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N
HISTORY OF MATHEMATICS
ANCIENT EGYPTIAN ARITHMETIC: 2/N BY
E. M. BRUINS (Communicated by Prof. L. E. J.
BROUWER
at the meeting of February 23, 1952)
Introduction The papyrus Rhind begins with a series of calculations concerning fractions from which a table of decompositions of the sum 1/N + 1/N in a sum of elementary fractions with different denominators can be deduced. This table is known as the 2/N-table. In this table the number N runs from 3 to 101 and for N = 3s the decomposition is § 1.
2/N = 1f2s
+ 1f6s.
As to the question how this table was constructed by the ancient Egyptian
mathematicians both 0. NEUGEBAUER and B. L. VAN DER WAERDEN concluded, that it was gradually developed, not excluding several centuries for the time needed. In our opinion this conclusion is inadmissible. The remaining numbers consist of 25 prime numbers and 9 products. As the table can be based on the decompositions of the primes: assuming that the table was develloped from 3400 B.C. till 1800 B.C., taking into account that the decomposition of the small primes are nearly evident, results in an average "speed of construction" of one single decomposition per century. This would mean that the grandfather attacked the problem for one special decomposition but failed to find the solution, that his son continued the research also failing to obtain a result and that the grandson just succeeded at the end of his life to add "one more" to the decompositions! It is clear that this must be considered as historically impossible. On the contrary even the assumption, that one and ·the same mathematician had used 25 years of his life to construct this table would force to admit that he obtained an average speed of (less than) one decomposition pro year, and this must be rejected. In the present paper we shall show, that an Egyptian mathematician, in accordance with the arithmetic of his days, was able to construct the table within a very short time, in any case within a day! § 2. Nomenclature In order to give a concise description of the 2/N-table we use a special nomenclature considering first the problem from the modern point of view. Suppose p is a prime number and 2/p = lfx 6
Series A
+
1/y
+
lfz
+ ... ,
82
then
2xyz ...
28~ =
=
ps~-I,
where s~ denotes the elementary symmetrical function of degree f3 in a variables. We see that at least one of the x, y, z, ... , say x, is a multiple of p
kp.
X=
Then we have
(2k-1)yz ... = 1.
(2k-1)S~=~
=
kpS~=r;
if 2k- 1 = p we obtain a direct binomial decomposition
2/p
=
1/k
+
1fkp;
if 2k- 1 = lp, I an odd number, we obtain indirect trinomial decompositions; 3. if 2k- 1 ¢:. 0 (mod. p) we must have y = mp and
2.
(2km- k -m)i ...
=
(2km -k -m) S~=~
=
kmpS~=~.
In the last case we have to distinguish between 1.
2. 3.
• 2km - k - m = I p, I divisor of km, direct trinomial decomposition; 2km- k- m = lp, I not a divisor of km, indirect quadrinomiadecomposition; 2km- k- m ¢:. 0 (mod. p) z = np.
The last case gives for
2kmn- km- kn- mn = lp, nomial decompositions etc.
I divisor of kmn, the direct quadril
Example: p = 11 2k- 1 = 11 k = 6 1/6 + 1/66 binomial decomposition; 2k~ 1 = 33 k = 17, 3yz = 17(y + z) y > z 5 < z < 12 y = 102, z = 6 1/6 + 1/102 + 1/187, indirect trinomial decomposition; 2km-k-m, k = 8, m = 2, I= 2 1/8 + 1/22 + 1/88, direct trinomial decomposition; k = 5, m = 3, I = 2 2yz = 15(y + z}, y = 120, z = 8; y = 45, z = 9; y = 30, z = 10; y = 20, z = 12, indirect quadrinomial decompositions; 2kmn- km ~ kn- mn, k = 4, m = 3, n = 2, I = 2 1/12 + 1/22 + 1/33 + 1/44, direct quadrinomial decomposition. The necessary and sufficient condition for a non trivial direct decomposition is, that all but one of the denominators are multiples of p. We call a decomposition trivial if all the denominators are multiples of p. Only one at most quadrinomial trivial decomposition exists:
2fp Indeed:
2
=
1/k
=
+
Ifp 1/m
+ +
1f2p 1/n
+
+
1/3p
+
1f6p.
