Real Numbers

Exercise 1.1

1. (i) 135 and 225
   Since 225 > 135, we apply the division lemma to 225 and 135 to obtain
   225 = 135 × 1 + 90
   Since remainder 90 ≠ 0, we apply the division lemma to 135 and 90 to obtain
   135 = 90 × 1 + 45
   We consider the new divisor 90 and new remainder 45, and apply the division lemma to obtain
   90 = 2 × 45 + 0
   Since the remainder is zero, the process stops.
   Since the divisor at this stage is 45,
   Therefore, the HCF of 135 and 225 is 45.
   (ii)196 and 38220
   Since 38220 > 196, we apply the division lemma to 38220 and 196 to obtain
   38220 = 196 × 195 + 0
   Since the remainder is zero, the process stops.
   Since the divisor at this stage is 196,
   Therefore, HCF of 196 and 38220 is 196.
   (iii)867 and 255
   Since 867 > 255, we apply the division lemma to 867 and 255 to obtain
   867 = 255 × 3 + 102
   Since remainder 102 ≠ 0, we apply the division lemma to 255 and 102 to obtain
   255 = 102 × 2 + 51
   We consider the new divisor 102 and new remainder 51, and apply the division lemma to obtain
   102 = 51 × 2 + 0
   Since the remainder is zero, the process stops.
   Since the divisor at this stage is 51,
   Therefore, HCF of 867 and 255 is 51.
   
2. Let abe any positive integer and b = 6. Then, by Euclid’s algorithm,
   a = 6q + rfor some integer q ≥ 0, and r = 0, 1, 2, 3, 4, 5 because 0 ≤ r < 6.
   Therefore,a = 6q or 6q + 1 or 6q + 2 or 6q + 3 or 6q + 4 or 6q + 5
   Also, 6q+ 1 = 2 × 3q + 1 = 2k₁ + 1, wherek₁ is a positive integer
   6q+ 3 = (6q + 2) + 1 = 2 (3q + 1) + 1 = 2k₂+ 1, where k₂ is an integer
   6q+ 5 = (6q + 4) + 1 = 2 (3q + 2) + 1 = 2k₃+ 1, where k₃ is an integer
   Clearly, 6q + 1, 6q + 3, 6q + 5 are of the form 2k+ 1, where k is an integer.
   Therefore, 6q + 1, 6q + 3, 6q + 5 are not exactly divisible by 2. Hence, these expressions of numbers are odd numbers.
   And therefore, any odd integer can be expressed in the form 6q + 1, or 6q + 3,
   or 6q+ 5
   
3. HCF (616, 32) will give the maximum number of columns in which they can march.
   We can use Euclid’s algorithm to find the HCF.
   616 = 32 × 19 + 8
   32 = 8 × 4 + 0
   The HCF (616, 32) is 8.
   Therefore, they can march in 8 columns each.
   
4. Let abe any positive integer and b = 3.
   Then a= 3q + r for some integer q ≥ 0
   And r= 0, 1, 2 because 0 ≤ r < 3
   Therefore,a = 3q or 3q + 1 or 3q + 2
   Or,
   
   Where k₁,k₂, and k₃ are some positive integers
   Hence, it can be said that the square of any positive integer is either of the form 3m or 3m + 1.
5. Let abe any positive integer and b = 3
   a =3q + r, where q ≥ 0 and 0 ≤ r < 3
   
   Therefore, every number can be represented as these three forms. There are three cases.
   Case 1: When a = 3q,
   
   Where mis an integer such that m =
   Case 2: When a = 3q + 1,
   a³= (3q +1)³
   a³= 27q³ + 27q² +9q + 1
   a³= 9(3q³ + 3q² + q) + 1
   a³= 9m + 1
   Where mis an integer such that m = (3q³ +3q² + q)
   Case 3: When a = 3q + 2,
   a³= (3q +2)³
   a³= 27q³ + 54q² +36q + 8
   a³= 9(3q³ + 6q² +4q) + 8
   a³= 9m + 8
   Where mis an integer such that m = (3q³ +6q² + 4q)
   Therefore, the cube of any positive integer is of the form 9m, 9m + 1,
   or 9m + 8.
   Exercise 1.2

1. 
   
2. 
   
   
   Hence, product of two numbers = HCF × LCM
   
   Hence, product of two numbers = HCF × LCM
   
   Hence, product of two numbers = HCF × LCM
3. 
   
   
   
   
   
    
4. 
   
