Generalized Cantor Expansions
By, Joseph Galante
1 Introduction:
In the beginning of our mathematical lives, we learn about numbers through
concrete examples. At this stage, a number is simply a collection of digits. Some
numbers have only a few digits, whereas others have “a lot”. By the time we get through
a real analysis course, numbers stop being digits and start being entities in their own
right. However we need not throw away information about numerical representations
simply because we have more refined notations of mathematics. Digit Related Number
Theory studies these representations and attempts to extract useful information about the
numbers from their representations. In this paper, we will examine the various types of
representations for numbers. The simplest and most familiar is base-10, which is
commonly used in everyday life. However a more uncommon way to represent a number
is the so called cantor expansion of the number. This system uses factorials rather than
exponentials as the basis for the representation, and it can be shown that this produces a
unique expansion for every natural number. However, if you view factorials as products,
then it becomes natural to ask what happens if you use other types of products as bases.
It can be shown that there is an uncountable number of representations which can be used
to represent the natural, and real numbers.
2 Motivation:
2.1 Definition: The cantor expansion of the natural number n is
n  ak * k! a ( k  1) * (k  1)!....  a 2 * 2! a1 *1!
where for all ai, 0  ai  i
n 1
The identity n!  1   i (i! ) is crucial since it allows “carries” to occur. When adding
i 0
numbers, it provides a meaning to advance to the next term in the base.
2.2 Example:
23 = 3*3!+2*2!+1*1!
24 = 23 + 1 = 3*3!+2*2!+1*1! + 1 = 4!
It is reasonably easy to show via induction that all natural numbers have a unique cantor
expansion. This will not be proven at this point. Instead we will offer a generalization of
this concept, and then show that the regular cantor expansion is a special case.
3 Generalization:
Knowing that the identity is the key, it is conceivable that a more general identity will
yield a more general result. By recognizing that the original expansion relied upon
factorials and their properties, if a more general product is considered, this should
provide the necessary generalization.
3.1 Lemma:
Let S be an ordered sequence of integers such that S  {1, x1, x 2,... | where xi  1, xi  N }
Let p(n) = the nth element in the set S, starting the counting at n=0
n
Let P ( n )   p(i )
i 0
Our generalized identity then becomes:
n
P(n  1)  1   ( p(i  1)  1) P(i )
(Equation 3.1.1)
i 0
Proof/ By Induction:
Basis: n=0
p(1)=p(1)
p(1)*1=p(1)-1 +1
P(1) = 1 + (p(1)-1)*1 since P(1)=1*p(1)
P(0+1) = 1 + (p(0+1)-1)*P(0) since P(0)=p(0)=1
Thus the basis case is established
We assume that equation 3.1.1 works for the first n natural numbers, then we need to
show it works for n+1 as well:
P( (n+1) +1 ) = P(n+2) = p(n+2)*P(n+1) ( from definition of P(n) )
= (p(n+2)-1)*P(n+1) + P(n+1)
n
= (p(n+2)-1)*P(n+1) + ( 1   ( p(i  1)  1) P(i ) ) ( By the Inductive Hypothesis )
i 0
n 1
P(n  2)  1   ( p(i  1)  1) P(i )
i 0
n
 By the principle of induction, P(n  1)  1   ( p(i  1)  1) P(i )
i 0
QED
n  0, n  N
Thus we have shown that there is a more generalized identity which, we hope, can be
used to extend the concept of a cantor expansion. Now we need a new definition to make
it rigorous.
3.2 Definition: The generalized cantor expansion (GCE) of the natural number n with
respect to sequence S is:
n  ak * P( k )  a ( k  1) * P(k  1)  ....  a1 * P(1)  a 0 * P(0)
0  ai  p(i  1), 0  i  k
where S, p, and P are as defined above in Lemma 3.1. S is referred to as the base set or
base sequence. It is important to note that the order of the elements in S matters.
Note the additional term a0*P(0) is not present in normal cantor expansion since it would
be the equivalent of adding the term 0*0! which does not effect the value of the
expansion. With this definition we are ready to prove that each natural can be written
uniquely in the generalized cantor expansion.
3.3 Theorem: All natural numbers n can be written in generalized cantor expansions,
as defined above in 3.2.
Proof/ By Induction
Basis:
0=0*1=0*P(0)
1=1*1=1*P(0)
Thus the basis case is established.
