SCERT AP 7th Class Maths Solutions Pdf Chapter 1 Integers Ex 1.3 Textbook Exercise Questions and Answers.

## AP State Syllabus 7th Class Maths Solutions 1st Lesson Integers Exercise 1.3

Question 1.

Identify the laws in the following statements :

(i) – 3 + 5 = 5 + (- 3)

Answer:

a + b = b + a (Additive commutative property)

– 3 + 5 = 5 – 3

2 = 2

(ii) – 2 × 1 = 1 × (-2) = – 2

Answer:

a × 1 = 1 × a = a

Multiplicative identity property

– 2 × 1 = 1 × (-2) = – 2

– 2 = – 2

(iii) [(-5) × 2)] × 3 = (-5) × [(2 × 3)]

Answer:

(a × b) × c = a × (b × c)

Multiplicative associative property.

(- 10) × 3 = (- 5) × (6)

– (10 × 3) = – (5 × 6)

– 30 = – 30

(iv) 18 × [7 +(- 3)] = [18 × 7] + [18 × (-3)]

Answer:

a × (b + c) = (a × b) + (a × c)

Distributive over addition property

18 × (7 – 3) = (126) + (- 54)

18 × 4= 126 – 54

72 = 72

(v) – 5 × 6 = – 30 .

Answer:

(- a) × b = – (ab)

Multiplicative closure property

– 30 = – 30

(vi) – 3 + 0 = 0 + (- 3) = – 3

Answer:

a + 0 = 0 + a = a

Additive identity property

-3 = 0 – 3 = -3

-3 = – 3 = – 3.

Question 2.

What will be the sign of the product of the following

(i) 24 times of negative integer?

(ii) 35 times of negative integer?

Answer:

(- 1) × (- 1) = + 1

(- 1) × (- 1) × (- 1) = – 1

(- 1) × (- 1) × (- 1) × (- 1) = + 1

(- 1) × (- 1) × (- 1) × (- 1) × (- 1) = – 1

If negative integer multiplied even number of times, the product is positive integer. If negative integer multiplied odd number of times, the product is negative integer.

(i) (-a) × (-a) × ………….. 24 times

= + a (Positive integer)

(ii) (-a) × (-a) × ……………. 35 times

= – a (Negative integer)

Question 3.

Write the suitable numbers in the blanks by using appropriate law.

(i) – 3 + ________= – 3

Answer:

a + 0 = a

– 3 + 0 = – 3

(ii) 2 × (-3) = (-3) × ________

Answer:

a × b = b × a

2 × (-3) = (-3) × 2

(iii) – 6 + [3 + (-2)] = [(- 6) + ________] + ________

Answer:

a + (b + c) = (a + b) + c

= [(-6) + 3] + (- 2)

(iv) – 6 × ________ = – 6

Answer:

a × 1 = a = 1 × a

-6 × ________ = – 6

(v) 5 × [(- 6) + 9] = ________ × (-6) + 5 × ________

Answer:

a × (b + c) = (a × b) + (a × c)

5×[(-6)+9] = 5 × (-6) + [(5 × 9 )]

Question 4.

State true or false. Write the reasons,

(i) 2 is the multiplicative identity of – 2.

Answer:

False.

2 is not the multiplicative identity of – 2. Multiplicative identity of – 2 is – \(\frac{1}{2}\).

2 is the additive inverse of – 2.

(ii) Integers are commutative under subtraction.

Answer:

False.

5 – 3 ≠ 3 – 5

2 ≠ – 2, i.e., a – b ≠ b – a.

Integers are not commutative under subtraction.

(iii) For any two integers a and b,

a × b = b × a

Answer:

True.

4 × (- 5) = (- 5) × 4

– (4 × 5) = -(5 × 4)

– 20 = – 20

So, for any two integers a and b,

a × b = b × a.

(iv) The division of integers by zero is not defined.

Answer:

True.

5 ÷ 0 is not defined.

So, the division of integers by zero is not defined.

(v) 6 + (-6) = (-6) + 6 = 0 indicates additive identity property. .

Answer:

False.

(-6) is the additive inverse of 6.

So, it indicates additive inverse property. But, not additive identity property.

Question 5.

Simplify the following using suitable laws.

(i) – 11 × (- 25) × (- 4)

Answer:

– 11 × (- 25) × (- 4)

a × (b × c)

= (- 11) × [(- 25) × (- 4)]

= (- 11) × (25 × 4)

= – 11 × 100

= – (11 × 100)

= – 1100

(ii)

3 × (- 18) + 3 × (- 32)

Answer:

3 × (- 18) + 3 × (- 32)

(a × b) + (a × c) = a × (b + c)

= 3 × [(- 18) + (- 32)]

= 3 × (- 18 – 32) .

= 3 × (- 50)

= – (3 × 50)

= – 150

Question 6.

Are the integers are associative under subtraction ? Explain by an example.

Answer:

No, integers are not associative under subtraction.

(a – b) – c ≠ a – (b – c)

(6 – 4) – 5 ≠ 6 – (4 – 5)

2 – 5 ≠ 6-( – 1)

– 3 ≠ 6 + 1

– 3 ≠ + 7

Question 7.

Verify [(-5) × 2)] × 3 = (- 5) × [(2 × 3)].

Answer:

[(- 5) × 2)] × 3 = (- 5) × [(2 × 3)]

(a × b) × c = a × (b × c)

[- (5 × 2)] × 3 = (- 5) × (6)

(- 10) × 3 = – (5 × 6)

– (10 × 3) = – 30

– 30 = – 30

L.H.S. = R.H.S.