Repeating Decimals - Kevin Carmody

REPEATING DECIMALS

A NUMERISTIC APPROACH

Kevin Carmody1000 University Manor Drive, Apt. 34

Fairfield, Iowa [email protected]

Fifth edition11 August 2020

Copyright c© 2010–2020 Kevin Carmody

Fifth edition 11 August 2020Fourth edition 13 December 2019

Third edition 5 June 2017Second edition 2 September 2016

First edition 24 March 2016

CONTENTS

Epigraph . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5

Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7

Definitions and notation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8

Fundamental theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11

Terminating decimals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13

Repeating decimals are multivalued . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15

Pure repeating decimals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17

Mixed repeating decimals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23

Nildecimals and nonidecimals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 25

Infinite left decimals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 28

Other bases . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 35

Infinity and zero . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 36

Decimal representations of infinity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 36

Unfolding infinity and zero . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 37

Infinite integers and rational numbers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 39

Decimal structure of real numbers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41

Folded real numbers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41

Unfolded real numbers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41

Acknowledgment . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 44

References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 45

Contents 3

4 Repeating Decimals

Epigraphs

C’est de l’Inde que nous vient l’ingenieuse methode d’exprimer tous lesnombres avec dix caracteres, en leur donnant a la fois une valeur absolue et une valeurde position, idee fine et importante, qui nous parait maintenant si simple que nousen sentons a peine le merite. Mais cette simplicite meme et l’extreme facilite qui enresulte pour tous les calculs placent notre systeme d’Arithmetique au premier rangdes inventions utiles, et l’on appreciera la difficulte d’y parvenir, si l’on considerequ’il a echappe au genie d’Archimede et d’Apollonius, deux des plus grands hommesdont l’antiquite s’honore.

It is from India that the ingenious method comes from expressing allnumbers with ten characters, giving them both an absolute value and a posi-tion value, a fine and important idea, which now seems so simple to us that wehardly feel the merit. But this very simplicity and the extreme ease which re-sults from it for all calculations place our Arithmetic system at the forefront ofuseful inventions, and we will appreciate the difficulty of achieving this, if weconsider that it has escaped the genius of Archimedes and Apollonius, two ofthe greatest and most honored men of antiquity.—Pierre-Simon Laplace, [Lp,p. 404–405]

5


SUMMARY

This document derives several theorems on terminating and repeatingdecimals, along with illustrative examples. It is shown that every rational num-ber has a unique decimal representation, except that some rational numbershave two representations, such as 0.999 . . . = 1.

It is also shown which rational numbers result in terminating, pure re-peating, and mixed repeating decimals. Some results for the lengths of termi-nating and repeating portions are proved.

Using numeristics, developed in a separate document, an extension tothe usual decimal scheme is developed here: infinite decimals on the left sideof the decimal point, such as . . . 999 = −1.

Other number bases are briefly considered.

Another extension to the decimal scheme, equipoint analysis, is brieflyintroduced here. This theory uses multiple levels of sensitivity to extend dec-imal arithmetic to infinite and infinitesimal numbers, including infinite andinfinitesimal integers and rational numbers.

Summary 7

DEFINITIONS AND NOTATION

A terminating decimal is a fraction whose decimal representation has afinite number of digits. Examples include 0.25 and 3.728.

A repeating decimal (also called a recurring or a periodic decimal) hasan infinite number of digits, and the digits are periodic after a certain point.Examples include 0.333 . . . = 0.3 and 0.81090909 . . . = 0.8109.

A terminating decimal can be considered a special case of a repeatingdecimal, since, for instance, 0.25 = 0.25000 . . . = 0.250. However, as we usethe term, repeating decimal means only a non-terminating repeating decimal,so that terminating and repeating decimals are distinct classes.

A pure repeating decimal is a repeating decimal in which all the digitsare periodic, i.e. the perodicity starts at the decimal point. A mixed repeatingdecimal is any repeating decimal which is not pure, i.e. the digits after thedecimal point consist of a nonperiodic portion followed by a periodic portion.

A non-repeating infinite decimal has an infinite number of digits to theright of the decimal point, but the digits never become periodic.

A ten-pure or regular number is a positive integer whose prime factorsinclude only 2 or 5 or both. The first few ten-pure numbers are 2, 4, 5, 8, 10, 16,20, and 25.

A ten-free number is a positive integer whose prime factors are all diif-ferent from 2 or 5. The first few ten-free numbers are 3, 7, 9, 11, 13, 17, 19, and21.

A ten-mixed number is a positive integer whose prime factors include2 or 5 or both, and other prime factors. The first few ten-mixed numbers are 6,12, 14, 15, 18, 22, 24, and 26.

In the following, uppercase letters denote integers, e.g. A, B, N, whichare positive unless otherwise noted.

Whenever we write a fraction MN

, we assume 0 < M < N and N ≥ 2,unless otherwise noted.


Subscripted lowercase letters denote decimal digits, e.g. d1 or dn. Astring such as d1d2d3 . . . dn denotes an integer or decimal represented by thosedigits. For example, if d1 = 1, d2 = 2, and d3 = 5, then d1d2d3 = 125, and0.d1d2d3 = 0.125.

For any terminating decimal 0.d1d2d3 . . . dn, the integer d1d2d3 . . . dn iscalled the terminant. For example, 125 is the terminant of 0.125.

Similarly, for any repeating decimal 0.d1d2d3 . . . dne1e2e3 . . . em, the inte-gere1e2e3 . . . em is called the repetend. (Articles in MathWorld, e.g. [WRD], use theword “reptend” instead, but this almost certainly originated as a typo, since nostandard references published prior to MathWorld used this spelling.)

When we consider fractions MN

, we usually consider only those that arein lowest terms, i.e. in which M and N are relatively prime, also known ascoprime. We call such fractions reduced.

We should note that if MN

is not reduced and the denominator N is ten-free, then in the reduced form J

K, the denominator K will still be ten-free, since

any prime factor in K must also be in N, and N contains no ten-pure primefactors. Similarly, if M

Nreduces to J

Kand N is ten-pure, then K is ten-pure.

However, if MN

reduces to JK

and N is ten-mixed, then K may be ten-mixed,ten-pure, or ten-free, depending on which prime factors M and N have incommon.

The stint of a fraction MN

is the number of digits between the decimalpoint and the end of the non-repeating portion of the decimal representationof M

N. The stint is the number of digits in the terminant, including leading

zeros but not trailing zeros, unless they are followed by a nonzero repetend.For example, the stint of 3

4 is 2, since 34 = 0.75 and there are two non-repeating

places in the decimal; it is not 3 even though 34 = 0.750. The stint of 1

12 is 2

because 112 = 0.083 and there are two non-repeating digits 08. We denote the

stint of 1N

as Λ(N).