We assume the first n natural numbers can be written in generalized cantor expansion,
and now want to show that (n+1) can be written this way as well.
For convenience, let us assume that ak  0 , so that n  P(k ) . If the first term was zero,
all that would be required would be to consider a smaller number of terms and re-labeling
subscripts accordingly. We can break the inductive step up into three cases.
(See Example 3.5 for more examples of the cases)
Case I: i  k st ai  p(i  1)  1
This case covers the addition of one to a number n without any arithmetic carries into the
kth place. The idea is that the number n has some digit before the kth place which when
one is added to it, the operation does not produce a carry. (To make this concrete, an
example would be n=32378 in regular base 10, and n+1 = 32378 + 1 = 32379. The
addition of the number one does not produce a carry which changes the digit in the
leftmost place).
Using this idea we see that the digits of n satisfy:
n  ai * P(i )  ....  a 0 * P(0) = y
y will always be strictly less than n since n will always have an additional nonzero term
which y does not, namely the term ak*P(i). (This would be like saying in the example
n=32378 that 32378 > 02378.)
When adding one, the strict inequality, becomes only an inequality.
n  ai * P(i )  ....  a 0 * P(0)  1
(n=32378 ≥ 02378+1 -- The idea is that we will never, in case I, have 9999, and the
addition of one will yield 10000, which will change the kth digit in the number.)
By the inductive hypothesis, we see that there exists a valid generalized cantor expansion
for y, since n≥ y ,so we can rewrite it as such:
y = ai * P(i )  ....  a 0 * P(0)  1  ( ai ' ) * P(i )  ....  (a 0' ) * P(0)
Then:
n  1  ak * P( k )  ....  ( ai  1) P(i  1)  ( ai * P(i )  ....  a 0 * P(0)  1)
n  1  ak * P( k )  ....  ( ai  1) P(i  1)  ( ai ' ) * P(i )  ....  ( a 0' ) * P(0)
which is a valid generalized cantor expansion.
Case II: i  k st ai  p(i  1)  1, ak  p(n  1)  1
This case covers a series of carries that terminates at the greatest digit place of the
number. (In another concrete example with base 10, n=39999, and n+1 = 39999 = 40000)
In this case n has the form: n  ak * P( k )  ( p(k )  1) * P( k  1)  ....  ( p(1)  1) * P(0)
Then n+1 is:
n  1  ak * P( k )  ( p(k )  1) * P( k  1)  ....  ( p(1)  1) * P(0)  1
n  1  ak * P( k )  P( k ) from Lemma 3.1
n  1  (ak  1) * P(k )  0 * P(k  1)  ...  0 * P(0)
which is a valid generalized cantor expansion since (ak  1)  ( p(n  1)  1)  1  p(n  1)
Case III: i  k ai  p( n  1)  1
This case covers a series of carries that results in advancing the number of digits in the
number representation by one. This is slightly different from the previous case, since we
advance to a new position, with a new P(x). (Concretely this would be like in base 10 :
99+1 = 100. In base 10, the next place is always 10 times as much as the previous one,
but in the generalized cantor expansion, this may not be true.)
Numbers in this case have the form:
n  ( p(k  1)  1) * P( k )  ( p(k )  1) * P( k  1)  ....  ( p(1)  1) * P(0)
So,
n  1  ( p(k  1)  1) * P( k )  ( p(k )  1) * P(k  1)  ....  ( p(1)  1) * P(0)  1
n  1  1 * P(k  1)  0 * P(k )  ....  0 * P(0) from Lemma 3.1
which is a valid generalized cantor expansion.
 Since in all possibilities are covered by the three cases, and each case yields that the
number can be represented as a generalized cantor expansion, then by the principle of
induction, we can conclude that all natural numbers can be put into generalized cantor
expansion.
QED
Note that the careful way in which we defined the set S is what makes this proof work.
The only term in S which is allowed to be one is the 0th term. Every other term must be 2
or greater. If we allowed other ones into the base set then case III could possibly fail.
Note that if p(k+1)=1, then ak must be zero, in order to satisfy definition 3.2. If we broke
the one’s rule, we could construct a base set with an infinite number of ones in a row, for
example S={1,1,1,1…}. Then each ai must be zero after a certain point (all ai in with
S={1,1,1,1…} ). This would mean that numerical expression of certain numbers is
impossible, as its representation with the base set S would require an infinite sequence of
zeros and yet the number itself would not be zero. If we allowed S={1,1,1,1…} as a valid
representation, then it would mean that all numbers would have the same expansion as
zero. This would not be very useful.
