INEQUALITIES FOR THE NUMBER
OF INTEGERS 111 A SUM OF
SETS OF GAUSSIAN INTEGERS
by
BETTY LOU KVARDA
A THESIS
submitted to
OREGON STATE UNIVERSITY
in partial fulfillment of
the requirements for the
degree of
DOCTOR OF PHILOSOPHY
June, 1.962
APPROVED
Redacted for privacy
Associate rofei;or of
in Charge of
hematics
or
Redacted for privacy
Chairman of Depa ment of Mathematic8
Redacted for privacy
Chairman of
ho
o
Science Graduate Committee
Redacted for privacy
Dean of Graduate School
Date thesis is presented
Typed by Jo1n Eriss
Mav 17. 1962
ACKNOWLEDGMENT
The author wishes to thank Professor
Stalley for his assistance during the
writing of this thesi
Robert D
TABLE OF CONTENTS
INTRODUCT1ON.......................
CHAPTER I.
A "OHEO-TYPE" THEOREM.. ...........
CHAPTER II.
ANALOGUES OF TWO THEOREMS CW
SCHNIRELMANN AND LANDAU FOR
GAUSSIAN INTEGERS.
CHAPTER III.
.............
. 15
ANALOGUE Of A THEOREM OF MANN
FOR GAUSSIAN INTEGERS...........
BIBLIOGRAPHY......
APPENDIX........
.......
26
4..
49
51
INEQUALITIES FOR THE NUMBER
OF INTEGERS IN A SUM OF
SETS OF GAUSSIAN INTEGERS
Let A be a set of positive integers and for any
positive integer n denote by A(n) the number of integers of A which are not greater than n Then the
Schnirelmann density of A is defined (15, p.65) to be
the quantity
a = gib 411)
n
Thus the set of 1 positive integers would have
Schnirelmann density 1, the set of all odd positive integers would have Schnirelmann density Jlis and the set of
all even positive integers would have Schnirelmann density
0,
B,scovitch (1, p. 246) introduced the density
glb A(n)
n41
and Erdcis (
p
66) the density
a = gib
1
where k
n>k
is the smallest positive integer not contained
2
in A.
but we omit this require
(Eras assumed k > 1
ment.)
For any two sets of positive integers A
define the sum Id
C
MB
b, a+b
a
I
a G A, b
B
Schnirelmann (15 p. 652) proved that if a,
Schnirelmann densities of
yZa+P- ap,
are the
C respectively, then
p. 57) proved that
A,
and Landau
p a 1 implies
and
y
1
In a famous paper by Mann it
526) that
y4 in
13)
This result is usually referred to as the
u+
Theorem.
In the same paper is proved (10, p. 526-527) a result
which implies that for any positive integer n not in
e have
and this inequality can be strengthened, by application
of a result in a later paper (11, p. 250-252) to the
relation
(1
B(n)
for any positive integer n not in C.
In a still more recent paper 12, p. 9
409 2) Mann
3
proved a theorem which implies, in our notation, that for
any positive integer n either C(n) r n or there exists
a positive integer
< n and
not in C such that
in
C(n)+1
(2)
611/..:411/...±-1
n+1
+
This in uality is
C(n)
(2.1)
strengthening of the inequality
A(m)+11( )
n
m
Theorem.
n order to obtain the
In this thesis we will be concerned w
attempts
to extend the above definitions and theorems to sets of
Gaussian inteciers, that is, numbers of the form x + yi
Where x and y are real integers
The sets discussed
above were subsets of the set of positive integers; in
our work we will consider subsets of the et
which Mann proved
yi
x
and
y
real integers, x + y
are nonnegative
Oj
For two subsets A and
of Q we will let A(S) denote the number of Gaussian integers in A CIS. In parti
cular, then,
is just the number of elements in S
Whenever we use the notation A(S) the set S will consist
of all Gaussian integers in a given bounded region of the
complex plane and therefore will be finite. For any two
subsets A and B of Q we define the sum set C = A+B
as we did for sets of real integers. The notation AB
Q(S)
4
will occasionally be used to denote the set of all
which are not in B.
Very little work has been done in this area;
the author% knowledge, Cheo (3, p.2) is the only one to
have extended the concept of Schnirelmann density to the
Gaussian integers. His definition is * follows; Let
+ yoi be any number in Q, and let S be the set
ments of
A
of all x +yi in
such that
Q
Then for any subset
Y
Yo
Q,
A
st gib
We will refer to
the ghig,dpnsitv of
A
a
discuss the theorems obtained by Cheo for this density,
as well as prove some similar theorems, in Chapters 1
and 2. These theorems also involve modifications of the
Cheo density.
Cheo was able to show by means of an example that
the u+p Theorem is not valid for his density, but a result
analogous to (2.1) for subsets of Q may still be true.
Cheo's example at east does not furnish a counterexample. This result would be the statement that if
+ yoi is any element of Q which is not in C and
S
is the set of all x + yi
yo
in
Q
h
then there exists an element x1 + y i in
S
which is not in C such that if
x + yi in Q with x / xi and
s the set of all
yl
then
+ B(T)
(T)
A(T
(2.2)
T
Inequality (2.1) implies the
p Theorem for sets 0
real positive integers, but the analogous inequality
(2.2) does not imply the a+p Theorem for subsets of g.
While Cheois definition of density is very natural and simple it proves to be a somewhat difficult one
with which to work. Accordingly, we will modify it in
the following way.
Definition 1.
numbers in
which 0 <
Q,
and y>
S
1,
for which
Q
and let
Than for any subset
A
t
1
of all x + yi in
yti
t
A
R
of
Q
to be the quantity
u = gib
R$
and
**
we de
be
y
(see Figure
ha density of
We will look upon this density as the extension of
the Schnirelmann density to subsets of Q. It will also
be necessary to have an extension of the ErdOs density,
which we define as follows.
Definition 2. If A is any subset of
modified density of A is the quantity
gib
taken over all sets
of the type described in Definition 1 for which A(R) < Q(R)
R
Whether or not the (4.13 Theorem is valid for the
density of Definition 1 is still a matter of conjecture;
however, it is shown in Chapter 3 that
MR)
for every set
R
for which the
x
1]
B(R)
of the type described in Definition 1
yli,
Xt
This extension of the inequality
t1 are not in C.
he main result
of the thesis.