1ft , k > m > n > t
83 requires: 2 <4ft
< 3/n i < 2/m 1
t = 1 n=2 m = 3
1 = lfk
l
=
1/k
.k
=
6,
+ ljm + + 1/m,
ljn,
and along the same line one shows that no binomial and trinomial trivial decompositions exist. With this nomenclature we can state the following properties of the 2/N-table: 1.
2. 3.
None of the denominators exceeds 1000; None of the decompositions is more than quadrinomial; All decompositions of primes N are direct.
The first property may be expressed by:
1a. The table is a three-symbol-table. Only the symbols 1, 10, 100 are present, higher units do not appear. It is moreover easy to check: A non trivial at most quadrinomial (direct) decomposition for primes N > 100 is only possible for 103, 107, 109, 139, with greatest denominator < 1000. Indeed: 10 > k > m > n > t gives
1. 2.
2k--;- 1 ~ 17. 2km- k- m
k= 9 k= 9 k = 8
m
8
127
m = 6 m = 7
97
=
93
whereas the series of valpes for gradually increasing k, m, n gives for
3.
2kmn- km- kn- mn k = 4 22; k = 5 29, 42, 70; k = 6 36, 52, 90, 68, 117, 166; k = 7 43, 62, 107, 81, 139, 197, 100, 171, 242, 313; k = 8 50, 72, 124, 94, 161, 228, 116, 198, 280, 362, 138, 235, 332, 429, 526; k = 9 57, 82, 141, 107, 183, 259, 132, 225, 318, 411, 157, 267, 377, 487, 597, 182, 309, 436, 563, 690, 817.
The decomposition of 101 must therefore be trivial! It has thus not to be considered as an addition given by an unskilled copyist, as NEUGEBAUER 888umed, but as the only possible decomposition with denominators smaller than 1000.
§ 3. Construction of the table In order to master the calculations with fractions the Egyptian mathematician uses different units of which one is a known multiple
84 of the other. The different units are distinguished by writing them in different colours, black ahd red. The red unit is not necessarily the smaller one! In problem 23 of the Rhind papyrus we find: 1 )
+ 8 + 10 + 30 + 45 114 + 528 + 42 + 12 + 1 4
and here the red unit is 45 times smaller than the black one. The calculations concerning the 2/N-table for N = 31 show:
20 = 1 + 2 + 10 and here the red unit is greater! In order to find a decomposition of p + p the Egyptian mathematician shall have to introduce a smaller unit, say N times as small as the original one; then the total number of 2 N smaller units must be decomposed in a sum of terms: p and simple fractions of N. By these fractions p must be multiplied in order to obtain a decomposition:
p+ p=
N
+ kp + mp + np.
In this way he will obtain only direct decompositions. Evidently N must be chosen thus, that 2N > p. On the other hand N will not be chosen greater than p! From the modern point of view this is immediately clear. For all decompositions we have
N (2 - 1/k - 1/m - 1/n) = p
so
2 N > p.
On the other hand the trinomial decompositions require k )! 3 11f )! 2, so N(2- 5/6) < p
N
the quadrinomial decompositions, with the exception of k = 4, m n = 2; k = 5, m = 3, n = 2 require N(2- 19/20)
~
p
=
3,
N
Multiplication is effected by duplications and additions. The Egyptian mathematician will for this reason meet with "bad" and "beautiful" multipliers. 32 e.g. is a beautiful multiplier because it leads to 5 duplications; 31 on the other hand is a bad multiplier as it requires 4 duplications and 4 additions. Starting the scheme of beautiful multipliers we obtain:
2 4 8 16
3 6 12 24
20 30 40 60 80 120
In the first column we have the multipliers 2k, the second column contains the series of the sum of two adjacent numbers of the first column viz 1)
In the following we indicate lfn by ii.