Now that we know every natural number has a generalized cantor expansion, the question
of uniqueness arises.
3.4 Theorem: The generalized cantor expansion of a natural number is unique.
Proof/ Suppose Not.
Then there exists two different GCE representations under the same set S for some n
such that:
n  ak * P( k )  a ( k  1) * P(k  1)  ....  a1 * P(1)  a 0 * P(0) and
n  ( ak )'*P( k )  ( a ( k  1))'*P( k  1)  ....  ( a1)'*P(1)  ( a 0)'*P(0)
n - n = 0 so, then (ak  ak ' ) * P(k )  ...  (a 0  a 0' ) * P(0)  0
since ai  p (i  1) and ai '  p(i  1) , then
(ai  ai ' )  p(i  1) => (ai  ai ' )  P(i  1)
We denote the difference in terms ( ai  ai ' )  di
And let d  dk * P( k )  ...  d 0 * P(0)
There are several cases to consider.
Case I: All the di are greater than or equal to zero.
So, 0≤ di < p(i+1). Then d is a GCE since each of the terms satisfy the criteria.
Consequently, d≥0. Also, by the way in which we defined d, we have
n  ( ak )'*P( k )  ( a ( k  1))'*P( k  1)  ....  ( a1)'*P(1)  ( a 0)'*P(0)
+ d  ( dk ) * P( k )  (dk  1) * P(k  1)  ... .  (d 1 )*P( 1 )  (d 0) * P(0)
-----------------------------------------------------------------------------------n  ( ak ) * P( k )  (a ( k  1)) * P(k  1)  ....  (a1) * P(1)  ( a 0) * P(0)
So n+d=n. But this implies d=0, and d cannot be zero if it has any nonzero coefficients,
so each di is zero as well, and ai=ai’
Case II: All the di are less than or equal to zero. We can apply a similar argument as
before, only we switch the order with which we subtract the representations. We can
make ( ai ' ai )  ci and all the ci will then be greater than or equal to zero, and we can
apply the case I argument.
Case III: Each di is zero so ai=ai’.
Case IV: The only remaining possibilities is if the di’s have mixed signs; some are
positive, some are negative, some are zero. Our objective in this case is to reduce to one
of the previous cases. We do this by employing a standard arithmetic procedure, the
carry. If a given di <0, then making the following alternation to the sum will not change
its value:
d  dk * P( k )  ...  (di  1  1) * P(i  1)  (di  p(i  1)) * P(i )  ...  d 0 * P(0)
(We are using the fact that P(i+1)=p(i+1)*P(i) )
We can apply this idea to each of the terms, except the kth, (multiple times if necessary)
so that all the coefficients are now positive, and meet the criteria for a GCE. We can do
this since by adding a negative to p(i+1), we produce a term, less than p(i+1), which can
satisfy the criteria for a GCE coefficient. Note that multiple applications to single
element might be required. (For example if S={1,10,2…}, and
d=3*P(1)+(-9)*P(1)+0*P(0), then the first carry will yield
d=2*P(1)+(-9+2)*P(1)+0*P(0). We would need multiple applications for the coefficient
in front of P(1) to be positive.)
The kth term, however, cannot fixed by carrying, since there is no greater term to take
from. (If we used a zero from the k+1 term, then carrying would make that negative, and
the next carry would present troubles). After carrying, we will re-label all our coefficients
to indicate that they are not the same from the original difference.
d  d ' k * P( k )  ...  d '0 * P(0) , with d’k any sign, and all other coefficients either zero or
positive.
If d’k ≥ 0, then the series ( d ' k ) * P( k )  ... .  (d '1 )*P( 1 )  ( d '0) * P(0) is a valid GCE,
and falls under case I.
So lets look at if d’k<0. From Lemma 3.1, we see that
( d ' k  1) * P( k  1)  ... .  (d '1 )*P( 1 )  ( d '0) * P(0)  P( k )
So ( d ' k ) * P( k )  ( d ' k  1) * P( k  1)  ... .  (d '1 )*P( 1 )  (d '0) * P(0)  ( d ' k  1) * P( k )
And (d’k+1)≤0, so then d≤0. But n-n=d≤0 and so d=0. But once again d cannot be zero,
if it has nonzero terms, so ai=ai’.
