IJMMSInternational Journal of Mathematics and Mathematical Sciences1687-04250161-1712Hindawi Publishing Corporation86570510.1155/2009/865705865705Research ArticleEngel Series and Cohen-Egyptian Fraction ExpansionsLaohakosolVichian1ChaichanaTuangrat2, 3Rattanamoong Jittinart2, 3KanasriNarakorn Rompurk4LouboutinStéphane1Department of MathematicsKasetsart UniversityBangkok 10900Thailandku.ac.th2Department of MathematicsFaculty of ScienceChulalongkorn UniversityBangkok 10330Thailandchula.ac.th3The Centre of Excellence in MathematicsCHESi Ayutthaya RoadBangkok 10400Thailandmahidol.ac.th4Department of MathematicsKhon Kaen UniversityKhon Kaen 40002Thailandkku.ac.th200908122009200911092009021220092009Copyright © 2009This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

Two kinds of series representations, referred to as the Engel series and the Cohen-Egyptian fraction expansions, of elements in two different fields, namely, the real number and the discrete-valued non-archimedean fields are constructed. Both representations are shown to be identical in all cases except the case of real rational numbers.

1. Introduction

It is well known  that each nonzero real number can be uniquely written as an Engel series expansion, or ES expansion for short, and an ES expansion represents a rational number if and only if each digit in such expansion is identical from certain point onward. In 1973, Cohen  devised an algorithm to uniquely represent each nonzero real number as a sum of Egyptian fractions, which we refer to as its Cohen-Egyptian fraction (or CEF) expansion. Cohen also characterized the real rational numbers as those with finite CEF expansions. At a glance, the shapes of both expansions seem quite similar. This naturally leads to the question whether the two expansions are related. We answer this question affirmatively for elements in two different fields. In Section 2, we treat the case of real numbers and show that for irrational numbers both kinds of expansion are identical, while for rational numbers, their ES expansions are infinite, periodic of period 1, but their CEF expansions always terminate. In Section 3, we treat the case of a discrete-valued non-archimedean field. After devising ES and CEF expansions for nonzero elements in this field, we see immediately that both expansions are identical. In Section 4, we characterize rational elements in three different non-archimedean fields.

2. The Case of Real Numbers

Recall the following result, see, for example, Kapitel IV of , which asserts that each nonzero real number can be uniquely represented as an infinite ES expansion and rational numbers have periodic ES expansions of period 1.

Theorem 2.1.

Each A{0} is uniquely representable as an infinite series expansion, called its Engel series (ES) expansion, of the form A=a0+n=11a1a2an, where a0={[A],if  A,A-1if  A,a12,an+1an(n1). Moreover, A if and only if an+1=an(2) for all sufficiently large n.

Proof.

Define A1=A-a0, then 0<A11. If An0  (n1) is already defined, put an=1+[1An],An+1=anAn-1. Observe that an is the least integer >1/An and 1an<An1an-1. We now prove the folowing.Claim 2.

We have 0<An+1AnA2A11.

Proof.

First, we show that An>0 for all n1 by induction. If n=1, we have seen that A1>0. Assume now that An>0 for n1. By (2.3), we see that an. Since An+1=anAn-1=(An-1an)an and 1/an<An, we have An+1>0. If there exists k such that Ak+1>Ak, then akAk-1=Ak+1>Ak and so ak-1>1/Ak, contradicting the minimal property of an and the Claim is proved.

