CHAPTER 1 Real Numbers [N.C.E.R.T. Chapter 1] POINTS FOR QUICK REVISION l Euclid s Division Lemma: Given two positive integers a and b, there exist unique integers q and r satisfying a = bq + r, 0 r < b. l Euclid s Division Algorithm: To obtain the HCF of two positive integers, say c and d, with c > d, we follow the steps given below: Step 1: Apply Euclid s division lemma, to c and d. So, we find whole numbers, q and r such that c = dq + r, 0 r < d. Step : If r = 0, d is the HCF of c and d. If r 0, apply the division lemma to d and r. Step : Continue the process till the remainder is zero. The divisor at this stage will be the required HCF. l Fundamental Theorem of Arithmetic: Every composite number can be expressed (factorised) as a product of primes, and this factorisation l l is unique, apart from the order in which the prime factors occur. If a and b are any two positive integers, then, HCF (a, b) LCM (a, b) = a b Theorem: Let p be a prime number. If p divides a, then p divides a, where a is a positive integer. l,, 5, 7, 11 are irrational numbers. l The sum or difference of a rational number and an irrational number is an irrational number. l The product or quotient of a non-zero rational number and an irrational number is an irrational number. l Let x be a rational number whose decimal expansion terminates. Then x can be expressed in the form p q, where p and q are co-prime, and the prime factorisation of q is of the form n 5 m, where n, m are non-negative integers. 1
MATHEMATICS X l Let x = p q be a rational number, such l If p, q, r are any three positive integers, then, l that the prime factorisation of q is of the form n 5 m, where n, m are nonnegative integers. Then x has a decimal expansion which terminates. Let x = p be a rational number, such q that the prime factorisation of q is not of the form n 5 m, where n, m are nonnegative integers. Then, x has a decimal expansion which is nonterminating repeatiy (recurring). HCF (p, q, r) = x LCM (p, q, r) p q r However, HCF(p, q, r) = p. q. r. LCM ( p, q, r) LCM ( pq, ). LCM ( qr, ). LCM ( pr, ) LCM (p, q, r) = p. q. r. HCF ( p, q, r). HCF ( pq, ). HCF ( qr, ). HCF ( pr, ) NCERT TEXT BOOK QUESTIONS (SOLVED) [NCERT Exercise 1.1] (Page 7) Example 1. Use Euclid s division algorithm to find the HCF of : (i) 15 and 5 (ii) 196 and 80 (iii) 867 and 55. Sol. (i) 15 and 5. Start with the larger integer, that is, 5. Apply the division lemma to 5 and 15, to get 5 = 15 1 + 90 Since the remainder 90 0, we apply the division lemma to 15 and 90, to get 15 = 90 1 + 45 We consider the new divisor 90 and the new remainder 45, and apply the division lemma to get 90 = 45 + 0 The remainder has now become zero, so our procedure stops. Since the divisor at this stage is 45, the HCF of 5 and 15 is 45. (ii) 196 and 80. Start with the larger integer, that is, 80. Apply the division lemma to 80 and 196, to get 80 = 196 195 + 0 The remainder is zero, so our procedure stops. Since the divisor is 196, the HCF of 80 and 196 is 196. (iii) 867 and 55. Start with the larger integer, that is, 867. Apply the division lemma to get 867 = 55 + 10 Since the remainder 10 0, we apply the division lemma to 55 and 10, to get 55 = 10 + 51 We, consider the new divisor 10 and the new remainder 51, and apply the division lemma, to get 10 = 51 + 0 The remainder has now become zero, so our procedure stops. Since the divisor at this stage is 51, the HCF of 867 and 55 is 51. Example. Show that any positive odd integer is of the form 6q + 1, or 6q +, or 6q + 5, where q is some integer.
