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Continued Fractions and the Euclidean AlgorithmWilliam F. HammondLecture notes prepared for MATH 326, Spring 1997 Department of Mathematics and Statistics University at AlbanyZGL&TableOfContentsFile;
&SecRef-1;IntroductionContinued 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)
&SecRef-2;The continued fraction expansion of a real numberEvery 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 $n\left[x\right]$ 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 $1u{}_{}$ of u1 satisfies $1u{}_{}1$ 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 $n{}_{}, n{}_{},, n{}_{}$ and real numbers $u{}_{}, u{}_{},, u{}_{}$ that are members of the unit interval I with $u{}_{}, u{}_{},, u{}_{}$ 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, $n{}_{}, n{}_{},$ of integers This sequence is called the continued fraction expansion of x AssertLabel-1conventionConventionWhen $\left[n{}_{}, n{}_{}, ...\right]$ is called a continued fraction, it is understood that all of the numbers nj are integers and that $n{}_{}1$ for j 2
&SecRef-3;First examples 1511 1 4111 11141 12 341 12 1431 12 11 13[1, 2, 1, 3] 10 3 111033 11033 16 111033 16 11033 16 16 1[3, 6, 6, 6, ] [2, 3, 5, 2 ] 2 1[3, 5, 2]2 13 1[5, 2]2 13 15 122 13 11122 13 2112 135112 11358135 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 $\left[n{}_{}, n{}_{},\right]$ 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
&SecRef-4;The case of a rational numberThe 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 $u{}_{}0$ is the case where x is an integer Otherwise u1 0 and the mod one decomposition of $1u{}_{}$ 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 $n{}_{}, n{}_{},$ 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 AssertLabel-2theoremsTheorem1The 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
&SecRef-5;The symbol [t1, t2, , tr]For arbitrary real numbers $t{}_{}, t{}_{},, t{}_{}$ with each $t{}_{}1$ for $j2$ the symbol $\left[ t{}_{}, t{}_{},, t{}_{}\right]$ is defined recursively by: [ t1 ] t1 t-receqn[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="t-rec"t-rec) is nonzero The condition $t{}_{}1$ for $j2$ guarantees, in fact, that [t2,,tr ] 0 as one may prove using inductionIt is an easy consequence of mathematical induction that the symbol $\left[t{}_{}, t{}_{},, t{}_{}\right]$ is a rational number if each tj is rational In particular, each finite continued fraction is a rational number (Note that the symbol $\left[t{}_{}, t{}_{},, t{}_{}\right]$ 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 $\left[t{}_{},, t{}_{}\right]$ 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, , tr] Let, in fact, $t{}_{}, t{}_{},$ be any (finite or infinite) sequence of real numbers One uses the double recursion pdefeqnpj tj pj1 pj2 , j 1 , p0 1 , p1 0 to define the sequence pj , j 1 The double recursion, differently initialized, qdefeqnqj 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 Mr-defeqnMj rrpj qjpj1qj1for j 0 Thus, for example, M0 rr1 001 , and M1 rrt1 110 It is easy to see that the matrices Mj satisfy the double recursion Mr-receqnMj rrtj 110 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 MrcompeqnMr rrtr 110 rrt2 110 rrt1 110 , r 1 This is summarized by: AssertLabel-3propositionsProposition1For any sequence $t{}_{},j1$ of real numbers, if $p{}_{}$ and $q{}_{}$ are the sequences defined by the double recursions (iref="pdef"pdef) and (iref="qdef"qdef), then one has the matrix identity prqr-roweqnrrpr qrpr1qr1 rrtr 110 rrt2 110 rrt1 110 for each integer r 1 pqidencorollariesCorollary1One has the identity $p{}_{}q{}_{}q{}_{}p{}_{}\left(1\right)r$ for each integer r 1Proof. The number $p{}_{}q{}_{}q{}_{}p{}_{}$ is the determinant of the matrix Mr From the formula (iref="Mrcomp"Mrcomp) the matrix $M{}_{}$ 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 videncorollariesCorollary2One has the vector identity prqr-veceqnrpr qr rrt1 110 rrt2 110 rrtr 110 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 bcd 