![JK9972397103
4.Use Euclid’s division lemma to show that the square of any positive integer is either of form
3m or 3m + 1 for some integer m. [Hint: Let x be any positive integer then it is of the form 3q,
3q + 1 or 3q + 2. Now square each of these and show that they can be rewritten in the form
3m or 3m + 1.]
Let a be any positive integer and b = 3.
Then a = 3q + r for some integer q ≥ 0
And r = 0, 1, 2 because 0 ≤ r < 3
Therefore, a = 3q or 3q + 1 or 3q + 2
Or,
a2 = (3q)2 or (3q+1)2 or (3q+2)2
a2 = (9q2) or (9q2 + 6q + 1) or (9q2 +12q+4)
= 3 x (3q2) or 3 x (3q2+2q) + 1 or 3 x (3q2+4q+1) + 1
= 3k1 or 3k2+1 or 3k3 +1
Where k1, k2, and k3 are some positive integers
Hence, it can be said that the square of any positive integer is either of the form 3m or 3m + 1.](https://image.slidesharecdn.com/realnumberspptbyjk-200505105352/85/Real-numbers-ppt-by-jk-7-320.jpg)
Real numbers ppt by jk
- 1. JYOTHI KUMAR.M, KARNATAK PUBLIC SCHOOL, SANTHEBENNURU, CHANNAGIRI-Tq, DAVANAGERE-Dt. 9972397103
- 2. JK9972397103 An algorithm is a series of well defined steps which gives a procedure for solving a type of problem. A lemma is a proven statement used for proving another statement. The word algorithm comes from the name of the 9th century Persian mathematician al-Khwarizmi. In fact, even the word ‘algebra’is derived from a book, he wrote, called Hisab al-jabr w’al-muqabala. Euclid’s division algorithm,as the name suggest,has to do with divisibility of integers.It says any positive integer “a” can be divided by any another positive integer “b” in such a way that it leaves a remainder “r” that is smaller than “b”. EUCLID’S DIVISION LEMMA(THEOREM NO:-1.1) Given positive integers “a” and “b” ,there exist unique integers “q” and “r” satiafying a=bq+r,0≤r<b.
- 3. JK9972397103 EXERCISE NO: 1.1 1. Use Euclid’s division algorithm to find the HCF of: (i) 135 and 225 135 and 225 Since 225 > 135, we apply the division lemma to 225 and 135 to obtain 225 = 135 × 1 + 90 Since remainder 90 ≠ 0, we apply the division lemma to 135 and 90 to obtain 135 = 90 × 1 + 45 We consider the new divisor 90 and new remainder 45, and apply the division lemma to obtain 90 = 2 × 45 + 0 Since the remainder is zero, the process stops. Since the divisor at this stage is 45, Therefore, the HCF of 135 and 225 is 45.
- 4. JK9972397103 (ii)196 and 38220 Since 38220 > 196, we apply the division lemma to 38220 and 196 to obtain 38220 = 196 × 195 + 0 Since the remainder is zero, the process stops. Since the divisor at this stage is 196, Therefore, HCF of 196 and 38220 is 196. (iii)867 and 255 Since 867 > 255, we apply the division lemma to 867 and 255 to obtain 867 = 255 × 3 + 102 Since remainder 102 ≠ 0, we apply the division lemma to 255 and 102 to obtain 255 = 102 × 2 + 51 We consider the new divisor 102 and new remainder 51, and apply the division lemma to obtain 102 = 51 × 2 + 0 Since the remainder is zero, the process stops. Since the divisor at this stage is 51, Therefore, HCF of 867 and 255 is 51.
