Non-archimedean analysis on the extendedhyperreal line d and some transcendenceconjectures over field .

Jaykov Foukzonjaykovfoukzon@list.ru

Center for Mathematical Sciences, Israel Institute of Technology,Haifa,Israel

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Abstract.

In this paper the importent application of the Dedekind completiond of theRobinson non-archimedean field , constructed by Dedekind sections [6,7] isconsidered. Given any analytic function of one complex variable f z, weinvestigate the arithmetic nature of the values of fz at transcendental pointsen,n . Main result are the both numbers e and e are irrational.

Keywords: Non-archimedean analysis,Robinson non-Archimedian field,Dedecind completion, Dedecind

hyperreals,Wattenberg embeding,Gosnor idempotent theory.

AMS Subject Classification:

1.Introduction.In 1873 French mathematician, Charles Hermite, proved that e is

transcendental. Coming as it did 100 years after Euler had established thesignificance of e, this meant that the issue of transcendence was onemathematicians could not afford to ignore.Within 10 years of Hermitesbreakthrough,his techniques had been extended by Lindemann and used to add to the list of known transcendental numbers. Mathematician then tried to prove thatother numbers such as e and e are transcendental too,but these questions

were too difficult and so no further examples emerged till todays time. Thetranscendence of e has been proved in1929 by A.O.Gelfond.

Conjecture 1. The both numbers e and e are irrational?Conjecture 2. The numbers e and are algebraically independent?However, the same question with e and has been answered:Theorem.(Nesterenko, 1996 [1]) The numbers e and are algebraicallyindependent.During of XX th century,a typical question: is whether f is a transcendental

number for each algebraic number has been investigated and answered manyauthors.Modern result in the case of entire functions satisfying a linear differentialequation provides the strongest results, related with Siegels E-functions [1],[2],ref[1] contains references to the subject before 1998, including Siegel E and Gfunctions.

Theorem.(Siegel C.L.) Suppose that , 1,2, . . . , 0.

z n0 zn

1 2 n . 1.1

Then is a transcendental number for each algebraic number 0.Let f be an analytic function of one complex variable f z.Conjecture 3.Is whether f is a irrational number for giventranscendental number .Conjecture 4.Is whether f is a transcendental number for giventranscendental number .In this paper we investigate the arithmetic nature of the values of fz at

transcendentalpoints en,n .Definition 1.1. Let gx : be any real analytic function such that

gx n0

anxn, |x| r,nan . 1.2

We will call any function given by Eq.(1.2) -analytic function and denoted bygx.

Definition 1.2.[3],[4]. Arbitrary transcendental number z is called#-transcendental number over field , if there no exist -analytic functiongx such that gz 0, i.e. for every -analytic function gx theinequality gz 0 is satisfies.Definition 1.3.[3],[4].Arbitrary transcendental number z called w-transcendentalnumber over field ,if z is not #-transcendental number over field ,i.e.

exist -analytic function gx such that gz 0.Example. Number is transcendental but number is not#-transcendental number over field as(1) function sinx is a -analytic and(2) sin 2 1, i.e.

1 2 3

233! 5255!

7277! . . .

12n12n122n12n 1! . . . 0. 1.3

Main result are.Theorem 1.1.[3],[4].Number e is #-transcendental over field .From theorem 1.1 immediately follows.Theorem 1.2.[3],[4]. The both numbers e and e are irrational.

2. Preliminaries.Short outline of Dedekindhyperreals and Gonshor idempotent theory

Let be the set of real numbers and a nonstandard model of [5]. is notDedekind complete.For example, 0 and are bounded subsets of whichhave no suprema or infima in .Possible completion of the field can beconstructed by Dedekind sections [6],[7]. In [6] Wattenberg constructed theDedekind completion of a nonstandard model of the real numbers and applied theconstruction to obtain certain kinds of special measures on the set of integers.Thus was established that the Dedekind completion d of the field is astructure of interest not for its own sake only and we establish further importentapplications here. Importent concept was introduce Gonshor [7] is that of theabsorption number of an element a d which, roughly speaking, measures thedegree to which the cancellation law a b a c b c fails for a.

