## 1.2 PROPERTIES OF INTEGERS:

**1.2 PROPERTIES OF INTEGERS:**

The sum of any two integers is always an integer. **Integers are closed under addition**.

For any two integers a and b, **a + b is an integer.**

If a and b are any two integers, then **a + b = b + a**

If a,b,c are any three integers, then **(a + b) + c = a + (b + c)**

If ‘a’ is any integer, then **a + 0 = a or 0 + a.** Zero is additive identity for integers.

The sum of an integer and its opposite is 0. Thus , if ‘a’ is an integer, then** a + (−a) = 0**

If ‘a’ and ‘b’ are any two integers and** a − b = c,** then ‘c’ is also an integer.

If ‘a’ is an integer. then **(a−1) is its predecessor.**

If ‘a’ is any integer , then** a−o =a,** i.e., zero subtracted from any integer is the integer itself.

**Exercise 1.2 **

**Write down a pair of integers whose:****a) Sum is – 7****b) Difference is****– 10 c) Sum is 0**

**Solution: **a) We can have (– 3) + (– 4) =– 7

The required pair of integers is – 3 and – 4

b) We can have – 16 – (– 6) = – 16 + 6 = – 10

The required pair of integers is – 16 and – 6

c) We can have – 8 + 8 = 0

**The required pair of integers is – 8 and 8**

**2.****a) Write a pair of negative integers whose difference gives 8 ****b) Write a negative integer and a positive integer whose sum is –5 ****c) Write a negative integer and a positive integer whose difference is – 3****.**

** Solution: **a) Since,** –**7 –( –15) = –7 + 15 = 8

So, ** – 7** and 15 is a pair of negative integers such that their difference is 8.

b) Since, (**– **6) + 1 = ** –**5

So, ** – **6 and 1 is a pair of integers such that their sum is** – **5 and one of them is a negative integer.

c) Since, –9 –( – 6) = –9 + 6 = – 3

**So, – 9 and – 6 are a pair of integers such that their difference is – 3 and one of them is positive and other is negative.**

**3.In a quiz, team A scored – 40, 10, 0 and team B scored 10, 0, – 40 in three successive rounds. Which team scored more? Can we say that we can add integers in any order?**

**Solution:- **Total score obtained by team A = – 40 + 10 + 0

= – 40 + 10 = – 30

Total Score of team B = 10, 0, −40

Total score obtained by team B = 10 + 0 + (– 40)

= 10 + 0 – 40 = – 30

**Therefore, scores of the both A team and B team are same. – 30.**

**Yes, we can say that we can add integers in any order.**

**4.Fill in the blanks to make the following statements true:**

**(i) (–5) + (– 8) = (– 8) + (…………)**

**(ii) –53 + ………… = –53**

**(iii) 17 + ………… = 0**

**(iv) [13 + (– 12)] + (…………) = 13 + [(–12) + (–7)]**

**(v) (– 4) + [15 + (–3)] = [– 4 + 15] +…………**

**Solution:- **

i) Since, integers can be added in any order, (– 5) + (– 8) = (– 8) + (– 5)

ii) If we add zero to any integer, we get the same integer. – 53 + 0 = – 53

iii) We know that the sum of an integer and its additive inverse is zero. 17 + (– 17) = 0

iv) Since, the addition of integers is associative. For three integers, a, b, and c, we have: ( a + b) + c = a + (b + c) Thus, [13 + (– 12)] + (– 7) = 13 + [(–12) + (–7)]

v) (– 4) + [15 + (–3)] = [– 4 + 15] + (–3)

( –4) + [15 –3] = [ 11] – 3

– 4 +12 = 11 – 3

8 = 8

**Try these: (page 10) **

Q.1) **Using number line, find:**

(i) 4 x ( –8) (ii) 8 x (–2) (iii) 3 (iv) 10

**Solution: **

From the number line, we have:

(8) (8) (8) (8) = 32

∴ 4 x (–8) = 32

From the number line, we have:

From the number line, we find:

(7) + (7) + (7) = 21

From the number line, we have:

