C OMPOSITIO M ATHEMATICA
A. B IJLSMA
On the simultaneous approximation of a, b and a
bCompositio Mathematica, tome 35, n
o1 (1977), p. 99-111
<http://www.numdam.org/item?id=CM_1977__35_1_99_0>
© Foundation Compositio Mathematica, 1977, tous droits réservés.
L’accès aux archives de la revue « Compositio Mathematica » (http:
//http://www.compositio.nl/) implique l’accord avec les conditions géné- rales d’utilisation (http://www.numdam.org/conditions). Toute utilisation commerciale ou impression systématique est constitutive d’une infrac- tion pénale. Toute copie ou impression de ce fichier doit contenir la présente mention de copyright.
Article numérisé dans le cadre du programme Numérisation de documents anciens mathématiques
http://www.numdam.org/
99
ON THE SIMULTANEOUS APPROXIMATION OF
a, b AND ab
A.
Bijlsma
Publishing
Printed in the Netherlands
1. Introduction
Let a and b be
complex
numbers witha ~ 0,
a ~ 1 andb ~ Q.
Letab
denoteexp(b log a )
for some fixed branch of thelogarithm.
Theproblem
is to determine whether it ispossible
that all three numbers a, band a’
can be wellapproximated by algebraic
numbers ofbounded
degree.
We take d fixed and considertriples (a, (3, y)
ofalgebraic
numbers ofdegree
at mostd ;
H will denote the maximum of theheights
of a,j3
and y.After initial results of Ricci
[8]
and Franklin[5],
Schneider[9]
stated that for any E >
0,
there existonly finitely
manytriples (a, (3, y)
with
Bundschuh
[2]
remarked that in Schneider’sproof
a condition like{3 ~ Q
is needed and tried to prove a theorem without such a restric- tion. His assertion is that for any E >0,
there areonly finitely
manytriples (a, (3, y)
withwhere
log2
meansloglog. However,
there is an error at thebeginning
of the
proof
of his Satz2a,
so that hisresult,
too, isonly
valid if onemakes some extra
assumption.
Earlier
Smelev [10]
hadproved
thatonly finitely
manytriples
(a, (3, y)
with{3 ~ Q
have theproperty
Cijsouw
and Waldschmidt[4] recently improved
upon the above resultsby showing
that for anyE > 0,
there areonly finitely
manytriples (a, 03B2, y)
with03B2 ~ Q
andThe main purpose of this paper is to show that from all these
theorems,
the condition{3 É 0
cannot be omitted. Moreprecisely,
thefollowing
will beproved:
THEOREM 1: For any
fixed
natural number K, there exist irrational numbers a, b E(0, 1)
such thatfor infinitely
manytriples (a, (3, 03B3) of
rational numbers
where H denotes the maximum
of
theheights of
a,{3
and y.In the theorems cited
above,
it would seem more natural toplace
a restriction upon thegiven
number b than upon/3.
Forinstance,
it isquite
easy to see that the estimate ofCijsouw
and Waldschmidt holds forarbitrary triples (a, (3, y)
if one assumesthat,
for realb,
the convergentspnlqn
of the continued fractionexpansion
of bsatisfy
(Note
that the real numbers b for which this condition is notfulfilled,
are U*-numbers
(see [9],
III§3)
and thus form a set ofLebesgue
measure
zero.)
Asharper
result in the same direction isgiven by
thenext theorem.
THEOREM 2:
Suppose E > 0, dEN,
a,bEC, a~0, b ~ Q ,
1 abranch
of
thelogarithm
withl(a) ~
0.If b e R,
orif
b ~ R such thatthe convergents
Pnl qn of
the continuedfraction expansion of
bsatisfy
there are
only finitely
manytriples (a, (3, y) of algebraic
numbersof
degree
at most d withwhere H denotes the maximum
of
theheights of
a,i3
and y anda b
= exp(bl(a)).
In the
proofs
of Theorems 1 and2,
thefollowing
notations will beemployed:
if a is analgebraic number, [a[ (
denotes the maximum of the absolute values of theconjugates
of a,h(a)
theheight
of a,dg(a)
the
degree
of a andden(a)
the denominator of a. We shall make useof the relations
den(a) h(a) and |a| ~ h(a)
+ 1.2. Two sequences of rational numbers
LEMMA 1: Let K, À E N be
given.
There is a sequence({3nY:=l of
rational numbers in
(0, 1)
such thatfor
all n E N thefollowing inequalities
hold:PROOF: Let
{31
be any rational number in(0, 1)
withh({31)
> À. The sequence({3n)~n=l
will be definedinductively; suppose 6,, already
chosen.