2.1 The Dedekind hyperreals dDefinition 2.1. A Dedekind hyperreal d is ordered pairU,V P P satisfied the next conditions:1.xyx U y V.2. U V .3.xx U yy V x y.

4. xx V yy V y x.5. xyx y x U y V.Designation 2.1. Let U,V d. We designate in this paper

U cut,V cut

cut,cutDesignation 2.2. Let .We designate in this paper

# cut,# cut

#,#Remark 2.1. The monad of it is the set: x | x is denoted by

.Supremum of 0 is denoted by d. Supremum of is denoted by d.Note

that [6]

d , 0 0,

d n

,n.

Let A be a subset of is bounded above then supA exists in d [6].Example 2.1. (i) d sup d\, (ii) d sup 0 d\.Remark 2.2. Anfortunately the set d inherits some but by no means allof the algebraic structure on .For example,d is not a group withrespect to addition since if x d y denotes the addition in d then:d d d d d 0 d d.Thus d is not iven a ring but pseudo-ring only.Definition 2.2We define:1.The additive identity (zero cut) 0 d , often denoted by 0# or simply 0 is0 d x | x 0 .2.The multiplicative identity 1 d , often denoted by 1# or simply 1 is1 d x | x 1 .Given two Dedekind hyperreal numbers d and d we define:3. Addition d of and often denoted by is x y| x ,y .It is easy to see that d 0 d for all d.It is easy to see that d is again a cut in and d d .

Another fundamental property of cut addition is associativity: d d d d .This follows from the corresponding property of .4.The opposite d of , often denoted by # or simply by , is x | x ,x is not the least element of \5.We say that the cut is positive if 0# or negative if 0#.The absolute value of ,denoted ||, is || , if 0 and || , if 06.If , 0 then multiplication d of and often denoted is z | z x y for some x ,y with x,y 0 .In general, 0 if 0 or 0, || || if 0, 0 or 0, 0, || || if 0, 0,or 0, 0.7. The cut order enjois on d the standard additional properties of:(i) transitivity: .(ii) trichotomy: eizer , or but only one of the three(iii) translation: d d .

2.2 The Wattenberg embeding into dDefinition 2.3.[6]. Wattenberg hyperreal or #-hyperreal is a nonepty subset

such that:(i) For every a and b a, b .(ii) , .(iii) has no greatest element.

Definition 2.4.[6].In paper [6] Wattenberg embed into d by following way:if the corresponding element, #, of d is

# x x 2.1Remark 2.3.[6]. In paper [6] Wattenberg note that: condition (iii) above is

included only to avoid nonuniqueness. Without it # would be represented by both#and # .

Remark 2.4.[7]. However in paper [7] H. Gonshor pointed out that the definition(2.1) in Wattenberg paper [6] is technically incorrect. Note that Wattenberg [6]defines in general by

a a . 2.2If d i.e. d\ has no mininum, then there is no any problem with definitions(2.1) and (2.2). However if a# for some a , i.e. # x x a thenaccording to the latter definition (2.2)

# x x a 2.3whereas the definition of d requires that:

# x x a , 2.4but this is a contradiction.

Remark 2.5.Note that in the usual treatment of Dedekind cuts for the ordinaryreal numbers both of the latter sets are regarded as equivalent so that no seriousproblem arises [7].

Remark 2.6.H.Gonshor [7] defines # by# x bb a b a , 2.5

Definition 2.5. (Wattenberg embeding) We embed into d of the followingway: (i) if , the corresponding element # of d is

# x |x 2.6

and# a a . 2.7

or in the equivalent way,i.e. if the corresponding element # of d is

# x |x 2.8

Thus if then # A|B where

A x |x ,B y |y . 2.9

Such embeding into d Such embeding we will name Wattenberg embedingand to designate by # d

Lemma 2.1.[6].(i) Addition d is commutative and associative ind.(ii) d : d 0 d .(iii) , : # d # #.Remark 2.7. Notice, here again something is lost going from to d since

a doesnot imply since 0 d but 0 d d d d.Lemma 2.2.[6].

