CC BY
43
UDC 511.11 DOI 10.24147/2222-8772.2021.1.46-59
EGYPTIAN FRACTIONS AS APPROXIMATORS
Olga Kosheleva
Ph.D. (Phys.-Math.), Associate Professor, e-mail: olgak@utep.edu
Vladik Kreinovich Ph.D. (Phys.-Math.), Professor, e-mail: vladik@utep.edu
University of Texas at El Paso, El Paso, Texas 79968, USA
Abstract. In ancient Egypt, fractions were represented as the sum of inverses to natural numbers. Processing fractions in this representation is computationally complicated. Because of this complexity, traditionally, Egyptian fractions used to be considered an early inefficient approach. In our previous papers, we showed, however, that the Egyptian fractions actually provide an optimal solution to problems important for ancient Egypt - such as the more efficient distribution of food between workers. In these papers, we assumed, for simplicity, that we know the exact amount of food needed for each worker - and that this value must be maintained with absolute accuracy. In this paper, we show that the corresponding food distribution can become even more efficient if we make the setting more realistic by allowing "almost exact" (approximate) representations.
Keywords: Egyptian fractions, approximate representation of fractions, optimal representation, representation accuracy.
1. Formulation of the Problem
Egyptian fractions: reminder. In ancient Egypt, fractions were represented as a
sum of fractions of the type —, with the smallest possible number of terms; see,
n
5
e.g., [1-4,11] and references therein. For example, — was represented as
A = i + i ID
12 3 12 w
Traditional history-of-mathematics view on Egyptian fractions. Dealing with such fractions is not computationally easy: e.g., multiplying two 3-term Egyptian fractions would generate the sum of 3 ■ 3 = 9 term-by-term products, and we face a complex problem of how to find the representation of this product that uses the smallest number of terms. Because of this complexity, books on history of mathematics usually dismiss Egyptian fractions as a not-very-effective approach to representing fractions; see, e.g., [1,4].
What we showed in our previous paper: a reminder. In our paper [6], we showed that Egyptian fractions actually provide an optimal solution to a practical problem described in many papyri from ancient Egypt: how to divide loaves of bread between multiple workers; see also [5,7-10].
Indeed, it is easy and fast to divide a rectangular-shaped bread into n equal pieces along its long side: this requires n — 1 cuts. So, if want to give each worker 5/12 of a loaf, a straightforward idea is to divide each loaf into 12 parts and give 5 such parts to each worker. In particular, if we have 12 workers, then we take 5 loaves and divide each of them into 12 parts. This requires 11 cuts per loaf, to the total of 5 ■ 11 = 55 cuts.
We can do all this cutting much faster if we take into account representation (1). For 12 workers, this means that we can:
• take 12■ - = 4 loaves and divide each loaf into 3 pieces; this requires 3 —1 = 2
3
cuts per loaf, to the total of 4 ■ 2 = 8 cuts, and
• take 12 • — = 1 loaf and divide it into 12 — 1 = 11 cuts.
12
Overall, we need 11 + 8 = 19 cuts, which is much fewer than the original 55 cuts.
P
A general description of what is optimal. Each fraction - can be represented
Q
in many ways as a sum of different fractions, and each fraction — in this sum can
n
be represented as
m 1 1 / . . s
— = —+ ... +— (m times). n n n
p 1 Thus, we can represent each fraction - as a sum of fractions of the type —:
q n
P 111
*- = — + — + ... + —. (2) q ni n2 nk
Without losing generality, we can assume that the integers ni are listed in increasing order, i.e., that ni ^ n2 ^ ... ^ nk.
Let N be the least common multiple of all the values ni. Then, the representation (2) means that to feed N workers, for each i from 1 to k, we take N ■ —
Hi
loaves and cut each into ni pieces. This requires ni — 1 cuts per loaf, to the total of
N ■ - ■ (m — 1) = N ■ (1 — . ni V niJ
Thus, the overall number of cuts is
N ■ (1 — — ) + N ■ (1 — — ) + ... + N ■ (1 — — ) = V nij V n2J V nkJ
n,
So, the number of cuts per worker is
4 -1 =Nik - s)
k j i+i+...+1)=k - *
\n 1 n2 nkJ q
Thus, to minimize the number of cuts, we need to minimize the number k of terms in the representation (2) - which is exactly what the Egyptian fractions do!
