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ALGEBRAIC NUMBERS AND TOPOLOGICALLY
EQUIVALENT MEASURES
DISSERTATION
Presented to the Graduate Council of the
North Texas State University in Partial
Fulfillment of the Requirements
For the Degree of
DOCTOR OF PHILOSOPHY
By
Kuoduo Huang, B.S., M.S,
Denton, Texas
December, 1983
y
Huang, Kuoduo J., Algebraic Numbers and Topologically
Equivalent Measures in the Cantor Set.
Doctor of Philosophy
(Mathematics), December, 1983, 66 pp., bibliography, 23
titles.
A set-theoretical point of view to study algebraic
numbers has been introduced.
We extend a result of
Navarro-Bermudez concerning shift invariant measures in
the Cantor space which are topologically equivalent to
shift invariant measures which correspond to some algebraic
integers.
It is known that any transcendental numbers and
rational numbers in the unit interval are not binomial.
We
proved that there are algebraic numbers of degree greater
than two so that they are binomial numbers.
Algebraic inte-
gers of degree 2 are proved not to be binomial numbers.
A
few compositive relations having to do with algebraic numbers
on the unit interval have been studied; for instance,
rationally related, integrally related, binomially related,
-related relations.
A formula between binomial numbers
and binomial coefficients has been stated.
A generalized
algebraic equation related to topologically equivalent
measures has also been stated.
PREFACE
This dissertation consists of three chapters.
chapter has its own introduction.
Each
The topologically equi-
valent measures in the Cantor set have been studied in
Chapter I.
The binomial numbers and Pisot numbers have
been discussed in Chapter II.
In Chapter III, a set-theo-
retic approach to algebraic numbers has been developed.
I hope that through this point of view, many nice theorems
can be discovered in the future.
At least the following theorems in this article are
original.
rems:
Some of them are generalizations of known theo-
Theorem 1.2, Theorem 1.7, Theorem 1.11, Theorem 2.2,
Theorem 2.3, Theorem 2.4, Theorem 3.1 and its corollary,
Theorem 3.3, Theorem 3.5, Theorem 3.6, Theorem 3.7, Theorem
3.8, and a relation between binomial numbers and binomial
coefficients is stated.
I wish to thank Dr. F.J. Navarro-Bermudez and Dr. Helge
Tverberg for generously allowing me to make use of some of
their unpublished manuscripts.
TABLE OF CONTENTS
Page
i
PREFACE
Chapter
I.
TOPOLOGICALLY EQUIVALENT MEASURES
IN THE CANTOR SET
§1.
§2.
§3.
II.
BINOMIAL NUMBERS IN THE UNIT INTERVAL
§1.
§2.
§3.
III.
Introduction
An Improvement of Navarro-Bermudez1s
Theorem
C-pairs
Introduction
Existence of Binomial Numbers
Existence of Non-binomially Algebraic
Numbers of Degree _> 2
ALGEBRAIC NUMBERS
§ 1.
§2.
§3.
§4.
§5.
§6.
§7.
23
41
Introduction
Equivalence Relations on I
Homogeneous Functions in Two Variables
Rationally Related Relation on I
Integrally Related Relation on I
Bi-related Relation
Binomial Numbers and Binomial
Coefficients
BIBLIOGRAPHY
65
11
CHAPTER I
TOPOLOGICALLY EQUIVALENT MEASURES
IN THE CANTOR SET
§ 1. Introduction
Let X be a topological space.
Two Borel measures y
and v are said to be topologically equivalent whenever y = vh
for some homeomorphism h of X onto itself.
This notion sets
up an equivalence relation which partitions the family of
Borel measures into disjoint classes each one consisting of
those measures which are mutually equivalent.
By restricting
attention, if necessary, to a suitably defined subfamily of
measures, one can ask for the number of classes there.
One
can also try to uncover necessary conditions, as well as
sufficient conditions, which intrinsically characterize measures which belong to the same class.
Topologically equivalent measures in the n-dimensional
unit cube, the space of irrational numbers in the unit interval, and the Hilbert cube have been studied, respectively, by
Oxtoby and Ulam [1], Oxtoby [2], and by Oxtoby and Prasad [3].
In [4], F.J. Navarro-Bermridez studied the topologically equivalent measures in the Cantor space.
his result.
I will be able to extend
Before getting into details, I prefer to define
a few necessary notations and terms.
(X,y) is a P-pair provided X is a space of the form
oo
n xn
n=li
x =
with the product topology.
Each factor X^ is finite or
countably infinite and carries discrete topology and each
X
has at least two elements.
]i is a Borel measure in X.
n
A P-pair is of type C in case each factor X^ is finite (In
[ 5 ], F.J. Navarro-Bermudez called aP-pair of type C a
"C-pair").
Every P-pair of type C is homeomorphic to the
Cantor space, which can be realized as X with X = {0,1} for
n
n=l,2,...,
and its topology is compatible with the metric d
such that for any two points x = ( x n ) a n d X'
=
( x ^) is given
by the formula
00
1
d(X/X ) =
^
-n
(%) d (x , x'),
n=l
each d
n
is the metric on X
n
n
n
n
which takes only
the values 0
u
or 1.
A P-pair is of type F-K0 in case there are infinitely
many X n having cardinality ^ , and infinitely many of X^
having finite elements.
Denote
[i1,i z,...,im] = {(x n)€X: x.=i.€X.
j j j for i=l,2,...,m}
to be a basic open set.
The set of all basic open sets forms
a base for the topology of X.
These sets are both open and
closed and will be referred to as the special closed-open
sets of X or a basic clopen set.
A P-pair is of type ( p 1 , p 2 , — , P n ) provided that the
following three conditions hold:
(i)
p
+ p 2 + ... + p n = 1 and p l ,p 2 ,...,P n are non-negative
real numbers.
(ii)
For each k, X
is a set consisting of n elements,
K.
for instance, may be taken to be integers 0,1,2,...,n-l.
00
(iii)
y (j) = p. for all n and j, and y =
n
=
y n (j)
3
Pj f° r
n
an
II
n=l
y
with
n
^
Similarly a P-pair is of type (p^,p2,...) provided the
following conditions hold:
oo
(a)
E p = 1, where p ,p ,... are non-negative real
1 2
n=l n
numbers and infinitely many of them are positive.
(b)
For each k, X
is a set consisting of K0 elements,
iv
for instance, may be taken to be integers 1,2,3,... .
00
(c)
y n (j)
=
y n (j)
=
Pj f° r
Pj f° r a-'-^ n
anc
a
H
n
an<
^ 3r
anc
^ U
=
n
y^ with
^ j*
Any two P-pairs of type (p^,p2,...) are topologically
equivalent which is an immediate result from [2] by John
C. Oxtoby.
He proved the following theorem in [2].
Theorem. - A topological measure space (X,y)
is homeo-
morphic to (J,A) if and only if X is homeomorphic to J and
y is an everywhere positive, non-atomic, normalized Borel
measure in X.
In particular, any such measure in J is topo-
logically equivalent to X.
Recall:
(1)
A topological measure space is a pair (X,y), where
X is a topological space and y is a measure on the class of
Borel subsets of X.
(2)
A measure y is everywhere positive if y(G) > 0 for
every non-empty open set G, non-atomic if y({x}) = 0 for each
x £ X , and normalized if y(X) = 1.
(3)
J denotes the set of irrational numbers in I = [0,1]
and A denotes the restriction of Lebesgue measure m to the
Borel subsets of J.
So we are only interested in a P-pair which is not of
type (plfp2,...)«
Especially we are interested in a P-pair of
type (p^/P2/.../Pn)•
From now on we let (X,y) be a fixed
P-pair of type (p1,P2,...,p ) so that for each positive integer
m, X = {0,1,2,...,n-1}.
m
basic open sets.
Every open set U of X is a union of
If U is also closed, hence compact, then
only finitely many basic open sets enter into this union.
It can be easily seen that the closed-open set U can be expressed as the union of finitely many disjoint basic open sets
each of which has the form
{i }x{i }x...x{i }xx
xx
x...(i =0,1,2,...,n-l)
1 2
m
m+1 m+2
k
(1.1)
for k=l,2,...,m.
The set (1.1) will be denoted as [i,,i„,...,i ]. Two
1 z
m
of these sets [i,,...,i ] and [j,,j„,...,j ] are equal or
1
m
i /
m
disjoint, depending on whether the n-tuples (i^,i , ...,i ) and
(jifj^/---fj )
1 2
m
are
identical or not.
The sets (1.1) have been
named differently by many other authors to be thin cylinder,
intervals, special closed-open sets, basic open sets, etc.
The truth of the following theorem is obvious.
Theorem 1.1. - Let U be a closed-open set of the Cantor
space X.
There is a positive integer n such that U can be
expressed as the disjoint union of finitely many special
closed-open sets of length n.
The family f of probability measures in X to which attention will be restricted consists of product measures
oo
y = II ym subject to the condition
m=i
1
y (i) = y
(i) for all m and i=0,1,2,...,n-l,
m
m+1
(1.2)
the factors y are all normalized; that is, y (0)+...+y (n) = l .
m
m
m
For convenience, we denote y = y(p,,p 0 ,
1 2
p.^ = V (i) f° r i=0,1, . . .,n-1.
,p ), where
m
Among the product measures,
there are precisely those which are invariant under the
Bernoulli shift transformation T of X defined by
T(x,x,...,x
1
z
,...) — ( x , x , . . . )
m
1 5
for each point x = (x^,x^,x ,...) in X.
X is invariant under T
A Borel measure y on
(or T is y-measure preserving) if
y ( T _ 1 ( U ) ) = y(U) for every Borel set U of X.
n
For each n triple (p 1 ,...,p11 ) in n
[0,1] with
i=l
p +...+P = 1 , there is a measure y(p_,...,p ). Since the
i
n
I
n
correspondence
( P w . - w P ) -> y(p_,...,p ) is one-to-one, the
i
n
I
n
family has continuum many measures. Each y(p , . . . , p ) is a
1
n
normalized Borel measure which is everywhere positive and
n
non-atomic for (p ,...,p ) € n (0,1).
n
i=l
If U = [i.,i0,...,i ], then
1 1
m
y(px,p2,...,pn)(u) = y 1 (i T )y 2 (i 2 )...y m (i m ).
It follows that
m. m
m
y(P1,. . .,Pn) (U) = P x P 2 •••PI1n
(1*3)
where m. is the number of times of i,=j ( l £ k £ n ) .
It is evident that the measure y(px,p2,...,pn) and
y(p^(l) f . . . 'Pfi(n))
are
always topologically equivalent, where
7T is any permutation of l,2,...,n.
Because y(pj,...,p ) =
oo
y(p
p . x)h, where h is the homeomorphism h = n h
m=l m
with h : X -* X given by h (i) = tt (i) i=l,2,... .
m
m
m
m
Theorem 1.2. - If the measure y(p^,...,p ) and
y(qj/.../q ) are topologically equivalent, then there are
positive integers r and s, and integers
. . .