The period of a fraction MN

, is the number of digits in the repeatingportion of the decimal representation of M

N. The period is the number of digits

in the repetend, including leading and trailing zeros in the repeating portion.For example, the period of 3

7 = 6, since 37 = 0.428571. The period of 1

11 is 2

because 111 = 0.09. The period is always derived from the minimum number of

Definitions and notation 9

decimal places necessary to represent MN

. For example, the period of 13 is 1 even

though 13 = 0.33, and the period of 4

5 is 0 even though 45 = 0.80. We denote the

period of 1N

as λ(N).


FUNDAMENTAL THEOREM

The decimal expansion of any fraction MN is either a terminat-

ing decimal or a repeating decimal.

PROOF. MN

is converted to a decimal expansion by long division of Mby N. In long division by N, each remainder is less than N, and the first re-mainder is M. The remainder at any point determines all digits of the quotientthat follow. At some point, the remainder will either be 0, in which case the ex-pansion terminates, or a nonzero number which has occurred before, in whichcase the expansion repeats periodically. In the latter case, the period is thenumber of steps between the recurring remainders and is less than N, sinceonly remainders from 1 to N − 1 are possible. �

We will see below that the converse is also true: any terminating orrepeating decimal represents some fraction M

N.

Figure 1 shows a typical long division yielding a repeating decimal.The final remainder of 1 is the same as the first remainder (the original divi-dend M), so continued division will repeat the digits of the quotient peridi-cally.

0.142857

7 1.0000007302820146056

403550491

FIG. 1:Computation of repetend of 1

7

Fundamental theorem 11

Table 2 shows stints and periods for the first unit fractions.

N Λ(N) λ(N) 1N

2 1 0 0.53 0 1 0.34 2 0 0.255 1 0 0.26 1 1 0.167 0 6 0.1428578 3 0 0.1259 0 1 0.1

10 1 0 0.111 0 2 0.0912 1 1 0.08313 0 6 0.07692314 1 6 0.071428515 1 1 0.0616 4 0 0.062517 0 16 0.058823529411764718 1 1 0.0519 0 18 0.05263157894736842120 2 0 0.05

TABLE 2: Stints and periods of1N

for N = 1 to 20, where

Λ(N) = stint of 1N

,

λ(N) = period of 1N


TERMINATING DECIMALS

The decimal representation of a fraction MN is terminating if

and only if N is ten-pure.

PROOF. Let MN

= 0.d1 . . . dQ for someQ and T = d1 . . . dQ. Then MN

= T10Q ,

and N | 10Q. Thus N is ten-pure. The converse follows by reversing thesesteps. �

The stint of 1N , Λ(N), where N is ten-pure and N = 2A5B , is

max(A,B).

PROOF. max(A,B) is the minimum Q such that 2A5B | 10Q. �

Given any reduced fraction MN , the length of the terminant is

independent of M and is thus equal to Λ(N), and the termi-nant of M

N is M times the terminant of 1M .

PROOF. Let P = the stint of 1N

and T its terminant, and let Q = the stint

of MN

and U its terminant. Then T = 10PN

and U = M10QN

. P is the minimum valuefor which N | 10P and likewise Q is the minimum value for which N |M10Q.Since M and N are coprime, they have no factors in common, so N | 10Q. Qmust be the minimum value for which this holds, since if we had S < Q andN | 10S, then we would haveN |M10S andQ would not be the minimum. ButP is also the minimum value for which N | 10P . Hence P = Q and U =MT .

For ten-pure numbers H and N, Λ(HN) ≤ Λ(H) + Λ(N).

PROOF. If H and N are ten-pure, they are of the form H = 2A5B andN = 2C5D . We then have Λ(HN) = Λ(2A+C5B+D) = max(A + C,B + D) ≤max(A,B) + max(C,D) = Λ(H) + Λ(N). �

Terminating decimals 13

For ten-pure N, Λ(NK) = KΛ(N).

PROOF. As before, N = 2C5D . Then Λ(HK) = Λ(2KC5KD) =max(KC,KD) = Kmax(C,D) = KΛ(N). �

Table 3 demonstrates this last theorem for N = 2 and K from 1 to 10.The terminants are powers of 5, since 2N5N

2N = 5N .

K 2K 12K Λ(2K)

1 2 0.5 12 4 0.25 23 8 0.125 34 16 0.0625 45 32 0.03125 56 64 0.015625 67 128 0.0078125 78 256 0.00390625 89 512 0.001953125 9

10 1024 0.0009765625 10

TABLE 3: Decimals and stints of1

2K for K from 1 to 10


REPEATING DECIMALSARE MULTIVALUED

A repeating decimal is a convergent geometric series. A finite geomet-ric series

x =N∑

K=M

aK = aM + aM+1 + aM+2 + aM+3 + . . . + aN

can easily be summed through a recurrence formula. When we multiply bothsides by a:

xa = aM+1 + aM+2 + aM+3 + aM+4 + . . . + aN+1,

we observe that xa + aM = x + aN+1. We therefore have x − xa = aM − aN+1,which yields

x =aM − aN+1

1 − a ,

the well-known resultN∑

K=m

aK = aM + aM+1 + aM+2 + aM+3 + . . . + aN =aM − aN+1

1 − a .

For the case of an infinite geometric series, the numeristic theory ofinfinite series postulates adding a single infinite element∞ to the real numbers.The resulting number system is called the projectively extended real numbersand is described in detail in [CN].∞ is simultaneously greater and less than allfinite real numbers. We also have:

∞ + k =∞, where k is any finite number

k∞ =∞k

=∞, where k is any nonzero finite number

10 =∞1∞ = 0∞ = −∞

Since +∞ = −∞, a∞ has two real values for most a, a−∞ = 0 and a+∞ =∞. This in turn means that most convergent series have two sums, one finiteand one infinite:

∞∑K=M

aK = aM + aM+1 + aM+2 + aM+3 + . . . ={∞, a

M

1 − a

}.

Repeating decimals are multivalued 15

For example,

0.3 = 3∞∑N=1

10−N

= 30.1 − 10∞+1

1 − 0.1

= 3{

0.1 −∞0.9

,0.1 − 0

0.9

}= 3{∞, 1

9

}={∞, 1

3

}

The infinite value can also be justified by comparison with ∞ · 0 =∑∞N=1 0, which may take any finite or infinite value, and yet each term of which

is infinitely smaller than the corresponding term of any geometric series withnonzero a.

Decimals with zero repetends, called nildecimals, are an exception tothe two-value rule:

0.0 = 0 · 0.1 = 0{∞, 1

9

}= {R, 0} = R.

