Continued fractions
Yann BUGEAUD
Let x0 , x1 , . . . be real numbers with x1 , x2 , . . . positive. A finite continued fraction
denotes any expression of the form
1
[x0 ; x1 , x2 . . . , xn ] = x0 +
,
1
x1 +
1
x2 +
... +
1
xn
and, more generally, we call any expression of the above form or of the form
1
[x0 ; x1 , x2 , . . .] = x0 +
x1 +
= lim [x0 ; x1 , x2 . . . , xn ]
1
x2 +
n→+∞
1
...
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 =
[x0 ; x1 , x2 , . . .], where x0 is an integer and xn a positive integer for any n ≥ 1. 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 (an )n≥0 , with an ≥ 1 for n ≥ 1, and we
show that the sequence of rational numbers [a0 ; a1 , a2 . . . , an ] 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 [1],
Dajani & Kraaikamp [3], Hardy & Wright [4], Perron [5], and Schmidt [7].
Lemma 1.1. Any rational number r has exactly two different continued fraction expansions. These are [r] and [r − 1; 1] if r is an integer and, otherwise, one of them has the
form [a0 ; a1 , . . . , an−1 , an ] with an ≥ 2, and the other one is [a0 ; a1 , . . . , an−1 , an − 1, 1].
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 = 1, then r = u = [u], and if r = [a0 ; a1 , . . . , an ] with
n ≥ 1, we have r = a0 + 1/[a1 ; a2 , . . . , an ]. Since a1 ≥ 1 and r is an integer, we deduce
that n = 1 and a1 = 1, thus r = a0 + 1 = [r − 1; 1].
We assume v ≥ 2 and that the lemma holds true for any rational of denominator
positive and at most equal to v − 1. Performing the Euclidean division of u by v, there
exist integers q and c with u = qv + c and 1 ≤ c ≤ v − 1. 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 2010.
1
which we denote by [a1 ; a2 , . . . , an−1 , an ] with an ≥ 2, and [a1 ; a2 , . . . , an−1 , an − 1, 1].
Setting a0 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 [1] if p/q = 1, and that one which does not terminate in 1 if p/q 6= 1.
The following algorithm allows us to associate to any irrational real number ξ an
infinite sequence of integers. Let define the integer a0 and the real number ξ1 > 1 by
a0 = [ξ]
and
ξ1 = 1/{ξ}.
We then have ξ = a0 + 1/ξ1 . For any positive integer n, we define inductively the integer
an and the real number ξn+1 > 1 by
an = [ξn ]
ξn+1 = 1/{ξn },
and
and we observe that ξn = an + 1/ξn+1 . 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 a0 , a1 , a2 , . . . with an positive for all n ≥ 1.
If ξ is rational, the same algorithm terminates and associates to ξ a finite sequence
of integers. Indeed, the ξj ’s are then rational numbers and, if we set ξj = uj /vj , with
uj , vj positive and gcd(uj , vj ) = 1, an easy induction shows that we get uj > uj+1 ,
for any positive integer j with vj 6= 1. Consequently, there must be some index n for
which vn−1 6= 1 and ξn is an integer. Thus, an+1 , an+2 , . . . are not defined. We have
ξ = [a0 ; a1 , . . . , an ], and this corresponds to the Euclidean algorithm.
Definition 1.2. Let ξ be an irrational number (resp. a rational number). Let a0 , a1 , . . .
(resp. a0 , a1 , . . . , aN ) be the sequence of integers associated to ξ by the algorithm defined
above. For any integer n ≥ 1 (resp. n = 1, . . . , N ), the rational number
pn
:= [a0 ; a1 , . . . , an ]
qn
is called the n-th convergent of ξ and an is termed the n-th partial quotient of ξ. Further,
for any integer n ≥ 1 (resp. n = 1, . . . , N − 1), there exists a real number ηn in ]0, 1[ such
that,
ξ = [a0 ; a1 , . . . , an−1 , an + ηn ].
