REAL NUMBERS 1not©toNbCeErReTpublished 1REAL NUMBERS1.1 IntroductionIn Class IX, you began your exploration of the world of real numbers and encounteredirrational numbers. We continue our discussion on real numbers in this chapter. Webegin with two very important properties of positive integers in Sections 1.2 and 1.3,namely the Euclid’s division algorithm and the FundamentalTheorem of Arithmetic. Euclid’s division algorithm, as the name suggests, has to do with divisibility ofintegers. Stated simply, it says any positive integer a can be divided by another positiveinteger b in such a way that it leaves a remainder r that is smaller than b. Many of youprobably recognise this as the usual long division process. Although this result is quiteeasy to state and understand, it has many applications related to the divisibility propertiesof integers. We touch upon a few of them, and use it mainly to compute the HCF oftwo positive integers. The Fundamental Theorem ofArithmetic, on the other hand, has to do somethingwith multiplication of positive integers. You already know that every composite numbercan be expressed as a product of primes in a unique way —this important fact is theFundamental Theorem ofArithmetic. Again, while it is a result that is easy to state andunderstand, it has some very deep and significant applications in the field of mathematics.We use the Fundamental Theorem of Arithmetic for two main applications. First, weuse it to prove the irrationality of many of the numbers you studied in Class IX, such as2 , 3 and 5 . Second, we apply this theorem to explore when exactly the decimal pexpansion of a rational number, say q (q ≠ 0) , is terminating and when it is non-terminating repeating. We do so by looking at the prime factorisation of the denominatorq of p . You will see that the prime factorisation of q will completely reveal the nature q pof the decimal expansion of q . So let us begin our exploration.
not©toNbCeErReTpublished2 MATHEMATICS1.2 Euclid’s Division LemmaConsider the following folk puzzle*. A trader was moving along a road selling eggs. An idler who didn’t havemuch work to do, started to get the trader into a wordy duel. This grew into afight, he pulled the basket with eggs and dashed it on the floor. The eggs broke.The trader requested the Panchayat to ask the idler to pay for the broken eggs.The Panchayat asked the trader how many eggs were broken. He gave thefollowing response: If counted in pairs, one will remain; If counted in threes, two will remain; If counted in fours, three will remain; If counted in fives, four will remain; If counted in sixes, five will remain; If counted in sevens, nothing will remain; My basket cannot accomodate more than 150 eggs. So, how many eggs were there? Let us try and solve the puzzle. Let the numberof eggs be a. Then working backwards, we see that a is less than or equal to 150: If counted in sevens, nothing will remain, which translates to a = 7p + 0, forsome natural number p. If counted in sixes, a = 6 q+ 5, for some natural number q. If counted in fives, four will remain. It translates to a = 5 w + 4, for some naturalnumber w. If counted in fours, three will remain. It translates to a = 4s + 3, for some naturalnumber s. If counted in threes, two will remain. It translates to a = 3t + 2, for some naturalnumber t. If counted in pairs, one will remain. It translates to a = 2 u + 1, for some naturalnumber u. That is, in each case, we have a and a positive integer b (in our example,b takes values 7, 6, 5, 4, 3 and 2, respectively) which divides a and leaves a remainderr (in our case, r is 0, 5, 4, 3, 2 and 1, respectively), that is smaller than b. The* This is modified form of a puzzle given in ‘Numeracy Counts!’by A. Rampal, and others.