REAL NUMBERS Sol. Let us start with taking a, where a is any positive odd integer. We apply the division algorithm, with a and b = 6. Since 0 r < 6, the possible remainders are 0, 1,,, 4, 5. That is, a can be 6q, or 6q + 1, or 6q +, or 6q +, or 6q + 4, or 6q + 5, where q is the quotient. However, since a is odd, we do not consider the cases 6q, 6q + and 6q + 4 (since all the three are divisible by ). any positive odd integer is of the form 6q + 1, or 6q +, or 6q + 5. Example. An army contingent of 616 members is to march behind an army band of members in a parade. The two groups are to march in the same number of columns. What is the maximum number of columns in which they can march? Sol. By applying the Euclid s division lemma, we can find the maximum number of columns in which an army contingent of 616 members can march behind an army band of members in a parade. HCF of 616 and is equal to maximum number of columns in which 616 and members can march. Since 616 >, we apply the division lemma to 616 and, to get 616 = 19 + 8 Since the remainder 8 0, we apply the division lemma to and 8, to get = 8 4 + 0 The remainder has now become zero, so our procedure stops. Since the divisor at this stage is 8, the HCF of 616 and is 8. the maximum number of columns in which an army contingent of 616 members can march behind an army band of members in a parade is 8. Example 4. Use Euclid s division lemma to show that the square of any positive integer is either of the form m or m + 1 for some integer m. [Hint: Let x be any positive integer then it is of the form q, q + 1 or q +. Now square each of these and show that they can be rewritten in the form m or m + 1.] Sol. Let a be any odd positive integer. We apply the division lemma with a and b =. Since 0 r <, the possible remainders are 0, 1 and. That is, a can be q, or q + 1, or q +, where q is the quotient. Now, (q) = 9q which can be written in the form m, since 9 is divisible by. Again, (q + 1) = 9q + 6q + 1 = (q + q) + 1 which can be written in the form m + 1 since 9q + 6q, i.e., (q + q) is divisible by. Lastly, (q + ) = 9q + 1q + 4 = (9q + 1q + ) + 1 = (q + 4q + 1) + 1 which can be written in the form m + 1, since 9q + 1q +, i.e., (q + 4q + 1) is divisible by. the square of any positive integer is either of the form m or m + 1 for some integer m. Example 5. Use Euclid s division lemma to show that the cube of any positive integer is of the form 9m, 9m + 1 or 9m + 8. Or If n is a positive integer, then prove that n can be expressed as 9m, 9m + 1 or 9m + 8. Sol. Let a be any positive integer. We apply the division lemma, with a and b =. Since 0 r <, the posible remainders are 0, 1 and. That is, a can be q, or q + 1, or q +, where q is the quotient. Now, (q ) = 7q which can be written in the from 9m, since 9 is divisible by 9. Again, (q + 1) = 7q + 7q + 9q + 1 = 9(q + q + q ) + 1 which can be written in the form 9m + 1, since 7q + 7q + 9q, i.e., 9(q + q + q ) is divisible by 9. CHAPTER 1
4 MATHEMATICS X Lastly, (q + ) = 7q + 54q + 6q + 8 = 9(q + 6q + 4q ) + 8 which can be written in the form 9m + 8, since 7q + 54q + 6q, i.e., 9(q + 6q + 4q ) is divisible by 9. the cube of any positive integer is of the form 9m, 9m + 1 or 9m + 8. [NCERT Exercise 1.] (Page 11) Example 1. Express each number as a product of its prime factors: (i) 140 (ii) 6 (iii) 85 (iv) 5005 (v) 749. Sol. (i) 140 140 70 So, 85 = 5 5 17 = 5 17 (iv) 5005 5005 5 1001 7 14 5 7 So, 140 = 5 7 = 5 7 (ii) 6 6 78 9 1 So, 6 = 1 = 1 (iii) 85 85 175 11 1 So, 5005 = 5 7 11 1. (v) 749 749 17 47 19 So, 749 = 17 19. Example. Find the LCM and HCF of the following pairs of integers and verify that LCM HCF = product of the two numbers. (i) 6 and 91 (ii) 510 and 9 (iii) 6 and 54. Sol. (i) 6 and 91 6 45 5 85 1 So, 6 = 1 5 17
REAL NUMBERS 5 91 7 1 So, 91 = 7 1 LCM (6, 91) = 7 1 = 18 HCF (6, 91) = 1 Verification: LCM HCF = 18 1 = 66 and 6 91 = 66 i.e., LCM HCF = Product of two numbers. (ii) 510 and 9 6 168 84 4 1 So, 6 = 7 = 4 7 7 CHAPTER 1 510 55 So, 510 = 5 51 5 9 46 So, 9 = LCM (510, 9) = 5 51 = 460 HCF (510, 9) = Verification: LCM HCF = 460 = 4690. and 510 9 = 4690 i.e., LCM HCF = Product of two numbers (iii) 6 and 54 51 54 7 9 So, 54 = = LCM (6, 5) = 4 7 = 04. HCF (6, 5) = = 6. Verification: LCM HCF = 04 6 = 18144 and 6 54 = 18144 i.e., LCM HCF = Product of two numbers. Example. Find the LCM and HCF of the following integers by applying the prime factorisation method. (i) 1, and 1 (ii) 17, and 9 (iii) 8, 9 and 5. Sol. (i) 1, and 1