Comment. Ancient Egyptians had additional restrictions on representation of type (2): e.g., they required that all denominators ni be different. This additional requirement is not related to our optimality result, so we do not consider it in this paper.
By the way, it would be interesting to understand the motivation behind this additional requirement.
What we did in our previous paper and what are the remaining problems. In
our previous paper, we describe an algorithm for computing the shortest possible representation of a fraction in the form (2). From the theoretical viewpoint, this solves the problem - although, of course, it is always desirable to look for a more efficient algorithm.
The remaining problem is that for some fractions, the number of cuts is still too large. The only reason why we need this many cuts is that we wanted to exactly represent the original fraction. But from the practical viewpoint, it may be beneficial to make slightly larger bread portions - we will spend slightly more money on bread but save on cuts. In other words, in addition to the problem of representing fractions in the form (2), we need to consider the problem of approximating the fractions by expressions of type (2).
Let us formulate this approximation problem in precise terms.
Formulating the approximation problem in precise terms. We are given two positive numbers:
• the price b of a loaf of bread, and
• the per-cut wages c that we need to pay the bread cutters.
We are also given the fractional part f e (0,1) of a loaf that needs to be given to each worker. What we then need to do is minimize the overall extra expenses: i.e., the cutting expenses plus extra-bread expenses per worker. In other words, we need to find, among the tuples (n1,..., nk) for which
/ ^- + - +... + -, (3)
n1 n2 Uk
the tuple for which the overall expenses
c (k- (1 + ! + +1)) + 6.(1 +1 + ... +1 - A (4)
V W n2 nk) ) \ni n2 Uk J
are the smallest possible.
What we do in this paper. In this paper, we describe an algorithm for solving the above optimal approximation problem.
2. Solution to the Problem
Let us first simplify the objective function. The above expression (4) can be
represented as
c ■ k + (b - c) ■( — + — + ... + — - A + c ■ f. (4a)
\ni n2 nk J
Adding a constant c ■ f to all the values of the objective function does not change which values are larger and which are smaller. Thus, minimizing the expression (4a) is equivalent to minimizing the expression
c ■ k + (b - c) ■( — + — + ... + — - A. (4b)
\ni n2 nk J
Similarly, dividing all the values of the objective function by the same constant b-c does not change wich value is larger and which value is smaller. So, minimizing the expression (4b) is equivalent to minimizing the expression (4b) divided by b - c, i.e., the expression
r ■ k + ( — + — + ... + — - A, (4c)
\ni n2 nk J
where we denoted r =f ° .
b - c
To solve the corresponding problem, it is sufficient to solve an auxiliary problem for several values k: idea. Our idea is that for each value k = 1,2, ■, we find the values n1,...,nk that minimize the expression (4c). Then, we find k for which the corresponding minimum is the smallest.
For each k, the term r ■ k does not depend on the choice of ni, so minimizing the expression (4c) is equivalent to finding the values n1,... ,nk that minimize the difference
- + - +... + - - /; (5)
Hi U2 nk
(provided that this difference is non-negative).
Let dk denote the smallest possible value of this difference corresponding to the given value k (we assume that for the given value k, there exist integers ni that satisfy the condition (3)). Then, the smallest possible value of the expression (4c) for given k is equal to
ek =f r ■ k + dk. (6)
Thus, once we know all the values dk, we must find the value k that minimizes the expression (6).
Each value f can be approximated, with any given accuracy, by a rational number, and each rational number can be represented, for some k, in the form (2). Thus if we approximate f with approximation error < r, then, for some k, we will get dk < r and thus, ek < r • k + r = r • (k + 1). Once we reach the value k for which dk < r, considering larger values k does not make sense: already the first term in the expression (6) will be larger than the current value ek. So, we arrive at the following reduction algorithm.
Reduction: algorithm. To solve the original optimization problems, for k = 0,1, 2,... we compute dk and ek. We stop when dk < r. We then:
• select the value k for which ek is the smallest, and
• for the selected value k, we find the values ni,...,nk that minimize the difference (5).