In each of the cases, we reduced to the fact that ai=ai’. But if this true for all the terms,
then there are not really two different expansions. This is a contradiction.
 The generalized cantor expansion of a natural number is unique.
QED
3.5 Example
Let S=set of squares of natural numbers={1,4,9,16,…}
p(0)=1, p(1)=4, p(2)=9
P(0) =1, P(1)=4*1, P(2) = 9*4*1
Case 1 used:
Convert 137 into general cantor expansion for the given S
137 = 3*P(2) + 7*P(1)+1*P(0) = 3*(9*4*1) + 7*(4*1) + 1*(1)
137 => (3 7 1)S
137+1=138= 3*P(2) + (7*P(1) + 1*P(0)+1)= 3*(9*4*1) + 7*(4*1) + 2*(1)
138 => (3 7 2)S
Case 2 used:
Convert 71 into general cantor expansion for the given S
71 = 1*P(2) + 8*P(1)+3*P(0) = 1*(9*4*1) + 8*(4*1) + 3*(1)
71 => (1 8 3)S
71 + 1 = 72 = 2*P(2) + 0*P(1) + 0*P(0) = 2*(9*4*1)
72 => (2 0 0)S
Case 3 used:
Convert 575 into general cantor expansion for the given S
575 = 15*P(2) + 8*P(1)+3*P(0) = 15*(9*4*1) + 8*(4*1) + 3*(1)
575 => (15 8 3)S
575 + 1 = 576 = 1*P(3) + 0*P(2) + 0*P(1) + 0*P(0) = 1*(16*9*4*1)
576 => (1 0 0 0)S
These examples additionally illuminate the fact that representations using GCE need not
cover every possible combination of digits used. We note that if we start counting
upwards using the base set used in example 3.5, our representations leap from (1 8 3)S to
(2 0 0)S . Thus the range of digits from (1 8 4)S to (1 9 9)S is off limits since it breaks the
rules of our representation. If we choose to break the rules, then we lose uniqueness of
representations.
It is interesting to consider the cardinality of the set of all possible generalized cantor
expansions.
3.6 Theorem: There are an uncountable number of base sets S which can be used to
make generalized cantor expansions.
Proof/
Each expansion has a unique sequence S which characterizes the expansion.
S  {1, x1, x 2,... | where xi  1, xi  N }
We can construct another base set S’ with a smaller range of terms such that
S '  {1, x1 mod 10, x 2 mod 10,... | where xi  S}
There is a one to one correspondence between the elements of S’ and the digits of a real
number x = 1. x1 x2 x3.... (standard notion of base 10 representation for reals used here).
Since every value of every variable will be reached if we consider the set of all S’s, then
every real number on the continuous interval [1,2) will be reached. Thus the cardinality
of the set of all S’ is the same as that of the reals, and thus is uncountable. The set of all
S, which is a larger set, is also then uncountable.
Therefore there are an uncountable number of base sets S which can be used to make
generalized cantor expansions.
QED
Now with powerful new facts about generalized cantor expansions, we can examine how
specific number systems fit into this definition. At this stage, we can see how the cantor
expansion works, and also how our regular base 10 number system fits into the picture.
3.7 Factorials
Factorials and the original cantor expansion, from which the generalization evolved, are a
simple case of the generalized theorems where:
S = {1,2,3,4,5….}
p(i) = i+1
P(i) = (i+1)!
And the identity then resolves as:
n 1
n 1
i 0
i 0
n
1   ( p(i  1)  1) * P(i )  1   (i  1)( i  1)!
= 1   i * (i! )
i 1
n
 1   i * (i! )
i 0
=(n+1)!
Which, is actually the original identity shifted up by one iteration;
n 1
n!  1   i (i! )
i 0
The coefficients for each term will be between 0  ai  p(i  1)  0  ai  i
which will yield the original cantor expansions:
n  ak * k! a ( k  1) * (k  1)!....  a 2 * 2! a1 *1!