85
3. 2k. By adding the two columns the same scheme in a tenfold is obtained. These multipliers are forming the first set to start with. In order to obtain as small a greatest denominator as possible in the final result the smallest term in the additional decomposition of 2N must be as great as possible. This process gives straightforward and by very simple operations, purely additive and nearly to effect by heart a series of decompositions for the prime numbers shown in Table I. The first TABLE I
I
N
N
3
2
4=
5
3
6= 5+ 1
Fractions
Sums
2 3 4 6
3+1
7
4
11
6
12 = 11
13
8
16 = 13 +
2 +
1
17
12
24 = 17 +
4 +
3 2
8=7+1
+
1
19
12
24 = 19 +
3 +
23
12
24 = 23 +
1
29
24
48
31
20
40 = 31 +
=
4 +
5 +
4
37
24
48 = 37 +
8 +
3
41
24
48 = 41 +
4 +
3
43
24
48 = 43 +
3 +
2
47
30
60 = 47 + 10 +
3
53
30
60 = 53 +
2
5 +
59
*
61
40
80 = 61 + 10
67
40
80 = 67 +
8 +
5
80 = 71 +
5 +
4
+
3
4
79
60
120 = 79 + 20 + 15 +
83
60
120 = 83 + 15 + 12 + 10 120 = 89 + 15 + 10 +
*
6 6
+
66 52+ 104 51 + 76
68
+ 114
24 +
4+ 8+ 5+ 8
+ 232
24 + 111 + 296 24 + 246 + 328 24 + 344 + 516 30 + 141 + 470 30 + 318 + 795 10
3+ 4+ 5 3 + 4 + 10
4+ 5+ 6 4 + 6 + 10
* *
58 + 174
20 + 124 + 155
8 + 10
120 = 73 + 20 + 15 + 12
*
2+ 6+ 8
*
5 +
60
* *
12
28
12 + 276
4+ 5 3+ 8 6+ 8 8 + 12 3 + 10 6 + 15
40
97
12 +
4+ 6
71
101
4+ 6+ 8+
4+ 8 3+ 4
73
89 . 60
2+ 6 3 + 15
12
29 + 12 +
*
Decompositions
40
+ 244
* + 488 + 610 '
4() + 335 + 536 40 + 568 + 710 60 + 219 + 292 + 365
+ 316 + 790 332 + 415 + 498
60 + 237 60 +
60 + 356 + 534 + 890
*
column contains the primes N, the second the multiplier N, the third the additional decomposition of 2N, the fourth the fractions of N, the last column gives the final result. So we obtained exactly the decompositions of the 2/N-table [with the
86 exception of 43 which is different ; 59, 97, 101 which cannot be found by the simplest multipliers] by one and the same direct method. We obtain a confirmation of the limitation of denominators from the decomposition of 101. It would have been very easy to take
+ 1515 !!
N
=
101
N
=
60
120
=
101 + 15 + 4
60 + 404
N
=
97
N
=
60
120
=
97 + 15 + 5 + 3
60 + 388 + 1164 + 1940.
The fact that the Egyptian mathematician aimed at a smallest maximumdenominator is confirmed by 29, 37 for:
N = 29_ N = 20
40 = 29 + 5 + 4 + 2 20 + 116 + 145 + 290; k = 10!
leads to a quadrinomial decomposition with a greater maximum denominator; N = 37
N= 20
40 = 37 + 2
+1
20 + 370 + 740
again, has a much greater maximum denominator. As we have already shown that tl;le only possible decomposition for N = 101 is a quadrinomial trivial one the mathematician is now forced to fill up only two gaps 59, 97. Let us consider first N = 97, so N > 48 and because of k ~ 10 2N must be greater than 102. Now we have
N= 52
104 = 97 + 7
N= 53
106 = 97 + 9
N= 54
108 = 97·+ 11
N= 55
110 = 97
N= 56
112 = 97 + 8 + 7
+ 13
and we have already· obtained the decomposition 97 + 97 =56+ 679 + 776. It is however impossible to demonstrate this to be the way in which the decomposition was obtained as the result is unique. The same holds for N = 59, N ~ 30, k < 18, in which case the series
N= 30
60 =59+ 1
N= 31
62 =59+ 3
N= 32
64 =59+ 5
N= 33
66 =59+ 7
N= 34
68 =59+ 9
N= 35
70 =59+ 11
N= 36
72 =59+ 9 + 4
leads to the unique decomposition 59 +59= 36 + 236 + 531. In these two cases the mathematician is forced to use other multipliers than those from the initial scheme: 2, 3, 4, 6, 8, 12, 20, 24, 30, 40, 60. Moreover the binomial decompositions, which are possible with
87
denominators smaller than 1000 make it clear that at first no other multipliers were used:
N= 3
N=2
4=3+1
N = 19
N= 10
20 = 19 + 1
N= 5
N=3
6=5+1
N = 23
N= 12
24 = 23 + 1
N= 7
N=4
8=7+1
N = 29
N= 15
30 = 29 + 1
N=ll
N=6
12=11+1
N = 31
N= 16
32 = 31 + 1
N = 13
N=7
14 = 13 + 1
N = 37
N= 19
38 = 37 + 1
N = 17
N=9
18 = 17 + l
N = 41
N= 21
42 = 41 + 1.