- 5. JK9972397103 2.Show that any positive odd integer is of the form, 6q + 1 or 6q + 3, or 6q + 5 , where q is some integer. Let a be any positive integer and b = 6. Then, by Euclid’s algorithm, a = 6q + r for some integer q ≥ 0, and r = 0, 1, 2, 3, 4, 5 because 0 ≤ r< 6. Therefore, a = 6q or 6q + 1 or 6q + 2 or 6q + 3 or 6q + 4 or 6q + 5 Also, 6q + 1 = 2 × 3q + 1 = 2k1 + 1, where k1 is a positive integer 6q + 3 = (6q + 2) + 1 = 2 (3q + 1) + 1 = 2k2 + 1, where k2 is an Integer 6q + 5 = (6q + 4) + 1 = 2 (3q + 2) + 1 = 2k3 + 1, where k3 is an Integer Clearly, 6q + 1, 6q + 3, 6q + 5 are of the form 2k + 1, where k is an integer. Therefore, 6q + 1, 6q + 3, 6q + 5 are not exactly divisible by 2. Hence, these expressions of numbers are odd numbers. And therefore, any odd integer can be expressed in the form 6q + 1, or 6q + 3, or 6q + 5
- 6. JK9972397103 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? HCF (616, 32) will give the maximum number of columns in which they can march. We can use Euclid’s algorithm to find the HCF. 616 = 32 × 19 + 8 32 = 8 × 4 + 0 The HCF (616, 32) is 8. Therefore, they can march in 8 columns each.
- 7. JK9972397103 4.Use Euclid’s division lemma to show that the square of any positive integer is either of form 3m or 3m + 1 for some integer m. [Hint: Let x be any positive integer then it is of the form 3q, 3q + 1 or 3q + 2. Now square each of these and show that they can be rewritten in the form 3m or 3m + 1.] Let a be any positive integer and b = 3. Then a = 3q + r for some integer q ≥ 0 And r = 0, 1, 2 because 0 ≤ r < 3 Therefore, a = 3q or 3q + 1 or 3q + 2 Or, a2 = (3q)2 or (3q+1)2 or (3q+2)2 a2 = (9q2) or (9q2 + 6q + 1) or (9q2 +12q+4) = 3 x (3q2) or 3 x (3q2+2q) + 1 or 3 x (3q2+4q+1) + 1 = 3k1 or 3k2+1 or 3k3 +1 Where k1, k2, and k3 are some positive integers Hence, it can be said that the square of any positive integer is either of the form 3m or 3m + 1.
- 8. JK9972397103 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. Let a be any positive integer and b = 3 a = 3q + r, where q ≥ 0 and 0 ≤ r < 3 ∴a=3q or 3q+1 or 3q+2 Therefore, every number can be represented as these three forms. There are three cases. Case 1: When a = 3q, a 3=(3q)3= 27q3 =9(3q3)= 9m, Where m is an integer such that m=3q3 Case 2: When a = 3q + 1, a3 = (3q +1)3 a3 = 27q3 + 27q2 + 9q + 1 a3= 9(3q3 + 3q2 + q) + 1 a3 = 9m + 1, Where m is an integer such that m = (3q3 + 3q2 + q) Case 3: When a = 3q + 2, a3 = (3q +2)3 a3 = 27q3 + 54q2 + 36q + 8 a3= 9(3q3 + 6q2 + 4q) + 8 a3 = 9m + 8, Where m is an integer such that m = (3q3 + 6q2 + 4q) Therefore, the cube of any positive integer is of the form 9m, 9m + 1, or 9m + 8.
- 9. JK9972397103 THE FUNDAMENTAL THEOREM OF ARITHMETIC (THEOREM NO:-1.2) An equivalent version of Theorem 8.2 was probably first recorded as Proposition 14 of Book IX in Euclid’s Elements, before it came to be known as the Fundamental Theorem of Arithmetic. However, the first correct proof was given by Carl Friedrich Gauss in his Disquisitiones Arithmeticae. Every composite number can be expressed (factorized)as a product of primes,and this factorization is unique,apart from the order in which the prime factors occur. The Fundamental Theorem of Arithmetic says that every composite number can be factorised as a product of primes. Actually it says more. It says that given any composite number it can be factorised as a product of prime numbers in a ‘unique’ way, except for the order in which the primes occur. That is, given any composite number there is one and only one way to write it as a product of primes,as long as we are not particular about the order in which the primes occur. So, for example, we regard 2 × 3 × 5 × 7 as the same as 3 × 5 × 7 × 2, or any other possible order in which these primes are written.
