Continued fractions. Yann BUGEAUD. x 1 + and, more generally, we call any expression of the above form or of the form


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1 Continued fractions Yann BUGEAUD Let x 0, x,... be real numbers with x, x 2,... positive. A finite continued fraction denotes any expression of the form [x 0 ; x, x 2..., x n ] = x 0 + x + x x n, and, more generally, we call any expression of the above form or of the form [x 0 ; x, x 2,...] = x 0 + x + x = lim n + [x 0; x, x 2..., x n ] a continued fraction, provided that the limit exists. The aim of this Section is to show how a real number ξ can be expressed as ξ = [x 0 ; x, x 2,...], where x 0 is an integer and x n a positive integer for any n. We first deal with the case of a rational number ξ, then we describe an algorithm which associates to any irrational ξ an infinite sequence of integers (a n ) n 0, with a n for n, and we show that the sequence of rational numbers [a 0 ; a, a 2..., a n ] converges to ξ. For more results on continued fractions or/and different points of view on this theory, the reader may consult e.g. a text of Van der Poorten [6] and the books of Cassels [], Dajani & Kraaikamp [3], Hardy & Wright [4], Perron [5], and Schmidt [7]. Lemma.. Any rational number r has exactly two different continued fraction expansions. These are [r] and [r ; ] if r is an integer and, otherwise, one of them has the form [a 0 ; a,..., a n, a n ] with a n 2, and the other one is [a 0 ; a,..., a n, a n, ]. Proof. Let r be a rational number and write r = u/v with v positive and u and v coprime. We argue by induction on v. If v =, then r = u = [u], and if r = [a 0 ; a,..., a n ] with n, we have r = a 0 + /[a ; a 2,..., a n ]. Since a and r is an integer, we deduce that n = and a =, thus r = a 0 + = [r ; ]. We assume v 2 and that the lemma holds true for any rational of denominator positive and at most equal to v. Performing the Euclidean division of u by v, there exist integers q and c with u = qv + c and c v. Thus, u/v = q + c/v and, by our inductive hypothesis, the rational v/c has exactly two expansions in continued fraction, École du CIMPA, Bamako, novembre 200.
2 which we denote by [a ; a 2,..., a n, a n ] with a n 2, and [a ; a 2,..., a n, a n, ]. Setting a 0 equal to q, the desired result follows for u/v. Unless otherwise explicitly stated, by the continued fraction expansion of a rational number p/q, we mean [] if p/q =, and that one which does not terminate in if p/q. The following algorithm allows us to associate to any irrational real number ξ an infinite sequence of integers. Let define the integer a 0 and the real number ξ > by a 0 = [ξ] and ξ = /{ξ}. We then have ξ = a 0 + /ξ. For any positive integer n, we define inductively the integer a n and the real number ξ n+ > by a n = [ξ n ] and ξ n+ = /{ξ n }, and we observe that ξ n = a n + /ξ n+. We point out that the algorithm does not stop since ξ is assumed to be irrational. Thus, we have associated to any irrational real number ξ an infinite sequence of integers a 0, a, a 2,... with a n positive for all n. If ξ is rational, the same algorithm terminates and associates to ξ a finite sequence of integers. Indeed, the ξ j s are then rational numbers and, if we set ξ j = u j /v j, with u j, v j positive and gcd(u j, v j ) =, an easy induction shows that we get u j > u j+, for any positive integer j with v j. Consequently, there must be some index n for which v n and ξ n is an integer. Thus, a n+, a n+2,... are not defined. We have ξ = [a 0 ; a,..., a n ], and this corresponds to the Euclidean algorithm. Definition.2. Let ξ be an irrational number (resp. a rational number). Let a 0, a,... (resp. a 0, a,..., a N ) be the sequence of integers associated to ξ by the algorithm defined above. For any integer n (resp. n =,..., N), the rational number p n := [a 0 ; a,..., a n ] is called the nth convergent of ξ and a n is termed the nth partial quotient of ξ. Further, for any integer n (resp. n =,..., N ), there