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Class –x Weekend assignment sheet -3 REAL nUMBERS Name: _______________________ 1. Find the HCF of the integers 455 and 42 using Euclid’s division algorithm. 2. Prove that any positive odd integer is of the form 4p+ 1 or 4p+ 3, where p is an integer. 3. Use Euclid’s division lemma to show that the square of any positive integer is either of the form 3p or 3p+ 1, where p is an integer. 4. The numbers 1424, 2096, and 1844 when divided by a number N give the same remainder 10. Find the highest such number N. 5. Prove that for any prime number p, is an irrational number. 6. Prove that there is no natural number n for which 4n ends with the digit 0. 7. Find the highest number less than 1000 that is divisible by each of the numbers 2, 3, 4, 5, 6, and 7. 8. Prove that is irrational. 9. State without actual division, whether the decimal expansions of the following rational numbers are terminating or non-terminating repeating. Give reasons to justify your answer. (a) (b) 10. The decimal expansions of some real numbers are given below. For each number, state whether it is rational or irrational. If the number is rational, then assuming it to be of the form , what can be said about the prime factors of its denominator? Give reasons to justify your answer. (a) 34.98 (b) 23.353355333555… (c) Solutions Question 1 ( 2.0 marks) Find the HCF of the integers 455 and 42 using Euclid’s division algorithm. Solution: The two numbers are 455 and 42. Applying Euclid’s division lemma on 455 and 42, we get: 455 = 42 × 10 + 35 The remainder obtained is 35 ≠ 0. Applying Euclid’s division lemma on the divisor 42 and the remainder 35, we get: 42 = 35 × 1 + 7 The remainder obtained is 7 ≠ 0. Applying Euclid’s division lemma on the divisor 35 and the remainder 7, we get: 35 = 7 × 5 + 0 The remainder obtained is 0. Thus, the HCF of 455 and 42 is the divisor at this stage, that is, 7. Question 2 ( 2.0 marks) Prove that any positive odd integer is of the form 4p+ 1 or 4p+ 3, where p is an integer. Solution: Let a be a positive odd integer. We apply Euclid’s division algorithm with a and b = 4 Since 0 r < 4, the possible remainders will be 0, 1, 2, and 3. This means that a can be 4p, 4p+ 1, 4p+ 2, or 4p+ 3, where pis the quotient. However, we know that a is odd. Therefore, a cannot be 4por 4p+ 2 since they are both divisible by 2, i.e., they are even. Therefore, any odd integer is of the form 4p+ 1 or 4p+ 3. Question 3 ( 2.0 marks) Use Euclid’s division lemma to show that the square of any positive integer is either of the form 3p or 3p+ 1, where p is an integer. Solution: Let a be a positive integer. Then, a will be of the form 3m, 3m + 1, or 3m + 2. Case I: a = 3m On squaring both sides, we get a2 = (3m)2 = 9m2 ⇒ a2 = 3(3m2) = 3p, where p = 3m2 Case II: a = 3m + 1 On squaring both sides, we get a2 = (3m + 1)2 = 9m2 + 6m + 1 ⇒ a2 = 3(3m2 + 2m) + 1 = 3p + 1, where p = 3m2 + 2m Case III: a = 3m + 2 On squaring both sides, we get a2 = (3m + 2)2 = 9m2 + 12m + 4 ⇒ a2 = 3(3m2 + 4m + 1) + 1 = 3p + 1, where p = 3m2 + 4m + 1 Hence, a is the form of 3p or 3p + 1. Question 4 ( 2.0 marks) The numbers 1424, 2096, and 1844 when divided by a number N give the same remainder 10. Find the highest such number N. Solution: It is given that the three numbers give a remainder of 10 when divided by N. This means that (1424 − 10), (2096 − 10), and (1844 − 10) are divisible by N. Hence, N is the HCF of the numbers 1414, 2086, and 1834. 