- 10. JK9972397103 EXERCISE NO: 1.2 1.Express each number as product of its prime factors: (i) 140 140 = 2 × 2 × 5 × 7 = 22 × 5 × 7 (ii) 156 156 = 2 × 2 × 3 × 13 = 22 × 3 × 13 (iii) 3825 3825 = 3 × 3 × 5 × 5 × 17 = 32 × 52 × 17 (iv) 5005 5005 = 5 ×7× 11 × 13 (v) 7429 7429 = 17 × 19 × 23
- 11. JK9972397103 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 26 = 2 × 13 91 = 7 × 13 HCF = 13 LCM = 2 × 7 × 13 = 182 Product of the two numbers = 26 × 91 = 2366 HCF × LCM = 13 × 182 = 2366 Hence, product of two numbers = HCF × LCM (ii) 510 and 92 510 = 2 × 3 × 5 × 17 92 = 2 × 2 × 23 HCF = 2 LCM = 2 × 2 × 3 × 5 × 17 × 23 = 23460 Product of the two numbers = 510 × 92 = 46920 HCF × LCM = 2 × 23460 = 46920 Hence, product of two numbers = HCF × LCM
- 12. JK9972397103 (iii) 336 and 54 336 = 2 × 2 × 2 × 2 × 3 × 7 336 = 24× 3× 7 54 = 2 × 3 × 3 × 3 54 = 2 × 33 HCF = 2 × 3 = 6 LCM = 24 × 33 × 7 = 3024 Product of the two numbers = 336 × 54 = 18144 HCF × LCM = 6 × 3024 = 18144 Hence, product of two numbers = HCF × LCM 3. Find the LCM and HCF of the following integers by applying the prime factorisation method. (i) 12,15 and 21 12 = 22× 3 15 = 3 × 5 21 = 3 × 7 HCF = 3 LCM = 22× 3 × 5 × 7 = 420
- 13. JK9972397103 (ii) 17,23 and 29 17 = 1 × 17 23 = 1 × 23 29 = 1 × 29 HCF = 1 LCM = 17 × 23 × 29 = 11339 (iii) 8, 9 and 25 8 = 2 × 2 × 2 9 = 3 × 3 25 = 5 × 5 HCF = 1 LCM = 2 × 2 × 2 × 3 × 3 × 5 × 5 = 1800
- 14. JK9972397103 4.Given that HCF (306, 657) = 9, find LCM (306, 657). HCF (306, 657) = 9 We know that, LCM × HCF = Product of two numbers ∴LCM × HCF = 306 × 657 LCM = 306 × 657 HCF = 306 × 657 9 LCM = 22338 5.Check whether 6n can end with the digit 0 for any natural number n. If any number ends with the digit 0, it should be divisible by 10 or in other words, it will also be divisible by 2 and 5 as 10 = 2 × 5 Prime factorisation of 6n = (2 ×3)n It can be observed that 5 is not in the prime factorisation of 6n. Hence, for any value of n, 6n will not be divisible by 5. Therefore, 6n cannot end with the digit 0 for any natural number n.
- 15. JK9972397103 6.Explain why 7 × 11 × 13 + 13 and 7 × 6 × 5 × 4 × 3 × 2 × 1 + 5 are composite numbers. Numbers are of two types - prime and composite. A prime number can be divided by 1 and only itself, where as a composite number have factors other than 1 and itself. It can be observed that 7 × 11 × 13 + 13 = 13 × (7 × 11 + 1) = 13 × (77 + 1) = 13 × 78 = 13 ×13 × 6 The given expression has 6 and 13 as its factors. Therefore, it is a composite number. 7 × 6 × 5 × 4 × 3 × 2 × 1 + 5 = 5 ×(7 × 6 × 4 × 3 × 2 × 1 + 1) = 5 × (1008 + 1) = 5 ×1009 1009 cannot be factorised further. Therefore, the given expression has 5 and 1009 as its factors. Hence, it is a composite number.