Clearly
there areinfinitely
many rationalnumbers 03B2*
E(0, 1)
with the
property
Only finitely
many rational numbers haveheights
boundedby
exp
(h K+2({3n)),
so there exists a03B2n+l with
bothh (03B2n+1)
> exp(h K+2(03B2n))
and
LEMMA 2: Let K, À EN be
given
and let({3n)::’=l
be a sequenceof
rational numbers in
(0, 1)
such that(1), (2)
and(3)
aresatisfied for
alln E N. Put
(3n
=vn/wn,
where Vn, Wn E N,(vn, wn)
= 1.If
À iss uffi cien tly large,
there is a sequence(an)1lJ=1 of
rational numbers in(0, 1),
suchthat
for
all n E N thefollowing
assertions hold :PROOF: Choose
al:= 2-wl.
The sequence(an)~n=1
will be definedinductively;
suppose a i, ..., an havealready
been chosen and possess the desiredproperties. By
Bertrand’s Postulate([6],
Theorem418),
there is a
prime
number Un+l withNotice
that,
if À issufficiently large,
Consider the
partition
of the interval
(0, 1).
Take t E{0,...,
Un+l -1}.
Thentherefore the width of the
partition
D does not exceedwn+ilun+i. By (8)
the interval{x E (0,1): lan - xl exp (-logK+lh(an))}
has alength greater
thanwn+ilun+i,
so that this interval contains at least one of thepoints
of D. This proves the existence of a tn+1 E{l, ...,
Unll -1}
withand
3. Proof of Theorem 1
1. Take À E N and let
({3n)~n=l
be a sequence of rational numbers in(0, 1)
such that(1)-(3)
are satisfied.Put On
=vn/wn,
where vn, wn E N,(vn, wn)
= 1. We may suppose that À issufliciently large
in the sense ofLemma
2;
let(an)n-=,
be a sequence of rational numbers in(0, 1)
suchthat
(4)-(7)
are satisfied.Put an
=(tnlun)Wn,
where tn, Un EN, (tn, Un) =
1. Define yn : =
a’-;
we haveso
yn E
Q andh(yn)
= u,,- u,,- =h(an).
ThereforeII. The sequence
(an):=l
has theproperty
if wi is
sufficiently large,
so x EIk-,.
III. From
(9)
we see that(an):=l
is aCauchy
sequence; it convergesto a
limit,
which we shall call a. Thenwhich, by
Theorem 186 of[6], implies
that a is irrational.In the same way one can prove that
({3n);=1
is aCauchy
sequence and that its limit b satisfiesso
that b,
too, is irrational.IV. The function xy is
continuously
differentiable on everycompact
subset K of(0, 1)
x(0, 1),
so that a constantCK, only depending
onK,
can be found with
From this follows the existence of a constant
Ca,b, depending only
ona and
b,
such thatfor
sufficiently large
n. 0It may be of interest to note that a
p -adic analogue
of Theorem 1can be
proved
withconsiderably
lessdifficulty. Indeed,
it suffices toconstruct a sequence
({3nr:=1
of natural numbers with theproperties /{3n - {3n+llp
exp(_{3:+1)
and/{3n/p
=p-2.
If b is thep-adic
limit of this sequence anda = b + 1, infinitely
manytriples (a, (3, y)
of naturalnumbers
satisfy
where H = max
(a, (3, y)
andab
is definedby
means of thep-adic logarithm
andexponential
function.4. A result on
vanishing
linear formsLEMMA 3:
Suppose
d E tBI, K a compact subsetof
thecomplex plane
not
containing 0, It
and12
branchesof
thelogarithm, defined
onK,
such that11
does not take the value 0. Thenonly finitely
manypairs
(a, y)
E K X Kof algebraic
numbersof degree
at most d have theproperty that a
8
El 0 exists withand
where H = max
(h (a ), h(y)).
PROOF: I.
Suppose
a, y EK, (3
EQ,
such that the conditions of the lemma are fulfilled.By
CI, c2, ... we shall denote natural numbersdepending only on d, K, Il
and12;
we suppose that H isgreater
than such anumber,
which will lead to a contradiction.P ut B : =
h(3);
thenDefine
L: = [2dB log-1/3B] -1.
We introduce theauxiliary
functionwhere aÀ1z = exp
(À izli(a )) , 03B303BB2z
= exp(À2z12(’Y))
and wherep(À1, À2)
are rational
integers
to be determined later. We haveNow
put a : = den(a ), b:=den({3), c :=den(y), S:=[logl/3B],
T : =
[B21og-1B]
and consider thesystem
of linearequations
These are ST
equations
in the(L
+1)2
unknownsp(Ài, À2);
the coeffi- cients arealgebraic integers
in the number fieldQ(a, y)
ofdegree
atmost d2. The absolute values of the
conjugates
of the coefficients areless than or
equal to
(here (10)
isused).
As
(L+1)2 ~ d2B2 log-2/3 B ~ d2ST,
Lemme 1.3.1 of[11] ]
states that there is a non-trivial choice for thep(Ài,À2),
such thatwhile
II. For kEN U
101
weput Tk :
=2kT;
suppose2k ~ logl/6 B. Then,
forour
special
choice of thep(À1, À2),
we haveThis is
proved by induction;
for k = 0 the assertion isprecisely (11).