3 Ji»'••']o
b
"
. . .
i l l ,
with 0 < a
inJ . y • • • 9 in ,
x
i
.
.
2 Ji >•••>] n
and 0 < b
<
m , 5 • • • jm ,
i
Ql
'/
<i
1
Z
t
2
3
1
l
i
< . . . < i £ n, l_<j < j
K.
x
<
k
m,
1
,
m
k
*/
m . 5 x• • • > in #
.
.
m,
x
x
i
i where
k
• • • • m.
•
• • • •in, •
J
l£i
a
J
Z
£
< . . . < j £ n , m. +...+m. = r
36
1X J.
k
and nu 1 +...+m_.Z = s, and l £ i , j £ n , such that (1.4) and (1.5)
3
J
hold:
. .
q. =
1
I
i1=l
a
r
1
p.
X
1
+
. .
.
Z
a
2
lli1<i2<n
m
m
i1' i2
m.
p.
X
1
m.
p.
2
+...
(1.4)
x
1
2
m.
-\ s • s a s
s i* / —. m • jfli. * • • • 5 m.
1—<1,<i
n<. . .<i- <n
1 2
k—
1x- 2 x0
kxn 1
m.
i-i
k
I-.
1
m1»*
p. =
J
J JlI T
b q.
Z
3
Jj-1
+
1
£
-J1
Proof:
3
b
^
J
2
V
i<^<i2<n
.
+
m. m.
3 Ji * J o
] i Jo
b
q.
q.
+ . . .
y
^
£-n
.
.
(1.5)
32
m.
Ji
q
m. . . .m.
3
J
I
£
m.
- i n
Jn
J 1
...q.
+...+ b
i,
l.
m
. i >11 1,
q....q
m
x
n
Let h be a homeomorphism of X onto itself such
that y(p,, . . . ,p ) = y(q,,...,q
1
ii
1
n
set
%
J l
[j] where 1<. j<_n.
)h, and let V be the clopen
So we have
y (P L , . . . ,p n ) (V) = p .
On the other hand, h(V), being a closed-open set, is, according to Theorem 1.1, a disjoint union U of finitely many thin
cylinders of length s.
By
(1.3) the y(q,,...,q )-measure of
1
II
[ t / . - . / t ] is
1
s
m. m.
m.
31 J o
^k
q.
q.
...q.
where 1 < i < j 2n <
J
J
1
2
k
"" 1
... < j, < n,
k -
{t , . . . ,t } = { j w . - . / j , } /
the number of times of
m. +...+m.
3
3
1
k
If
b
k is a positive integer and m.
3
i
repeating in {tj,...,t }, thus
= s.
is the number of clopen sets in U
in « ) m , $ • • • ) m ,
3
3
3
1
2
k
m.
m.
J1
Jv
whose y(q,/...,q )-measure is q.
...q
(hence
1
]
j
i
k
3 jljjoj'-'jji
cI
z
k
0<
b 1
<
•/
,
_
,), then y(q ,...,q )(h(V))
m. • • • m.
—
m. I . . • m. !
1
n
J
3
3
3
1
k
1
k
is equal to
m
n
j j
l r
E b q + ... +
3
ji =1
1
Z
j J,-•-j
h 1
U
v k
b 1
q. . . .q.
J
<m m j •""j, J 1
k
k
j 1>2
nm
m
+ ... + b
q. ...q n .
m, , . . . ,m 1
n
1
n
Formula (1.5) follows by substituting into the equation
y(P1,...,Pn) = y(qx,...,qn)(h(V)), the values obtained for
y(Pj,...,Pn)(V) and y(qx,...,q )(h(V)).
Formula (1.4) can be proven similarly by considering
1
h
instead of h.
Q.E.D.
This elementary theorem will be a very useful tool in
proving other interesting theorems.
The following corollary
will be obtained immediately from Theorem 1.2.
Corollary 1.3. - If y(pj,...,p ) is topologically equivalent to y(q^,...,q ), then the following statements hold:
(i)
If p ,...,p
i
n
9
#
*3^ ' ^2 ' * * *
(ii)
q
r
are all rationals, then so are
If p ^ . . . ^
are all algebraic numbers, so are
If p^,...,p
are all algebraic integers, so are
. ..,qn.
(iii)
^3 j f • • • f
•
The following corollary 1.4 is a special case of Theorem
1.2, but plays an important role in this paper.
It was first
proved by F.J. Navarro-Bermrtdez in [4].
Corollary 1.4. - Given r,s€ [0,1].
If y(r,l-r), y(s,l-s)
are topologically equivalent, then there are positive integers
m,n and integers a Q/ a 1 ,... f a n ; b ^ b ^ . . . , ^ with 0 £ a
< (J)
and 0 < b. < (?) such that
~ J - J
s = a r m +a rm_1(l-r)+...+a
r(l-r) m_1 +a (l-r)m
u
m-1
m
r = b
(1.6)
0sn+bisn_1(1-s)+*••+bn-lS(1~S)n~1+bn(1~S)n *
(1
*7)
Let C(p1/...,p ) be the class of measures in f which are
topologically equivalent to y(p,,...,p ).
I
We have mentioned
n
previously that C(pj,...,p ) always contains the measures
is any permutation on {l,2,...,n}.
Corollary 1.5. - Each class C(p ,...,p ) contains at
l
n
most countably many measures of the family f and the cardinality °f classes C(pj,...,p ) is of power the continuum.
Proof:
If y(q , — , q ) belongs to the class C(p.,
,p ),
i
II
i
n
then the numbers q ^ . . . ^ must satisfy formula (1.4) for
some choice of integers r,
• •
•
i ii > • • • »in
a
with 1 < i < n
m. 5 • • • *ni.
—
—
^i
i
'
r!
<
^m. ! . . .m. !' 1 - """1 < i 2 <
"*"£ — n '
X
X
x
1
£
1
£
i i
i
m. +...+m
= r. For each fixed r , i a s O < a ^
^
m
1
£
,1
£
1i i
,
r!i
, th
»• •i
• j-'-n
by var in
m I...m !'
Y g
e integers a
* , the
n ,•..,n.
X
1
*
1
right hand side of (1*4) generates only finitely many differ—
ent numbers
. It follows that the number of measures
-
a
m.
—
y (c[•,/•••/q
) is at most countable,
j11) in the class C(p i,...,p
n
10
Since there are continuum many measures in f and each
class C(pj,...,p ) has only countably many measures, the
cardinality of classes C(p^,...,p ) is of continuum.
Q.E.D.
§ 2. An Improvement
of Navarro-Bermudez 1 s Theorem
In [4], F.J. Navarro-Bermudez has proved the following
interesting theorem.
Theorem 1.6. - If y
= y(r,l-r) is topologically equiv-
alent to y g = y(s,l-s), and if s is rational or transcendental
in the unit interval, then either r = s or r = l - s .
In other words, there are only two (except s = h ) elements in the equivalent class C(s,l-s).
In this section, we
are going to show the same result if s is a real algebraic
integer of degree 2 in the unit interval (There are infinitely
many algebraic integers of degree 2 in the unit interval).
Actually we prove slightly more in Theorem 1.7.
Theorem 1.7. - If r is an algebraic integer of degree
n ^ 2 , and y
is topologically equivalent to y , and s = p+qr
S
XT
for some rational numbers p,q, then s = r or s = 1-r.
We denote irr(0) to be the irreducible polynomial of
an algebraic number 0 in Z[x] with relatively prime integer
coefficients and positive leading coefficient.
In this
chapter, r and s are always in the unit interval.
11
Theorem 1.8. - If y is topologically equivalent to y
TC
S
and s is a real algebraic number of degree n, then there
exists a unique representation of r in terms of s.
r = £
n-1
+ £ 0 s n - 2 + ...+£, s +
n-2
1 0
for some rationals
Furthermore, if s is an
algebraic integer, then
, are integers.
U 1
n-1
Let irr(s) = x 4 - a ^ x
+ . . . + a 1 x + a Q . Upon using
n 1
n
s
=
-a
n_^
s
-...-a^s-a
and (1.6) of Corollary 1.4, it
follows that there is a representation for (2.1) . Since
l,s,...,sn
1
form a basis of Q(s) over a, so the represen-
tation of (2.1) is unique.
Theorem 1.9. - The sum and product of two algebraic
integers are algebraic integers.
Also, the set of all
algebraic numbers forms a field.
Refer to [ 8 ].
Theorem 1.10. - If r is an algebraic number of degree n
and y is topologically equivalent to y , then s is an algeS
"JC
braic number of degree n.
Furthermore, in addition, if r is
an algebraic integer, then s is an algebraic integer.
Proof:
degree of r.
We are going to prove this by recursion on the
If n = l , then this theorem holds according to
Theorem 1.6 and Theorem 1.9.
Assume this theorem is true
for all n_<k and r is an algebraic number of degree k + 1 ,
and y is topologically equivalent to y . According to
s
x*
(2.1)
12
•n
Theorem 1.8, s =
^0''—'^n-1'
1
+ ... + £^r + £Q for some rationals
and this
representation is unique.
It follows
that s€Q(r), where Q(r) is the minimal field extension over
Q so that r€Q(r).
It is obvious Q(s) £ Q(r), and it follows
that the dimension of Q(s) over Q is less than or equal to
the dimension of Q(r) over Q.
= k+1.
This implies deg(s) £ deg(r)
But deg(s) cannot be less than k + 1 ; otherwise, by
our inductive hypothesis, the degree of r will be less than
k+1.
This contradicts deg(r) = k + 1 .
the algebraic number s is k + 1 .
Hence the degree of
In addition, if r is an
algebraic integer then, according to Theorem 1.8 and Theorem
1.9, s is an algebraic integer.
Recall Theorem 1.7.
Q.E.D.
If r is an algebraic integer of
degree n>_2, and yg is topologically equivalent to y r , and
s = p+qr for some rational numbers p,q, then s = r or s = 1-r.
Proof:
As r is an algebraic integer of degree n> 2 and
y s is topologically equivalent to y^, according to Theorem
1.10, it follows that s is an algebraic integer of degree n.
Furthermore, owing to Theorem 1.8, we have
s = £
r11"1 + ... + £ r + £
n-1
1 0
for some integers
unique.
,£^,...,£n_^ and this representation is
This forces p and q to be integers and q ^ 0.
r = (l/q)s-p/q.
So, using the same argument as above, we
can have 1/q and p/q are integers.
or -1.
Notice
Thus q can be either 1
13
Case 1.
q = 1.
As 0 < s = p + qr = p + r < 1 and p is an integer and
0 < r
<1, it follows that, p = 0; that is, s = r.
Case 2.
q = -1.
Since 0 < s = p + qr = p - r < 1 and p is an integer
and 0 < r < 1, so p can only be 1; that is, s = 1 - r.
