CBSE students can refer to NCERT Solutions for Class 7 Maths Chapter 1 Integers InText Questions and Answers are provided by experts in order to help students secure good marks in exams.

## Class 7 Maths NCERT Solutions Chapter 1 Integers InText Questions

Try These (Page 2)

Question 1.

A number line representing integers is given below :

– 3 and – 2 are marked by E and F, respectively. Which integers are marked by B, D, H, J, M, and O?

Solution:

We know that on an integer number line, numbers right to 0 is positive integers, and numbers left to 0 are negative integers. We have to identify on the given number line which integers are marked by the letters on the number line.

For it, we have to first find the letter which represents the integer 0. Since, – 3 and – 2 are marked by E and F respectively, counting forward, we get

Clearly, H represents 0.

Let, we first find the numbers for the letters to the left of H. It will be done by counting backward from 0.

Clearly, B represents – 6 and D represents – 4.

Let us find numbers for J, M, and O. It can be done by counting right to 0, i.e., counting forward.

Clearly, J represents 2, M represents 5 and O represents 7.

Question 2.

Arrange 7, – 5,4,0, and – 4 in ascending order and then mark them on a number line to check your answer.

Solution:

Arranging the given numbers 7, – 5,4,0 and -4 in the ascending order, we have – 5,- 4,0,4 and 7 …(1)

In order to represent these integers on the number line, we draw a line and mark a point on it almost in the middle of it as shown. Now, we set off equal distances on the right-hand side as well as on the left-hand side to 0.

Starting from 0 and proceeding 4 units on the right of it, we obtain 4 as marked by A, and further proceeding 3 units from A to the right, we obtain 7 marked by B.

Again, starting from 0 and proceeding 4 units to the left of it, we obtain – 4 marked by C, and further proceeding 1 unit from C to its left, we obtain – 5 marked by D.

Thus, looking at the numbers, we find that the order of numbers written in eq. (1) is correct.

Try These (Page 2)

Question 1.

State whether the following statements are correct or incorrect. Correct those which are wrong:

(i) When two positive integers are added, we get a positive integer.

Solution:

Correct Five examples :

(a) 46 + 83 = 129

(b) 103 + 92 = 195

(c) 26 + 53 = 79

(d) 120 + 402 = 522

(e) 217 + 13 = 230

(ii) When two negative integers are added, we get a positive integer.

Solution:

Incorrect

Correct statement: When two negative integers are added, we get a negative integer.

Five examples :

(a) (- 46) + (- 83) = – 129

(b) (- 103) + (- 92) = – 195

(c) (- 26) + (- 53) = – 79

(d) (-120) + (- 403) = – 523

(e) (- 217) + (- 13) = – 230

(iii) When a positive integer and a negative integer are added, we always get a negative integer.

Solution:

Incorrect

Correct statement: When one positive integer and one negative integer are added, we determine the difference of their absolute values and assign the sign of addend having greater absolute value.

Five examples :

(a) 26 + (- 33) = – 7

(b) (- 66) + 83 = 17

(c) (- 133) + 102 = – 31

(d) 145 + (- 102) = 43

(e) (- 220) + 325 = 105

(iv) Additive inverse of an integer 8 is (- 8) and additive inverse of (- 8) is 8.

Solution:

Correct

Five Example:

Integer | Its additive inverse |

10 | -10 |

-10 | 10 |

76 | -76 |

-76 | 76 |

18 | -18 |

-18 | 18 |

20 | -20 |

-20 | 20 |

39 | -39 |

-39 | 39 |

(v) For subtraction, we add the additive inverse of the integer that is being subtracted to the other integer.

Solution:

Incorrect

Correct statement For subtracting two integers, we take the additive inverse (or negative) of the integer to be subtracted and add it to the other integer.

Example: Subtract

(a) 5 from 8 = 8 + (- 5) = 3

(b) – 3 from 9 = 9 + (+3) = 9 + 3 = 12

(c) 4 from – 9 = (- 9) + (- 4) = – 13

(d) -8 from – 6 = (- 6) + (+8) = (- 6) + 8 = 2

(e) 2 from – 4 = (- 4) + (- 2) = – 6

(vi) (-10) + 3 = 10 – 3

Solution:

Incorrect

Correct statement: (-10) + 3 = – 10 + 3

(vii) 8 + (- 7) – (- 4) = 8 + 7 – 4

Solution:

Incorrect

Correct statement: 8 + (- 7) – (- 4) = 8 – 7 + 4

Try These (Page 3)

Question 1.

Can you find a pattern for each of the following? If yes, complete them:

(a) 7, 3, – 1, – 5, _______ , _______ , _______

Solution:

We see that 7 – 4 = 3, 3 – 4 = -1, -1 – 4 = – 5, so the next numbers are – 5 – 4 = -9, – 9 – 4 = -13 and -13 – 4 = – 17. Thus, 7, 3, -1, -5, -9, -13, -17.

