Class 8 RD Sharma Chapter 1 Rational Numbers  Exercise 1.2 (original) (raw)

Last Updated : 10 Sep, 2024

Exercise 1.2 in Chapter 1 of RD Sharma's Class 8 mathematics textbook focuses on rational numbers and their properties. This exercise builds upon the fundamental concepts of rational numbers, exploring their representation on a number line, comparison, and basic operations. Students will learn to identify, compare, and perform calculations with rational numbers, enhancing their understanding of this crucial mathematical concept.

**Question 1: Verify the commutativity of the addition of rational numbers for each of the following pairs of rational numbers

****(i) -11 / 5 and 4 / 7**

**Solution:

Commutativity property is verified if two rational numbers a and b, a + b = b + a

In this case a = -11 / 5 and b = 4 / 7

a + b = (-11) / 5 + 4 / 7

LCM of 5 and 7 is 35

= (-11 × 7 + 4 × 5) / 35

= (-77 + 20) / 35

= (-57) / 35

b + a = 4 / 7 + (-11) / 5

LCM of 5 and 7 is 35

= (4 × 5 + (-11 × 7)) / 35

= (20 - 77) / 35

= (-57) / 35

Hence, a + b = b + a

Therefore, commutativity is verified in this case.

****(ii) 4 / 9 and 7 / -12**

**Solution:

Here a = 4 / 9, b = (-7) / 12

a + b = 4 / 9 + (-7) / 12

LCM of 9 and 12

9 = 3*3

12 = 3 * 2 * 2

LCM = 3 * 3 * 2 * 2 = 36

= (4 * 4 + (-7 * 3)) / 36

= (16 - 21) / 36

= (-5) / 36

Now, b + a = (-7) / 12 + 4 / 9

= ((-7 * 3) + 4 * 4) / 36

= (-21 + 16) / 36

= (-5) / 36

Hence, a + b = b + a

Therefore, commutativity is verified in this case.

****(iii) -3 / 5 and -2 / -15**

**Solution:

Here a = -3 / 5, b = -2 / -15 = 2 / 15

a + b = (-3) / 5 + 2 / 15

LCM is 15

= (-3 * 3 + 2 * 1) / 15

= (-9 + 2) / 15

= (-7) / 15

Now,

b + a = 2 / 15 + (-3) / 5

= (2 × 1 + (-3 * 3)) / 15

= (2 - 9) / 15

= (-7) / 15

Hence, a + b = b + a

Therefore, commutativity is verified.

****(iv) 2 / -7 and 12 / -35**

**Solution:

a = 2 / -7 = -2 / 7

b = 12 / -35 = -12 / 35

a + b = -2 / 7 + (-12) / 35

LCM of 7 and 35 is 35

= (-2 × 5 + (-12 × 1)) / 35

= (-10 - 12) / 35

= (-22) / 35

b + a = (-12) / 35 + (-2) / 7

LCM of 7 and 35 is 35

= (-12 × 1 + (-2 × 5)) / 35

= (-12 - 10) / 35

= (-22) / 35

a + b = b + a

Therefore, commutativity is verified.

****(v) 4 and -3 / 5**

**Solution:

a = 4 / 1

b = -3 / 5

a + b = 4 / 1 + (-3) / 5

LCM of 1 and 5 is 5

= (4 * 5 + (-3 * 1)) / 5

= (20 - 3) / 5

= 17 / 5

b + a = (-3) / 5 + 4 / 1

= (-3 + 4 * 5) / 5

= (-3 + 20) / 5

= (17) / 5

a + b = b + a

Therefore, commutativity is verified.

****(vi) -4 and 4 / -7**

**Solution:

a = -4 / 1

b = -4 / 7

a + b = (-4) / 1 + (-4) / 7

LCM of 1 and 7 is 7

= (-4 * 7 + (-4 * 1)) / 7

= (-28 - 4) / 7

= (-32) / 7

b + a = (-4) / 7 + (-4) / 1

= (-4 * 1 + (-4 * 7)) / 7

= (-4 - 28) / 7

= (-32) / 7

a + b = b + a

Therefore, commutativity is verified.