See Nildecimals and nonidecimals (p. 25).

Since only finite values are usually of interest with repeating decimals,and since almost all of the proofs in this document use calculations which arevalid only for finite values, we will usually ignore the infinite value of repeat-ing decimals and use only the finite values. We also usually ignore the indeter-minate nature of nildecimals and use only the single value 0.


PURE REPEATING DECIMALS

For any pure repeating decimal, 0.d1 . . . dP = R10P−1 , whereR =

d1 . . . dP is the repetend and P is the period.

PROOF. A repeating decimal is a convergent geometric series. 0.d1 . . . dPis a convergent geometric series whose first term is 0.d1 . . . dP and whose termratio is 10−P . The sum is therefore

0.d1 . . . dP1 − 10−P

=R

10−P

1 − 10−P=

R

10P − 1. �

The decimal expansion of a reduced fraction MN is a pure re-

peating decimal if and only if N is ten-free.

PROOF. Let MN

be the reduced form of R10P−1 which is given above. Since

M and N have no common factors, N | 10P − 1. Each prime factor of N alsodivides 10P − 1. But 2 and 5 can never divide 10P − 1, since the last digit of10P − 1 is always 9. N is therefore ten-free. These steps are reversible, so theconverse easily follows. �

Given any reduced fraction MN , the length of the repetend is

independent ofM and is thus equal to λ(N), and the repetendof M

N is M times the repetend of 1M .

PROOF. Let P be the period of 1N

and R its repetend, and let Q be the

period of MN

and S its repetend. Then R = 10P−1N

and S = M(10Q−1)N

. P is theminimum value for which N | 10P − 1, and likewise Q is the minimum valuefor which N |M(10Q −1). Since M and N are coprime, they have no factors incommon, so N | 10Q − 1. Q must be the minimum value for which this holds,since if we had T < Q and N | 10T − 1, then we would have N | M(10T − 1)and Q would not be the minimum. But P is also the minimum value for whichN | 10P − 1. Hence P = Q and S =MR.

Pure repeating decimals 17

The period λ(N) of MN never exceeds N − 1.

PROOF. All remainders in the long division of M by N are in the range1 to N − 1. Whenever any of them occurs a second time, a repetend is estab-lished. Hence the period is always within this range. �

We now define a repnine of length N as 9192 . . . 9N = 10N − 1, a repunit

of length N as 1112 . . . 1N = 10N−19 , and a repdigit of length N as d1d2 . . . dN =

d9 (10N − 1) for any digit d. We denote a repunit of length N as U(N).

Every prime M, except 2, 3, and 5, divides a repunit U(P) forsome P < M.

PROOF. Since 1M

= R9U(P) for some P < M, where R is the repetend, we

have RM = 9U(P). Since M 6= 3, M | U(P). �

For M = 3, P = M, since 3 | 111. Since repdigits are multiples ofrepunits, the same is true for every repdigit.

If the repetend of MN isR and λ(N) =N−1, then the repetend

of JN is a cyclic permutationof R.

PROOF. Since 0 < M, J < N and λ(N) = N − 1, every integer from1 to N − 1, including M and J , must occur as one of the remainders in thecalculation of the decimal expansion of M

N. The numerator is always the first

remainder in any such calculation, so J will occur as a remainder at some laterpoint in the calculation of the digits of M

N. When it does, the digit in M

Nat

that point will be the first digit of JN

. Since subsequent digits depend only on

the current remainder, the digits of JN

will repeat those of MN

but starting at a

different point, i.e. the digits of JN

will be a cyclic permutation of the digits ofMN

. �

If λ(N) =N − 1, then N is prime.


PROOF. If N were composite, then MN

would reduce to JK

for some M

and K < N. Since λ(N) can never exceed N − 1, λ(K) could not exceed K − 1,which is less than N − 1. But by the previous theorem, λ(K) = λ(N) = N − 1.Hence N is prime. �

The converse of this theorem is not true. Two conterexamples are λ(3) =1 and λ(11) = 2.

If λ(N) < N −1, then there may be multiple cycles of length λ(N). Thiscan be expressed more precisely as follows.

λ(N) | φ(N), where φ(N) is the Euler totient function.

PROOF. The Euler totient function φ(N) is the number of integers that

are relatively prime to N. It is thus the number of integers M < N such that MN

is reduced. The number of cycles of length λ(N) is φ(N)λ(N) . �

Tables 4 through 7 show the example N = 21, where φ(N) = 12 andλ = 6.

M M21 M M

21 M M21

1 0.047619 8 0.380952 15 0.7142852 0.095238 9 0.428571 16 0.7619043 0.142857 10 0.476190 17 0.8095234 0.190476 11 0.523809 18 0.8571425 0.238095 12 0.571428 19 0.9047616 0.285174 13 0.619048 20 0.9523807 0.333333 14 0.666666 21 0.999999

TABLE 4: Decimal representations of all M21


M M21 M M

21

1 0.047619 2 0.09523810 0.476190 20 0.95238016 0.761904 11 0.52380913 0.619048 5 0.2380954 0.190476 8 0.380952

19 0.904761 17 0.809523

TABLE 5: The two cycles for which M21 is reduced

M M21

3 0.1428579 0.4285716 0.285174

18 0.85714212 0.57142815 0.714285

TABLE 6: The cycle for which M21 reduces to M

7

M M21

7 0.314 0.621 0.9

TABLE 7: The cycle for which M21 reduces to M

3

λ is not multiplcative for ten-free integers, i.e. for ten-free Hand N, λ(HN) is not necessarily equal to λ(H)λ(N), andλ(NK) is not necessarily equal to Kλ(N).


PROOF. Tables 8 and 9 show some examples where multiplicativitydoes not hold, and some where it does. �

λ(H)·H N 1

H1N λ(H) λ(N) λ(N) λ(HN) 1

HN HN

3 11 0.3 0.09 1 2 2 2 0.03 333 13 0.3 0.076923 1 6 6 6 0.025641 397 11 0.142857 0.09 6 2 12 6 0.012987 77

11 13 0.09 0.076923 2 6 12 6 0.006993 143

TABLE 8: λ(H)λ(N) compared to λ(HN)for select H and N

K·N 1

N λ(N) K λ(N) λ(NK) 1NK NK

3 0.3 1 2 2 1 0.1 93 0.3 1 3 3 3 0.037 277 0.142857 6 2 12 42 0.0204081632653061224489795 49

918367346938775513 0.3 1 4 4 10 0.0123456789 81

11 0.09 2 2 4 22 0.0082644628099173553719 12113 0.076923 6 2 12 78 0.0059171597633136094674556 169

21301775147928994082840236686390532544378698224852071

TABLE 9: λ(NK) compared to Kλ(N)for select N and K

In the decimal expansion of MN , if N is a ten-free prime and

λ(N) is even, then the sum of the two halves of the repetendis a repnine of the length of the halves, e.g. 1

7 = 0.142857 and142 + 857 = 999.