From the Claim and (2.3), we deduce that a12 and an+1an  (n1). Iterating (2.4), we get A1=1a1+1a1a2++1a1a2an+An+1a1a2an. To establish convergence, let Bn=1a1+1a1a2++1a1a2an(n1). Since An>0 and an for all n1, the sequence of real numbers (Bn) is increasing and bounded above by A1. Thus, limnBn exists and so 1a1a2an0(n). By the Claim, 0<An+1a1a2an1a1a2an0(n), showing that any real number has an ES expansion. To prove uniqueness, assume that we have two infinite such expansions such that a0+n=11a1a2an=b0+n=11b1b2bn, with the restrictions a0,  a12,  an+1an  (n1) and the same restrictions also for the bn's. From the restrictions, we note that 0<A1:=n=11a1a2an1. If A1=1, then by (2.12) we also have n11/b1b2bn=1, forcing a0=b0. If 0<A1<1, then (2.12) shows that 0<n11/b1b2bn<1, forcing again a0=b0. In either case, cancelling out the terms a0,b0 in (2.12) we get A1:=n=11a1a2an=n=11b1b2bn. Since an+1an, then a1A1-1=1a2+1a2a3+1a2a3a4+1a1+1a1a2+1a1a2a3+=A1, so 0<a1-1/A11. But there is exactly one integer a1 satisfying these restrictions. Thus, a1=b1. Cancelling out the terms a1 and b1 in (2.14) and repeating the arguments we see that ai=bi for all i.

Concerning the rationality characterization, if its ES expansion is infinite periodic of period 1, it clearly represents a rational number. To prove its converse, let A=a/b{0}. Since

A1=A-a0=a-ba0b, we see that A1 is a rational number in the interval (0,1] whose denominator is b. In general, from (2.4), we deduce that each An  (n1) is a rational number in the interval (0,1] whose denominator is b. But the number of rational numbers in the interval (0,1] whose denominator is b is finite implying that there are two least suffixes h,k such that Ah+k=Ah. Thus, by (2.3), we have ah+k=ah. From (2.2), we know that the sequence {an} is increasing. We must then have k=1 and the assertion follows.

Remark 2 s.

In passing, we make the following observations.

For n1, we have an+1=anAn+1=AnanAn-1=Anan=1+1An1An.

If Ac, then Anc and so 1/An for all n1.

If A, then its ES expansion is A=A-1+12+122+123+.

To construct a Cohen-Egyptian fraction expansion, we proceed as in  making use of the following lemma.

Lemma 2.2.

For any y(0,1), there exist a unique integer n2 and a unique r such that 1=ny-r,0r<y.

Proof.

Let n=1/y and r=ny-1. Put 1/y:=n-1/y[0,1) and so r=ny-1=y1y[0,y). To prove uniqueness, assume ny-r=1=my-s so that 1+1y>n=1+ry1y. Since there is only one integer with this property, we deduce n=m and consequently, r=s proving the lemma.

Theorem 2.3.

Each A{0} is uniquely representable as a CEF expansion of the form A=n0+k=11n1n2nk, subject to the condition n0,n12,nk+1nk(k1), and no term of the sequence appears infinitely often. Moreover, each CEF expansion terminates if and only if it represents a rational number.

Proof.

To construct a CEF expansion for A{0}, define r0=A-n0[0,1). If r0=0, then the process stops and we write A=n0. If r00, by Lemma 2.2, there are unique n1 and r1 such that 1=n1r0-r1,0r1<r0,n12. Thus, A=n0+r0=n0+1n1+r1n1. If r1=0, then the process stops and we write A=n0+1/n1. If r10, by Lemma 2.2, there are n2 and r2 such that 1=n2r1-r2,0r2<r1,n2n1, the last inequality being followed from n1=1/r0,  n2=1/r1, and r1<r0. Observe also that A=n0+1n1+1n1n2+r2n1n2. Continuing this process, we get A=n0+1n1+1n1n2++1n1n2nk+rkn1n2nk, with 1=niri-1-ri,1>ri-1>ri0,2nini+1(i=1,2,). If some rk=0, then the process stops, otherwise the series convergence follows at once from |rkn1n2nk|0(k).