The reader will notice that Theorem 1.1 and the
theorem of Cheo's quoted in Chapter I both require that
all numbers ji, j = 1, 2,
shall be contained in
the sets A and B. and that this same condition on
is needed in the theorem of Cheo's which is quoted at the
end of Chapter 2. The arguments used to prove these theorems are essentially one-dimensional, and these strong
hypotheses are required in order to carry out these arguments. The proofs of Theorems 2.1, 2.2, and 3.1 are
two-dimensional and no such restrictions need be placed
on
A
and
B.
t appears that the methods used to prove the
theorems of Chapters 2 and 3 can be equally well applied
to sets of lattice points in n-dimensional space if density is suitably defined. The amount of exposition required is then greatly increased, of course, and we have
omitted all such work from this thesis.
The method used to prove inequality (3) above can
also be used to give a new proof of (I) This proof, plus
a second new proof, is presented in the Appendix.
CHAPTER
EO-TYPE" THEOREM
A
t
be two sub
and let the Cheo density of C be ya
and
A
titan, p04)
Let
B
(see Introduc.
be the set of all numbers
A
such that a + ji is in A, j 0,
A (x) be the number of elements
2,
ii
and
f
AA
with
a < x where x is any non-negative integer. Similarly
define B B (x), C
and C (x). Let S be the set
f all
ith
y
x
for an
nt
Cheo
p. 10)
proved the following theorem:
=
= glb
(1)
and
Yc
B (x)
for
+ pi
ji is in
0
gib
and
gm 0, 1, 2,
A, B
Po+
pn
where
gib
a,
(2) 1
tp 2 po
then
2,
+p
The method used by Cho° to prove this
heorem
9
(among other things) using one.dimensional arguments on the sets Aj
and C and then summing the
involve
inequalities thus obtained over all j such that ji is
contained in a set S of the type described above. Theo em 1.1 below is somewhat similar to Chem's; our proof
uses the inequality (2) in the Introduction and is simpler
than his
Theo
f there is a positive integer not
contained in C
k be the smallest such integer,
taken over all sets S of the type
described above for which
f there is no positive integer not con-
min (Y0 y*)
tamed in
'f'
C
A
and
(2)
yf
mit 0, 1
2,
gib
gib
0
For all j
gib
glb x +
(1)
let y'
and
+ Pi 5. 1 and a1
ji
A, B
po s 1 for all 3Itg 0,1,2,
for all
then
2,
p
f
Pro
$
y0i be n element of
the set of all x + yi in
Q
with
Q
with
10
and
y<
0.
be the set of all positive in
Le
+ ji is in A . Similarly defi
tegers a such that
R(13.1]and
RDCi]
if j > 0
we have
and
REA0) x
Since
(Aj)
) * A0( 0)
ji is in Ay B
) a Aj(
KB )
j>0
1
with like relationships for the
is in A
Also, the fact that
i andC j
B, for j > 0 and Vitt C 0
A
o
RfC 1 D l[A4]
and C
and ICJ
and
implies that
for
A01 4. Rtfy
all jm. 0, 1, 2,
j > 0, applying Mann'
(see Introduction, page ) either C (
Then for
or there
that
1
a positive integer m not in
RCA
(m
and
such
and
)44
or
.960 theorem
)+1
ewise
Therefore, in either case,
(4)
and noting that
Applying the same theorem to CO3
>k
p.
0(
(x )
we have
o)
A
80(m) 4. 1
m+11
xo +
( 5)
Po
We now add the inequalities (4) and
C
(
C0(x0) Cl(x )
)(y0+1)
S) + 1) (c0
to obtain
= C(S)
12
Therefore,
+f3
for all sets
$
of the type used in defining
+p
Y*
Now for sets
we clearly have
ya,
and
.
of the type used in defining
S
.e*
a' +13
Suppose
x0+ yo
(4) obtained f
is in
< lc.
Q,
j > 0 is still valid.
The inequality
j
For
0
now hay.
4. a0 + A
)
.
From (4),
+1
Ci(x0)
2
(ao
p0+ ai + 13j)
j > 0. By adding these inequalities we obtain
C (xo) +
+ Cy
(x)at
C(S)
)(y
2
(y0+1)..2
2
(
*Po)
we
13
Q(S) (a' +
for all sets
then C(S)
S
y
)
.
such that
Q(S)
a
and
This gives us
and, therefore,
>a
which completes the proof.
It is obvious that the
of Theorem 1.1 is al
ways less than or equal to the p of Cheo's heor
We will now show that this is also true of the 09
Accordingly, let S be any set of the type described in
the first paragraph of this chapter.
Then
14
al
where the
are those defined in the
rue for all
statement of Theorem 1.1. Since this
of Cheo s theothese sets S, it follows that the
of Theorem 1.1.
rem is greater than or equal to the
Consequently, if yc y' then Theorem 1 cannot give
a stronger result than Cheo's theor
The two theorems are not really comparable when
ye but the following example illustrates the re suits for one such
Example:
and
j
3k+1+ji
Let
.2 1,2,.1
I
that
k
A
B=
3k+1. 3k+Ji
0, 1, 2,
1,0 = Qt
011.
and
0,1.2."
k
Then C
[3k+1. 3k+2 3k+j
2,
and j
1/2.
For Cheo $ theorem V*
ye > U 4For Theorem 1.1 the
25 2/3
I
We see
y
4*
3
and
= 1/4, and
CHAPTER II
ANALOGUES OF TWO THECREMS
OF SCHNIRELMANN AND LANDAU FOR
GAUSSIAN INTEGERS
and
B
t was mentioned in the Introduction that if A
are two sets of positive integers C . A + B.
p, y the Schnirelmann densities of A B, C respectively, then a + p 1 implies y =1, and, in any
case, y > a 4, p 4. Cheo (3, p. 6) was able to establish the first of these theorems for his density, and
we will now prove both of them for the density defined
in Definition 1 of the Introduction. The proof of
Theorem 2.1 is essentially that given by Cho°.
A and
Theorem 2.1
Q, C
A+B,
a,
and
Now
y = gib
be two subsets of
, y the densities of A, B C,
respectively.