- 16. JK9972397103 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? It can be observed that Ravi takes lesser time than Sonia for completing 1 round of the circular path. As they are going in the same direction, they will meet again at the same time when Ravi will have completed 1 round of that circular path with respect to Sonia. And the total time taken for completing this 1 round of circular path will be the LCM of time taken by Sonia and Ravi for completing 1 round of circular path respectively i.e., LCM of 18 minutes and 12 minutes. 18 = 2 ×3 ×3 And, 12 = 2 ×2 ×3 LCM of 12 and 18 = 2 × 2 × 3 × 3 = 36 Therefore, Ravi and Sonia will meet together at the starting point after 36 minutes.
- 17. JK9972397103 REVISITING IRRATIONAL NUMBERS A number ‘s’ is called irrational if it cannot be written in the form 𝑝 𝑞 ,where p and q are integers and q ≠ 0. The sum or difference of a rational and an irrational number is irrational. The product and quotient of a non-zero rational and irrational number is irrational. THEOREM NO:-1.3 Let p be a prime number. If p divides a2, then p divides a, where a is a positive integer.
- 18. JK9972397103 EXERCISE NO: 1.3 1.Prove that 5 is irrational. Let 5 is a rational number. Therefore, we can find two integers a, b (b ≠ 0) such that 5 = 𝑎 𝑏 Let a and b have a common factor other than 1. Then we can divide them by the common factor, and assume that a and b are co-prime. a = 5 b squaring on both sides,then a2=5b2⇒ a2 5 =b2 Therefore, a2 is divisible by 5 and it can be said that a is divisible by 5.⟶(1) Let a 5 =k,then a= 5k, where k is an integer.then we know that a2=5b2 by a value ,we will get (5k)2=5b2 ⇒25 k2=5b2 ⇒k2= b2 5 This means that b2 is divisible by 5 and hence, b is divisibleby 5. ⟶(2) From (1) & (2), implies that a and b have 5 as a common factor. And this is a contradiction to the fact that a and b are co-prime. Hence, 5 cannot be expressed as 𝑝 𝑞 or it can be said that 5 is irrational.
- 19. JK9972397103 2.Prove that is irrational 3+2 5. Let is 3+2 5 rational. Therefore, we can find two integers a, b (b ≠ 0) such that 3+2 5 = 𝑎 𝑏 2 5 = 𝑎 𝑏 -3⇒ 5 = 1 2 𝑎 𝑏 −3 Since a and b are integers, 1 2 𝑎 𝑏 −3 will also be rational and therefore, 5 is rational. This contradicts the fact that 5 is irrational. Hence, our assumption that 3+2 5 is rational is false. Therefore, 3+2 5 is irrational. 3.Prove that the following are irrationals: (i) 1 2 Let 1 2 is rational. Therefore, we can find two integers a, b (b ≠ 0) such that 1 2 = 𝑎 𝑏 ⇒ 2= 𝑏 𝑎 But 𝑏 𝑎 is rational as a and b are integers. Therefore, 2 is rational which contradicts to the fact that 2 is irrational. Hence, our assumption is false and 1 2 is irrational.
- 20. JK9972397103 (ii) 7 5 Let 7 5 is rational. Therefore, we can find two integers a, b (b ≠ 0) such that 7 5 = 𝑎 𝑏 5 = 𝑎 7𝑏 Since a and b are integers, 𝑎 7𝑏 will also be rational and therefore, 5 is rational. This contradicts the fact that 5 is irrational. Hence, our assumption that 7 5 is rational is false. Therefore, 7 5 is irrational. (iii) 6+ 2. Let is 6+ 2 rational. Therefore, we can find two integers a, b (b ≠ 0) such that 6+ 2 = 𝑎 𝑏 2 = 𝑎 𝑏 -6,Since a and b are integers, 𝑎 𝑏 -6will also be rational and therefore, 2 is rational. This contradicts the fact that 2 is irrational. Hence, our assumption that 6+ 2 is rational is false. Therefore, 6+ 2 is irrational.