Q.E.D.
Theorem 1.11. - If r is an algebraic integer of degree 2
and y is topologically equivalent to y / then s = r or
s
x*
s = 1 - r.
Proof:
As r is an algebraic integer of degree 2 and y
S
is topologically equivalent to y , it follows from Theorem
1.10 and Theorem 1.8, that s = p + q r for some rationals p
and q.
Hence, according to Theorem 1.7, s = r or s = 1 - r .
Q.E.D.
§3. C-pairs
This section consists of some results which have to do
with Topologically Equivalent Measures in the Cantor Space.
They were done by F.J. Navarro-Bermudez in [5 ] which is an
unpublished article.
In order to make a complete exposure
of the topologically equivalent measures in the Cantor Space,
1 decided to put it here as a section.
By a C-pair, we mean a pair (X,y) where X is a space of
the form
x =
n sn
n =il
14
with the product topology.
Each factor S
n
carries the discrete topology.
is finite and
y is a Borel measure in X.
It is well known, of course, that whenever (X,y) is a
C-pair then X is homeomorphic to the Cantor space of infinite
sequences of zeros and ones, and its topology is compatible
with the metric d which, for any two points x = (x ) and
n
x' = (x'n), is given by the formula
00
d(x,x') =
r»
i (%) d (x ,x1 ) .
,
n=l
By d
n
n
is meant the metric on S
n
0 or 1.
n
n
which takes only the values
Furthermore, it can readily be seen that a countable
basis for the topology of X consists of sets of the form
[i ,i ,...,i ] = { (x ) £ X: x.=i. for j=l, 2, . . . ,m} .
I
z
ni
n
3
3
These sets are both open and closed and will be referred to
as the special closed-open sets of X.
They are obtained by
fixing the first coordinate, the second coordinate, and so
on, up to a finite number of coordinates.
Definition. - Let t be an integer, t>2, and let
P1,P2/...,Pt be non-negative real numbers such that
P 1 + P2 + '-- + P t ~
A
C-pair (X,y) is said to be of type
(t?P1/P2'•••'Pt) if the following two conditions hold:
(i)
For each n,
is a set consisting of t elements
which, for convenience, may be taken to be the integers
1,2,...,t.
15
(ii)
p is a shift invariant product measure y =
n
n u
,pn
=l
with y (j) = cp. for all n and j.
n
2
Let (X,y) be a C-pair of type (t;p x ,p 2 ,...,p ) .
x can
be expressed as a disjoint union
X =
U [j]
j=l
where y([j]) — p_. and diam([j]) =
Each one of the special
closed-open sets [i ] can in turn be expressed as a disjoint
union
[i,] =
U [i,,j]
j=l
where y ( [ i p j ] ) = p ± p
and diamffi^j]) = %.
this process, a sequence of covers ^
By continuing
^ , % ,... of X can be
constructed with the following properties:
n
^ U iji ...i : l<i.<t, l<j<n^*
12
The members of ^
n
— j—
(3.1)
-J-
are mutually disjoint non-empty closed-
open sets of diameter less than 1/n.
For fixed i , i , — , i ,
1 I
II
•
. = U {U. .
. .€ #
1
1 2",;Ln
j=l 1 1 1 2 " * " 1n:'
}
» = Pi Pi •••Pi •
J- l
n
1 2
(3.2)
<3-3)
n
Theorem 1.12. - Let X be a complete metric space and y
a Borel measure in X.
If there exists an integer t ^ 2 , non-
negative real numbers p 1 ,p 2 ,...,p t with p 1 +p 2 +...+p t = 1, and
16
a sequence of covers of X with properties (3.1), (3.2) and
(3.3), then (X,y) is isomorphic with a C-pair of type
(t;
, p^ , . . . , p^ ) .
00
Proof: Put Y = n
, where S = {l,2,...,t}, and let
n
oo
n =l
v = II v be the product measure in Y given by v (j) = p
n
n=l n
i
for all n and j. For y = ( ^ , , i , . . .) in Y, define
h(y) = o
n
o
1
nu
n...
1
12
<3.4)
1 2 3
Since X is a complete space and lim (diam U,
) =
1 1
n->oo
l 2*'' : L n
o,
the
intersection on the right hand side of (2.4) consists of
exactly one point.
to X.
\
Thus, h is a well defined function from Y
It is onto on account of the fact that the families
are covers of x
' it: i s one-to-one because each
cover %n consists of disjoint sets.
and h~ 1 (U
=u
12* * * n
12' '
family of all sets U
.
12
_
Clearly, h([i ,i ,...,i ])
1 2
n
) = [i ,i ,...,i ]. As the
1
2
n
n
is a basis for the topology of
n
X, and the family of special closed-open sets is a basis for
the topology of Y, both functions h and h ^ are continuous.
Finally, for U = [i
i
1
2
..,i ], it is the case that v(U) =
n
. ) = y(h(U)). But every open set
P. P* •••?. = y( u . .
1 2
n
1 2"'Xn
of Y is a disjoint union of countably many special closedopen sets.
Hence, the equation v(U) = y(h(U)) holds for
every open set U of Y, and, consequently, for every Borel
set as well.
17
The purely topological content of Theorem 1.12, namely,
that every compact metric space is the continuous image of
the Cantor space, is a well known result.
Theorem 1.12 was
inspired by the proof that A.H. Schoenfeld
[ 9 ] has supplied
for that well known result.
It is obvious to observe that the converse statement of
Theorem 2.1 is nearly true.
isomorphic with a C-pair
Y to X.
(3.2) and
Indeed, suppose that
(X,y) is
(Y,v) through a homeomorphism h from
As Y has a sequence of covers with properties
(3.1),
(3.3), h carries these covers to covers of X with
the same properties, except that there is no guarantee that
a set of diameter less than 1/n in Y is carried by h to a set
of diameter also less than 1/n in X.
Some Consequences of Theorem 1.12
A C-pair of a given type may well be isomorphic with a
C-pair of a different type.
In particular instances, this
can be established by appealing to Theorem 1.12.
For example,
00
the pair
II {1,2} , y )
n
n= 1
(X=
of type
(2;k,h)
and the pair
00
(Y =
n {1,2,3} ,v) of type
n_i
(3;h,h,h) are isomorphic.
X
prove this, a sequence of covers ^ ' ^ 3 ' * * *
constructed satisfying properties
t=3,
P1=h, P
=
2
P
=
3
^'
L e t
the special closed-open sets
U3 =
[2,2].
t
(3.1),
(3.2) and
^ le f i r s t cover %
=
[1],
=
To
will be
(3.3) with
consist of
[2,1] and
These sets are mutually disjoint, of diameter
less than 1, and u(U^) = h, y ( U 2 ) = y(U^)
=
In
general,
18
if covers
\ > " t \ have been constructed satisfying
properties (3.1), (3.2) and (3.3), then the cover ^
is
n+1
constructed as follows. Let U
= H
i
i 1 ho
i i
.i
LJ
12
n
any of the special closed-open sets in % . Put U.
n
i i . . .i i
12
n
J 2 , • • •, js/l] / U ± ±
2 - [ jx , j 2 , . . ., jg ,2,1] and
12*
* * iin
JU
/ j»2,2].
i i ...i 3 =
j2/« • j
is defined
s / 2, 2 The cover U
1 2
n
n+i
to consist of the special closed-open sets U
1 1
1 2 * * * """n+l
These sets are clearly disjoint and of diameter less than
l/n+l«
Property (3.1) is satisfied, and the same is easily
seen to be true of properties (3.2) and (3.3).
The function
h from Y to X defined by (3.4) using the covers
establishes an isomorphism between (X,y) and (Y,v).
In
fact, it is even possible to describe the action of h on the
points (i^i^...) Of Y.
Put f (1) = 1; f (2) = 2,1; f (3) = 2,2
Then h(i1,i2,...) = ( f d ^ ,f(i2)
.
Theorem 1.13. - Let (X,y) be a C-pair of type
' • • • '*3^ •
-*-n or der for (X,y) to be isomorphic with
a C-pair of type (t;Pl,p2,...,pt), it is sufficient that there
exist disjoint special closed-open sets U ,U ,...,u
J.
that x = U U U
U... U U
such
t
and y (U.) = p. for all j.
t
Proof:
z.
J
J
Write U
= [i ,i
.
i
1 fnr -i—i o
+•
j
1 J2
-"nCj)
D-J-f • • • ,t.
Then p = y(U ) = g q ...q
. Let ^ be the cover of X
J
J
1
. .
1 J2
n (j)
consisting of the special closed-open sets U ,U , . . . ,u
1 2
t
Suppose that covers
, \ of X have been constructed
satisfying properties (3.1), (3.2) and (3.3), each cover
19
consisting of special closed-open sets.
Let U
be
l±2 • • • i
any member of %
say U
= [k ,k ,...,k ]. Fo? fixed
11
12***
x
k
put " i i l 2 . . . v - l V 2
kr,jl>j2
jn(j)1,
i
and define the cover % ,, to consist of the sets U
n+1
i1 i2
i
n+l
These sets are disjoint special closed—open sets of diameter
less than 1/n+l.
a
y
„ d s i „ ~ ^
Properties (3.1) and (3.2) are satisfied,
l l V
- V
" V
V
"
\
V
V
V
«
•
(Ui i
property (3.3) is satisfied
i )P 1 = P. P. ...p. p.,
1 2* * n 3
2
n3
as well.
Thus, it is possible to construct a sequence of
covers of X with the requirements of Theorem 2.1.
Hence,
(X,u) is isomorphic with a C-pair of type (t;p ,p ,...,p ).
1 2
t
To illustrate Theorem 1.13, consider a C—pair (X,y) of
type (2; 1/3,2/3) . Put ^
= [1], U £ = [2,1], U 3 = 3*2,2%.
X
is the disjoint union of these special sets and since
y(U^) = 1/3, y(U2) = 2/9, li(U^) = 4/9, (X,y) is isomorphic
with a C-pair of type (3;1/3,2/9,4/9).
put Vj = [1,1], V 2 = [1,2],
u(^2) = 2/9, y(V^) = 2/3.
= [2].
On the other hand,
This time yfV^) = 1/9,
Thus, (X,y) is also isomorphic
with a C-pair of type (3;1/9,2/9,2/3).
Therefore, two C-pairs
of types (3;1/3,2/9,4/9) and (3;1/9,2/9/2/3), respectively,
are always isomorphic.
Note that the condition in Theorem 1.13 that the closedopen sets U_. be special cannot be dropped.
be a C-pair of type (4jh,h,k,h).
u
=
2
[2] U [3] U [4].
U 2 , and y (U^) =
Put
Indeed, let (X,y)
= [1] and
Then X is the disjoint union of u
y(U2) = 3/4.
However, (X,y) cannot be
and
20
isomorphic with a C-pair of type (2;l/4,3/4), for indeed, by
an easy application of Theorem 1.13, (X,y) can be seen to be
isomorphic with a C-pair of type (2;h,h).