The fact that N = 31 was not solved by the multiplier N = 16 might easily be explained by the greater maximum-denominator (31 + 31 = 16 + 496), but this argumentation would fail for N = 13 N = 7 13 + 13 = 7 + 91! Now one can easily check that according to the principle of the smallest maximum denominator the table contains already the "best" decompositions for 3, 5, 17, 19, 31, 37, 41, 59, 67, 73, 79, 83, 97, 101,
<
whereas for multipliers N been obtained
N= 7 N = 13
N=6
p the following ameliorations could have
12=7+3+2
6 + 14 + 21 7 + 91
-
N=7
14 = 13 + 1
N= 10
20 = 13 + 5 + 2
lO + 26 + 65
N= 12
24 = 13 + 6 + 3 + 2
12 + 26 + 52 + 78
!1
=
43
N= 42
84 = 43 + 21 + 14 + 6
42 + 86 + 129 + 301
N
=
47
N= 36
72 = 47 + 12 + 9 + 4
36 + 141 + 188 + 423
N= 40
80 = 47 + 20 + 8 + 5
40 + 94 + 235 + 376
N =53
N= 36
72 = 53 + 9 + 6 + 4
36 + 212 + 318 + 477
N = 61
N= 45
90
45 + 183 + 305 + 549
N=71
N= 42
84 = 71 + 7 + 6
N = 89
N= 63
=
61 + 15 + 9 + 5
126 = 89 + 21 + 9 + 7
42 + 426 + 497
---
63 + 267 + 623 + 801.
From this it is clear that only N = 13 N = 7 would give a better decomposition with a smaller N. The resulting decompositions for 7, 43, 4 7, 53 contain more terms than those of the table. Nevertheless the table contains this last decomposition of N = 43. All the others are absent! Indeed the solution given for N = 24, k = 12 gives one of the greatest values for k, which is pushed ·down by N = 42 to k = 7! In our opinion one should also consider the decompossition of 11, 23, 29
88
as the best ones because of the fact that they can only be ameliorated by exorbitant values of N, N > p! We give the results for all other N > p: N=ll
N= 12 N= 60
12 + 22 + 33 + 44 +6 +4 +3 ---55 + 33 + 20 + 12 12 + 20 + 33 + 55 * 92 + 12 + 10 + 6 15 + 115 + 138 + 230 207 + 115 + 20 + 18 20 + 36 + 207 + 230 * 29 + 15 + 10 + 6 30 + 58 + 87 + 145 ---203 + 145 + 42 + 30 30 j- 42 + 145 + 203 * 159 + 53 + 13 + 9 39 + 117 + 477 + 689 * 689 + 159 + 52 + 36 36 + 156 + 477 + 689 *.
24 = 11 120
=
N= 60 120 = N= 180 360 = N=29 N= 30 60 = N= 210 420 = N= 53 N= 117 234 = N=23
N= 468 936 =
The indirect quadrinomial decompositions, marked with *, are excluded by the "Egyptian method" as indicated above which leads immediately to direct decompositions. The remaining decompositions for N = 11 and N = 23 which would be quadrinomial instead of binomial can evidently not be considered as an improvement of the decomposition contained by the table. · Thus, finally, we are inclined to consider the decompositions of 3, 5, 7, ll, 17, 19, 23, 29, 31, 37, 41, 47, 59, 67, 73, 79, 83, 97, 101
as the best ones, given straightforward by the construction of the table; to conclude that N = 43 was replaced by a better, and the best decomposition and that N = 13, 53, 61, 71, 89 were not altered for a reason which might be found from the analysis of the compound numbers N.
§ 4. Compound numbers Finally we have to consider the decompositions of the 9 compound numbers, which are not divisible by 3: 25, 35, 49, 55, 65, 77, 85, 91, 95. We stated already the well known fact that all multiples of 3 were obtained according to the process 2)
N= 3
2
+
N=9=3x3
6
+ 18
N = 15
=
5
X
3
10
+
6
30
etc. 2) The arrangement of the fractions in BM 10250 decidedly contradicts a systematical division of the fractions of the decomposition of a prime N by s = 2, 3, 4, 5,. .. . They seem to us to have been ordered in groups belonging to the same value of N. e.g. N = 25
+ 8 + 8 + 1, +8 +1 ' + 6 + 3 + 1,
25 = 8 25 ~= 16 25 = 15
+ + +
+ + +
+
f/8 = 1/25 1/25 1/25 1f200, 1/16 = [l/25 1/50 1/400] = 1/50 1/6 = 1/10 1/25 lf50 1/150.
