William F. HammondLecture notes prepared for MATH 326, Spring 1997 Department of Mathematics and Statistics University at AlbanyZGL

Introduction

Continued fractions offer a means of concrete representation for arbitrary real numbers The continued fraction expansion of a real number is an alternative to the representation of such a number as a (possibly infinite) decimalThe reasons for including this topic in the course on Classical Algebra are: (i)The subject provides many applications of the method of recursion(ii)It is closely related to the Euclidean algorithm and, in particular, to Bezouts Identity(iii)It provides an opportunity to introduce the subject of group theory via the 2dimensional unimodular group GL₂(Z)

The continued fraction expansion of a real number

Every real number x is represented by a point on the real line and, as such, falls between two integers For example, if n is an integer and n x n 1 , x falls between n and n1 and there is one and only one such integer n for any given real x In the case where x itself is an integer, one has n x The integer n is sometimes called the floor of x and one often introduces a notation for the floor of x such as n [x] Examples: 1. 2 [1.5] 2. 3 [] For any real x with $n[x]$ the number $uxn$ falls in the unit interval I consisting of all real numbers u for which 0 u 1Thus, for given real x there is a unique decomposition x n u where n is an integer and u is in the unit interval Moreover, u 0 if and only if x is an integer This decomposition is sometimes called the mod one decomposition of a real number It is the first step in the process of expanding x as a continued fractionThe process of finding the continued fraction expansion of a real number is a recursive process that procedes one step at a time Given x one begins with the mod one decomposition x n₁ u₁ , where n₁ is an integer and 0 u₁ 1If u₁ 0 which happens if and only if x is an integer, the recursive process terminates with this first step The idea is to obtain a sequence of integers that give a precise determination of xIf u₁ 0 then the reciprocal $1u$₁ of u₁ satisfies $1u$₁ 1 since u₁ is in I and, therefore, u₁ 1 In this case the second step in the recursive determination of the continued fraction expansion of x is to apply the mod one decomposition to 1u₁ One writes 1u₁ n₂ u₂ , where n₂ is an integer and 0 u₂ 1 Combining the equations that represent the first two steps, one may write x n₁ 1n₂ u₂ Either u₂ 0 in which case the process ends with the expansion x n₁ 1n₂ , or u₂ 0 In the latter case one does to u₂ what had just been done to u₁ above under the assumption u₁ 0 One writes 1u₂ n₃ u₃ , where n₃ is an integer and 0 u₃ 1 Then combining the equations that represent the first three steps, one may write x n₁ 1n₂ 1n₃ u₃ After k steps, if the process has gone that far, one has integers $n$₁, n₂, , n_k and real numbers $u$₁, u₂, , u_k that are members of the unit interval I with $u$₁, u₂, , u_k1 all positive One may write x n₁ 1n₂ 1n₃ 1 1n_k u_k Alternatively, one may write x [ n₁, n₂, n₃, , n_k u_k ] If u_k 0 the process ends after k steps Otherwise, the process continues at least one more step with 1u_k n_k1 u_k1 In this way one associates with any real number x a sequence, which could be either finite or infinite, $n$₁, n₂, of integers This sequence is called the continued fraction expansion of x conventionConventionWhen $[n$₁, n₂, ...] is called a continued fraction, it is understood that all of the numbers n_j are integers and that $n$_j 1 for j 2

First examples

1511 1 411 1 1114 1 12 34 1 12 143 1 12 11 13 [1, 2, 1, 3] 10 3 11103 3 1103 3 16 11103 3 16 1103 3 16 16 1 [3, 6, 6, 6, ] [2, 3, 5, 2 ] 2 1[3, 5, 2] 2 13 1[5, 2] 2 13 15 12 2 13 1112 2 13 211 2 13511 2 1135 8135 Let x 11213121312 In this case one finds that x 1 1y , where y 213121312 Further reflection shows that the continued fraction structure for y is selfsimilar: y 2131y This simplifies to y 7y23y1 and leads to the quadratic equation 3y² 6y 2 0 with discriminant 60 Since y 2 one of the two roots of the quadratic equation cannot be y and, therefore, y 3 153 Finally, x 1512 The idea of the calculation above leads to the conclusion that any continued fraction $[n$₁, n₂, ] that eventually repeats is the solution of a quadratic equation with positive discriminant and integer coefficients The converse of this statement is also true, but a proof requires further consideration