But, by Theorem
3.3 of [4], two C-pairs of types (2;l/4,3/4) and (2;h,h),
respectively, cannot be isomorphic, and this in spite of the
fact that both measures take values on closed-open sets which
are dyadic rationals.
Theorem 1.14. - If the requirement that the closed-open
sets Uj be special is dropped, then the sufficient condition
for a C—pair to be isomorphic with a C—pair of a given type
as stated in Theorem 1.13 is also necessary.
Proof:
Let h: X •+ Y establish an isomorphism between
(X,y) and a C-pair (Y,v) of type (t;p 1 ,p 2 ,...,p ).
V
j
=
[j3 for j = l,2,...,t.
These are mutually disjoint
closed-open sets of Y such that Y = V
1
and v (Vj) = p_. .
The sets
Put
U V
2
U
U V
t
= h ^ (V ) are mutually disjoint
closed-open sets of X such that X = U
y(U ) = v(V.) = p..
J
J
3
U U U ... U U and
•L
Z
t
Observe that while the sets V are
j
special, the same is not necessarily true of the sets U .
3
Let (X,y) and (Y,v) be two C-pairs.
Denote by C the
Cantor space of infinite sequences of zeros and ones, and
let f and g be homeomorphisms from C to X and Y, respectively.
Put y^ (B) = y(f(B)) and
B of C.
Clearly, y^ and
(B) = v(g(B)) for every Borel set
are Borel measures which are
topologically equivalent in C if and only if the C-pairs
21
(X,y) and (Y,v) are isomorphic.
Thus, the results of Theorem
1.13 and Theorem 1.14 can be carried to measures in C of the
form yf where f is a homeomorphism from C to some product
space X and y is a shift invariant product measure in X.
CHAPTER BIBLIOGRAPHY
Oxtoby, J.C. and S.M. Ulaiti, "Measure preserving
Homeomorphisms and Metrical Transitivity,"
Annals of Mathematics (2) 42 (1941), 874-920.
Oxtoby, J.C., "Homeomorphic Measures in Metric
Space," Proceedings of the American Mathematical Society 24 (1970), 419-423.
Oxtoby, J.C. and V. Prasad, "Homeomorphic Measures
in the Hilbert cube," Pacific Journal of
Mathematics 77 (1978), 483-497.
Navarro-Bermudez, F.J., "Topologically equivalent
measures in the Cantor space," Proceedings
of the American Mathematical Society (2)
77 (1979), 229-236.
, "Topologically equivalent
measures in the Cantor space II," preprint.
Prasad, V.S., "A survey of Homeomorphic Measures,"
Measure Theory Conference, Oberwolfach, 1981,
Springer-Verlag Lecture Notes Series.
Von Neumann, J., "Comparison of cells," Collected
Works, Vol. II, 558.
Hungerford, T.W., Algebra, New York, Holt, Rinehart
and Winston, Inc., 1973.
Schoenfeld, A.H., "Continuous Surjections from
Cantor Sets to Compact Metric Space,"
Proceedings of the American Mathematical
Society 46 (1974), 141-142.
22
CHAPTER II
BINOMIAL NUMBERS IN THE UNIT INTERVAL
§ 1. Introduction
In view of Theorem 1.2 and Corollary 1.4 of Chapter I,
we will say that two numbers r and s in the unit interval
[0,1] are binomially related whenever (1.6) and (1.7) of
Corollary 1.4 hold.
Furthermore (p1,...,pn) and (q1,...fq )
are said to be multinomially related in case (1.4) and (1.5)
of Theorem 1.4 hold.
In this chapter, we are only interested
in binomial relationship.
The following definition plays
an important role in this chapter.
Definition. — A number r in the unit interval is said
to be a binomial number provided that there exists s£ [0,1]
such that s is binomially related to r and s ^ r, s#l-r.
One can ask whether there are any binomial numbers.
this chapter we give the affirmative answer.
In
According to
[1], the rational and transcendental numbers are not binomial
numbers.
I will give a delicate proof to show that the alge-
braic integers of degree 2 are not binomial numbers.
It is
evident that binomially related relation is an equivalence
relation in the unit interval.
In other words, a binomial
number is a number whose equivalence class has more than two
elements.
23
24
Definition. - A Selmer's number r is an algebraic integer
of degree n_>3 so that r11 + r-1 = 0 or r11 - r-1 = 0.
In this chapter, we are going to show any Selmer's number
in the unit interval is a binomial number.
Let us have a
brief review of Selmer's number.
Through a study of generalized continued fractions, Ernst
S. Selmer has been led to certain questions concerning the
irreducibility of polynomials.
Let us first consider the
general problem:
The ordinary algorithm for a continued fraction may be
described as a systematic replacement of two non-negative
numbers, a^ ^ and
(k+1) _ a (k) a n d ,a (k)=
a
l
~ 2
2
division a x (k) /a^ k) .
by two other numbers, usually
the
(positive) remainder by the
Formulated in this way, a generalization
to n dimensions is immediate.
Such a division algorithm has
been extensively studied by Perron [2].
An alternative pro-
cedure, using subtraction instead of division, was introduced
by Brun [3]:
a
At each step, with the n numbers
ik) - a2k) -
- a n k ) - °'
(1-D
we replace a^ k ) by the difference a* k ) - a ^ k ) , and rearrange
the numbers according to magnitude.
Of particular interest are periodic expansions, characterized by
a^
k+£
a (k>
1
)
a (k+£)
=
a(k)
2
a (k + £)
=
••• =
V )
a
„
=
A
"
d-2)
25
In any procedure for generalized continued fractions, the
ratio X is determined by an equation of degree n, and this
equation is irreducible if and only if the given n numbers
are linearly independent.
Irreducibility and independence
are then, of course, related to the same basic field of
rationality.
In what follows, we shall always assume this
to be the ordinary field of rational numbers.
The simplest period consists of one step only; that is,
1 = 1 in (1.2).
a
ik^ ~ a 2 ^
= a
In Brun's algorithm, we must then have
nk+1^'
since
otherwise the smallest number
would not be involved at all in the process.
The equations
(1.2) then take the form
£ £
(k)
,(k)
a< k) -a 2 (k)
n
(k)
a<k>
n
= A.
n
Showing that A is a characteristic root of the n x n matrix
0
1 0
0
0
1 0
0
0
0
1 - 1 0
0
0
0
0
0
0
0
1
0
The corresponding equation for A takes the form
An + A-l = 0,
(1.3)
26
This naturally led E.S. Selmer to study the irreducibility
of the polynomials x n + x ± 1 which have very small discriminants.
Note for discriminants.
Let K be a field.
If f€ K[x]
is a monic polynomial of degree n, with roots x, ,...,x we
1
'n
define the discriminant of f by:
d(f) =
1)/2
n (x -x ) = (-1)
1
i<j
J
n (x.-x.).
.
/ . 1J J
If J
Thus D(f) € K and D(f) ^0 if and only if f has distinct roots.
It is important to compute the discriminant of an irreducible polynomial f = xn+a1xI1"1 + . . .+an without knowing a
priori its roots.
= x^+...+xk (where x . , . . . , x
If p
k.
i
n
i
are the
n
roots of f) for k=0,1,2,..., then pQ = n, pl = -ax and
?2'?3'"* * m a ^
be com
P u t e d recursively by the well-known
Newton formulas:
(a)
P
If k_<n, then
k-Pk-l S l + Pk-2 S 2---- +( - 1)k " 1 Pl s k -l +( - 1)kkS k
(b)
=
°"
If k > n, then
P
k-Pk-lSl+---+<-1)nPk-nSn = °'
where Sj,...^^ are the elementary symmetric polynomials on
the indeterminants x.
,x
I
,. Then
n-1
n-1
n
D(f) =
n-1
n
2n- 2
27
In some cases, the computation of the above determinant
is rather awkward.
However, it is possible to compute the
discriminant of f by the more direct formula:
D ,£, = < - D n < n - l ) / 2 N k ( x ) / k ( f ' ( x ) )
where x is any root of f and f 1 is the derivative of the
polynomial f.
N
k(x)/k
is the norm of f 1
(relative to k).
In [4], E.S. Selmer obtained
D (x n -x-l) = ( - l ) ^ ^ " ^ ( n - 2 ) [n n +(-l) n (n-l) n - 1 ]
D (x n +x+l) = (_i ) i 2 n ( n- 1 ) [ n ^ - D 1 1 - 1 (n-l) n _ 1 ] .
Furthermore, he proved that x n ± x - 1 are irreducible for all
n.
It happens that there exists a real number y
that y ii is a zero of x +x-l.
numbers in this chapter.
We can see y
n
e
[0,1], such
are binomial
That is an interesting result.
Let
us state the following theorem since we will use it in this
article.
Theorem 2.1.
x n - x - 1 are irreducible for all positive
integers n>_2.
By a trinomial is meant a polynomial of the form x n ± x ± 1.
Due to the fast development of computer science, the irreducibility of trinomials over
theory.
play some role in the coding
There are quite a few articles studying irreducibility
or irreducible factors of trinomials.
would be an interesting one.
For instance, [5]
28
We need Gauss' Lemma, in this article.
Before stating the
lemma, let D be a unique factorization domain and f =
.
E a.x1
1=0 1
.
a non-zero polynomial in D[x].
A greatest common divisor of
the coefficients aQ,a^,...,a^ is called a content of f and is
denoted by C(f).
Strictly speaking, the notation C(f) is
ambiguous since greatest common divisors are not unique.
an
But
y two contents of f are necessarily associates and any
associate of a content of f is also a content of f.
write b ~ c whenever b and c are associates in D.
We shall
Now ~ is
an equivalence relation on D and since D is an integral domain,
b ~ c if and only if b = cu for some unit u £ D by Theorem
3.2(vi) of Chapter III in [12].
C(af) ~ aC(f).
If f€ D[x] and C(f) is a unit in D, then f
is said to be primitive.
9
=
If a€ D and f€ D[x], then
Clearly for any polynomial g6 D[x],
C(g)g^ with g^ primitive.
(Gauss' Lemma).
If D is a unique factorization domain
and g,f€D[x], then C(fg) s C(f)C(g).
In particular, the
product of primitive polynomials is primitive.
Eisenstein s Criterion. — Let D be a unique factorization
domain with quotient field F.
If f =
E a. x* € D[x], deg f> 1
i = 0
1
and p is an irreducible element of D, such that
PVV
P|a. for i=0,1, . . .,n-1; p
then f is irreducible in F[x],
irreducible in D[x],
For a proof, refer to [5].
2
^,
If f is primitive, then f is
29
§2. Existence of Binomial Numbers
In view of [1,6], one may ask the existence of binomial
numbers.
tively.
The following theorem answers the question affirmaMoreover, the theorem gives binomial numbers for
each degree n > 3.