3.8 Base-b
In general base-b numbers can be represented using:
S = {1,b,b,b,….} for b>1, b a natural number
p(0)=1, p(i) = b i>0
P(i) = bi
Making the identity:
n
P( n  1)  1   (b  1) * b i  b ( n 1)
i 0
Additionally the coefficients will be between:
0  ai  p(i  1)  0  ai  (b  1)
and n  ak * bk  a( k  1) * b( k 1)  ...  a0
which is the standard definition of a number in base-b notation
3.9 Other Interesting Examples of Natural Numbers
There are several other interesting cases to consider for the base set S.
Letting S  {1,2,3,5,7,11,13...( primes of increasing order )...}
The consecutive products become what are known as primorials. (See reference 7.9 for
some more interesting discussion and properties of primorials.)
By picking this base set, we can write numbers as sums of primorials.
17 = 2*(3*2*1) + 2*(2*1) + 1*(1) => (2 2 1)S
42 = 1*(5*3*2*1) + 2*(3*2*1)+0*(2*1)+0*(1) => (1 2 0 0) S
We can also go the other direction:
(1 0 0 0 0 0 0 0 0 0 0 1) S =>
1*(31*29*23*19*17*13*11*7*5*3*2*1) + 1*1=200560490131
(9 8 7 6 5 4 3 2 1) S => 9*(19*17*13*11*7*5*3*2*1) +
8*(17*13*11*7*5*3*2*1)+7*(13*11*7*5*3*2*1)+6*(11*7*5*3*2*1)+5*(7*5*3*2*1)+
4*(5*3*2*1)+3*(3*2*1)+2*(2*1)+1*(1) = 91606553
both of which are actually prime numbers.
4 The Leap to The Reals
Having considered several cases with the natural numbers, it becomes logical to question
whether Generalized Cantor Expansions can be generalized to the real numbers. With a
little work, and some motivational cases, it can be done.
4.1 The New Identity
Just as in the case with the natural numbers, we need an identity to make the
generalization possible. Additionally, we want the identity to look as much alike as
possible to the one used above, for consistency and with the hope of reusing some ideas.
Let S  {1, x1, x 2,... | where xi  1, xi  N }
Let p(n) = the nth element in the set S, starting the counting at n=0
n
Let P( n )  1 /[  p(i )]
i 0
Our generalized identity then becomes:
n
1   ( ( p(i )  1) * P(i ) )  P( n ) (equation 4.1.1)
i 0
Proof/ By Induction
Basis:
n=0
1=0*1+1=(1-1)*1+1=(p(0)-1)*P(0) + P(0)
n=1
1=p(1)/p(1)
1=(p(1)-1)/p(1) + 1/p(1)
and since P(1) =1/(p(1)*p(0))=1/(p(1)*1) = 1/p(1) and (p(0)-1)*P(1) =0
1= ( (p(0)-1)*P(0) + (p(1)-1)*P(1) ) + P(1)
Thus the basis case has been established.
We assume equation 4.1.1 is true for n and we want show it is true for (n+1).
n 1
1   ( ( p(i )  1) * P(i ) )  P( n  1)
i 0
n
1   ( ( p(i )  1) * P(i ) )  ( p( n  1) - 1) P( n  1)  P( n  1)
i 0
n
1   ( ( p(i )  1) * P(i ) )  p( n  1) P( n  1)
i 0
n
1   ( ( p(i )  1) * P(i ) )  p( n  1) * ( P( n ) / p( n  1))
i 0
n
1   ( ( p(i )  1) * P(i ) )  P( n )
i 0
which by the inductive hypothesis is assumed to be true.
Therefore the identity is true for all n≥1.
QED
Now that we have an identity, which in some respects is similar to that above for natural
numbers, we can create a definition.
4.2 Definition
A number x, 0≤x<1 can be represented in a Fractional Generalized Cantor Expansion
with respect to a set S iff

x  c1 * P(1)  c 2 * P( 2)  ...   (ci * P(i ))
i 1
where 0≤ci<p(i), with p, P, and S are defined immediately above.
We can write in short hand: x=( . c1 c2 c3 ….)S
It would be nice if all FGCE’s converge so that our definition is well defined, but first we
must know some of the properties of the function P.
4.3 Lemma: P convergences to zero as n approaches infinity.
Proof/
Since xi  2 for all i≥1, then
1
0<P(n) = ---------------- ≤
x1*x2*...*xn
1
----------- = (1/2)n for all n
2*2*...*2
As n approaches infinity, (1/2)n approaches zero, and thus P converges to zero.
QED
With this nice property of P, we can continue.
