The theory of continued fractions and the approximations of irrational numbers

Transcription

1 교육학석사학위청구논문 연분수이론과무리수의근사 The theory of continued fractions and the approximations of irrational numbers 202 년 2 월 인하대학교교육대학원 수학교육전공 정미라

2 교육학석사학위청구논문 연분수이론과무리수의근사 The theory of continued fractions and the approximations of irrational numbers 202 년 2 월 지도교수김재문 이논문을석사학위논문으로제출함.

3 본논문을정미라의석사학위논문으로인준함. 202 년 2 월 주심 ( 인 ) 부심 ( 인 ) 부심 ( 인 )

4 Contents Abstract 국문요약 i ii Introduction 2 Preliminaries 3 3 Approximations of irrational numbers 0 References 5

5 ABSTRACT In this paper, we study approximations of a given irrational number by rational numbers. Section is introductory. We briefly explain what the continued fraction is about. In section 2, we discuss infinite continued fractions and study several theorems that are used in section 3. Finally, in section 3, we examine approximations of a given irrational number by rational numbers by using the theory of continued fractions.

6 국문초록 이논문에서는주어진무리수를유리수로근사하는것에대해논해보고자한다. 장에서는연분수에대해간단히언급하고연분수의형태를알아본다. 2장에서는무한연분수를논의하기위해필요한내용을정리하고, 3장을논의하기전에알아야할정리와따름정리들을소개하면서그정리들을증명해본다. 마지막으로 3장에서는 2장에서살펴본연분수이론을이용하여무리수를유리수로근사하는것에대해연구해본다.

7 Introduction The aim of this paper is to study approximations of a given irrational number by rational numbers. For example, as we will see later in section 3, the rational numbers 3, , and 06 give good approximations of π with errors only about 0.459, , and , respectively. Our approach is based on the theory of continued fractions. The theory has a variety of applications. For instance, the Pell s equation x 2 dy 2 = can be solved by using the theory. It can also be applied to the theory of cryptography. In this paper, however, we do not discuss them. We only focus on the application of the theory to the approximations of irrational numbers. A continued fraction is a real number of the form < x 0, x, x 2, >= x 0 + x + x , where x i > 0 for i. When all the x i s are integers, it is called a simple continued fraction. A finite simple continued fraction <a 0, a,, a n > obviously represents a rational number. On the other hand, for an infinite simple continued fraction < a 0, a, a 2, >, we need to make it clear what it means. The most natural way to define < a 0, a, a 2, > would be

8 < a 0, a, a 2, > = lim n c n, where c n = <a 0, a,, a n > if this limit exists. We will see that the limit indeed exists, and that the value is, in fact, an irrational number. All of these are justified in section 2. We also discuss how to find the infinite simple continued fraction of a given irrational number in the same section. In section 3, we examine approximations of a given irrational number α by rational numbers, which is the main topic of this paper. We first represent α as an infinite simple continued fraction α = <a 0, a, a 2, >. Then we have a sequence {c n } of rational numbers, where c n = <a 0, a,, a n >. We will investigate this sequence {c n }, and see that it gives good (best in some sense) approximations of α. 2

9 2 Preliminaries In this section, we briefly review some basic concepts of continued fractions. Definition 2.. For real numbers x 0, x,, x n with x i > 0 for i, we define a finite continued fraction < x 0, x,, x n > by < x 0, x,, x n >= x 0 + x x n. When all the entries are integers, it is called a simple continued fraction. Clearly, a finite simple continued fraction represents a rational number. For example, < 2, 3,, 4 >= = Conversely, every rational number r has exactly two representations as a finite simple continued fraction, one with an even number of terms and the other with an odd number. This happens because <a 0, a,, a n > = <a 0, a,, a n, > if a n >. To discuss an infinite simple continued fraction, we need to define sequences {p k } and {q k } from a given sequence {a k }. Let a 0, a, a 2, be an infinite sequence of integers, all of them positive except possibly a 0. We define s- 3

10 equences {p k } and {q k } of integers in a recursive way as follows: i) p 2 = 0, p =, p k = a k p k + p k 2, k 0. ii) q 2 =, q = 0, q k = a k q k + q k 2, k 0. Then Theorem 2.2 below can be easily proved by induction. Theorem 2.2. For each positive real number x and for each nonnegative integer k, we have < a 0, a, a 2,, a k, x > = xp k + p k xq k + q k. Corollary 2.3. For each nonnegative integer k, we have < a 0, a,, a k > = p k q k. Proof. Substitute a k+ for x in Theorem 2.2 to obtain < a 0, a,, a k, a k+ >= a k+p k + p k a k+ q k + q k = p k+ q k+. The sequences {p k } and {q k } satisfy the following two important and interesting properties. Theorem 2.4. For each integer k, we have p k q k p k q k = ( ) k. () Proof. We use induction on k. It is easy to check the equality when k = or 0. Assume p k q k p k q k = ( ) k. Then p k+ q k p k q k+ = (a k+ p k + p k )q k p k (a k+ q k + q k ) = a k+ p k q k + p k q k a k+ p k q k p k q k 4