PROOF. Let L = 12λ(N) be the length of each half, and let R be the

repetend. The decimal expansion is a pure repetend since N is ten-free. MN

isreduced for allM sinceN is prime. We then have M

N= R

102L−1 , andN | 102L−1 =(10L − 1)(10L + 1). Since N is prime, either N | 10L − 1 or N | 10L + 1. But ifN | 10L − 1, then λ(N) would be L, whereas we have assumed λ(N) = 2L. SoN | 10L + 1.

ComputeA = MN(10L+1)−1, which is an integer in the range 0 to 10L−1,

since N | 10L +1 and M < N. A then has at most L digits. If it has fewer than Ldigits, left pad it with zeros to L places. A then has exactly L digits in all cases.Denote A as a1a2 . . . aL.

Next compute B = 10L − 1−A, which is also an integer in the range 0 to10L − 1, and denote it b1b2 . . . bL.

Finally compute a1a2 . . . aLb1b2 . . . bL = A10L+B =[MN(102L + 1) − 1

]10L+

10L − 1−[MN(10L + 1) − 1

]= M

N(10L + 1)10L − M

N(10L − 1) = M

N(10L + 1)(10L − 1) =

MN(102L − 1) = R.

A and B are thus the two halves of the repetend, and their sum A +B =10L − 1 = 9U(L). �


MIXED REPEATING DECIMALS

For any fraction MN

, we have shown that the decimal expansion is aterminating decimal if and only if N is ten-pure, and a pure repeating decimalif and only if N is ten-free. By elimination then, we have established that thedecimal expansion is a mixed repeating decimal if and only if N is ten-mixed.We will show this more positively below.

For a ten-mixed integer N, we will find it helpful to factor N into ten-pure and ten-free components. We define the ten-pure part of a ten-mixedinteger N as the product of all the ten-pure prime factors of N (the 2’s and 5’sin the prime factorization of N), and similarly the ten-free part of a ten-mixedinteger N as the the product of all the ten-free prime factors of N (everythingbut the 2’s and 5’s in the prime factorization of N). For example, the ten-purepart of 12 is 4, and the ten-free part of 12 is 3. Obviously, for any ten-mixed N,if P is the ten-pure part and F is ten-free part, then N = PF.

The decimal expansion of a fraction MN is a mixed repeating

decimal if and only if N is ten-mixed.

PROOF. We begin by factoring N into its ten-pure part P and ten-freepart F. Since P and F have no factors in common, they are coprime.

We can express MN

as the sum of a pure repeating decimal and a termi-nating decimal by setting M

N= A

P+ B

Fand solving for A and B. We then have

AP+ B

F= AF+BP

N, so M = AF + BP . From elementary number theory it is known

that if P and F are coprime, then there exist nonzero integers C and D suchthat CF +DP = 1. Then A =MC, B =MD, and M

N= MC

P+ MD

F.

MCP

can be represented by a terminating decimal, plus an integer if|MC| ≥ |P |. MD

Fcan be represented by a pure repeating decimal, plus an inte-

ger if |MD| ≥ |F|. The sum is a mixed repeating decimal. Since these steps caneasily be reversed, the converse follows. �

A few examples of this process:16 = 2

3 −12 = 0.666 . . . − 0.5 = 0.166 . . . = 0.16

Mixed repeating decimals 23

56 = 1

3 + 12 = 0.333 . . . + 0.5 = 0.833 . . . = 0.83

112 = 1

3 −14 = 0.333 . . . − 0.25 = 0.083 . . . = 0.083

1112 = 2

3 + 14 = 0.666 . . . + 0.25 = 0.916 . . . = 0.916

For any ten-mixed N, Λ(N) = Λ(P) and λ(N) = λ(F), whereP is the ten-pure part of N and F is the ten-free part of N.

PROOF. We compute the decimal representation of an arbitrary reducedfraction M

N. Let C = 10Λ(P). Since C

Pis the terminant of P , it is an integer, P | C,

and MCP

is an integer. Since MN

is reduced, M and N have no common factors,and since C and P are ten-pure, MC

Pand F have no common factors. Therefore,

MC/PF

= MCN

is a reduced fraction. Since F is ten-free, the decimal representationof MC

Nis a pure repeating decimal whose period is λ(F). But since MC

N= CM

N

and C is the Λ(P) power of 10, the decimal representation of MCN

is derivedfrom the decimal representation of M

Nby shifting the decimal point left by Λ(P)

places. The integer to the left of the decimal point is the terminant of MN

, andthe pure repeating decimal to the right of the decimal point is the repetend ofMN

. Therefore, Λ(N) = Λ(P) and λ(N) = λ(F). �


NILDECIMALS AND NONIDECIMALS

We define the nilrepetend as the repetend 0, and a nildecimal as anydecimal representation with an explicit nilrepetend. An example of a nildeci-mal is 0.50.

Nildecimals and decimal representations ending with a finite numberof trailing zeros, while they can be considered as distinct, duplicate represen-tations of the decimal without the trailing zeros, are such trivial duplicationsthat we usually ignore trailing zeros and do not consider decimals with themto be distinct.

We define the nonirepetend (the Latin prefix noni- means 9) as the repe-tend 9, and a nonidecimal as any decimal representation with the nonirepe-tend. Nonidecimals are special since they are nontrivial duplicate representa-tions, as we shall soon see.

There is one pure nildecimal, 0.000 . . . = 0.0, and one pure nonidecimal,0.999 . . . = 0.9. There are many mixed nildecimals and nonidecimals, e.g. e.g.0.35000 . . . = 0.350 and 0.24999 . . . = 0.249.

Nildecimals are obviously duplicates of terminating decimals, but thefollowing establishes a similar result for nonidecimals.

0.9 = 1.

PROOF. 0.9 = 0.91−0.1 = 1. �

0.0102 . . . 0P9 = 10−P .

PROOF. 0.0102 . . . 0P9 = 0.0102 ...0P 91−10P = 0.9·10−P

0.9 = 10−P . �

These theorems mean that nonidecimals are nontrivial duplicate dec-imal representations of nildecimals and terminating decimal representations,e.g. 3.9 = 4 and 0.349 = 0.34 + 0.009 = 0.34 + 0.01 = 0.35.

Nildecimals and nonidecimals 25

The second theorem implies that we can convert a nonidecimal to a ter-minating decimal by adding 1 to the terminant and deleting the nonirepetend,and reverse it by subtracting 1 from the terminant and adding the nonirepe-tend.