We observe that the real numbers ηn occurring in Definition 1.2 are exactly the real
numbers 1/ξn+1 given by the algorithm.
In all what follows until the end of Theorem 1.7 included, unless otherwise explicitly
stated, we assume that ξ is a real irrational number and we associate to ξ the sequences
(an )n≥0 and (pn /qn )n≥1 as given by Definition 1.2. However, the statements below remain
true for rational numbers ξ provided that the an ’s and the pn /qn ’s are well-defined.
The integers pn and qn can be easily expressed in terms of an , pn−1 , pn−2 , qn−1 , and
qn−2 .
2
Theorem 1.3. Setting
p−1 = 1, q−1 = 0, p0 = a0 , and q0 = 1,
we have, for any positive integer n,
pn = an pn−1 + pn−2
and
qn = an qn−1 + qn−2 .
Proof. We proceed by induction. Since p1 /q1 = a0 + 1/a1 = (a0 a1 + 1)/a1 , the definitions
of p−1 , q−1 , p0 , and q0 show that the theorem is true for n = 1. Assume that it holds
true for a positive integer n and denote by p00 /q00 , . . . , p0n /qn0 the convergents of the rational
number [a1 ; a2 , . . . , an+1 ]. For any integer j with 0 ≤ j ≤ n + 1 we have
0
qj−1
1
pj
= [a0 ; a1 , . . . , aj ] = a0 +
= a0 + 0 ,
qj
[a1 ; a2 , . . . , aj ]
pj−1
thus
0
pj = a0 p0j−1 + qj−1
and qj = p0j−1 .
(1)
It follows from (1) with j = n + 1 and the inductive hypothesis applied to the rational
[a1 ; a2 , . . . , an+1 ] that
0
0
pn+1 = a0 (an+1 p0n−1 + p0n−2 ) + an+1 qn−1
+ qn−2
0
0
= an+1 (a0 p0n−1 + qn−1
) + (a0 p0n−2 + qn−2
)
and
qn+1 = an+1 p0n−1 + p0n−2 ,
whence, by (1) with j = n and j = n − 1, we get qn+1 = an+1 qn + qn−1 and pn+1 =
an+1 pn + pn−1 , as claimed.
Theorem 1.4. For any non-negative integer n, we have
qn pn−1 − pn qn−1 = (−1)n
(2)
qn pn−2 − pn qn−2 = (−1)n−1 an .
(3)
and, for all n ≥ 1,
Proof. These equalities are clearly true for n = 0 and n = 1. It then suffices to argue by
induction, using Theorem 1.3.
Lemma 1.5. For any irrational number ξ and any non-negative integer n, the difference
ξ − pn /qn is positive if, and only if, n is even.
Proof. We easily check that this is true for n = 0, 1, and 2 and we proceed by induction.
Let n ≥ 4 be an even integer. Then ξ = [a0 ; a1 , [a2 ; a3 , . . . , an + ηn ]] and the inductive
hypothesis implies that [a2 ; a3 , . . . , an + ηn ] > [a2 ; a3 , . . . , an ]. Since [a0 ; a1 , u] > [a0 ; a1 , v]
holds for all positive real numbers u > v, we get that ξ > [a0 ; a1 , [a2 ; a3 , . . . , an ]] = pn /qn .
We deal with the case n odd in exactly the same way.
As a corollary of Lemma 1.5, we get a result of Vahlen.
3
Corollary 1.6. Let pn /qn and pn+1 /qn+1 be two consecutive convergents of the continued
fraction expansion of an irrational number ξ. Then at least one of them satisfies
p
ξ − < 1 .
(4)
q 2q 2
In particular, there are infinitely many rational numbers p/q satisfying (4).
Proof. We infer from Lemma 1.5 that ξ is an inner point of the interval bounded by pn /qn
and pn+1 /qn+1 . Thus, using (2) and the inequality a2 + b2 > 2ab, valid for any distinct
real numbers a and b, we get
pn
p
p
1
1
1
p
n
n+1
n+1
= ξ −
+ ξ −
,
+
>
=
−
2
qn
2qn2
2qn+1
qn qn+1
qn+1 qn qn+1 and the claimed result follows.