exists a real number η n in ]0, [ such that, ξ = [a 0 ; a,..., a n, a n + η n ]. We observe that the real numbers η n occurring in Definition.2 are exactly the real numbers /ξ n+ given by the algorithm. In all what follows until the end of Theorem.7 included, unless otherwise explicitly stated, we assume that ξ is a real irrational number and we associate to ξ the sequences (a n ) n 0 and (p n / ) n as given by Definition.2. However, the statements below remain true for rational numbers ξ provided that the a n s and the p n / s are welldefined. The integers p n and can be easily expressed in terms of a n, p n, p n 2,, and 2. 2
3 Theorem.3. Setting we have, for any positive integer n, p =, q = 0, p 0 = a 0, and q 0 =, p n = a n p n + p n 2 and = a n + 2. Proof. We proceed by induction. Since p /q = a 0 + /a = (a 0 a + )/a, the definitions of p, q, p 0, and q 0 show that the theorem is true for n =. Assume that it holds true for a positive integer n and denote by p 0/q 0,..., p n/q n the convergents of the rational number [a ; a 2,..., a n+ ]. For any integer j with 0 j n + we have thus p j q j = [a 0 ; a,..., a j ] = a 0 + [a ; a 2,..., a j ] = a 0 + q j, p j p j = a 0 p j + q j and q j = p j. () It follows from () with j = n + and the inductive hypothesis applied to the rational [a ; a 2,..., a n+ ] that and p n+ = a 0 (a n+ p n + p n 2) + a n+ q n + q n 2 = a n+ (a 0 p n + q n ) + (a 0 p n 2 + q n 2) + = a n+ p n + p n 2, whence, by () with j = n and j = n, we get + = a n+ + and p n+ = a n+ p n + p n, as claimed. Theorem.4. For any nonnegative integer n, we have and, for all n, p n p n = ( ) n (2) p n 2 p n 2 = ( ) n a n. (3) Proof. These equalities are clearly true for n = 0 and n =. It then suffices to argue by induction, using Theorem.3. Lemma.5. For any irrational number ξ and any nonnegative integer n, the difference ξ p n / is positive if, and only if, n is even. Proof. We easily check that this is true for n = 0,, and 2 and we proceed by induction. Let n 4 be an even integer. Then ξ = [a 0 ; a, [a 2 ; a 3,..., a n + η n ]] and the inductive hypothesis implies that [a 2 ; a 3,..., a n + η n ] > [a 2 ; a 3,..., a n ]. Since [a 0 ; a, u] > [a 0 ; a, v] holds for all positive real numbers u > v, we get that ξ > [a 0 ; a, [a 2 ; a 3,..., a n ]] = p n /. We deal with the case n odd in exactly the same way. As a corollary of Lemma.5, we get a result of Vahlen. 3
4 Corollary.6. Let p n / and p n+ /+ be two consecutive convergents of the continued fraction expansion of an irrational number ξ. Then at least one of them satisfies ξ p q < 2q 2. (4) In particular, there are infinitely many rational numbers p/q satisfying (4). Proof. We infer from Lemma.5 that ξ is an inner point of the interval bounded by p n / and p n+ /+. Thus, using (2) and the inequality a 2 + b 2 > 2ab, valid for any distinct real numbers a and b, we get 2q 2 n + 2q 2 n+ > and the claimed result follows. + = p n p n+ = ξ p n + ξ p n+, The next two theorems show that the real irrational numbers are in a onetoone correspondence with the set of integer sequences (a i ) i 0 with a i positive for i. Theorem.7. The convergents of even order of any real irrational ξ form a strictly increasing sequence and those of odd order a strictly decreasing sequence. The sequence of convergents (p n / ) n 0 converges to ξ, and we set + ξ = [a 0 ; a, a 2,...]. Any irrational number has a unique expansion in continued fraction. Proof. It follows from (3) that for any integer n with n 2 we have p n 2 2 p n = ( )n a n 2, and, since the a n s are positive, we deduce that the convergents of even order of the real irrational ξ form a strictly increasing sequence and those of odd order a strictly decreasing sequence. To conclude, we observe that, by Lemma.5, we have p 2n /q 2n < ξ < p 2n+ /q 2n+ for all n 0, and, by (2), the difference p 2n /q 2n p 2n+ /q 2n+ tends to 0 when n tends to infinity. Unicity is clear. Indeed, if (b i ) i 0 is a sequence of integers with b i positive for i and such that lim n + [b 0 ; b, b 2,...] exist, then this limit cannot be equal to lim n + [a 0 ; a, a 2,...] as soon as there exists a nonnegative integer i with a i b i. Theorem.8. Let a 0, a,... be integers with a, a 2,... positive. Then the sequence of rational numbers [a 0 ; a,..., a i ], i, converges to the irrational number whose partial quotients are precisely a 0, a,... Proof. For any positive integer n, denote by p n / the rational number [a 0 ; a,..., a n ]. The recurrence relations obtained in Theorems.3 and.4 hold true in the present context. As 4 +
