CHAPTER – 1 NUMBER SYSTEM EXERCISE 1.1

 Introduction : 

• Euclid’s Division Lemma (or Euclid’s Division Algorithm):  Let a and b be any two integers and b>0. Then there exists unique integers q and r such that a = bq + r and 0 ≤ r < b.
• Lemma: A lemma is a provable statement used in proving another statement.
• Algorithm: An algorithm is a well defined sequence of steps forming a process of solving given problem.
• Euclid’s Algorithm for finding HCF of two given positive integers:

1.Find the quotient and remainder of the division of the greater number by the smaller.
2.If the remainder is zero, then the divisor is the HCF.
3.Else, taking the previous remainder as the new divisor and the previous divisor as the new dividend, find the quotient and remainder.
4.Continue the process till the remainder is zero. 

5.The last divisor is the required HCF.
Theorem: If a = bq + r, then (a, b) = (b, r).
Note: The symbol (a, b) denotes the HCF of two positive integers a and b.

EXERCISE 1.1

 
1.Using Euclid’s algorithm find the HCF of
 (i) 1240 and 1984
Solution:          We have
1984 = 1240 ×1+744
1240 = 744 ×1+496
744 = 496 ×1+248
496 = 248 ×2+0
∴ (1984,1240) =248

(ii) 348 and 504
Solution:     We have
504 = 348 ×1+156
348 = 156 ×2+36
156 = 36 ×4+12
36 = 12 ×3+0
∴ (504,348) =12
 
(iii) 986 and 899
Solution:     We have
986 = 899 ×1+87
899 = 87 ×10+29
87 = 29 ×3+0
∴ (986,899) =29
 
(iv) 4216 and 1240
Solution:     We have
4216 = 1240 ×3+496
1240 = 496 ×2+248
496 = 248 ×2+0
∴ (4216,1240) =248
 
(v)10605 and 5256
Solution:-
We have
10605 = 5256 ×2+93
5256 = 93 ×56+48
93 = 48 ×1+45
48 = 45 ×1+3
45 = 3 ×15+0
∴ (10605, 5256) =3

(vi) 10005 and 9269
Solution:-
We have
10005 = 9269 ×1+736
9269 = 736 ×12+437
736 = 437 ×1+299
437 = 299 ×1+138
299 = 138 ×2+ 23
138 = 23 ×6+ 0
∴ (10005, 9269) =23

2.Show that the product of two consecutive integers is divisible by 2.
Solution:
Let a and a+1 be any two consecutive integers.
The integer a is of the form 2q or 2q + 1 for some integer q. 

If a = 2q, a(a + 1) = 2q(2q + 1)  which is divisible by 2. 

 If a = 2q + 1, a(a + 1) = (2q + 1)(2q + 1 + 1)
= (2q + 1)(2q + 2)
= (2q + 1) × 2(q + 1)
= 2(2q + 1)(q + 1) which is divisible by 2.
Thus, the product of two consecutive integers is divisible by 2.

3.Show that the product of two consecutive even integers is divisible by 8.
Solution:
Let 2a and 2a + 2 be any two consecutive even integers. 

The integer a is of the form 2q or 2q + 1 .
If a = 2q, 2a(2a + 2) = 2 × 2q(2 × 2q + 2)
= 4q(4q + 2)
= 4q × 2(2q + 1)
= 8q(2q + 1), which is divisible by 8.
 
If 

a = 2q + 1, 2a(2a + 2) = 2(2q + 1){2(2q + 1) + 2}
= 2(2q + 1)(4q + 2 + 2)
= 2(2q + 1)(4q + 4)
= 2(2q + 1) × 4(q + 1)
= 8(2q + 1)(q + 1), which is divisible by 8.
Thus, the product of two consecutive even integers is divisible by 8.

4.Show that every integer is of the form 4q, 4q+1, 4q+2 or 4q- 1.
Solution:
Let a be any integer.
Taking b = 4 

and applying Euclid’s division lemma 

we get
a = 4q + r, where r = 0, 1, 2 or 3 

If r = 0, a = 4q + 0 = 4q

If r = 1, a = 4q + 1 

If r = 2, a = 4q + 2 

If r = 3, a = 4q + 3

= 4q + 4 − 1
= 4(q + 1) − 1
= 4k − 1 , where k = q + 1 is an integer.
Hence, every integer is of the form 4q, 4q + 1, 4q + 2 or 4q − 1.

5.Show that the product of three consecutive integers is divisible by 6.
Solution:-
Let a − 1, a and a + 1 be three consecutive integers.
The integer a is of the form 6q, 6q + 1, 6q + 2, 6q + 3, 6q − 2 or 6q − 1. 

If a = 6q, (a − 1)a(a + 1) = (6q − 1) × 6q(6q + 1)
= 6q(6q − 1)(6q + 1) which is divisible by 6.
If a = 6q + 1, (a − 1)a(a + 1) = (6q + 1 − 1)(6q + 1)(6q + 1 + 1)
= 6q(6q + 1)(6q + 2) which is divisible by 6.
If 

a = 6q + 2, (a − 1)a(a + 1) 

= (6q + 2 − 1)(6q + 2)(6q + 2 + 1)
= (6q + 1)(6q + 2)(6q + 3)
= (6q + 1) × 2(3q + 1) × 3(2q + 1)
= 6(6q + 1)(3q + 1)(2q + 1) which is divisible by 6.
If a = 6q + 3, (a − 1)a(a + 1) 

= (6q + 3 − 1)(6q + 3)(6q + 3 + 1)