+
+ 1/30 + 1/150 + 1/400,
89
0. NEUGEBAUER stated in his "Vorlesungen tiber vorgriechische Mathematik" page 159: 5 + 5 = 3 + 15. Wie man sieht ist dies die kanonische Zerlegung der 2/N-Tabelle und gilt entsprechend fiir die ganze Folge der Zerlegungen fiir . ungerade durch 5 teilbare N (sofern sie noch nicht durch die von 3 = 2 + 6 ausgehenden Zerlegungen erfaszt sind)." This statement is incorrect. The numbers 25, 35, 55, 65, 85, 95 give rise to decompositions according to the Rhind papyrus: N
=
65
39 + 195
N= 35
+ 75 30 + 42
N
=
85
51+ 255
N= 55
30 + 330
N
=
95
60 + 380 + 570 '
N=
15
~5
and only the decompositions of 25, 65, 85 can be obtained by 5s + 5s
=
3s + 15s;
the other three are treated differently! Considering first N = 35 N = 6 12 = 5 + 7. 30 + 42 is evident. Secondly N = 55 gives different possibilities:
N =55
N=3
33 + 165'
the subdivision of the decomposition for N
N= 55
N = 6
12 = 11
+1
=
5, is not in the table.
1 = 330
30 + 330
is the decomposition given by the Rhind papyrus, whereas
N= 55
N= 8
16 = 11
+5
40
+ 88,
the "best" one according to the maximum-denominator is not given. The second, contained in the Rhind papyrus will, indeed, be the best for practical use, because of the greater number of divisors of 30!! But now N = 95 would result in
N= 95
N= 12
24 = 19
+5
60 + 228 !!
This shows, that this more complicated factorising of N = 95 = 5 x 19 has escaped the attention of the constructors of the 2/N-table! The Rhind papyrus shows the reducible trinomial decomposition 95 + 95
=
60 + 380 + 570 .
This fact allows us to conclude that the ·greater N were automatically decomposed according to the general scheme with gradually increasing multipliers! Indeed:
N = 95
N = 60
120 = 95 + 25 = 95 + 15 + 10
90 leads to a trinomial decomposition as 25 is not a divisor of 60! And where even such a simple compound number N was not an object of further research in order to diminish the number of components and the greatest denominator we can safely assume that for the same reason N = (47), 53, 61, 71, 89 were not changed too. The two remaining compound numbers N = 49, N = 77 can be treated by N = 4 8 = 7 + 1 resulting in a subdivision of the decomposition of
N= 7. § 5.
The operations of the Rhind-papyrus
Having shown how the decompositions from the Rhind papyrus can be obtained straightforward from one and the same principle we wish to make a few remarks which make clear the reason for the more complicated, in our opinion impossible interpretations of B. L. VAN DER WAERDEN and 0. NEUGEBAUER. The first operations on fractions, which lead VAN DER W AERDEN to his hypothesis that the table was obtained by simple dividing out 2 : N, the operations being given only in a highly abbreviated form, are nothing but the check of the table given in full. Indeed, here are the results given before the "division", according to VAN DER WAERDEN, could have begun. In a scheme as e.g. for 1
N = 31 l
31
20
1
124
4
155
5
2
2
20
5
4
20
we have nothing of a division resulting in 2 : 31 = 20 + 124 merely a complete check! 2 + 4 + 5 + 20 = 1 being evident [10 + 1 = 20]. Moreover a division 2 : 7 would be easier by 2:7 = 6
+ 155 but +5+4+
+ 14 + 21;
again a completion of 1 + 2 + 12, having been obtained as 1/60 of 95 would be more easily effected by 3 + 12 than by 4 + 6. The choice of this last one is only justified by the "denominator < 1000"-condition! As to the theory of 0. NEUGEBAUER: he started with the number 1 +;, found in the calculations of the Rhind papyrus and supposed, that by gradually complicating the numbers 1 + ~ and calculating 7f = 1 - ~ the table was constructed with 25 different combinations ;;, {j, such that (1 + ~) N =iii, which is a rather complicated problem.
