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1 CHAPTER 1 Points to Remember : REAL NUMBERS 1. Euclid s division lemma : Given positive integers a and b, there exists whole numbers q and r satisfying a = bq + r, 0 r < b.. Euclid s division algorithm : This is based on Euclid s division lemma. According to this, the HCF of any two positive integers a and b, with a > b, is obtained as follows : Apply Euclid s division lemma to find q and r where a = bq + r, 0 r < b. If r = 0, the HCF is b. If r 0, apply the Euclid s lemma to b and r. Continue the process till the remainder is zero. The divisor at this stage will be HCF (a, b).. The fundamental theorem of arithmetic : Every composite number can be expressed (factorised) as a product of primes, and this factorisation is unique, except for the order in which the prime factors occur. 4. For any two positive integers a and b, HCF (a, b) LCM (a, b) = a b. a b HCF (a, b) and LCM( a, b) a b LCM ( a, b) HCF( a, b). For any three positive integes a, b and c, we have a b c LCM ( a, b, c) HCF (a, b, c) LCM ( a, b) LCM ( b, c) LCM ( a, c) a b c HCF ( a, b, c) and LCM (a, b, c) HCF ( a, b) HCF ( b, c) HCF ( a, c) 6. Let a be a positive integer and p be a prime number such that p/a, then p/a. 7. If p is a positive prime, then p is an irrational number. 8. Let x be a rational number, whose decimal expansion terminates. Then we can express x in the form p q, where p and q are co-prime and the prime factorisation of q is of the form n m, where n, m are nonnegative integers. 9. Let x p q be a rational number, such that the prime factorisation of q is of the form n m, where n, m are non-negative integers. Then x has a decimal expansion which terminates. m are non-negative integers. Then x has a decimal expansion which is non terminating repeating (recurring). 10. Let x p q be a rational number, such that the prime factorisation of q is not of the form n m, where n, MATHEMATICS X REAL NUMBERS 1

2 ILLUSTRATIVE EXAMPLES Example 1. Use Eculid s division algorithm to find the HCF of 1 and. Solution. Since > 1. Let a =, b = 1. Applying Eculid s division algorithm, we get = Since the remainder So, we apply the euclid s division algorithm to 1 and 90, to get 1 = Again, since the remainder is non-zero, so we apply the euclid s division algorithm to 90 and 4, to get 90 = We observe that remainder at this stage is zero. the divisor at this stage i.e. 4 is the HCF of 1 and. Symbolically we write HCF (1, ) = 4 Ans. 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? [NCERT] Solution. Clearly, the maximum number of columns in which members can march is the HCF of 616 and. So, let us find the HCF of 616 and by Euclid s division algorithm Since, 616 >. Let a = 616, b = Applying Euclid s division algorithm, we get = Since the remainder is non-zero, so we apply the Euclid s division algorithm to and 8, to get = We observe that the remainder is zero at this stage. The divisor at this stage i.e. 8 is the HCF of 616 and. Hence, the required maximum number of columns is 8. REAL NUMBERS MATHEMATICS X Example. Use Eculid s division demma, to show that the cube of any positive integer is of the form 9m, 9m + 1 or 9m + 8. [NCERT] Solution. Let x be any positive integer. Then, it is of the form q or, q + 1 or, q +. So, we have the following cases : Case I : When x = q. then, x = (q) = 7q = 9 (q ) = 9m, where m = q. 4

3 Case II : When x = q + 1 then, x = (q + 1) = 7q + 7q + 9q + 1 = 9 q (q + q + 1) + 1 = 9m + 1, where m = q (q + q + 1) Case III. When x = q + then, x = (q + ) = 7 q + 4q + 6q + 8 = 9q (q + 6q + 4) + 8 = 9 m + 8, where m = q (q + 6q + 4) Hence, x is either of the form 9 m or 9 m + 1 or, 9 m + 8. Example 4. Express each of the following number as a product of its prime factors : (i) 140 (ii) 8 (iii) 00 [NCERT] Solution. (i) Using the factor tree for prime factorisation, we have = 7 = 7 (ii) Using the factor tree for prime factorisation, we have MATHEMATICS X REAL NUMBERS = 17 = 17 (iii) Using the factor tree for prime factorisation, we have =