5 in the proof of Theorem.7, we deduce from (2) and (3) that the sequences (p 2n /q 2n ) n and (p 2n+ /q 2n+ ) n are adjacent. Hence, they converge to the same limit, namely to the irrational number [a 0 ; a, a 2,...], whose partial quotients are precisely a 0, a,..., by Theorem.7. We observe that for any irrational number ξ the sequences (a n ) n 0 and (ξ n ) n given by the algorithm defined below Lemma. satisfy ξ n = [a n ; a n+, a n+2,...] for all positive integers n. Theorem.9. Let n be a positive integer and ξ = [a 0 ; a, a 2...] be an irrational number. We then have ξ = [a 0 ; a,..., a n, ξ n+ ] = p nξ n+ + p n ξ n+ + and ξ p n = ( ) n ξ n+ + = ( )n ξ n+ + [0; a n, a n,..., a ]. Furthermore, the set of real numbers having a continued fraction expansion whose n + first partial quotients are a 0, a,..., a n is precisely the closed interval bounded by (p n + p n )/( + ) and p n /, which are equal to [a 0 ; a,..., a n, ] and [a 0 ; a,..., a n ], respectively. Proof. We proceed by induction, using Theorem.3 and noticing that we have ξ n = a n + /ξ n+ and / = [a n ; a n,..., a ] for all positive integers n. The last assertion of the theorem follows immediately, since the admissible values of ξ n+ run exactly through the interval ], + [. Corollary.0. For any irrational number ξ and any nonnegative integer n, we have ( + + ) < ξ p n <. + Proof. Writing ξ = [a 0 ; a, a 2,...] and ξ n+ = [a n+ ; a n+2,...], we observe that a n+ < ξ n+ < a n+ +, and we get from Theorem.9 that ( ) < (an+ + ) + q ξ p n < n (a n+ + ). The corollary follows then from Theorem.3. The following result of Legendre provides a partial converse to Corollary.6. Theorem.. Let ξ be a real number. Any nonzero rational number a/b with ξ a b < 2b 2 is a convergent of ξ. Proof. We assume that ξ a/b and we write ξ a/b = εθ/b 2, with ε = ± and 0 < θ < /2. By Lemma., setting a n = if necessary, we may write a/b = [a 0 ; a,..., a n ], with 5
6 n given by ( ) n = ε, and we denote by p /q,..., p n / the convergents of a/b. Let ω be such that ξ = p n ω + p n 2 ω + 2 = [a 0 ; a,..., a n, ω]. We check that whence εθ b 2 = ξ a b = ξ p n = (ξ p n ) = ω = θ 2 ( ) n ω + 2, and ω >. Denote by [a n ; a n+, a n+2,...] the continued fraction expansion of ω. Since a j for all j, we have ξ = [a 0 ; a,..., a n, ω] = [a 0 ; a, a 2,...], and Theorem.8 yields that a/b = p n / is a convergent of ξ. A lessknown result of Fatou provides a satisfactory complement to Theorem.. It asserts that if the real number ξ and the rational number a/b satisfy ξ a/b < /b 2, then there exists an integer n such that a b belongs to { pn, p n+ + p n, p } n+2 p n Definition.2. The irrational real number ξ is said to be badly approximable if there exists a positive constant c(ξ) such that ξ p q c(ξ) q 2 for any rational p/q distinct from ξ. The set of badly approximable real numbers is denoted by B. Quadratic real numbers are badly approximable (see Exercise ), and many other real numbers share this property, as follows from the next theorem. Theorem.3. An irrational real number ξ is badly approximable if, and only if, the sequence of its partial quotients is bounded. Consequently, the set B is uncountable. Proof. Assume that ξ is badly approximable, and let c be a positive real number such that ξ p/q > c/q 2 for any rational p/q. Let n be a positive integer. It follows from Corollary.0 that /c. Since Theorem.3 yields that a n, we get a n /c, and the sequence of partial quotients of ξ is bounded. Conversely, if the partial quotients of ξ are not greater than a constant M, we then have + (M + ) for any nonnegative integer n. Furthermore, if a/b satisfies 6