91 In our opinion 0. NEUGEBAUER also started from the wrong side. Our Table I shows immediately, that the fractions 1f are those of the fourth column; in our scheme, however, the 7f are given by one and the same procedure, the fractions 1 + ~ the completions of {i to 2, need not to be calculated, the decomposition being obtained straightforward. Conclusions The operations in the papyrus Rhind are merely the check by the pupil of the 2/N-table. 2. The 2/N-table is a three symbol table; all denominators are less than 1000. 3. The decompositions for prime numbers are direct, at most quadrinomial. 4. The trivial decomposition of 2/101 is the only possible one according to the property stated in 2. 5. The greatest denominator should be· as small as possible. 6. According to the introduction of a smaller unit by simple multipliers, the table can be constructed in a very short time. 7. The decomposition of 2/95 is reducible to a binomial one. This proves that for greater values of N the table obtained by simple multipliers was not improved any more. 8. The value N = 43 has been replaced by the "best" one according to the principle of smallest maximum denominator at the cost of one . more term in the decomposition. 9. The immediately obtained decompositions of § 6. 1.
3, 5, 7, 11, 17, 19, 23, 29, 31, 37, 41, 47, 59, 67, 73, 79, 83, 97, 101 are unique or the best ones. Only 13, 53, 61, 71, 89couldhave been improved. 10. For higher prime numbers only 103, 107, 109, 139 could give rise to decompositions in a non trivial way.
ANCIENT EGYPTIAN ARITHMETIC: 2/N BY
E. M. BRUINS (Communicated by Prof. L. E. J.
BROUWER
at the meeting of February 23, 1952)
Introduction The papyrus Rhind begins with a series of calculations concerning fractions from which a table of decompositions of the sum 1/N + 1/N in a sum of elementary fractions with different denominators can be deduced. This table is known as the 2/N-table. In this table the number N runs from 3 to 101 and for N = 3s the decomposition is § 1.
2/N = 1f2s
+ 1f6s.
As to the question how this table was constructed by the ancient Egyptian
mathematicians both 0. NEUGEBAUER and B. L. VAN DER WAERDEN concluded, that it was gradually developed, not excluding several centuries for the time needed. In our opinion this conclusion is inadmissible. The remaining numbers consist of 25 prime numbers and 9 products. As the table can be based on the decompositions of the primes: assuming that the table was develloped from 3400 B.C. till 1800 B.C., taking into account that the decomposition of the small primes are nearly evident, results in an average "speed of construction" of one single decomposition per century. This would mean that the grandfather attacked the problem for one special decomposition but failed to find the solution, that his son continued the research also failing to obtain a result and that the grandson just succeeded at the end of his life to add "one more" to the decompositions! It is clear that this must be considered as historically impossible. On the contrary even the assumption, that one and ·the same mathematician had used 25 years of his life to construct this table would force to admit that he obtained an average speed of (less than) one decomposition pro year, and this must be rejected. In the present paper we shall show, that an Egyptian mathematician, in accordance with the arithmetic of his days, was able to construct the table within a very short time, in any case within a day! § 2. Nomenclature In order to give a concise description of the 2/N-table we use a special nomenclature considering first the problem from the modern point of view. Suppose p is a prime number and 2/p = lfx 6
Series A
+
1/y
+
lfz
+ ... ,
82
then
2xyz ...
28~ =
=
ps~-I,
where s~ denotes the elementary symmetrical function of degree f3 in a variables. We see that at least one of the x, y, z, ... , say x, is a multiple of p
kp.
X=
Then we have
(2k-1)yz ... = 1.
(2k-1)S~=~
=
kpS~=r;
if 2k- 1 = p we obtain a direct binomial decomposition
2/p
=
1/k
+
1fkp;
if 2k- 1 = lp, I an odd number, we obtain indirect trinomial decompositions; 3. if 2k- 1 ¢:. 0 (mod. p) we must have y = mp and
2.
(2km- k -m)i ...
=
(2km -k -m) S~=~
=
kmpS~=~.
In the last case we have to distinguish between 1.
2. 3.
• 2km - k - m = I p, I divisor of km, direct trinomial decomposition; 2km- k- m = lp, I not a divisor of km, indirect quadrinomiadecomposition; 2km- k- m ¢:. 0 (mod. p) z = np.
The last case gives for
2kmn- km- kn- mn = lp, nomial decompositions etc.
