CONTINUED FRACTIONS. 1. Finite continued fractions A finite continued fraction is defined by [a 0,..., a n ] = a 0 +

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1 CONTINUED FRACTIONS Abstract Continued fractions are used in various areas of number theory and its applications They are seen in the extended Euclidean algorithm; in approximations (of numbers and functions); in Pell s equation; and more This chapter is largely based on Chapters X and XI of [HW], using matrices and the action of PGL (Z) wherever possible Finite continued fractions A finite continued fraction is defined by [a 0,, a n ] = a 0 + [a,,a n ] Our first main theorem is that an infinite continued fraction is always well defined Proposition We have the equality [a 0,, a m, a m,, a n ] = [a 0,, a m, [a m+,, a n ]] Proof It is enough to prove the case m = 0, which follows from definition There is a strong connection to the Euclidean algorithm, which we do not elaborate on here Convergence of the continued fraction Let a sequence a 0, a, be given; we assume a n for n The purpose of this section is to study infinite continued fractions, thereby showing ( that )[a 0, a, ] is always a well defined real number We set A n = an 0 and note that det(a n ) = Definition Let p =, q = 0; p 0 = a 0, q 0 = ; and for n, () Setting the definition can be rephrased as p n = a n p n + p n, = a n + Q n = pn p n q, n Notes for a course in elementary number theory, Uzi Vishne, 006, Bar-Ilan University

2 CONTINUED FRACTIONS Corollary but Q 0 = A 0, so that Q n = A n Q n ; () Q n = A n A 0 The main equality: Corollary 3 The determinant p n p n = det Q n is equal to ( ) n Corollary 4 From Corollary 3 it follows that p n, are co-prime, so p n is always reduced In a similar manner, we obtain Corollary 5 Computing p n q n p n q by substituting () for the first n line, we get ( ) n a n We now turn to study the finite sections of the continued fraction Theorem 6 Let x n = [a 0,, a n ] Then p n = x n Proof By induction on n For n = 0, this is clear If the claim holds for fractions of length n, then compute for x n = [a 0,, a n + that it is equal to (a n + an )p n +p n 3 (a n + an ) + 3 = p n + an p n + an = = p n Corollary 7 Dividing the equations in Propositions 3 and 5 by and, respectively, we get (3) (4) x n x n = ( )n ; x n x n = ( )n a n Corollary 8 The sequence x m is decreasing; the sequence x m+ is increasing Proof Directly from Equation (4) Corollary 9 x m > x k+ always (otherwise apply (3)) Remark 0 F n where F n are the Fibonachi numbers (induction) Theorem The sequence x n converges a n ]

3 CONTINUED FRACTIONS 3 Proof Three proofs Combining Corollaries 8 and 9, the sequences converge, and by (3) they have a joint limit Also, by (3) we have x n = x 0 + n ( ) n k=, so converges by Leibnitz theorem Alternatively, the sequence absolutely converges by comparison test, using Remark 0 Remark Now that infinite continued fractions makes sense, we can repeat Remark for this situation: [a 0, a, ] = [a 0, [a, ]] Now, for the other direction, namely existence Given a real number y R, we define y 0 = y; a n = [y n ]; y n = y n a n The definition breaks down if some y n Z, but this is easily seen to happen iff y Q Theorem 3 (No proof) [a 0, a, ] = y Theorem 4 (No proof) Every x R has a unique expression as a continued fraction Example 5 5+ = [,, ] To conclude, there is a one-to-one correspondence between R and the sequences [a 0, ] with a n > 0 for n, assigning every number its expansion as a continued fraction We need to define the action of PGL (Z) on R Recall that GL (Z) is the set of matrices in M (Z) which are invertible, with inverse in the same set Those are precisely the matrices whose determinant in ± (by Cramer s formula) The center of this group is the group of order, ±I, and by definition PGL (Z) = GL (Z)/ I is the quotient group By an action of a group on a set we mean that every element of the group defines a function on the set, such that the action of gh is the composition of those of g and h The action is defined by the same action as B ( a b ) z = az+b Notice that B induces cz+d Exercise 6 Verify that for every B, B GL (Z), (BB ) z = B (B z) Exercise 7 Verify that the vector B z is projectively equivalent to B z

