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Chapter - 1 Real Numbers Euclids Division Lemma All rational and irrational numbers are called real numbers. E.g. 3,4, v5 t 6, The set of real numbers is denoted by R. Real Numbers can be represented on a number line. Classification of Real Numbers: Rational Numbers [ irrational Numbers ] — Whole Numbers == Natural Numbers: Natural Numbers is a set of counting numbers. They are denoted by N. N= (1, 2,3, 4ocs.00} Whole Numbers: Whole numbers is a set of natural numbers plus zero. Integers: Integers is a set of whole numbers and negative of all natural numbers. Z={-3,-2,-1,0,1, 2,3} Rational Numbers: All the numbers that can be written in the form , where and are integers and # 0 are called rational numbers. 8 -3 Eg. 11, 17 Irrational Numbers: All the numbers that cannot be written in the form, are called irrational numbers. All the non-terminating and non-repeating decimal numbers are irrational numbers. E.g. V5, V3, V5 + V3, v ¢ Now, consider ¥ . When we calculate the value of V3 we get V3 =1.7320508075688 here the decimal number do not repeat and do not terminate. Hence, it is an irrational number. e V3 is an irrational number, but ¥ is not an irrational number because V9 = 3, which is a rational number. Hence, the square roots of all the numbers do not give an irrational number. e The value of the famous irrational number 7 is 3.145926...... Again the decimal number is neither repeating nor terminating. Euclid’s Division Lemma A lemma is a proven statement used for proving another statement. Euclid’s Division Lemma states that any positive integer can be divided by another positive integer in such a way that it leaves a remainder which is smaller than and this is also known as the long division process. According to Euclid’s Division Lemma, if a and b are two positive Now,0=r<2 Therefore, the possible remainders are 0 and 1. When ,r=0,a=2q+0 a=2q Whenr =1,a=2q+1 If a = 2q, then is an even integer. If a = 2q, then is an odd integer. Let us consider an even positive integer, 6 and an odd positive integer, 9. Then, 6 =2 x3 and9=2x4+1 Clearly, the positive even integer is of the form and odd positive integer is of the form 2q +1. Example: Show that any positive odd integer is of the form 6q +1, or 6q + 3 or 6q + 5, where is any integer. Let be any odd positive integer andb=6. According to Euclid’s division lemma there exist q and r such that a=bq+r,where 0=r (i) Putting b= 6 in Eq.(i) we get, a=6q+r, where is any integer and 0 6 As 0 =r <6, ..r = 0,1,2,3,4,5 When,r=0 ,a= 6q+ 0 r= 1,a=6q+ 1 r= 2,a= 6q+ 2 r= 3,a=6q+ 3 r= 4, a= 6q+ 4 r=5,a=6q+ 5 Assuming q=2, then a=6q=6x2=12, an even integer a=6q+2=6x2+2=14, an even integer a=6q+4=6x2+4=16. an even integer. Since, a is an odd integer a # 6q, a * 6q4+2, a * 6q+4 Any positive odd integer is of the form 6q+1,6q+3,6q+5 Now, the value of can q be any integer. Let us assume that the value of q is 2. Then,6q + 1 = 6x2+1=13, positive odd integer 6q+3=6x2+3=12+3=15, positive odd integer 6q+5 = 6x2+5=12+5=17, positive odd integer Example: Show that the square of any positive odd integer is of the form 4q + 1 for some integer q. Let a be any positive integer According to Euclid’s division lemma there exist q andr such that a=bq+tr,whereO0 =r 225 . By using Euclid’s Division Lemma, a = bq + r, 0 =r 176. By using Euclid’s Division Lemma, a = bq + r, 0 504 By using Euclid’s Division Lemma we get, 735 =504x1+ 231 Now,r = 23170. We will consider new divisor 504 and new remainder 231 and apply Euclid’s Division Lemma to get, 504 = 231x2 +42 Again, r = 42 #0. We now consider new divisor 231 and new remainder 42, and apply Euclid’s Division Lemma to get, 231=42x5=21 r= 21#0. We now consider new divisor 42 and new remainder 21, and apply Euclid’s Division Lemma to get, 42 =21x2+0 The remainder is equal to zero, the divisor at this stage is 21. Hence the HCF of 504 and 735 is 21. Therefore, the maximum capacity of the required container is 21L. Example : Three pieces of timber 42 m, 49 m and 56 m long have to be divided into planks of the same length. What is the greatest possible length of each plank? The greatest possible length of the plank will be the HCF of 42m, 49 m and 56 m. Now, 56 > 49 > 42 We will first find the HCF of 56 and 49. By using Euclid’s Division Lemma we get, 56=49x1+4+7 Now,r =7 #0. We will consider new divisor 49 and new remainder 7 and apply Euclid’s Division Lemma to get, 49=7x7+ 0 The remainder is equal to zero, the divisor at this stage is 7. Hence the HCF of 56 and 49 is 7. Now, using Euclid’s Division Lemma for 42 and 7 we get, 42=7x6+0 As the remainder is 0, the divisor in this case is the HCF of 42 and 7. ©. © 8) Aa @ 270 =2xX3° X'S 270=5 X 33x 2 Here, in the prime factorization of 270, the prime numbers appearing in both the cases are the same only the order in which they appear are different. Therefore, the prime factorization of 270 is unique except for the order in which the primes occur. Example: Check whether 15n can end with digit zero for any natural number n. If a number ends with the digit 0, then it is divisible by both 2 and 5. But prime factors of 15 are 3 and 5. 