**Question 2: Verify associativity of addition of rational numbers i.e (x + y) + z = x + (y + z) when

****(i) x = 1 / 2, y = 2 / 3, z = -1 / 5**

**Solution:

To verify associativity solving the LHS

(1 / 2 + 2 / 3) + (-1 / 5)

Bracket needs to be solved first

LCM of 2 and 3 is 6

= (1 * 3 + 2 * 2) / 6 + (-1 / 5)

= (3 + 4) / 6 + (-1 / 5)

= 7 / 6 + (-1 / 5)

Now solving these,

LCM of 5 and 6 is 30

= (7 * 5 + (-1 * 6)) / 30

= (35 - 6) / 30

= 29 / 30

Now solving the RHS

1 / 2 + (2 / 3 + (-1 / 5))

Solving the bracket first, LCM of 3 and 5 is 15

= 1 / 2 + (2 * 5 + (-1 * 3)) / 15

= 1 / 2 + (10 - 3) / 15

= 1 / 2 + 7 / 15

LCM of 2 and 15 is 30

= (1 * 15 + 7 * 2) / 30

= (15 + 14) / 30

= 29 / 30

Hence LHS = RHS, associativity property is verified

****(ii) x = -2 / 5, y = 4 / 3, z = -7 / 10**

**Solution:

According to the property x + (y + z) = (x + y) + z

Solving LHS

-2 / 5 + (4 / 3 + (-7 / 10))

Solving bracket first

LCM of 3 and 10 is 30

= -2 / 5 + (4 * 10 + (-7 * 3)) / 30

= -2 / 5 + (40 - 21) / 30

= -2 / 5 + 19 / 30

Now, LCM of 5 and 30 is 30

= (-2 * 6 + 19 * 1) / 30

= (-12 + 19) / 30

= 7 / 30

Now, considering RHS

(-2 / 5 + 4 / 3) + (-7 / 10)

LCM of 5 and 3 is 15

= (-2 * 3 + 4 * 5) / 15 + (-7 / 10)

= (-6 + 20) / 15 + (-7 / 10)

=(14) / 15 + (-7) / 10

LCM of 15 and 10 is 30

= (14 * 2 + (-7 * 3)) / 30

= (28 - 21) / 30

= 7 / 30

Hence, LHS = RHS

Therefore, associativity property is verified

****(iii) x = -7 / 11, y = 2 / -5, z = -3 / 22**

**Solution:

According to the property,

LHS is (-7 / 11 + (-2 / 5)) + (-3 / 22)

Solving bracket first

LCM of 11 and 5 is 55

= (-7 * 5 + (-2 * 11)) / 55 + (-3 / 22)

= (-35 - 22) / 55 + (-3 / 22)

= (-57) / 55 + (-3 / 22)

LCM of 55 and 2 is 110

= (-57 * 2 + (-3 * 5)) / 110

= (-114 - 15) / 110

= (-129) / 110

Now, solving RHS

= -7 / 11 + (-2 / 5 + (-3) / 22)

LCM of 22 and 5 is 110

= -7 / 11 + (-2 * 22 + (-3 * 5)) / 110

= -7 / 11 + (-44 - 15) / 110

= -7 / 11 + (-59) / 110

LCM of 11 and 110 is 110

= (-7 * 10 + (-59 * 1)) / 110

= (-70 - 59) / 110

= (-129) / 110

Hence, LHS = RHS, associativity is verified.

****(iv) x = -2, y = 3 / 5, z = -4 / 3**

**Solution:

According to the property,

LHS is (-2 / 1 + 3 / 5) + (-4 / 3)

LCM of 1 and 5 is 5

= (-2 * 5 + 3 * 1) / 5 + (-4 / 3)

= (-10 + 3) / 5 + (-4 / 3)

= (-7) / 5 + (-4 / 3)

LCM of 5 and 3 is 15

= (-7 * 3 + (-4 * 5)) / 15

= (-21 - 20) / 15

= (-41) / 15

Now solving RHS

-2 / 1 + (3 / 5 + (-4 / 3))

LCM of 5 and 3 is 15

= -2 / 1 + (3 * 3 + (-4 * 5)) / 15

= -2 / 1 + (9 - 20) / 15

= -2 / 1 + (-11) / 15

LCM of 1 and 15 is 15

= (-2 * 15 + (-11 * 1)) / 15

= (-30 - 11) / 15

= (-41) / 15

Hence, LHS = RHS

Therefore, associativity is verified.