- 21. JK9972397103 REVISITING RATIONAL NUMBERS AND THEIR DECIMAL EXPANSIONS THEOREM NO:-1.5 Let x be a rational number whose decimal expansion terminates.Then x can be expressed in the form 𝑝 𝑞 ,where p and q are coprime, and the prime factorisation of q is of the form 2n5m, where n, m are non-negative integers. THEOREM NO:-1.6 Let x = 𝑝 𝑞 be a rational number, such that the prime factorization of q is of the form 2n5m, where n, m are non-negative integers. Then x has a decimal expansion which terminates. THEOREM NO:-1.7 Let x = 𝑝 𝑞 be a rational number, such that the prime factorization of q is not of the form 2n5m ,where n, m are non-negative integers. Then, x has a decimal expansion which is non-terminating repeating (recurring).
- 22. JK9972397103 EXERCISE NO: 1.4 1. 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) 13 3125 3125 = 55 The denominator is of the form 5m. Hence, the decimal expansion of 13 3125 is terminating. (ii) 17 8 8=23 The denominator is of the form 2m. Hence, the decimal expansion of 17 8 is terminating. (iii) 64 455 455 = 5 × 7 × 13 ,Since the denominator is not in the form 2m× 5n, and it also contains 7 and 13 as its factors, its decimal expansion will be non-terminating repeating.
- 23. JK9972397103 (iv) 15 1600 1600 = 26 × 52 The denominator is of the form 2m× 5n. Hence, the decimal expansion of 15 1600 is terminating. (v) 29 343 343=73 The denominator is not in the form 2m× 5n, and it also contains 7 as its factor, its decimal expansion will be non-terminating repeating. (vi) 23 2352 The denominator is of the form 2m× 5n. Hence, the decimal expansion of 23 2352 is terminating. (vii) 129 225775 The denominator is not in the form 2m× 5n, and it also contains 7 as its factor, its decimal expansion will be non-terminating repeating.
- 24. JK9972397103 (viii) 6 15 = 23 35 = 2 5 The denominator is of the form 5n. Hence, the decimal expansion of 6 15 is terminating. (ix) 35 50 = 75 105 = 7 25 The denominator is of the form 2m× 5n.Hence the decimal expansion of 35 50 is terminating. (x) 77 210 = 711 307 = 11 30 30=235 The denominator is not in the form 2m× 5n, and it also contains 3 as its factor, its decimal expansion will be non-terminating repeating.
- 25. JK9972397103 2.Write down the decimal expansions of those rational numbers in Question 1 above which have terminating decimal expansions. (i) 13 3125 13 3125 = 13 2055 Multiplying and dividing by25 13 2055 25 25= 1325 205525= 1332 15525= 416 (10)5=0.00416 (ii) 17 8 17 8 = 17 2350 Multiplying and dividing by53 17 2350 53 53= 1753 235053= 17125 12353= 2125 (10)3=2.125
- 26. JK9972397103 (iii) 15 1600 3 320 = 3 2651 Multiplying and dividing by55 3 2651 55 55= 355 2656= 33125 2656 =9375=0.009375 (iv) 23 2352 23 2352 Multiplying and dividing by51 23 2352 51 51= 235 2353= 115 (10)3=0.115 (v) 6 15 = 3 5 =0.4 (vi) 35 50 = 7 10 =0.7
- 27. JK9972397103 3.The following real numbers have decimal expansions as given below. In each case, decide whether they are rational or not. If they are rational, and of the form 𝑝 𝑞 , what can you say about the prime factor of q? (i) 43.123456789 Since this number has a terminating decimal expansion, it is a rational number of the form 𝑝 𝑞 and q is of the form2m5n i.e., the prime factors of q will be either 2 or 5 or both. (ii) 0.120120012000120000 … The decimal expansion is neither terminating nor recurring. Therefore, the given number is an irrational number.
- 28. JK9972397103 (iii) 43.123456789 Since the decimal expansion is non-terminating recurring, the given number is a rational number of the form and q is not of the form 𝑝 𝑞 and q is of the form2m5n i.e., the prime factors of q will also have a factor other than 2 or 5.