The case of a rational number

The process of finding the continued fraction expansion of a rational number is essentially identical to the process of applying the Euclidean algorithm to the pair of integers given by its numerator and denominatorLet x ab, b 0 be a representation of a rational number x as a quotient of integers a and b The mod one decomposition ab n₁ u₁ , u₁ a n₁ bb shows that u₁ r₁b where r₁ is the remainder for division of a by b The case where $u$₁ 0 is the case where x is an integer Otherwise u₁ 0 and the mod one decomposition of $1u$₁ gives br₁ n₂ u₂ , u₂ b n₂ r₁r₁ This shows that u₂ r₂r₁ where r₂ is the remainder for division of b by r₁ Thus, the successive quotients in Euclids algorithm are the integers $n$₁, n₂, occurring in the continued fraction Euclids algorithm terminates after a finite number of steps with the appearance of a zero remainder Hence, the continued fraction expansion of every rational number is finite theoremsTheoremThe continued fraction expansion of a real number is finite if and only if the real number is rationalProof. It has just been shown that if x is rational, then the continued fraction expansion of x is finite because its calculation is given by application of the Euclidean algorithm to the numerator and denominator of x The converse statement is the statement that every finite continued fraction represents a rational number That statement will be demonstrated in the following section

The symbol [t₁, t₂, , t_r]

