representation of real numbers

0.1 Introduction

It is well-known that there are several methods to introduce the real numbers. We shall follow an inductive method which is instructive as well as elementary. Apart from that such treatment is modern, interesting and is obtained through two theorems and a lemma, which are relatively easy to understand. So that our aim will be to prove the above propositionsPlanetmathPlanetmath. Our starting point is the following theorem.

Theorem 1.

Let {ai} be a sequenceMathworldPlanetmath of positive integers such that ai>1, for all i1. Then any real number ρ is uniquely expressible by

ρ=b0+i=1bij=1iaj, (1)

where the bi are integers satisfying the inequalitiesMathworldPlanetmath 0biai-1 for all i1, and bi<ai-1 for infinitely many i.


Let {bi}0 be a sequence of integers and {ρi}1 a sequence of real numbers defined by the equations

b0=[ρ],ρ1=ρ-b0, (2)

and for all i1

bi=[aiρi],ρi+1=aiρi-bi, (3)

denoting [] the integral part function. Clearly ρi+1 is the fractional part of aiρi, therefore for all i1 we have,

0ρi<1. (4)

Next we multiply (4) by ai whence 0aiρi<ai, but bi=[aiρi], so that


an inequality required by the theorem.
Now, by (3) and (4), and applying inductionMathworldPlanetmath on ρi, we can establish that

==b0+i=1nbij=1iaj+ρn+1j=1naj. (5)

Now we define


thus from (4), (5) and by the hypothesisMathworldPlanetmath ai2, we arrive to


because the fractional part ρn+1<1. Then if we let n grows beyond of any bound, ρ-dn will be so close to zero as we want and such a observation implies the representation (1).
We still need to prove the another inequality of the theorem, i.e. bi<ai-1 for infinitely many i, but also the uniqueness of representation (1). To do that we need to make use of the identityPlanetmathPlanetmath

i=1an+i-1j=1ian+j=1. (6)

(It is legitimate to consider this identity as a lemma, as we need it to prove this theorem as well as the next one).
We shall prove later this identity. Let us prove the inequality bi<ai-1 by tertio excluso; thus suppose that there is a fixed integer n such that bi=ai-1 for all i>n. From (1) and (6) we get

ρ= b0+i=1nbij=1iaj+i=n+1ai-1j=1iaj
= b0+i=1nbij=1iaj+1j=1naji=1an+i-1j=1ian+j
= b0+i=1nbij=1iaj+1j=1naj,

and comparing this with (5) one realizes that ρn+1=1, contradicting (4).
Finally we must prove the uniqueness of the representation (1). So that we suppose


Here the integers ci satisfy the same conditions as do the bi. It is necessary (and sufficient!) to show that ci=bi for every i. The condition ci<ai-1, for infinitely many i, altogether with the identity (6) imply that


so we see that c0 is the integral part of ρ, i.e. c0=[ρ], but also from (2) b0=[ρ], therefore c0=b0. Next we shall again use tertio excluso. On this way let us suppose that for some n1 the pair bn and cn are unequal. There is no loss of generality in assuming that n is the smallest integer with this property (which is justified by a simple inductive argumentMathworldPlanetmath), and that bn>cn, so that bn-cn1. Thus we have


(There is no contradictionMathworldPlanetmathPlanetmath at all in this equality, as it is easily seen in the next below equation).
It is obvious that these series are absolutely convergent, so we may rearrange terms to obtain


But then recalling that ci<ai-1, we see tat ci-bi<ai-1. From this fact and (6), we can write


This is a contradiction with respect to the inequality found before, thus the proof of this theorem is completePlanetmathPlanetmathPlanetmathPlanetmath. ∎

0.2 Some implications

First we remarked that Theorem 1 is a generalizationPlanetmathPlanetmath of the standard decimal expansion for a real number ρ. This may be seen by taking all the integers ai=10. Thus, if ρ>0, (1) gives the decimal representation

ρ=b0+i=1bi10i=b0.b1b2. (7)

Second, if ρ<0, we must write its decimal representation on the form (7)and then changing all signs. Third, any real ρ could have an ambiguous decimal representation, e.g. ρ=1.5699, having an infiniteMathworldPlanetmath successive sequence of 9’s, which also involves a geometric seriesMathworldPlanetmath in 10-i in turns implying (at the limit) that also ρ=1.57. For that reason, (7) represents that number with an infinite succession of 0,s, that is, ρ=1.57=1.5700. The reason for this resides in that an infinite succession of 9’s is ruled out by the condition of the theorem that bi<ai-1 for infinitely many i, a condition that in the present example takes the form bi<9 for infinitely many i.
Now we prove (6).

Lemma 1.

For the integers sequence {ai}, where ai>1 for every i1, we have


Let us take the partial sum


since when in the operator an+j-1an+j, then the index value of j, at its upper limitMathworldPlanetmath, m+1m, but its lower limit does not. Thus, central sums cancel and the last term vanishes because, by hypothesis, we have


and the lemma is proved. ∎

Theorem 1 also represents an irrational number whenever we add a couple of additional conditions. Thus we have the following important theorem.

Theorem 2.

Let us consider the same integers sequence {ai} described in the preceding theorem, and that the integers bi satisfying the inequalities of that result. In additionPlanetmathPlanetmath, let us assume that infinite integers bi are positive, and that each prime numberMathworldPlanetmath divides infinitely many ai. Then ρ is irrational.


We contradict the thesis by supposing ρ=p/q is rational (p, q, coprimeMathworldPlanetmath). By the last hypothesis in the preceding theorem, we can choose an integer n sufficiently large in order to q be a divisorMathworldPlanetmathPlanetmath of j=1naj. Now we may use (1) replacing the LHS by our rational numberPlanetmathPlanetmath assumptionPlanetmathPlanetmath, next multiplying both side by the latter product, and rearranging terms we get (we do the partitionPlanetmathPlanetmathPlanetmath sum i=1,,n;n+1,)

j=1naj(p-b0q)q-i=1nb0j=1najj=1iaj=i=1bn+ij=1ian+j. (8)

By hypothesis, the LHS of (8) is obviously an integer. However, we have proved already that the last inequality of theorem 1 requires that bn+i<an+i-1 for infinitely many i, so that, from the RHS of (8) and the lemma we see that


a clear contradiction, proving the theorem. ∎

0.3 Example

e is irrational.
All we know there are different ways to prove the irrationality of e. In particular, it results illustrative if we adapt our theorems and its hypotheses (all of which are true in this case) to this problem. From Taylor’s expansion


Let us use (1), by setting ρ=e, b0=2, bi-1=1 for all i2, and aj=j+1, 0ji-1. Thus,


So far our discussion on real numbers. An interesting approach on transcendental numbersMathworldPlanetmath as well as and extensive bibliography on real numbers are given, for instance, in [1].


Title representation of real numbers
Canonical name RepresentationOfRealNumbers
Date of creation 2013-03-22 19:10:36
Last modified on 2013-03-22 19:10:36
Owner perucho (2192)
Last modified by perucho (2192)
Numerical id 10
Author perucho (2192)
Entry type Topic
Classification msc 11A63
Related topic UniquenessOfDigitalRepresentation