(b) – 2, – 4, – 6, – 8, _______ , _______ , _______

Solution:

Clearly, – 2 – 2 = – 4, – 4 – 2 = – 6, – 6 – 2 = – 8, so the next numbers are – 8 – 2 = -10, -10 – 2 = – 12 and – 12 – 2 = -14. Thus, – 2, – 4, – 6, – 8, – 10, – 12, – 14.

(c) 15, 10, 5, 0 _______ , _______ , _______

Solution:

Clearly, 15 – 5 = 10, 10 – 5 = 5, 5 – 5 = 0, so the next numbers are 0 – 5 = – 5, -5 – 5 = – 10 and -10 – 5 = – 15. Thus, 15, 10, 5, 0, – 5, -10, – 15.

(d) – 11, – 8, – 5, – 2, _______ , _______ , _______

Solution:

Clearly, – 11 + 3 = – 8, – 8 + 3 = – 5, – 5 + 3 = – 2, so the next numbers are – 2 + 3 =1,1 + 3 = 4 and 4 + 3 = 7. Thus, -11, – 8, – 5, – 2, 1, 4, 7.

Make some more such patterns and ask your friends to complete them:

Some more patterns are :

(a) 9, 7, 5, 3,________ ,________ , ________

(b) – 7, – 14, – 21, – 28,________ ,_______ , ________

(c) 9,18,27,36,________ , _______ , _______

(d) -11,-1,9,19, _______ , _______ , _______

Try These (Page 5)

Properties of Addition and Subtraction of Integers Closure under Addition Following are some pairs of integers. Observe the following table and complete it

Statement | Observation |

17 + 23 = 40 | Result is an integer |

(- 10) + 3 = ___________ | ______________ |

(- 75) + 18 = ________ | ______________ |

19 + (- 25) = – 6 | Result is an integer |

27 + (- 27) = _________ | ______________ |

(- 20) + 0 = __________ | ______________ |

(- 35) + (- 10) =________ | _______________ |

What do you observe? Is the sum of two integers always an integer? Did you find a pair of integers whose sum is not an integer?

Solution:

Statement | Observation |

17 + 23 = 40 | Result is an integer |

(- 10) + 3 = – 7 | Result is an integer |

(- 75) + 18 = – 57 | Result is an integer |

19 + (- 25) = – 6 | Result is an integer |

27 + (- 27) = 0 | Result is an integer |

(- 20) + 0 = 20 | Result is an integer |

(- 35) + (- 10) = – 45 | Result is an integer |

We observe that for any two integers a and b, a + b is always an integer. Yes, the sum of two integers is always an integer. No, we cannot find a pair of integers whose sum is not an integer. Since the addition of integers gives integers, we can say that integers are closed under addition.

Try These (Page 6)

Question 1.

Closure under Subtraction

What happens when we subtract an integer from another integer? Can we say that their difference is also an integer? Observe the following table and complete it:

Statement | Observation |

7 – 9 = – 2 | Result is an integer |

17- (- 21) = _________ | _____________ |

(- 8) – (- 14) = 6 | Result is an integer |

(- 21) – (-10) = _____ | ______________ |

32 – (-17) = ________ | ______________ |

(-18) – (-18) = _____ | ______________ |

(- 29) – 0 = ________ | _____________ |

What do you observe? Is there any pair of integers whose difference is not an integer? Can we say integers are closed under subtraction?

Solution:

On completion of the table, we have,

Statement | Observation |

7 – 9 = – 2 | Result is an integer |

17 – (- 21) = 17 + 21 = 38 | Result is an integer |

(- 8) – (-14) = 6 | Result is an integer |

(- 21) – (-10) = – 21+10 = – 11 | Result is an integer |

32 – ( -17) = 32 +17 = 49 | Result is an integer |

(-18) – (-18) = – 18 +18 = 0 | Result is an integer |

(-29) – 0 = -29 | Result is an integer |

When we subtract two integers, the result obtained is also an integer. No, there is no pair of integers whose difference is not an integer. Since subtraction of integers always gives an integer, we can say that integers are closed under subtraction.

Question 2.

Do the whole numbers satisfy this property?

Solution:

Whole numbers do not satisfy this property, for example, 5 – 7 = – 2, which is not a whole number.

Try These (Page 7)

Question 1.

Try this with five other pairs of integers. Do you find any pair of integers for which the sums are different when the order is changed?