(i) d a linear ordering on d often denoted ,which extends the usualordering on

.(ii) d d d d d .(iii) d d d d d .(iv) is dense in d.That is if d in d there is an a then

d a# d .(v) Suppose that A d is bounded above then supA

Asup A cut

exist in d.(vi) Suppose that A d is bounded below then infA

Ainf A cut

exist in d.Remark 2.8.Note that in general case infA

Ainf

Acut. In particular

the formula for infA given in [6] on the top of page 229 is not quite correct [7], seeExample 2.2. However by Lemma 2.2 (vi) this is no problem.

Example 2.2.[7].The formula infA Ainf

Acut says

Ainf a dd 0 a d

Acut

Let A be the set A a d where d runs through the set of all positive numbers in, then infA a x|x a. However

Acut x|x a.

Lemma 2.3.[6].(i) If then d # #.(ii) dd .(iii) d d d d .(iv) d d d d d d .(v) a : a# d d d a# d .(vi) d d d 0 d .Proof.(v) By (iv): a# a# .(1) Suppose now c a# this means(2) c1c c1 a# and therefore(3) c1 a# .(4) Note that: c a (since c a and a c c1 a# implyc1 a c c1 c a a# but this is a contradiction)(5) Thus c a and therefore c a .(6) By similar reasoning one obtain: c1 a .(7) Note that: a c1 c a# and therefore

c a c1 c c1 a a# .

Lemma 2.4.(i) a , d, , 0 : a# # #a# ,(ii) a , d, , 0 : a# # #a# .Proof.(i) For 0 the statement are clear. Suppose now without loss of

generality 0. By Lemma 2.3.(iv): a# # #a# #.(1) Suppose c #a# and therefore c a# , but this means(2) c1 c c1 a# and therefore(3) c1 a# .(4) Note that: c a (since c a and a c c1 a# imply c1 a c c1 c a a# but this is a contradiction)(5) Thus c a and therefore c a #.(6) By similar reasoning one obtain: c1 a #.(7) Note that: a c1 c #a# and thereforec a c1 c c1 a a# #.(ii) Immediately follows from (i) by Lemma 2.3.Definition 2.6.Suppose d. The absolute value of written ||is defined as follows:

|| if d 0 dd if d 0 d

Definition 2.7.Suppose , d.The product d , is definedas follows: Case (1) , d 0 d :

d a b|0 d d a# d 0 d d b# d , 0#.

2.10

Case (2) d 0 d d 0 d : d 0 d .Case (3) d 0 d d 0 d d 0 d d 0 d

d || d || iff d 0 d d 0 d ,

d d || d || iff d 0 d d 0 d .2.11

Lemma 2.5.[6]. (i) a,b : a b# a# d b#.(ii) Multiplication is associative and commutative:

d d d d , d d . 2.12

(iii) 1 d d ; 1 d d d , where 1 d 1 #.(iv) || d || || d ||.(v)

0 0 0 d d d d d . 2.13

(vi)

0 d d d , 0 d d d d d d . 2.14

Lemma 2.6.Suppose and , d. Then

# 0 0 # d d # d d # d . 2.15

Proof. We choose now: (1) a such that: a# 0.(2) Note that # # #a# #a#.Then from (2) by Lemma 2.4.(ii) one obtain(3) # # a# #a#.Therefore(4) # # a# #a#.(5) Then from (4) by Lemma 2.5.(v) one obtain(6) # # # a# #a#.Then from (6) by Lemma 2.4.(ii) one obtain(7) # # # a# # #a# # #.

Definition 2.8. Suppose d, 0 d then 1d isdefined as follows:(i) 0 d d : 1d infa1 |a ,(ii) d 0 : 1d d d 1d .