Nonidecimals may occur through certain operations on decimals, e.g.when we multiply the decimal representation of 1

Nby N for ten-free or ten-

mixed N. Examples:

3 · 0.3 = 0.9 = 1

7 · 0.142857 = 0.999999 = 0.9 = 1

12 · 0.083 = 0.96 + 0.03 + 0.009 = 0.999 = 0.9 = 1

Terminating decimals and corresponding nonidecimals arethe only nontrivial duplicate forms among all decimal repre-sentations.

PROOF. A duplicate can occur only if adding or subtracting zerochanges the digits in a nontrivial way. We will show the conditions underwhich this occurs by adding or subtracting zero in the form of 0.9 − 1. First welook at some examples.

0.5 + 0.9 − 1 = 1.49 − 1 = 0.49

0.49 + 1 − 0.9 = 1.49 − 0.9 = 0.5

Next consider 311 = 0.27. We have

0.27 + 0.99 = 1.2627

0.27 + 0.9999 = 1.272627

0.27 + 0.999999 = 1.27272627

. . .

0.27 + 0.99 = 1.27

and so3

11= 0.27

= 0.27 + 0.9 − 1

= 0.27 + 0.99 − 1


= 1.27 − 1= 0.27

Now consider π = 3.14159 . . . 502 . . ..3.14159 . . . 502 . . . + 0.9 = 4.04159 . . . 502 . . .

3.14159 . . . 502 . . . + 0.99 = 4.13159 . . . 502 . . .3.14159 . . . 502 . . . + 0.999 = 4.14059 . . . 502 . . .

3.14159 . . . 502 . . . + 0.9999 = 4.14149 . . . 502 . . .3.14159 . . . 502 . . . + 0.99999 = 4.14158 . . . 502 . . .

3.14159 . . . 502 . . . + 0.99999 . . . 9 = 4.14159 . . . 402 . . .3.14159 . . . 502 . . . + 0.99999 . . . 99 = 4.14159 . . . 492 . . .

3.14159 . . . 502 . . . + 0.99999 . . . 999 = 4.14159 . . . 501 . . .so

π = 3.14159 . . . 502 . . .= 3.14159 . . . 502 . . . + 0.9 − 1= 4.14159 . . . 502 . . . − 1= 3.14159 . . . 502 . . .

These examples should make it clear that only when we start with aterminating decimal, i.e. when we can append the nilrepetend and thus havean infinite number of consecutive zeros, will the digits be nontrivially affectedby the subtraction of zero.

0 is the only terminating decimal representation which does not havea nonidecimal duplicate, as 0 + 0.9 − 1 = 0.9 − 1 = −0. The minus form is theduplicate in this case. �

Nildecimals and nonidecimals 27

INFINITE LEFT DECIMALS

We now explore extending decimal places infinitely far to the left ofthe decimal point. We define an infinite left decimal as a decimal expansionextending infinitely far to the left. More generally, we use the term left alongwith any of our previous terms to denote a corresponding decimal represen-tation to the left of the decimal point. For example, a a left repeating decimalhas a repetend which extends infinitely far to the left, such as . . . 333 = 3. Nat-urally, we use the term right to refer to the decimal representations extendingto the right that we considered above. For example, we now call a decimalrepresentation such as 0.333 . . . = 0.3 a right repeating decimal.

Just as a right repeating decimal is a convergent geometric series, soa left repeating decimal is a divergent geometric series. Divergent series ingeneral are sometimes controversial, but numeristics also provides a coherentframework for understanding them, which is explored in detail in [CD]. Aswe did with right repeating decimals, for left repeating decimals, we use theprojectively extended real numbers and the identities

+∞ = −∞a∞ = {∞, 0} for a 6= 0, 1,∞

N∑K=M

aK = aM + aM+1 + aM+2 + aM+3 + . . . + aN =aM − aN+1

1 − a∞∑

K=M

aK = aM + aM+1 + aM+2 + aM+3 + . . . ={∞, a

M

1 − a

}

A left repeating decimal therefore has two values, one infinite and onefinite. While the infinite value may seem more natural for an infinite left dec-imal and a finite value for an infinite right decimal, both values are valid forleft and right. Following this principle leads to an easy and consistent theory.

However, as we did with infinite right decimals, in the theorems andproofs below for infinite left decimals, we will usually not be interested in theinfinite value, and the processes we use will usually be valid only for finitevalues. Therefore, we will usually ignore the infinite value and work only withthe finite values.

. . . 999 = −1.


PROOF. . . . 999 = 9 = 9(1 + 10 + 100 + 1000 + . . .) = 9(100 + 101 + 102 +103 + . . .) = 9

1−10 = − 99 = −1. �

Left nonidecimals can be used as an alterative to right decimal repre-sentations:

98 = 9 − 1 = −2927.5 = 9 − 71.5 = −72.5

Other left repetends produce negative numbers: 3 = . . . 333 = 3 · 101 +3 · 102 + 3 · 103 + . . . = 3

1−10 = − 13 . This is the negative of the corresponding right

repeating decimal 0.3 = 13 . This generalizes as follows.

A pure repeating decimal consisting of the same repetend Ron both the left and the right is always zero, e.g. 3.3 = 0.

PROOF. Let Q be the period. The first term of the left repeating decimalis R, and the first term of the right repeating decimal is R

10Q . The value of theleft repeating decimal is R

1−10Q , and the value of the right repeating decimal isR/10Q

1−10−Q = R10Q−1 . The value of the total decimal is therefore zero. �

Every real number has an infinite number of decimal repre-sentations with left repetends.

PROOF. If we choose any repetend R and use it in a pure repeat-ing decimal on both the left and right, the value is zero. Zero thus has aninfinite number of decimal representations with left repetends, for example3.3 = 147.147 = 09.09 = 0. If we add the usual decimal representation of a realnumber r to any of these forms of zero, then we obtain an alternate decimalrepresentation of r. Since there are an infinite number of possible repetends,there are an infinite number of alternate representations for r. �

Some examples:12 = 0.5 = 0.5 + 1.1 = 1.61

= 3.83 = 09.590 = . . .

Infinite left decimals 29

13 = 0.3 = 0.3 + 1.1 = 1.4

= 3.6 = 09.42 = 142857.476190 = . . .

2 = 2 + 1.1 = 13.1

= 35.3 = 0911.09 = . . .

π = 3.14145265 . . . + 1.1 = 14.25270376 . . . = . . .

The signum of a decimal representation is the minus or plus symbolthat indicates sign at the front of the representation. We have already seen thatinfinite left decimals allow us to represent some negative numbers without asignum. We will find that this is true of all negative numbers.

