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CHAPTER 1, , REAL NUMBERS, , (A) Main Concepts and Results, , •, •, , •, •, , 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., Euclid’s Division Algorithm to obtain the HCF of two positive integers, say c and d,, c > d., Step 1 : Apply Euclid’s division lemma to c and d, to find whole numbers q and r,, such that c = dq + r, 0 ≤ r < d., Step 2 : If r = 0, d is the HCF of c and d. If r ≠ 0, apply the division lemma to, d and r., Step 3 : Continue the process till the remainder is zero. The divisor at this stage, will be the required HCF., Fundamental Theorem of Arithmetic : Every composite number can be expressed, as a product of primes, and this expression (factorisation) is unique, apart from the, order in which the prime factors occur., Let p be a prime number. If p divides a2, then p divides a, where a is a positive, integer., , •, •, •, , The sum or difference of a rational and an irrational number is irrational., , •, , For any two positive integers a and b, HCF (a, b) × LCM (a, b) = a × b., , 2,, , 3,, , 5 are irrational numbers., , The product or quotient of a non-zero rational number and an irrational number is, irrational., , 03/05/18

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2, , •, , EXEMPLAR PROBLEMS, , p, Let x = q , p and q are co-prime, be a rational number whose decimal expansion, terminates. Then, the prime factorisation of q is of the form 2m.5n; m, n are, non-negative integers., , •, , p, Let x = q be a rational number such that the prime factorisation of q is not of the, form 2m.5n; m, n being non-negative integers. Then, x has a non-terminating, repeating decimal expansion., , (B) Multiple Choice Questions, Choose the correct answer from the given four options:, Sample Question 1 : The decimal expansion of the rational number, , 33, will, 2 2.5, , terminate after, (A) one decimal place, (B) two decimal places, (C) three decimal places, (D) more than 3 decimal places, Solution : Answer (B), Sample Question 2 : Euclid’s division lemma states that for two positive integers a, and b, there exist unique integers q and r such that a = bq + r, where r must satisfy, (A) 1 < r < b, (B) 0 < r ≤ b, (C) 0 ≤ r < b, (D) 0 < r < b, Solution : Answer (C), , EXERCISE 1.1, Choose the correct answer from the given four options in the following questions:, 1. For some integer m, every even integer is of the form, (A) m, (B) m + 1, (C) 2m, (D) 2m + 1, 2. For some integer q, every odd integer is of the form, (A) q, (B) q + 1, (C) 2q, (D) 2q + 1, , 03/05/18

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REAL NUMBERS, , 3, , 3. n2 – 1 is divisible by 8, if n is, (A) an integer, (B) a natural number, (C) an odd integer, (D) an even integer, 4. If the HCF of 65 and 117 is expressible in the form 65m – 117, then the value of m is, (A) 4, (B) 2, (C) 1, (D) 3, 5. The largest number which divides 70 and 125, leaving remainders 5 and 8,, respectively, is, , 6., , 7., , 8., , 9., , (A) 13, (B) 65, (C) 875, (D) 1750, If two positive integers a and b are written as, a = x3y2 and b = xy3; x, y are prime numbers, then HCF (a, b) is, (C) x 3 y 3, (D) x 2 y 2, (A) xy, (B) xy 2, If two positive integers p and q can be expressed as, p = ab2 and q = a3b; a, b being prime numbers, then LCM (p, q) is, (C) a 3 b 2, (D) a 3 b 3, (A) ab, (B) a 2 b 2, The product of a non-zero rational and an irrational number is, (A) always irrational, (B) always rational, (C) rational or irrational, (D) one, The least number that is divisible by all the numbers from 1 to 10 (both inclusive) is, (A) 10, (B) 100, (C) 504, (D) 2520, , 10. The decimal expansion of the rational number, (A), (C), , one decimal place, three decimal places, , (B), (D), , 14587, will terminate after:, 1250, two decimal places, four decimal places, , (C) Short Answer Questions with Reasoning, Sample Question 1: The values of the remainder r, when a positive integer a is, divided by 3 are 0 and 1 only. Justify your answer., Solution : No., , 03/05/18

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4, , EXEMPLAR PROBLEMS, , According to Euclid’s division lemma,, a = 3q + r, where 0 ≤ r < 3, and r is an integer. Therefore, the values of r can be 0, 1 or 2., Sample Question 2: Can the number 6n, n being a natural number, end with the, digit 5? Give reasons., Solution : No, because 6n = (2 × 3)n = 2n × 3n, so the only primes in the factorisation, of 6n are 2 and 3, and not 5., Hence, it cannot end with the digit 5., , EXERCISE 1.2, 1. Write whether every positive integer can be of the form 4q + 2, where q is an, integer. Justify your answer., 2. “The product of two consecutive positive integers is divisible by 2”. Is this statement, true or false? Give reasons., 3. “The product of three consecutive positive integers is divisible by 6”. Is this statement, true or false”? Justify your answer., 4. Write whether the square of any positive integer can be of the form 3m + 2, where, m is a natural number. Justify your answer., 5. A positive integer is of the form 3q + 1, q being a natural number. Can you write its, square in any form other than 3m + 1, i.e., 3m or 3m + 2 for some integer m? Justify, your answer., 6. The numbers 525 and 3000 are both divisible only by 3, 5, 15, 25 and 75. What is, HCF (525, 3000)? Justify your answer., 7. Explain why 3 × 5 × 7 + 7 is a composite number., 8. Can two numbers have 18 as their HCF and 380 as their LCM? Give reasons., , 987, will have terminating, 10500, or non-terminating (repeating) decimal expansion. Give reasons for your answer., , 9. Without actually performing the long division, find if, , 10. A rational number in its decimal expansion is 327.7081. What can you say about, , p, the prime factors of q, when this number is expressed in the form q ? Give reasons., , 03/05/18

