Exact Computer Calculations With Infinitely Repeating Decimals
May, 2001
Dr. Michael J. Bossé*
Department of Mathematics
Indiana University of Pennsylvania
Indiana, PA 15705
Dr. Fengshan Liu*
Department of Mathematics
Delaware State University
1200 N duPont Hwy
Dover, DE 19901
fliu@dsc.edu
(302) 857-6646
Dr. N. R. Nandakumar*
Department of Mathematics
Delaware State University
1200 N duPont Hwy
Dover, DE 19901
nnandaku@dsc.edu
(302) 857-6647
* Authors’ names appear in alphabetical order.
Exact Computer Calculations With Infinitely Repeating Decimals
ABSTRACT
Both computers and calculators are limited by architecture, operating system, and software, to some
predetermined level of precision within decimal number presentation and calculation. Rational fractions
produce either terminating or repeating decimal numbers. In decimal form, rational fractions which
produce repeating decimals cannot be represented exactly on computers or calculators. At some decimal
place value, the decimal number must be truncated. This process leads to approximations, rather than
exact numeric representation and calculation. Exact calculations can only be performed by computers
and calculators on terminating decimal numbers. This paper demonstrates that for any rational fraction
which produces a repeating decimal, there exists a base into which that number can be converted so that it
produces a terminating decimal. Therefore, through converting to appropriate bases computers and
calculators can provide exact representations and operations on newly formed terminating fractions.
1
Exact Computer Calculations With Infinitely Repeating Decimals
Preliminaries
Assignment #1:
Use a calculator or a computer to perform the following addition. Provide an exact
answer, not an approximation (no rounding or truncating of the decimal is allowed).
(Remember that 1.2345  1.234523452345... )
1.2345  1.23456
Discussion: In decimal form, it is impossible to enter decimal numbers with infinitely repeating cycles
into a calculator or computer. At some point, the number must either be rounded or truncated to some
predetermined decimal place value. However, this process will only produce an estimate of the exact
answer accurate to one digit less than the decimal place value to which the original numbers were
truncated. While computer and calculators may inadequately tackle such calculations, some paper and
pencil techniques are available to perform such tasks.
Technique 1: This technique simply expands each infinitely repeating cycle until the cycles in each
number commonly meet at the end of a cycle. In order to consider this we must introduce new notation
and operations. Let cl(x) = n be an operation which enumerates the cycle length of a repeating decimal


 
LCM cl 1.2345; cl 1.23456  LCM 4,5  20 .
number. Thus cl 1.248  3 and cl 3.5798  4 . In our example,
Therefore, in order to accurately calculate the sum presented above, both 1.2345 and 1.23456 must be
written to 20 decimal places in order for the cycles to terminate at a common decimal place. This
produces:
1.23452345234523452345...
 1.23456234562345623456...
. Thus, 1.2345  1.23456  2.46808579796869075801 .
2.46908579796869075801...
Notably, this sum is again an infinitely repeating decimal cycle and is exact.
Technique 2: The numbers discussed within this investigation are of the form of either terminating
or repeating decimals. Let us review the two definitions of the rational numbers.
1. Q = {p/q: p, q  Z, q  0}, which means that all rational numbers can be expressed as a fraction
of integers, with a non-zero denominator.
2. Q = {x| x is a terminating or repeating decimal}. Terminating decimals terminate at some
2
decimal place value. Repeating decimals include some infinitely repeating cycle of digits.
A simple procedure allows repeating decimals to be represented as fractions and provides a method
which can produce an exact answer to our original problem: 1.2345  1.23456 . We shall first convert
these repeating decimals to fractional form.
10,000 x  12,345.2345
x 
1.2345
.
9,999 x  12,344.000
100,000 y  123,456.23456
y 
1.23456
.
99,999 y  123,455.0000
Thus, x 
12,344
.
9,999
Thus, y 
123,455
99,999
Since 9,999 = 9(1,111) and 99,999 = 9(11,111), the LCM(9,999; 99,999) = 9(1,111)(11,111).
Thus,
12,344 123,455  12,344 11,111   123,455 1,111  274,312,689




.