Now, all we need to do is to show how to compute dk.
Proposition 1. There exists an algorithm that, given f and k, computes the smallest possible value dk.
Proof. Since ni * n2 * ... * nk, we have
1 1 1
— * ... * — * —, nk n2 ni
so the inequality (3) implies that
f < k--, ni
which is equivalent to
k
ni * 7'
Thus, we need to consider only finitely many values ni. For each of these values ni, (3) implies that
i - ^ * ^ + + ^
ni n2 nk
hence
f -1 * (k -1)^ ^ ni n2
and
k -1 n2 * —
f -1
ni
(If the difference is 0, then di = 0, so further computations are not needed.) Thus, for each ni, we only need to consider finitely many values n2.
In general, once we have selected the values ni,... ,ni; then (3) implies that
hence
and
(1+- + -1)
\ni m)
(
l l
« — +... + —
ni+1 nk
f — ( 1 +... + y « №-0
ni+i
ni+1
«
k — i
f— I ± +... + -i
ni ni
So overall, we need to consider a finite set of possible tuples (ni,... ,nk). We then select the tuple for which the difference (4) is the smallest. This is all we need to do to compute the desired smallest value dk. The proposition is proven.
How accurately can we thus represent a number? A natural question is: what accuracy can we achieve by such an approximation? In this paper, we provide the answer to this question for k =1, k = 2, and k = 3.
Definition 1. For each positive integer k, by an accuracy ak of approximation by a ^ fc-term Egyptian fraction, we mean the smallest number ak for which: for every number f e (0,1), there exist integers ni,...,nj, j ^ k for which
f « — + ... + — « f + ak,
ni
ni
i.e., for which
0 « — + ... + — - f « ak.
ni
n^
Proposition 2. For k = l, the accuracy a1 of approximation by a « l-term Egyptian fraction is a1 = -.
smallest possible value n
Proof. The largest possible value of the expression 1 < l corresponds to the
n
2. Thus, all values f > - have to be approximated from above by the number 1 (in this case, no cuts are needed). For each e > 0, for the value f = - + e, the difference l — f is equal to - — e, thus a1 cannot be
smaller than ^.
2
The value a1 = - satisfies the desired condition:
• for values f ^ 1, we take ni worked half of a loaf), and
2 (i.e., we cut each loaf in half and give each
l
• for values f > 1, we take ni = 1 (i.e., we give each worker the whole loaf and do not cut anything at all).
The proposition is proven.
Proposition 3. For k = 2, the accuracy a2 of approximation by a * 2-term
Egyptian fraction is a2 = -.
6
Proof. Let us first find the largest possible value f < 1 that can be represented as
f = ^ + ^
Ui U2
In general, we could have ni = 2 or ni ^ 3.
1 1 1 1
• For ni = 2, to have--1--< 1, we must have — < -, i.e., we must have
ni U2 U2 2
n2 > 2. The largest value of this fraction corresponds to the smallest value
of n2 that satisfies this inequality, i.e. to the value n2 = 3. In this case,
1 1 _ 1 1 _ 5
ni n2 2 3 6
• For ni ^ 3, due to ni * n2, we have
11112 — + — * o + 0 = 0 .
ni n2 3 3 3
25 The upper bound - for cases ni ^ 3 is smaller than the value - achievable for
36 ni = 2. Thus the largest values f < 1 that can be represented by a * 2-term
Egyptian fraction is -.
6
5
So, each value f which is larger than - has to be approximated by number 1.
6
Hence, the value a2 cannot be smaller than the differences
1 - (5+«)=6 - -
for any e > 0. Thus, we must have a2 ^ 1.
6
To complete the proof, we need to show that every number f from the interval (0,1) can indeed be approximated by a * 2-term Egyptian fraction with accuracy
-. Indeed:
6
• values v e (0, -6
can be approximated by 1;
6
values v e -, -
1 1
6, 3
• values v e -,-
1 1 3, 2
values v e -, -
1 2 2, 3
• values v e -,-
2 5
3, 6
5
can be approximated by -;
3
can be approximated by -;
can be approximated by -
3
1 1 ;
2 + 6;
can be approximated by 5 = 1 + 1;
• values v e ( -, 1 ) can be approximated by 1.