**Question 3: Write the additive inverse of each of the following

****(i) -2 / 17**

Additive inverse is a number which when added to the given number gives 0. Therefore, it is negative of the number given.

Additive inverse of -2 / 17 = -(-2 / 17)

= 2 / 17

****(ii) 3 / -11**

Additive inverse of -3 / 11 is 3 / 11

****(iii) -17 / 5**

Additive inverse of -17 / 5 is 17 / 5

****(iv) -11 / -25**

It can be written as 11 / 25

Additive inverse is -11 / 25

**Question 4: Write the negative (additive) inverse of each of the following

****(i) -2 / 5**

Negative (Additive) inverse is 2 / 5

****(ii) 7 / -9**

It can be written as -7 / 9

Negative(Additive) inverse is 7 / 9

****(iii) -16 / 13**

Negative(Additive) inverse is 16 / 13

****(iv) -5 / 1**

Negative(Additive)inverse is 5

****(v) 0**

0 is neutral number

Negative(Additive) inverse is 0

****(vi) 1**

Negative(Additive) inverse is -1

****(vii) -1**

Additive inverse is 1

**Question 5: Using commutativity and associativity of the addition of rational numbers, express each of the following as rational numbers

****(i) 2 / 5 + 7 / 3 + -4 / 5 + -1 / 3**

**Solution:

According to commutativity order of numbers can be changed, so writing numbers with same denominators together.

(2 / 5 + -4 / 5) + (7 / 3 + -1 / 3)

= (2 - 4 / 5) + (7 - 1 / 3)

= -2 / 5 + 6 / 3

LCM of 5 and 3 is 15

= (-2 × 3 + 6 × 5) / 15

= (-6 + 30) / 15

= 24 / 15

3 is a common number so can be simplified further

= 8 / 5

****(ii) 3 / 7 + -4 / 9 + -11 / 7 + 7 / 9**

**Solution:

According to the commutativity order of numbers can be changed, so writing numbers with the same denominators together.

= (3 / 7 + -11 / 7) + (-4 / 9 + 7 / 9)

= (3 - 11) / 7 + (-4 + 7) / 9

= -8 / 7 + 3 / 9

LCM of 7 and 9 is 63

= (-8 × 9 + 3 × 7) / 63

= (-72 + 21) / 63

= (-51) / 63

3 is the common number so can be simplified further

= -17 / 21

****(iii) 2 / 5 + 8 / 3 + -11 / 15 + 4 / 5 + -2 / 3**

**Solution:

According to the commutativity order of numbers can be changed, so writing numbers with the same denominators together.

= (2 / 5 + 4 / 5) + (8 / 3 + -2 / 3) + (-11 / 15)

= (6) / 5 + 6 / 3 + (-11 / 15)

Applying associativity

LCM of 5 and 3 is 15

= (6 × 3 + 6 × 5) / 15 = () + (-11) / 15

= (18 + 30) / 15 + (-11) / 15

= 48 / 15 + (-11) / 15

= 37 / 15

****(iv) 4 / 7 + 0 + -8 / 9 + -13 / 7 + 17 / 21**

**Solution:

According to the commutativity order of numbers can be changed, so writing numbers with the same denominators together.

(4 / 7 + -13 / 7) + -8 / 9 + 17 / 21

= -9 / 7 + (-8 / 9) + 17 / 21

LCM of 7, 9 and 21

7 = 7×1

9 = 3×3

21 = 3×7

LCM is 3 × 3 × 7 = 63

= (-9 × 9 + (-8 × 7) + 17 × 3) / 63

= (-81 - 56 + 51) / 63

= (-86) / 63

**Question 6: Rearrange suitably and find the sum in each of the following

****(i) 11 / 12 + -17 / 3 + 11 / 2 + -25 / 2**

**Solution:

According to the commutativity order of numbers can be changed, so writing numbers with the same denominators together.