For arbitrary real numbers $t$₁, t₂, , t_r with each $t$_j 1 for $j2$ the symbol $[\; t$₁, t₂, , t_r ] is defined recursively by: [ t₁ ] t₁ treceqn[t₁,t₂,,t_r ]t₁1[t₂,,t_r ] In order for this definition to make sense one needs to know that the denominator in the righthand side of (iref="trec"trec) is nonzero The condition $t$_j 1 for $j2$ guarantees, in fact, that [t₂,,t_r ] 0 as one may prove using inductionIt is an easy consequence of mathematical induction that the symbol $[t$₁, t₂, , t_r] is a rational number if each t_j is rational In particular, each finite continued fraction is a rational number (Note that the symbol $[t$₁, t₂, , t_r] is to be called a continued fraction, according to the convention of the first section, only when each t_j is an integer.)Observe that the recursive nature of the symbol $[t$₁, , t_r] suggests that the symbol should be computed in a particular case working from right to left Consider again, for example, the computation above showing that [2, 3, 5, 2] 8135 Working from right to left one has: [2] 2 [5, 2] 51[2] 512 112 [3, 5, 2] 31[5, 2] 3211 3511[2, 3, 5, 2] 21[3, 5, 2] 21135 8135There is, however, another approach to computing [t₁, t₂, , t_r] Let, in fact, $t$₁, t₂, be any (finite or infinite) sequence of real numbers One uses the double recursion pdefeqnp_j t_j p_j1 p_j2 , j 1 , p₀ 1 , p₁ 0 to define the sequence p_j , j 1 The double recursion, differently initialized, qdefeqnq_j t_j q_j1 q_j2 , j 1 , q₀ 0 , q₁ 1 defines the sequence q_j , j 1 Note that p₁ t₁ p₂ t₁t₂ 1 and q₁ 1 q₂ t₂ q₃ t₂t₃ 1 One now forms the matrix MrdefeqnM_j rrp_j q_j p_j1 q_j1 for j 0 Thus, for example, M₀ rr1 0 0 1 , and M₁ rrt₁ 1 1 0 It is easy to see that the matrices M_j satisfy the double recursion MrreceqnM_j rrt_j 1 1 0 M_j1 , j 1 as a consequence of the double recursion formulas for the p_j and q_j Hence, a simple argument by mathematical induction shows that MrcompeqnM_r rrt_r 1 1 0 rrt₂ 1 1 0 rrt₁ 1 1 0 , r 1 This is summarized by: propositionsPropositionFor any sequence $t$_j, j 1 of real numbers, if $p$_j and $q$_j are the sequences defined by the double recursions (iref="pdef"pdef) and (iref="qdef"qdef), then one has the matrix identity prqrroweqnrrp_r q_r p_r1 q_r1 rrt_r 1 1 0 rrt₂ 1 1 0 rrt₁ 1 1 0 for each integer r 1 pqidencorollariesCorollaryOne has the identity $p$_r q_r1 q_r p_r1 (1)^r for each integer r 1Proof. The number $p$_r q_r1 q_r p_r1 is the determinant of the matrix M_r From the formula (iref="Mrcomp"Mrcomp) the matrix $M$_r is the product of r matrix factors, each of which has determinant 1 Since the determinant of the product of matrices is the product of the determinants of the factors, it is clear that det(M_r) (1)^r videncorollariesCorollaryOne has the vector identity prqrveceqnrp_r q_r rrt₁ 1 1 0 rrt₂ 1 1 0 rrt_r 1 1 0 r1 0 for each integer r 1Proof. First recall (i) that the product of a matrix and a (column) vector is defined by the relation rra b c d rx y rax by cx dy , (ii) that the transpose of a matrix is the matrix whose rows are the columns of the given matrix, and (iii) that the transpose operation reverses matrix multiplication One tranposes both sides of the relation (iref="prqrrow"prqrrow) to obtain: prqrcoleqnrrp_r p_r1 q_r q_r1 rrt₁ 1 1 0 rrt₂ 1 1 0 rrt_r 1 1 0 To this relation one applies the principle that the first column of any 2 2 matrix is the product of that matrix with the column r1 0 in order to obtain the column identity (iref="prqrvec"prqrvec) theoremsTheoremFor any sequence $t$_j, j 1 of real numbers, if $p$_j and $q$_j are the sequences defined by the double recursions (iref="pdef"pdef) and (iref="qdef"qdef), and if $t$_j 1 for j 2 then the value of the symbol $[t$₁, , t_r] is given by the formula tcompeqn[t₁, t₂, , t_r] p_rq_r for r 1 Proof. What is slightly strange about this important result is that while the $p$_r and the $q$_r are defined by the front end recursions, albeit double recursions, (iref="pdef"pdef) and (iref="qdef"qdef), the symbol $[t$₁, , t_r] is defined by the back end recursion (iref="trec"trec) The proof begins with the comment that the righthand side of (iref="tcomp"tcomp) does not make sense unless one can be sure that the denominator q_r 0 One can show easily by induction on r that $q$_r 1 for $r1$ under the hypothesis $t$_j 1 for j 2The proof proceeds by induction on r If r 1 the assertion of the theorem is simply the statement t₁ p₁q₁ and, as noted above, $p$₁t₁ and q₁1 Assume now that r 2 By induction we may assume the correctness of the statement (iref="tcomp"tcomp) for symbols of length r1 and, therefore, for the symbol [t₂, , t_r] That case of the statement says that $[t$₂, , t_r] must be equal to ac where by corollary iref="viden"viden ra c rra b c d r1 0 with rra b c d rrt₂ 1 1 0 rrt_r 1 1 0 Now by (iref="trec"trec) [t₁, t₂, , t_r] t₁ 1ac t₁ ca at₁ ca But by corollary iref="viden"viden again rp_r q_r rrt₁ 1 1 0 rra b c d r1 0 rrat₁ c bt₁ d a b r1 0 rat₁ c a Hence, p_rq_r at₁ ca [t₁, t₂, , t_r]