(i) (- 8) + (- 9) and (- 9) + (- 8)

Solution:

(- 8) + (- 9) = – 8 – 9 = – 17

and, (- 9) +(- 8) = – 9 – 8 = – 17

=> (- 8) +(- 9) = (- 9) +(- 8)

(ii) (- 23) + 32 and 32 + (- 23)

Solution:

(- 23)+ 32 = – 23 + 32 = 9

and, 32+ (- 23) = 32 – 23 = 9

=> (- 23) + 32 = 32 + (- 23)

(iii) (- 45) + 0 and 0 + (- 45)

Solution:

(- 45) + 0 = – 45

and, 0 + (- 45) = – 45

=> (- 45) + 0 = 0 + (- 45)

Let us take five other pairs of integers :

(a) 5 + (- 8) = – 3

and, (- 8) + 5 = – 3

∴ 5 + (- 8) = (- 8) + 5

(b) (- 7) +(- 9) = – 16

and, (- 9) +(- 7) = – 16

∴ (- 7) +(- 9) = (- 9) +(- 7)

(c) 3 + (- 4) = -1

and, (- 4) + 3 = -1

∴ 3 +(- 4) = (- 4)+ 3

(d) 5 + (- 9) = – 4

and, (- 9) + 5 = – 4

∴ 5 +(- 9) = (- 9) + 5

(e) (-12)+ 8 = – 4

and, 8 + (-12) = – 4

∴ (-12) + 8 = 8 + (- 12)

From the above examples, we observe that there Is no pair of integers for which the sum is different when the order is changed.

Thus, we conclude that the addition is commutative for integers also. In general, if a and b are any two integers, then a + b = b + a, that is the sum of two whole integers remains the same even if the order of integers is changed.

We know that the subtraction of whole numbers is not commutative, that is, if a and b are two whole numbers, then in general (a – b) is not equal to (b – a).

Since, 9 – 5 = 4, but 5 – 9 is not possible. Thus, for two whole numbers a and b (a > b), then a – b is a whole number but b – a is not possible, and if b > a, then b – a is a whole number but a – b is not possible.

Question 2.

Take at least five different pairs of integers to conclude that subtraction is not commutative for integers.

Solution:

Let us consider five different pairs of integers :

(i) 10 and 6

10 – 6 = 4

and 6 -10 = – 4

∴ 10 – 6 ≠ 6 – 10

(ii) – 9 and 5

(- 9) – 5 = – 9 – 5 = -14

and, 5 – (- 9) = 5 + 9 =14

∴ (- 9) – 5 ≠ 5 – (- 9)

(iii) 12 and 125

12-125 = -113

and, 125-12 =113

∴ 12-125 ≠ 125-12

(iv) 28 and 0

28 – 0 = 28

and 0 – 28 = – 28

∴ 28 – 0 ≠ 0 – 28

(v) – 7 and – 9

-(7) – (- 9) = – 7 + 9 = 2

and, (- 9) – (- 7) = – 9 + 7 = – 2

∴ (- 7) -(- 9) ≠ (- 9) – (- 7)

We observe that the result is not the same in any case. Hence, we conclude that the subtraction of integers is not commutative. i.e., for any two integers a and b, a – b ≠ b – a.

Try These (Page 7)

Question 1.

Associative Property

If a, b,c are any three integers, then

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

That is, the addition of integers is associative.

Consider three integers: 3,1 and – 7

We have, (- 3) + [1 + (- 7)] = (- 3) + (- 6) = – 9

and, [(- 3) +1] + (- 7) = (- 2) + (- 7) = – 9

∴ (- 3) + [1 + (- 7)] = [(- 3) +1] + (- 7)

Let us verify it with five more examples :

(i) Consider three integers : – 9,3 and – 4

We have, [(- 9) + 3] + (- 4) = (- 6) + (- 4) = – 10

and, (- 9) + [3 + (- 4)] = – 9 + (- 1) = – 10

∴ [(- 9) + 3] + (- 4) = (- 9) + [3 + (- 4)]

(ii) Consider three integers: 6, – 4 and – 3

We have, [(- 6) + (- 4)] + (-3) = (-10) + (-3) = -13

and, (- 6) + [(- 4) + (- 3)] = (- 6) + (+ 7) = -13

∴ [(- 6) + (- 4)] + (- 3) = (- 6) + [(- 4) + (- 3)]

(iii) Consider three integers: 8, 2 and – 5

We have, [(- 8) + 2] + (- 5) = (- 6) + (- 5) = -11

and, (- 8) +[2 +(- 5)] = (- 8) + (- 3) = – 11

∴ [(- 8) + 2] + (- 5) = (- 8) +[2 +(- 5)]

(iv) Consider three integers: 5, – 9 and 2

We have, [5 + (- 9)]+ 2 = (- 4) + 2 = – 2

and, 5 + [(- 9) + 2] = 5 +(- 7) = – 2

∴ [5 +(- 9)+ 2] = 5+ [(- 9) + 2]

(v) Consider three integers: -8,-6 and -5

We have, [(- 8) + (- 6)] + (- 5) = (- 14) + (- 5) = -19

and, (- 8) + [(- 6) + (- 5)] = (- 8)+ (-11) = -19

∴ [(- 8) + (- 6)] + (- 5) = (- 8) + [(- 6) + (- 5)]

We see that there is no example for which the sums are different. Thus, we conclude that addition is associative for integers.

Try These (Page 8)

Question 1.

Additive Identity

We know that if a is a whole number, then a + 0 = a = 0 + a

That is the sum of any whole number and zero is the number itself. The whole number 0 (zero) is called the additive identity or the identity element for the addition of whole numbers. In particular, we can also say that zero is an additive identity for positive integers. Is zero an additive identity for negative integers also? Observe the following table and check it.