Lemma 2.7.[6].(i) a : a#1d w a1 #.(ii) 1 1 .(iii) 0 d d d 1d d 1d .(iv) 0 d d 0 d d

1d d 1d d d 1d(v) a : a 0 a#1d d 1d a# d 1d .(vi) d 1d d 1 d .Lemma 2.8.[6]. Suppose that a ,a 0,, d, 0, 0.Thena# d d a# d d a# d .Theorem 2.1.Suppose that S is a non-empty subset of d which isbounded from above,i.e. supS exist and suppose that , 0.Then

xSsup # x #

xSsup x # supS. 2.16

Proof.Let B supS.Then B is the smallest number such that, for any x S,x B.Let T # x|x S.Since # 0,# x # B for any x S.Hence T isbounded above by # B.Hence T has a supremum CT s-supT. Now we have topruve that CT # B # supS.Since # B # supS is an apperbound for Tand C is the smollest apper bound for T,CT # B.Now we repeat theargument above with the roles of S and T reversed. We know that CT is thesmallest number such that, for any y T,y CT.Since # 0 it follows that#1 y #1 CT for any y T.But S #1 y|y T .Hence #1 CTis an apper bound for S.But B is a supremum for S.Hence B #1 CT and# B CT.We have shown that CT # B and alsothat # B CT.Thus# B CT.

2.3 Absorption numbers in d.One of standard ways of defining the completion of involves restricting

oneself to subsets a which have the following property 0xx yy y x .It is well known that in this case we obtain a field. In fact the proof is essentially thesame as the one used in the case of ordinary Dedekind cuts in the development ofthe standard real numbers, d,of course, does not have the above propertybecause no infinitesimal works.This suggests the introduction of the concept ofabsorption part ab.p. of a number for an element of d which, roughlyspeaking, measures how much a departs from having the above property [7].

Definition 2.9.[7]. Suppose d, then

ab.p. d 0|xxx d . 2.17

Example 2.5.(i) : ab.p. # 0,(ii) ab.p. d d,(iii) ab.p. d d,(iv) : ab.p. # d d,(v) : ab.p. # d d.Lemma 2.9.[7].(i) c ab.p. and 0 d c d ab.p. (ii) c ab.p. and d ab.p. c d ab.p. .Remark 2.9. By Lemma 2.7 ab.p. may be regarded as anelement of d by adding on all negative elements of d to ab.p. .Of course if the condition d 0 in the definition of ab.p. is deleted weautomatically get all the negative elements to be in ab.p. sincex y x .The reason for our definition is that the real interest liesin the non-negative numbers. A technicality occurs if ab.p. 0. Wethen identify ab.p. with 0. [ab.p. becomes x|x 0 which by ourearly convention is not in d].Remark 2.10. By Lemma 2.7(ii), ab.p. is additive idempotent.Lemma 2.10.[7].(i) ab.p. is the maximum element d such that .(ii) ab.p. for 0.(iii) If is positive and idempotent then ab.p. .Lemma 2.11.[7]. Let d satsify 0. Then the following areequivalent. In what follows assume a,b 0.(i) is idempotent,(ii) a,b a b ,(iii) a 2a ,(iv) nna n a ,(v) a r a , for all finite r .Theorem 2.2.[7]. ab.p. .Theorem 2.3.[7]. ab.p. ab.p. .Theorem 2.4.[7].(i) ab.p. .(ii) ab.p. ab.p. .Theorem 2.5.[7].Suppose , d, then(i) ab.p. ab.p. ,(ii) ab.p. maxab.p. ,ab.p. Theorem 2.6.[7]. Assume 0. If absorbs then absorbs .

Theorem 2.7.[7]. Let 0 d. Then the following are equivalent(i) is an idempotent,(ii) ,(iii) .Theorem 2.8.[7].(i) Let 1 and 2 be two positive idempotents such that 2 1.Then 2 1 2.