REAL NUMBERS 3moment we write down such equations we are using Euclid’s division lemma,which is given in Theorem 1.1. Getting back to our puzzle, do you have any idea how you will solve it? Yes!Youmust look for the multiples of 7 which satisfy all the conditions. By trial and error(using the concept of LCM), you will find he had 119 eggs.not©toNbCeErReTpublished In order to get a feel for what Euclid’s division lemma is, consider the followingpairs of integers:17, 6; 5, 12; 20, 4 Like we did in the example, we can write the following relations for each suchpair:17 = 6 × 2 + 5 (6 goes into 17 twice and leaves a remainder 5)5 = 12 × 0 + 5 (This relation holds since 12 is larger than 5)20 = 4 × 5 + 0 (Here 4 goes into 20 five-times and leaves no remainder) That is, for each pair of positive integers a and b, we have found whole numbersq and r, satisfying the relation: a = bq + r, 0 ≤ r < bNote that q or r can also be zero. Why don’t you now try finding integersq and r for the following pairs of positiveintegers a and b?(i) 10, 3; (ii) 4, 19; (iii) 81, 3 Did you notice that q and r are unique? These are the only integers satisfying theconditions a = bq + r, where 0 ≤ r < b. You may have also realised that this is nothingbut a restatement of the long division process you have been doing all these years, andthat the integers q and r are called the quotient and remainder, respectively.A formal statement of this result is as follows :Theorem 1.1 (Euclid’s Division Lemma) : Given positive integers a and b,there exist unique integers q and r satisfying a = bq + r, 0 ≤ r < b. This result was perhaps known for a long time, but was first recorded in Book VII of Euclid’s Elements. Euclid’s division algorithm is based on this lemma.
4 MATHEMATICSAn algorithm is a series of well defined stepswhich gives a procedure for solving a type ofproblem. The wordalgorithm comes from the nameof the 9th century Persian mathematicianal-Khwarizmi. In fact, even the word ‘algebra’is derived from a book, he wrote, called Hisabal-jabr w’al-muqabala.not©toNbCeErReTpublishedA lemma is a proven statement used for Muhammad ibn Musa al-Khwarizmiproving another statement. (A.D. 780 – 850) Euclid’s division algorithm is a technique to compute the Highest Common Factor(HCF) of two given positive integers. Recall that the HCF of two positive integers aand b is the largest positive integer d that divides both a and b. Let us see how the algorithm works, through an example first. Suppose we needto find the HCF of the integers 455 and 42. We start with the larger integer, that is,455. Then we use Euclid’s lemma to get 455 = 42 × 10 + 35 Now consider the divisor 42 and the remainder 35, and apply the division lemmato get 42 = 35 × 1 + 7 Now consider the divisor 35 and the remainder 7, and apply the division lemmato get 35 = 7 × 5 + 0 Notice that the remainder has become zero, and we cannot proceed any further.We claim that the HCF of 455 and 42 is the divisor at this stage, i.e., 7. You can easilyverify this by listing all the factors of 455 and 42. Why does this method work? Itworks because of the following result. So, let us state Euclid’s division algorithm clearly. To obtain the HCF of two positive integers, say c and d, with c > d, followthe steps below:Step 1 : Apply Euclid’s division lemma, to c and d. So, we find whole numbers, q and r such that c = dq + r, 0 ≤ r < d.Step 2 : If r = 0, d is the HCF of c and d. If r ≠ 0, apply the division lemma to d and r.Step 3 : Continue the process till the remainder is zero. The divisor at this stage will be the required HCF.
REAL NUMBERS 5not©toNbCeErReTpublished This algorithm works because HCF (c, d) = HCF (d, r) where the symbol HCF (c, d) denotes the HCF of c and d, etc. Example 1 : Use Euclid’s algorithm to find the HCF of 4052 and 12576. Solution : Step 1 : Since 12576 > 4052, we apply the division lemma to 12576 and 4052, to get 12576 = 4052 × 3 + 420 Step 2 : Since the remainder 420≠ 0, we apply the division lemma to 4052 and 420, to get 4052 = 420 × 9 + 272 Step 3 : We consider the new divisor 420 and the new remainder 272, and apply the division lemma to get 420 = 272 × 1 + 148 We consider the new divisor 272 and the new remainder 148, and apply the division lemma to get 272 = 148 × 1 + 124 We consider the new divisor 148 and the new remainder 124, and apply the division lemma to get 148 = 124 × 1 + 24 We consider the new divisor 124 and the new remainder 24, and apply the division lemma to get 124 = 24 × 5 + 4 We consider the new divisor 24 and the new remainder 4, and apply the division lemma to get 24 = 4 × 6 + 0 The remainder has now become zero, so our procedure stops. Since the divisor at this stage is 4, the HCF of 12576 and 4052 is 4. Notice that 4 = HCF (24, 4) = HCF (124, 24) = HCF (148, 124) = HCF (272, 148) = HCF (420, 272) = HCF (4052, 420) = HCF (12576, 4052). Euclid’s division algorithm is not only useful for calculating the HCF of very large numbers, but also because it is one of the earliest examples of an algorithm that a computer had been programmed to carry out. Remarks : 1. Euclid’s division lemma and algorithm are so closely interlinked that people often call former as the division algorithm also. 2. Although Euclid’s Division Algorithm is stated for only positive integers, it can be extended for all integers except zero, i.e., b ≠ 0. However, we shall not discuss this aspect here.