(i) (- 8) + 0 = – 8

Solution:

(- 8) + 0 = – 8

(ii) 0 + (- 8) = – 8

Solution:

0 + (- 8) = – 8

(iii) (- 23) + 0 =

Solution:

(- 23) + 0 = – 23

(iv) 0 + (- 37) = – 37

Solution:

0 + (- 37) = – 37

(v) 0 + (-59) =

Solution:

0 + (- 59) = – 59

(vi) 0 + = – 43

Solution:

0 + (- 43) = – 43

(vii) – 61 + = – 61

Solution:

– 61 + 0 = – 61

(viii) + 0 =

Solution:

– 15 + 0 = -15

We observe that the sum of any negative integer and zero is the integer itself. Thus, for any integer a

a + 0 = a = 0 + a The integer 0 (zero) is called the additive identity.

Try These (Page 8)

Question 1.

Write a pair of integers whose sum gives

(a) a negative integer.

Solution:

Consider the pair of integers – 6 and – 5 such that – 6 + (- 5) = – 6 – 5 = – 11

(b) zero

Solution:

Consider the pair of integers 8 and – 8 such that 8 – 8 = 0.

(c) an integer smaller than both the integers.

Solution:

Consider the pair of integers – 9 and – 8 such that – 9 + (- 8) = -17, where -17 is smaller than – 9 and – 8.

(d) an integer smaller than only one of the integers.

Solution:

Consider the pair of integers 8 and – 3 such that 8 + (- 3) = 5, where 5 is greater than – 3.

(e) an integer greater than both the integers.

Solution:

Consider the pair of integers 9 and 8 such that 9 + 8 =17, where 17 is greater than both the integers.

Question 2.

Write a pair of integers whose difference gives

(a) a negative integer.

Solution:

A pair of integers whose difference gives a negative may be taken as – 8 and 5 as (- 8) – 5 = – 8 – 5 = -13.

(b) zero.

Answer:

A pair of integers whose difference gives zero may be taken as – 3 and – 3 as (- 3) – (- 3) = – 3 + 3 = 0.

(c) an integer smaller than both the integers.

Solution:

A pair of integers whose difference gives an integer smaller than both the integers may be taken as 5 and 4 as 5 – 4 = 1, clearly 1 < 5 and 1 < 4.

(d) an integer greater than only one of the integers.

Solution:

A pair of integers whose difference gives an integer greater than only one of the integers may be taken as 8 and 2 as 8 – 2 = 6 and 2 < 6 < 8.

(e) an integer greater than both the integers.

Solution:

A pair of integers whose difference gives an integer greater than, both the integers may be taken as 7 and – 2 as 7 – (-2) = 7 + 2 = 9 clearly 9 > 7 and. 9 > – 2.

Try These (Page 10)

Question 1.

Find 4 x (- 8), 8 x (- 2), 3 x (- 7), 10 x (- 1) using number line. Sol. We can write 4 x (- 8) as 4 x (- 8) = (- 8) +(- 8) +

(- 8) +(- 8) In can be represented on the number line as given below :

We have, 8 x(- 2) = (- 2) + (- 2) + (- 2) + (- 2) + (- 2) + (- 2) + (- 2)+(- 2) It cab be represented on the number line as given below :

We have, 3 x (- 7) = (- 7) + (- 7) + (- 7)

It can be represented on the number line as given below :

We have,

10 x (- 1) = (- 1) + ( – 1) + (- 1) + (- 1)+(- 1) + (- 1) +(- 1) + (- 1) + (- 1) + (-1) It can be represented on the number line given below :

Question 2.

To find the product of a positive integer and a negative integer without using a number line. To find 3 × (- 5). First, find 3×5 and then put minus sign (-) before the product obtained as under Find in a similar way

(i) 4 × (- 8) = =

Solution:

4 × (- 8) = – (4 × 8) = – 32

(ii) 3 × (- 7) = =

Solution:

3 × (- 7) = – (3 × 7) = – 21

(iii) 6 × (- 5) = =

Solution:

6 × (- 5) = – (6 × 5) = – 30

(iv) 2 × (- 9) = =

Solution:

2 × (- 9) = – (2 × 9) = -18

Thus, we find that while multiplying a positive integer and a negative integer, we multiply them as whole numbers and put a minus sign (-) before the product.

Try These (Page 10)

Question 1.

Find:

(i) 6 × (-19)

Solution:

(i) 6 × (-19) = -(6 × 19) = -114

(ii) 12 × (-32)

Solution:

(ii) 12 × (-32) = – (12 × 32) = – 384

(iii) 7 × (- 22)

Solution:

(iii) . 7 × (- 22) = – (7 x 22) = – 154

Try These (Page 11)

Question 1.

Using patterns, find (- 4) × 8, ( – 3) × 7, (- 6) × 5 and (- 2) × 9 Check whether, (- 4) × 8 = 4 × (- 8), (- 3) × 7, = 3 × (- 7), (- 6) × 5 = 6 × (- 5) and (- 2) × 9 = 2 × (- 9).