2.4 The -part of absorption numberDefinition 2.10.[4].Suppose ab.p. d i.e. has type 1, i.e. a# d,a .Then ,where 0, , is the -part of iff:

y a# y a# y y # d . 2.18

Theorem 2.9.[4]. 0 d # d .Proof. Immediately follows from definitions.

2.5 The pseudo-ring of Wattenberg hyperintegers dLemma 2.12. [6].Suppose that d.Then the following two conditions on

are equivalent:(i) sup #| ,(ii) inf #| .

Definition 2.11.[6].If satisfies the conditions mentioned above is said to bethe Wattenberg hyperinteger. The set of all Wattenberg hyperintegers is denotedby d.

The set of all positive Wattenberg hyperintegers is called the Wattenberghypernaturals and is denoted by d.

Definition 2.12.Suppose that (i) , d, (ii) #, # and (iii) |.If d and d satisfies these conditions is said is divisible by and

that is denoted by#|#.Definition 2.13.Suppose that (i) d and (ii) there exist # d such that(1) sup #| | or(2) inf #| | .If satisfies the conditions mentioned above is said is divisible by # and that

is denoted by#|.Theorem 2.10. Let p , Mp, be a prime hypernaturals such that (i)

p Mp. Let d be a Wattenberg hypernatural such that (i) p|. Then

|Mp | 1. 2.19Proof. Immediately follows from definitions (2.12)-(2.13).

2.6 External summ of the countable infinite series ind

Definition 2.14.[4]. Let be snn1 an countable sequence sn : .such that(i) nsn 0 or (ii) nsn 0 or(iii) snn1 sn1n11 sn2n22 ,n1 n1 1 sn1 0,n2 n2 2 sn2 0, 1 2.Then external sum (#-sum) of the countable sequence sn : is defined by

i nsn 0 :

#Ext-nsn#

ksup

nksn# ,

ii nsn 0 :

#Ext-nsn#

kinf

nksn#

k sup

nksn# ,

iii n1n1 1sn1 0,

n2n2 2sn2 0, 1 2 :

#Ext-nsn# #Ext-

n11

sn1# #Ext- n22

sn2# .

2.20

Theorem 2.11.(i) Let be snn1 an countable sequence sn : such thatnn sn 0, n1 sn and infinite seriesn1 sn absolutely convergesto in .Then

#Ext-nsn#

ksup

nksn# # d d, 2.21

(ii) Let be snn1 countable sequence sn : such that nn sn 0,n1 sn and infinite seriesn1 sn absolutely converges to in .Then

#Ext-nsn#

kinf

nksn# # d d, 2.22

(iii) Let be snn1 countable sequence sn : such that (1)snn1 sn1n11 sn2n22 ,n1 n1 1 sn1 0,n2 n2 2 sn2 0, 1 2 and (2)

n11

sn1 1 , n22

sn2 2 . Then

#Ext-nsn#

#Ext- n11

sn1# #Ext- n22

sn2# 1# 2# d d.2.23

Theorem 2.12.Let be ann1 countable sequence an : d such thatnan 0 and infinite seriesn1 sn absolutely converges in .Let #Ext-

nan# be

external sum of the countable sequence ann1 denoted by s.Let be bnn1countable sequence where bn amn any rearrangement of terms of the sequenceann1 . Then external sum #Ext-

nbn of the countable sequence bnm1 has

the same value s as external sum of the countable sequence an, i.e. s d.

Theorem 2.13.(i) Let be ann1 countable sequence an : d,such thatnan 0 and let #Ext-

nan# be external sum of the sequence an#n1 . Then for

any c the equality is satisfied

c# #Ext-nan #Ext-

nc# an 2.24

(ii) Let be ann1 countable sequence an : d, such that nan 0 andlet

#Ext-nan# be external sum of the sequence an#n1 . Then for any c the

equality is satisfied:

c# #Ext-nan #Ext-

nc# an 2.25

(iii) Let be snn1 countable sequence sn : d such that

snn1 sn1n11 sn2n22 ,n1 n1 1 sn1 0,n2 n2 2 sn2 0, 1 2.Then the equality is satisfied:

c# #Ext-nsn# c# #Ext-

n11

sn1# c# #Ext- n22

sn2# . 2.26

3. The proof of the #-transcendene of the numbersek,k .

In this section we will be proved of the #-transcendene of the numbersek,k .Let M0n,p,Mkn,p and kn,p be the Shidlovsky quantities [8]:

M0n,p 0

xp1x 1. . . x npex

p 1! dx 0, 3.1

Mkn,p ek k

xp1x 1. . . x npex

p 1! dx,k 1,2, . . . 3.2

kn,p ek 0

kxp1x 1. . . x np ex

p 1! dx,k 1,2, . . . 3.3

where p this is any prime number.Using Eqs.(3.1)-(3.3.) by simple calculationone obtain:

Mkn,p kn,p ekM0n,p 0,k 1,2, . . . . 3.4

and consequently

ek Mkn,p kn,pM0n,p

k 1,2, . . .3.5

Lemma 3.1.[8]. Let p be a prime number. ThenM0n,p 1nn!p p1,1 .

Proof. ([8], p.128) By simple calculation one obtain the equality

xp1x 1. . . x np 1nn!nxn1 p1

n1pc1x1,

c , p,p 1, . . . , n 1 p 1,

3.6

By using equality 0 x1exdx 1!,where , from Eq.(3.1) and

(3.6) one obtain

M0n,p 1nn!p pp 1! p1n1p

c1

p 1!

1nn!p cpp cn1pp 1 . . .

1nn!p p 1,1 .

3.7

Thus

M0n,p 1nn!p p 1n,p,1n,p . 3.8

Lemma 3.2.[8]. Let p be a prime number. Then Mkn,p p 2n,p,2n,p , k 1,2, . . . .

Proof.([8], p.128) By subsitution x k u dx du from Eq.(3.3) one obtain

Mkn,p 0

u kp1u k 1 . . .u . . .u k npeup 1! du

k 1,2, . . .

3.9

By using equality

u kp1u k 1 . . .u . . .u k np p1

n1pd1u1,

d , p,p 1, . . . , n 1 p 1,

3.10

and by subsitution Eq.(3.10) into RHS of the Eq.(3.9) one obtain

Mkn,p 1p 1! 0p1

n1pd1u1du p 2n,p,

2n,p ,k 1,2, . . . .

3.11

Lemma 3.3.[8]. (i) There exists sequences an,n and gn,n such that

|kn,p| n gn anp1

p 1! , 3.12

where sequences an,n and gn,n does not depend on number p. (ii) Forany n : kn,p 0 if p .

Proof.([8], p.129) Obviously there exist sequences an,n andgn,k ,n such that an,n and gn,n does not depend on numberp

|xx 1. . . x n| an, 0 x n 3.13

and

|xx 1. . . x nexk | gn, 0 x n,k 1,2, . . . . 3.14

Substitution inequalities (3.13)-(3.14) into RHS of the Eq.(3.3) by simple calculationgives

kn,p gn anp1

p 1! 0k

dx n gkn anp1

p 1! . 3.15

Statement (i) follows from (3.15). Statement (ii) immediately follows from astatement (ii).

Lemma 3.4.[8]. For any k n and for any such that 0 1 there exist p such that

ek Mkn,pM0n,p . 3.16

Proof.From Eq.(3.5) one obtain

ek Mkn,pM0n,p |kn,p|M0n,p . 3.17

From Eq.(3.5) by using Lemma 3.3.(ii) one obtain (3.16).Remark 3.1.We remind now the proof of the transcendence of e following

Shidlovsky proof is given in his book [8].Theorem 3.1. The number e is transcendental.Proof.([8], pp.126-129) Suppose now that e is an algebraic number; then it

satisfies some relation of the form

a0 k1

n

akek 0, 3.16

where a0,a1, . . . ,an are an integers and where a0 0.Having substituted RHSof the Eq.(3.5) into Eq.(3.16) one obtain

a0 k1

n

akMkn,p kn,p

M0n,p a0 k1n

akMkn,pM0n,p k1

n

akkn,pM0n,p 0.