To prove uniqueness, let

n0+k=11n1n2nk=A=m0+k=11m1m2mk, with the restrictions (2.23) on both digits ni and mj. Now k11n1n2nkk112k=1. It is clear that the restrictions (2.23) imply the strict inequality in (2.33). This also applies to the right-hand sum in (2.32). Equating integer and fractional parts in (2.32), we get n0=m0,k=11n1n2nk=k=11m1m2mk=:w,  say. Since nk+1nk, then n1w-1=1n2+1n2n3+1n2n3n4+1n1+1n1n2+1n1n2n3+=w, so 0<n1-1/w1. But there is exactly one integer n1 satisfying these restrictions. Then n1=m1 and k21n2nk=k21m2mk. Proceeding in the same manner, we conclude that ni=mi for all i.

Finally, we look at its rationality characterization. If A, then r0, say r0:=p/q, where p,q. From (2.30), we see that each ri is a rational number whose denominator is q. Using this fact and the second inequality condition in (2.30), we deduce that rj=0 for some jp, that is, the expansion terminates. On the other hand, it is clear that each terminating CEF expansion represents a rational number. Now suppose that A is irrational and there is a j and integer n such that ni=n for all ij. Then

A=n0+k=1j1n1n2nk+1n1n2njk=11nk. Since k11/nk=1/(n-1), it follows that A is rational, which is impossible.

The connection and distinction between ES and CEF expansions of a real number are described in the next theorem.

Theorem 2.4.

Let A{0} and the notation be as set out in Theorems 2.1 and 2.3.

(i) If A, then its ES expansion is infinite periodic of period 1, while its CEF expansion is finite. More precisely, for A, let its ES and CEF expansions be, respectively,

A=a0+n=11a1a2an=n0+k=11n1n2nk. If m is the least positive integer such that 1/Am, then a0=n0,a1=n1,,am-1=nm-1,am=nm+1,am=am+i(i1), and the digits ni terminate at nm.

(ii) If A, then its ES and its CEF expansions are identical.

Proof.

Both assertions follow mostly from Theorems 2.1, 2.3, and Remark (b) except for the result related to the expansions in (2.38) which we show now.

Let A and let m be the least positive integer such that 1/Am. We treat two seperate cases.

Case 1 (<inline-formula><mml:math xmlns:mml="http://www.w3.org/1998/Math/MathML" id="M170"><mml:mi>m</mml:mi><mml:mo>=</mml:mo><mml:mn>1</mml:mn></mml:math></inline-formula>).

In this case, we have 1/A1 and a1=1+[1/A1]=1+1/A1. Since r0=A-n0=A-[A]=A-a0=A1, we get n1=1/A1 and so a1=n1+1. We have r1=n1r0-1=0, and so the CEF expansion terminates. On the other hand, by Remark (a) after Theorem 2.1, we have a1=ai  (i2).

Case 2 (<inline-formula><mml:math xmlns:mml="http://www.w3.org/1998/Math/MathML" id="M178"><mml:mi>m</mml:mi><mml:mo>></mml:mo><mml:mn>1</mml:mn></mml:math></inline-formula>).

Thus, 1/A1 and A1=r0. By Lemma 2.2, we have a1=n1. For 1im-2, assume that Ai=ri-1 and ai=ni. Then Ai+1=aiAi-1=niri-1-1=ri. Since 1/Ai+1, again by Lemma 2.2, ai+1=ni+1. This shows that a1=n1,,am-1=nm-1. Since 1/Am, we have am=1+[1/Am]=1+1/Am and thus Am=am-1Am-1-1=nm-1rm-2-1=rm-1. From the construction of CEF, we know that nm=1/rm-1. Thus, nm=1/Am showing that am=nm+1. Furthermore, rm=nmrm-1-1=0, implying that the CEF terminates at nm, and by Remark (a) after Theorem 2.1, am=am+i  (i1).