Thus indeed, a2
-. The proposition is proven. 6
Proposition 4. For k = 3, the accuracy a3 of approximation by a * 3-term Egyptian fraction is a3 = —.
Proof. Let us first find the largest possible value f e (0,1) that can be represented
111
as the sum--1---1--of three inverses.
Hi U2 n3 In general, we can have ni = 2 or ni > 2.
If ni = 2, then we cannot have n2 = 2 - this would lead to the sum equal to 1. Thus, we must have n2 ^ 3.
• If n2 = 3, then to get the sum smaller than 1, we must have n3 > 6. The smallest such value is n3 = 7 for which
1 1 1 41
- + - + - = — = 0.976...
2 3 7 42
By the way, for n3 = 8, we get
1 1 1 23
- + - + - = — = 0.958...
2 3 8 24
• If n2 = 4, then to get the sum smaller than 1, we must have n3 > 4. The smallest such value is n3 = 5 for which
1 1 1 19
2 + 4 + 5 = 20 = 0
This is smaller than the previous value, so the largest representable f cannot be reached for this value of n2.
• If n2 ^ 5, then we have n3 ^ n2 ^ 5, hence
1 1 11119 — + — + — * - + - + - = — = 0.9, ni n2 n3 2 5 5 10
which is also smaller than what we had for n2 = 3.
If ni = 3, then we can have n2 ^ 3.
• If n2 = 3, then, to get the sum smaller than 1, we must have n3 > 3. The smallest such value is n3 = 4 for which
1 1 1 11 - + - + - = — = 0.916 ...; 3 3 4 12 '
this is smaller than what we had earlier;
• If n2 ^ 4, then n3 ^ 4, hence
1 1 1 1115 - + - + - ^ - + - + - = - = 0.833..., ni n2 n3 3 4 4 6
also smaller.
Finally, if we have ni ^ 4, then n2 ^ 4, n3 ^ 4, hence
1 1 1 1113
— + — + — +7+7=7= 0.75, ni n2 n3 4 4 4 4
still smaller.
41
So, the largest number representable as a sum of three Egyptian terms is —.
Thus, every largest number has to be approximated by 1, so a3 cannot be smaller than
41 1
1 - — = — = 0.0238...
42 42
Let us prove that can indeed approximate any value f e (0,1) with this accuracy. Most of the numbers can be approximated with an even better accuracy —:
60
• values v e (0, — 60
can be approximated by ^;
• values v e ( —, — 60' 60
can be approximated by — = —; FF J 60 30
• values v e ( —, — 60' 60
can be approximated by — = —; FF J 60 20
• values v e ( —, — 60' 60
41
can be approximated by — = —;
60 15
• values v e ( —, — 60' 60
can be approximated by 5 = 1;
56
• values v e —, — 60' 60
can be approximated by — = —;
60 10
• values v e
• values v e
• values v e
• values v e
• values v e
• values v e
• values v e
• values v e
• values v e
• values v e
• values v e
• values v e
• values v e
• values v e
• values v e
• values v e
• values v e
6 7
60 60
7 8
60 60
8 9
60 60
9 10
60 60
10 11
60 60
11 12
60 60
12 13
60 60
13 14
60 60
14 15
60 60
15 16
60 60
16 17
60 60
17 18
60 60
18 19
60 60
19 20
60 60
20 21
60 60
21 22
60 60
22 23
60 60
7 60 = 1 1 ; 10 + 60 '
8 2 1 1 " 10 + 30
60 = 15 =
9 3 1 1 " 10 + 20
60 = 20 =
10 1 '
60 = 6 '
11 60 = 1 6 + 1 ' 60 '
12 1 '
60 = 5 '
13 60 = 1 5 + 1 ' 60 '
14 7 1 = 5 + 1 '
60 = 30 = 30 '
15 1 '
60 = 4 '
16 4 1 = 4 + 1 '
60 = 15 = 60 '
17 60 = 1 4 + 1 ' 30 '
18 3 1 = 4 + 1 '
60 = 10 = 20 '
19 60 = 1 4 + 1 ' 15 '
20 1 ;
60 = 3 '
21 7 1 = 3 + 1 '
60 = 20 = 60 '