4 4 CONTINUED FRACTIONS Let us say that two infinite sequences are equivalent if they become equal after removing a finite header The following theorem essentially follows from Remark Theorem 8 (No proof) The expansions of x, y R are equivalent iff x, y are in the same orbit of PGL (Z) 3 Approximations We say that α R is approximable of order n and power C, if there are infinitely many pairs (p, q) such that 0 α p q < Cq n Remark 3 Notice that when n > 0, infinitely many pairs and infinitely many fractions above, give equivalent definitions Remark 3 Not being approximable of any order > n is equivalent to having ɛ > 0 such that α p q ɛ Remark 33 Every number is trivially approximable to order and power / ɛ Proposition 34 Rational numbers are not approximable to order > Theorem 35 (Dirichlet) Every irrational number is approximable to order and power Proof Fix T, and consider the T + values {(qα) : q = 0,, T } [0, ), which are all different Dividing the segment to T sub-segments, the piegone hole principle implies some (qα) (q α) < /T This provides one approximation To get new ones, choose the new T such that /T is smaller than all differences in previous numbers Theorem 36 (Liouville) If [Q[α]:Q] = d, then α is not approximable of order > d Proof We may assume f(α) = 0 for f(λ) Z[λ], and replacing by a prime factor, we may assume f (α) 0 Let γ be such that [α γ, α+γ] has no zero of f (thus no zero of f, other than α) Let ɛ = inf f (y) on that segment Compute that in the segment, f(p/q) q d Then use the mean value theorem to conclude f(x) f(α) = f (y)(x α) so α p q ɛq d for p/q in the segment

5 CONTINUED FRACTIONS 5 Corollary 37 Lioville s number α = n= 0 n! is approximable to any order, and so transcendental This is the first explicit non-algebraic number constructed, although cardinality arguments easily show existence 4 Continued fractions and approximations Continued fractions can be used to improve the power of nd order approximations The results obtained in this way are the best possible Write x n = [a n+, a n+, ] Remark can be read to say that (5) x n = A n x n Proposition 4 We have that Proof By definition of x, we have x = Q nt x n (6) x = [a 0,, a n, a n + /x n] Setting p n = (a n + /x n)p n + p n = p n + p n /x n and q n = (a n + /x n) + = + /x n as in the first definition, we know from Theorem 6 that x = p n/q n = pnx n+p n x n + = Q n t x n Remark 4 We have < q n = + /x n < + + By (3) applied to (6), we have (7) x p n = < <, qn a n qn showing that we have approximations of order by continued fractions, which are of power except for conjugates of 5+ But the same inequality also implies x p n = /qn By the remark we have < x p n < + Corollary 43 The series x p n is decreasing (this is stronger than x p n being decreasing) Theorem 44 If q < and p/q p n /, then qx p > x p n pn p Proof Assume q Let u, v Z such that n uv = q n pq Since q = u + v, we see that v 0 and uv 0 Moreover x p n and x p n have different signs, so qx p = u x p n + v x p n > x p n

6 6 CONTINUED FRACTIONS Example 45 π = [3, 7, 5,, 93, ], so p 4 /q 4 = 355/ is the best approximation with q < = 330 Theorem 46 One of every two consecutive p n / satisfies (8) x p n < qn In fact, x p n < a n a n+ qn Proof of theorem Otherwise a n a n+ q n + q n = x n x n = x n x + x x n, and f( / ) f(a n ) for f(t) = t t + ; but f is decreasing, and / > a n contradiction Corollary 47 Every irrational number is approximable of order and power ; for all numbers except conjugates of 5+, the power increases to 5/ It should be remarked that (9) characterizes convergents to continued fractions: Theorem 48 (No proof) If p/q satisfies x p q <, then p/q = q p n / for some n Theorem 49 (No proof; see [HW, Thm 95]) One of every three consecutive p n / satisfies (9) x p n < 5q n This is the best possible result, in light of the following: Theorem 40 ([HW, Thm 84]) Every fraction p/q satisfying (9) is a convergent Proof (Sketch) Based on the following: if x = a b y where y >, a b PGL (Z) and c > d > 0, then c/d = p n /, a/b = p n / and y = x n for x On the other hand ([HW, Thm 94]), we have: Proposition 4 ζ = 5 is not approximable to any power larger than 5

