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1 1 REAL NUMBERS Exercise 1.1 Q.1. Use Euclid s division algorithm to find the HCF of: (i) 135 and 225 (ii) 196 and (iii) 867 and 255 Solution. (i) In 135 and 225, 225 is larger integer. Using Euclid s division algorithm, [Where 135 is divisor, 90 is remainder] Since, remainder 90 0, by applying Eudid s division algorithm to 135 and Again since, remainder 45 0, by applying Eudid s division algorithm to 90 and Now, the remainder is zero so, our procedure stops. Hence, HCF of 135 and 225 is 45. (ii) In 196 and 38220, is larger integer. Using Euclid s division algorithm, remainder 0 The remainder is zero so our procedure stops. Hence, HCF of 196 and is 196. (iii) In 867 and 255, 867 is larger integer. Using Euclid s division algorithm [Where 255 is divisor, 102 is remainder] Since, remainder 102 0, by applying Eudid s division algorithm to 255 and Since, remainder 51 0, by applying Eudid s division algorithm 102 and 51 So, remainder 0 The remainder is zero so our procedure stops. Hence, HCF of 867 and 255 is 51. Q.2. Show that any positive odd integer is of the form or or 6 + 5, where is some integer. 1 P A G E C L A S S 1 0 N C E R T S O L U T I O N S D E E P A K S I R
2 Solution. Let a be any positive odd integer. We apply Euclid s division algorithm with a and b 6. Since 0 r 6, the positive remainders are 0, 1, 2, 3, 4 and 5. i.e., a can be 6 or 6 1, or 6 2, or 6 3, or 6 4, or 6 5 where is the uotient. But, a is odd so a cannot be eual to 6, 6 2, 6 4 numbers. (divisible by 2) Any odd integer is of the form 6 1 or 6 3 or 6 5. which are even Q.3. An army contingent of 616 members is to march behind an army band of 32 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? Solution. Total members in army are 616 and 32, 616 is larger integer i.e. 616 > 32 Using Euclid s division algorithm, [Where 32 is divisor, 8 is remainder] Since, remainder 8 0, by applying Eudid s division algorithm, we get The remainder is zero, Hence, the maximum numbers of columns in which both 616 members and 32 members can march is 8 columns. Q.4. Use Euclid s division lemma to show that the suare of any positive integer is either of the form 3m or 3m + 1 for some integer m. Solution. Let x be any positive integer, then it is of the form 3, 3 1 or 3 2. So x 3 (x) 2 (3) 2 (Suaring both sides) x (3 2 ) 3m, (where m 3 2 ) Hence If x 2 3m x 3 1 x 2 (3 1) 2...(i) (Suaring both sides) where, x x 2 3(3 2 2) 1 x 2 3m 1 m (ii) 2 P A G E C L A S S 1 0 N C E R T S O L U T I O N S D E E P A K S I R
3 From euation (i) and (ii), we get x 2 3m, 3m 1 3 P A G E C L A S S 1 0 N C E R T S O L U T I O N S D E E P A K S I R
4 Suare of any positive integer is either of the form 3m integer m. or 3m 1 for some Q.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. Solution. Let a be any positive integer, then it is of the form 3, 3 1 or 3 2. If a 3 then a (cubing both sides) a m...(i) where m (3 3 ) If a (3 1) a 3 (3 1) 3 (cubing both sides) [We know (a b) 3 a 3 b 3 3a 2 b 3b 2 a ] a 3 (3) 3 (1) 3 3 (3) a a a 3 9( ) 1 a 3 9m 1...(ii) Where m If a 3 2 a 3 (3 2) 3 (cubing both sides) a 3 (3) 3 (2) 3 3 (3) (2) 2 3 a a a 3 9( ) 8 a 3 9m 8...(iii) 4 P A G E C L A S S 1 0 N C E R T S O L U T I O N S D E E P A K S I R
5 Where m Hence, from euation (i), (ii) and (iii) the cube of any integer is of the form 9m, 9m 1 or 9m 8. Exercise 1.2 Q.1. Express each number as a product of its prime factors: (i) 140 (ii) 156 (iii) 3825 (iv) 5005 (v) 7429 Solution. (i) Hence, (ii) Hence, (iii) Hence, P A G E C L A S S 1 0 N C E R T S O L U T I O N S D E E P A K S I R
6 (iv) 5005 Masters Tuition Center Hence, (v) Hence, Q.2. Find the LCM and HCF of the following pairs of integers and verify that LCM HCF Product of the two numbers. (i) 26 and 91 (ii) 510 and 92 (iii) 336 and 54 Solution. (i) 26 and HCF 13 LCM Verification: LCM HCF Product of two numbers Product of two numbers LCM HCF (ii) 510 and 92 6 P A G E C L A S S 1 0 N C E R T S O L U T I O N S D E E P A K S I R