7 ξ a/b < /(2b 2 ), then, by Theorem. and Corollary.0, there exists a nonnegative integer n such that a/b = p n / and ξ a b > b(b + + ) (M + 2)b 2. This shows that ξ is badly approximable. The last assertion of the theorem follows from Theorem.8. Quadratic numbers are badly approximable numbers, thus, by Theorem.3, their partial quotients are bounded. However, much more is known: the sequence of their partial quotients is ultimately periodic. Theorem.4. The real irrational number ξ = [a 0 ; a, a 2,...] has a periodic continued fraction expansion (that is, there exist integers k 0 and n such that a m+n = a m for all m k) if, and only if, ξ is a quadratic irrationality. The only if part is due to Euler, and the (more difficult) if part was established by Lagrange in 770, see Exercise 2. The next result dates back to (at least) 877 and can be found in the book of Serret. Theorem.5. Let ξ = [a 0 ; a, a 2,...] and η = [b 0 ; b, b 2,...] be two irrational numbers. There exist integers u and v such that a u+n = b v+n for every positive integer n if, and only if, there exist integers a, b, c, and d such that ad bc = and η = (aξ + b)/(cξ + d). Two real numbers satisfying the equivalent conditions of Theorem.5 are called equivalent. Hurwitz improved the last assertion of Corollary.6. Theorem.6. For any real irrational number ξ, there exist infinitely many rationals p/q satisfying ξ p q <. 5q 2 Further, the constant 5 cannot be replaced by a larger real number. We observe that the Golden Section ( + 5)/2 = [;,,...,,...] is, up to equivalence, the irrational number which is the most badly approximable by rational numbers. As shown by Hurwitz, Theorem.6 can be improved if, besides the rationals, we also exclude the numbers which are equivalent to the Golden Section. Theorem.7. For any irrational real number ξ which is not equivalent to ( + 5)/2, there exist infinitely many rationals p/q satisfying ξ p q <. 8q 2 The constant 8 cannot be replaced by a larger real number. In other words, the assumptions of Theorem.7 are satisfied by any irrational ξ having infinitely many partial quotients greater than or equal to 2. The limiting case is obtained with real numbers equivalent to ( + 2)/2. 7
8 More generally, for a real ξ we define ν(ξ) = lim inf q + q qξ, where denotes the function distance to the nearest integer. Clearly, ν(ξ) = 0 holds for any rational ξ. The set of values taken by the function ν is called the Lagrange spectrum. By Theorem.6, it is included in the interval [0, / 5]. Theorem.8. There exists a sequence of numbers ν = 5 > ν 2 = 8 > ν 3 > ν 4 >... tending to /3 such that, for all ν i, there is, up to equivalence, a finite number of real numbers ξ satisfying ν(ξ) = ν i. More results on the Lagrange spectrum can be found in the book of Cusick & Flahive [2]. We continue with some complements. It is sometimes important to take the matrix point of view, that is, to note that the recurrence relations given in Theorem.3 imply that, if ξ = [a 0 ; a,...] is a positive real irrational number, with convergents p n / = [a 0 ; a,..., a n ], then, for n, we have ( ) pn p M n := n = ( a0 0 ) ( ) a... 0 ( ) an. (5) 0 Theorem.9. Let n 2 be an integer and a,..., a n be positive integers. For j =,..., n, set p j /q j = [0; a,..., a j ]. Then, we have Proof. We get from Theorem.3 that = [0; a n, a n,..., a ]. + = a n+ +. Theorem.9 then follows by induction. Alternatively, we can conclude by setting a 0 = 0 in (5) and then by taking the transpose of both sides of (5). Definition.20. Let m and a,..., a m be positive integers. The continuant of a,..., a m, usually denoted by K m (a,..., a m ), is the denominator of the rational number [0; a,..., a m ]. Theorem.2. For any positive integers a,..., a m and any integer k with k m, we have K m (a,..., a m ) = K m (a m,..., a ), (6) 8