1414 = 2 × 7 × 101 2086 = 2 × 7 × 149 1834 = 2 × 7 × 131 The prime factors common to all these numbers are 2 and 7. Therefore, the HCF of the numbers 1414, 2086, and 1834 is 2 × 7 = 14 Thus, the highest such number N is 14. Question 5 ( 2.0 marks) Prove that for any prime number p, is an irrational number. Solution: Let us assume that is rational. Then, there exist positive co-primes a and b such that On squaring both sides and rearranging: ⇒ b2p = a2 This implies that p is a factor of a2. Hence, p is a factor of a. We can hence, write a = pc for some positive integer c. Now, b2p = a2 ∴b2p = p2c2 ⇒ b2 = pc2 This means that b2 is a multiple of p. Hence, b is a multiple of p, i.e., p is a factor of b. It is seen that p is a factor of both and b. This cannot happen if a and b are co-primes. Thus, our assumption is incorrect. Hence, is irrational. Question 6 ( 2.0 marks) Prove that there is no natural number n for which 4n ends with the digit 0. Solution: We know that any positive integer ending with the digit 0 must be divisible by the number 5. In other words, 5 must be one of the prime factors of the numbers ending with 0. We have 4n = = (2)2n Therefore, the only prime factor of 4n is 2. Since the Fundamental Theorem of Arithmeticsays that prime factorisation is unique, the number 5 cannot occur in the prime factorisation of 4n for any value of n. Hence, the number 4n does not end with 0 for any natural number n. Question 7 ( 2.0 marks) Find the highest number less than 1000 that is divisible by each of the numbers 2, 3, 4, 5, 6, and 7. Solution: The LCM of 2, 3, 4, 5, 6, and 7 is 420. Hence, 420 and every multiple of 420 is divisible by each of these numbers. Hence, the numbers 420, 840, 1260, and 1680 are all divisible by each of these numbers. We can see that 840 is the highest number less than 1000, which is a multiple of 420. Hence, the highest number less than 1000 that is divisible by each of the numbers 2, 3, 4, 5, 6, and 7 is 840. Question 8 ( 2.0 marks) Prove that is irrational. Solution: Let be a rational number equal to . Then, Since a and b are integers, the RHS is rational and so, However, this contradicts the fact that is a rational number is an irrational number. This contradiction has arisen because of our incorrect assumption that is irrational. is rational. Hence, Question 9 ( 2.0 marks) State without actual division, whether the decimal expansions of the following rational numbers are terminating or non-terminating repeating. Give reasons to justify your answer. (a) (1 mark) (b) (1 mark) Solution: (a) The numerator and the denominator of the rational number are respectively 189 and 160. 189 = 33 × 7 160 = 25 × 5 Thus, the numerator and the denominator are co-prime. Now, the prime factorization of the denominator is of the form 2 a × 5b, where a and b are positive integers, and there is no other prime factor. This implies that the decimal expansion of is terminating. (b) The given rational number is . This can be simplified as The numerator and the denominator of the rational number are now co-prime. Since the denominator is not of the form 2a× 5b, where a and b are positive integers, and it also has 3 as its factor, the decimal expansion of or is non-terminating repeating. Question 10 ( 3.0 marks) The decimal expansions of some real numbers are given below. For each number, state whether it is rational or irrational. If the number is rational, then assuming it to be of the form factors of its denominator? Give reasons to justify your answer. , what can be said about the prime (a) 34.98 (1 mark) (b) 23.353355333555… (1 mark) (c) (1 mark) Solution: (a) The given number is 34.98 Since this number has a terminating decimal expansion, it is a rational number of the form form 2a× 5b, where a and b are positive integers. , where q is of the This implies that the prime factors of the denominator, i.e., q will either be 2 or 5 or both. (b) The given number is 23.353355333555… The decimal expansion is neither terminating nor recurring. This implies that the number is an irrational number. (c) The given number is Since this number has a non-terminating recurring decimal expansion, it is a rational number of the form However, q is not of the form 2a× 5b, where a and b are positive integers. . This implies that the prime factors of the denominator, i.e., q will comprise a prime number other than 2 or 5.