= (6q + 2)(6q + 3)(6q + 4)
= 2(3q + 1) × 3(2q + 1) × 2(3q + 2)
= 6 × 2(3q + 1)(2q + 1)(3q + 2) which is divisible by 6.
If 

a = 6q − 2, (a − 1)a(a + 1) 

= (6q − 2 − 1)(6q − 2)(6q − 2 + 1)
= (6q − 3)(6q − 2)(6q − 1)
= 3(2q − 1) × 2(3q − 1)(6q − 1)
= 6 × (2q − 1)(3q − 1)(6q − 1) which is divisible by 6.
If 

a = 6q − 1, (a − 1)a(a + 1)

 = (6q − 1 − 1)(6q − 1)(6q − 1 + 1)
= (6q − 2)(6q − 1) × 6q
= 6q(6q − 2)(6q − 1) which is divisible by 6.
Thus, the product of three consecutive integers is divisible by 6.

6.Show that the square of an odd integer is of the form 8k + 1.
Solution:-
Let a be any odd integer. Then a is of the form 4q ± 1. Now, a2 = (4q ± 1)2
= (4q)2 ± 2 × 4q × 1 + 12
= 16q2 ± 8q + 1
= 8(2q2 ± q) + 1
= 8k + 1 , where k = 2q2 ± q is an integer. Thus, the square of an odd integer is of the form 8k + 1.

7.If a is divisible by neither 2 nor 3, show that a2 − 1 is divisible by 24.
Solution:-
As a is divisible by neither 2 nor 3, 

a is of the form 12q ± 1 or 12q ± 5. 

If a = 12q ± 1, a2 − 1 = (12q ± 1)2 − 1
= (12q)2 ± 2 × 12q × 1 + 12 − 1
= 144q2 ± 24q + 1 − 1
= 24(6q2 ± q), which is divisible by 24.
 
If a = 12q ± 5, a2 − 1 = (12q ± 5)2 − 1
= (12q)2 ± 2 × 12q × 5 + 52 − 1
= 144q2 ± 120q + 24
= 24(6q2 ± 5q + 1) , which is divisible by 24. 

Thus if a is divisible by neither 2 nor 3, then a2 − 1 is divisible by 24.

8.Show that any square number cannot be put in the form 4k + 2.
Solution:-
We have,
4k+2=2(2k+1).
For this product to be a perfect square, the number 2k+1 must have 2 as its prime factor, which is impossible as the number 2k+1 is odd. Therefore, the number 4k+2 cannot be a perfect square.
Hence, any square number cannot be put in the form 4k+2.

9.Show that any square number is of the form 3n or 3n+1.
Solution:-
Let a be any integer. 

Then a is of the form 3q, 3q +1 or 3q −1 If a = 3q, a2 = (3q)2
= 3 × 3q2
= 3n where n = 3q2 is an integer.
If a = 3q + 1, a2 = (3q + 1)2
= (3q)2 + 2 × 3q × 1 + 12
= 9q2 + 6q + 1
= 3(3q2 + 2q) + 1
= 3n + 1, where n = 3q2 + 2q is an integer.
If a = 3q − 1, a2 = (3q − 1)2
= (3q)2 − 2 × 3q × 1 + 12
= 9q2 − 6q + 1
= 3(3q2 − 2q) + 1
= 3n + 1, where n = 3q2 − 2q is an integer.
Thus, any square integer is of the form 3n or 3n + 1.

10.Show that one of three consecutive odd integers is a multiple of 3.
Solution:-
Let 2a+1, 2a+3 and 2a+5 be any three consecutive odd integer.

 Then a is of the form 3q, 3q + 1 or 3q + 2.
If a = 3q, 2a + 3 = 2 × 3q + 3
= 6q + 3
= 3(2q + 1) which is a multiple of 3.
If a = 3q + 1, 2a + 1 = 2(3q + 1) + 1
= 6q + 2 + 1
= 6q + 3
= 3(2q + 1) which is a multiple of 3.
If a = 3q + 2, 2a + 5 = 2(3q + 2) + 5
= 6q + 4 + 5
= 6q + 9
= 3(2q + 3) which is a multiple of 3 .

Thus, one of three consecutive odd integers is a multiple of 3.

11.Show that the product of any three consecutive even integers is divisible by 48.
Solution:-
Let a − 2, a and a + 2 be any three consecutive even integers. Then a is of the form 2q.
Now,   (a − 2)a(a + 2) = (2q − 2) × 2q × (2q + 2)
= 2(q − 1) × 2q × 2(q + 1)
= 8(q − 1)q(q + 1)
= 8 × 6k where 6k = (q − 1)q(q + 1) ∈ Z for the product of three
consecutive integers is divisible by 6.
= 48k which is divisible by 48.
Thus, the product of any three consecutive even integers is divisible by 4

Extra Question

Q.Show that one of three consecutive even integers is a multiple of 3.
Solution:-
Let 2a, 2a + 2 and 2a + 4 be any three consecutive even integers. The integer a is of the form 3q, 3q + 1 or 3q + 2 for some integer q.
If a = 3q, 2a = 2 × 3q
= 3 × 2q which is a multiple of 3.
If a = 3q + 1, 2a + 4 = 2(3q + 1) + 4
= 6q + 2 + 4
= 6q + 6
= 6(q + 1)
= 3 × 2(q + 1) which is a multiple of 3.
If a = 3q + 2, 2a + 2 = 2(3q + 2) + 2
= 6q + 4 + 2
= 6q + 6
= 6(q + 1)
= 3 × 2(q + 1) which is a multiple of 3. Thus, one of three consecutive even integers is a multiple of 3.

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