9,999 99,999  9,999 11,111   99,999 1,111  111,098,889
Discussion: Interestingly, due to the length of the repeating cycles of decimals in Technique 1 and the
precision of the fractions in Technique 2, most calculators and computers will not accurately produce
these results. However, cumbersome paper and pencil techniques often successfully accomplish tasks
difficult or impossible for technology. The question then arises, can technology be modified to accurately
handle infinitely repeating decimals? The response is a resounding “yes.” We consider this as follows.
Terminating Fractions
We define terminating fractions as fractions (rational numbers) which can be rewritten as decimals which
terminate. For the remainder of this investigation, we will consider
n
, where n and d are relatively
d
prime positive integers. Since this study will consider numeration systems in different bases, b will be a
positive integer. In base 10, the portion of the number to the right of the decimal point is commonly
named the “decimal” part of the number. To allow generic discussions of base b, this investigation will
title the portion of the number to the right of the “fraction point” the “bessimal” part of the base b
number. All exponents will be recognized as non-negative integers.
Base 10.
In base 10, a rational number is defined as
3
n
 i w 10 w  i w1 10 w1  ...  ii 10 j  ...  i 0  f 1 10 1  f 2 10  2  ...  f k 10 k  ... ,
d
where n and d are relatively prime non-negative integers, and i j and f k are integers less than 10. Can
it be predicted, in base 10, which fractions will produce terminating and repeating decimals? We offer
here a theorem for base 10 which will be proven in discussions regarding generic base b.
Theorem 1.
For
n
m
m
in lowest form in base 10, if d  2 1  5 2 , where m1 and m2 are nond
negative integers, then
n
d
will produce a terminating decimal.
All other
denominators will result in decimal forms with infinitely repeating cycles.
Proof: Special case of Theorem 3.
Examples: The following rational numbers produce terminating decimals:
3
3
 2
 0.15
20 2  5
97
97
 4 3
2000 2  5
197 197

 0.3152
625 5 4
The following rational numbers produce decimal numbers with infinitely repeating
decimal cycles:
37
37
 2
 0.123
300 2  3  5 2
Theorem 2.
For
n
d
61
61

 0.17428571
350 2  5 2  7
9
 0.81
11
m
m
in lowest form in base 10, if d  2 1  5 2 , where m1 and m2 are non-
negative integers, the number of decimal places within the decimal number produced
by
n
is max[m1, m2].
d
Proof: Special case of Corollary 1.
Examples:
49
49
 4 2  0.1225 ; max(4, 2) = 4; four decimal places.
400 2  5
17
17
 3 6  0.000136 ; max(3, 6) = 6; six decimal places.
125000 2  5
Base b.
Definition 1.
Let integer b > 0 be the base of a number system. Then a rational number is
n
 i w  b w  i w1  b w1  ...  ii  b j  ...  i 0  f 1  b 1  f 2  b  2  ...  f k  b k  ...
d
4
where n and d are relatively prime non-negative integers, i j and f k are integers less than b.
Definition 2.
A fraction is called a terminating fraction in base b if there is an integer T such that
f k = 0 for all k > T in Definition 1. T is called the fraction length.
Example:
For the hexidecimal number 4.AF36C200000..., 2 is in the position k = 6. However,
for all positions such that k > 6, f k = 0. Thus, this number terminates.
The following conditions hold for each of the ensuing theorems:
n
will denote a fraction within a
d
becimal system where n and d are relatively prime positive integers.
Lemma 1. A positive integer multiple of a terminating fraction is also a terminating fraction.
Proof obvious.
Lemma 2. Given any fraction
n
, if it terminates in a base b, then it terminates in bases of positive
d
integer multiple of b.
Proof: Since
n
n
terminates in base b, b T  must be an integer, where T is the fraction length in
d
d
base b. Let k be a positive integer. Thus kb 
T
therefore,
n
n