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REAL NUMBERS, , 5, , (D) Short Answer Questions, Sample Question 1: Using Euclid’s division algorithm, find which of the following, pairs of numbers are co-prime:, (i) 231, 396, , (ii) 847, 2160, , Solution : Let us find the HCF of each pair of numbers., (i), , 396 = 231 × 1 + 165, 231 = 165 × 1 + 66, 165 = 66 × 2 + 33, 66 = 33 × 2 + 0, , Therefore, HCF = 33. Hence, numbers are not co-prime., (ii), , 2160 = 847 × 2 + 466, 847 = 466 × 1 + 381, 466 = 381 × 1 + 85, 381 = 85 × 4 + 41, 85 = 41 × 2 + 3, 41 = 3 × 13 + 2, 3, , = 2×1 + 1, , 2, , = 1×2 + 0, , Therefore, the HCF = 1. Hence, the numbers are co-prime., Sample Question 2: Show that the square of an odd positive integer is of the form, 8m + 1, for some whole number m., Solution: Any positive odd integer is of the form 2q + 1, where q is a whole number., Therefore, (2q + 1)2 = 4q2 + 4q + 1 = 4q (q + 1) + 1,, , (1), , q (q + 1) is either 0 or even. So, it is 2m, where m is a whole number., Therefore, (2q + 1)2 = 4.2 m + 1 = 8 m + 1., Sample Question 3: Prove that, , 2 + 3 is irrational., , Solution : Let us suppose that, , 2 + 3 is rational. Let, , [From (1)], , 2 + 3 = a , where a is, , rational., , 03/05/18

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6, , EXEMPLAR PROBLEMS, , Therefore,, , 2 = a− 3, , Squaring on both sides, we get, 2 = a2 + 3 – 2a 3, Therefore,, , 3=, , number while, , a 2 +1, , which is a contradiction as the right hand side is a rational, 2a, , 3 is irrational. Hence,, , 2 + 3 is irrational., , EXERCISE 1.3, 1. Show that the square of any positive integer is either of the form 4q or 4q + 1 for, some integer q., 2. Show that cube of any positive integer is of the form 4m, 4m + 1 or 4m + 3, for, some integer m., 3. Show that the square of any positive integer cannot be of the form 5q + 2 or, 5q + 3 for any integer q., 4. Show that the square of any positive integer cannot be of the form 6m + 2 or, 6m + 5 for any integer m., 5. Show that the square of any odd integer is of the form 4q + 1, for some integer q., 6. If n is an odd integer, then show that n2 – 1 is divisible by 8., 7. Prove that if x and y are both odd positive integers, then x2 + y2 is even but not, divisible by 4., 8. Use Euclid’s division algorithm to find the HCF of 441, 567, 693., 9. Using Euclid’s division algorithm, find the largest number that divides 1251, 9377, and 15628 leaving remainders 1, 2 and 3, respectively., 10. Prove that, , 3 + 5 is irrational., , 11. Show that 12n cannot end with the digit 0 or 5 for any natural number n., 12. On a morning walk, three persons step off together and their steps measure 40 cm,, 42 cm and 45 cm, respectively. What is the minimum distance each should walk so, that each can cover the same distance in complete steps?, , 03/05/18

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REAL NUMBERS, , 7, , 257, in the form 2m × 5n, where, 5000, m, n are non-negative integers. Hence, write its decimal expansion, without actual, division., , 13. Write the denominator of the rational number, , 14. Prove that, , p +, , q is irrational, where p, q are primes., , (E) Long Answer Questions, Sample Question 1 : Show that the square of an odd positive integer can be of the, form 6q + 1 or 6q + 3 for some integer q., Solution : We know that any positive integer can be of the form 6m, 6m + 1, 6m + 2,, 6m + 3, 6m + 4 or 6m + 5, for some integer m., Thus, an odd positive integer can be of the form 6m + 1, 6m + 3, or 6m + 5, Thus we have:, (6 m +1)2 = 36 m2 + 12 m + 1 = 6 (6 m2 + 2 m) + 1 = 6 q + 1, q is an integer, (6 m + 3)2 = 36 m2 + 36 m + 9 = 6 (6 m2 + 6 m + 1) + 3 = 6 q + 3, q is an integer, (6 m + 5)2 = 36 m2 + 60 m + 25 = 6 (6 m2 + 10 m + 4) + 1 = 6 q + 1, q is an integer., Thus, the square of an odd positive integer can be of the form 6q + 1 or 6q + 3., , EXERCISE 1.4, 1. Show that the cube of a positive integer of the form 6q + r, q is an integer and, r = 0, 1, 2, 3, 4, 5 is also of the form 6m + r., 2. Prove that one and only one out of n, n + 2 and n + 4 is divisible by 3, where n is, any positive integer., 3. Prove that one of any three consecutive positive integers must be divisible by 3., 4. For any positive integer n, prove that n3 – n is divisible by 6., 5. Show that one and only one out of n, n + 4, n + 8, n + 12 and n + 16 is divisible, by 5, where n is any positive integer., [Hint: Any positive integer can be written in the form 5q, 5q+1, 5q+2, 5q+3,, 5q+4]., , 03/05/18