22 11 1 = 3 + 1 '
60 = 30 = 30 '
23 60 = 1 3 + 1 ' 20 '
• values v e
• values v e
• values v e
• values v e
• values v e
• values v e
• values v e
• values v e
• values v e
• values v e
• values v e
• values v e
• values v e
• values v e
• values v e
• values v e
• values v e
23 24
60 60
24 25
60 60
25 26
60 60
26 27
60 60
27 28
60 60
28 29
60 60
29 30
60 60
30 31
60 60
31 32
60 60
32 33
60 60
33 34
60 60
34 35
60 60
35 36
60 60
36 37
60 60
37 38
60 60
38 39
60 60
39 40
60 60
24 2 1 1 ' 3 + 15 '
60 = 5 =
25 5 1 1 ' " 5 + 12 '
60 = 12 =
26 13 1 1 ' " 3 + 10'
60 = 30 =
27 9 11 = 4 + 5 '
60 = 20 =
28 7 1 1 " 3 + 10
60 = 15 =
29 60 = 1 3 + 1 1 ' 10 + 20 '
30 1 '
60 = 2 '
31 60 = 1 2 + 1 ' 60 '
32 8 1 1 ' " 2 + 30'
60 = 15 =
33 11 1 1 ' " 2 + 20'
60 = 20 =
34 17 1 1 ' " 2 + 15'
60 = 30 =
35 7 1 1 ' " 2 + 12 '
60 = 12 =
36 3 1 1 ' 2 + 10 '
60 = 5 =
37 60 = 1 2 + 1 1 ' 10 + 60'
38 19 1 1 " 2 + 10 '
60 = 30 =
39 13 1 1 " 2 + 10 '
60 = 20 =
40 2 1 1 ' 2 + 6 '
60 = 3 =
• values v e
• values v e
• values v e
• values v e
• values v e
• values v e
• values v e
• values v e
• values v e
• values v e
• values v e
• values v e
• values v e
• values v e
• values v e
• values v e
• values v g
40 41
60
41 60
42 60
43 60
44 60
45 60
46 60
47 60
48 60
49 60
50 60
51 60
52 60
53 60
54 60
55 60
56 60
60
42' 60
43' 60
44' 60
45' 60
46' 60
47' 60
48' 60
49' 60
50' 60
51' 60
52' 60
53' 60
54' 60
55' 60
56' 60
60
can be approximated by
can be approximated by
41 1
- — - +
60 2
42 7
60 = 10 =
43 1
— = - +
60 2
44 11
60 = 15 =
45 3
60 = 4 =
46 23
60 = 30 =
47 1
— — - +
60 2
48 4
60 = 5 =
49 1
— — - +
60 2
50 5
60 = 6 =
51 17
60 = 20 =
52 13
60 = 20 =
53 1
— — - +
60 2
54 9
60 = 10 =
55 11
60 = 12 =
56 14
60 = 15 =
57 19
60 — 20 =
4 _ 1 1 1 '
5 = 2 + 5 + 10 '
^ ^ '
. 57 23' •values (60 ,24
• . , U 23 1 1 1 can be approximated by — = - + - + - ;
24 2 3 8
/23 41' • values v e — , — 24 42
41 1 1 1 can be approximated by — = - + - + - ;
/41 \
• values v e ( — , 1 1 can be approximated by 1.
The proposition is proven.
3. Conclusion
Ancient Egyptians represented each fraction as a sum of inverses of integers,
e.g., - was represented as - + -. In our previous paper, we showed that such repre-
6 2 3
sentations correspond to the optimal solution to a problem that is mentioned several
times in Egyptian papyri: how to divide bread loaves between workers. Egyptian
fractions correspond to the smallest number of cuts needed for this division.
Sometimes, however, the attempt to provide the exact amount of bread to each
worker leads to too many cuts. In many such cases, if we give every worker a
little bit more bread, we will spend more on bread but our overall expenses will
be lower, since we will need fewer cuts and thus, we will need to hire fewer bread
cutters. In this paper, we show how to find the solution that minimizes the overall
expenses. Depending on how many cuts per worker we allow, we can make sure
that the resulting portion of a loaf is close to the original one. For example, if we
allow 3 cuts, we can get the accuracy of — « 2.5%.