7 CONTINUED FRACTIONS 7 Proof Write ζ = p + δ q q 5 = ζ +, we have for 5 δ < Thus δ = q(qζ p), and since δ q δ 5 = (p qζ)(p + qζ + q) = p + pq q (ζ + ζ) = p + pq q in which the left hand side is less than in absolute value for q large, a contradiction We remark that any number not PGL (Z)-equivalent to ζ, is approximable of order and power ; again this is best possible Finally, to put Equation (7) in better perspective, we quote Theorem 4 ([HW, Thm 96]) The set of x R for which {a n } is bounded, has measure zero Much more is known about approximations and the average behavior of a n : see [K] 5 Periodic fractions and Pell s equation The equation was extensively studied by the Indian mathematician and astronomer Brahmagupta, in the 7th century (see History of Dickson, about the origin of the name) The minimal solution for (0) x Dy = can be quite large, for example for D = 6, it is x = , y = Easy theorem: if the fraction is (ultimately) periodic, then x is quadratic (since x k = x k+j ) The converse is due to Lagrange: Theorem 5 (No proof; see [HW, p 44, Thm 77]) Every quadratic number has a periodic continued fraction A hint that continued fractions are involved is obtain by dividing (0) by y : then D x y = < D+ x y y 3 Dy, so applying Theorem 40, the only possible solutions are convergents in continued fraction approximation of D Write x = D = [a 0,, a k, a k+,, a k+j, a k+j ], where j is the length of the period, and k + is its starting point Periodic fractions are particularly amenable to the matrix analysis, as we shall see Moreover, since the fraction for x is periodic, we have x k = x k+j Proposition 5 Let M = Q k+j t Q k t, then x = M x

8 8 CONTINUED FRACTIONS Proof By Proposition 4, x = Q t k+j x k+j = Q k+j t x k = Q k+j t Q t k x Theorem 53 Writing M = a b, a Dc = ( ) j is a non-trivial solution Proof Clearly M is a rational matrix Writing M = a b, the equality M D = D implies d = a and b = Dc, so in fact M = a Dc c a, and a Dc = det(m) = ( ) j We only need to prove that M 0 0, but M = I will imply Q k+j = Q k, contradiction to q k+j > q k Remark 54 Computing M from the definition, the explicit equation proved here is p k+j q k p k+j q k = q k+j p k q k+j p k, p k+j p k p k+j p k = D(q k+j q k q k+j q k ) Quite frequently k = 0, and then M = ( ) j pj p j a 0 p j q j q j a 0 q j When this is the case, the semi-symmetry of M translates to pj q = a0 D pj j a 0 q, j and p j Dq j = ( ) j Example 55 Compute 9 = [4,,, 3,,, 8, ] (so in the previous notation, k = 0 and j = 6) We then have n a n p n p n 9qn ** Remark 56 We could replace j by any multiple mj, and get similar results; thus producing infinitely many (and in fact all) solutions

9 CONTINUED FRACTIONS 9 Remark 57 It seems that k = 0 for every D Also, it was noticed by Meyer Goldberg that a k+j = a 0, and also that a k+,, a k+j is palindromic I do not know a proof of these claims Exercise 58 Using 43 = [6,,, 3,, 5,, 3,,, ], solve a 43c = References [HW] Hardy and Wright, An Introduction to the Theory of Numbers [K] Khinchin, Continued Fractions