 k T   b T   is an integer and,
d
d

n
also terminates in base kb.
d
Theorem 1.Given any fraction
n
n
n
n
, let d  p1 1 p2 2  ps s , where p1, p2, ..., ps are prime factors
d
with powers n1 , n 2 , n s  1respectively. Then the fraction
n
is a terminating fraction in a base of a
d
positive integer multiple of p1  p2  ps . The minimal base for this is p1  p2  ps .
Proof: Case 1. n = 1.
Let N=max{ n1 , n 2 , n s }.
( p1  p2  ps ) N  d  p1
N n
N n
N n1
p2
N n2
 ps
N ns
N  ns
1 p1 1 p2 2  ps

d
( p1  p2  ps ) N
5
Since p1
p1
N  n1
N n1
p2
p2
N  n2
N n2
 ps
 ps
N  ns
N ns
 ( p1  p2  ps ) N , we can write
 f1  ( p1  p2  ps ) N 1    f N 1  ( p1  p2  ps )  f N
where integers f1 ,, f N 1 , f N are between 0 and
p1  p2  ps .
Therefore,
1 f1  ( p1  p 2  p s ) N 1    f N 1  ( p1  p 2  p s )  f N

d
( p1  p 2  p s ) N

f N 1
fN
f1


.
N 1
p1  p 2  p s
( p1  p 2  p s )
( p1  p 2  p s ) N
Case 2. General Case, numerator n is not necessary 1.
By Lemma 1,
n
is also a terminating fraction. By Lemma 2, is also a terminating fraction in bases of
d
integer multiple of p1 , p2 ,, ps .
Examples:
bases
1
1
is a terminating fraction in bases 2  3 and 2 2  3 . 3 1 is a terminating fraction in
1
2 3
2 3
1
23  3
and
25  32 .
1
is a terminating fraction in bases
2  3  52  7 3
3
2
2  3  5  7 11and
23  32  52  73  11 .
Corollary 1.
For terminating fraction
n
in lowest form in base d  p1 , p2 ,, ps , when
d
d  p1 1 , p2 2 ,, ps s , the number of becimal places within the number in becimal form is
m
m
m
max[ m1 , m2 ,, ms ] . See proof above.
Corollary 2.
If there exists some ps  qs , then
n
, produces a bessimal with an infinitely
d
repeating cycle. Proof: Obvious.
Corollary 3.
The minimal base for a terminating fraction,
bmin  p1  p2  p3  ps .
6
n
m
m
m
, where d  p1 1 , p2 2 ,, ps s , is
d
Theorem 2.
Proof:
If
n
is a terminating bessimal in base b, then d = kb, where k  N .
d
n
 iw  b w  iw 1  b w 1    i j  b j    i0  f1  b 1  f 2  b  2    f k  r  k
d
Multiplying both sides by b k produces,
bk 
n
d
 b k  (iw  b w  iw 1  b w 1    i j  b j    i0  f1  b 1  f 2  b  2    f k  r  k ) .
bk 
n
d
 iw  b w  k  iw 1  b w  k 1    i j  b j  k    i0  b k  f1  b k 1  f 2  b k  2    f k
Since the right hand side of this equation is an integer, b k 
divisible by d  p1 1 , p2 2 ,, ps
m
m
ms
n
must also be an integer. Notably, b k is
d
. Since p1 , p2 ,, ps are all distinct prime numbers, b k must be
divisible by p1 , p2 ,, ps . Finally, b must be a multiple of p1 , p2 ,, ps .
1