11 = p k q k p k q k = ( ) k = ( ) k. Therefore we have p k q k p k q k = ( ) k. Theorem 2.5. For each nonnegative integer k, we have p k q k 2 p k 2 q k = ( ) k a k. Proof. By the definitions of the sequences {p k }, {q k } and Theorem 2.4, we have p k q k 2 p k 2 q k = (a k p k + p k 2 )q k 2 p k 2 (a k q k + q k 2 ) = a k p k q k 2 + p k 2 q k 2 a k p k 2 q k p k 2 q k 2 = a k (p k q k 2 p k 2 q k ) = ( ) k 2 a k = ( ) k a k. Now we return to the discussion about the infinite simple continued fraction <a 0, a, a 2, >. We define <a 0, a, a 2, > by < a 0, a, a 2, >= lim k c k, where c k = <a 0, a,, a k >. For this to be meaningful, the limit must exist, which is guaranteed by the next theorem. The kth term c k will be called the kth convergent. 5

12 Theorem 2.6. Let c k = <a 0, a,, a k > = p k q k. Then (i) The even convergents form a strictly increasing sequence, that is, c 0 < c 2 < c 4 <. (ii) The odd convergents form a strictly decreasing sequence, that is, c > c 3 > c 5 >. (iii) Every odd convergent is greater than every even convergent. (iv) lim k c k exists. Proof. By Theorem 2.5, we have p k q k p k 2 q k 2 = ( )k a k q k 2 q k. Since the q s are all positive and a k > 0, we obtain (i) and (ii). To prove (iii), let us observe that if t is a nonnegative integer, from the equation () with k = 2t +, we find that or Certainly, then, p 2t+ q 2t+ p 2t q 2t = c 2t+ = c 2t + q 2t+ q 2t q 2t+ q 2t. (2) c 2t+ > c 2t (3) Now let r and s be arbitrary nonnegative integers and let us establish part (iii) of the theorem by showing that c 2r+ > c 2s. If r = s, we have the desired result by the equation (3). If r > s, then the equation (3) and part (i) of the present theorem show that c 2r+ > c 2r > c 2s. 6

13 Similarly, if r < s, by using part (ii) of the present theorem, we find that c 2r+ > c 2s+ > c 2s. Hence, in every case, c 2r+ > c 2s, and this completes the proof of (iii). By (i) and (iii), the sequence {c 2k } is an increasing sequence which is bounded above by any odd convergent, and therefore lim k c 2k exists and is greater than each even convergent. In like manner, using (ii) and (iii), we know that lim k c 2k+ exists and is less than each odd convergent. Moreover, from the equation (2), we have lim c 2k+ = lim c 2k + lim. k k k q 2k+ q 2k Since q k k, this last limit is zero, and we conclude that lim c 2k+ = lim c 2k. k k We have shown that the sequence of odd convergents has the same limit as the sequence of even convergents, and this implies that Thus, lim k c k exists. lim c k = lim c 2k = lim c 2k+. k k k We finish this section with a brief explanation of how to obtain an infinite simple continued fraction of a given irrational number α. First of all, notice that < a 0, a, >= a 0 + < a, a 2, >, where a 0 is the integral part of <a 0, a, >. Now suppose that α is an irra- 7

14 tional number which, for notational convenience, we shall also denote by α 0. Let a 0 = [α], the integral part of α 0. Then, since α 0 is irrational, we have 0 < α 0 a 0 <. We define α = /(α 0 a 0 ) and observe that α is irrational since α 0 is irrational, and that α >. We may remark that the equation defining α may also be written in the form α 0 = a 0 +. α We may define an irrational number α 2 in terms of α in the same way that α was defined in terms of α 0. In general, by recursion, we define an infinite sequence a k as follows, in which we use the notation a k = [α k ]: α 0 = α, α k = a k +, (k =, 2, ). α k Then since α k an irrational number, so is α k ; hence by induction all α k are irrational and called the complete quotients. For each positive integer k we must have 0 < α k a k <, and we see that α k > and hence that a k > 0 for every k > 0. And then we find that for each nonnegative integer k, α =< a 0, a,, a k, α k > =< a 0, a, a 2, >. Example. In this example, we find the infinite simple continued fraction of α = 5. For this, we compute a k and α k successively. a 0 = [ 5] = 2, α = a = [α ] = 4, α 2 = α a = = =