The remaining theorems in this chapter illustrate the similarities anddifferences between left and right decimals. Each theorem has two parts: aresult for right decimals that we have demonstrated in previous chapters, fol-lowed by a result for left decimals which we prove here.

A right terminating decimal without a signum (which usesonly a right terminant) can represent any positive fraction ofthe form M

N , where 0 < M < N, and N is ten-pure. This rep-resentation is unique.

A left terminating decimal without a signum (which uses onlya left terminant) can represent any positive integer. This rep-resentation is unique.

PROOF. A left terminating decimal is simply the conventional represen-tation of a positive integer. �

A pure right repeating decimal without a signum (which usesonly a right repetend) can represent any positive fraction ofthe form M

N , where 0 ≤ M ≤ N, and N is ten-free. When

M = 0, the representation is the pure right nildecimal 0.0,and when M = N, the representation is the pure right non-

idecimal 0.9. This representation is unique.


A pure left repeating decimal without a signum (which usesonly a left repetend) can represent any negative fraction −MN ,where 0 ≤ M ≤ N, and N is positive and ten-free. WhenM = 0, the representation is the pure left nildecimal 0, andwhen M =N, the representation is the pure left nonidecimal9. Within this type of representation, the decimal is unique.

PROOF. The representation of the positive fraction MN

is a pure rightrepeating decimal. The same repetend on the left gives a representation of itsnegative. �

Examples:

− 13 = −0.333 . . . = −0.3

= . . . 333 = 3

− 23 = −0.666 . . . = −0.6

= . . . 666 = 6

− 33 = −1 = −0.999 . . . = −0.9

= . . . 999 = 9

− 111 = −0.090909 . . . = −0.09

= . . . 090909 = 09

− 311 = −0.272727 . . . = −0.27

= . . . 272727 = 27

A mixed right repeating decimal without a signum (whichuses only a right terminant and a right repetend) can repre-sent any positive fraction M

N , where 0 < M < N, and N isten-mixed. This representation is unique.

A mixed left repeating decimal without a signum (which usesonly a left repetend and a left terminant) can represent anypositive or negative fraction M

N , whereN is ten-free. This rep-resentation is unique.


PROOF. The terminant portion of a mixed left repeating decimal repre-sents a positive integer J . If we extend the left repetend right to the decimalpoint, this becomes a pure left repeating decimal, which can represent − L

N,

where 0 ≤ L ≤ N and N is ten-free. If we subtract the pure repeating decimalfrom the original mixed decimal, the repetend cancels, and we are left with anintegerK, which may be positive or negative. Thus the mixed decimal isK− L

N.

Since −1 ≤ − LN≤ 0, i.e. it is somewhere within the negative unit interval, K − L

N

relocates − LN

to the interval between two integers K − 1 and K.

Given any MN

, we compute the integer K such that K − 1 ≤ MN< K and

L such that −1 ≤ − LN< 0, and add the left decimal representations of K and L

N

to obtain the left decimal representation of MN

.

Fractions and left decimals are thus mapped uniquely and completelyin both directions. �

To illustrate this process, we consider examples from the following fourcases. The first two occur when M is a multiple of N and M

Nis an integer.

• Any positive integer added to the pure left nonidecimal re-sults in a mixed left nildecimal, i.e. a left terminant precededby the nilrepetend. This is a trivial duplicate of the ordinarydecimal representation of that integer. Examples:

4 = −1 + 5 = . . . 999 + 5 = . . . 0004 = 04

15 = −1 + 16 = . . . 999 + 16 = . . . 0015 = 015

• Any negative integer subtracted from the pure left nonideci-mal results in a left terminant preceded by the nonirepetend.Examples:

−4 = −1 − 3 = . . . 999 − 3 = . . . 996 = 96

−15 = −1 − 14 = . . . 999 − 14 = . . . 985 = 985

• Any integer added to a left repeating decimal other thanthe left nonidecimal eventually stops carrying and leaves therepetend intact after that point. The result is a left terminantpreceded by a left repetend. Examples:

323 = 11 − 1

3 = . . . 333 + 11 = . . . 344 = 34412211 = 12 − 10

11 = . . . 909090 + 12 = . . . 909102 = 09102


• Any integer subtracted from a left repeating decimal eventu-ally stops borrowing and leaves the repetend intact after thatpoint. The result is a left terminant preceded by a left repe-tend. Examples:

− 323 = −10 − 2

3 = . . . 666 − 10 = . . . 656 = 656

− 12211 = −11 − 1

11 = . . . 090909 − 11 = . . . 090898 = 09898

A mixed right repeating decimal plus a left terminating deci-mal without a signum can represent any positive fraction M

N ,where N is ten-mixed. This representation is unique.

A mixed left repeating decimal plus a right terminating dec-imal without a signum can represent any positive or nega-tive fraction M

N , where N is ten-mixed. This representationis unique.

PROOF. We begin by factoring N into its ten-pure part P and ten-freepart F. Since P and F have no factors in common, they are coprime.

We can express MN

as the sum of a mixed left repeating decimal and aright terminating decimal by setting M

N= A

P+ B

Fand solving for A and B. We

then have AP+ B

F= AF+BP

N, so M = AF + BP . As before, we use the theorem that

if P and F are coprime, then there exist nonzero integers C and D such thatCF +DP = 1. Then A =MC, B =MD, and M

N= MC

P+ MD

F.

MCP

can be represented by a right terminating decimal, plus an integerif |MC| ≥ |P |. MD

Fcan be represented by a mixed left repeating decimal. The

sum is a mixed left repeating decimal plus a right terminating decimal. Sincethese steps can easily be reversed, the converse follows. �

Some examples:16 = 2

3 −12 = 0.666 . . . − 0.5 = 0.166 . . . = 0.16

= − 13 + 1

2 = . . . 333.5 = 3.5

− 16 = −0.16

= − 23 + 1

2 = . . . 666.5 = 6.5


56 = 1

3 + 12 = 0.333 . . . + 0.5 = 0.833 . . . = 0.83

= − 23 + 3

2 = . . . 666 + 1.5 = ..667.5 = 67.5

− 56 = −0.83

= − 13 −

12 = . . . 333 − 0.5 = . . . 332.5 = 32.5

112 = 1

3 −14 = 0.333 . . . − 0.25 = 0.083 . . . = 0.083

= − 23 + 3

4 = . . . 666.75 = 6.75

− 112 = −0.083

= − 13 + 1

4 = . . . 333.25 = 3.251112 = 2

3 + 14 = 0.666 . . . + 0.25 = 0.916 . . . = 0.916

= − 13 + 5

4 = . . . 333 + 1.25 = . . . 334.25 = 34.25

− 1112 = −0.916

= − 23 −

14 = . . . 666 − 0.25 = . . . 665.75 = 65.75

To summarize:

A left terminating decimal plus a right terminating or repeat-ing decimal without a signum can represent any positive ra-tional number, with a terminating decimal being duplicatedby a nonidecimal.