(–1)+ (–1) +(–1) +(–1)+ (–1) +(–1) +(–1)+ (–1) +(–1) +(–1) = –10

**Try these: **** (page 10)**

**i) 6 ×–19 ****ii) 12 ×–32 **** iii) 7 × –22**

**Solution:**

i) 6 **×** **–**19= **– **[6 × 19]** = –114**

ii) 12 **×–**32 = **– **[12 × 32] **= –384**

iii) 7 **×–**22 = **– **[7 × 22]** = –154**

** Try these: (page 11)**

**a) 15 × ( –16) ****b) 21× ( –32) **** c) (–42****) **** × 12 d) (–55****)× 15**

**Solution: **

**While multiplying a positive integer and a negative integer, we multiply them first and put a minus sign (–****) before the product.**

a) 15 **×** (**–** 16) = [15 **×** **–**16] = **–240**

b) 21 **×****–**32 = [21 **×–**32 ] =** –672**

c) (**–**42)**×** 12 = **–**[42 ×12] = **–504**

d) (**–**55) ×15 = **–**[55 ×15]**= –82**

**Try these:**** (page 12)**

**1) Find: (–****31)****(–****100), (–****25)****(–****72), (–****83)****(–****28)**

**Solution: ****We multiply the two negative integers and put the positive sign(+) before the product.**

(**–**31)**×**(**–**100) = [ **–**31**×–**100] = **+ 3100**

(**–**25)**×**(**–**72) = [**–**25 **×**–72] = **+1800**

(–83)**×**(–28) = [–83 **×**–28] = **+2324**

**Try these:**** (page 18) **

**i) Is 10 x [6 + (−2)] = 10 x 6 + 10 x (−2)**

**ii) Is (−15****) x [(−7) + (−1)] = (-15) x (-7) + (−15) x (−1) **

**Solution:**

i) [* a *x (b + c) = *a* x b + *a *x c]

10 x [ 6 + (−2) ] = 10 x [ 6 − 2 ]

= 10 x 4** = 40**

And, 10 x 6 + 10 x (−2) = 60 − 20 **= 40.**

**Thus, 10 x [6 + ( –2)] = 10 x 6 + 10 ( –2)**

(ii) [ *a *x ( b + c ) = *a* x b + *a *x c]

( **–**15) x [ **–**(7) + ( **–**1) = ( **–**15 ) x [ **–**7 **–**1 ]

= ( **–**15 ) x ( **–**8 )

= ( + ) ( 15 x 8 ) **= 120.**

And, ( **–** 15 ) x ( **–** 7 ) + ( **–** 15 ) x ( **–**1 )

= (+) ( 15 x 7 ) + ( + ) ( 15 x 1 )

= 105 + 15 **= 120.**

**Thus, ( – 15 ) x (–7) + ( – 1 ) =(–15) x (7) + (–15) x (–1)**

**Try these :**

**(Page 19)**

**By using distributive property find: i) (– 49 ) x 18; ii) (–25) x (–31); iii) 70 x (–19) + (–1) x 70 **

**Solution:**

**i) ( – 49 ) x 18:**

18 = 10 + 8

( **–**49 ) x 18

= ( **–** 49 ) x [10 x 8 ]

= ( **–**49 ) x 10 + ( **–**49 ) x 8

Using Distributive Property we get

= **–** 490 + ( **–**49 ) [ 10 **–** 2 ] because [ 8 = 10 **–**2 ]

= − 490 + ( **–** 49 ) x 10 **–** ( **–**49 ) x 2

= 490 + (**–** 490 ) + 98

= **–**980 + 98 **= 882**

**(ii) ( –25 ) x (–31):**

**–**31 = ( **–**30 ) + ( **–** 1 )

( **–**25 ) x ( **–**31 )

= ( **–** 25 ) x [ ( **–**30 ) + (**–**1) ]

= (**–**25 ) x ( **–**30 ) + ( **–**25) x (**–**1)

using Distributive Property

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

= 750 + 25 **= 775**

**(iii) 70 x ( –19 ) + ( –1 ) x 70 :**

*a ×* b + *a ×* c = *a ×* [ b + c ]

70 *× *( **–**19 ) + (**–** 1 ) *×* 70

= 70 *×* [ ( 19 ) + (**–**1) ]

= 70 *×* [ **–** 20 ] = [ 70 *×* 20 ] ** = –1400**