Proof
B
1
then y
We know y < 1. Hence, assume y < 1
C
1
implies there exists a set
of the type used in defining the density such that
C(R0) < Q(
which in turn implies that there exists
a numberxo + yoi contained in
Q
but not in C.
We
16
may let ao be the set of all
y<y
x
x + yi is in
some
b1+ b5i
Sint
x0+ y
B
either
Then for any x + yi in a
or x + yi
A,
(bl+boi) for
(xo+y
or neither, but never both.
in B ri Re,
t cannot be in
not in
(b + bet) for any
Pi (X0 yoi)
since
yi in Q such that
x
A.
Also,
boi
does not contain 0. Hence, we have
A(a ) +
)<
)
and
which is a c
C
0
A
and
B
be two subsets of
y the densities of
A, B C,
Q,
rifts
Then
Proof:
and
t
Theo em 2
A + B, and
pectively.
a
radiction, Therefore,
If
1
or
i is missing
then
A
and the theorem is obvious. Hence,.w assume
i are in
A.
Also
if
AmQ
then y
a
nd
again the theorem is obvious, We will, therefore, assume
that there exists a set a of the type used in defining
the density such that A(R0) < Q(Ro
17
et H be the set of all x yi which are in
but not in A By the choice of Re H is not
empty.
We want to partition
empty subsets
into disjoint, non..
of the following
H
I.
there exists a
type: For each
nA
number
in
Ro
contained
and a
of the type used in defining
such that
the density, and
4' Yi e
I
will speak of
being "based on" the n
aj + a3i. Note that a1 + aji is not in
Li
In order to effect this partition of
H
be the smallest real integer for which there exis
a real integer
such that
x
either (x +1)
a'i
x
or x
in AflR,
2
Let
+1)j
(
ti
be the smallest such x. (The existence of
follows from the fact that 1
x + yi is in Ro and ei he
x<
at'
then x + vi
be that number in
(2) for any
and ay,
x
H
are in A. If
or y = al and
and
y<a
is in
A.
but
Nnw let
such that
<
le
Yi
x
in
for any
41,1
+ n)i in
m)
0,
su
Q
max(m,n)
00 there exists x+yi in
with
x < 41
a1
1minimal.
(
I have
2,1
3.
does not axis
x+yi not in
i
or
0
and
1
in
H
for which
which contradicts the hoice o
al
with
the choice of
41
al S
(t1
t4
4,1
1
1
Also,
1 have
WO
for the same reason. We continue, des
in this
4,1
manner as long as possib e, and
numbers
m)m)+(4
n = 0, and we have a contradiction on
nating numbers 41
(x
and
4,1
)
1,t ither 41..1+ 4.1
> 0, n 0, and there
n)i with
(4.4
and
be that
satisfies the con-
Then
since
41
nal i
such tha
H
ditions (1) (2),
44
pick that one
H
inima
number in
H
<y
more than one such 41
which 44
but not in
Q
n 1,1 + 4,1n
4-
x yi G Q x+yi
Let the set
the last of the
Lk
=
19
Then the set
UL
L
same type as the
L
alt
isofthe
U Ln
U
described above.
f H Ll is not empty let a be the smallest real
integer for which there exists a real integer x such
but either (x+1) + a;i
that x + ai is in A (IR
a be the
or x + (a' +
is in H LI
smallest such x and form the set L0 based on
in the same way as was done for L1 substitutin
in this
for 4. We continue forming sets L1, Lst,
manner as long as possible. Call the last set formed
Suppose there exists a number
(LIU " U L ).
We know that
a1
M
x+yilx+yi e Q, x+yi A a j
s' <y<h) E
A.
ail
in
so there
based on
Let
1-11
Because of the way
hai
hs
are two possibilities:
i) There exists a set
such that
111
was formed, the s
i.
J
must contain at least one point
not, we wou d have h1 + h5i
in Lj)
Let
be the smallest real integer for which there exists
al integer
x
such that x +
m
20
and for any x'
y'i
x+a
< y/ < ha
have
x'+ y'i in
smallest such
the sets
in Q with
Let
H.
would lead us to form a setL
L1,
which would include
we would not have
h
0 We have
h5i in
H
and
UL
(
j > h1 whenever
11
the aj + aji upon which the
exists a number x+yi in A n
fi,
be the
Then eventually our method of forming
based upon ak
y<h
ak
ased. Then there
such that
hit
for example, since ha > 0 by choice of
ak + ai Just as
Again, among these numbers we can find
in (i) which must eventually be chosen as the base for a
set Lk which would include h + hal.
Hence
the set LI U "6 U Lr exhausts
Figure 2 illustra
and
A(RØ) < Q(R0)
s
Ro
and A such that 1 i GA
for which the set
H
has been parti-
tioned in the manner described above. The dots represent
points of A and the circles points of H.
21,
a8+ati
9
.8
0
6--9 a
1)
aV.
+X'/ifc:
,
6,,
Figure 2
Now, for each j
1,
we note that the
x+yi lE Lj), and that
set L'= [(x-a3) +
(Ys*
Q(L)= Q(
bl+ b5i is in
(a3
not in
3)i
I
L,
B
ali) + (b1 + bgi) is in C n L
A.
C(Ro
and, therefore,
Hence,
4.
A(R
13(
> A(R
o have
Q(R0)
Re
so we substitute to get
C(R0)
A Ro
zg A R
+p
Ro) - A(110)]
R
a[A(R0))
We divide both members of this
nequali
obtain
c(R0)
Urj
such that
Since this relationship holds for any
A(R0)
such that
it must also hold for any
Q(R0)
C(R0) < Q(R0),
Or
implies (1- )(1-p)
Hence, if C(R0) = Q(110) than
Ms*,
1>a+
and
C(no)
1
P
o,
(LP
CITITITT
Since this relationship holds for every set
type used in defining the density
y>a+p
Although Cheo did not prove
for all sets A, B, C
theorem:
and lot
Let
Let
Pc
A
and
gib
2,
Yc
he did prove the following
be subsets of Q C A +
MB,
B
and yebe the Cheo densities of
1 real integers in
be the set of
A
must have
A (x)
If
then
B
B
A,
contains all numbers
and C.
and
2
Ye
ce
implies
I this conclusion
and
Yc
o
ap .