A left terminating or repeating decimal plus a right terminat-ing decimal without a signum can represent any positive ornegative rational number, with no duplication.

A left terminating or repeating decimal plus a right terminat-ing or repeating decimal without a signum can represent anypositive or negative rational number, with an infinite numberof duplicate representations for every real number.


OTHER BASES

Entirely analogous results ensue in any other number base, also calleda radix. What is called a decimal point in base 10 is called a radix point in anarbitrary base, and the decimal representation can be called a radix represen-tation.

For example, consider base 2, the familiar binary system. We can definea two-pure number as any power of 2, a two-mixed number as an even numberthat is not a power of 2, and a two-free number as an odd number. We alsodefine a unirepetend representation as one with a repetend of 1, such as 0.12 or0.012.

Then the radix-two representation of a fraction is terminating orunirepetend if and only if the denominator is two-pure, mixed repeating if andonly if the denominator is two-mixed, and pure repeating non-unirepetend ifand only if the denominator is two-free.

Examples in this base:12=

1102

= 0.12 = 0.012

15=

11012

= 0.00112

λ2(5) = 4

1012 · 00112 = 5 · 3 = 15 = 2λ2(5) − 1

12 = −1

0.12 = 1

Other bases 35

INFINITY AND ZERO

Decimal representations of infinity

There is one number in our extended number system for which we havenot yet developed a right or left decimal, namely ∞. To do this, we need toreturn to our decision to use only the finite value of a∞ = a−∞. Now we use theinfinite value. ∞ can be thus represented by any right or left repeating decimal,such as:

1 = 101 + 102 + 103 + . . . + 10∞

=10∞+1 − 101

10 − 1=

10∞ − 109

=∞− 10

9=∞9

=∞

0.1 = 10−1 + 10−2 + 10−3 + . . . + 10−∞

=10−∞+1 − 10−1

10−1 − 1=

10−∞ − 0.10.9

=∞− 0.1

9=∞0.9

=∞

In previous chapters, we used the projectively extended real numbersto obtain finite results for infinite left decimals. We now examine another ex-tended real number system, the affinely extended real number. Both of thesesystems are explored in detail in [CN]. The affinely extended system adds twodistinct infinite elements, +∞ and −∞. In this system, a+∞ = +∞ and a−∞ = 0,so we do not have any choice between finite and infinite values of a±∞. Infiniteleft decimals are always equal to +∞, and infinite right decimals are alwaysfinite, so the sum of an infinite left and an infinite right decimal is always infi-nite. +∞ can be represented by any infinite left decimal without a signum, and−∞ can be represented by any infinite left decimal with a minus signum.

While the affinely extended real number system may initially appearmore palatable than the projectively extended system, there are two problemswith using the affinely extended system: (1) it leads to a more difficult theorywith various conflicting approaches, and (2) the results are not fully consistentwith quantum renormalization. Renormalization, which has been repeatedlyverified by physical experiment, is a mathematical procedure which uses as-sumptions similar to those of the projectively extended system. These consid-erations are examined in more detail in [CD].


Unfolding infinity and zero

To understand infinity more deeply, we extend the number systemagain. This time we unfold infinity and zero into multiple infinite and infinites-imal values. To do this we first develop a way of handling multiple valuessimultaneously.

A numeristic class is the simultaneous presence of zero or more values.It is potentially multivalued number or other numeric or number-like construc-tion. Classes have a flat structure: Every number is a single valued class; a classcontaining a single number is identical to the number.

Classes are somewhat similar to sets, but the main difference is thatclasses are flat, whereas sets introduce extra structure on top of their elements.Since we don’t use sets here, we borrow set theoretical notation for class con-cepts. For example, {+1,−1} is a class containing these two numbers. Theflatness property of classes means that a = {a} for any single number a.

A class containing multiple numbers distributes operations on it andstatements about it over each element. For example, {+1,−1}+5 = {+1+5,−1+5} = {+6,+4}.

0−2 −1 1 2

................................................

............................

.......................

....................

.........................................................................................................................................................................................................................................

...............................

.......................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................

...........................................................................

0 · 0′−2 · 0′ −0′ 0′ 2 · 0′

FIG. 10:Real number line with

microscope view of unfolded 0

Infinity and zero 37

0 · ∞′−2 · ∞′ −∞′ ∞′ 2 · ∞′

................................................

............................

.......................

....................

.........................................................................................................................................................................................................................................

...............................

.......................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................

...........................................................................

0−2 −1 1 2

FIG. 11:Line of infinities with microscope view of

real number line within 0 · ∞′

We extend the number system into multiple infinite and infinitesimalvalues using unfoldings or sensitivity levels. The ordinary real numbers arethe folded real numbers. Every number becomes a multivalued class when itis unfolded at a greater sensitivity level, and the class of all unfoldings is theunfolded real numbers. For 0 and ∞, we denote representative units withintheir unfoldings as 0′ and∞′.

Figure 10 shows the real number line with the unfolding of 0 in an in-finitely expanded space, where the single element 0 becomes a multivaluedclass of unfolded multiples of 0′, the class R0′. The bubble in this diagram iscalled a microscope. Figure 11 shows the unfolding of ∞ into a multivaluedclass consisting of multiples of∞′.

Equality is relative to the sensitivity level: if two elements a and b areidentical at the unfolded level, we denote this as a =′ b (say “a unfolded equalsb” or “a equals prime b”). If they are members of the unfolding of the samereal number, we write a = b (“a equals b” or “a folded equals b”). For example:

0′ = 2 · 0′ (folded)0′ 6=′ 2 · 0′ (unfolded)

0′2 =′ 0′ (unfolded)There may be multiple unfoldings: an unfolded number 0′ may itself be un-folded into a second unfolding. If two expressions are identical at all unfold-ings, we write a ≡ b (“a is equivalent to b”), e.g.

(a + b)2 ≡ a2 + 2ab + b2.


The unfolding of ∞ includes both positive and negative multiples of∞′, such as 2 · ∞′ and −3 · ∞′. At the folded level, such numbers are equal andboth positive and negative, but at the unfolded level, they are distinct and areeither positive or negative but not both, e.g.:

2 · ∞′ = −3 · ∞′

2 · ∞′ 6=′ −3 · ∞′

2 · ∞′ >′ 0

−3 · ∞′ <′ 0.