3.17

From Eq.(3.17) one obtain

a0M0n,p k1

n

akMkn,p a0M0n,p k1

n

akMkn,p k1

n

akkn,p 0. 3.18

We rewrite the Eq.(3.18) for short in the form

a0M0n,p k1

n

akMkn,p k1

n

akkn,p

a0M0n,p n,p k1

n

akkn,p 0,

n,p k1

n

akMkn,p.

3.19

We choose now the integers M1n,p,M2n,p, . . . ,Mnn,p such that:

p|M1n,p,p|M2n,p, . . . ,p|Mnn,pwhere p |a0 |

3.20

and p M0n,p. Note that p| n,p.Thus one obtain

p a0M0n,p n,p 3.21

and therefore

a0M0n,p n,p ,

a0M0n,p n,p 0.3.22

By using Lemma 3.4 for any such that 0 1 we can choose a prime numberp p such that:

k1

n

akkn,p k1

n

|ak | 1. 3.23

From (3.23) and Eq.(3.19) we obtain

a0M0n,p n,p 0. 3.24

From (3.24) and Eq.(3.22) one obtain the contradiction.This contradiction finalizedthe proof.The proof of the #-transcendene of the numbersek,k . We will divide the proof into four partsPart I.The Robinson transfer

1.Using Robinson transfer principle [4],[5], from Eq.(3.8) one obtain directly

M0n,p 1nn!p p 1n,p,

1n,p ,n,p .3.25

From Eq.(3.11) using Robinson transfer principle one obtain kk :

Mkn,p p 2n,p ,

2n,p ,k 1,2, . . . ,k ,n,p .3.26

Using Robinson transfer principle from inequality (3.13) one obtain kk :

kn,p n gkn anp1

p 1! ,

k 1,2, . . . ,k ,n,p .3.27

Using Robinson transfer principle, from Eq.(3.5) one obtain kk :

ek ek Mkn,p kn,p

M0n,p ,

k 1,2, . . . ,k ,n,p .3.28

Lemma 3.5. Let n , then for any k and for any 0, thereexist p such that

ek Mkn,pM0n,p . 3.29

Proof. From Eq.(3.28) we obtain kk :

ek Mkn,pM0n,p

|kn,p||M0n,p| ,

k ,n,p .3.30

From Eq.(3.30) and (3.27) we obtain (3.29).

Part II.The Wattenberg imbedding ek into d1.By using Wattenberg imbedding # d, from Eq.(3.28.) one obtain

ek# e# k Mkn,p# kn,p#

M0n,p# ,

k 1,2, . . . ,k ,n,p .3.31

2.By using Wattenberg imbedding # d, from Eq.(3.25) one obtain

M0n,p# 1n # n!p # p# 1n,p#

1#n# n!# p# p# 1n,p#,

1n,p ,d,n,p .

3.32

3.By using Wattenberg imbedding # #, from Eq.(3.26) one obtain

Mkn,p# p# 2n,p #,

2n,p# ,d,

k 1,2, . . . ,k ,n,p .

3.33

Lemma 3.6. Let n , then for any k and for any 0, thereexist p such that

ek# Mkn,p#M0n,p#

#. 3.34

Proof. Inequality (3.34) immediately from inequality (3.34) by using Wattenbergimbedding # d.