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="prqr-row"prqr-row) to obtain: prqr-coleqnrrpr pr1qrqr1 rrt1 110 rrt2 110 rrtr 110 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="prqr-vec"prqr-vec) AssertLabel-6theoremsTheorem2For any sequence $t{}_{},j1$ of real numbers, if $p{}_{}$ and $q{}_{}$ are the sequences defined by the double recursions (iref="pdef"pdef) and (iref="qdef"qdef), and if $t{}_{}1$ for j 2 then the value of the symbol $\left[t{}_{},, t{}_{}\right]$ is given by the formula t-compeqn[t1, t2, , tr] prqr for r 1 Proof. What is slightly strange about this important result is that while the $p{}_{}$ and the $q{}_{}$ are defined by the front end recursions, albeit double recursions, (iref="pdef"pdef) and (iref="qdef"qdef), the symbol $\left[t{}_{},, t{}_{}\right]$ is defined by the back end recursion (iref="t-rec"t-rec) The proof begins with the comment that the righthand side of (iref="t-comp"t-comp) does not make sense unless one can be sure that the denominator qr 0 One can show easily by induction on r that $q{}_{}1$ for $r1$ under the hypothesis $t{}_{}1$ 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, $p{}_{}t{}_{}$ and q11 Assume now that r 2 By induction we may assume the correctness of the statement (iref="t-comp"t-comp) for symbols of length r1 and, therefore, for the symbol [t2, , tr] That case of the statement says that $\left[t{}_{},, t{}_{}\right]$ must be equal to ac where by corollary iref="viden"viden ra c rra bcd r1 0 with rra bcd rrt2 110 rrtr 110 Now by (iref="t-rec"t-rec) [t1, t2, , tr] t1 1ac t1 ca at1 ca But by corollary iref="viden"viden again rpr qr rrt1 110 rra bcd r1 0 rrat1 c bt1 dab r1 0 rat1 c a Hence, prqr at1 ca [t1, t2, , tr]
&SecRef-6;Application to Continued FractionsRecall that $\left[n{}_{}, n{}_{},\right]$ is called a continued fraction only when each nj is an integer and $n{}_{}1$ for j 2 The sequence $n{}_{}, n{}_{},$ may be finite or infinite The symbol $c{}_{}\left[n{}_{}, n{}_{},, n{}_{}\right]$ formed with the first r terms of the sequence, is called the rth convergent of the continued fraction Associated with a given sequence $n{}_{}, n{}_{},$ are two sequences $p{}_{}, p{}_{},$ and $q{}_{}, q{}_{},$ that are given, according to the double recursions (iref="pdef"pdef), (iref="qdef"qdef) of the previous section with tj nj AssertLabel-7propositionsProposition2If $\left[n{}_{}, n{}_{},\right]$ is a continued fraction, then the integers pr and qr are coprime for each r 1Proof. 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 AssertLabel-8propositionsProposition3The difference between successive convergents of the continued fraction $\left[n{}_{}, n{}_{},\right]$ is given by the formula convgtdiffeqncr cr1 (1)rqr qr1 for r 2 Proof. According to the theorem (formula iref="t-comp"t-comp) 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.) AssertLabel-9remarksRemark1The formula (iref="convgtdiff"convgtdiff) remains true if $c{}_{}\left[t{}_{},, t{}_{}\right]$ where the tj are real numbers subject to the assumption $t{}_{}1$ for j 1 AssertLabel-10AssertDefaultLemmaThe sequence $q{}_{}$ is a strictly increasing sequence for j 2Proof. This is easily proved by induction from the recursive definition (iref="qdef"qdef) of the sequence AssertLabel-11theoremsTheorem3If $\left[n{}_{}, n{}_{},\right]$ is an infinite continued fraction, then the limit limr prqr always existsProof. 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 $q{}_{}q{}_{}$ has infinite limit, and so the sequence of reciprocals $1q{}_{}q{}_{}$ 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 AssertLabel-12definitionsDefinition1The limit of the sequence of convergents of an infinite continued fraction is called the value of that continued fraction AssertLabel-13theoremsTheorem4If $\left[n{}_{}, n{}_{},\right]$ 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 numberProof. For each $r1$ the continued fraction expansion $\left[n{}_{}, n{}_{},\right]$ of x is characterized by the identity xconfraceqnx [n1, n2, , nr ur] , where ur is a real number with 0 ur 1 The sequences of ps and qs for the symbol $\left[n{}_{}, n{}_{},, n{}_{}u{}_{}\right]$ agree with those for the symbol $\left[n{}_{}, n{}_{},, n{}_{}\right]$ except for the rth terms One has by (iref="t-comp"t-comp) [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