9 2 (m )/2 K m (a,..., a m ) ( + max{a,..., a m }) m, (7) and K k (a,..., a k ) K m k (a k+,..., a m ) K m (a,..., a m ) 2 K k (a,..., a k ) K m k (a k+,..., a m ). (8) Proof. The first statement is an immediate consequence of Theorem.9. For the second statement, letting a be the maximum of a,..., a m, it follows from Theorem.3 that K m (,..., ) K m (a,..., a m ) K m (a,..., a) (a + ) m. Since K () =, K 2 (, ) = 2 and 2 l/2 2 (l )/2 + 2 (l 2)/2 for l 2, an immediate induction gives the first inequality of (7). Combining K m (a,..., a m ) = a m K m (a,..., a m )+K m 2 (a,..., a m 2 ) with (6), we get K m (a,..., a m ) = a K m (a 2,..., a m ) + K m 2 (a 3,..., a m ), which implies (8) for k =. Let k be in {, 2,..., m 2} be such that K m := K m (a,..., a m ) =K k (a,..., a k ) K m k (a k+,..., a m ) + K k (a,..., a k ) K m k+ (a k+2,..., a m ), (9) with the convention that K 0 =. We then have K m = K k (a,..., a k ) (a k+ K m k+ (a k+2,..., a m ) + K m k+2 (a k+3,..., a m ) ) + K k (a,..., a k ) K m k+ (a k+2,..., a m ) = ( a k+ K k (a,..., a k ) + K k (a,..., a k ) ) K m k+ (a k+2,..., a m ) + K k (a,..., a k ) K m k+2 (a k+3,..., a m ), giving (9) for the index k +. This shows that (9) and, a fortiori, (8) hold for k =,..., m. 9
10 Exercises Exercise. Prove that every real quadratic number a + b d is badly approximable (introduce a b d). Exercise 2. (Proof of Theorem.4). Show that any irrational real number having a periodic continued fraction expansion is a quadratic irrationality. Prove the converse, following Steinig. Let ξ be a quadratic real number and define the sequences (a n ) n 0 and (ξ n ) n by the algorithm given above. Set ξ 0 := ξ. ) Show, by induction, that for each nonnegative integer n there is an integer polynomial f n (x) := A n x 2 + B n x + C n, with Bn 2 4A n C n positive and not a square, such that f n (ξ n ) = 0. Prove that A n+, B n+ and C n+ are given by A n+ = a 2 n A n + a n B n + C n, B n+ = 2a n A n + B n, and C n+ = A n. (0) 2) Observe that D := B 2 0 4A 0 C 0 = B 2 n 4A n C n, for any n 0, () and deduce from (0) that there exists an infinite set N of distinct positive integers such that A n A n is negative for any n in N. 3) Deduce from () that for any n in N. Conclude. B n < D, A n D/4, and C n D/4, Exercise 3. (Proof of Hurwitz Theorems.6 and.7, following Forder.) Let ξ be an irrational number and assume that its convergents p n /, p n /, and p n+ /+ satisfy ξ p j /q j /( 5qj 2 ) for j = n, n, n +. ) Prove that we have ( 5 qn 2 + ) qn 2, and deduce that / and / belong to the interval ]( 5 )/2, ( 5 + )/2[. Prove that the same conclusion also holds for /+ and + /. Conclude. 2) With a suitable adaptation of the above proof, show that if the convergents p n / of an irrational number ξ satisfy ξ p n / /( 8q 2 n) from some integer n 0 onwards then + < 2 + holds for any sufficiently large integer n. Conclude. 0
11 References [] J. W. S. Cassels, An introduction to Diophantine Approximation. Cambridge Tracts in Math. and Math. Phys., vol. 99, Cambridge University Press, 957. [2] T. W. Cusick and M. E. Flahive, The Markoff and Lagrange spectra. Mathematical Surveys and Monographs, 30. American Mathematical Society, Providence, RI, 989. [3] K. Dajani and C. Kraaikamp, Ergodic theory of numbers. Carus Mathematical Monographs 29, Mathematical Association of America, Washington, DC, [4] G. H. Hardy and E. M. Wright, An introduction to the theory of numbers. 5th. edition, Clarendon Press, 979. [5] O. Perron, Die Lehre von den Ketterbrüchen. Teubner, Leipzig, 929. [6] A. J. van der Poorten, An introduction to continued fractions. In: Diophantine analysis (Kensington, 985), 99 38, London Math. Soc. Lecture Note Ser. 09, Cambridge Univ. Press, Cambridge, 986. [7] W. M. Schmidt, Diophantine Approximation. Lecture Notes in Math. 785, Springer, Berlin, 980. Yann Bugeaud Université de Strasbourg Mathématiques 7, rue René Descartes STRASBOURG (FRANCE)
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