would be a terminating fraction in any of the following bases:
3
2
4
3
 2 3 5 7 
Examples: 
23  34  54  7 4
11  23  32  54  73
500  25  35  55  7 4
Applications
It has been shown that for any rational number, there exists a base b in which
fraction. For instance, in base 10 the rational number
n
will be a terminating
d
1
 0.3 is a non-terminating decimal, whereas
3
1
   0.13 , which is a terminating bessimal. While this paper does not consider techniques for
 3 10
converting fractions from one base to another, this investigation will demonstrate how to make the
selection of bases for which any
n
can be translated into a terminating fraction.
d
Base Selection for Terminating Fractions.
Example. Consider the rational number
5
 0.416 .
12
demonstrate that any of the following bases would convert
7
2
Since 12  2  3 , the previous theorems
5
into a terminating fraction:
12
23  6;
2 2  3  12 ;
2 2  32  36 ;
22  3  7  84 ;
or
k  2 m  3n , where k, m, n are positive integers.
Assignment #2a.
Consider various bases through which adding or multiplying
17
 0.283 with
60
69
 0.19714285 produces exact results.
350
Since
60  22  3  5
and
350  2  52  7 ,
bases
k1  2m1  3n1  5 p1
( ka , ma , na , pa , qa are positive integers) are appropriate conversions for
Thus, an appropriate base for converting both
17
60
and
and
k2  2m2  5 p 2  7 q 2
69
respectively.
350
17
69
and
to terminating fractions could be
60
350
k3  2m3  3 p3  5 p3  7 q3 , where m3  min( m1 , m2 ), n3  n1, p3  min( p1, p2 ) , and q3  q2 .
Assignment #2b.
Find the smallest base in which both
17
69
and
terminate.
60
350
The smallest base which would be appropriate for this problem would be 2  3  5  7  210 .
Computer Applications.
Clearly, non-terminating decimal numbers are problematic for computers and calculators.
All
truncating of decimals produces approximations rather than exact results. However, terminating decimals
allow for computers and calculators to produce exact mathematical calculations. This investigation
demonstrates that any rational number may be converted into another base in which the bessimal will be
terminating. Additionally, although not in the general case, Assignment 2 evidences that an appropriate
base can be found to perform exact calculations on multiple rational numbers.
While this all seems quite simplistic, the nature of computer architecture introduces additional caveats
and concerns. Computer architecture allows for convenient interchange between bases 2, 8, and 16 (and
with some additional but easy transformations into any base 2 n , where n is a positive integer). However,
algorithms for conversions to other bases are somewhat more complex. Furthermore, while conversions
to other bases are possible, they are extremely artificial. Although computers may mimic arithmetic in
bases other than 2 n , current technological architecture requires that all mathematics within computers
and calculators actually be performed in base 2. Hence, to fully accomplish the conversion of rational
numbers into terminating bessimals in other bases, this discussion assumes the existence of a virtual
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computer which can perform operations in any base.
Example. Quadratic equations ax 2  bx  c  0 with rational coefficients a, b, and c have rational
roots when the discriminant b 2  4ac is a perfect square over rationals. However, since many rational
numbers have non-terminating decimal expansions, when calculators and computers are employed to
compute the roots using the quadratic formula, derived results are often approximations rather than exact
results. Through the discussions above, expressing the coefficients of a quadratic expression in a suitable
base and computing the roots in this base using quadratic formula yields exact values.


Consider the equation x 2   7 
1
 x  1  0 . When the roots are computed using a calculator we obtain
7
x=7, 0.143 rounded to 3 decimal places (an approximation). Certainly, the smaller root is not exact.
Increasing the accuracy to more decimal place values does not produce exact results and only continues to
produce approximations.
However, by expressing the coefficients in base 7 we obtain,
x 2  10.17 x  17  0 . Thus,
 b  b 2  4ac
2a
 10.1  10.12  4 




2


x
7
 10.1  102.1  4 

 

2

7
 10.1  65.01 

 

2


 10.1  6.6 


2

7
Hence,
 10.1  6.6   10.1  6.6 
x
 ,

2
2

7 
7
 20.0   0.2 

 ,

 2 7  2 7
 107 , 0.17
1
 7 10 ,  
 7 10


1
7
Therefore, although a computer could not calculate exactly the roots of x 2   7   x  1  0 , exact
9
roots were found for x 2  10.17 x  17  0 . Thus, by converting from one base to another, exact
mathematics can be readily performed and inherent technological weaknesses can be averted.
Background Reading
Dickson, L. E. (1934). History of the Theory of Numbers, Volume I, Divisibility and Primality. New
York: G. E. Stechert & Co.
Mathews, J. H. & Fink, K. D. (1999). Numerical Methods Using MATLAB (3rd Ed.). Upper Saddlr
River, NJ: Prentice Hall
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