Now suppose that
(12)
holds forsome k,
while2k+l ~ logl/6B. By
Lemma 7 of
[3]
we haveHere
and
Substitution in
(13) gives
so
from which we conclude
However, l-;t(a )tP(t)(s)
isalgebraic
and formula(1.2.3)
of[11]
states that every non-zeroalgebraic number e
has theproperty
Now
so either
in the latter case
Combining (15)
and(17) gives 0("(s)
= 0 for s =0,...,
,S -1,
t =0, ..., T,,,, -
1. Thiscompletes
theproof
of(12).
III. Now let k be the
largest
natural number with2k ~ logl/6 B.
From(12)
it follows thatOnce more
apply
Lemma 7 of[3];
thisgives (13) again
and(14)
alsoremains
unchanged,
but from themaximality
of k we now getso
whence
Conclusion:
For these values of t we have
so
according
to 1in the latter case
Combining (18)
and(19) gives
As the
p(Ài,À2)
are not all zero, it follows that the coefficient matrix of the system(20),
which is of the Vandermondetype,
must besingular.
From this we deduce the existence ofkl, À2, Ai, À2E {0,
...,LI
with À +À2{3
=À1
+À2{3,
orThis
gives
so we
get
a contradiction forsufficiently large
H(and B).
D5. Proof of Theorem 2
1. The case
b ~
R istrivial;
we shall therefore suppose that b is real and that its continued fractionexpansion
has theproperty
described in the theorem. Let(a, (3, y)
be atriple fulfilling
the conditions of thetheorem;
wesuppose H
to begreater
than a certain bounddepending only
on E,d, a, b
and 1. This will lead to a contradiction.As a ~ 0 and
a ~ 0,
we may assume a ~ 0 andy ~ 0.
Forsuitably
chosen branches
li
and12
of thelogarithm
we havefrom
l (a ) ~
0 we thusget li(a) ~
0. As a consequence of(21), (22)
andwe have
If it were the case that
(311(a) -12( ’Y) ~ 0,
Theorem 1 of[4]
wouldimply
which is a contradiction. Therefore
(311(a) -12( ’Y)
= 0.II. We have
just proved
that1,(a)
andl2(y)
arelinearly dependent
over the field of all
algebraic numbers; using
Theorem 1 of[1]
we findthat these numbers must also be
linearly dependent
over 0. In otherwords,
there aree, 17
EQ,
not both zero, such that03BEll(a)
+~l2(y)
= 0.Here q 0
0 becauseI1(a) ~ 0,
sousing
Lemma 3 above we see thath({3) log
H.Put q :
=den({3);
then qlog H,
soAs q must tend to
infinity
withH,
we may assumeand
thus, by
Satz 2.11 of[7], {3
is aconvergent of b,
say{3
=Pnl qn. By (12) in § 13
of[7],
wehave,
for some constant c,which contradicts
(23).
DREFERENCES
[1] A. BAKER: Linear forms in the logarithms of algebraic numbers (II). Mathematika 14 (1967) 102-107.
[2] P. BUNDSCHUH: Zum Franklin-Schneiderschen Satz. J. Reine Angew. Math. 260 (1973) 103-118.
[3] P. L. CIJSOUW: Transcendence measures of exponentials and logarithms of algebraic numbers. Compositio Math. 28 (1974) 163-178.
[4] P. L. CIJSOUW and M. WALDSCHMIDT: Linear forms and simultaneous ap-
proximations (to appear).
[5] P. FRANKLIN: A new class of transcendental numbers. Trans. Amer. Math. Soc.
42 (1937) 155-182.
[6] G. H. HARDY and E. M. WRIGHT: An introduction to the theory of numbers, 4th edition. Oxford University Press, 1960.
[7] O. PERRON: Die Lehre von den Kettenbrüchen. Band I, 3. Auflage. B. G. Teubner, Stuttgart, 1954.
[8] G. RICCI: Sul settimo problema di Hilbert. Ann. Scuola Norm. Sup. Pisa (2) 4 (1935) 341-372.
[9] TH. SCHNEIDER: Einführung in die transzendenten Zahlen. Springer-Verlag, Ber- lin, 1957.
[10] A. A.
0160MELEV:
A. O. Gelfond’s method in the theory of transcendental numbers (in Russian). Math. Zametki 10 (1971) 415-426. English translation: Math. Notes 10 (1971) 672-678.[11] M. WALDSCHMIDT: Nombres transcendants. Lecture Notes in Mathematics 402.
Springer-Verlag, Berlin, 1974.
(Oblatum 27-VIII-1976 & 29-IX-1976) Mathematisch Instituut Universiteit van Amsterdam Amsterdam
The Netherlands