Application to Continued Fractions

Recall that $[n$₁, n₂, ] is called a continued fraction only when each n_j is an integer and $n$_j 1 for j 2 The sequence $n$₁, n₂, may be finite or infinite The symbol $c$_r [n₁, n₂, , n_r] formed with the first r terms of the sequence, is called the r^th convergent of the continued fraction Associated with a given sequence $n$₁, n₂, are two sequences $p$₁, p₂, and $q$₁, q₂, that are given, according to the double recursions (iref="pdef"pdef), (iref="qdef"qdef) of the previous section with t_j n_j propositionsPropositionIf $[n$₁, n₂, ] is a continued fraction, then the integers p_r and q_r are coprime for each r 1Proof. By Corollary iref="pqiden"pqiden of the previous section p_r q_r1 q_r p_r1 (1)^r Hence, any positive divisor of both p_r and q_r must divide the lefthand side of this relation, and, therefore, must also divide (1)^r propositionsPropositionThe difference between successive convergents of the continued fraction $[n$₁, n₂, ] is given by the formula convgtdiffeqnc_r c_r1 (1)^rq_r q_r1 for r 2 Proof. According to the theorem (formula iref="tcomp"tcomp) at the end of the last section the convergent c_r is given by c_r p_rq_r Hence, c_r c_r1 p_rq_r p_r1q_r1 p_r q_r1 p_r1 q_rq_r q_r1 (1)^rq_r q_r1 (The last step is by Corollary iref="pqiden"pqiden above.) remarksRemarkThe formula (iref="convgtdiff"convgtdiff) remains true if $c$_r [t₁, , t_r] where the t_j are real numbers subject to the assumption $t$_j 1 for j 1 LemmaThe sequence $q$_j is a strictly increasing sequence for j 2Proof. This is easily proved by induction from the recursive definition (iref="qdef"qdef) of the sequence theoremsTheoremIf $[n$₁, n₂, ] is an infinite continued fraction, then the limit lim_r p_rq_r always existsProof. As one plots the convergents c_r on the line of real numbers, one moves alternately right and left The formula (iref="convgtdiff"convgtdiff) for the difference between successive convergents elucidates not only the fact of alternate right and left movement but also the fact that each successive movement is smaller than the one preceding Therefore, one has c₁ c₃ c₅ c₆ c₄ c₂ Since any strictly increasing sequence of positive integers must have infinite limit, the seqence $q$_j q_j1 has infinite limit, and so the sequence of reciprocals $1q$_j q_j1 must converge to zero Hence, the sequences of odd and evenindexed convergents must have the same limit, which is the limit of the sequence of all convergents definitionsDefinitionThe limit of the sequence of convergents of an infinite continued fraction is called the value of that continued fraction theoremsTheoremIf $[n$₁, n₂, ] is the continued fraction expansion of an irrational number x then x lim_r p_rq_r ; that is, the value of the continued fraction expansion of a real number is that real numberProof. For each $r1$ the continued fraction expansion $[n$₁, n₂, ] of x is characterized by the identity xconfraceqnx [n₁, n₂, , n_r u_r] , where u_r is a real number with 0 u_r 1 The sequences of ps and qs for the symbol $[n$₁, n₂, , n_r u_r] agree with those for the symbol $[n$₁, n₂, , n_r] except for the r^th terms One has by (iref="tcomp"tcomp) [n₁, n₂, , n_r u_r] P_rQ_r , where by (iref="qdef"qdef) q_r n_r q_r1 q_r2 Q_r (n_r u_r) q_r1 q_r2 Hence, Q_r q_r u_r q_r1 Therefore, the displacement from c_r1 to x is by (iref="convgtdiff"convgtdiff) (1)^rQ_r q_r1 (1)^r(q_r q_r1 u_r q_r1²) , which is in the same direction but of smaller magnitude than the displacement from c_r1 to c_r Therefore, x must be larger than every oddindexed convergent and smaller than every evenindexed convergent But since all convergents have the same limit, that limit must be x