6 MATHEMATICS X 1 5 6 So, 1 = = 5 So, = 5 1 7 So, 1 = 7 HCF (1,, 1) = LCM (1,, 1) = 5 7 = 40 (ii) 17, and 9 17 = 17 = 9 = 9 HCF (17,, 9) = 1 LCM (17,, 9) = 17 9 = 119 (iii) 8, 9 and 5 8 4 So, 8 = = 9 So, 9 = = 5 5 So, 5 = 5 5 = 5 HCF (8, 9, 5) = 1 LCM (8, 9, 5) = 5 = 1800. Example 4. Given that HCF (06, 657) = 9, find LCM (06, 657). Sol. LCM (06, 657) 06 t 657 = HCF (06, 657) 06 t 657 = = 8. 9 Example 5. Check whether 6 n can end with the digit 0 for any natural number n. Sol. If the number 6 n, for any natural number n, ends with digit 0, then it would be divisible by 5. That is, the prime factorisation of 6 n contain would the prime 5. This is not possible because 6 n = ( ) n = n n ; so the only primes in the factorisation of 6 n are and and the uniqueness of the Fundamental Theorem of Arithmetic guarantees that there are no other primes in the factorisation of 6 n. So, there is no natural number n for which 6 n ends with the digit zero. Example 6. Explain why 7 11 1 + 1 and 7 6 5 4 1 + 5 are composite numbers. Sol. (i) 7 11 1 + 1 = (7 11 + 1) 1 = (77 + 1) 1 = 78 1 = ( 1) 1 Q 78 = 1 = 1 78 9 1
REAL NUMBERS 7 Since, 7 11 1 + 1 can be expressed as a product of primes, therefore, it is a composite number. (ii) 7 6 5 4 1 + 5 = (7 6 4 1 + 1) 5 = (1008 + 1) 5 = 1009 5 = 5 1009 Since, 7 6 5 4 1 + 5 can be expressed as a product of primes, therefore, it is a composite number. Example 7. There is a circular path around a sports field. Sonia takes 18 minutes to drive one round of the field, while Ravi takes 1 minutes for the same. Suppose they both start at the same point and at the same time, and go in the same direction. After how many minutes will they meet again at the starting point? Sol. By taking LCM of time taken (in minutes) by Sonia and Ravi, we can get the actual number of minutes after which they meet again at the starting point after both start at same point and at the same time, and go in the same direction. 18 = = 18 9 1 = = 1 6 LCM (18, 1) = = 6. both Sonia and Ravi will meet again at the starting point after 6 minutes. CHAPTER 1 [NCERT Exercise 1.] (Page 14) Example 1. Prove that 5 is irrational. Sol. Let us assume, to the contrary, that 5 is rational. So, we can find coprime integers a and b ( 0) such that 5 a b 5 b = a Squaring on both sides, we get 5 b = a 5 divides a. 5 divides a So, we can write a = 5 c for some integer c. Substituting for a, we get 5 b = 5c b = 5c This means that 5 divides b, and so 5 divides b. a and b have at least 5 as a common factor. But this contradicts the fact that a and b have no common factor other than 1. This contradiction arose because of our incorrect assumption that 5 is rational. So, we conclude that 5 is irrational. Example. Prove that + is irrational. Sol. Let us assume, to the contrary, that + is rational. That is, we can find coprime integers a and b (b 0) such that + = a b a b = a b = b
8 MATHEMATICS X a b = b 5 a = 5 b Since a and b are integers, we get a b is rational, and so 5 is rational. But this contradicts the fact that 5 is irrational. This contradiction has arisen because of our incorrect assumption that + is rational. So, we conclude that + is irrational. Example. Prove that the following are irrationals: (i) 1 (iii) 6 +. Sol. (i) 1. (ii) 7 5 Let us assume, to the contrary, that 1 is rational. So, we can find coprime integers a and b ( 0) such that 1 a = b b a Since, a and b are integers, b is rational, a and so is rational. But this contradicts the fact that is irrational. So, we conclude that 1 is irrational. (ii) 7 5. Let us assume to the contrary, that 7 5 is rational. So, we can find coprime integers a and b ( 0) such that 5 7 5 a b a 7b Since, a and b are integers, a is rational, 7b and so, 5 is rational. But this contradicts the fact that 5 is irrational. So, we conclude that 7 5 is irrational. (iii) 6 +. Let us assume to the contrary, that is rational. Then, 6 is rational. So, we can find coprime integers a and b ( 0) such that 6 + a b 6 a b Since, a and b are integers, we get a b is rational and so, 6 a is rational and so, is b rational. But this contradicts the fact that is irrational. So, we conclude that 6 + is irrational.