Theorem 2.2. - For each positive integer n>^3, there are
algebraic integer x and algebraic fractional y of degree n
such that x,y are binomial numbers.
Proof:
(I)
For any positive integers n^3,
First show that there are algebraic integers of
degree n which are binomial numbers.
Case 1.
n is even.
Let f(x) = x n + x - l .
As f(X) is irreducible (according
to Theorem 2.1) and f(0) = -1 < 0, f(i) = 1 > 0, it follows
that there exists an algebraic integer r of degree n such
that
r11 + r - 1 = 0
that is,
r = l-r 11 = 1- (r2 ) n ^2
which is equivalent to
r
m
= 1 - sm _
=
T
K
,c
-,m
m-k.,1
V( l - s*
) Kk
k
k=l
vz
(2.1)
where m = n/2 and
s = r2
(2.2)
30
As s = r or s = 1-r will imply that r is an algebraic
number of degree <2 which contradicts r is an algebraic
integer of degree nj>3, it follows that s ^ r and s^l-r.
Furthermore, (2.1) and (2.2) imply that r is binomially
related to s.
Case 2.
Hence r is a binomial number of degree n.
n is odd, say n=2m+l, m>_l.
Considering f(x) = x2m+1 +x2m - 1, which is irreducible
(according to Theorem 2.1, Gauss' Lemma, and substituting —
for x).
Moreover f(0) = —1 < 0 and f(1) = 1 > o.
Therefore,
there is an algebraic integer r of degree n in the unit interval so that
2m+l
2m
r
+ r - 1 = n0.
(2.3)
Multiplying both sides of equation (2.3) by (r-1), we obtain
r = l - r2m + r 2 m + 2
(2.4)
s = r2.
(2.5)
Set
Then we have
y~
— i
I
m
m+1
m n N
s +s
= 1 - s (1-s);
.
(2.6)
that is,
m+1
_ / y /-.m+1 m-k+1 .1 ,k, . m,, .
r - ( h C
s
(1-s) ) +ms (1-s).
k=0
k#l
r
(2.7)
31
Since s = r or s = 1-r implies the degree of r is less than
or equal to 2, we have a contradiction.
s^l-r.
Therefore s # r and
Equations (2.5) and (2.7) imply that r is a binomial
number.
(II)
Next we must prove that there are algebraic frac-
tional of degree n ^ 3 , which are binomial numbers.
The following lemma will simplify the original proof of
(II), which is also done by a Norwegian mathematician, Helge
Tverberg, at the University of Bergen in Norway.
Lemma.
2x 2n + 2x 2 n _ 1 - x 2 - x - 1 is irreducible over
Z[x] for all positive integers n>_2.
Proof:
According to Gauss' Lemma and substituting ^
for x, it suffices to prove
f(x) = x 2 n + x 2 n _ 1 + x 2 n ~ 2 - 2x - 2
to be irreducible.
Assume the contrary that f(x) = g(x)h(x) where g,h€ Z [x]
and deg g >_ 1, deg h
1.
As f(x) 5 x 2 n + x 2 n _ 1 + x 2 n " 2 E x 2n " 2 (x 2 +x+l) (mod 2), and
2
x +x+l is irreducible in 7>1 [x] , it follows that
g(x) = (x2+x+l)xr+ 2gx(x)
h(x) = x s + 2h (x)
where r + 2 = deg(g) > degfg^, s = deg(h) > degd^) and
gj,h e z[x].
Thus
32
„2n
x
2n-l
+x
2n-2
+x
~
„
- 2x - 2
(2.8)
= x 2 n + x 2 n _ 1 + x 2 n " 2 + 2[ (x2+x+l)xrh1+xsg1] +4g 1 h 1>
Now s> 0. If also r> 0, then the right hand side of (2.8)
would have a constant term divisible by r. Thus r=0, and so
g(x) = (x2+x+l) +2(ax+b),
where a,b€ Z.
As g(1) divides f(l), g(-1) divides f(-l), g(0) divides
f(0), it follows that
either g(x) = x2+x-l or g(x) =x2-x-l.
(2.9)
If g(x) = x 2 + x - l , then
f(x) = ( x 2 n + x 2 n _ 1 - x 2 n - 2 ) + 2(x2n_2-x-l)
which is equivalent to
(x2+x-l)h(x) = (x2+x-l)x2n_1 + 2(x2n-2-x-l).
This implies that x 2 n " 2 - x - 1 has factor x 2 + x - 1 which
2 n~ 2
contradicts x
— x — 1 is irreducible for n > 2 (according
to Theorem 2.1).
If g(x) = x 2 + x - l , then
f(x) = (-x 2 n +x 2 n _ 1 +x 2 n ~ 1 ) +2(x2n-x-l).
A similar argument shows that x 2 n - x - 1 has a factor x 2 - x - 1,
which contradicts x 2 n - x -1 is irreducible.
Q.E.D.
33
Continuing to prove Theorem 2.2:
(II)
To show that there are algebraic fractionals of
degree n>_3 (for each n) , which are binomial numbers.
Case 3.
n is odd, say n = 2m+1.
f(x) = 2x 2 m + 1 - 2x 2m + 2xm - 1 is irreducible over ]R ,
according to Eisenstein's Criterion.
Using the fact that
f(0) = -1 < 0, f(l) = 1 > 0, we find that there exists an
algebraic fractional r of degree 2m + 1 in the unit interval
so that
2r 2 m + 1 - 2r 2m + 2rm - 1 = 0
which is equivalent to (2.10), if we multiply (1-r) to both
sides of the previous equation.
r = 1 - 2rm(1-r) + 2[rm(1-r)]2
(2.10)
s = rm(1-r).
(2.11)
Set
Then (2.10) can be transformed into
r = 1 - 2s + 2s2
which is equivalent to
r = s 2 + (1-s)1.
Using the fact that r ^ s , r^l-s, (2.11) and (2.12), we
conclude that r is a binomial number of degree 2m +1.
(2.12)
34
Case 4.
n is even, say n = 2m, m> 1.
Considering f(x) = 2x 2m + 2x 2 m _ 1 - x 2 - x - 1, according to
the previous lemma, f(x) is irreducible over Z[x].
Upon using
the fact that f(0) < 0, f(1) > 0, we find an algebraic fractional r of degree n in the unit interval such that (2.13)
holds.
o„2m 0 2m-l
2
.
2r + 2r
- r - r - 1 = 0;
(2.13)
that is,
1+ r + r 2 = 2r 2m_1 (1+r).
Multiplying both sides of the above equation by (1-r), we
obtain
1 - r 3 = 2r 2m ~ 1 (1-r 2 )
which is equivalent to
1 = r 3 + 2r 2m - 1 (l-r 2 ).
MultiplyinQ" both sid.Gs of th.B cibovG scjiicition by ITF WG obtain
r = r 4 + 2r2nl(l-r2).
(2 . 1 4)
Set
s = rl
-
(2.15)
Then (2.15) becomes
r = s 2 + 2sm(1-s)
(2.16)
35
which is equivalent to
r
= (m+1)sm(1-s) +
m-1
z ( m ~ )s
k=0 k
k^l
(l-s)k.
Similarly reasoning, we have s / r , s ^ 1 - r .
(2 17)
Therefore
(2.17), (2.15), s ^ r, and s i1 1-r imply that r is a binomial
number.
Q.E.D.
§3. Existence of Non-binomially
Algebraic Numbers of Degree > 2
Iri this section, we are going to prove that any algebraic
integer of degree 2 is not a binomial number.
Let [x] be the largest
the fractional part of x.
integer £ x and <x) = x - [x] be
Then (3.1), (3.2), (3.3), (3.4) are
evidently true.
If 0 € (0,1) and a,b€ B , then
either
or
(3.1)
<(a+b)0) = <a0} + <b0),
((a+b)0) = <a0) + <b0) - 1.
If s > r, s,r are positive integers, O < 0 < 1 , then
<(s-r)6) =
f<se)-<r0)
<
|^<s0) - (l-<r0))
((q(r0)) = l - < q e )
if <S0> > <r0)
if <r0) > <s0)
if q € JSF , O < 0 < 1
(-n0) = 1 - <n0)
(3.2)
if n€ u .
The following theorem is very useful.
(3.3)
(3.4)
36
Theorem 2.3. - Let s be an irrational number and m an
integer with |m| > 1 .
If r = <ms>, then s^<kr> for any
integer k.
Proofs
It is obvious that k 5^ 0. Assume the contrary
that s = <kr) and r = <ms) with |m| > 1, k # 0.
Notice that
kms = k ([ms] + <ms >)
= (k[ms] + [k<ms>]) + <k<ms>)
= (k[ms] + [k<ms)]) + s.
Therefore, we have (km-l)s as an integer.
By using the fact
that s is irrational and (km - l)s is an integer.
have km - 1 = 0.
This contradicts |km| > 1.
So we must
Hence s#<kr)
for any non-zero integer k.
Q.E.D.
Theorem 2.4. — If r is an algebraic integer of degree 2
and s is binomially related to r, then either s= r or s=1-r.
Proof:
Since s is binomially related to r so that there
are non-negative integers m,n; a , a a
; bn,b1f...,b so
u i
n
u i
m
that 0 < a. < (^), 0 < b < ( m ) for i=0,1,2,...,n;
•J
J
=
j 0/1/2,...,m, and (3.5) holds.
s
=
a rI1
0
+ a 1 rI1 ~ 1 d-r) + ... + a^ (1-r) n
r = b s m + b sm_1(1—s) + ... + b (l-s)m
1
Jm
Using the fact that r is an algebraic integer of degree 2,
there exist integers a,b so that
2
r + ar + b = 0;
(3
' 5)
37
that is, (3.6) holds.
r2 = - a r - b .
(3.6)
Repeatedly applying (3.5) and (3.6), we obtain s = <pr)
and r = (qs) for some integers p,q.
According to Theorem 2.3,
we must have p = 0,1,-1 and q = 0,1,-1.
These possible combi-
nations of values of p and q force either s = <r) or s = <-r).
In other words, according to (3.4), s = r or s = 1-r.
Q.E.D.
It follows from Theorem 2.4 that any algebraic integer
of degree 2 is not a binomial number.
§4. Binomial Numbers and Pisot Numbers
We shall denote by ||a|| the absolute value of the
difference between a and the nearest integer.
Thus,
I I a| | = min |a-n|, n=0,±l,±2,... .
If m is the integer nearest to a, we shall also write
a = m ±e
where 0 < e £
so that ||a|| =
e.
If n is an integer, then
||a|| = ||a+n||.
Definition. — Let 0 > 0 be a real algebraic integer such
that all its conjugates (not 0 itself) have moduli strictly
less than 1.
Then we shall say that 0 is a Pisot number.
Refer to [7, 8, 9, 10].
38
Theorem 2.5. - If 0 is a Pisot number, then | | 0 n | | -> o
as n
oo.
