Continued Fractions and the Euclidean Algorithm William F. Hammond Lecture notes prepared for MATH 326, Spring 1997 Department of Mathematics and Statistics University at Albany Z GL
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) decimal The 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 u x n falls in the unit interval I consisting of all real numbers u for which 0 u 1 Thus, 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 fraction The 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 1 If 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 x If u1 0 then the reciprocal 1u1 of u1 satisfies 1u1 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 n1, n2, , nk and real numbers u1, u2, , uk that are members of the unit interval I with u1, u2, , uk1 all positive One may write x n1 1n2 1n3 1 1nk uk Alternatively, one may write x [ n1, n2, n3, , nk uk ] If uk 0 the process ends after k steps Otherwise, the process continues at least one more step with 1uk nk1 uk1 In this way one associates with any real number x a sequence, which could be either finite or infinite, n1, n2, of integers This sequence is called the continued fraction expansion of x conventionConvention When [n1, n2, ...] is called a continued fraction, it is understood that all of the numbers nj are integers and that nj 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 [n1, 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 denominator Let 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 u1 0 is the case where x is an integer Otherwise u1 0 and the mod one decomposition of 1u1 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 n1, 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 theoremsTheorem The continued fraction expansion of a real number is finite if and only if the real number is rational Proof. 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, , tr]
For arbitrary real numbers t1, t2, , tr with each tj 1 for j 2 the symbol [ t1, t2, , tr ] is defined recursively by: [ t1 ] t1 treceqn [t1,t2,,tr ]t11[t2,,tr ] 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 tj 1 for j 2 guarantees, in fact, that [t2,,tr ] 0 as one may prove using induction It is an easy consequence of mathematical induction that the symbol [t1, t2, , tr] is a rational number if each tj is rational In particular, each finite continued fraction is a rational number (Note that the symbol [t1, t2, , tr] is to be called a continued fraction, according to the convention of the first section, only when each tj is an integer.) Observe that the recursive nature of the symbol [t1, , tr] 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 8135 There is, however, another approach to computing [t1, t2, , tr] Let, in fact, t1, t2, be any (finite or infinite) sequence of real numbers One uses the double recursion pdefeqn pj tj pj1 pj2 , j 1 , p0 1 , p1 0 to define the sequence pj , j 1 The double recursion, differently initialized, qdefeqn qj tj qj1 qj2 , j 1 , q0 0 , q1 1 defines the sequence qj , j 1 Note that p1 t1 p2 t1t2 1 and q1 1 q2 t2 q3 t2t3 1 One now forms the matrix Mrdefeqn Mj rr pj qj pj1 qj1 for j 0 Thus, for example, M0 rr 1 0 0 1 , and M1 rr t1 1 1 0 It is easy to see that the matrices Mj satisfy the double recursion Mrreceqn Mj rr tj 1 1 0 Mj1 , j 1 as a consequence of the double recursion formulas for the pj and qj Hence, a simple argument by mathematical induction shows that Mrcompeqn Mr rr tr 1 1 0 rr t2 1 1 0 rr t1 1 1 0 , r 1 This is summarized by: propositionsProposition For any sequence tj, j 1 of real numbers, if pj and qj are the sequences defined by the double recursions (iref="pdef"pdef) and (iref="qdef"qdef), then one has the matrix identity prqrroweqn rr pr qr pr1 qr1 rr tr 1 1 0 rr t2 1 1 0 rr t1 1 1 0 for each integer r 1 pqidencorollariesCorollary One has the identity pr qr1 qr pr1 (1)r for each integer r 1 Proof. The number pr qr1 qr pr1 is the determinant of the matrix Mr From the formula (iref="Mrcomp"Mrcomp) the matrix Mr 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(Mr) (1)r videncorollariesCorollary One has the vector identity prqrveceqn r pr qr rr t1 1 1 0 rr t2 1 1 0 rr tr 1 1 0 r 1 0 for each integer r 1 Proof. First recall (i) that the product of a matrix and a (column) vector is defined by the relation rr a b c d r x y r ax 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: prqrcoleqn rr pr pr1 qr qr1 rr t1 1 1 0 rr t2 1 1 0 rr tr 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 r 1 0 in order to obtain the column identity (iref="prqrvec"prqrvec) theoremsTheorem For any sequence tj, j 1 of real numbers, if pj and qj are the sequences defined by the double recursions (iref="pdef"pdef) and (iref="qdef"qdef), and if tj 1 for j 2 then the value of the symbol [t1, , tr] is given by the formula tcompeqn [t1, t2, , tr] prqr for r 1 Proof. What is slightly strange about this important result is that while the pr and the qr are defined by the front end recursions, albeit double recursions, (iref="pdef"pdef) and (iref="qdef"qdef), the symbol [t1, , tr] 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 qr 0 One can show easily by induction on r that qr 1 for r 1 under the hypothesis tj 1 for j 2 The proof proceeds by induction on r If r 1 the assertion of the theorem is simply the statement t1 p1q1 and, as noted above, p1t1 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, , tr] That case of the statement says that [t2, , tr] must be equal to ac where by corollary iref="viden"viden r a c rr a b c d r 1 0 with rr a b c d rr t2 1 1 0 rr tr 1 1 0 Now by (iref="trec"trec) [t1, t2, , tr] t1 1ac t1 ca at1 ca But by corollary iref="viden"viden again r pr qr rr t1 1 1 0 rr a b c d r 1 0 rr at1 c bt1 d a b r 1 0 r at1 c a Hence, prqr at1 ca [t1, t2, , tr]
Application to Continued Fractions
Recall that [n1, n2, ] is called a continued fraction only when each nj is an integer and nj 1 for j 2 The sequence n1, n2, may be finite or infinite The symbol cr [n1, n2, , nr] formed with the first r terms of the sequence, is called the rth convergent of the continued fraction Associated with a given sequence n1, n2, are two sequences p1, p2, and q1, q2, that are given, according to the double recursions (iref="pdef"pdef), (iref="qdef"qdef) of the previous section with tj nj propositionsProposition If [n1, n2, ] is a continued fraction, then the integers pr and qr are coprime for each r 1 Proof. By Corollary iref="pqiden"pqiden of the previous section pr qr1 qr pr1 (1)r Hence, any positive divisor of both pr and qr must divide the lefthand side of this relation, and, therefore, must also divide (1)r propositionsProposition The difference between successive convergents of the continued fraction [n1, n2, ] is given by the formula convgtdiffeqn cr cr1 (1)rqr qr1 for r 2 Proof. According to the theorem (formula iref="tcomp"tcomp) at the end of the last section the convergent cr is given by cr prqr Hence, cr cr1 prqr pr1qr1 pr qr1 pr1 qrqr qr1 (1)rqr qr1 (The last step is by Corollary iref="pqiden"pqiden above.) remarksRemark The formula (iref="convgtdiff"convgtdiff) remains true if cr [t1, , tr] where the tj are real numbers subject to the assumption tj 1 for j 1 Lemma The sequence qj is a strictly increasing sequence for j 2 Proof. This is easily proved by induction from the recursive definition (iref="qdef"qdef) of the sequence theoremsTheorem If [n1, n2, ] is an infinite continued fraction, then the limit limr prqr always exists Proof. As one plots the convergents cr 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 qj qj1 has infinite limit, and so the sequence of reciprocals 1qj qj1 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 definitionsDefinition The limit of the sequence of convergents of an infinite continued fraction is called the value of that continued fraction theoremsTheorem If [n1, n2, ] is the continued fraction expansion of an irrational number x then x limr prqr ; that is, the value of the continued fraction expansion of a real number is that real number Proof. For each r 1 the continued fraction expansion [n1, n2, ] of x is characterized by the identity xconfraceqn x [n1, n2, , nr ur] , where ur is a real number with 0 ur 1 The sequences of ps and qs for the symbol [n1, n2, , nr ur] agree with those for the symbol [n1, n2, , nr] except for the rth terms One has by (iref="tcomp"tcomp) [n1, n2, , nr ur] PrQr , where by (iref="qdef"qdef) qr nr qr1 qr2 Qr (nr ur) qr1 qr2 Hence, Qr qr ur qr1 Therefore, the displacement from cr1 to x is by (iref="convgtdiff"convgtdiff) (1)rQr qr1 (1)r(qr qr1 ur qr12) , which is in the same direction but of smaller magnitude than the displacement from cr1 to cr 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 jth stage one performs long division of rj2 by rj1 to obtain the integer quotient nj and the integer remainder rj that satisfies 0 rj rj1 Thus, remreceqn rj rj2 nj rj1 The Euclidean algorithm admits an additional stage if rj 0 Since 0 rj rj1 r2 r1 r0 b , there can be at most b stages One may use the sequence of successive quotients nj (j 1) to form sequences pj and qj as in the previous section, according to the double recursions: npdefeqn pj nj pj1pj2 ,j 1 ;p0 1 ,p1 0 nqdefeqn qj nj qj1qj2 ,j 1 ;q0 0 ,q1 1 It has already been observed that qj 1 for j 1 and [n1, n2, , nj]pjqj ,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 theoremsTheorem If the application of the Euclidean algorithm to a and b (b 0) ends with the mth long division, i.e., rm 0 then bezouteqn rj (1)j1 qj a pj b , 1 j m Proof. One uses induction on j For j 1 the statement is r1 q1 a p1 b Since by (iref="npdef"npdef, iref="nqdef"nqdef) q1 1 and p1 n1 this statement is simply the case j 1 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 rj rj2 nj rj1 In this equation one may use (iref="bezout"bezout) to expand the terms rj2 and rj1 to obtain: rj (1)j3(qj2a pj2b) nj (1)j2(qj1a pj1b) (1)j1(qj2a pj2b) nj (1)j1(qj1a pj1b) (1)j1(qj2a pj2b) nj (qj1a pj1b) (1)j1(qj2 nj qj1)a (pj2 nj pj1)b (1)j1 qj a pj b corollariesCorollary The greatest common divisor d of a and b is given by the formula gcdbezouteqn d (1)m (qm1a pm1 b) , where m is the number of divisions required to obtain zero remainder in the Euclidean algorithm Proof. One knows that d is the last nonzero remainder rm1 in the Euclidean algorithm This formula for d is the case j m1 in (iref="bezout"bezout) corollariesCorollary checkbezouteqn pm ad , qm bd Proof. The last remainder rm 0 The case j m in (iref="bezout"bezout) shows that ab pmqm Since, by the first proposition of the preceding section, pm and qm have no common factor, this corollary is evident
The action of GL2(Z) on the projective line
If a b c d are real numbers with ad bc 0and M rr a b c d is the matrix with entries a b c and d then M z for z real, will denote the expression glacteqn M z a z bc z d One calls M z the action of M on z M z is a perfectly good function of z except for the case z dc where the denominator cz d vanishes If it were also true that az b 0 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 M w as w increases without bound, then one may regard the expression M z 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 propositionsProposition If [n1, n2, ] is any continued fraction, then cfgleqn [n1, n2, , nr, nr1, ] M [nr1, ] where M rr n1 1 1 0 rr nr 1 1 0 Proof. Let z [nr1, ] Then [n1, n2, , nr, nr1, ] [n1, n2, , nr, z ] The statement of the proposition now becomes [n1, n2, , nr, z ] M z This may be seen to follow by multiplying both sides in formula (iref="prqrcol"prqrcol), after replacing tj with nj, by the column r z 1 The matrix M in the preceding proposition is an integer matrix with determinant 1 The notation GL2(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 GL2(Z) is a member of GL2(Z) and that the matrix inverse of a member of GL2(Z) is a member of GL2(Z) Thus, GL2(Z) forms what is called a group The formula (iref="glact"glact) defines what is called the action of GL2(Z) on the projective line One says that two points z and w of the projective line are rationally equivalent if there is a matrix M in GL2(Z) for which w M z Since (i) GL2(Z) is a group, (ii) M1 (M2 z) (M1 M2) z and (iii) w M z 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 Terminology The 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) examplesExample The set of real numbers rationally equivalent to the point is precisely the set of rational numbers examplesExample The 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 fraction Note that a periodic continued fraction cannot represent a rational number since the continued fraction expansion of a rational number is finite theoremsTheorem Every periodic continued fraction is the continued fraction expansion of a real quadratic irrational number Proof. 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 square Numbers of this form with fixed m but varying integers a b and c 0 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 GL2(Z) the number M z will be a number of this form or if and only if z is in the same class Since 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, , nr, n1, , nr, n1, , nr, ] be a purely periodic continued fraction By the proposition of the preceding section, x M x 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 rr a b c d Then x ax bcx d , or, otherwise said, x is a solution of the quadratic equation cx2 (ad) x b 0 remarksRemark It is conversely true that the continued fraction expansion of every real quadratic irrationality is periodic This 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