The next two theorems show that the real irrational numbers are in a one-to-one
correspondence with the set of integer sequences (ai )i≥0 with ai positive for i ≥ 1.
Theorem 1.7. 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 (pn /qn )n≥0 converges to ξ, and we set
ξ = [a0 ; a1 , a2 , . . .].
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
pn−2
pn
(−1)n−1 an
−
=
,
qn−2
qn
qn qn−2
and, since the an ’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 1.5, we have p2n /q2n < ξ <
p2n+1 /q2n+1 for all n ≥ 0, and, by (2), the difference p2n /q2n − p2n+1 /q2n+1 tends to 0
when n tends to infinity. Unicity is clear. Indeed, if (bi )i≥0 is a sequence of integers with
bi positive for i ≥ 1 and such that limn→+∞ [b0 ; b1 , b2 , . . .] exist, then this limit cannot
be equal to limn→+∞ [a0 ; a1 , a2 , . . .] as soon as there exists a non-negative integer i with
ai 6= bi .
Theorem 1.8. Let a0 , a1 , . . . be integers with a1 , a2 , . . . positive. Then the sequence of
rational numbers [a0 ; a1 , . . . , ai ], i ≥ 1, converges to the irrational number whose partial
quotients are precisely a0 , a1 , . . .
Proof. For any positive integer n, denote by pn /qn the rational number [a0 ; a1 , . . . , an ]. The
recurrence relations obtained in Theorems 1.3 and 1.4 hold true in the present context. As
4
in the proof of Theorem 1.7, we deduce from (2) and (3) that the sequences (p2n /q2n )n≥1
and (p2n+1 /q2n+1 )n≥1 are adjacent. Hence, they converge to the same limit, namely to
the irrational number [a0 ; a1 , a2 , . . .], whose partial quotients are precisely a0 , a1 , . . ., by
Theorem 1.7.
We observe that for any irrational number ξ the sequences (an )n≥0 and (ξn )n≥1 given
by the algorithm defined below Lemma 1.1 satisfy ξn = [an ; an+1 , an+2 , . . .] for all positive
integers n.
Theorem 1.9. Let n be a positive integer and ξ = [a0 ; a1 , a2 . . .] be an irrational number.
We then have
pn ξn+1 + pn−1
ξ = [a0 ; a1 , . . . , an , ξn+1 ] =
qn ξn+1 + qn−1
and
qn ξ − pn =
(−1)n
(−1)n
1
=
·
.
qn ξn+1 + qn−1
qn
ξn+1 + [0; an , an−1 , . . . , a1 ]
Furthermore, the set of real numbers having a continued fraction expansion whose n + 1
first partial quotients are a0 , a1 , . . . , an is precisely the closed interval bounded by (pn−1 +
pn )/(qn−1 + qn ) and pn /qn , which are equal to [a0 ; a1 , . . . , an , 1] and [a0 ; a1 , . . . , an ], respectively.
Proof. We proceed by induction, using Theorem 1.3 and noticing that we have ξn =
an + 1/ξn+1 and qn /qn−1 = [an ; an−1 , . . . , a1 ] for all positive integers n. The last assertion
of the theorem follows immediately, since the admissible values of ξn+1 run exactly through
the interval ]1, +∞[.
Corollary 1.10. For any irrational number ξ and any non-negative integer n, we have
p
1
1
n
<
< ξ −
.
qn (qn + qn+1 )
qn
qn qn+1
Proof. Writing ξ = [a0 ; a1 , a2 , . . .] and ξn+1 = [an+1 ; an+2 , . . .], we observe that an+1 <
ξn+1 < an+1 + 1, and we get from Theorem 1.9 that
1
1
p
n
<
< ξ −
.
qn
qn (an+1 qn + qn−1 )
qn (an+1 + 1)qn + qn−1
The corollary follows then from Theorem 1.3.