For a proof, refer to [7, 8].
It is worthy to mention an important unsolved problem
here.
Suppose that 0 > 1 is a number such that | |0n| |
o as
n + oo (or, more generally, that 0 is such that there exists
a real number A such that | |X0n| | + 0 as n -* oo) .
C an
we
assert that 0 is a Pisot number?
The following theorem is a well-known good theorem.
Theorem 2.6. - The set of all Pisot numbers is a closed
set.
For a proof, refer to [8].
In view of Theorem 2.6, we
have the following definition.
Definition. - A real number 0 > 1 is a binomial number if
1/0 is a binomial number in [0,1].
It is very interesting to determine the relationship
between Pisot numbers and binomial numbers or other similar
questions.
According to the previous definition and [1], we know
that the positive integers 2,3,4,— are trivially non-binomial
Pisot numbers.
Using Theorem 2.4, we can determine some of
the Pisot numbers of degree 2 that are non-binomial.
instance, if a is an integer and a <-2, then
(-1 + /aJ^4) /2
is a non-binomial Pisot number.
For
39
According to Theorem 2.2 and Case 3, the binomially
algebraic integer 0 of degree 3 such that
I 3 - 2 0 2 + 20 - 2 =
and it is obvious 1 < 0 < 1.5.
0ab =
0
Let a,b be its conjugates.
2 and 1 < 0 < 1.5, it follows that |ab| > 1.
words, at least one of a,b has modulus > 1 .
and non-Pisot number.
m
As
other
So 0 is a binomial
For more details of binomial numbers
and Pisot numbers, refer to [11].
CHAPTER BIBLIOGRAPHY
1.
Navarro-Bermudez, F.J., "Topologically equivalent measures
in the Cantor space," Proceedings of the American
Mathematical Society (2) 77 (1979), 229-236^
2.
Perron, 0., "Grundlagen fur eine Theorie des Jacubischen
Kettenbruchalgorithms," Mathematische Annalen 64
(1907), 1-76.
3.
Brun, V. "Engeneralisation av kjedebroken I, II" (avec
des resumes en fransais), Vid, Selsk, Skrifter,
Mat.-Nat. Kl., Kristiania 1919, Nr. 6 1-29 1920,
1-24.
4.
Selmer, Ernst S., "On the irreducibility of certain
trinomials," Mathematica Scandinavica 4 M9R61.
287-302.
5.
Marsh, Richard W., W.H. Mills, Robert L. Ward, Howard
Ramsey, Jr., and Lloyd R. Welch, "Round Trinomials,"
Pacific Journal of Mathematics, Vol. 96.
1
'
1981, 175-192.
6.
Navarro-Bermudez, F.J., "A Number Theoretic Result with
Applications to Measure Theory," preprint.
7.
Meyer, Yves, Algebraic Numbers and Harmonic Analysis,
Amsterdam, North-Holland Publishing Co., 1972.
8.
Salem^ Raphael, Algebraic Numbers and Fourier Analysis,
Belmont, California, Wadsworth Mathematics Series,
1963.
'
Pisot, C. and J. Dufresnoy, Annales Scientifigues de
l'Ecole Normale Superieure, Vol. 70 (1953), 105-133.
10.
Pisot, C., Annali di Pisa, Vol. 7 (1938), 205-248.
11.
Huang, K., "Binomial Numbers and Pisot Numbers," to
appear.
12.
Hungerford, T.W., Algebra, New York, Holt, Rinehart
and Winston, Inc., 1973.
40
CHAPTER III
ALGEBRAIC NUMBERS
§1. Introduction
There are many points of view from which one may
approach algebraic number theory.
The most favorite point
of view, possibly, is the algebraic point of view.
Study-
ing algebraic numbers through algebraic point of view has
to do with groups, rings, fields, ideals, quotient fields,
homeomorphisms and isomorphisms, modules and vector spaces,
field extensions, Galois theory, Dedkind rings, localization,
the ideal function, ideles and adeles, class field theory,
etc.
There is another favorite point of view; namely, the
analytic point of view.
For instance, there are Liouville's
Theorem, Functional Equation ([1]), Diophantine Approximation
([2]), Fourier Analysis ([3]), Harmonic Analysis ([4]), etc.
There are some interrelations between these two viewpoints.
There are numerously unsolved problems related to algebraic
numbers.
Neither algebraic point of view nor analytic point
of view can solve all the problems.
There are many popular
theorems having to do with algebraic numbers.
Both points of
view have developed a lot of concepts and techniques to solve
problems.
They also have developed their own theories and
have proved many nice theorems, respectively.
41
42
The study of topologically equivalent measures has not
only led us to study the binomial numbers but also intrinsically led us to have a different point of view for looking at
algebraic numbers.
§2. Equivalence Relations on I
Let I be the unit interval on the real line, and S be
the set of all the functions from I into I.
If T is a subset
of S, we give the following definition.
Definition. - Let r and s be two elements of I.
We say
that r is T-related to s provided that there exist f,g in T
so that r = f(s) , s = g(r).
If T is closed under the operation of composition of two
functions, then T-related relation is an equivalence relation
on I.
From now on, we are only interested in this type of
relation.
We shall call it compositive relation.
In other
words, if T-related relation is a compositive relation, then
it is an equivalence relation on I.
For instance, if T = {id}, where id is the identity
function on I, then each equivalence class has only one element.
itself,
If T = S, then there is only one equivalent class - I
if T = {f: f(x) = 1-x or f = {id}}, then each equi-
valent class is of the form {r,l-r} for r£ I.
It is obvious
that the three relations appearing above are compositive
relations on I.
43
Let
Z[x] - the set of all polynomials with integer
coefficients,
Q[x] = the set of all polynomials with rational
coefficients,
£ ak x k (1-x) n - k
k=0
a
° ± k l (£) for k-0,1,... ,n}.
B [x] = {f 6 Z[x]: f(x) =
Bq[x]-related equivalence relation has been called binomially
related in Chapter II.
Let us call Z[x]-related relation the
integrally related relation and Q[x]-related relation the
rationally related relation.
It is obvious that integrally related and rationally
related relations are compositive relations.
It is not triv-
ial that Bq[X]-related relation is a compositive relation.
The next section will prove that Bq[x]-related relation is a
compositive relation.
§3. Homogeneous Functions in Two Variables
Let H[x,y] be the set of all homogeneous functions in
two variables x,y with non-negative coefficients.
that
n
f(x / Y ) =
Z a x iy n-i,
i=0
g(x,y) =
are two arbitrary elements of H[x,y].
f <g
11
n
Z b.x j y n ~ j
j=0 J
We define
if a. <b. for all i=0,l,2, .. . ,n.
Assume
44
One can easily prove the following two statements (3.1),
(3.2):
If f,g,h are in H[x,y] and if f < g, then f«h < g-h.
(3.1)
If f < g, then f n •< g11 for n=l,2,
(3.2)
Let B [x,y] = {f 6 H[x,y] : f(x,y) =
n
O i a . l t . ) for i=0,1,2,..., n} .
.
E a.x1y11
i=
0 1
1
and
Lemma. - Let A — a x m + a x m ^y+ ... + a y m and
u
i
m
B = b x m + b 1 x m _ 1 y + ... + b y m
U
I
ill
where
°-ak^(k}' ° <
b
k
^O'
and
°^ak+bk^(k)'
If f(A,B) = C q A £ + C 1 A A _ 1 B + ... + C £ _ 1 A B £ _ 1 + C £ B £
= e xp + e x p _ 1 y + • • • + e
u
I
p-1
1
u
x y p - 1 + e2 y p
p-
and 0 £ C i £ ( £ ) , then p = m£ and 0 < e± <_ ( p ) for i=0,l, . P.
..,
Proof:
First, it is easy to see p = m£, and
e x p + e x p _ 1 y + . . . + e .1x y p _ 1 + e
yp
u
i
p-1 pJ
= f(A,B)
= C Q A £ + C 1 A £ _ 1 B + ... + C £ _ 1 AB £ ~ 1 + C^B £
< (£)A^ + ( £ ) A £ _ 1 B + . . . + ( * )AB £ _ 1 + ( £ )B £
= (A + B) £
=
[ (a
n + b n ) x m + ( a i + b i )xin 1y + • • • + (a +b )y m ] Z
u u
1 1
m m
"< [(™)x m + ("1)xm~1y+ ... + ( m )y m ] £
u
1
m
= (x + y)
= ( um n £ )x m £ + (n,8-)xm<!-1y+ ... + ( m J)y m £ ;
1
m£
45
that is,
e
0 X ^ + e ^x p
y + ... + e _. xy p ^ + e y p
P
p
•< ( m 0 £ ) x m £ + ( m 1 A )x m A - 1 y+ ... + ( ^ ) y m £ .
Uisng the fact that p = m£ / we find
° -
8
i -
*
Q.E.D.
If f(x,y) 6 B[x,y], and
f(x,y) = a 0 x n + a i x n _ 1 y + ... + a n y n
where 0 < a. < (J) for i=0,1,2,...,n.
Then we define the
conjugate of f, denoted by f, by
f(x,y) = (l-a 0 )x n + ((J)-a )xn"1y+
... + ( ( n ) -a )y n .
n
n
From the previous lemma, we obtain the following theorem.
Theorem 3.1. - if f(x,y) 6 B[x,y], then f(g,g) 6 B[x,y]
for any g 6 B[x,y].
Corollary. - B q [x]-related relation is a compositive
relation on I.
Proof:
and g(x) =
Suppose g , f 6 B
[x] and f(x) =
m
z bk x k (l-x) m ~ k .
k=0
E a x k (l-x) n ~ k ,
f
-K.= rj k
Define
F(x,y) =
E akxkyn~k,
k=0
G (x,y)
=
? b xkym"k.
k=0 k
According to Theorem 3.1, F(G,G) €B[x,y] and
46
F(G(x,1-x),G (x,1-x))
n
=
k=0
=
aK lG(x,l-x))k(G(x,l-x))n-k
, l a k ( j „ V £ ( 1 - x ) m " £ ) k ( ? ((®)-b£)xil(l-x)in-£)n-k.
k=0
£=0
£=o
Using the fact that 1 = (x+(l-x))m =
in
/Hmix)x
__kk(l-x)
, ^ vm m-k
£ (
~k,
we
k=
o k
find that
h(x) = F(G(x,l-x),G(x,l-x))
n
=
Z a (g(x)) (l-g(x)) n_k = f(g(x)).
k=0
Since F(G,G) € B[x,y], so h(x) 6 B q [x].
Hence f(g(x)) is an
element in B Q [X], that is, Bq[x]-related relation is a compositive relation on I.
§4. Rationally Related Relation on I
In this section, we are going to investigate the rationally related relation on I.
Although it does little classi-
fication of algebraic numbers, it certainly gives a very
clear-cut equivalence classes for the unit interval.