Part III.To prove that e is #-transcendental number we must show that e does not

w-transcendental, i.e., there is does not exist real -analytic functiongx

n0

anxn, 0 |x| r e with rational coefficients a0,a1, . . . ,an, . . .

such that

n0

aken 0. 3.35

Suppose that e is w-transcendental, i.e., there is exist an -analytic functionx

n0

nxn,with rational coefficients:

0 k0m0 ,1 k1m1 , . . . ,n knmn , . . . ,

0 03.36

such that the equality is satisfied:

n0

nen 0. 3.37

Hence there exist sequences nii0 and njj0 such that

nieni 0,ilim

n0

nninen ,njenj 0,

j0

j

njenj . 3.38

FromEq.(3.35)-Eq.(3.23) by using Definition 2.14. and Theorem2.11 (seesubsection 2.6) one obtain the equality [4]

0# #Ext-n

n# en# d. 3.39

Theorem 3.2.[4] The equality (3.39) is a contradictory.Proof.Let us considered hypernatural number defined by countable

sequence

m0,m0 m1, . . . ,m0 m1 . . .mn, . . . 3.40

From Eq.(3.39) and Eq.(3.40) one obtain

# 0# # #Ext-n

n# en# # d. 3.41

From inequality (3.27) one obtain

kn,p# n# gkn# anp1 #

p 1!# ,

k 1,2, . . . ,k ,n,p .3.42

Substitution Eq.(3.31) into Eq.(3.41) gives

0# #Ext-n

k# en#

0# #Ext-k

k# Mkn,p# kn,p#

M0n,p# # d.

k# # k#,k ,0# # 0#.

3.43

Multiplying Eq.(3.43) by Wattenberg hyperinteger (see subsection 2.5)M0n,pw# d one obtain

0# M0n,p# #Ext-k

k# Mkn,p# k# kn,p#

# M0n,p# d.3.44

By using inequality (3.42) for a given , 0 we will choose prime hyperinteger p such that:

#Ext-k

k# kn,p# # M0n,p# # d. 3.45

Now using inequality (3.42) we willing to choose prime hyper integer p and# #p 0 in the Eq.(3.45) for a given , 0 such that:

# M0n,p# #p #. 3.46

Hence from Eq.(3.45) and Eq.(3.46) we obtain

#Ext-k

k# kn,p# # d. 3.47

Therefore from Eq.(3.44) and (3.47) we obtain

0# M0n,p# #Ext-k

k# Mkn,p# # M0n,p# d. 3.48

We will choose now prime hyper integer p in Eq.(3.48) p p such that

p# max|0# |,n#. 3.49

Hence from Eq.(3.32) follows

p# M0n,p#. 3.50

Note that M0n,p# 0#.Using Eq.(3.50) and Eq.(3.40) one obtain:

p# M0n,p# 0#. 3.51

Using Eq.(3.33) one obtain

p# Mkn,p#,k 1,2, . . . . 3.52

From Eq.(3.48) by Definition 2.10 and Theorem 2.9 (see subsection 2.4) we obtain

0# M0n,p#

#Ext-k

k# Mkn,p##

# M0n,p# d # # # M0n,p# d .

3.53

where # #p 0.But for a sufficiently small # #p 0 the equality (3.40)by Theorem 2.10 (see subsection 2.5) is a contradictory. This contradictioncompleted the proof.

References

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[3] Foukzon J. 2006 Spring Central Sectional Meeting Notre Dame,IN,April 8-9,2006 Meeting #1016 The solution of one very old problem intranscendental numbers theory. Preliminary report.http://www.ams.org/meetings/sectional/1016-11-8.pdf

[4] Foukzon J. Non-archimedean analysis on the extended hyperreal line dand some transcendence conjectures over field and .http://arxiv.org/abs/0907.0467

[5] Goldblatt,R.,Lectures on the Hyperreals. Springer-Verlag, New York,NY,1998.

[6] Wattenberg,F. 0,-valued, translation invariant measures on andthe Dedekind completion of .Pacific J. Math. Volume 90,Number 1 (1980), 223-247.

[7] Gonshor,H. Remarks on the Dedekind completion of a nonstandardmodel of the reals.Pacific J. Math. Volume 118, Number1 (1985),117-132.

[8] Shidlovsky, A.B. "Diophantine Approximations and TranscendentalNumbers", Moscov, Univ.Press,1982 (in Russian).http://en.bookfi.org/book/506517http://bookre.org/reader?file506517&pg129