4.4 Theorem: All FGCE series are convergent, with respect to a given base set S.
Proof/
From Lemma 4.3, which used the largest values of ci’s available under the definition, we
see that:
n
n
i 1
i 1
1   ( ( p(i )  1) * P(i ) )  P(n ) → 1   ( ( p(i )  1) * P(i ) )  0
Thus the largest FGCE is bounded for any n.
Since P(n) is bounded above by (1/2)n we can write:
n
0  (1   ( ( p(i )  1) * P(i ) ) )  P( n )  (1 / 2) n
i 1
Thus the sum converges, since it is bounded by a larger convergent sequence.
We then note that:
n
n
i 1
i 1
0   ( ci * P(i )))   ( ( p(i )  1) * P(i )) since all coefficients ci are so that 0≤ ci ≤p(i)-1
So all of the terms of any FGCE series have corresponding terms which are positive and
less than the terms of the larger convergent sum. Since the larger sum converges, so does
the smaller one.
Therefore all FGCE series converge.
QED
4.5 Definition: A terminating FGCE of length n is an FGCE that contains only a finite
number of nonzero terms such that all the nonzero terms occur before the n+1 term, for
some nonnegative integer n.
Example: In base 10, 0.1742 would have a terminating FGCE of length 4, since all the
nonzero terms occur before the 5th place. (Remember 0 counts as the 0th place.)
Example: In base 10, 1/3=0.3333333… would not have a terminating FGCE
4.6 Lemma: Terminating FGCE’s of length n divide the interval [0,1) up into
increments of P(n) for a given n and given base set S.
Proof/
Note: In this proof, we will be using both the GCE and FGCE, so we will denote the GCE
base sequence as S’ and the GCE P and p functions as P’ and p’.
Let S={1, x1, x2 ,….,xn} ( We do not care about terms after xn )
Let S’={1, xn, x(n-1) ,….,x1}. (The strange indexing becomes apparent later).
Let l = m/P(n), and where 0≤m≤P’(n)-1. We can see that as m varies between 0 and
P(n)-1, that the l’s divide up [0,1) into increments of length P(n).
We now want to write m as a GCE:
m = cn + cn-1 (xn)+ cn-2 (xn *xn-1)+ … + c2 (xn*…*x4*x3 ) + c1 (xn*…*x3*x2)
With 0≤ ci< p’(i+1) (Note that we are counting down from n, with our ci’s)
Also we omit the term c0 since it is zero, as m<P’(n)= x1*x2*...*xn
So l = m/P(n) = m/( x1*x2*...*xn)
After some shuffling of terms, we get:
c1 x2*x3*...*xn c2 x3*x4*…*xn
cn
l= ----------------- + ------------------ +….+ ---------------x1*x2*...*xn
x1*x2*...*xn
x1*x2*...*xn
After simplification of the fractions, we get:
c1
c2
cn
l= --- + ------ +….+ ---------------x1 x1*x2
x1*x2*...*xn
At this point, we notice that l has been put into a terminating FGCE of length n since the
coefficients satisfy 0≤ ci< p’(i+1) = xi = p(i)
n
Thus, l   ( ci * P(i )) c1 * P(1)  ...  cn * P( n ) , with the all the ci ≥ 0.
i 1
So the l’s, which are actually terminating FGCE’s, divide up the interval [0,1) into
increments of length P(n).
QED
Next we show that each real number has an FGCE associated with it.
4.7 Theorem: Each real number 0≤x<1 has a FGCE, with respect to base set S.
Proof/
This figure nicely illustrates the process which we will be employing:
Figure 4.7.1 - P’s dividing up [0,1) with an x in between (S={1,2,3,5…} shown)
We notice that for a given x, and a fixed n, we can pick ci’s so that:
c1*P(1)+c2*P(2)+…+cn*P(n) ≤ x ≤ c1*P(1)+c2*P(2)+…+(cn+1)*P(n) (eqn. 4.7.2)
We are picking the ci’s so that the x’s location in [0,1) can be specified within an interval
of length P(n). The coefficients simply help to determine the start and end points of this
interval.