3. The Non-Archimedean Case

We recapitulate some facts about discrete-valued non-archimedean fields taken from [3, Chapter 4]. Let K be a field complete with respect to a discrete non-archimedean valuation |·| and 𝒪:={AK;|A|1} its ring of integers. The set 𝒫:={AK;|A|<1} is an ideal in 𝒪 which is both a maximal ideal and a principal ideal generated by a prime element πK. The quotient ring 𝒪/𝒫 is a field, called the residue class field. Let 𝒜𝒪 be a set of representatives of 𝒪/𝒫. Every AK{0} is uniquely of the shape

A=n=Nbnπn(bn𝒜,bN0) for some N, and define the order v(A) of A by |A|=2-v(A)=2-N, with v(0):=+. The head part A of A is defined as the finite series

A=n=v(A)0bnπnif  v(A)0,  and  0  otherwise. Denote the set of all head parts by

S:={A;AK}. The Knopfmachers' series expansion algorithm for series expansions in K  proceeds as follows. For AK, let a0:=AS. Define A1:=A-a0. If An0  (n1) is already defined, put

an=1An,An+1=(An-1an)snrn if an0, where rn and snK{0} which may depend on a1,,an. Then for n1

A=a0+A1==a0+1a1+r1s11a2++r1rn-1s1sn-11an+r1rns1snAn+1. The process ends in a finite expansion if some An+1=0. If some an=0, then An+1 is not defined. To take care of this difficulty, we impose the condition

v(sn)-v(rn)2v(an)-1. Thus

A=a0+1a1+n=1r1rns1sn·1an+1. When rn=1,  sn=an, the algorithm yields a well-defined (with respect to the valuation) and unique series expansion, termed non-archimedean Engel series expansion. Summing up, we have the following.

Theorem 3.1.

Every AK{0} has a finite or an infinite convergent non-archimedean ES expansion of the form A=a0+n=11a1a2an, where the digits an are subject to the restrictions a0=AS,anS,v(an)-n,v(an+1)v(an)-1(n1).

Now we turn to the construction of a non-archimedean Cohen-Egyptian fraction expansion, in the same spirit as that of the real numbers, that is, by way of Lemma 2.2. To this end, we start with the following lemma.

Lemma 3.2.

For any yK{0} such that v(y)1, there exist a unique nS such that v(n)-1 and a unique rK such that 1=ny-r,v(r)v(y)+1(i.e.,0|r|<|y|).

Proof.

Let n=1/y. Then v(n)=v(1/y)=-v(y)-1. Putting r=ny-1, we show now that v(r)v(y)+1. Since n=1/y, we have 1y=n+k1ckπk, where ck𝒜, and so ny-1=-yk1ckπk. Thus v(r)=v(ny-1)=v(-y)+v(k1ckπk)v(y)+1>v(y). To prove the uniqueness, assume that there exist n1S such that v(n1)-1 and r1K such that 1=n1y-r1,0|r1|<|y|. From ny-r=1=n1y-r1, we get (n-n1)y=r-r1. If nn1, since n,n1S we have |n-n1|1. Using |y|>|r-r1|, we deduce that |r-r1|<|y||n-n1||y|=|r-r1|, which is a contradiction. Thus, n=n1 and so r=r1.

For a non-archimedean CEF expansion, we now prove the following.

Theorem 3.3.

Each yK{0} has a non-archimedean CEF expansion of the form y=n0+k=11n1n2nk, where nkS,v(nk)-k,v(nk+1)v(nk)-1(k1). This series representation is unique subject to the digit condition (3.19).

Proof.