not©toNbCeErReTpublished6 MATHEMATICS Euclid’s division lemma/algorithm has several applications related to finding properties of numbers. We give some examples of these applications below: Example 2 : Show that every positive even integer is of the form 2q, and that every positive odd integer is of the form 2q + 1, where q is some integer. Solution : Let a be any positive integer and b = 2. Then, by Euclid’s algorithm, a = 2q + r, for some integer q ≥ 0, and r = 0 or r = 1, because 0 ≤ r < 2. So, a = 2q or 2q + 1. If a is of the form 2q, then a is an even integer. Also, a positive integer can be either even or odd. Therefore, any positive odd integer is of the form 2q + 1. Example 3 : Show that any positive odd integer is of the form 4q + 1 or 4q + 3, where q is some integer. Solution : Let us start with taking a, where a is a positive odd integer. We apply the division algorithm with a and b = 4. Since 0 ≤ r < 4, the possible remainders are 0, 1, 2 and 3. That is, a can be 4q, or 4q + 1, or 4q + 2, or 4q + 3, where q is the quotient. However, since a is odd, a cannot be 4q or 4q + 2 (since they are both divisible by 2). Therefore, any odd integer is of the form 4q + 1 or 4q + 3. Example 4 : A sweetseller has 420 kaju barfis and 130 badam barfis. She wants to stack them in such a way that each stack has the same number, and they take up the least area of the tray. What is the number of that can be placed in each stack for this purpose? Solution : This can be done by trial and error. But to do it systematically, we find HCF (420, 130). Then this number will give the maximum number of barfis in each stack and the number of stacks will then be the least. The area of the tray that is used up will be the least. Now, let us use Euclid’s algorithm to find their HCF. We have : 420 = 130 × 3 + 30 130 = 30 × 4 + 10 30 = 10 × 3 + 0 So, the HCF of 420 and 130 is 10. Therefore, the sweetseller can make stacks of 10 for both kinds of barfi.
REAL NUMBERS 7 EXERCISE 1.11. Use Euclid’s division algorithm to find the HCF of :(i) 135 and 225 (ii) 196 and 38220 (iii) 867 and 2552. Show that any positive odd integer is of the form 6q + 1, or 6q + 3, or 6q + 5, where q is some integer.not©toNbCeErReTpublished3. 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?4. Use Euclid’s division lemma to show that the square of any positive integer is either of the 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 squareeach of these and show that they can be rewritten in the form 3m or 3m + 1.]5. Use Euclid’s division lemma to show that the cube of any positive integer is of the form 9m, 9m + 1 or 9m + 8.1.3 The Fundamental Theorem ofArithmeticIn your earlier classes, you have seen that any natural number can be written as aproduct of its prime factors. For instance, 2 = 2, 4 = 2 × 2, 253 = 11 × 23, and so on.Now, let us try and look at natural numbers from the other direction. That is, can anynatural number be obtained by multiplying prime numbers? Let us see. Take any collection of prime numbers, say 2, 3, 7, 11 and 23. If we multiplysome or all of these numbers, allowing them to repeat as many times as we wish,we can produce a large collection of positive integers (In fact, infinitely many).Let us list a few :7 × 11 × 23 = 1771 3 × 7 × 11 × 23 = 53132 × 3 × 7 × 11 × 23 = 10626 23 × 3 × 73 = 8232 22 × 3 × 7 × 11 × 23 = 21252and so on. Now, let us suppose your collection of primes includes all the possible primes.What is your guess about the size of this collection? Does it contain only a finitenumber of integers, or infinitely many? Infact, there are infinitely many primes. So, ifwe combine all these primes in all possible ways, we will get an infinite collection ofnumbers, all the primes and all possible products of primes. The question is – can weproduce all the composite numbers this way? What do you think? Do you think thatthere may be a composite number which is not the product of powers of primes?Before we answer this, let us factorise positive integers, that is, do the opposite ofwhat we have done so far.