Acknowledgments
This work was supported in part by the National Science Foundation grants 1623190 (A Model of Change for Preparing a New Generation for Professional Practice in Computer Science), and HRD-1834620 and HRD-2034030 (CAHSI Includes).
References
1. Boyer C.B. and Merzbach U.C. A History of Mathematics. Wiley, New York, 1991.
2. Eppstein D. Egyptian fractions website. URL: http://www.ics.uci.edu/$\ sim$eppstein/numth/egypt/.
3. Gardner M. Puzzles and number-theoretic problems arising from the curious fractions of Ancient Egypt. Scientific American, October 1978.
4. Hoffman P. The main who loved only numbers: the story of Paul Erdos and the search for mathematical truth. Hyperion, New York, 1998.
5. Kosheleva O. and Kreinovich V. Egyptian fractions revisited. Abstracts of the 2005 Meeting of the Southwestern Section of the Mathematical Association of America (MAA), April 1-2, 2005, p. 6.
6. Kosheleva O. and Kreinovich V. Egyptian fractions revisited. Informatics in Education, 2009, vol. 8, no. 1, pp. 35-48.
7. Kosheleva O., Kreinovich V., and Zapata F. Egyptian fractions re-revisited. Russian Digital Libraries Journal, vol. 22, no. 6, pp. 763-768.
8. Kosheleva O. and Lyublinskaya I. Teaching Fractions with the Help of Egyptian Papyrus and Technology. Abstracts of the Teachers Teaching with Technology T3 Regional Conference "Using Technology to Engage Students in Discovery Learning", Staten Island, New York, November 3-4, 2006.
9. Kosheleva O. and Lyublinskaya I. Can Egyptian papirus enrich our students' understanding of fractions? Abstracts of the Annual Meeting of the National Council of Teachers of Mathematics NCTM "Mathematics: Representing the Future", Atlanta, Georgia, March 21-24, 2007, p. 40.
10. Kosheleva O. and Lyublinskaya I. Using innovative fraction activities as a vehicle for examining conceptual understanding of fraction concepts in pre-service elementary teachers mathematical education. In: T. Lamberg and L. R. Wiest (Eds.), Proceedings of the 29th Annual Meeting of the North American Chapter of the International Group for the Psychology of Mathematics Education PME-NA 2007, Stateline (Lake Tahoe), Nevada, October 25-28, 2007, University of Nevada Publ., Reno, 2007, pp. 36-38.
11. Streefland L. Fractions in Realistic Mathematics Education: A Paradigm of Developmental Research. Kluwer Academic Publishers, Dodrecht, The Netherlands, 1991.
ЕГИПЕТСКИЕ ДРОБИ КАК ПРИБЛИЖЕНИЯ
О. Кошелева
к.ф.-м.н., доцент, e-mail: olgak@utep.edu
В. Крейнович
к.ф.-м.н., профессор, e-mail: vladik@utep.edu
Техасский университет в Эль-Пасо, Эль-Пасо, Техас 79968, США
Аннотация. В Древнем Египте дроби представляли как сумму обратных натуральных чисел. Но работа с дробями в таком представлении вычислительно сложна. Из-за этой сложности египетские дроби традиционно считались ранним неэффективным подходом. Однако, в наших предыдущих статьях мы показали, что египетские дроби на самом деле обеспечивают оптимальное решение задач, важных для Древнего Египта - таких как более эффективное распределение еды между рабочими. В этих работах мы для простоты предположили, что знаем точное количество еды, необходимое для каждого рабочего, и что это значение должно поддерживаться с абсолютной точностью. В этой статье мы показываем, что соответствующее распределение еды может стать еще более эффективным, если мы сделаем обстановку более реалистичной, допустив <почти точное> (приблизительное) представление.
Ключевые слова: египетские дроби, приближенное представление дробей, оптимальное представление, точность представления.
Дата поступления в редакцию: 21.10.2020