sove the requirement that ji
B
by means of an argument like that
for j 1, 2.
used to establish Theorem 2.2, since it would be necessary that the sets Li be of the type S used in dewould have to be
fining the Cheo density, and thus
such that for each
partitioned into sets Li,
We Ca
Li
there exists a number
i in H for
number
A fl R0
cI
and a
<
and
x+yi
Li
f a1+ai)
This is not always possible as the reader can easily verify by means of the example shown in Figure 3. Again,
the dots are points of A while the circles are points
of H. The sit Ro is the set of all rfYi in Q with
and
) A +
v ha
2
k
d(A1+
some for
integer
a5 a1
-
1
that
d(At
Assume
1.
or
as
d(Al+
A
implies 2.2
density 'the
If Proofs
The 2 = n
of density the be
Al
Introduction.) the in defined set
sum the to equivalent is 2 = n if and associative, and
commutative clearly is addition (This n).
d(Al
1010 U Ai
= An
4.
5iC
A
Let
each
set
sum
0$
the define and
al
2,
I
> n
Q
of sets
his. as
the essentially are proofs Our 655). p. (15, gers
int.. positive of sets for 2.2 Corollary of statement the
prove to result this used then below, 2.1 Corollary in VOn
gi- that to integers positive of sets for ap
result the extended 652-653) p. (15, elmann Schn
same
Then
d
ie.
Ak+l
d Al+
1
and the prcof s complete.
at k be a real integer,
A a basic
t of Q of order k
k
summands (or kA) equals
k
1
We will call
+A
A
Q who
k
is minimal.
Coro1.arv 2,2. If the density a of A is posi-
tive then
A
Proofs
(1-q
16
is a basic set of
Q.
There exists an integer
such that
Then Corollary 2,1 implies that
1 - d(nA) <
n<
and
From Theorem 2.1, d(nA) + d(nA) > 1 implies d(nA + nA
or 2nA m Q. Note that the order of A is less than or
equal to 2n.
26
CHAPTER III
ANALOGUE OF A THEOREM OF MANN
FOR GAUSSIAN INTEGERS
are two sets of positive integers,
C=A
and if n is any positive integer not in Co
then Mann has proved a result that implies that
C(n)
a1(n+1) + B(n) where al is the Erd;s density
A
of
A
and
B
(see page 2). We will now extend this theorem to
sets of Gaussian integers. Accordingly, throughout the
remainder of this chapter it is to be understood that A
A + B. Also, al is
and B are subsets of Q and C
the modified density of A described in Definition 2 of
the Introduction. Our theorem is then the following:
Theorem
to define
a
1.
Let
R
be any set of the type used
for which the numbers
are not in C.
C(R) >
Then
[Q(R)
1]
R(R)
We will need to make frequent reference to sets
of two special types. For convenience we will define
them here.
27
Definiton 3. A set S will be said to be of
type S1 if it satisfies the following conditions 2,
and either 1,
y
There exist in Q numbers X
x + y1
such that
X
X3
X/
=x
= Y1
Y1
m=2
+x,
then yXt. 1
equivalently, if for any
thenx
> yri
that xu
xm
1
+
yui be in
x'n-1 ). Not that the requirement
implies that if u = I then
Q
max (x3, Ye) > 0. Let T = [x+yi
y
ix+yi
T4
i,x+yi
1
t
If v > 1
Ye]
x+yi E Q, xl
I
x+yi E Q,
1
= (x+yi
1
<
x+yi e Q, x4.1 < x
If u > 1 let T3= [x+yi(x+yi e Q,
(x+yi
Tues..
(x+yi
x+yi E Q, x
1
1
o
yti
with 2 < n < v such that
n
there exists xt
Ya
I>Y11,->- Y*1
Yu
such that
1
1,
Y2,and if for
yli with
u, there exists
X2
any
Yi
xiv + yo1
V
Yli,
Yui0 xi
+yi e Q,
1.10,
x+yi
Q, X<
28
u =v=1 5= 1. Ifu
Then if
S
I - (.1" U e*
T - (T1U
T . (T V 4-,
ith
then
Li T:.f),If u > 1, v 411, 1 then
v > 1, then
Tu./ )
If u
UT
u
Li Tt Li ..
Yi. Xs
such that
[a+yi j x+yi e Q0
X1 <
with Yi > Ys such that
S
X
Yli x
y
[x+yilx yi e Q, x
Y
2.
B(8)
3
in
Yi
x5)
a
There exist numbers
Yi
U I'
There exist numbers
xl
Q
v
S
C(s) >
in B n s and
If bl+ ba
but not in C then (9
in Q. That is,
and
(b
95i)
bs
ba
with strict in-
equality in at least one of these cases.
Figure 4 illustrates a set S of the type described in requirement 1 above S being those points
Q
which are on and within the solid lines. The set I
consists of all points of Q in the entire rectangle.
The reader will note that S is a set R R' where
R' C: R and R',R are sets of the type used to define
the density.
Figure 4
will be said to be o
2,
if it satisfies the following conditions
+y
in
There exist numbers x+ y
Pefinitkon A. A set
type S°
Q9
S
such that if t > I then
1,
yi
Let Sr
x+yi
1
and
+yi C Qa
X
Yrb
Y
(In other words,
.
Then
S
x
is of the same type as the set R used to define
a
and
(S) - C(S)
3.
S
but no
if
b1+ NS.
C
is in
S
then
The reader will note that a set
in
and
in Q.
S
may satisf
30
or 1",2,3 of Definition 3 and a
tions 1, 2, 3 of Definition 4. Thus S may be
conditions
ondi
both
type S° and type
We will now prove three lemmas which wi
needed for the proof of Theorem 3.1.
be a set of type
fying requirement 1 of Definition 3, and le
S
set of all
in
x+yi
°**, u.