not©toNbCeErReTpublished8 MATHEMATICS We are going to use the factor tree with which you are all familiar. Let us take some large number, say, 32760, and factorise it as shown : 32760 2 16380 2 8190 2 4095 3 1365 3 455 5 91 7 13 So we have factorised 32760 as 2 × 2 × 2 × 3 × 3 × 5 × 7 × 13 as a product of primes, i.e., 32760 = 23 × 32 × 5 × 7 × 13 as a product of powers of primes. Let us try another number, say, 123456789. This can be written as 32 × 3803 × 3607. Of course, you have to check that 3803 and 3607 are primes! (Try it out for several other natural numbers yourself.) This leads us to a conjecture that every composite number can be written as the product of powers of primes. In fact, this statement is true, and is called the Fundamental Theorem of Arithmetic because of its basic crucial importance to the study of integers. Let us now formally state this theorem. Theorem 1.2 (Fundamental Theorem of Arithmetic) : Every composite number can be expressed ( factorised ) as a product of primes, and this factorisation is unique, apart from the order in which the prime factors occur.
REAL NUMBERS 9An equivalent version of Theorem 1.2 was probably firstrecorded as Proposition 14 of Book IX in Euclid’sElements, before it came to be known as the FundamentalTheorem of Arithmetic. However, the first correct proofwas given by Carl Friedrich Gauss in his DisquisitionesArithmeticae .not©toNbCeErReTpublishedCarl Friedrich Gauss is often referred to as the ‘Prince ofMathematicians’ and is considered one of the threegreatest mathematicians of all time, along withArchimedesand Newton. He has made fundamental contributions to Carl Friedrich Gaussboth mathematics and science. (1777 – 1855) The Fundamental Theorem of Arithmetic says that every composite numbercan be factorised as a product of primes. Actually it says more. It says that givenany 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 anycomposite 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, forexample, we regard 2 × 3 × 5 × 7 as the same as 3 × 5 × 7 × 2, or any otherpossible order in which these primes are written. This fact is also stated in thefollowing form: The prime factorisation of a natural number is unique, except for the orderof its factors.In general, given a composite number x, we factorise it as x = p1p2 ... pn, wherep , p ,..., p are primes and written in ascending order, i.e., p ≤ p12 n 12≤ . . . ≤ pn. If we combine the same primes, we will get powers of primes. For example,32760 = 2 × 2 × 2 × 3 × 3 × 5 × 7 × 13 = 23 × 32 × 5 × 7 × 13 Once we have decided that the order will be ascending, then the way the numberis factorised, is unique. The Fundamental Theorem of Arithmetic has many applications, both withinmathematics and in other fields. Let us look at some examples.Example 5 : Consider the numbers 4n, where n is a natural number. Check whetherthere is any value of n for which 4n ends with the digit zero.Solution : If the number 4n, for any n, were to end with the digit zero, then it would bedivisible by 5. That is, the prime factorisation of 4n would contain the prime 5. This is