( In the figure 4.7.1 with S={1,2,3,5…}
for n=1 c1=1 since 1*P(1) ≤ x≤ (1+1)*P(1)
for n=2 c1=1 and c2=0 since 1*P(1)+0*P(2) ≤ x ≤ 1*P(1)+(0+1)*P(1)
for n=3 c1=1, c2=0, and c3=2
since 1*P(1)+0*P(2)+2*P(3)≤ x ≤ 1*P(1)+0*P(1)+(2+1)*P(3)
)
Rewriting equation 4.7.2, we see that our choice of n’s actually makes partial sums of
FGCE’s. So for a given S, and an x in the interval [0,1), we can pick an n and ci’s that
satisfies, for x a real number:
n
n
i 1
i 1
 (ci * P(i ))  x   (ci * P(i ))  P(n)
Then,
n
0  x   ( ci * P(i ))  P( n )
i 1
n
As n approaches infinity, x   ( ci * P(i )) is squeezed in between zero and P(n), which
i 1
n
is known to converge to zero. So, x   ( ci * P (i ))  0 as n approaches infinity, which is
i 1
n
to say that
 (c * P(i ))  x as n approaches infinity.
i
i 1
Thus for the given x, there is an FGCE series which converges to it.
QED
Unlike the natural number expansions, we cannot show that FGCE’s for a given x and S
are unique. Since FGCE’s for real numbers are infinite series, they may have different
coefficients, provided that they simply converge to x. All that is needed is that the
expansions agree enough so that the difference between it and x is arbitrarily small. For
example in decimal, 1=0.9999999999999…
As a useful convention, we can define the proper FGCE as the terminating expansion
which is exactly equal to the number it represents after a finite number of terms. The
proper FCGE may not exist for some numbers, since the numbers may have an infinite
number of digits in the expansion to begin with, regardless of the base set S. For example
in decimal (S={1,10,10,10…}), the number π would not have a proper FGCE since there
is no finite sum of fractions which will equal π exactly.
4.8 Example
S={1,2,2,2,2…}
1/2= ( . 1,0,0,0….)S or 1/2=( . 0,1,1,1,1…) S
In this case ( . 1,0,0,0….)S is the proper expansion since after one term, the expansion
when evaluated, will give back the exact number it is meant to represent.
In regular decimal notation,
1 is a proper FGCE, as opposed to 0.9999999…
Armed with the definition of a proper FGCE, we can notice that the FGCE’s produces
some very interesting properties with rational numbers.
4.9 Examples:
Consider S = {1,2,3,4,5…}

e-2=(1 1 1 1..) S which is expected since e  1 / k!
k 0
Log(2) = (1 1 0 3 1 0 3 6 2 5 4 6 11..) S (pattern not apparent)
24/25 = (1 2 3 0 1 1 3 1 8 0 0 0 0..) S
1/7 = (0 0 3 2 0 6 0 0 0 0..) S
These although we saw only two example of rational numbers with terminating
expansions, we can still raise the natural question of whether all rational numbers have
terminating expansions for a specific base set.
4.10 Theorem: A rational number x=a/b with 0≤x<1 has a proper FGCE if and only if
there exists an i such that b divides x1*x2*...*xi with respect to the specific base set S.
Proof/
We take x=a/b with a and b in reduced form, so that a and b are relatively prime.
<=:
Assume there exists an i such that b divides x1*x2*...*xi
Then b*d= x1*x2*...*xi
a a*d
a*d
- = ----- = -------------- < 1
b b*d
x1*x2*...*xi
By lemma 4.5 , a FGCE divides the unit interval up into fractions with denominators of
x1*x2*...*xi for a given i, and a/b is a fraction of such a form. Therefore a/b has a finite,
terminating expansion of length i, and thus it has proper FGCE.
=>:
Assume that x has a proper FGCE.
Then
a
c1
c2
ci
x= - = -- + ------ +…. +------------b
x1 x1*x2
x1*x2*...*xi
By simply putting all the fractions in common denominator form and adding them, we
get
a
t
x = -- = -------------- < 1
b
x1*x2*...*xi
If t / ( x1*x2*...*xi) is in reduced form, then a=t and b=( x1*x2*...*xi), and so b divides
x1*x2*...*xi, since b divide itself.
If t / ( x1*x2*...*xi) is not in reduced form, then both t and ( x1*x2*...*xi) have some
common factor c>1. If we divide both t and ( x1*x2*...*xi) by this common factor, then
(t/c) and (( x1*x2*...*xi)/c) become relatively prime. But then this would mean that a=t/c
and b=(( x1*x2*...*xi)/c).
b=(( x1*x2*...*xi)/c) => b*c =( x1*x2*...*xi) => b divides ( x1*x2*...*xi)
In either situation b divides x1*x2*...*xi.