The following result of Legendre provides a partial converse to Corollary 1.6.
Theorem 1.11. Let ξ be a real number. Any non-zero rational number a/b with
a
ξ − < 1
b 2b2
is a convergent of ξ.
Proof. We assume that ξ 6= a/b and we write ξ−a/b = εθ/b2 , with ε = ±1 and 0 < θ < 1/2.
By Lemma 1.1, setting an−1 = 1 if necessary, we may write a/b = [a0 ; a1 , . . . , an−1 ], with
5
n given by (−1)n−1 = ε, and we denote by p1 /q1 , . . . , pn−1 /qn−1 the convergents of a/b.
Let ω be such that
pn−1 ω + pn−2
ξ=
= [a0 ; a1 , . . . , an−1 , ω].
qn−1 ω + qn−2
We check that
a
εθ
pn−1
1
1
(−1)n−1
=
ξ
−
=
ξ
−
=
(ξq
−
p
)
=
·
,
n−1
n−1
b2
b
qn−1
qn−1
qn−1 ωqn−1 + qn−2
whence
ω=
1 qn−2
−
θ qn−1
and ω > 1. Denote by [an ; an+1 , an+2 , . . .] the continued fraction expansion of ω. Since
aj ≥ 1 for all j ≥ 1, we have
ξ = [a0 ; a1 , . . . , an−1 , ω] = [a0 ; a1 , a2 , . . .],
and Theorem 1.8 yields that a/b = pn−1 /qn−1 is a convergent of ξ.
A less-known result of Fatou provides a satisfactory complement to Theorem 1.11. It
asserts that if the real number ξ and the rational number a/b satisfy |ξ − a/b| < 1/b2 , then
there exists an integer n such that
a
b
belongs to
pn pn+1 + pn pn+2 − pn+1
,
,
qn qn+1 + qn qn+2 − qn+1
.
Definition 1.12. The irrational real number ξ is said to be badly approximable if there
exists a positive constant c(ξ) such that
ξ −
p c(ξ)
≥ 2
q
q
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 1), and many other real
numbers share this property, as follows from the next theorem.
Theorem 1.13. 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
1.10 that qn ≤ qn−1 /c. Since Theorem 1.3 yields that qn ≥ an qn−1 , we get an ≤ 1/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 qn+1 ≤ (M + 1)qn for any non-negative integer n. Furthermore, if a/b satisfies
6
|ξ − a/b| < 1/(2b2 ), then, by Theorem 1.11 and Corollary 1.10, there exists a non-negative
integer n such that a/b = pn /qn and
a
1
1
ξ − >
.
≥
b b(b + qn+1 )
(M + 2)b2
This shows that ξ is badly approximable. The last assertion of the theorem follows from
Theorem 1.8.
Quadratic numbers are badly approximable numbers, thus, by Theorem 1.13, their
partial quotients are bounded. However, much more is known: the sequence of their partial
quotients is ultimately periodic.
Theorem 1.14. The real irrational number ξ = [a0 ; a1 , a2 , . . .] has a periodic continued
fraction expansion (that is, there exist integers k ≥ 0 and n ≥ 1 such that am+n = am 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 1770, see Exercise 2.
The next result dates back to (at least) 1877 and can be found in the book of Serret.
Theorem 1.15. Let ξ = [a0 ; a1 , a2 , . . .] and η = [b0 ; b1 , b2 , . . .] be two irrational numbers.
There exist integers u and v such that au+n = bv+n for every positive integer n if, and
only if, there exist integers a, b, c, and d such that |ad − bc| = 1 and η = (aξ + b)/(cξ + d).
Two real numbers satisfying the equivalent conditions of Theorem 1.15 are called
equivalent.
Hurwitz improved the last assertion of Corollary 1.6.