4 Example. Find the HCF of 96 and 404 by prime factorisation method. Hence, find their LCM. Solution. We have, 96 = and 404 = 101 HCF = = 4. Now, HCF LCM = LCM = 9696 Ans. HCF 4 Example 6. Check whether 6 n can end with the digit 0 for any natural number n. Solution. Example 7. Prove that [NCERT] We have, 6 n = ( ) n = n n. Therefore, prime factorisation of 6 n does not contain as a factor. Hence, 6 n can never end with the digit 0 for any natural number n. is an irrational number. Solution. Let us assume, to the contrary, that is rational. So, we can find integers r and s ( 0) such that r. Suppose r and s have a common factor other than 1. Then, we divide by the common s factor to get a, where a and b are co-prime. b So, b a. Squaring on both sides, we get b = a. Therefore, divides a divides a. So, we can write a = c for some integer c. Substituting for a, we get b = 4c, or b = c. This means that divides b, and so divides b. Therefore, a and b have atleast as a common factor. But this contradicts the fact that a and b have no common factors other than 1. This contradiction has arisen because of our incorrect assumption that So, we conclude that Example 8. Prove that + is irrational. is irrational. Hence proved. is rational. [NCERT] Solution. Let us assume on the contrary that is rational. Then there exists co-prime integers a and b such that, a b a b a b b is rational. [ a, b are integers, This contradicts the fact that Hence, is an irrational number. a b is a rational] b is irrational. So, our supposition is incorrect. 4 REAL NUMBERS MATHEMATICS X

5 Example 9. Without actually performing the long division, state whether the following rational number will have a terminating decimal expansion or a non-terminating repeating decimal expansion : 1 (i) 1 (ii) 17 8 (iii) (iv) 1600 (v) 9 4 Solution. (i) So, the denominator is of the form m n. Hence, it has terminating decimal expansion. (ii) So, the denominator is of the form m n. Hence, it has terminating decimal expansion. [NCERT] (iii) Clearly, the denominator is not of the form m n. Hence, it has non-terminating repeating decimal expansion. 1 (iv) So, the denominator is of the form m n. Hence, it has terminating decimal expansion. (v) Clearly, the denominator is not of the form m n. Hence, it has non-terminating repeating decimal expansion. Example 10. The following real numbers have decimal expression as given below. In each case decide whether p they are rational or not. If they are rational, and of the form, what can you say about the q prime factors of q? (i) (ii) (iii) [NCERT] Solution. (i) is a rational number of the form p q and q is of the form m n. (ii) (iii) is not a rational number since it has non-recurring and non-terminating decimal and q is not of the form m n is a rational number having recurring and non-terminating decimal expansion since q is not of the form m n. PRACTICE EXERCISE 1. Using Euclid s division algorithm, find the HCF of : (i) 10 and (ii) 117 and 6 (iii) 40 and 104 (iv) 91 and 4 (v) 188 and 7 (vi) 1 and 18. Find the largest number which exactly divides 80 and 14 leaving remainders 4 and, respectively.. Find the HCF of 00, 40, 890 by using Euclid s division algorithm. MATHEMATICS X REAL NUMBERS

6 4. A merchant has 10 litres of oil of one kind, 140 litres of another kind and 17 litres of third kind. He wants to sell the oil by filling the three kinds of oil in tins of equal capacity. What should be the greatest capacity of such a tin?. The length, breadth and height of a room are 8 m cm, 6 m 7 cm and 4 m 0 cm respectively. Determine the length of largest rod which can measure the three dimensions of the room exactly. 6. In a seminar, the number of participants in Hindi, English and Mathematics are 60, 84 and 108, respectively. Find the minimum number of participants are to be seated and all of them being in the same subject. 7. Show that every positive even integer is of the form q, and that every positive odd integer is of the form q + 1, where q is some integer. 8. Show that any positive integer is of the form q or, q + 1, or q + for some integer q. 9. Show that any positive odd integer is of the form 4q + 1, or 4q +, where q is some integer. 10. Show that the square of any positive integer is of the form m or, m + 1 for some integer m. 11. Show that one and only one out of n, n + or, n + 4 is divisible by, where n is any positive integer. 1. Show that one and only one out of n, n +, n + 6, n + 9 is divisible by Express each of the following positive integers as the product of its prime factors : (i) 60 (ii) 186 (iii) 64 (iv) 184 (v) 808 (vi) Explain why and are composite numbers? 1. Check whether 8 n can end with the digit zero for any natural number n. 16. Find the LCM and HCF of the following pairs of integers and verify that LCM HCF = Product of the integers. (i) 6 and 168 (ii) 144 and 160 (iii) 10 and 9 (iv), Find the LCM and HCF of the following integers by applying the prime factorisation method : (i) 1, 1 and 1 (ii) 1, 4 and 6 (iii), 6 and 60 (iv) 40, 104, The HCF and LCM of two numbers are 1 and 40 respectively. If one of these numbers is 48, find the other number. 19. The product of two numbers is 076 and their HCF is 4. Find their LCM. 0. The LCM of two numbers is 19 and their product is 07. Find their HCF. 1. Prove that is an irrational number.. Show that the following numbers are irrational : (i) (ii) (iii) 4 (iv) (v) (vi) (vii) 1 (viii). (i) If p is a prime number, prove that p is irrational. 7 (viii) (ii) Show that n 1 n 1 is irrational for every natural number n. 6 REAL NUMBERS MATHEMATICS X