7 So, HCF 2 LCM Verification: LCM HCF Product of two numbers Product of two numbers LCM HCF (iii) 336 and HCF LCM Verification: LCM HCF Product of two numbers Product of two numbers LCM HCF Q.3. Find the LCM and HCF of the following integers by applying the prime 7 P A G E C L A S S 1 0 N C E R T S O L U T I O N S D E E P A K S I R
8 factorisation method. (i) 12, 15 and 21 (ii) 17, 23 and 29 (iii) 8, 9 and 25 Solution. (i) 12, 15 and 21 Prime factorisation of Prime factorisation of Prime factorisation of HCF of 12, 15 and 21 3 LCM of 12, 15 and (ii) 17, 23 and 29 Prime factorisation of Prime factorisation of Prime factorisation of HCF of 17, 23 and 29 1 LCM of 17, 23 and (iii) 8, 9 and Prime factorisation of Prime factorisation of Prime factorisation of HCF of 8, 9 and 25 1 LCM of 8, 9 and Q.4. Given that HCF (306, 657) = 9, find LCM (306, 657). Solution. We know that, HCF LCM Product of the two numbers 8 P A G E C L A S S 1 0 N C E R T S O L U T I O N S D E E P A K S I R
9 HCF (306, 657) 9 (given) 9 LCM LCM Q.5. Check whether 6 n can end with the digit 0 for any natural number n. Solution. Let number 6 n end with the digit 0 for any n N. Then, 6 n will be divisible by 5. But, prime factors of 6 are 2 and 3. Prime factor of (6) n are (2 3) n So, it is clear that in prime factorisation of 6 n there is no 5. 9 P A G E C L A S S 1 0 N C E R T S O L U T I O N S D E E P A K S I R
10 By the uniueness of the Fundamental Theorem of Arithmetic, every composite number can be expressed as a product of primes and this factorisation is uniue, apart from the order in which the prime factors occur. Hence, there exists no any natural number n for which 6 n ends with the digit zero. Q.6. Explain why and are composite numbers. Solution , which is not a prime number because it has more than two factors. So, it is a composite number. And, , which is not a prime number because it has more than two factors. So, it is also a composite number. Q.7. There is a circular path around a sports field. Sonia takes 18 minutes to drive one round of the field, while Ravi takes 12 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? Solution. Time taken by Sonia to drive one round of the field 18 minutes. Time taken by Ravi to drive one round of the field 12 minutes. 2 18, , 6 3 9, 3 3 3, 1 1, 1 So, LCM of 18 and Hence, they will meet again at the starting point after 36 minutes. Exercise 1.3 Q.1. Prove that 5 is irrational. Solution. Let 5 is rational. r s (where, r and s are integers and s 0) Where r and s have no common factor except 1. Now by suaring both sides, we get 10 P A G E C L A S S 1 0 N C E R T S O L U T I O N S D E E P A K S I R
11 ( 5) 2 r 2 s r 2 5 s 2 11 P A G E C L A S S 1 0 N C E R T S O L U T I O N S D E E P A K S I R
12 Masters Tuition Center r 2 5s 2 5 is a factor of r 2...(i) Since, 5 is a prime number, if 5 divides r 2, then, 5 also divides r. Let Put r 5n, where n is an integer. r 5n in euation (i), we get, 5 (5n) 2 5s 2 25n 2 5s 2 s 2 5n 2 5 is a factor of s 2. As, 5 is a prime number, 5 also divides s. So, 5 is a common factor of r and s. But it contracts the facts that r and s has only one common factor i.e. 1. Hence, is an irrational number. Q.2. Prove that is irrational. Solution. Let is a rational number. Then we can find coprime numbers a and b. (where b 0 ) Such that a b 2 a 3 b 1 a 3 is rational 2 b So, 5 is rational. But it contracts the fact that Hence, is irrational. 5 is irrational Q.3. Prove that the following are irrationals: 1 (i) 2 (iii) 6+ 5 Solution. (i) Let P A G E C L A S S 1 0 N C E R T S O L U T I O N S D E E P A K S I R
13 (ii) 7 So, Masters Tuition Center 1 2 a, where a and b are coprime and b 0. b 13 P A G E C L A S S 1 0 N C E R T S O L U T I O N S D E E P A K S I R