Solution:

Try These (Page 11)

Question 1.

Find:

(a) 15 × (-16)

Solution:

15 × (-16) = – (15 × 16) = – 240

(b) 21 × ( -32)

Solution:

21 × (- 32) = – (21 × 32) = – 672

(c) (- 42) x 12

Solution:

(- 42) × 12 = – (42 × 12) = – 504

(d) – 55 × 15

Solution:

(- 55) × 15 = – (55 × 15) = – 825

Question 2.

Check if

(a) 25 × (- 21) = (- 25) × 21

Solution:

L.H.S. = 25 × (- 21)

=-(25 × 21) = – 525

R.H.S. = (-25) × 21

= – (25 × 21) = – 525

∴ L.H.S. = R.H.S.

(b) (- 23) × 20 = 23 × (- 20)

Solution:

(b) L.H.S. = (- 23) × 2.0

= – (23 × 20) = – 460

R.H.S. = 23 × (-20)

= – (23 × 20) = – 460

∴ L.H.S. = R.H.S.

Five examples are given below :

(i) 15 × (-17) = (-15) × 17 = – (15 × 17)

(ii) 55 × (- 21) = (- 55) × 21 = – (55 × 21)

(iii) 19 × (-11) = (-19) × 11 =- (19 × 11)

(iv) 35 × (-12) = (- 35) × 12 = – (35 × 12)

(v) 95 × (- 25) = (- 95) × 25 = – (95 × 25)

Thus, we can say that for any two intgers a and b, a × (- b) = (- a) × b = -(a × b).

Try These (Page 11)

Question 1.

Multiplication of Two Negative Integers Can you find the product (- 3) × (- 2)?

Solution:

Observe the following :

– 3 × 4 = – 12

– 3 × 3 = – 9 = -12 – (- 3)

– 3 × 2 = – 6 = – 9 – (- 3)

– 3 × 1 = – 3 = – 6 – (- 3)

– 3 × 0 = 0 = – 3 – ( – 3)

– 3 × – 1 = 0 – (- 3) = 0 + 3 = 3

– 3 × – 2 = 3 – (- 3) = 3 + 3 = 6

Try These (Page 11)

Question 1.

Do you see any pattern? Observe how the products change. Based on this observation, complete the following :

(i) – 3 x – 3 =

Solution:

– 3 × 4 = – 12

– 3 × 3 = – 9 = – 12 – (- 3)

– 3 × 2 = – 6 = – 9 – (-3)

– 3 × 1 = – 3 = – 6 – (- 3)

– 3 × 0 = 0 = – 3 – (- 3)

– 3 × -1 = 0 – (- 3) = 0 + 3

= 3 – 3 × – 2 = 3 -(- 3)

= 3 + 3 = 6 – 3 ×-3 = 6 – (- 3) = 6 + 3 = 9

So, – 3 × – 3 = 9

(ii) – 3 × 4 =

Solution:

– 3 × 4 = -12 – 3 × 3 = – 9 = -12 – (-3)

– 3 × 2 = – 6 = – 9 – (- 3)

– 3 × 1 = – 3 = – 6 – (- 3)

– 3 × 0 = 0 = – 3 – (- 3)

– 3 × – 1 = 0 – (- 3) = 0 + 3 = 3 – 3 × – 2 = 3 – (- 3) = 3 + 3 =6 – 3 × – 3 = 6 -(- 3) = 6 + 3 = 9 – 3 × – 4 = 9 – (- 3) = 9 + 3 =12

So, – 3 × – 4 = 12

Try These (Page 12)

Question 1.

Now observe these products and fill in the blanks :

(i) – 4 × 4 = 16

Solution:

– 4 × 4 = -16

(ii) – 4 × 3 = – 12 = 16 + 4

Solution:

– 4 × 3 = – 12 = -16 + 4

(iii) – 4 × 2 = _______ = -12 + 4

Answer:

– 4 × 2 = – 8 = – 12 + 4

(iv) – 4 × 1= _______

Solution:

– 4 × 1 = – 4 = – 8 + 4

(v) – 4 × 0 = _______

Solution:

– 4 × 0 = 0 = 4 + 4

(vi) – 4 × (-1) = _______

Solution:

– 4 × (-1) = 0 – (- 4) = 0 + 4 = 4

(vii) – 4 x (- 2) = _______

Solution:

– 4 × (- 2) = 4 -(- 4) =4 + 4 = 8

(viii) – 4 × ( – 3) = _______

Solution:

– 4 × (- 3) = 8 – (- 4) = 8 + 4 =12

From these patterns we observe that

(- 4) × (- 2) = 4 × 2 = 8

(- 4) × (- 3) = 4 × 3 = 12

Thus, we can say that the product of two negative integers is a positive integer. We multiply the two negative integers as whole numbers and put the positive sign before the product.