7 4. Without actually performing the long division, state whether the following rational numbers will have terminating decimal expansion or a non-terminating repeating decimal expansion : (i) 1 11 (iv) (ii) 7 18 (v) (iii) 19 4 (vi) Write down the decimal expansions of the following rational numbers by writing their denominators in the form m n, where m, n are non-negative integers : (i) 8 (iv) 1 6 (ii) 17 1 (v) (iii) (vi) 000 HINTS TO SELECTED QUESTIONS. Required number = HCF (80 4, 14 ) = HCF (76, 14). 4. Greatest capacity = HCF (10 l, 140 l, 17 l).. Length of largest rod = HCF (8 cm, 67 cm, 40 cm) 6. Number of participants in each room = HCF (60, 84, 108) = 1. Now, total number of participants = =. number of rooms required Let a be any positive integer. Then, it is of the form q or, q + 1 or, q +. So, we have the following cases : Case I : When a = q, here a ( q) 9q ( q) m, where m q. Case II : When a = q + 1, a = (q + 1) = 9q + 6q + 1 = q (q + ) + 1, where m = q (q + ). Case III : When a = q +, here, a (q ) 9q 1q 4 (q 4q 1) 1 m 1, where m q 4q 1. Hence, a is of the form m or m = 1 ( ) = 1 78, which is a composite no. also, = ( ) = 1009, which is a composite no. m. (i) Let p be a prime number and if possible, let p is rational. Let its simplest form be p, n where m and n are integers having no common factor other than 1, and n 0. m m Now, p p pn m n n p divides m p divides m. MATHEMATICS X REAL NUMBERS 7...(i)

8 Let m = pq for some integer q. putting m = pq in (i), we get pn = p q n = pq p divides n p divides n....(ii) from (i) and (ii), we observe that p is a common factor of m and n, which contradicts our assumption. Hence, p is irrational. a (ii) Let n 1 n 1 is rational. Let n 1 n 1, where a, b are positive integers. b a Now, n 1 n 1 b...(i) b 1 b n 1 n 1 n 1 n 1 a n 1 n 1 a ( n 1) ( n 1) b n 1 n 1 a...(ii) adding (i) and (ii) and subtracting (ii) from (i), we get 1 a b and 1 a n n b ab ab n 1 and n 1 are rationals. [ a, b are integers, a b and a b ab ab are rationals.] (n + 1) and (n 1) are perfect squares of positive integers, which is not possible as any two perfect squares differ atleast by. Hence, n 1 n 1 is irrational. MULTIPLE CHOICE QUESTIONS Mark the correct alternative in each of the following : 1. The product of three consecutive positive integers is always divisible by : (a) 4 (b) (c) 6 (d) 1. The HCF of 048 and 960 is : (a) (b) 64 (c) 18 (d) none of these. The HCF of 67and 96 is : (a) (b) 6 (c) 9 (d) none of these 4. The largest number which divides 61 and 96 leaving remainder 6 in each case is : (a) 9 (b) 87 (c) 116 (d) none of these. The HCF of 40, 840 and 960 is (a) 9 (b) 1 (c) 4 (d) none of these 6. The GCD of 144, 418 and 11 is ; (a) (b) 4 (c) 8 (d) 1 7. The prime factorisation of 468 is : (a) 1 (b) 1 (c) 1 (d) none of these 8 REAL NUMBERS MATHEMATICS X