(xi
yi)
S=S
such that
$' be the sot of all (x+yi)
is in
St =S U .00
for which x+yi
+yii
x+y
Let
S
U
Let
US .)
sati
Si bathe
Then
Si
and
(Clearly,
Q(S
I then S' s just a translation
Q(S)
S except one, and Q(S')
Assume that for some k > 1 we have Q(S9 < Q(S)
k+1 (see Figure 5)
whenever u < k. and consider u
If
of all points of
Proofs
FIGURE 5
Define sets
k+1
+yi
1
**
k+1
follows:
x+yi E So
[(x+Yi)
(x+yi)
(Yk- Yk+1)
<Y<
I "14
(xk
Then
Q(S'k)
CgSk
> Q(Sk
Q(Si USJ4.1
and
Q(W
rvyi
32
k=I
Q SI
k > 1 then
St
then
St
U
(Si
Q(11/)
Q(Slu
St
(S'.
I,J
+Q(S
Us
U
Q(S
the induction hypothesis
U SI) < Q(S1 U
Q(S1 U
Q(S)
Q(W)
Therefore,
o(s)
Q SI
U
+ Q(W) + 1
Q(S')
or
1:1(s9
Q( S)
Lemma 3.2.
C(S)
Proof:
A
is any set of type
S
then
[Q(S)) + B(S).
satisfies requIrement I of Deft
nition 3.
We know
n S is not empty.
the largest real integer for which there exists s
b°a
be
integer
such that b(i) + bi
bi°
bo
t + b°a = max (b I
if
fl
S
and
baiEB 5). (That
° + b°i,
b'1
a
b1
ba
B
+ b'i
C Bes b'fbail
a
max tb1 + bsj, then b: > ID:
b
3.
Likewise, let g:
the largest real, integer for which there exists a real
integer g such that gi g:i is in S but not in
Define the set
r > I, Q(S) c(s)
The set of all
ba
C 5, g+ gsi
gl +
g2 + g° = max (gl
and
C
Let B(S)
in Lemma 3.1.
S
9c1, and
Q(S')
A(S
i) with
(
gA.
b
b
numbers in Q (Definition S,
rement 3) which are not in A4 since suppose
in
B
fl
S
(g2 + g:i)
gives
r
asi e A.
(31 + bsi)
(s1 + Alai) + (bs,
Then
qui-
g2
bsi) EC which is a contradiction.
Since
We also show that these r numbers are in
bl+bsi is in S. there exists an Sj as defined in
i is in Si. This Implies
Lemma 3.1 such that b
j < bi, y
b2.
But
g: + g:i is also in
+ (yj + n)i
so
(go.. m)
Then
(g:
bl
j
92
Let
bt
and
bI
bs < g:,
so
+ ba
j
nce
b -n=y
g:i)
(g:
(b1
which is in
Lik
and, therefore, in
SI
with g+gi
in
a
S
1
921 1g(1.4 g21 gives
and
-m)+(
se, the possibly empty) set of
(be+
b01)
2
1
C
=I
)
gc
1
but not in
numbers which
are in St but not in A. We must show that these two
sets are disjoint. Hence, suppose that for some gl+ gsi
Pt%
b+
b
1
g:i (and, therefor
10(b°+
b i)
bi
(1i)
a
we have
gx+91i)-(b
i).
Then
b°
We add these
uslities to obtain
g:)
(b!+ b:)
i
The method of choosing
that we must have
But this,
Hence,
g2
g1
implies
a
and
ga
(b1+
b°+ bi implies
d
> g and
b°+
bo
a > b2.
d we have obtained a contra..
diction.
Therefore, the two sets are disjoin
35
Q(s)
and
Q(s)..ge
Q(S )
gA+ Q(S)
4,r0
We recall that 0(S).Q(S1).1 > 0 (Lemma 3 1), and note
that S' is a set of the type R used to define al
Hence, we may write
C(S)
A(S1)+N(S)-Q(S')-1) B(S)
o1N(S')+1) + atiA(S)-Q(S')*1)
* allQ(S))
B(S)*
Thus, the Lemma is established for any set
type Si which satisfies requirement 1. If S
fies requirement 11 we may let uml and xl+y
in the above, or if
lot uml and x
S
of
Xl+Yi
satisfies requirement 1" we may
Nisi
The proof will then be
the same, except that now CAS
Lemma 3.1 is not needed.
3.3. If
S)
$
c(s)
1
0 and
is any set of type
(S)
Prgmit( ) Suppose
Q(S )
)
then
B(S).
B(S)0. Then C(S)=A(S)
a1[Q(s) + 1) + a(s).
(ii)
Suppose
S(S)
g:+ g:i as in the proof of L
Then define
bp,
3.2. Let Q(S)4k(S)*VA.
36
B(B) = r
1,
191)
(b1 +bai)
[(914.
4(S)
scg
C(
b1
Again, the two sets
1
baiEBrIS)
-(1)14 b:i 191+ 9
i
ES, gl+
gl+
give us gc -1 + r distinct numbers not in
gc:+
A which now wtli be in S.
o4-Rns.
in the first of these
assume
not
The number
(bf+ bp.)
g:i = ..(43
nod in the
cont
second set, then
and
+ b°+
b°*
s
gi is in neither
which is impossible. There
of these two sets. However,
in
C,
S
but not
hence not in A.
Since
fine
is in
S
t of the
a
R
pE
a1 we have
A
9c
Q(S-d`gc
C(S)
Q
A(S)
A 41
B(S)
r.
used to
37
B(S)
Q(s)
e are now ready to prove Theorem 3.1.
any set of the type described in the statement of the theWe will use induction on the number of elements of
such elements being referred to
R which are not in C,
in the following as nu of C in R.
(i) Suppose there is just one gap of C in R.
orem.
and
Then we must have
Figure 6)
is in
If bl+ b2i e B rl R then (x141 i)-(
Therefore.
Q
is that gap (see
x
R
is a set of type
S0
and we
may apply Lemma 3.3.
Figure 6
(it) Suppose there are two gaps of
in
C
R.
(Consideration of this case is not necessary for
proof, but is included for greater clarity.) Then we
may have t=1, in which case x14.y1i is one of the gaps
and the othei
Figure 7a)
x fyli
and
umber
we may have
x2415i
g
gni in
R
R
(see
t=2 and the two gaps are
see Figure 7b).
38
Figure 7a
R
Figure lb
the case illustrated in Figure 7a
and let W
B(W) = 0
is of type
W
type
Sl
and
S
(W)
0
V
is of type
Se
hen
V=
+yilx+yi 64,
then
set of
Hence,
C(V) + C(W
COO
al[Q(v) + I
B(V) 4'
allQ(R)
B(R).