Let Q(r) be the field obtained by adjoining r to Q.
Q(r)
a
= deg r.
vector space over Q and the dimension [Q(r): r]
The following theorem is needed in this section.
Theorem 3.2. - Let W be a subspace of a vector space V
over a division ring D.
(i)
(ii)
Then the following statements hold.
dim W £ dim V;
if dim W = dim V and dim V is finite, then W = V;
47
(iii)
dim V = dim W + dim (V/W).
The following theorem is the main theorem of this section.
Theorem 3.3. - The equivalence classes for rationally
related relation on I are either of the form (4.1) or of
the form (4.2).
n-1
A(r) = {s: s 6 i , s = a„+a
1r+...- + a
l
0* ret
l" * * n-1 r
'
where deg s= deg r, a ,a , ...,a
are
U 1
n-1
rational numbers}
if
r
(4.1)
i-s a n algebraic number of degree n and r is in the unit
interval.
T(r) = {a+br€l: a,b are rationals, b ^ O }
(4.2)
if r is a transcendental number and r€ I.
In order to prove Theorem 3.2, we need the following
famous classical theorem in algebra.
Theorem 3.4. - Let k be a field, u an element of a larger
field, and suppose that u is algebraic over k.
Let f be a
monic polynomial with coefficients in k of least degree such
that f(u) = 0, and let this minimal degree be n.
Then
(a)
f is unique;
(b)
f is irreducible over k;
2
n__ ^
l,u,u ,...,u
form a vector space basis of K(u)
(c)
over k;
48
(d)
[k(u) : k] = n;
(e)
A polynomial g with coefficients in k satisfies
g(u) = 0 if and only if g is a multiple of f.
For a proof, refer to [6].
Proof of Theorem 3.3:
Case 1.
Suppose r€ [0,1].
Suppose r is transcendental.
Let s be rationally related to r.
f(x) = aQ + a ] [ x+ ... + a x
n
Then there are f,g, say
(a ^0), g(x) = b + b. x + . .. + b x m
11
u
i
m
(bm^0) being two elements in Q[x], such that r = f(s) and
s = g(r).
It is easy to see that r = f(g(r)), which is equi-
valent to
r = a
+ a
0
i ( b 0 + b i r + * • • + b m r m ) + ••• + a n ( b o+...+ b m r m ) n
Claim.
(4.3)
m = l and n = 1.
Suppose the contrary - that m ^ l or n # l .
Since m = 0 ,
n - 0 would imply that r is rational, so we can assume m> 1
or n> 1.
Then the degree of f(g(x)) is nm and a b n is its
n m
leading coefficient.
As the left hand side and right hand
side of (4.3) are equal, it follows that a b n = 0.
That
n m
contradicts
S = b
0+bir*
a
n
^ 0 and
I n ot
Hence m = l , n = l ; that is,
her words, if s is rationally related to
r, then s = b + b r.
Conversely, if s = a + b r € I and a,b are rational numbers
with b ^ O , then we take f(x) = a + bx and g(x) = ^ + I x
b
Then
b
s = f(r) and r = g(s); that is, r and s are rationally related.
Hence, the ecjuiv&lence eXciss of T is of the form (4.2)
49
Case 2.
r is algebraic over Q.
Suppose s is rationally related to r.
Then there exist
f,g£Q[x] so that r = f(s), s = g(r) . Let
irr(r) = x n + a ^ x 1 1 - 1 + ...+ a Q , where a. 6 Q, for i=0,1,...,n-l.
So we have r n + a ^ r 1 1 " 1 + ... + a Q = 0; that is,
n
r
= _a
rn-l
n-1
" *'* ~ V
(4.4)
Applying (4.4) t o s = g ( r ) finitely many times, we obtain
s6A(r).
Furthermore deg s < deg r.
r = f(s) which would imply deg r < n.
Conversely, suppose s€A(r).
If deg s < deg r, then
Therefore, deg s = deg r.
We want to show s is
rationally related to r.
s€A(r) implies there are rational numbers a ,a ,...,a
0' 1'
' n-1
so that (4.5) holds.
S = a0 + a l r + • • • + V l 1 " " 1 Let f (x)
a Q + ajX + . . . + ^ ^ x 1 1
1
.
s€A(r) also implies deg s = deg r.
A
H.s)
So we have s = f (r) .
Using the fact that
(r) £ Q(r), s^A(r), we find that Q(s) £Q(r).
Moreover,
Q(r) is a finite dimensional vector space, Q(s) is a subspace
of Q(r), and dim Q(s) = dim Q(r).
Therefore Q(r) = Q(s).
That is, r = g(s) for some g(x) € Q [ X ] .
Q.E.D.
§5. Integrally Related Relation on I
After investigating the rationally related relation on I,
the next composition relation on I to be studied is integrally
related relation.
We have the following theorem for the inte-
grally related relation on I.
50
Theorem 3.5. - Let r be any number in the unit interval.
Then the following statements are true.
(a)
If r is transcendental and s is integrally related
to r, then either s = r o r s = l - r.
(b)
If r is an algebraic integer of degree 2 and s is
integrally related to r, then either s = r or s = 1 — r.
(c)
If r is an algebraic integer, then s is integrally
related to r if and only if z [ s ] = Z[r].
Proof:
(a)
Suppose r is transcendental and s is integrally
related to r.
Then there exist f,g€ Z [x] , say
f (x) = a + a.x+ ... + a x n ;
u
i
n
g(x) = b + b x + ... + b x n (b #0) ,
u
i
n
n
so that r = f(s) and s = g(r).
r =
f(g(r)).
r =
a
0
+ a
Claim.
it is easy to see that
r = f(g(r)) is equivalent to
l
<
b
+ b
0
1
r+. •
•+bmrTn) +
... +
a n (b Q +. . .+b m r m ) n .
(5.1)
m = l and n = l .
Suppose the contrary that m # l or n # l .
Since m = 0 ,
n — 0 would imply that r is rational, so we can assume m > 1
or n > 1.
and a
nbm
Since a n # 0 , k>m^0, the degree of f(g(x)) is nm
is i t s lead
i n g coefficient.
As the left hand side
and the right hand side of (5.1) are equal and l,r,r^,...
form a basis of Q[r] over Q, it follows that a b n = 0. This
n m
a
contradicts n ^ 0 and fc>m^0. Hence m = l , n = l; that is,
51
s
r
=
b
=
a
0
+ b^r
(5.2)
0
+ alS
(5.3)
From (5.2) and (5.3), we obtain
r = a
So we have
0 + alb0 + alblr-
= 1.
Then either
= b^ = 1 or
and (5.2) becomes either s = b Q + r or s = b Q - r.
= b =-1
Using the
fact that r € [0,1] and b^ is an integer, we conclude that
either s = r o r s = l - r.
(b)
If r is an algebraic integer of degree 2 and s is
integrally related to r, applying Theorem 1.7, Theorem 1.8,
we will have r,s satisfy (5.2) and (5.3) for some integers
b
0' b l ,a 0' a l*
With
a
similar argument, we reach s = r or
s = 1 - r.
(c) is obviously true.
Z[r] is a larger set than the equivalent class of r under
integrally related relation.
For instance, s = nr is not
integrally related to r for any integer |n| > 1.
§6. B^-related Relation
Let Bj[x] be the set of all polynomials f(x) of Z[x]
so that there are non-negative integers a.,a.,...,a with
U 1
n
(aQ = 0 or 1) and (an = 0 or 1) , and
fix) = a 0 x n + a 1 x n - 1 (l-x) + ... +a n _ 1 x(l-x) n - 1 +a n (l-x) n .
It is obvious that B1[x]-related relation is a compositive relation on I.
From now on, we will call Bx[x]-related
(6.1)
52
relation the B^related relation.
Since (s is B -related to
r) implies (s is integrally related to r), the following
theorem is an immediate result of Theorem 3.5.
Theorem 3.6. - Let r be a number in the unit interval
and s Bj-related to r.
And if r is an algebraic integer of
degree 2 or if r is transcendental, then either s = r or
s = 1-r.
The following Theorem 3.7 was inspired by the proof that
F.J. Navarro-Bermudez has supplied for his main theorem in
[7].
Lemmas (6.1), (6.2), (6.3), and (6.4) are based on his
proof in [7].
Theorem 3.7. - Let r and s be two rational numbers in
the unit interval.
The numbers r and s are B^-related if
and only if s = r or s = 1-r.
Proof:
The sufficient condition is clear.
the necessary condition proceeds as follows.
The proof for
Let r = a/b and
s - c/d, where b and d are positive integers and a and c are
non-negative integers.
in reduced form.
Assume that the fractions are written
Using the fact that r and s are B^-related,
we find (6.2), (6.3) are true for some non-negative integers
a Q ,a 1 ,...,a n ; b Q ,^,...,b m ;
a
0 /b 0 ,a n ,b m
are either 0 or 1, and
n,m are positive integers.
s = V
1
r
m
=
V
"!''"
+
1
!
1
" ! + ... + a n _ 1 r ( l - r ) n - 1 + a n ( l - r ) n
V m " 1 < l - = > + - . . + b B _ 1 s < l - s ) m - 1 + b (i-sl"1.
(6.2)
(6-3)
53
Since Bj related relation is an equivalence relation, we
can interchange the roles of r and s, if necessary, and assume,
without loss of generality, that b < d.
If r = 0 or 1 and applying (6.2), then s = 0 or 1.
now on, we can suppose that neither r nor s is 0 or 1.
From
in
other words b > 1, 0 < a < b and 0 < c < d .
Suppose the contrary that s ^ r and s # 1-r.
We are going
to show that (6.3) cannot be true.
Note that if c = d-c, then s = c/d = 1/2.
and since l<b<_d, then b = 2 also.
Hence d = 2,
But then a = l and r = %.
Thus c £ d-c and, by interchanging the roles of s and 1-s, if
necessary, it may be assumed, without loss of generality,
that c < d-c.
It is being assumed throughout that b £ d .
From now on,
the proof splits into two cases; either b = d or b < d.
Case 1.
b = d.
If it were true that a = b-a, then r = a/b = 1/2.
Hence,
d = b = 2, and since 0 < c < d, it would follow that c = 1 and
therefore s=h.
Thus a ^ b - a and, by interchanging the roles
of r and 1-r, if necessary, it may be assumed that a < b-a.
Also, if it were true that d - c < a , the d-c < b-a, and as
c< d-c <a, it would follow that d = c + (d-c) < a + (b-a) = b.
But b = d.
Consequently, a<d-c.
Finally, observe that under
the assumption b = d, it is not possible to have a = c, as this
would force s = r.
54
Lemma 6.1. - I f b - d , c < d - c and a < d - c, there
cannot exist non-negative integers b ,b , — , b
(m>l) with
u i
m
—
b Q = 0 or 1 such that
^ = b (—)m + b f
b
Vd' + Vd'
Proof:
(6.4) holds.
ad
—
i
+
,i /d-c»m
• * • + bm < — ) •
(6.4)
Assume that, for some choice of integers, line
Then, upon clearing denominators,
b Q c +b 1 c m
1
(d-c) + .. . + b^ (d-c)m.