10 MATHEMATICSnot possible because 4n = (2)2n; so the only prime in the factorisation of 4n is 2. So, theuniqueness of the Fundamental Theorem of Arithmetic guarantees that there are noother primes in the factorisation of 4n. So, there is no natural number n for which 4nends with the digit zero. You have already learnt how to find the HCF and LCM of two positive integersusing the Fundamental Theorem of Arithmetic in earlier classes, without realising it!This method is also called the prime factorisation method. Let us recall this methodthrough an example.not©toNbCeErReTpublishedExample 6 : Find the LCM and HCF of 6 and 20 by the prime factorisation method.Solution : We have : 6 = 21 × 31 and 20 = 2 × 2 × 5 = 22 × 51.You can find HCF(6, 20) = 2 and LCM(6, 20) = 2 × 2 × 3 × 5 = 60, as done in yourearlier classes.Note that HCF(6, 20) = 21 = Product of the smallest power of each common prime factor in the numbers.LCM (6, 20) = 22 × 31 × 51 = Product of the greatest power of each prime factor, involved in the numbers. From the example above, you might have noticed that HCF(6, 20) × LCM(6, 20)= 6 × 20. In fact, we can verify that for any two positive integers a and b,HCF (a, b) × LCM (a, b) = a × b. We can use this result to find the LCM of twopositive integers, if we have already found the HCF of the two positive integers.Example 7 : Find the HCF of 96 and 404 by the prime factorisation method. Hence,find their LCM.Solution : The prime factorisation of 96 and 404 gives : 96 = 25 × 3, 404 = 22 × 101Therefore, the HCF of these two integers is 22 = 4.Also, LCM (96, 404) = 96 × 404 = 96 × 404 = 9696 HCF(96, 404) 4Example 8 : Find the HCF and LCM of 6, 72 and 120, using the prime factorisationmethod.Solution : We have : 6 = 2 × 3, 72 = 23 × 32, 120 = 23 × 3 × 5Here, 21 and 31 are the smallest powers of the common factors 2 and 3, respectively.
REAL NUMBERS 11So, HCF (6, 72, 120) = 21 × 31 = 2 × 3 = 6 23, 32 and 51 are the greatest powers of the prime factors 2, 3 and 5 respectivelyinvolved in the three numbers.So, LCM (6, 72, 120) = 23 × 32 × 51 = 360not©toNbCeErReTpublishedRemark : Notice, 6 × 72 × 120 ≠ HCF (6, 72, 120) × LCM (6, 72, 120). So, theproduct of three numbers is not equal to the product of their HCF and LCM. EXERCISE 1.21. Express each number as a product of its prime factors:(i) 140 (ii) 156 (iii) 3825 (iv) 5005 (v) 74292. 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 543. Find the LCM and HCF of the following integers by applying the prime factorisation method.(i) 12, 15 and 21 (ii) 17, 23 and 29 (iii) 8, 9 and 254. Given that HCF (306, 657) = 9, find LCM (306, 657).5. Check whether 6n can end with the digit 0 for any natural number n .6. Explain why 7 × 11 × 13 + 13 and 7 × 6 × 5 × 4 × 3 × 2 × 1 + 5 are composite numbers.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?1.4 Revisiting Irrational NumbersIn Class IX, you were introduced to irrational numbers and many of their properties.You studied about their existence and how the rationals and the irrationals togethermade up the real numbers. You even studied how to locate irrationals on the numberline. However, we did not prove that they were irrationals. In this section, we willprove that 2 , 3 , 5 and, in general, p is irrational, where p is a prime. One ofthe theorems, we use in our proof, is the Fundamental Theorem of Arithmetic.Recall, a number ‘s’ is called irrational if it cannot be written in the form p, qwhere p and q are integers and q ≠ 0. Some examples of irrational numbers, with