1
ILQ(W)]
illustrated in Figure 7b let
(x+yilx+yi C Q, x <x2, y< 1/1)
In the
[x+yilx+yi EQ, xs< x
x1, y< Yi)
fx+yilx+yi CQ, x <x2, yl< y
If
B(W
B(W5)
0
where
is of type
2.
= 0 thenR is
J=2
Se
or
where
then
ype S.
type St
and
9
Assume the theorem i true for any set
the proper type in which there are less than k gaps
i
a set of the
for an integer k > 3
proper type in which there are k gaps of Co of which
xt yt ,i< t < k
are the points x
of
C
If
B(R)
0
then
R
is a
we are done. Hence. assume B(R) >
bi
Figure 8b
Figure
Let
b2
there exists a
be the largest real integer for which
integer
such that x+bai E B
and let bi be the largest such
Ro
The set
Ex+yilx+yiERxb
then contains precisely one element of
fl
y > b5)
B.
Also,
b+bi
40
is
RU
of the In
oduc
such that
1
(for notation, see Definition I
UR
on); hence, there exists an integer
and be biki is in
implies that x54y5i is in
Case 1
or all
b
then
(R
I
e done.
and we
b*+ b*i
in
*
ith
nR
B
R
k
is a set of type
b+
ba
1
b* < ba
gi be the smallest real integer for which there
Let
exists a real integera such that
of
1.
then there exist numbers
b*i
*
b
Q(W)-C(W)
gaps of C
This implies that bl
k,
Q(W)C(W)
Figure 8a)
and
that is,
W,
This
Rs.
C
in
If
W.
gsi
(31
is a gap
for all. b*+
b*i
in
2
1
b*1 S g 1
B (*II R
is
set of type S° and we are done.
b: be the smallest real integer
Otherwise,
for which there exists an integer b; such that
then again
R
b'+ bi ER rl B
bl >
and
fx+yi
is of type
91+
,
I
(Note that
i, so B(W) > 1
Q(Wi
Then the set
x yi E R. x
W.
co tai
(w4) >
There is at least one gap of
xt+YI
Let
g
C
in
namely
be the smallest real integer for which
41
there x
R -W1.
such thai
If we have
g:i
for all b
a
gl
b
b:i in
is of type S. If not, let
then
Br(R
be the
b
smallest real integer for which there exists 4 real
ger bsa such that bw + bi E B fl (R-W
and b"
1)
I
S
Then the set
= [x yi
of type
x yi E'.W4,
(Note that
W5
UW
U
of type Si.
c( w)
and
X
such that
C(
)
+ 11 +
al[Q(
Case 2.
C
BW)
N(R)+1] B(
w)
C(W) <k
(
Figure
)
least one but no more than k-1
conta
Let
)]
h'
be the largest real integer for
which there exists a real integer
is a gap of
b'i
of type Si Ws,
n'
1[Q(
gaps of C.
bl
Then
C(R)
R-W
contains
disjoint sets
Constructed
Then
bflj
i
may continue this process until we him
)
cq,
1
in RW, and let
x
hi,
such that
x+
be the largest
42
hi =
x. (Not. that
such
be the largast real in
h
a gap of C in
Continue this proce
gaps of
C
(If
m,
=
is
be the largest such x*
h
in RW, h
hl
hm+ 11;i
and let
andh <
> h1
i
yit then
x yi I x+y
RJ
x
as long as possible. We will then
such that
1
such that
h
and let
R-W
Let
bt>
f one exists,
and there exists
<h
/111
if
Yti
yli
y < hp,
h
R
U 111:1.
Since
of the type described in the statement of Theoin which the number of gaps of C is less than
rem
have
(R') >
by the induction hypothesis.
.(W U Ri).
Let
the
h+
h1+ h'
gaps of
C
in V.
ready a set
into
n
Because of the way in which
were chosen, there can be no
if any exis
Elements
B flV will have b< b
of typo
disjoint subsets
R')
[Q(R') + 1
Hence
either
WUV
is al
or we may partition WUV
oh of typeSI,
43
n precisely the same manner as was done for the set
in Case 1
Th
gives us
C(R)
C(R')
c( w)
C(
%MR') 1)+B(R')
IN(
1[Q(
))+B(W
)) + B(Wn
1]
Ql[Q(R)
R
B(R
which completes the proof.
The hypothesis of Theorem
that the numbers
xt+y i be gaps of C
analogous to the
requirement in Mann's theorem that
n
be a gap of Co
We can now see that this is somewhat more than is neces-
sary, however; in constructing the disjoint subsets
R
in Case 1
i
we had a point
C
in
B
a point
b(3711
4)1.of
v(n-s.)(n-1)
i
Let
S
and a gap
a point bl+ bli
and a gap girt
that bln-1)
of
and of W U V in Case 2
B
(110.1
i
in
W,
gi+ gai of
and a gap
(n.
2
of
point
C
(wit
91
b(n
could not be included in
be the smallest subscript such that
44
W,
W
1 <$St
From the method of construction of
is in
we know that
x>
s b "1)
Wn
and
Thus the requirement that
Yti
x + y i be gap* of
is then sufficient to in-
C
sure obtaining a last
of type
would be
require that
enough farourpurposes, howev
R
be
so chosen that for any point b1+ b5i in B ri R there
exist a gap
b
g
in
R
of
g
in R such that
C
and that there
at least one gap of C
even if B(R) = 0.
This last requirement is ne-
cessary, since if B(R) = 0 then
cannot conclude that AU
C(R)
A(R)
and we
a1[Q(R) + 1) unle
A(R) < Q(R).)
Corollarx 3.1,. Define
for the set B in the
same manner as a wes defined or the set A. Let y: =
1
gib
taken over all R satisfying the hypotheses
of The
I
Proof,:
Then
For any set
R
of Theorem 3.1 we have
C(R)
[Q(R)+1]
satisfying the hypotheses
and
T heref o
Corollary 3.2.
be the density of
Let
(Definition 1, Introduction) and
taken over all
3.1.