Define n0=y and r0=y-n0. Then v(r0)1. If r0=0, the process stops and we write y=n0. If r00, by Lemma 3.2, there are n1S and r1K such that n1=1r0,r1=n1r0-1, where v(n1)-1 and v(r1)v(r0)+1. So y=n0+r0=n0+1n1+r1n1. If r1=0, the process stops and we write y=n0+1/n1. If r10, by Lemma 3.2, there are n2S and r2K such that n2=1r1,r2=n2r1-1, where v(n2)-1 and v(r2)v(r1)+1. So y=n0+1n1+1n1n2+r2n1n2. Continuing the process, in general, nk=1rk-1,rk=nkrk-1-1,y=n0+1n1+1n1n2++1n1n2nk+rkn1n2nk, where nkS,v(nk)-1,v(rk)v(rk-1)+1(k1). Thus, v(nk+1)=-v(rk)-v(rk-1)-1=v(nk)-1(k1). We observe that the process terminates if rk=0. Next, we show that v(nk)-k  (k1). By construction, we have v(n1)-1. Assume that v(nk)-k, then v(nk+1)v(nk)-1-k-1. Regarding convergence, consider v(rkn1n2nk)=-v(n1)-v(n2)--v(nk)+v(rk)=-v(n1)-v(n2)--v(nk)-v(nk+1)1+2++k+(k+1)(k). It remains to prove the uniqueness. Suppose that xK{0} has two such expansions x=n0+j1n1n2nj=m0+i1m1m2mi. Since v(j1/n1n2nj)=v(1/n1)1 and n0S, we have n0=y. Similarly, m0=y yielding by uniqueness n0=m0 and j11/n1n2nj=i11/m1m2mi. Putting ω:=j11n1n2nj=i11m1m2mi, we have n1ω=1+j21/n2nj and so 1=n1ω-j21n2nj,v(j21n2nj)=v(1n2)=-v(n2)-v(n1)+1=v(ω)+1. By Lemma 3.2, since n1 is the unique element in S with such property, we deduce n1=m1. Continuing in the same manner, we conclude that the two expansions are identical.

It is clear that the construction of non-archimedean ES and CEF expansions is identical which implies at once that the two representations are exactly the same in the non-archimedean case.

4. Rationality Characterization in the Non-Archimedean Case

In the case of real numbers, we have seen that both ES and CEF expansions can be used to characterize rational numbers with quite different outcomes. In the non-archimedean situation, though ES and CEF expansions are identical, their use to characterize rational elements depend significantly on the underlying nature of each specific field. We end this paper by providing information on the rationality characterization in three different non-archimedean fields, namely, the field of p-adic numbers and the two function fields, one completed with respect to the degree valuation and the other with respect to a prime-adic valuation.

The following characterization of rational numbers by p-adic ES expansions is due to Grabner and Knopfmacher .

Theorem 4.1.

Let xpp{0}. Then x is rational, x=α/β, if and only if either the p-adic ES expansion of x is finite, or there exist an m and an sm, such that am+j=ps+j+1-γps+j(j=0,1,2,), where γβ.

Now for function fields, we need more terminology. Let 𝔽 denote a field and π(x) an irreducible polynomial of degree d over 𝔽. There are two types of valuation in the field of rational functions 𝔽(x), namely, the π(x)-adic valuation |·|π, and the degree valuation |·|1/x defined as follows. From the unique representation in 𝔽(x),

f(x)g(x)=π(x)mr(x)s(x),f(x),g(x),r(x),s(x)𝔽[x]{0};π(x)r(x)s(x),m, set |0|π=0,|f(x)g(x)|π=2-md;|0|1/x=0,|f(x)g(x)|1/x=2degf(x)-degg(x). Let 𝔽((π)) and 𝔽((1/x)) be the completions of 𝔽(x), with respect to the π(x)-adic and the degree valuations, respectively. The extension of the valuations to 𝔽((π)) and 𝔽((1/x)) is also denoted by |·|π and |·|1/x.

For a characterization of rational elements, we prove the following.

Theorem 4.2.

The CEF of y𝔽((π)) or in 𝔽((1/x)) terminates if and only if y𝔽(x).

Proof.

Although the assertions in both fields 𝔽((π)) and 𝔽((1/x)) are the same, their respective proofs are different. In fact, when the field 𝔽 has finite characteristic, both results have already been shown in  and the proof given here is basically the same.