&SecRef-7;Bezouts Identity and the double recursionIt 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, remreceqnrj 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 stagesOne may use the sequence of successive quotients nj (j 1) to form sequences $p{}_{}$ and qj as in the previous section, according to the double recursions: npdefeqnpj nj pj1pj2 ,j 1 ;p0 1 ,p1 0 nqdefeqnqj nj qj1qj2 ,j 1 ;q0 0 ,q1 1 It has already been observed that $q{}_{}1$ for $j1$ 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 AssertLabel-14theoremsTheorem5If the application of the Euclidean algorithm to a and b (b 0) ends with the mth long division, i.e., rm 0 then bezouteqnrj (1)j1 qj a pj 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) $q{}_{}1$ 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 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 AssertLabel-15corollariesCorollary3The greatest common divisor d of a and b is given by the formula gcdbezouteqnd (1)m (qm1a pm1 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 rm1 in the Euclidean algorithm This formula for d is the case $jm1$ in (iref="bezout"bezout) AssertLabel-16corollariesCorollary4checkbezouteqnpm ad , qm bd Proof. The last remainder rm 0 The case $jm$ 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
&SecRef-8;The action of $GL{}_{}\left(Z\right)$ on the projective lineIf a b c d are real numbers with $adbc0$and M rra bcd 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 AssertLabel-17propositionsProposition4If $\left[n{}_{}, n{}_{},\right]$ is any continued fraction, then cfgleqn[n1, n2, , nr, nr1, ] M [nr1, ] where M rrn1 110 rrnr 110 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="prqr-col"prqr-col), after replacing tj with nj, by the column rz 1 The matrix M in the preceding proposition is an integer matrix with determinant 1 The notation $GL{}_{}\left(Z\right)$ 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 $GL{}_{}\left(Z\right)$ is a member of $GL{}_{}\left(Z\right)$ and that the matrix inverse of a member of $GL{}_{}\left(Z\right)$ is a member of GL2(Z) Thus, $GL{}_{}\left(Z\right)$ forms what is called a group The formula (iref="glact"glact) defines what is called the action of $GL{}_{}\left(Z\right)$ 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 $GL{}_{}\left(Z\right)$ for which w M z Since (i) $GL{}_{}\left(Z\right)$ 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 AssertLabel-18AssertDefaultTerminologyThe 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) AssertLabel-19examplesExample1The set of real numbers rationally equivalent to the point is precisely the set of rational numbers AssertLabel-20examplesExample2The 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
&SecRef-9;Periodic continued fractionsIn 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, $\left[6,6,6,\right]103$ 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 AssertLabel-21theoremsTheorem6Every 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 $GL{}_{}\left(Z\right)$ 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, , nr, n1, , nr, n1, , nr, ] 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="prqr-col"prqr-col) of M in terms of convergents, let M rra bcd Then x ax bcx d , or, otherwise said, x is a solution of the quadratic equation cx2 (ad) x b 0 AssertLabel-22remarksRemark2It is conversely true that the continued fraction expansion of every real quadratic irrationality is periodicThis converse will not be proved here
&BibRef-KEY-1;KEY-1 G. Chrystal, Algebra: An Elementary Textbook (2 vols.), Chelsea &BibRef-KEY-2;KEY-2 G. Hardy E. Wright, An Introduction to the Theory of Numbers, Oxford Univ. Press &BibRef-KEY-3;KEY-3 S. Lang, Introduction to Diophantine Approximations, AddisonWesley &BibRef-KEY-4;KEY-4 O. Perron, Die Lehre von den Kettenbruchen, 2nd ed., Chelsea