15 a 2 = [α 2 ] = 4, α 3 = Therefore 5 =< 2, 4, 4, 4, >. α 2 a 2 = 5 2 =

16 3 Approximations of irrational numbers In this section, we discuss approximations of an irrational number by rational numbers by using the theory of continued fractions introduced in the previous section. Theorem 3.. Let α = <a 0, a, a 2, >, and {p k } and {q k } be the sequences as in section 2. Then for each positive integer n, we have α p n < α p n. q n q n Proof. We have α = <a 0, a,, a n, α n+ >, where α n+ denotes the appropriate complete quotient. It follows from Theorem 2.2 that which may be written in the form Now, dividing by α n+ q n, we find that Since α n+ > and q n and the proof is completed. α = α n+p n + p n, α n+ q n + q n α n+ (αq n p n ) = αq n + p n ( = q n α p ) n. q n α p n q n = q n α n+ q n ( α p ) n. q n q n, it follows from this equation that α p n < α p n, q n q n Example 2. The continued fraction of π is known to be π = < 3, 7, 5,, 0

17 >. Then the sequences {p k } and {q k } are as in the table below. k a k p k q k We see that the sequence p k q gives good approximations of π with errors about k π 3 = 0.459, π 22 7 = , π = The following theorem says that p k q k is the best approximation of a given irrational number α in the sense that among all the rational numbers whose denominators are less than or equal to q n, p n qn is the closest rational number to α. Theorem 3.2. Let p n /q n be the nth convergent to the infinite continued fraction representing the irrational number α. If c and d are integers with d > 0 and (c, d) = such that α c < d then d > q n. α p n, n > 0, q n

18 Proof. Of the two successive convergents p n /q n and p n /q n, one is even and the other is odd, and hence α lies between them. Moreover, the preceding theorem shows that α is nearer to p n /q n than to p n /q n. The hypothesis of the present theorem then implies that c/d lies between p n /q n and p n /q n, and hence that c d p n < q n p n p n q n. q n By using p n q n p n q n =, this implies that d > q n cq n dp n. Clearly, cp n dq n 0 since, otherwise, it would follow that c/d = p n /q n. Hence cq n dp n is a positive integer. Therefore d > q n. Theorem 3.3. Of any two consecutive convergents to α, at least one of them satisfies the inequality α p/q < /2q 2. Proof. α lies between p n /q n and p n+ /q n+ so that we have p n+ p n q n+ = p n α q n + p n+ α q n+. If the theorem is false, then q n q n+ = q n p n+ p n q n+ q n 2q 2 n giving (q n+ q n ) 2 0, which is impossible if n > 0. +, 2qn+ 2 Example 3. It may happen that α p n qn 2q 2 n for some n. For 2

19 instance, we have 5 = < 2, 4, 4, >. We then have the following table. k a k p k q k We see that is one of the convergents of the infinite continued fraction of 5. Since = and = , we have > However, it is known that if c and d are integers with 2 d > 0 and (c, d) = such that α c d < 2d 2, then c/d is one of the convergents in the infinite continued fraction representation of α. Theorem 3.4. Of any three consecutive convergents to α, at least one of them satisfies the inequality α p q <. 5q 2 3

20 Proof. Let β n+ = q n /q n, so that, by equation (4), p n α q n = q n (α n+ q n + q n ) = qn(α 2 n+ + β n+ ). We proceed to prove that α i + β i 5 (4) cannot hold for three consecutive values i = n, n, n+. Suppose that equation (5) holds for i = n and i = n. From α n = a n + /α n and β n = q n q n 2 = a n q n 2 + q n 3 q n 2 = a n + β n, we have giving α n + β n = α n + β n 5, = α n α n ( 5 β n ) ( 5 β n ). Thus β n + /β n 5, from which we deduce easily that β n > 2 ( 5 ). Similarly, if equation (5) holds for i = n and i = n+, then β n+ > 2 ( 5 ), and we arrive at a n = β n+ β n < 5 β n+ β n < 5 ( 5 ) =, which is impossible. The theorem is proved. 4

21 REFERENCES [] Neal H. McCoy, The Theory of Numbers, The Macmillan Company, 970. [2] Hua Loo-Keng, Introduction to Number Theory, Springer- Verlag,