= 11 / 12 + -17 / 3 + (11 - 25) / 2

= 11 / 12 + (-17) / 3 + (-14) / 2

LCM of 3 and 2 is 6

= 11 / 12 + (-17 × 2 + (-14 × 3)) / 6

= 11 / 12 + (-34 - 42) / 6

= 11 / 12 + (-76) / 6

LCM of 6 and 12 is 12

= (11 + (-76 × 2)) / 12

= (11 - 152) / 12

= (-141) / 12

****(ii) -6 / 7 + -5 / 6 + -4 / 9 + -15 / 7**

**Solution:

According to the commutativity order of numbers can be changed, so writing numbers with the same denominators together.

= (-6 / 7 + -15 / 7) + -5 / 6 + -4 / 9

= (-6 - 15) / 7 + -5 / 6 + -4 / 9

= -21 / 7 + -5 / 6 + -4 / 9

= -3 / 1 + -5 / 6 + -4 / 9

LCM of 6 and 9 is

6 = 3 × 2

9 = 3 × 3

LCM is 18

= (-3 × 18 + (-5 × 3) + (-4 × 2)) / 18

= (-54 - 15 - 8) / 18

= -77 / 18

****(iii) 3 / 5 + 7 / 3 + 9 / 5 + -13 / 15 + -7 / 3**

**Solution:

According to the commutativity order of numbers can be changed, so writing numbers with the same denominators together.

= (3 / 5 + 9 / 5) + (7 / 3 + -7 / 3) + -13 / 15

= (3 + 9) / 5 + (7 - 7) / 3 + -13 / 15

= 12 / 5 + 0 / 3 + -13 / 15

= 12 / 5 + 0 + -13 / 15

= 12 / 5 + -13 / 15

LCM of 5 and 15 is 15

= (12 × 3 + (-13 × 1)) / 15

= (36 - 13) / 15

= 23 / 15

****(iv) 4 / 13 + -5 / 8 + -8 / 13 + 9 / 13**

**Solution:

According to commutativity order of numbers can be changed, so writing numbers with same denominators together.

= (4 / 13 + -8 / 13 + 9 / 13) + -5 / 8

= (4 - 8 + 9) / 13 + -5 / 8

= (5) / 13 + -5 / 8

LCM of 13 and 8 is 104

= (5 × 8 + (-5 × 13)) / 104

= (40 - 65) / 104

= -25 / 104

****(v) 2 / 3 + -4 / 5 + 1 / 3 + 2 / 5**

**Solution:

According to the commutativity order of numbers can be changed, so writing numbers with the same denominators together.

= (2 / 3 + 1 / 3) + (-4 / 5 + 2 / 5)

= (2 + 1) / 3 + (-4 + 2) / 5

= (3) / 3 + (-2) / 5

LCM of 3 and 5 is 15

= (3 × 5 + (-2 × 3)) / 15

= (15 - 6) / 15

= 9 / 15

= 3 / 5

****(vi) 1 / 8 + 5 / 12 + 2 / 7 + 7 / 12 + 9 / 7 + -5 / 16**

**Solution:

According to the commutativity order of numbers can be changed, so writing numbers with the same denominators together.

= (5 / 12 + 7 / 12) + (2 / 7 + 9 / 7) + 1 / 8 + -5 / 16

= (5 + 7) / 12 + (2 + 9) / 7 + 1 / 8 + -5 / 16

= 12 / 12 + 11 / 7 + 1 / 8 + -5 / 16

= 1 / 1 + 11 / 7 + 1 / 8 + -5 / 16

LCM of 1, 7, 8 and 16 is 112

= (112 + 11 × 16 + 14 + (-5 × 7)) / 112

= (112 + 176 + 14 - 35) / 112

= (267) / 112

Summary

Rational numbers are numbers that can be expressed as the quotient of two integers, where the denominator is not zero. In Exercise 1.2, students explore various aspects of rational numbers, including their representation on a number line, comparison techniques, and arithmetic operations. Key concepts covered include identifying rational numbers, converting fractions to decimals and vice versa, comparing rational numbers using the number line or cross-multiplication, and performing addition, subtraction, multiplication, and division of rational numbers. The exercise also reinforces the properties of rational numbers, such as closure under arithmetic operations, the existence of additive and multiplicative inverses, and the density property.