Te
B
9
satisfying the hypotheses of Theorem
R
Then
Proof. We have
C(R)
IN(R) + 1] + B(R)
Q(R)1+ B(R),
for any set
R
satisfying the hypotheses of Theorem
Therefore,
Yi
The following example shows that
a+p
yl
may equal
and thus that the conclusion of Corollary 3.1 is
in a sense, the best possible.
46
Example 1.
Let A
l, 2,
B
in Q with x>5 or y>2
Then C asti, 2.
1+i, 2+1., 3+i, all x+yi in Q with x>5 or
4/5,
and we have
2/5,
P
41
.
The proof of Theorem 3.1 given here nvolved an
induction on the number of Gaussian integers in the set
R
which were not in the sum set C the induction hypothesis being applied only to a subset R' of R which
was known to contain fewer of these gaps of C than
did R. Mann's proof of the corresponding theorem for
sets of positive integers, C(n)
1(n+1) + B(n) for
not in C, is carried out by induction on the number
of gaps of C in the interval I 74 (x 1 < x < n) How
ever, his proof involves the application of the induction
hypothesis in two different ways; in one case it is
applied to a smaller interval which is known to contain
one less gap of C than I, and in the other case it is
applied to a new
C
which contains C as a proper
n
I
subset.
Mann's approach yields a quite simple and elegant
proof for the theorem in the one-dimensional case, and it
is only natural to attempt to apply similar methods to
the sets of Gaussian integers. Difficulties are soon encountered, however. It appears that the extension to the
47
C
Gaussian integers of Mann's method of transforming
the new set
x
C
y i of C
gl+
such that for any other gap
R wehave
in
consideration of sets
R
statement of Theorem 3.1
ourselves to sets
R
g5i
of
This precludes
yo
gi
of a gap
R
requires the existence in
CI
of the type described in the
t>1.
for which
If we restrict
for which tI we find that the
case in which Mann transformed
C
CI
to the new set
ion now yields to almost precisely the same
one dime
treatment, but the other case presents difficulties which
seem to the author to be insurmountable since apparently
use of the induction hypothesis will require
of sets
R'
for which / may be greaterconsiderstien
than 1.
Attempts to find other ways of transforming the
C
set
to a new set
Ci
to which the induction hypotheAmong these
sis could be applied were also unsuccessful,
attempts was one which led to a new proof of the theorem
for sets of positive integers
the second of the two
A trans-
proofs given in the Appendix to this thesis.
formation of the type described therein
can be success
fully carried out for sets of Gaussian integers whenever
the set
so
1
R
< s
such that
has been so chosen so that for some subscript
to
there exists a number
(or
b1
ba
in
)
and
48
and such that
Ys
of C not equal to
x
t is not necess
strict ourselves to sets a
in order to
use this transformation; however, the case in which the
in which t=1
transformation cannot be used has not yet been resolved.
There are, of course, other approaches to this
theorem. Artounting process" devised by Besicovitch (1,
Pe 246-248) has been used in several 'says to prove the
theorem for the one-dimensional case (9 pib 20..29) and
may perhaps be extendible to the Gaussian integers.
Attempts to apply a transformation developed by Dyson
(4, p. 8-14) to the theorem for the one-dimensional case
have so far not been assful
p. 30-37 but this
transformation cannot be completely dismissed as a pos-
sibility.
49
BIBLIOGRAPHY
On the density of the sum of two
sicovitch, A. S
sequences of integers. Journal of the London
Mathematical Society 10:246-248. 1935.
Chao, Luther P.
On the density of sets of Gaussian
Mathematical Monthly 58:
integers. American
618-620.
1951.
The density of the sum of sets of
Gaussian integers. Ph.D. thesis. Eugene,
University of Oregon 1951. 35 numb. leaves.
4.
Dyson, F. J. A theorem on the densities of sets
integers. Journal of the London Mathematical
Society 20:8-14. 1945.
Erdes, Paul. On the asymptotic density of the sum
of two sequences. Annals of Mathematics 43:
65-68. 1942.
Erd(4, Paul and Ivan N von. The a+P hypothesis
and related problems. American Mathematical
Monthly 53:314-317. 1946.
7.
Khi chin, A, Y. Three pearls of number theory.
Ti'. from 2d Russian ed. Rochester Greylock
Press, 1952. 64 ps
Landau, Edmund. Uber einige neuere Fortschritte der
additiven Zahlentheorie. London, Cambridge
University Press, 1937. 94 p. (Cambridge Tracts
in Mathematics and Mathematical Physics, No. 35
Lim, Yearn Seng.
An inequality for the number of integers in the sum of two sets of integers.
Master's thesis. Corvallis, Oregon State C 1
lege, 1962. 44 numb. leaves.
10.
Mann, Henry B
A proof of the fundamental theorem on
the density of sums of sets of positive integers.
Annals of Mathematics 43:523-527. 1942.
50
On the number of integers in the sum
Journal of Mathematics 1:249-253. 1951.
A refinement of the fundamental theorem on the density of the sum of two sets of
integers. Pacific Journal of Mathematics 0:
of two sets of positive integers. Pacific
909-915.
1960.
Niv n, Ivan. The asymptotic density of sequences.
Bulletin of the American Mathematical Society
57:420-434.
14,
1951.
Schexk, Peter. An inequality for sets of integer
Pacific Journal of Mathematics 5:585-587.
1955.
15.
Schn relmann, L. jber additive Eigen chaften von
Zahlen. Ma hematische Annalen 107:649-690.
1933.
APPENDIX
TWO NEW PROOFS OF A THEOREM
OF MANN
The investigations which led to the proof o
Theorem 3.1 produced two new proofs of the corresponding
theorem for sets of positive integers. The first of
these is just a specialization of the proof presented
in Chapter
however, this specialization is so much
simpler that it is of interest to have it presented separately. The second proof involves the use of a new
transfor tion on the set B. We restate thetheorem:
Let A
Theorem A.1.
tive integers, and
sity of A. Then
and
MB.
C
B
be two sets of posi
be the Erdos den-
Let
(n+l) +
C(
for any positive integer n not in C.
will apply our first new method of proof
to Theorem A.2 below, which is the theorem actually proved
by Mann, and then show that Theorem A.2 implies Theorem
Al.
Theorem A.2.
Let
negative integers with
and
A
0
in
Ay
B
be two sets of nonin
B.