14 Masters Tuition Center b a b a b is a rational i.e, a is rational. But it contracts the fact that 1 Hence, is irrational. 2 is irrational. (ii) Let 7 5 is a rational. 7 a where a and b are co  prime and b 0. b a 7b Since a and b are integers, a 7b is rational, and 5 is also rational. But, it contracts the fact that Hence, is irrational. is irrational. (iii) Let 6 2 is rational. 6 2 a where a and b are co  prime and b 0. b a 6 b Since, a b is rational so, a 6 is rational, also b contractation as Hence, 6 is irrational. is irrational. is rational. But it is a 14 P A G E C L A S S 1 0 N C E R T S O L U T I O N S D E E P A K S I R
15 Exercise 1.4 Q.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) (ii) 17 8 (iv) (v) (vi) P A G E C L A S S 1 0 N C E R T S O L U T I O N S D E E P A K S I R
16 129 (viii) 6 (vii) (ix) Solution. (i) Let x (x)...(1) Comparing euation (1) with x p, we get p 13 and 5 5 Here is of the form 2 n 5 m (n 0, m 5) Thus, x have a terminating decimal expansion. (ii) Let x (i) Comparing euation (2) with x p, we get p 17 and 2 3 Here is of the form 2 n 5 m (n 3, m 0) Thus, x 17 8 have a terminating decimal expansion. (iii) Let x (3) Comparing euation (3) with x p, we get p 64 and Here is not of the form 2 n 5 m Thus, x has a nonterminating repeating decimal expansion. (iv) Let x (4) P A G E C L A S S 1 0 N C E R T S O L U T I O N S D E E P A K S I R
17 Comparing euation (4) with x p, we get p 15 and Here is of the form 2 n 5 m (n 6, m 2) Thus, x having a terminating decimal expansion. (v) Let x (5) P A G E C L A S S 1 0 N C E R T S O L U T I O N S D E E P A K S I R
18 Comparing euation (5) with x p, we get p 29 and 7 3 Here is not of the form 2 n 5 m Thus, x having a nonterminating decimal expansion. (vi) Let x (6) Comparing euation (6) with x p, we get p 23 and Here is of the form 2 n 5 m (n 3, m 2) Thus, x have a terminating decimal expansion. (vii) Let x (7) Comparing euation (7) with x p, we get p 129 and Here is not of the form 2 n 5 m Thus, x have a nonterminating decimal expansion. (viii) Let x (8) 15 5 Comparing euation (8) with x p, we get p 2 and Here is of the form 2 n 5 m (n 0, m 1) Thus, x 6 have a terminating decimal expansion. 15 (ix) Let x (9) 18 P a g e
19 50 10 Comparing euation (9) with x p, we get p 7 and Here is of the form 2 n 5 m (n 1, m 1) Thus, x 7 have a terminating decimal expansion. 10 (x) Let x (10) P a g e
20 Comparing euation (10) with x p, we get p 11 and Here is not of the form 2 n 5 m Thus, x have a nonterminating decimal expansion. Q.2. Write down the decimal expansions of those rational numbers in Question 1 above which have terminating decimal expansions. Solution. From uestion 1, above, we know that only (i), (ii), (iv), (vi), (viii) and (ix) have terminating decimal expansions. (i) x x x (2 5) 5 x 416 (10) (Multiply and divide by 2 5 ) x (ii) x x x (2 5) 3 (Multiply and divide with 5 3 ) x (10) x x (iv) x
21 x (Multiply and divide by 5 4 ) 14
22 x x x (2 5) 6 So, x 9375 (10) x (vi) x So, 115 x (2 5) x (viii) x x (ix) So, 6 15 x 0.4 x x (2 5) 7 1 (10) So, x Q.3. The following real numbers have decimal expansions as given below. In each 15
23 case, decide whether they are rational or not. If they are rational, and of the form p, what can you say about the prime factors of? (i) (ii) (iii)
24 Solution. (i) Let x (i) It is clear from the number that x is a rational number. x x (ii) 10 9 From euation (ii) we see that x is a rational number and of the form p. where p and 10 9 = Prime factors of are (ii) Let x Since x neither terminating nor nonterminating repeating. So, is it is an irrational number. (iii) Let x (1) It is clear that the given number is a rational number because it is a nonterminating and repeating decimal. Multiply euation (i) with 10 9 on both sides, we get x = (2) Subtracting (i) from (2), we get x x x x Which is a rational number of the form p. x , where p and x ( ) Which is not of the form 2 n 5 m, n, m I. 17
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
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