12 MATHEMATICSwhich you are already familiar, are : 2, 3, 15 , π, − 2 , 0.10110111011110 . . . , etc. 3 Before we prove that 2 is irrational, we need the following theorem, whoseproof is based on the Fundamental Theorem of Arithmetic.not©toNbCeErReTpublishedTheorem 1.3 : Let p be a prime number. If p divides a2, then p divides a, wherea is a positive integer.*Proof : Let the prime factorisation of a be as follows :a = p p . . . p , where p ,p , . . ., p are primes, not necessarily distinct.12 n 12 nTherefore, a2 = ( p1p2 . . . pn)( p1p2 . . . pn) = p12 p22 . . . p2n.Now, we are given that p divides a2. Therefore, from the Fundamental Theorem ofArithmetic, it follows that p is one of the prime factors of a2. However, using theuniqueness part of the Fundamental Theorem of Arithmetic, we realise that the onlyprime factors of a2 are p1, p2, . . ., pn. So p is one of p1, p2, . . ., pn.Now, since a = p p . . . p , p divides a. 12 nWe are now ready to give a proof that 2 is irrational.The proof is based on a technique called ‘proof by contradiction’. (This technique isdiscussed in some detail in Appendix 1).Theorem 1.4 : 2 is irrational.Proof : Let us assume, to the contrary, that 2 is rational. rSo, we can find integers r and s (≠ 0) such that 2 = s .Suppose r and s have a common factor other than 1. Then, we divide by the commonfactor to get 2 = a , where a and b are coprime. bSo, b 2 = a.Squaring on both sides and rearranging, we get 2b2 = a2. Therefore, 2 divides a2.Now, by Theorem 1.3, it follows that 2 divides a.So, we can write a = 2c for some integer c.* Not from the examination point of view.
REAL NUMBERS 13Substituting for a, we get 2b2 = 4c2, that is, b2 = 2c2.This means that 2 divides b2, and so 2 divides b (again using Theorem 1.3 with p = 2).Therefore, a and b have at least 2 as a common factor.not©toNbCeErReTpublishedBut this contradicts the fact that a and b have no common factors other than 1.This contradiction has arisen because of our incorrect assumption that 2 is rational.So, we conclude that 2 is irrational. Example 9 : Prove that 3 is irrational.Solution : Let us assume, to the contrary, that 3 is rational.That is, we can find integers a and b (≠ 0) such that 3 = a⋅ bSuppose a and b have a common factor other than 1, then we can divide by thecommon factor, and assume that a and b are coprime.So, b 3 = a⋅Squaring on both sides, and rearranging, we get 3b2 = a2.Therefore, a2 is divisible by 3, and by Theorem 1.3, it follows that a is also divisibleby 3.So, we can write a = 3c for some integer c.Substituting for a, we get 3b2 = 9c2, that is, b2 = 3c2.This means that b2 is divisible by 3, and so b is also divisible by 3 (using Theorem 1.3with p = 3).Therefore, a and b have at least 3 as a common factor.But this contradicts the fact that a and b are coprime.This contradiction has arisen because of our incorrect assumption that 3 is rational.So, we conclude that 3 is irrational. In Class IX, we mentioned that :z the sum or difference of a rational and an irrational number is irrational andz the product and quotient of a non-zero rational and irrational number is irrational.We prove some particular cases here.