Thus, we have,

(- 10) × (- 12) = 120 Similarly, (- 15) × (- 6) = 90

In general, for any two positive integers a and b, (- a) × (- b) = a × b

Try These (Page 12)

Question 1.

(i) Starting from (- 5) × 4, find (- 5) × (- 6)

Solution:

– 5 × 4 = – 20

– 5 × 3 = -15 = – 20 – (- 5)

– 5 × 2 = – 10 = – 15 – (- 5)

– 5 × 1 = – 5 = – 10 – (- 5)

– 5 × 0 = 0 = – 5 – (- 5)

– 5 × -1 = 0 – (- 5) = 0 + 5 = 5

– 5 × – 2 = 5 – (- 5) = 5 + 5 = 10

– 5 × – 3 = 10 – (- 5) = 10 + 5 = 15

– 5 × – 4 = 15 – (- 5) = 15 + 5 = 20

– 5 × – 5 = 20 – (- 5) = 20 + 5 = 25

– 5 × – 6 = 25 – (- 5) = 25 + 5 = 30

(ii) Starting from (- 6) × 3, find (- 6) × (- 7)

Solution:

(- 6) × 3 = – 18

(- 6) × 2 = – 12 = -18 – (- 6)

(- 6) × 1 = – 6 = -12 – ( -6)

(- 6) × 0 = 0 = – 6 – (- 6)

(- 6) × – 1 = 0 – (- 6) = 0 + 6 = 6

(- 6) × – 2 = 6 -(- 6) = 6 + 6 = 12

(- 6) × – 3 = 12 – (- 6) =12 + 6 = 18

(- 6) × – 4 = 18 – (-6) = 18 + 6 = 24

(- 6) × – 5 = 24 – (- 6) = 24 + 6 = 30

(- 6) × – 6 = 30 – (- 6) = 30 + 6 = 36

(- 6) × – 7 = 36 – (- 6) = 36 = 6 = 42

Try These (Page 12)

Question 1.

Find : (- 31) × (- 100), (- 25) × (- 72), (- 83) × (- 28).

Solution:

(- 31) × (- 100) = 31 × 100 = 3100

(- 25) x (- 72) = 25 × 72 = 1800

(- 83) × (- 28) = 83 × 28 × 28

= 2324

Try These (Page 14)

Question 1.

What is the product of five negative integers in (d)? So, what will be the product of six negative integers?

Solution:

The product of five negative integers in (d) is a negative integer. So, the product of six negative integers will be a positive integer. Thus, we can say that “if the number of negative integers in a product is even, then the product is a positive integer; if the number of negative integers in a product is odd, then the product is a negative integer”.

Question 2.

Justify it by taking five more examples of each kind.

Solution:

When the number of negative integers is even :

(i) (- 2) × (- 4) = 8

(ii) (-1) × (- 2) × (- 3) × (- 4) = [(- 1) × (- 2)] × [(- 3) × (- 4)] = (2) × (12) = 24

(iii) (- 1) × (- 2) × (- 3) × (- 4) × (- 5) × (- 6) = [(- 1) × (- 2)] × ((- 3) x (- 4)] × [(- 5) × (- 6)] = (2) × (12) × (30) = [2 × 12] × 30 = 24 × 30 = 720

(iv) (- 2) × 3 × (-4 ) = [(- 2) × (- 4)] x 3= 8 × 3 = 24

(v) (- 2) × (- 3) × (- 4) × (- 5) × 6 = [(- 2) × (- 3)] × [(- 4) × (- 5)] × 6 = (6) × (20) × 6 = [6 × 20] × 6 = 120 × 6 = 720

Thus, we can say that if the number of negative integers is even, then the product is a positive integer. When the number of negative integers is odd:

(i) (-1) × (- 2) × (- 3)

= [(-1) × (- 2)] × (- 3)

= 2 × (- 3)

= – (2 × 3)

= – 6

(ii) (- 1) × (- 2) × (- 3) × (- 4) × (- 5)

= [(-1) × (- 2)] x [(- 3) × (- 4)] × (- 5)

= (2) × (12) × (- 5)

= [2 × 12] × (- 5)

= 24 × (- 5)

= – (24 × 5)

= -120

(iii) (- 2) × (- 3) × (- 4) x 5

= [(- 2) × (- 3)] × (- 4) x 5

= 1(6) × (- 4)] × 5

= – (6 × 4) × 5

= – 24 × 5

= – (24 × 5)

= – 120

(iv) (- 3) × (- 4) × (- 5) × 3

= [(- 3) × (- 4)] × (- 5) × 3

= [(12) × (- 5)] × 3

= – (12 × 5) × 3

= (- 60) × 3

= (- 60 × 3)

= – 180

(v) (- 1) × (- 2) × (- 3) × (- 4) × (- 5) × (- 6) × (- 7)

= [(-1) × (- 2)] × [(- 3) × (- 4)] × [(- 5) × (- 6)] × (- 7)

= (2) × (12) × (30) × (- 7)

= [2 × 12] × 30 × (- 7)

= 24 × 30 x (- 7)

= [24 × 30] × (- 7)

= 720 × (- 7)

= – (720 × 7)

= – 5040

Try These (Page 15)

Question 1.

