}

is

(\frac{15}{56})^{- 1} = \frac{56}{15}

(v)

- 1 \times \frac{- 2}{7} = \frac{2}{7}

The multiplicative inverse of

\frac{2}{7} = \frac{7}{2}

(vi) The multiplicative inverse of -1 is -1 .

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

Sol. (i) Existence of multiplicative identity. (ii) Commutative property of multiplication. (iii) Existence of multiplicative inverse.

Multiply

\frac{6}{13}

by the reciprocal of

\frac{- 7}{16}

. Sol.

\frac{6}{13} \times (

the reciprocal of

\frac{- 7}{16})

= \frac{6}{13} \times (\frac{- 7}{16})^{- 1}

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

.

Tell what property allows you to compute

\frac{1}{3} \times (6 \times \frac{4}{3})

as

(\frac{1}{3} \times 6) \times \frac{4}{3}

. Sol. Associative property of multiplication over rational numbers allows us to compute :

\frac{1}{3} \times (6 \times \frac{4}{3})

as

(\frac{1}{3} \times 6) \times \frac{4}{3}

.

Is

\frac{8}{9}

the multiplicative inverse of

- 1 \frac{1}{8}

? Why or why not? Sol. No,

\frac{8}{9}

is not the multiplicative inverse of

- 1 \frac{1}{8}

. Because

\frac{8}{9} \times (- 1 \frac{1}{8}) = \frac{8}{9} \times \frac{- 9}{8} = - 1 \neq = 1

.

Is 0.3 the multiplicative inverse of

3 \frac{1}{3}

? Why or why not? Sol. Yes, 0.3 is multiplicative inverse of

3 \frac{1}{3}

. Because

0.3 \times 3 \frac{1}{3} = \frac{3}{10} \times \frac{10}{3} = 1

.

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. Sol. (i) We know that there is no rational number which when multiplied with 0 , gives 1. Therefore, the rational number 0 has no reciprocal. (ii) We know that the reciprocal of 1 is 1 and the reciprocal of -1 is -1 . Therefore 1 and -1 are the only rational numbers which are equal to their reciprocals. (iii) The rational number 0 is equal to its negative.

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 \neq = 0

is

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

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

-. Sol. (i) No (ii)

1, - 1

(iii)

\frac{- 1}{5}

(iv) x (v) Rational number (vi) Positive

NCERT Class 8 Maths Chapter 1 : All Exercises	Number of Questions
NCERT Class 8 Maths Chapter 1 Exercise 1.1 Solutions	11
NCERT Class 8 Maths Chapter 1 Exercise 1.2 Solutions	7

NCERT Class 8 Maths Chapter 1 : All Exercises	Number of Questions
NCERT Class 8 Maths Chapter 1 Exercise 1.1 Solutions	11
NCERT Class 8 Maths Chapter 1 Exercise 1.2 Solutions	7

Frequently Asked Questions

How can rational numbers be represented on a number line?

Why is it important to convert a rational number into its standard form?

How do the NCERT Solutions help in understanding rational numbers?

Join ALLEN!

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NCERT Solutions Class 8 Maths Chapter 1 Rational Numbers

1.0NCERT Solutions for Class 8 Maths Chapter 1 PDF Download

2.0NCERT Solutions for Class 8 Maths Chapter 1 : All Exercises

3.0NCERT Questions with Solutions Class 8 Maths Chapter 1 - Detailed Solutions

Exercise : 1.1

EXERCISE : 1.2

4.0Quick Insights About the chapter - Rational Numbers

Frequently Asked Questions

How can rational numbers be represented on a number line?

Why is it important to convert a rational number into its standard form?

How do the NCERT Solutions help in understanding rational numbers?

Join ALLEN!

Related Articles:-

NCERT Solutions Class 8 Maths Chapter 4 Data Handling

NCERT Solutions for Class 8 Science Chapter 10 Reaching the Age of Adolescence

NCERT Solutions Class 8 Maths Chapter 6 Cubes and Cube Roots

NCERT Solutions Class 8 Science Chapter 11 Chemical Effects of Electric Current

NCERT Solutions Class 8 Science Chapter 9 Friction

NCERT Solutions Class 8 Science Chapter 6 Reproduction In Animals

NCERT Solutions Class 8 Maths Chapter 1 Rational Numbers

1.0NCERT Solutions for Class 8 Maths Chapter 1 PDF Download

2.0NCERT Solutions for Class 8 Maths Chapter 1 : All Exercises

3.0NCERT Questions with Solutions Class 8 Maths Chapter 1 - Detailed Solutions

Exercise : 1.1

EXERCISE : 1.2

4.0Quick Insights About the chapter - Rational Numbers