Bezouts Identity and the double recursion

It has already been observed that the process of finding the continued fraction expansion of a rational number ab (b 0), involves the same series of long divisions that are used in the application of the Euclidean algorithm to the pair of integers a and b Recall that at each stage in the Euclidean algorithm the divisor for the current stage is the remainder from the previous stage and the dividend for the current stage is the divisor from the previous stage, or, equivalently, the dividend for the current stage is the remainder from the second previous stage The Euclidean algorithm may thus be viewed as a double recursion that is used to construct the sequence of remainders One starts the double recursion with r₁ a and r₀ b At the j^th stage one performs long division of r_j2 by r_j1 to obtain the integer quotient n_j and the integer remainder r_j that satisfies 0 r_j r_j1 Thus, remreceqnr_j r_j2 n_j r_j1 The Euclidean algorithm admits an additional stage if r_j 0 Since 0 r_j r_j1 r₂ r₁ r₀ b , there can be at most b stagesOne may use the sequence of successive quotients n_j (j 1) to form sequences $p$_j and q_j as in the previous section, according to the double recursions: npdefeqnp_j n_j p_j1p_j2 ,j 1 ;p₀ 1 ,p₁ 0 nqdefeqnq_j n_j q_j1q_j2 ,j 1 ;q₀ 0 ,q₁ 1 It has already been observed that $q$_j 1 for $j1$ and [n₁, n₂, , n_j]p_jq_j ,j 1 Bezouts Identity says not only that the greatest common divisor of a and b is an integer linear combination of them but that the coefficents in that integer linear combination may be taken, up to a sign, as q and p theoremsTheoremIf the application of the Euclidean algorithm to a and b (b 0) ends with the m^th long division, i.e., r_m 0 then bezouteqnr_j (1)^j1 q_j a p_j b , 1 j m Proof. One uses induction on j For $j1$ the statement is r₁ q₁ a p₁ b Since by (iref="npdef"npdef, iref="nqdef"nqdef) $q$₁ 1 and p₁ n₁ this statement is simply the case $j1$ in (iref="remrec"remrec) Assume j 2 and that the formula (iref="bezout"bezout) has been established for indices smaller than j By (iref="remrec"remrec) one has r_j r_j2 n_j r_j1 In this equation one may use (iref="bezout"bezout) to expand the terms r_j2 and r_j1 to obtain: r_j (1)^j3(q_j2a p_j2b) n_j (1)^j2(q_j1a p_j1b) (1)^j1(q_j2a p_j2b) n_j (1)^j1(q_j1a p_j1b) (1)^j1(q_j2a p_j2b) n_j (q_j1a p_j1b) (1)^j1(q_j2 n_j q_j1)a (p_j2 n_j p_j1)b (1)^j1 q_j a p_j b corollariesCorollaryThe greatest common divisor d of a and b is given by the formula gcdbezouteqnd (1)^m (q_m1a p_m1 b) , where m is the number of divisions required to obtain zero remainder in the Euclidean algorithmProof. One knows that d is the last nonzero remainder r_m1 in the Euclidean algorithm This formula for d is the case $jm1$ in (iref="bezout"bezout) corollariesCorollarycheckbezouteqnp_m ad , q_m bd Proof. The last remainder r_m 0 The case $jm$ in (iref="bezout"bezout) shows that ab p_mq_m Since, by the first proposition of the preceding section, p_m and q_m have no common factor, this corollary is evident