14 MATHEMATICSExample 10 : Show that 5 – 3 is irrational.Solution : Let us assume, to the contrary, that 5 – 3 is rational.That is, we can find coprime a and b (b ≠ 0) such that 5 − 3 = a ⋅not©toNbCeErReTpublished bTherefore, 5 − a = 3 ⋅ bRearranging this equation, we get 3 = 5 – a = 5b − a ⋅ bbSince a and b are integers, we get 5 – a is rational, and so 3 is rational. bBut this contradicts the fact that 3 is irrational.This contradiction has arisen because of our incorrect assumption that 5 – 3 isrational.So, we conclude that 5 − 3 is irrational.Example 11 : Show that 3 2 is irrational.Solution : Let us assume, to the contrary, that 3 2 is rational.That is, we can find coprime a and b (b ≠ 0) such that 3 2 = a ⋅ bRearranging, we get 2 = a ⋅ 3b aSince 3, a and b are integers, is rational, and so 2 is rational. 3bBut this contradicts the fact that 2 is irrational.So, we conclude that 3 2 is irrational. EXERCISE 1.31. Prove that 5 is irrational.2. Prove that 3 + 2 5 is irrational.3. Prove that the following are irrationals :(i) 1 (ii) 7 5 (iii) 6 + 2 2
REAL NUMBERS 151.5 Revisiting Rational Numbers and Their Decimal ExpansionsIn Class IX, you studied that rational numbers have either a terminating decimalexpansion or a non-terminating repeating decimal expansion. In this section, we aregoing to consider a rational number, say p (q ≠ 0) , and explore exactly when the p qnot©toNbCeErReTpublisheddecimal expansion of q is terminating and when it is non-terminating repeating(or recurring). We do so by considering several examples. Let us consider the following rational numbers : (i) 0.375 (ii) 0.104 (iii) 0.0875 (iv) 23.3408.Now (i) 0.375 = 375 = 375 (ii) 0.104 = 104 = 104 1000 103 1000 103 (iii) 0.0875 = 875 = 875 (iv) 23.3408 = 233408 = 233408 10000 104 10000 104 As one would expect, they can all be expressed as rational numbers whosedenominators are powers of 10. Let us try and cancel the common factors betweenthe numerator and denominator and see what we get : (i) 0.375 = 375 = 3 × 53 = 3 (ii) 0.104 = 104 = 13 × 23 = 13 103 23 × 53 23 103 23 × 53 53 (iii) 0.0875 = 875 = 7 (iv) 23.3408 = 233408 = 22 × 7 ×521 104 24 × 5 104 54 Do you see any pattern? It appears that, we have converted a real numberwhose decimal expansion terminates into a rational number of the form p , where p qand q are coprime, and the prime factorisation of the denominator (that is, q) has onlypowers of 2, or powers of 5, or both. We should expect the denominator to look likethis, since powers of 10 can only have powers of 2 and 5 as factors. Even though, we have worked only with a few examples, you can see that anyreal number which has a decimal expansion that terminates can be expressed as arational number whose denominator is a power of 10. Also the only prime factors of 10are 2 and 5. So, cancelling out the common factors between the numerator and thedenominator, we find that this real number is a rational number of the form p , where qthe prime factorisation of q is of the form 2n5m, and n, m are some non-negative integers. Let us write our result formally:
16 MATHEMATICSTheorem 1.5 : Let x be a rational number whose decimal expansion terminates.Then x can be expressed in the form p, where p and q are coprime, and the qprime factorisation of q is of the form 2n5m, where n, m are non-negative integers.not©toNbCeErReTpublished You are probably wondering what happens the other way round in Theorem 1.5.That is, if we have a rational number of the form p , and the prime factorisation of q qpis of the form 2n5m, where n, m are non negative integers, then does q have aterminating decimal expansion? Let us see if there is some obvious reason why this is true. You will surely agreethat any rational number of the form a , where b is a power of 10, will have a terminating bdecimal expansion. So it seems to make sense to convert a rational number of theform p , where q is of the form 2n5m, to an equivalent rational number of the form a , qbwhere b is a power of 10. Let us go back to our examples above and work backwards.(i) 3 = 3 = 3 × 53 = 375 = 0.375 8 23 23 × 53 103(ii) 13 = 13 = 13 × 23 = 104 = 0.104 125 53 23 × 53 103(iii) 7 = 7 = 7 × 53 = 875 = 0.0875 80 24 × 5 24 × 54 104(iv) 14588 = 22 × 7 × 521 = 26 × 7 × 521 = 233408 = 23.3408 625 54 24 × 54 104 So, these examples show us how we can convert a rational number of the form p , where q is of the form 2n5m, to an equivalent rational number of the form a ,qbwhere b is a power of 10. Therefore, the decimal expansion of such a rational numberterminates. Let us write down our result formally.Theorem 1.6 : Let x = p be a rational number, such that the prime factorisation qof q is of the form 2n5m, where n, m are non-negative integers. Then x has adecimal expansion which terminates.