Let
C
be
52
the set of all naubers of
A,
is in
b
Let
B.
a1
form a+b where
h
is in
a
be the Erd6s density of
A.
Then
for any positive integer n not in
(Note: In the
definition
A(n) is still the number of positive
integers in A which are less than or equal to n. Also,
B(n) and C(n) are interpreted similarly.)
Proof:
Let
ger* and for any se
be the set of all positive nteS of positive integers let A(S)
denote the number of integer in A (1 5. Choose any positive integer no which is not in C.
1
b
be all the integers in
< bs
Let
r
be the maximum
j such that
(
)
I( J)
b1. Then there exists a gap
n of C such that br < n < b
(If B is a finite
set of r elements we have only br<
t
be the largest gap of C such that
where S
b<
r
n1
(If
br+1
let
St
<
o
br+1
let
co
whichever is
does not exist, then n1* no
1
x
e I, bx b,4.l}
If
no
fx
<
f
egers
The
>
in S.
A
gaps of
let
gives
These are each less than
A
the other
hence dist nc
gA
egers n-br
g p of C in
n
ps of
gc
1 number o
then the set of
C
such that
be the
gc
.
Letting
be the total number of gaps of A
thus sho
(1)
1.
k
then
k
is the smallest pos&tive integer not in
since nt -1 is not in A. Therefore,
k
s in
Hence
from (
have
I(S)
g
This is equivalent to
C(S1)
A(S1
B(S
LI(S;) +
= al[I(S )3
If
)
]+
Si
B(S
we are done.
a set Sas we did
Si,
using
can
instead of
b1
and
54
instead of br,t being the maximum
c(sh)
I(Sh) where
1
bt
such that
h
let
br+1
I
IxES 2 x > br i.
n the same
Proceeding
manner as before and
k<
r+i where n
noting that kES'2 since
is the gap of C in S2
2
corresponds to ni
we obtain
in
ye at
Eventually we mus
end point is
no
which
set
whose right
Then
*
We must now show that Theorem A.2 implies Theorem
at A and
A.1, which has been shown before 9, P. 9).
B
be two sets of positive integers, C is A + B,
A U CO)
B*
B U [Oh
set of all a b
Let
B
C*
where
a
be the set of
1
C U (0).
is in
A'
b 1 with
Then CI
and
b
b
is the
is in Bl.
in B'
and
the set of all a+b+1 with a in
If n is a gap in CI (therefore a gap in
A'
b+1
C),
C1
in B1.
then n+i
55
I
I
is a gap
in,. The sets
A',
satisfy the hypo*
Bs
theses of Theorem A.2, and so we must have
Ci(n+1)
al
where
A.
But
(n+1) + B(n)
is the Erdgs density
A'
or, equivalen
(n 1) vs C(n) + 1 and
(n
)
Y,
1.
=B
Therefore
n)
n+1)
C(
h implies Theorem A.I.
2. We will apply our second new method of proof
to Theorem A.1 directly. Let A and B be any two sets
of positive integers, C A + B. Let ki be the smallest positive integer not in A.
is a positive integer not in C such
n-I. Since A is a subset of C we must
Suppose
that C(n)
have
n
kl<
any
then either
si ve in
is in
B,
C(n)
n,.4
such that IS x< n
is of the form n-a
for some a in A and less than n or neither, but
not both. The integer n is neither in B nor of the
form n-a. Therefore,
x
or
x
A(n)
(M ) B(n)
Thus the theorem is established for the
a
gal)
of
C
and we proceed by induction on the number of
gaps of C which are less than or equal to a given gap.
To be more precise, we assume that if C(n) >
for some integer
2 and n not in Co then
C(n)a (n+1)
be the first r gaps o
Let n/
so that C(n )
i)
Suppose
We distinguish two cases*
nr. n
C(nr
(1
+1) 4 B(n
by the induction hypothesis. Also,
for any integer x then eit
of the form nra for some
T
in
is in 8, or
A
We have
x
such that
or neither, but not both.
r
neither.
x
Aga
1 > k , so w
write
(2)
>
Adding
(nr
r)
he inequalities (1) and (2) gives the desired
result
Ign
57
ii)
may assume that
r nr*
the largest set such that A + B = C in the interval from 1 to n. or, in other words, that B consists
Suppose
n
of all integers in this interval except those of the set
aeAU[0), a
1 ,S
then it will also be established for any set B' such
that A 4 B = C. since necessaril
is a subset
and B(n) B4
Let k1< k2<
be all the gaps o
k
All th gaps of C are also
which are less tha
gaps of
A,
t
so
1.
Ei(nr) a 0 then C
1 to
and C(
n
B(n
th.re ex
If
y
is in A
)
(
(
is in
x
)
positive integer
y
plies that the elements of
of the set [
k
j
1
y
+ I) 4, B(
B,
Hence,
1<x<n
then
such that x + y =
is in C
then x + y
not in C. Therefore,
exists j such that
in the interval from
A
However,
n
is a gap of A and there
and
B
y = kj.
This JAI..
will be found among those
tj The largest element
of this set is
1 call
< ki, so
are assuming
n1
r- then
xl
s not in
x
Then there exists
If
xx* k 1.
were in
x4
B
t
implies
is in
implies that
1
h <j.
C. Cons,'
and the assumption B(nr)1
Hence, there can be at most j
greater than
This implies
x
and we must have
ro.
know that each of the integers
is not in
let C
B.
A 4. B
For each
in 5,
A.
some i<r, then there exists
and k *
ISn Xf k
such that
C
B,
yl
less than n .)
largest element o
gaps of
iheref ore,
and
2, Let j be the smallest subscript such
k
in B (That is, xj is the
that
kh
Suppose
we would ha
not in
quently,
B.
such that
Yi
ri
yi
We
1 <h<
and hence there ex
have
u h that
nr.j
59
i <r,
a 2g0 or aE A, such thai
This implies ni >
so n.1
Since
We also know
can only be one of the gapsnrso,u.
n - a is in
in C
0
have added at least one eleme
Hence,
but not mor than
elements,
Thus we have
BI(n
and
(3)
C(
)
C
r
B(nr
By the induction hypothesis,
(4)
Adding
Ct
and (4), we obtain
C(nr
or aEA, we have
1) + B(n
in forming