Properties of Multiplication of Integers Closure under Multiplication Observe the following table and complete it:

Statement | Inference |

(-20) × (-5) = 100 | Product is an integer |

(-15) × 17 = -255 | Product is an integer |

(-30) × 12 = | ___________ |

(-15) × (-23)- | ___________ |

(-14) × (-13)- | ___________ |

12 × (-30) – | ___________ |

What do you observe? Can you find a pair of integers whose product is not an integer?

Solution:

Statement | Inference |

(- 20) × (- 5) = 100 | Product is an integer |

(- 15) × 17 = – 255 | Product is an integer |

(- 30) × 12 = – 360 | Product is an integer |

(-15) × (- 23) = 345 | Product is an integer |

(-14) × (-13) = -182 | Product is an integer |

12 × (- 30) = – 360 | Product is an integer |

We observe that if we multiply two integers, we get an integer. In other words, if a and b are two integers, then their product a × b is an integer. So, we can say that integers are closed under multiplication. Thus, we cannot find a pair of integers whose product is not an integer.

Question 2.

Find the product of five more pairs of integers and verify the above statement.

Solution:

Let us take five pairs of integers and multiply them as shown below in order to verify the statement

We find that the product is always an integer.

Try These (Page 16)

Question 1.

What are your observations?

Solution:

We observe that the multiplication of integers is commutative. In other words, if a and b are any two integers, then

a ×b = b × a

Question 2.

Write five more such examples and verify.

Solution:

In order, to verify this property let us take five pairs of integers and multiply these integers in different orders as shown below:

We find that, in whatever order we multiply two integers, the product remains the same.

Try These (Page 16)

Question 1.

What happens when we multiply any integer with -1? Complete the following:

(i) (- 3) × (-1) =

Solution:

(- 3) × (- 1) = 3

(ii) (-6) × (-1) =

Solution:

(- 6) × (- 1) = 6

(iii) (-1) × 13 =

Solution:

(- 1) × 13 = -13

(iv) (-1) × (- 25) =

Solution:

(-1) × (- 25) = 25

(v) 18 × (- 1) =

Solution:

18 × -1 = – 18

We observe that the product of any integer and the number – 1 is the additive inverse of the integer.

In other words, if a is any integer, then a × (-1) = (-1) × a = – a

No, we can’t say that -1 is a multiplicative identity of integers because for any integer a.

a × (- 1) = – a ≠ a

Try These (Page 17)

Question 1.

Associativity for Multiplication Look at this and complete the products :

[(7) × (- 6)] × 4 × 4 =

7 × [(- 6) × 4] = 7 × =

Is [7 × (- 6)] × (4) = 7 × [(- 6) × 4] ?

Solution:

[(7) × ( – 6)] × 4 = (- 42) × 4 = -168

7 × [(- 6) × 4] = 7 x (-24) = -168

So, [7 × (- 6)] × (4) = 7 × [(- 6) × (4)]

In general, for any three integers a, b and c (a × b) x c = a × (b × c)

Take any five values for a, b and c each and verify this property.

In order to verify this property, we take three integers, say a, b, c, and find the values of the expressions (a × b) x c and a × (b × c) as shown below :

We find that in each case (a × b) × c = a × (b × c).

Thus, the multiplication of integers is associative.

Distributive Property The multiplication of integers distributes over their addition. In other words, if a, b, c are any three integers, then a × (b + c) = a × b + a × c

Try These (Page 18)

Question 1.

Take at least five different values for each of a, b, and c and verify the above distributive property.

Solution:

In order to verify this property, we take any three integers, a, b, c and find the values of the expressions a × (b + c) and a × b + a × c as shown below:

We find that the expressions a × (b + c) and a × b + a × e are equal in each case.

Try These (Page 18)

(i) Is 10 × [6 + (- 2)] = 10 × 6 +10 × (- 2)?

Solution:

We have,

10 × [6 + (- 2)] =10 × 4 = 40

and, 10 × 6 + 10 × (- 2)= 60 – 20 = 40

∴ 10 × [6 + (- 2)] = 10 × 6 +10 × (- 2)

(ii) Is (- 15) × [(- 7) + (-1)] = (-15) × (- 7) + (-15) × (-1)

Solution:

We have,

(-15) × [(- 7) + (-1)] = (-15) × (- 8) = 120

and, (-15) × (- 7) +(-15) × (-1) = 105 + 15 = 120

(-15) × [(- 7) + (- 1)] = (-15) × (-7) + (-15) × (-1)

Look at the following :

(- 5) × [(- 4) – (- 6)] = (- 5) × 2 = – 10

[(- 5) × (- 4)] – [(- 5) × (- 6)] = 20 – 30 = -10

So, (- 5) × [(-4) – (- 6)] = [(- 5) × (- 4)] – [(- 5) × (- 6)]

Check this for (- 9) x [10 -(- 3)]

and [(- 9) × 10] -[(- 9) × (- 3)]

Now, (- 9) × [10 – (- 3)] = (- 9) × [10 + 3] = (- 9) × 13

= – (9 × 13) = -117

and, [( – 9) × 10] – [(- 9) × (- 3)] = [-(9 x 10)] – [9 x 3]

= – 90 – 27 = – 117

So, (- 9) × [10 – (- 3)] = [(- 9) × 10] – [(- 9) × (- 3)]

In general, for any three integers a, b and c, a × (b – c) = a × b – a × c

Question 2.