5. If any number ends with the digit 0, it should be divisible by 10 or in other words, it will also be divisible by 2 and 5 as 10 = 2 × 5
   Prime factorisation of 6ⁿ = (2 ×3)ⁿ
   It can be observed that 5 is not in the prime factorisation of 6ⁿ.
   Hence, for any value of n, 6ⁿ will not be divisible by 5.
   Therefore, 6ⁿ cannot end with the digit 0 for any natural number n.
6. Numbers are of two types - prime and composite. Prime numbers can be divided by 1 and only itself, whereas composite numbers have factors other than 1 and itself.
   It can be observed that
   7 × 11 × 13 + 13 = 13 × (7 × 11 + 1) = 13 × (77 + 1)
   = 13 × 78
   = 13 ×13 × 6
   The given expression has 6 and 13 as its factors. Therefore, it is a composite number.
   7 × 6 × 5 × 4 × 3 × 2 × 1 + 5 = 5 ×(7 × 6 × 4 × 3 × 2 × 1 + 1)
   = 5 × (1008 + 1)
   = 5 ×1009
   1009 cannot be factorised further. Therefore, the given expression has 5 and 1009 as its factors. Hence, it is a composite number.

It can be observed that Ravi takes lesser time than Sonia for completing 1 round of the circular path. As they are going in the same direction, they will meet again at the same time when Ravi will have completed 1 round of that circular path with respect to Sonia. And the total time taken for completing this 1 round of circular path will be the LCM of time taken by Sonia and Ravi for completing 1 round of circular path respectively i.e., LCM of 18 minutes and 12 minutes.
18 = 2 ×3 × 3
And, 12 = 2 × 2 × 3
LCM of 12 and 18 = 2 × 2 × 3 × 3 = 36
Therefore, Ravi and Sonia will meet together at the starting pointafter 36 minutes.

Exercise 1.3

1. Let is a rational number.
   Therefore, we can find two integers a, b (b ≠ 0) such that
   Let aand b have a common factor other than 1. Then we can divide them by the common factor, and assume that a and b are co-prime.
   
   Therefore,a² is divisible by 5 and it can be said that ais divisible by 5.
   Let a= 5k, where k is an integer
   
   This means that b² is divisible by 5 and hence, bis divisible by 5.
   This implies that a and b have 5 as a common factor.
   And this is a contradiction to the fact that a and b are co-prime.
   Hence,cannot be expressed as or it can be said thatis irrational.
   
2. Let is rational.
   Therefore, we can find two integers a, b (b ≠ 0) such that
   
   Since aand b are integers, will also be rational and therefore,is rational.
   This contradicts the fact thatis irrational. Hence, our assumption that is rational is false. Therefore, is irrational.
3. 
   Let is rational.
   Therefore, we can find two integers a, b (b ≠ 0) such that
   
   
   is rational as a and b are integers.
   Therefore, is rational which contradicts to the fact that is irrational.
   Hence, our assumption is false and is irrational.
   
   Let is rational.
   Therefore, we can find two integers a, b (b ≠ 0) such that
   for some integers a and b
   
   is rational as a and b are integers.
   Therefore, should be rational.
   This contradicts the fact thatis irrational. Therefore, our assumption that is rational is false. Hence, is irrational.
   
   Let be rational.
   Therefore, we can find two integers a, b (b ≠ 0) such that
   
   Since a and b are integers, is also rational and hence, should be rational. This contradicts the fact thatis irrational. Therefore, our assumption is false and hence, is irrational.
   
   Exercise 1.4

1. (i)
   
   The denominator is of the form 5^m.
   Hence, the decimal expansion ofis terminating.
   (ii)
   
   The denominator is of the form 2^m.
   Hence, the decimal expansion of is terminating.
   (iii)
   455 = 5 × 7 × 13
   Since the denominator is not in the form 2^m × 5ⁿ, and it also contains 7 and 13 as its factors, its decimal expansion will be non-terminating repeating.
   (iv)
   1600 = 2⁶ × 5²
   The denominator is of the form 2^m ×5ⁿ.
   Hence, the decimal expansion of is terminating.
   (v)
   
   Since the denominator is not in the form 2^m × 5ⁿ, and it has 7 as its factor, the decimal expansion of is non-terminating repeating.
   (vi)
   The denominator is of the form 2^m × 5ⁿ.
   Hence, the decimal expansion of is terminating.
   (vii)
   Since the denominator is not of the form 2^m × 5ⁿ, and it also has 7 as its factor, the decimal expansion of is non-terminating repeating.
   (viii)
   The denominator is of the form 5ⁿ.
   Hence, the decimal expansion of is terminating.
   (ix)
   
   The denominator is of the form 2^m × 5ⁿ.
   Hence, the decimal expansion of is terminating.
   (x)
   
   Since the denominator is not of the form 2^m × 5ⁿ, and it also has 3 as its factors, the decimal expansion of is non-terminating repeating.
2.  







(viii)


3. (i) 43.123456789
   Since this number has a terminating decimal expansion, it is a rational number of the formand q is of the form
   i.e., the prime factors of q will be either 2 or 5 or both.
   (ii) 0.120120012000120000…
   The decimal expansion is neither terminating nor recurring. Therefore, the given number is an irrational number.
   (iii)
   Since the decimal expansion is non-terminating recurring, the given number is a rational number of the form and q is not of the form i.e., the prime factors of q will also have a factor other than 2 or 5.