(6.5)
As d e c (mod d-c), it follows from (6.5) that
a
c m - 1 = b0cm
(mod d-c).
(6>6)
Since c/d is in reduced form, the integers c and d-c are
Line (6.6) can be divided by cm-''' to
relatively prime.
obtain
a e bQc
(mod d-c).
(6m7)
However, 0 < a < d-c, so that it is not possible for b Q to be
0.
Thus, b Q = 1, but then a E c (mod d-c) is not possible
either, as both a and c are less than d-c and a ^ c .
This
contradiction proves the lemma.
Case 2.
b < d.
Then the inequalities a > c and b-a > d-c cannot be both
true.
Thus, a < c or b-a < d-c, and there is no loss of gener-
ality to assume that a < c, provided that it is not assumed also
that c < d-c as was done earlier.
55
Lemma 6.2. - Let b < d and a < c.
If a E b m b (mod c) and
a = b Q b (mod d-c) with b m = 0 or 1, bQ = 0 or 1, then
(i) b
m
= 1
'
(ii) a < d - c
Proof:
k.
'
and
(iii) b Q = 1.
If b^ = 0, then a = kc for some positive integer
Hence a >_ c, which contradicts the hypothesis.
Therefore
b m = 1; thus proving (i).
It follows from (i) that b-a = jc for some non-negative
integer j.
As b ^ a , j must be positive.
Then b-a > c.
If
it were true that a _> d-c, it would follow that
b = a + (b-a) >_ (d-c) + c = d, or b > d, contrary to the hypothesis.
Therefore, a < d-c, thus proving (ii).
then a = p(d-c) for some positive integer p.
but this contradicts (ii).
If b^ = 0,
Hence a > d-c,
So, b Q = 1, and the lemma is
proved.
Lemma 6.3. - Let b < d and a < c.
Then it is not possible
to have a = b (mod c) and a = b (mod d-c) simultaneously.
Proof:
Suppose on the contrary that b = a + k c and that
b = a + j(d-c) for some non-negative integers k and j.
As
b # a , both k and j must be larger than or equal to 1.
Both
inequalities c < d/2 and d-c < d/2 cannot be true simultaneously.
Assume that c
d/2.
If k > 2, then b = a + k c > a + d
or b > d, which contradicts the hypothesis.
Hence, k = 1.
But then, c = kc = b-a = j(d-c), so that d-c divides c.
c and d-c are relatively prime and distinct.
But
Hence, d-c = 1,
which is impossible on account of Lemma 6.2(ii) and the fact
that a is positive.
d,
56
It must then be true that d-c > d/2.
b = a + j (d-c)
thesis.
a + d > d.
Thus, j = l.
c divides d-c.
distinct.
If j > 2, then
But this contradicts the hypo-
Then d-c = j(d-c) = b-a = kc, and
Again c and d-c are relatively prime and
So, c = 1, which is impossible as a < c by the
hypothesis and a is positive.
This contradiction proves the
lemma.
Lemma 6.4. - Let b < d and a < c.
Then there cannot
exist non-negative integers hQ/bj r • •. fbm (m> 1) with b Q = 0
or
1/ b m = 0 or 1, such that
f - V§>"
Proof:
(6.8) holds.
+
Vi>'"1<TS>
+
-"+VT
£
>".
Assume that, for some choice of integers, line
Then, upon clearing denominators,
d m a = b 0 bc m + b 1 bc m - 1 (d-c) + ... + bb(d-c) m .
As d E c
(6-8)
(6 .9)
(mod d-c) and d E d-c (mod c) , it follows from (6.9)
that
c
" 3. = b be
(mod d-c)
(6.10)
and
(d-c) a = b^bfd-c)"1
(mod c) .
Since c and d-c are relatively prime, line (6.10) can be
divided by c m and line (6.11) by (d-c)m to obtain
(6.11)
57
a = b b
(mod d-c)
a = b b
ni
(mod c) .
(6.12)
Then, by Lemma 6.2, b = b = 1 and thus a E b (mod c) and
u
m
a E b (mod d-c). However, by Lemma 6.3, these two congruences
cannot hold simultaneously.
This contradiction proves the
lemma.
Lemma 6.1 and Lemma 6.2 show that if s ^ r and s^ 1-r,
then Equation (6.3) does not hold.
Therefore, Theorem 3.7
has been proved.
Q.E.D.
As BQ [X] ^ B1[x], it follows that "s and r are binomially
related" implies "s and r are Bj[x]-related."
We will have
the following theorem.
Theorem 3.8. - Let n be a positive integer.
The follow-
ing two statements hold.
(a)
For n>_2, there are algebraic fractional numbers in
the unit interval so that each of their equivalence classes
under B -related has more than two elements.
(b) For n>_3, there are algebraic fractional numbers in
the unit interval such that each of their equivalence classes
under B^-related has more than two elements.
Proof:
Using the fact that "s,r are binomially related"
implies "s,r are B -related," and in Theorem 2.2 we find that
the only thing we need to prove here is to find the algebraic
fractional 0 of degree 2 so that the equivalence class of 0
under B^-related has at least three elements.
58
Look at f(x) = 4x(l-x) = 4x2(1-x) +4x(l-x)2.
r = (5 + /5)/8 and s = (5-/5)/8.
and f(x) 6
Let
Since s = f(r) and r = f(s)
[x], so the equivalence class of r has at least
four elements, namely r, s, 1-r, 1-s.
And it is obvious that
the degree of r is 2.
Note.
Q.E.D.
Let A be the set of all polynomials f(x) of Z[x]
so that there are integers a^ra^,...,a^ with (aQ = 0 or 1)
and (an = 0 or 1) , and 0 _< a^ £ (?)+l for i=2,3,...,n-l.
f (x) = a.x11 + a. x n ~ 1 (1-x) + . . . + a
U
I
And
. x (1-x) n ~ 1 + a (1-x) n .
n-1
n
Let B be one smallest subset containing A so that the
compositions of two functions B are closed in B.
All the
theorems true for B^-related relation are true for B-related
relation in this section.
§7. Binomial Numbers and Binomial Coefficients
Binomial coefficients are the combinations of n things,
k at a time.
The Pascal triangle identity (7.1), the
binomial theorem (7.2), and many other combinatorial identities have to do with the magic binomial coefficients.
(j) - I";1)+ <;:}>
(a + b)n =
? (")a k b n_k .
n.u
(7.2)
k=0
Borel measures on the Cantor set are naturally connected
with binomial coefficients through taking the measures on the
basic open sets.
59
In this section, we are going to expose an interesting
formula related to binomial numbers and binomial coefficients.
Let r be a binomial number in the unit interval and s is
binomially related to r.
So there exists f,gf BQ[x] such
that s = f(r) and r = g(s).
Hence r = g(f(r)).
According
to Theorem 3.1, g,f€B^[x]; that is, there exist integers
n,aQ,a1,...,an such that n > 1, 0 < a
£ (") for i=0,l,...,n;
and
r = aQrn +
(1-r) + . . . + a R (l-r)n.
(7.3)
Expanding the right hand side by using the binomial
theorem, we have
r = (ajj-ajt.. , + (-l) n a n )r n + (a1-(jj) a 2 + (J) a3-... + (-1) n+1 (") a^
+
••• +
+
---
(a
+
k" ( k k 1 ) a k+l + --- + ( - 1 ) n + k < k ) a „ ) r k
( a
n - l " C l
) a
n
) r +
<7"4»
O V
Define h(x) as follows:
h(x) = ( Z (-l) k a 1 )x I1 +( E (-l) k+1 (J) a. )x n_1 + . . .
k
k=0
+
1
k=l
k
(ky-l)k+£(k)ak)x^...
+ ((
Z
k=n-1
{?>5)
(-l) k + n _ 1 ( k )a )-l)x + ( n )a .
n-1
k
n
n
Now let irr(r) = b^ + b, x + . . . + b x m .
0
1
m
As h(r) = 0 and b_+b. r + . . . + b r m = 0, it follows that
0
1
irr(r) is a factor of h(x).
m
Therefore there exist integers
d ,d1,...,dn (dn^0) and t(x)€z[x], so that
60
h(x) = t(x) (irr (r)) = d Q + c^x + ... + d n x n ,
(7.6)
Comparing (7.5) with (7.6), we have (7.7).
d
n
= a
d
n-l
0-al+a2+---+(-1)nan
=
U
1 " (l)a2+ (l)a3"
+
<-l>n+1<?>an>
(7.7)
j
t
d
= a
k"
1
,k+l.
,
k )ak+l + ••• +
, , -i v n+k ,ni
k n
(_1)
n
d.1 + 1 = a n-1 - (n-1
)a n
d. = a .
0
n
In other words,
n
n-1
(7.7)
= A
d +1
n
where
(-1)n
1 - 1 1
0
1
(-l) n+1 (?)
(\)
A =
0
0
1
0
-Ci'
61
The determinant of A is 1 and the inverse matrix of A
exists, say B.
Using the orthogonal relation (7.8) (refer
to [8]), we obtain (7.9).
6
=
(7.8)
E ( - l ) ^ + k (?) (J)
3
k
j =k
where 6
is the kronecker delta.
mk
1
1
1
1
0
0
1 0
2
1
3
3
0
0
0
1
(7.9)
B =
( nJ
n-1
1 (*) .
1
1
1
Hence
• • • ©
n
(7.10)
= B
dl + 1
a
n —
According to3 (7.7) and i
(7.10), we have the following:
m
( k ) d , dm =
m k
k=m
k+m(k)a
m k
where d, = d. for k # l , d = d +1.
k
k
1
i
By using (7.10), we will be able to use the computer to
help us to study a specific binomial number with its given
irreducible polynomial.
62
In the end of Chapter III, I would like to mention two
unsolved problems which were thought to be important by many
mathematicians; for instance, Problem 1 by Stan Ulam and Dan
Mauldin, Problem 2 by R.D. Mauldin.
Problem 1. - Let r,s be any two elements in the unit
interval.
If r is binomially related to s, is it true that
u is topologically equivalent to us ?
r
Problem 2. - Are there any r£ [0,1] so that the equivalent class of r under binomially related relation (or y r under
topologically equivalent relation) has infinitely many elements?
Of course, these are not the only two unsolved problems
left in this article.
If you read this article carefully,
you will find many other interesting unsolved problems.
CHAPTER BIBLIOGRAPHY
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2.
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3.
Salem, Raphael, Algebraic Numbers and Fourier Analysis,
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Meyer, Yves, Algebraic Numbers and Harmonic Analysis,
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63
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65
Oxtoby, J.C. and V. Prasad, "Homeomorphic Measures in
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66
Announcement of Results
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