9 8. The least number that is divisible by all the numbers between 1 and 10 (both inclusive) is: (a) (b) 0 (c) 160 (d) none of these 9. The HCF of two numbers is 14 and their LCM is 17. If one number is 7, the other number is : (a) 07 (b) 870 (c) 87 (d) If the sum of two numbers is 11 and their HCF is 81, the total number of such pairs is : (a) (b) (c) 4 (d) 11. Which of the following number is rational : 1. (a) (b) (7 ) (7 ) 1, when expressed as a non-terminating and recurring decimal is given by : 7 MATHEMATICS X REAL NUMBERS 9 (c) (d) ( 7) (a) (b) (c) (d) none of these 1. Which of the following number is given by 0.16 : (a) 17 7 (b) 4 0 (c) Which of the following number have a terminating decimal representation : (a) 7 80 (b) 4 (c) Which of the following number do not have a terminating decimal representation : (a) (b) 1000 (c) 1 10 (d) 1 8 (d) 4 (d) VERY SHORT ANSWER TYPE QUESTIONS (1 MARK QUESTIONS) 1. State Euclid s division lemma.. State fundamental theorem of arithmetic.. Write the condition to be satisfied by q so that a rational number p q 4. Write the condition to be satisfied by q so that a rational number p q decimal expansion.. Find the HCF of 6 and If HCF (7, 10) = 4, then what is the LCM (7, 10)? has a terminating decimal expansion. has a non-terminating recurring 7. LCM of two numbers is 079 and their HCF is 7. If one of the numbers is 97, find the other number. 8. If LCM (4, 80) = 40, then what is the HCF (4, 80)? 9. Why the number 4 n, where n is a natural number, cannot end with 0? 10. Why is a composite number? 11. Express 16 as a product of its prime factors.

10 1. Find the missing number in the following factor tree: Give an example of two irrational numbers whose sum is a rational number. 14. Give an example of two irrational numbers whose product is a rational number. 10 REAL NUMBERS MATHEMATICS X? 1. Without actually performing long division, state why has a non-terminating repeating decimal 4 expansion. 16. Write down the decimal expansion of State whether x. is a rational number or not How many prime factors are there in the prime factorisation of By which smallest irrational number 8 be multiplied so as to get a rational number? 0. Which number should be multiplied to ( 7 ) to get a rational number? M.M : 0 General Instructions : PRACTICE TEST Q. 1-4 carry marks, Q. -8 carry marks and Q carry marks each. 1. Write 04 as a product of its prime factors.. Find the LCM and HCF of 6 and 91.. Is ( ) ( ) is a rational number? Justify. Time : 1 hour 4. Write the rational number in decimal form. 6. Use Euclid s division algorithm to find HCF of 867 and. 6. Prove that one of every three consecutive positive integers is divisible by. 7. Find the HCF and LCM of 144, 180 and 19 by prime factorisation method. 8. Two tankers contains 80 litres and 680 litres of petrol respectively. Find the maximum capacity of a container which can measure the petrol of either tanker in exact number of times. 9. Prove that is an irrational number. 10. Without actually performing the long division, state whether the following rational numbers will have a terminating decimal expansion or a non-terminating repeating decimal expansion : (i) (iv) 7 (ii) (v) (iii) 00

11 ANSWERS OF PRACTICE EXERCISE 1. (i) (ii) 1 (iii) 16 (iv) 17 (v) (vi) cm (i) (ii) 7 11 (iii) 11 1 (iv) 7 1 (v) 7 11 (vi) No 16. (i) LCM = 04, HCF = 1 (ii) LCM = 1440, HCF = 16 (iii) LCM = 460, HCF = (iv) LCM = 0744, HCF = (i) LCM = 40, HCF = (ii) LCM = 60, HCF = (iii) LCM = 00, HCF = (iv) LCM = 160, HCF = (i) non-terminating repeating (ii) terminating (iii) non-terminating repeating (v) non-terminating repeating (iv) terminating (vi) terminating. (i) 0.6 (ii) 0.16 (iii) 0.16 (iv) (v) 17. (vi) ANSWERS OF MULTIPLE CHOICE QUESTIONS 1. (c). (b). (c) 4. (b). (b) 6. (a) 7. (c) 8. (b) 9. (d) 10. (c) 11. (b) 1. (a) 1. (c) 14. (a) 1. (c) ANSWERS OF VERY SHORT ANSWER TYPE QUESTIONS. The prime factorisation of q should be of the form n m, where n, m are non-negative integers and 14. and rational ANSWERS OF PRACTICE TEST LCM = 18, HCF = 1. yes LCM = 880, HCF = litres 10. (i) non-terminating repeating (ii) non-terminating repeating (iii) Terminating (v) Terminating (iv) non-terminating repeating MATHEMATICS X REAL NUMBERS 11

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 t

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 t 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