Theorem 1.16. For any real irrational number ξ, there exist infinitely many rationals
p/q satisfying
p
ξ − < √ 1 .
q
5q 2
√
Further, the constant 5 cannot be replaced by a larger real number.
√
We observe that the Golden Section (1 + 5)/2 = [1; 1, 1, . . . , 1, . . .] is, up to equivalence, the irrational number which is the most badly approximable by rational numbers.
As shown by Hurwitz, Theorem 1.16 can be improved if, besides the rationals, we also
exclude the numbers which are equivalent to the Golden Section.
√
Theorem 1.17. For any irrational real number ξ which is not equivalent to (1 + 5)/2,
there exist infinitely many rationals p/q satisfying
p
ξ − < √ 1 .
q
8q 2
√
The constant 8 cannot be replaced by a larger real number.
In other words, the assumptions of Theorem 1.17 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 (1 + 2)/2.
7
More generally, for a real ξ we define
ν(ξ) = lim inf qkqξk,
q→+∞
where k · k 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 1.16, it is included in the interval [0, 1/ 5].
Theorem 1.18. There exists a sequence of numbers
1
1
ν1 = √ > ν2 = √ > ν3 > ν4 > . . .
5
8
tending to 1/3 such that, for all νi , there is, up to equivalence, a finite number of real
numbers ξ satisfying ν(ξ) = νi .
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 1.3 imply
that, if ξ = [a0 ; a1 , . . .] is a positive real irrational number, with convergents pn /qn =
[a0 ; a1 , . . . , an ], then, for n ≥ 1, we have
Mn :=
pn
qn
pn−1
qn−1
=
a0
1
1
0
a1
1
1
0
...
an
1
1
0
.
(5)
Theorem 1.19. Let n ≥ 2 be an integer and a1 , . . . , an be positive integers. For j =
1, . . . , n, set pj /qj = [0; a1 , . . . , aj ]. Then, we have
qn−1
= [0; an , an−1 , . . . , a1 ].
qn
Proof. We get from Theorem 1.3 that
qn+1
qn−1
= an+1 +
.
qn
qn
Theorem 1.19 then follows by induction.
Alternatively, we can conclude by setting a0 = 0 in (5) and then by taking the transpose of both sides of (5).
Definition 1.20. Let m ≥ 1 and a1 , . . . , am be positive integers. The continuant of
a1 , . . . , am , usually denoted by Km (a1 , . . . , am ), is the denominator of the rational number
[0; a1 , . . . , am ].
Theorem 1.21. For any positive integers a1 , . . . , am and any integer k with 1 ≤ k ≤ m−1,
we have
Km (a1 , . . . , am ) = Km (am , . . . , a1 ),
(6)
8
2(m−1)/2 ≤ Km (a1 , . . . , am ) ≤ (1 + max{a1 , . . . , am })m ,
(7)
and
Kk (a1 , . . . , ak ) · Km−k (ak+1 , . . . , am ) ≤ Km (a1 , . . . , am )
(8)
≤ 2 Kk (a1 , . . . , ak ) · Km−k (ak+1 , . . . , am ).
Proof. The first statement is an immediate consequence of Theorem 1.19. For the second
statement, letting a be the maximum of a1 , . . . , am , it follows from Theorem 1.3 that
Km (1, . . . , 1) ≤ Km (a1 , . . . , am ) ≤ Km (a, . . . , a) ≤ (a + 1)m .
Since K1 (1) = 1, K2 (1, 1) = 2 and 2`/2 ≥ 2(`−1)/2 + 2(`−2)/2 for ` ≥ 2, an immediate
induction gives the first inequality of (7).