REAL NUMBERS 9 Example 1. Without actually performing the long division, state whether the following rational numbers will have a terminating decimal expansion or a nonterminating repeating decimal expansion: (i) (iii) (v) (vii) 1 15 64 455 9 4 (ix) 5 50 19 7 5 (viii) 7 Sol. (i) 1 15 (ii) 17 8 (iv) 1600 (vi) (x) 6 1 15 = 1 5 5 77 10. Here, q = 5 5, which is of the form n 5 m (n = 0, m = 5). So, the rational number has a terminating decimal expansion. [NCERT Exercise 1.4] (Pages 17 18) 1 15 (ii) 17 8 17 8 = 17 Here, q =, which is of the form n 5 m (n =, m = 0). So, the rational number 17 8 has a terminating decimal expansion. 64 (iii) 455 64 455 = 64 5t7t1 Here, q = 5 7 1, which is not of the form n 5 m. So, the rational number 64 455 has a non-terminating repeating decimal expansion. (iv) 1600 1600 = 0 = 6 t 5 1 Here, q = 6 5 1, which is of the form n 5 m (n = 6, m = 1). So, the rational number has a terminating decimal expansion. (v) 9 4 1600 9 4 = 9 7 Here, q = 7, which is not of the form n 5 m. So, the rational number 9 has a nonterminating repeating decimal 4 expansion. (vi) Here, q = 5, which is of the form n 5 m (n =, m = ). So, the rational number has a terminating decimal expansion. (vii) 19 7 5 7 Here, q = 7 7 5, which is not of the form n 5 m 19. So, the rational number 7 5 7 has a non-terminating repeating decimal expansion. 6 (viii) 6 = Here, q = 5, which is of the form n 5 m ( n = 0, m = 1). So, the rational number 6 has a terminating decimal expansion. CHAPTER 1
10 MATHEMATICS X (ix) 5 50 5 50 = 7 10 = 7 t 5 Here q = 5, which is of the form n 5 m (n = 1, m = 1). So, the rational number 5 has a terminating decimal expansion. 50 77 (x) 10 77 10 = 11 0 = 11 tt5 Here, q = 5, which is not of the form n 5 m. So, the rational number 77 10 has a non-terminating repeating decimal expansion. Example. Write down the decimal expansions of those rational numbers in Question 1 above which have terminating decimal expansions. 1 Sol. (i) 15 = 1 5 5 = 1 t 5 5 5 t = 416 10 5 = 0.00416 17 (ii) 8 = 17 17 5 17 5 = = 5 10 = 15 10 =.15 (iv) 1600 = 4 t 10 4 t 5 = 4 4 t5 t10 = 975 10 6 = 0.00975 (vi) = t t t5 t = 100 10 5 = 0.1 (viii) (ix) 6 = = t = 4 5t 10 = 0.4 5 50 = 7 10 = 0.7. Example. The following real numbers have decimal expansions as given below. In each case, decide whether they are rational, or not. If they are rational, and of the form p q, what can you say about the prime factors of q? (i) 4.1456789 (ii) 0.10 100 1000 10000... (iii) 4.1456789. Sol. (i) 4.1456789 Since, the decimal expansion terminates, so the given real number is rational and therefore of the form p q. 4.1456789 = 41456789 1000 000 000 = 41456789 10 9 = 41456789 ( t 5) 9 = 41456789 9 9 Here, q = 9 5 9 The prime factorization of q is of the form n 5 m, where n = 9, m = 9. (ii) 0.10 100 1000 10000... Since, the decimal expansion is neither terminating nor non-terminating repeating, therefore, the given real number is not rational. (iii) 4.1456789 Since, the decimal expansion is nonterminating repeating, therefore, the given real number is rational and therefore of the form p q.