REAL NUMBERS 17We are now ready to move on to the rational numbers 0.1428571whose decimal expansions are non-terminating and recurring. 7 10Once again, let us look at an example to see what is going on. 7We refer to Example 5, Chapter 1, from your Class IX 30not©toNbCeErReTpublished1 28textbook, namely, . Here, remainders are 3, 2, 6, 4, 5, 1, 3, 20 7 142, 6, 4, 5, 1, . . . and divisor is 7. 60Notice that the denominator here, i.e., 7 is clearly not of 56the form 2n5m. Therefore, from Theorems 1.5 and 1.6, we 40 1 35know that will not have a terminating decimal expansion. 507 49Hence, 0 will not show up as a remainder (Why?), and the 10remainders will start repeating after a certain stage. So, we 7will have a block of digits, namely, 142857, repeating in the 301quotient of .71What we have seen, in the case of 7 , is true for any rational number not coveredby Theorems 1.5 and 1.6. For such numbers we have : pTheorem 1.7 : Let x = q be a rational number, such that the prime factorisationof q is not of the form 2n5m, where n, m are non-negative integers. Then, x has adecimal expansion which is non-terminating repeating (recurring). From the discussion above, we can conclude that the decimal expansion ofevery rational number is either terminating or non-terminating repeating. EXERCISE 1.41. 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: 13 17 64 15(i) 3125 (ii) 8 (iii) 455 (iv) 1600 29 23 129 6(v) 343 (vi) 2352 (vii) 22577 5 (viii) 15 35 77(ix) (x) 50 210
18 MATHEMATICS2. Write down the decimal expansions of those rational numbers in Question 1 above which have terminating decimal expansions.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 , what can qnot©toNbCeErReTpublishedyou say about the prime factors of q? (i) 43.123456789 (ii) 0.120120012000120000. . . (iii) 43.1234567891.6 SummaryIn this chapter, you have studied the following points: 1. Euclid’s division lemma : Given positive integers a and b, there exist whole numbers q and r satisfying a = bq + r, 0 ≤ r < b. 2. Euclid’s division algorithm : This is based on Euclid’s division lemma. According to this, the HCF of any two positive integers a and b, with a > b, is obtained as follows: Step 1 : Apply the division lemma to find q and r where a = bq + r, 0 ≤ r < b. Step 2 : If r = 0, the HCF is b. If r ≠ 0, apply Euclid’s lemma to b and r. Step 3 : Continue the process till the remainder is zero. The divisor at this stage will be HCF (a, b). Also, HCF(a, b) = HCF(b, r). 3. The Fundamental Theorem of Arithmetic : Every composite number can be expressed (factorised) as a product of primes, and this factorisation is unique, apart from the order in which the prime factors occur. 4. If p is a prime and p divides a2, then p divides a, where a is a positive integer.5. To prove that 2, 3 are irrationals.6. Let x be a rational number whose decimal expansion terminates. Then we can express x p in the form 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.7. Let x = p be a rational number, such that the prime factorisation of q is of the form 2n5m, q where n, m are non-negative integers. Then x has a decimal expansion which terminates.8. Let x = p be a rational number, such that the prime factorisation of q is not of the form q 2n 5m, where n, m are non- negative integers. Then x has a decimal expansion which is non-terminating repeating (recurring).
REAL NUMBERS 19 A NOTE TO THE READERYou have seen that :HCF ( p, q, r) × LCM (p, q, r) ≠ p × q × r, where p, q, r are positive integers(see Example 8). However, the following results hold good for three numbersp, q and r : p ⋅q ⋅r ⋅HCF(p, q, r) LCM (p, q, r) = HCF( p, q) ⋅ HCF(q,r) ⋅ HCF( p, r) p ⋅q ⋅r⋅ LCM(p, q, r) HCF (p, q, r) = LCM( p, q) ⋅ LCM(q, r) ⋅ LCM( p, r)not©toNbCeErReTpublished