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 38220 (iii) 867 and 255 Solution. (i) In 135 and 225, 225 is larger integer. Using Euclid s division algorithm, 225 135 1 90 [Where 135 is divisor, 90 is remainder] Since, remainder 90 0, by applying Eudid s division algorithm to 135 and 90 135 90 1 45 Again since, remainder 45 0, by applying Eudid s division algorithm to 90 and 45 90 45 2 0 Now, the remainder is zero so, our procedure stops. Hence, HCF of 135 and 225 is 45. (ii) In 196 and 38220, 38220 is larger integer. Using Euclid s division algorithm, 38220 196 195 0 remainder 0 The remainder is zero so our procedure stops. Hence, HCF of 196 and 38220 is 196. (iii) In 867 and 255, 867 is larger integer. Using Euclid s division algorithm 867 255 3 102 [Where 255 is divisor, 102 is remainder] Since, remainder 102 0, by applying Eudid s division algorithm to 255 and 102 255 102 2 51 Since, remainder 51 0, by applying Eudid s division algorithm 102 and 51 So, 102 51 2 0 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 6 + 1 or 6 + 3 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
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, 616 32 19 8 [Where 32 is divisor, 8 is remainder] Since, remainder 8 0, by applying Eudid s division algorithm, we get 32 8 4 0 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 2 9 2 3(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 2 9 2 1 2 3 1 x 2 3(3 2 2) 1 x 2 3m 1 m 3 2 2...(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
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
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 3 27 3 (cubing both sides) a 3 9 3 3 9m...(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) 2 1 31 3 a 3 27 3 1 27 2 9 a 3 27 3 27 2 9 1 a 3 9(3 3 3 2 ) 1 a 3 9m 1...(ii) Where m 3 3 3 2 If a 3 2 a 3 (3 2) 3 (cubing both sides) a 3 (3) 3 (2) 3 3 (3) 2 2 3 (2) 2 3 a 3 27 3 8 54 2 36 a 3 27 3 54 2 36 8 a 3 9(3 3 6 2 4) 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
Where m 3 3 6 2 4 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) 140 140 70 2 35 2 7 5 Hence, 140 2 2 5 7 2 2 5 7 (ii) 156 156 78 2 39 2 13 3 Hence, 156 2 2 313 2 2 313 (iii) 3825 3825 1275 425 17 85 Hence, 3825 3 3 5 517 3 2 5 2 17 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
(iv) 5005 Masters Tuition Center 5005 100 1 143 11 13 Hence, 5005 5 7 1113 (v) 7429 7429 17 437 19 23 Hence, 7429 17 19 23 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 91 26 2 13 91 7 13 HCF 13 LCM 2 7 13 182 Verification: LCM HCF 182 13 2366 Product of two numbers 26 91 2366 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
So, 510 2 3 517 92 2 2 23 HCF 2 LCM 2 2 3 517 23 23460 Verification: LCM HCF 2 23460 46920 Product of two numbers 510 92 46920 Product of two numbers LCM HCF 336 54 168 27 84 42 21 (iii) 336 and 54 336 2 2 2 2 3 7 54 2 3 3 3 HCF 2 3 6 LCM 2 2 2 2 3 3 3 7 3024 Verification: LCM HCF 6 3024 18144 Product of two numbers 336 54 18144 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
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 12 2 2 3 Prime factorisation of 15 3 5 Prime factorisation of 21 3 7 HCF of 12, 15 and 21 3 LCM of 12, 15 and 21 2 2 3 5 7 420 17 23 29 17 23 29 (ii) 17, 23 and 29 Prime factorisation of 17 117 Prime factorisation of 23 1 23 Prime factorisation of 29 1 29 HCF of 17, 23 and 29 1 LCM of 17, 23 and 29 117 23 29 11339 (iii) 8, 9 and 25 8 9 25 Prime factorisation of 8 1 2 2 2 Prime factorisation of 9 1 3 3 Prime factorisation of 25 1 5 5 HCF of 8, 9 and 25 1 LCM of 8, 9 and 25 2 2 2 3 3 5 5 1800 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
HCF (306, 657) 9 (given) 9 LCM 306 657 LCM 306 657 22338 9 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
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 7 11 13 + 13 and 7 6 5 4 3 2 1 + 5 are composite numbers. Solution. 7 1113 13 137 111 13 78, which is not a prime number because it has more than two factors. So, it is a composite number. And, 7 6 5 4 3 2 1 5 57 6 4 3 2 1 51009, 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, 12 2 9, 6 3 9, 3 3 3, 1 1, 1 So, LCM of 18 and 12 2 2 3 3 36 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
( 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