Combining Km (a1 , . . . , am ) = am Km−1 (a1 , . . . , am−1 )+Km−2 (a1 , . . . , am−2 ) with (6),
we get
Km (a1 , . . . , am ) = a1 Km−1 (a2 , . . . , am ) + Km−2 (a3 , . . . , am ),
which implies (8) for k = 1. Let k be in {1, 2, . . . , m − 2} be such that
Km := Km (a1 , . . . , am ) =Kk (a1 , . . . , ak ) · Km−k (ak+1 , . . . , am )
(9)
+ Kk−1 (a1 , . . . , ak−1 ) · Km−k+1 (ak+2 , . . . , am ),
with the convention that K0 = 1. We then have
Km = Kk (a1 , . . . , ak ) · ak+1 Km−k+1 (ak+2 , . . . , am ) + Km−k+2 (ak+3 , . . . , am )
+ Kk−1 (a1 , . . . , ak−1 ) · Km−k+1 (ak+2 , . . . , am )
= ak+1 Kk (a1 , . . . , ak ) + Kk−1 (a1 , . . . , ak−1 ) · Km−k+1 (ak+2 , . . . , am )
+ Kk (a1 , . . . , ak ) · Km−k+2 (ak+3 , . . . , am ),
giving (9) for the index k + 1. This shows that (9) and, a fortiori, (8) hold for k =
1, . . . , m − 1.
9
Exercises
√
Exercise 1.√Prove that every real quadratic number a + b d is badly approximable (introduce a − b d).
Exercise 2. (Proof of Theorem 1.14). 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 (an )n≥0 and
(ξn )n≥1 by the algorithm given above. Set ξ0 := ξ.
1) Show, by induction, that for each non-negative integer n there is an integer polynomial fn (x) := An x2 + Bn x + Cn , with Bn2 − 4An Cn positive and not a square, such that
fn (ξn ) = 0. Prove that An+1 , Bn+1 and Cn+1 are given by
An+1 = a2n An + an Bn + Cn , Bn+1 = 2an An + Bn ,
and Cn+1 = An .
(10)
2) Observe that
D := B02 − 4A0 C0 = Bn2 − 4An Cn ,
for any n ≥ 0,
(11)
and deduce from (10) that there exists an infinite set N of distinct positive integers such
that An An−1 is negative for any n in N .
3) Deduce from (11) that
|Bn | <
√
D, |An | ≤ D/4,
and |Cn | ≤ D/4,
for any n in N . Conclude.
Exercise 3. (Proof of Hurwitz’ Theorems 1.16 and 1.17, following Forder.)
Let ξ be an irrational number and
that its convergents pn−1 /qn−1 , pn /qn , and
√ assume
2
pn+1 /qn+1 satisfy |ξ − pj /qj | ≥ 1/( 5qj ) for j = n − 1, n, n + 1.
1) Prove that we have
1
1
≥√
qn−1 qn
5
1
2
qn−1
1
+ 2 ,
qn
√
√
and deduce that qn /qn−1 and qn−1 /qn belong to the interval ]( 5 − 1)/2, ( 5 + 1)/2[.
Prove that the same conclusion also holds for qn /qn+1 and qn+1 /qn . Conclude.
2) With a suitable adaptation of the above proof,
that if the convergents pn /qn
√ show
2
of an irrational number ξ satisfy |ξ − pn /qn | ≥ 1/( 8qn ) from some integer n0 onwards
then qn+1 < 2qn + qn−1 holds for any sufficiently large integer n. Conclude.
10
References
[1] J. W. S. Cassels, An introduction to Diophantine Approximation. Cambridge Tracts
in Math. and Math. Phys., vol. 99, Cambridge University Press, 1957.
[2] T. W. Cusick and M. E. Flahive, The Markoff and Lagrange spectra. Mathematical
Surveys and Monographs, 30. American Mathematical Society, Providence, RI, 1989.
[3] K. Dajani and C. Kraaikamp, Ergodic theory of numbers. Carus Mathematical
Monographs 29, Mathematical Association of America, Washington, DC, 2002.
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[5] O. Perron, Die Lehre von den Ketterbrüchen. Teubner, Leipzig, 1929.
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Yann Bugeaud
Université de Strasbourg
Mathématiques
7, rue René Descartes
67084 STRASBOURG (FRANCE)
bugeaud@math.unistra.fr
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