The action of $GL$₂(Z) on the projective line

If a b c d are real numbers with $adbc0$and M rra b c d is the matrix with entries a b c and d then M z for z real, will denote the expression glacteqnM z a z bc z d One calls $Mz$ the action of M on z$Mz$ is a perfectly good function of z except for the case $zdc$ where the denominator $czd$ vanishes If it were also true that $azb0$ for the same z then one would have ba dc in contradiction of the assumption ad bc 0 Thus, when z dc the value of M w increases beyond all bounds as w approaches z and it is convenient to say that M dc where is regarded as large and signless If further it is agreed to define M ac , which is the limiting value of $Mw$ as w increases without bound, then one may regard the expression $Mz$ as being defined always for all real z and for The set consisting of all real numbers and also the object (not a number) is called the projective line The projective line is therefore the union of the (ordinary) affine line with a single point propositionsPropositionIf $[n$₁, n₂, ] is any continued fraction, then cfgleqn[n₁, n₂, , n_r, n_r1, ] M [n_r1, ] where M rrn₁ 1 1 0 rrn_r 1 1 0 Proof. Let z [n_r1, ] Then [n₁, n₂, , n_r, n_r1, ] [n₁, n₂, , n_r, z ] The statement of the proposition now becomes [n₁, n₂, , n_r, z ] M z This may be seen to follow by multiplying both sides in formula (iref="prqrcol"prqrcol), after replacing t_j with n_j, by the column rz 1 The matrix M in the preceding proposition is an integer matrix with determinant 1 The notation $GL$₂(Z) denotes the set of all such matrices (The 2 indicates the size of the matrices, and the Z indicates that the entries in such matrices are numbers in the set Z of integers.) It is easy to check that the product of two members of $GL$₂(Z) is a member of $GL$₂(Z) and that the matrix inverse of a member of $GL$₂(Z) is a member of GL₂(Z) Thus, $GL$₂(Z) forms what is called a group The formula (iref="glact"glact) defines what is called the action of $GL$₂(Z) on the projective lineOne says that two points z and w of the projective line are rationally equivalent if there is a matrix M in $GL$₂(Z) for which w M z Since (i) $GL$₂(Z) is a group, (ii) M₁ (M₂ z) (M₁ M₂) z and (iii) $wMz$ if and only z M¹ w it is easy to see that every point of the projective line belongs to one and only one rational equivalence class and that two points rationally equivalent to a third must be rationally equivalent to each other TerminologyThe rational equivalence of points on the projective line is said to be the equivalence relation on the projective line defined by the action of GL₂(Z) examplesExampleThe set of real numbers rationally equivalent to the point is precisely the set of rational numbers examplesExampleThe proposition above shows that any continued fraction is rationally equivalent to each of its tails It follows that all tails of a continued fraction are rationally equivalent to each other

Periodic continued fractions

In one of the first examples of a continued fraction expansion, it was shown that 10 [3,6,6,6,] This is an example of a periodic continued fraction After a finite number of terms the sequence of integers repeats cyclically If a cyclic pattern is present from the very first term, then the continued fraction is called purely periodic Evidently, $[6,6,6,]103$ is an example of a purely periodic continued fractionNote that a periodic continued fraction cannot represent a rational number since the continued fraction expansion of a rational number is finite theoremsTheoremEvery periodic continued fraction is the continued fraction expansion of a real quadratic irrational numberProof. For clarity: it is being asserted that every periodic continued fraction represent a number of the form a bmc where a b c and m are all integers with m 0 c 0 and m not a perfect squareNumbers of this form with fixed m but varying integers a b and $c0$ may be added, subtracted, multiplied, and divided without leaving the class of such numbers (The statement here about division becomes clear if one remembers always to rationalize denominators.) Consequently, for M in $GL$₂(Z) the number $Mz$ will be a number of this form or if and only if z is in the same classSince a periodic continued fraction is rationally equivalent to a purely periodic continued fraction, the question of whether any periodic continued fraction is a quadratic irrationality reduces to the question of whether a purely periodic continued fraction is such Let x [n₁, , n_r, n₁, , n_r, n₁, , n_r, ] be a purely periodic continued fraction By the proposition of the preceding section, $xMx$ where M is notationally identical to the M in (iref="cfgl"cfgl) Ignoring the computation (iref="prqrcol"prqrcol) of M in terms of convergents, let M rra b c d Then x ax bcx d , or, otherwise said, x is a solution of the quadratic equation cx² (ad) x b 0 remarksRemarkIt is conversely true that the continued fraction expansion of every real quadratic irrationality is periodicThis converse will not be proved here G. Chrystal, Algebra: An Elementary Textbook (2 vols.), Chelsea G. Hardy E. Wright, An Introduction to the Theory of Numbers, Oxford Univ. Press S. Lang, Introduction to Diophantine Approximations, AddisonWesley O. Perron, Die Lehre von den Kettenbruchen, 2nd ed., Chelsea