Take at least five different values for each of a, b, and c and verify this property.

Solution:

In order to verify this property, we take any three integers a, b, c and find the values of the expressions a × (b – c) and a × b – a × c as shown below:

We find that in each case, the expressions a × (b – c) and a b x b- a × c are equal.

Try These (Page 18)

(i) Is 10 × [6 – (- 2)] = 10 × 6 -10 × (-2)?

Solution:

We have,

10 × [6 – (- 2)] = 10 × [6 + 2]

= 10 × 8 = 80 10 × 6 – 10 x (- 2) = 60 +10 x 2 = 60 + 20 = 80

10 × [6 – (-2)] = 10 × 6 -10 × (- 2)

(ii) Is (- 15) × [(- 7) – (- 1)] = (- 15) × (- 7) – (- 15) × (- 1)?

Solution:

We have,

(- 15) × [(- 7) – (-1)] = (- 15) × (- 7 + 1)

. = (-15) × (- 6) = 90

and, – 15 × (- 7) – (- 15) × (- 1) = 105 – 15 = 90

(- 15) × [(- 7) – (-1)] = (-15) × (- 7) – (- 15) × (- 1)

Try These (Page 19)

Question 1.

Find (- 49) x 18 ; (- 25) x (- 31); 70 x (-19) + (-1) x 70 using distributive property.

Solution:

We have, (- 49) x 18 = (- 49) x (20 – 2)

= (- 49) x 20 – (- 49) x (2)

= -980 + 98 = – 882

We have,

(- 25) x (- 31) = (- 25) x [(- 30) + (- 1)]

= (- 25) x (- 30) + (- 25) x (- 1) = 750 + 25 = 775

We have,

70 X (-19) + (- 1) X 70 = 70 X (- 19) + 70 x (- 1)

= 70 x [(- 19) + (- 1)]

= 70 x [(- 19) + (- 1)]

= 70 x (- 20) = – 1400

Try These (Page 22)

Question 1.

Find:

(a) (- 100) ÷ 5

Solution:

(-100) ÷ 5 = – 20

(b)(- 81) + 9

Solution:

(- 81) ÷ 9 = – 9

(c)(- 75) + 5

Solution:

(- 75) ÷ 5 = – 15

(d) (- 32) ÷ 2

Solution:

(- 32) ÷ 2 = -16

Thus, we find that when we divide a negative integer by a positive integer, we divide them as whole numbers and then put a minus sign (-) before the quotient. We, thus, get a negative integer.

Try These (Page 23)

Question 1.

Find:

(a) 126 ÷ (- 25)

Solution:

125 ÷ (- 25) = – 5

(b) 80 ÷ (- 5)

Solution:

80 ÷ (- 5) = -16

(c) 64 ÷ (-16)

Solution:

64 ÷ (-16) = – 4

Try These (Page 23)

Question 1.

Find:

(a) (- 36) ÷ (- 4)

Solution:

(-36) ÷ (-4) = 36 ÷ 4= 9

(b) (- 201) ÷ (- 3)

Solution:

(- 201) ÷ (- 3) = 201 ÷ 3 = 67

(c) (- 325) ÷ (- 18)

Solution:

(- 325) ÷ (-13) = 325 ÷ 13 = 25

Try These (Page 24)

(i) Is (- 9) ÷ 3 the same as 3 ÷ (- 9) ?

Solution:

We have, (- 9) ÷ 3 = – 3

and 3 ÷ (- 9) =

∴ (- 9) ÷ 3 ≠ 3 ÷ (- 9)

(ii) Is (- 30) ÷ (- 6) the same as (- 6) ÷ (- 30) ?

Solution:

We have, (- 30) + (- 6) =

and (- 6) ÷ (- 30) =

(- 30) ÷ (- 6) ≠ (- 6) ÷ (- 30)

Hence, commutative property does not hold good for integers. In other words, if a(≠ 0) and b(≠ 0) are any two integers, then a + b ≠ b + a, except when a = b.

Try These (Page 24)

Question 1.

(i) 1 + a = 1?

Solution:

Let us take a = 3 ,

∴ 1 ÷ a = 1 ÷ 3 = ≠ 1

so, 1 ÷ a ≠ 1

(ii) a ÷ (- 1) = – a?

Solution:

Let us take a = 5

. a ÷ (-1) = 5 ÷ (- 1) = – 5

So, a ÷ (-1) = – a is true.