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 3+ 2 5 is irrational. Solution. Let 3 2 5 is a rational number. Then we can find coprime numbers a and b. (where b 0 ) Such that 3 2 5 a b 2 a 3 b 1 a 3 is rational 2 b So, 5 is rational. But it contracts the fact that Hence, 3 2 5 is irrational. 5 is irrational Q.3. Prove that the following are irrationals: 1 (i) 2 (iii) 6+ 5 Solution. (i) Let 1 2 12 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
(ii) 7 So, Masters Tuition Center 1 2 a, where a and b are co-prime 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
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, 7 5 5 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
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) 13 3125 64 455 (ii) 17 8 (iv) 15 1600 (v) 29 343 (vi) 23 2 3 5 2 15 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
129 (viii) 6 (vii) 2 2 5 7 7 5 15 (ix) 35 50 Solution. (i) Let x 13 13 (x)...(1) 77 210 3125 5 5 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 13 3125 have a terminating decimal expansion. (ii) Let x 17 17 8 2 3...(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 64 64...(3) 455 5 7 13 Comparing euation (3) with x p, we get p 64 and 5 7 13 Here is not of the form 2 n 5 m Thus, x 64 455 has a non-terminating repeating decimal expansion. (iv) Let x 15 15...(4) 1600 2 6 5 2 16 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
Comparing euation (4) with x p, we get p 15 and 2 6 5 2 Here is of the form 2 n 5 m (n 6, m 2) Thus, x 15 1600 having a terminating decimal expansion. (v) Let x 29 29...(5) 343 7 3 17 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
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 29 343 having a non-terminating decimal expansion. (vi) Let x 23 2 3 5 2...(6) Comparing euation (6) with x p, we get p 23 and 2 3 5 2 Here is of the form 2 n 5 m (n 3, m 2) Thus, x 23 2 3 5 2 have a terminating decimal expansion. (vii) Let x 129 2 2 5 7 7 5...(7) Comparing euation (7) with x p, we get p 129 and 2 5 5 7 7 5 Here is not of the form 2 n 5 m Thus, x 129 2 2 5 7 7 5 have a non-terminating decimal expansion. (viii) Let x 6 2...(8) 15 5 Comparing euation (8) with x p, we get p 2 and 5 2 0 5 1 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 35 7...(9) 18 P a g e
50 10 Comparing euation (9) with x p, we get p 7 and 10 2 1 5 1 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 77 11...(10) 210 30 19 P a g e
Comparing euation (10) with x p, we get p 11 and 30 2 1 3 1 5 1 Here is not of the form 2 n 5 m Thus, x 77 210 have a non-terminating 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 13 13 3125 5 5 13 2 5 x 5 5 2 5 13 32 x (2 5) 5 x 416 (10) 5 416 100000 (Multiply and divide by 2 5 ) 13 3125 x 0.00416 (ii) x 17 17 8 2 3 17 5 3 x 2 3 5 3 17 125 x (2 5) 3 (Multiply and divide with 5 3 ) x 2125 2125 (10) 3 1000 x 2.125 17 x 2.125 8 (iv) x 15 1600 13
x 5 1 2 6 5 2 (Multiply and divide by 5 4 ) 14
15 5 4 x 2 6 5 2 5 4 15 625 x 2 6 5 6 9375 x (2 5) 6 So, x 9375 (10) 6 15 1600 x 0.009375 9375 1000000 0.009375 (vi) x 23 23 5 115 2 3 5 2 2 3 5 2 5 2 3 5 3 So, 115 x (2 5) 115 3 1000 0.115 23 2 3 5 2 x 0.115 (viii) x 6 2 15 5 2 2 1 4 x 2 1 5 1 10 0.4 (ix) So, 6 15 x 0.4 x 35 7 7 50 10 2 1 5 1 7 x (2 5) 7 1 (10) 0.7 1 35 So, x 0.7 50 Q.3. The following real numbers have decimal expansions as given below. In each 15
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) 43.123456789 (ii) 0.120120012000120000... (iii) 43.123456789 16
Solution. (i) Let x 43.123456789...(i) It is clear from the number that x is a rational number. x 43123456789 1000000000 x 43123456789...(ii) 10 9 From euation (ii) we see that x is a rational number and of the form p. where p 43123456789 and 10 9 = 2 5 5 9 Prime factors of are 2 9 5 9 (ii) Let x 0.120120012000120000... Since x neither terminating nor non-terminating repeating. So, is it is an irrational number. (iii) Let x 43.123456789...(1) It is clear that the given number is a rational number because it is a non-terminating and repeating decimal. Multiply euation (i) with 10 9 on both sides, we get 1000000000x = 43123456789.123456289 (2) Subtracting (i) from (2), we get 1000000000x 43123456789.123456789... x 43.123456789... 999999999x 43123456746 x 43123456746 999999999 Which is a rational number of the form p. x 4791495194, where 111111111 p 4791495194 and 111111111 x 4791495194 3 2 (12345679) Which is not of the form 2 n 5 m, n, m I. 17