We use the notation of the last section with added subscripts π or 1/x to distinguish their corresponding meanings.

If the CEF of y in either field is finite, then y is clearly rational. It remains to prove the converse and we begin with the field 𝔽((π)). Assume that y𝔽(x){0}. By construction, each rk𝔽(x) and so can be uniquely represented in the form

rk=π(x)v(rk)pk(x)qk(x), where pk(x),  qk(x)  (0)𝔽[x] with gcd  (pk(x),qk(x))=1,  π(x)pk(x)qk(x). Since nk=1/rk-1Sπ and v(nk)-k, it is of the form nk=sv(nk)(x)π(x)v(nk)+sv(nk)+1(x)π(x)v(nk)+1++s-1(x)π(x)-1+s0(x)=:mk(x)π(x)v(nk), where sv(nk)(x),,s0(x) are polynomials over 𝔽, not all 0, of degree <d and mk(x)𝔽[x]. Thus, |nk|1/xmax{|sv(nk)(x)π(x)v(nk)|1/x,|sv(nk)+1(x)π(x)v(nk)+1|1/x,,|s-1(x)π(x)-1|1/x,|s0(x)|1/x}2d-1, yielding |mk(x)|1/x2d-dv(nk)-1. By construction, we have rk=nkrk-1-1. Substituting (4.4) and (4.5) into (4.8) and using v(rk-1)=-v(nk) lead to π(x)-v(nk+1)pk(x)qk-1(x)=qk(x)(mk(x)pk-1(x)-qk-1(x)). Since gcd  (π(x)-v(nk+1)pk(x),qk(x))=1, it follows that qk(x)qk-1(x), and so successively, we have |qk(x)|1/x|qk-1(x)|1/x|q1(x)|1/x, which together with (4.9) yield |pk(x)|1/x|π(x)|1/xv(nk+1)max{|mk(x)pk-1(x)|1/x,|q1(x)|1/x}. Using (3.19) and (4.7), we consequently have |pk(x)|1/x2d(v(nk)-1)max{2d-dv(nk)-1|pk-1(x)|1/x,|q1(x)|1/x}max{12|pk-1(x)|1/x,|q1(x)|1/x2d(k+1)}.   This shows that |pk(x)|1/x(1/2)|pk-1(x)|1/x for all large k which implies that from some k onwards, pk(x)=0, and so rk=0, that is, the expansion terminates.

Finally for the field 𝔽((1/x)), assume that y=p(x)/q(x)𝔽(x){0}. Without loss of generality, assume degp(x)degq(x). By the Euclidean algorithm, we have

y=p(x)q(x)=N0(x)+R0(x)q(x):=n0+r0, where n0:=N0(x)=y1/x,R0(x)𝔽[x],0degR0<degq,r0=R0(x)q(x). From the Euclidean algorithm, q(x)R0(x)=N1(x)+R1(x)R0(x),N1(x),R1(x)𝔽[x],0degR1<degR0<degq, which is, in the terminology of Lemma 3.2, 1=r0N1+R1q=r0n1-r1. Again, from the Euclidean algorithm, -q(x)R1(x)=N2(x)+R2(x)R1(x),N2(x),R2(x)𝔽[x],0degR2<degR1<degR0<degq, which is, in the terminology of Lemma 3.2, 1=r1N2-R2q=r1n2-r2. Proceeding in the same manner, in general we have rj=(-1)jRjq  ,0degRj<degRj-1<<degR1<degq. There must then exist k such that degRk=0, that is, Rk𝔽{0}. Thus, the CEF of y is y=n0+1n1++1n1nk+rkn1nk=n0+1n1++1n1nk+1n1nknk+1, where nk+1=(-1)kRk-1q𝔽[x], which is a terminating CEF.

Acknowledgments

This work was supported by the Commission on Higher Education and the Thailand Research Fund RTA5180005 and by the Centre of Excellence In Mathematics, the Commission on Higher Education.

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