Of course this whole proof really depends upon the choice of S. A necessary condition
for S to give a specific fraction a terminating expansion is that it contains the factors of
the denominator of x. Thus an appropriate choice of S to make all fractions have
terminating expansions would be one in which all possible factors where included. Once
we have selected such a base, the above statements hold.
QED
This is an interesting result since it implies that there could be base set in which all
rational numbers between zero and one have a terminating expansion. We present such a
base set now. Consider S ={1,2,3,4,5…}. In this base set, there is always an i such that b
divides x1*x2*...*xi. This is so since all natural numbers are contained in the set and we
can pick the i such that xi = i =b. Then b divides x1*x2*...*xi, since b divides x1*x2*...*b.
Finally, we realize that every real number consists of an integer and decimal part. The
GCE allows all the integer parts to be represented, and the FGCE allows all decimal parts
to be represented. Thus the expansion of a real number can be thought of as a list of
coefficients for the GCE and FGCE with respect to two base sets. We can write for a real
number x:
x=( cn … c2 c1 . c1‘ c2‘ c3‘ ….)S, S’
with the S and numbers to the left of the decimal point representing a GCE and the S’ and
numbers to the right of the decimal point representing a FGCE. Since there are an
uncountable number of valid choices for the S’, then there are an uncountable number of
ways to represent the real numbers.
4.11 Example:
S=S’ = {1,2,3,5,7…primes of increasing order}
ππ = 36 + 0.46215960…
= ( 1 1 0 0 . 0 2 3 6 0 7 11 1 21 4 28 ….)S, S’ (no apparent pattern)
5 Conclusion
Having taken a brief excursion into the world of numerical representations, we recall that
some representation methods are common day, such as decimal, others are more exotic,
such as the factorial base, and still others have many practical applications, such as
binary. But question now becomes how general can we make our representations?
Donald Knuth (7.7) said that “π is 10 in base π”. Perhaps the generalized representations
presented here can be extended further to include rational and real numbers into the base
sets. These new representations could be very useful in digit related number theory. For
example, now that we have shown that there are bases in which all rational numbers have
terminating expansions, we can extend previously discrete and finite methods to include
rational numbers. The concepts of additive persistence and digital roots normally use
base 10 numbers, however, now since it has been shown there are an uncountable number
of bases to pick from, those concepts can be expanded. Such may be true of much in the
field of digit related number theory.
6 Thanks
Special Thanks to Ken McMurdy for constructive feedback, comments, and support that
helped make this paper possible.
7 References:
7.1 Friedberg, Insel, Spence. Linear Algebra. Third Edition. Prentice Hall. New Jersey
ISBN 0-13-233859-9
7.2 Johnsonbaugh, Richard. Discrete Mathematics. Second Edition. Macmillan
Publishing
Company. New York. ISBN 0-02-359690-2
7.3 Ore, Oystein. Number Theory and Its History. Dover Publications. Inc. New York.
ISBN 0-486-65620-9
7.4 Rosen, Kenneth H. Discrete Mathematics. Fourth Edition. McGraw Hill.
ISBN 0-07-289905-0
7.5 Spivak, Michael. Calculus. Third Edition. Publish or Perish Inc.
ISBN 0-914098-89-6
7.6 Weisstein, Eric. MathWorld. “Additive Persistence”
http://mathworld.wolfram.com/AdditivePersistence.html
7.7 Weisstein, Eric. MathWorld. “Base.” http://mathworld.wolfram.com/Base.html
Note: The reference to Knuth was taken off of this site, which in turn got the
reference from Knuth’s Book:
Knuth, D. E. "Positional Number Systems." §4.1 in The Art of Computer
Programming, Vol. 2: Seminumerical Algorithms, 3rd ed. Reading, MA: AddisonWesley, pp. 195-213, 1998
7.8 Weisstein, Eric. MathWorld. “Digital Root”
http://mathworld.wolfram.com/DigitalRoot.html
7.9 Weisstein, Eric. MathWorld. “Primorial”
http://mathworld.wolfram.com/Primorial.html
7.10 Wikipedia. “Dedekind Cut” http://www.wikipedia.org/wiki/Dedekind_cut