15"= (3 x 5)" = 3" x 5" Here, the prime factorization of 15" contains only 5 but not 2. The uniqueness of the Fundamental Theorem of Arithmetic guarantees that there are no other primes in the prime factorization of 15". Therefore,15" cannot end with digit zero for any natural number n. Example: Explain, why (7x6x5x4x3x2x1)+5and(3x5 x 13 x 46) + 23 is a composite number? Composite numbers are those numbers that have at least one factor other than 1 and the number itself. E.g.4, 6, 9 i)(7x6x5x4x3x2x1)+5=50404+5=5045 5045 =5 x 1009 As the factors of 5045 are 5 and 1009, it is a composite number. ii) (38 x 5 x 13 x 46) + 23 = 8970 + 23 = 8993 8993 =17 x 23 x 23 x2 As the factors of 8993 are 17, 23 and 2, it is a composite number. HCF and LCM by Prime Factorisation Method In this method, we first express the given numbers as a product of prime factors separately. Then, HCF is the product of the smaller power of each common prime factor in the numbers and LCM is the product of the greatest power of each prime factor involved in the numbers. For any two positive integers a and b, HCF(a, b) x LCM (a,b) =axXb Example: Find the LCM and HCF of 120 and 144 by the fundamental theorem of arithmetic. 120=23x3x5 144 = 24 x 32 Now, HCF is the product of the smallest power of each common prime factor in the numbers | Common Prime Factors Smallest Power of Prime Factor 2 = 3 3" HCF (120,144) = 23x3=8x3=24 Example: If HCF (253,440) = 11 and LCM (253, 440) = 253 x R. Find the value of R. We know that, HCF(a, b) x LCM (a,b) =a xb « HCF(253, 440) x LCM (253, 440) = 253 x 440 11 x 253 x R= 253 x 440 Example: Ravi and Shikha drive around a circular sports field. Ravi takes 16 min to complete one round, while Shikha completes the round in 20 min. If both start at the same point, at the same time and go in the same direction, then after how much time will they meet at the starting point? Time taken by Ravi to drive one round of the circular field = 16 min Time taken by Shikha to drive one round of the circular field =20 min The time after which they will again meet at the starting point will be equal to the LCM of 16 min and 20 min. 16 = 2+ 20=22x5 LCM(16,20) = 24x 5=16x5=80 Therefore, Ravi and Shikha will meet again at the starting point after 80 min. Revisiting Irrational Numbers Revisiting Irrational Numbers Pp Irrational numbers are those numbers which cannot be written in the form 4%, where p and qare integers and q # 0. E.g.V2,V3, v15 The square roots of all the numbers do not give an irrational number. For example, v2 is an irrational number but v4 = 2, which is rational. Therefore square roots of all prime numbers are irrational. If p is a prime number then VP is an irrational number. Theorem 1: If a prime number p divides a2, then p divides a, where a is a positive integer. Proof: Every positive integer can be expressed as the product of primes. Let a = pip2p3 .. +. +... Pn Where pip2ps «.. ... «+. Pn are all the prime numbers of a. a? = (p1p2p3........... pn) a? = (pip2ps «.. +. PN)(P1p2ps... + s+» Pn) a2 = (pi2p2?p2? ......... pn? ) It is given that p divides a2. According to the Fundamental Theorem of Arithmetic, we can say that p is one of the prime factors of a?. According to the Fundamental Theorem of Arithmetic, the prime factorization of a natural number is unique. Now the only prime factors of a2 are pip2ps .. Pi p2p? wwe ee PB . Pn. Therefore, p is one of So p is also a factor of a If p divides a2 then p also divides a. Let us consider a positive integer 12. Now, the factors of 12 are 2, 2 and 3. On squaring 12 we get, 122 = 144 Example: Show that 3v2 is an irrational number. Let us assume that 3V2 is rational. a Then 3v2 can be expressed in the form ‘b where a and b are integers and b # 0. Now, a and b have no common factor other than 1. a 3V2 =), where a and bare coprime integers a - V2 = 3b a Since, 3, a and b are integers, 3b is rational. Now, a rational number cannot be equal to an irrational number, that isV2 So, we conclude that 3v2 is irrational. Example: Show that 5 — V3 is irrational. Let us assume that 5 -Vv3 is rational. As it is a rational number, it can be a expressed in the form b where aand b are integers and b + 0. Now, aand b have no common factor other than 1. /q . 5 — V3 =ab, where a and b are coprime a. 5 -b= V3 5b-a b =v3 5b—a As, 5,a,andbareintegers, 6 isrational. Now, V3 is an irrational number and cannot be equal to a rational number. So, we conclude that 5 — V3 is irrational. Example: Prove that 2V3 + V5 is an irrational number. Let us assume that 2v3 + V5 is rational. As it is a rational number, it can be a expressed in the form D where aand b are integers and b + 0. Now,a andb have no common factor other than 1. a 2v3 ae V5 = >, where aand b are coprime a 2v3= 6 — V5 On squaring both sides, we get a 12=(b-V5)2 a 12 =a2/b2 — 2V5 b+ 5 2 a a B-2v5 b=7 2 a a BP-7=2v5b > ape a’ — 7h a — oF 22 =Vv5b 2 2 ae — The Since, 2,7,aand bare integers 2a) is rational A rational number cannot be equal to an irrational number. So, our assumption is wrong. Hence, 2v3 + V5 is irrational. Revisiting rational numbers and their decimal expansion Revisiting Rational Numbers Rational Numbers and their Decimal Expansion 