We add to this the postulate of the projectively extended real numbersthat +∞ = −∞. Thus we have

−∞′ = −∞ =∞ =∞′

10−∞′= 10−∞ = 10∞ = 10∞

′= {0,∞}

log 0′ = log 0 = log∞ = log∞′ =∞ ⊃′ {∞′′,−∞′′}

Infinite integers and rational numbers

Given any sensitivity unit 0′, the unfolded infinite numbersR

0′contain

unfolded integers, rationals, and irrationals. Similarly, the unfolded infinitesi-mal numbers R0′ contain unfolded rationals and irrationals.

If M is an infinite integer and n is a finite integer, thenM

nis an infinite

rational, andn

Mis an infinitesimal rational or zero rational.

If M and N are both infinite integers, thenM

Nmay be either rational or

irrational, and either infinite, zero, or perfinite (neither infinite nor zero).M

Ncan be irrational even though it is a ratio of integers, because M and N havean infinite number of digits.

A perfinite irrational can be unfolded into a class which includes suchratios. For example,

√2 = 1.414 . . ., when unfolded, includes elements of the

form 1.414 . . . d(∞′), where d(n) is the n-th digit:√

2 = 1.414 . . . d(∞′) =′ 1414...d(∞′)10∞′

≡ M1

N1

Infinity and zero 39

√2 = 1.414 . . . d(∞′ + 1) =′

1414...d(∞′ + 1)10∞′+1

≡ M2

N2

√2 = 1.414 . . . d(2∞′) =′ 1414...d(2∞′)

102∞′ ≡ M3

N3

A perfinite irrational can also be thought of as having a repetend ofinfinite length:

√2 =

M1

N1=′ 10 · 0.1414 . . . d(∞′)

≡ 1.414 . . . d(∞′)1414 . . . d(∞′)1414 . . . d(∞′)1414 . . .

Another way of finding such infinite integers is with a continued frac-tion:

√2 = 1 +

1

2 +1

2 +1

2 +. . .

.

The convergents (partial evaluations) of this continued fraction are11

,

32

,75

,1712

,4129

,9970, . . .

anbn, . . ., where bn are Pell numbers and an are one-half the

companion Pell numbers or Pell-Lucas numbers. These numbers satisfy therelations

an = 2an−1 + an−2 = an−1 + 2bn−1

bn = 2bn−1 + bn−2 = an−1 + bn−1

an =

(1 +√

2)n

+(

1 −√

2)n

2

bn =

(1 +√

2)n−(

1 −√

2)n

2√

2

So for M and N we have

M =

(1 +√

2)∞′

+(

1 −√

2)∞′

2

N =

(1 +√

2)∞′−(

1 −√

2)∞′

2√

2


DECIMAL STRUCTUREOF REAL NUMBERS

Folded real numbers

A folded real number r, identified with a decimal expansion, has theform

r =+∞∑

m=−∞dm10m,wheredm ∈ Z10 = 0, 1, . . . , 9.

For example, for √2 = 1.414 . . .

we havedn = 0 for n > 0.d0 = 1d−1 = 4d−2 = 1d−3 = 4

. . .

This form allows for infinite left decimals (p. 28). It also includes du-plicates, as discussed in Nildecimals and nonidecimals (p. 25) and Decimalrepresentations of infinity (p. 36).

Unfolded real numbers

In general, an unfolded real number is the sum of one or more of: (1)an infinite number, (2) a perfinite number, and (3) an infinitesimal number.

In the first unfolding, an arbitrary real number r takes this form:r ≡ r+1∞′ + r0 + r−1 0′

≡+∞′′−1∑m=−∞′′

d+1,m10m∞′ ++∞′′−1∑m=−∞′′

d0,m10m ++∞′′−1∑m=−∞′′

d−1,m10m 0′,

Decimal structure of real numbers 41

where ∞′ :≡ 10′ and r+1, r0, r−1 are folded real numbers, and we choose ∞′′ so

that this becomes a single sequence from smallest infinitesimal to largest infi-nite:

10−∞′′ ∞′ ≡ 10+∞′′ and 10−∞

′′ ≡ 10+∞′′ 0′ ≡ 10+∞′′ ∞′−1

∞′ ≡ 102∞′′

∞′′ ≡ 12

log10∞′ ≡ −1

2log10 0′.

Then

r ≡+∞′′−1∑m=−∞′′

d+1,m102∞′′+m + d0,m10m + d−1,m10−2∞′′+m

≡3∞′′−1∑m=−3∞′′

dm10m

where dm =

d+1,m−2∞′′ for +∞′′ ≤ m ≤ +3∞′′ − 1d0,m for −∞′′ ≤ m ≤ +∞′′ − 1d−1,m+2∞′′ for −3∞′′ ≤ m ≤ −∞′′ − 1

or dm = dk,m−2k∞′′ for (2k − 1)∞′′ ≤ m ≤ (2k + 1)∞′′ − 1 and k = +1, 0,−1.

In the second unfolding,

r ≡ r+2∞′2 + r+1∞′ + r0 + r−1 0′ + r−2 0′2

≡+∞′′−1∑m=−∞′′

d+2,m10m∞′2 + d+1,m10m∞′ + d0,m10m + d−1,m10m 0′ + d−2,m10m 0′2

≡+∞′′−1∑m=−∞′′

d+2,m104∞′′+m + d+1,m102∞′′+m + d0,m10m + d−1,m10−2∞′′+m + d−2,m10−4∞′′+m

≡+∞′′−1∑m=−∞′′

+2∑k=−2

dk,m102k∞′′+m

≡+5∞′′−1∑m=−5∞′′

dm10m.

In the n-th unfolding,

r ≡+∞′′−1∑m=−∞′′

+n∑k=−n

dk,m102k∞′′+m

≡+(2n+1)∞′′−1∑m=−(2n+1)∞′′

dm10m.


This includes the ultimate unfolding, where n =∞.

Decimal structure of real numbers 43

ACKNOWLEDGMENT

The author wishes to thank the organizers, participants, and supporters of theInvincible America Assembly at Maharishi International University in Fairfield, Iowafor creating one of the best research environments in the world.


REFERENCES

CD K. Carmody, Divergent series: A numeristic approach,http://kevincarmody.com/math/divergentseries.pdf

CN K. Carmody, Numeristics: A number-based foundational theory of mathe-matics, http://kevincarmody.com/math/numeristics.pdf

WRD E. Weisstein, Repeating Decimal, from MathWorld,http://mathworld.wolfram.com/RepeatingDecimal.html

References 45

http://mathworld.wolfram.com/RepeatingDecimal.html

http://kevincarmody.com/math/numeristics.pdf

http://kevincarmody.com/math/divergentseries.pdf

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Repeating Decimals - Kevin Carmody