I divisor of kmn, the direct quadril
Example: p = 11 2k- 1 = 11 k = 6 1/6 + 1/66 binomial decomposition; 2k~ 1 = 33 k = 17, 3yz = 17(y + z) y > z 5 < z < 12 y = 102, z = 6 1/6 + 1/102 + 1/187, indirect trinomial decomposition; 2km-k-m, k = 8, m = 2, I= 2 1/8 + 1/22 + 1/88, direct trinomial decomposition; k = 5, m = 3, I = 2 2yz = 15(y + z}, y = 120, z = 8; y = 45, z = 9; y = 30, z = 10; y = 20, z = 12, indirect quadrinomial decompositions; 2kmn- km ~ kn- mn, k = 4, m = 3, n = 2, I = 2 1/12 + 1/22 + 1/33 + 1/44, direct quadrinomial decomposition. The necessary and sufficient condition for a non trivial direct decomposition is, that all but one of the denominators are multiples of p. We call a decomposition trivial if all the denominators are multiples of p. Only one at most quadrinomial trivial decomposition exists:
2fp Indeed:
2
=
1/k
=
+
Ifp 1/m
+ +
1f2p 1/n
+
+
1/3p
+
1f6p.
1ft , k > m > n > t
83 requires: 2 <4ft
< 3/n i < 2/m 1
t = 1 n=2 m = 3
1 = lfk
l
=
1/k
.k
=
6,
+ ljm + + 1/m,
ljn,
and along the same line one shows that no binomial and trinomial trivial decompositions exist. With this nomenclature we can state the following properties of the 2/N-table: 1.
2. 3.
None of the denominators exceeds 1000; None of the decompositions is more than quadrinomial; All decompositions of primes N are direct.
The first property may be expressed by:
1a. The table is a three-symbol-table. Only the symbols 1, 10, 100 are present, higher units do not appear. It is moreover easy to check: A non trivial at most quadrinomial (direct) decomposition for primes N > 100 is only possible for 103, 107, 109, 139, with greatest denominator < 1000. Indeed: 10 > k > m > n > t gives
1. 2.
2k--;- 1 ~ 17. 2km- k- m
k= 9 k= 9 k = 8
m
8
127
m = 6 m = 7
97
=
93
whereas the series of valpes for gradually increasing k, m, n gives for
3.
2kmn- km- kn- mn k = 4 22; k = 5 29, 42, 70; k = 6 36, 52, 90, 68, 117, 166; k = 7 43, 62, 107, 81, 139, 197, 100, 171, 242, 313; k = 8 50, 72, 124, 94, 161, 228, 116, 198, 280, 362, 138, 235, 332, 429, 526; k = 9 57, 82, 141, 107, 183, 259, 132, 225, 318, 411, 157, 267, 377, 487, 597, 182, 309, 436, 563, 690, 817.
The decomposition of 101 must therefore be trivial! It has thus not to be considered as an addition given by an unskilled copyist, as NEUGEBAUER 888umed, but as the only possible decomposition with denominators smaller than 1000.
§ 3. Construction of the table In order to master the calculations with fractions the Egyptian mathematician uses different units of which one is a known multiple
84 of the other. The different units are distinguished by writing them in different colours, black ahd red. The red unit is not necessarily the smaller one! In problem 23 of the Rhind papyrus we find: 1 )
+ 8 + 10 + 30 + 45 114 + 528 + 42 + 12 + 1 4
and here the red unit is 45 times smaller than the black one. The calculations concerning the 2/N-table for N = 31 show:
20 = 1 + 2 + 10 and here the red unit is greater! In order to find a decomposition of p + p the Egyptian mathematician shall have to introduce a smaller unit, say N times as small as the original one; then the total number of 2 N smaller units must be decomposed in a sum of terms: p and simple fractions of N. By these fractions p must be multiplied in order to obtain a decomposition:
p+ p=
N
+ kp + mp + np.
In this way he will obtain only direct decompositions. Evidently N must be chosen thus, that 2N > p. On the other hand N will not be chosen greater than p! From the modern point of view this is immediately clear. For all decompositions we have
N (2 - 1/k - 1/m - 1/n) = p
so
2 N > p.
On the other hand the trinomial decompositions require k )! 3 11f )! 2, so N(2- 5/6) < p
N
the quadrinomial decompositions, with the exception of k = 4, m n = 2; k = 5, m = 3, n = 2 require N(2- 19/20)
~
p
=
3,
N
Multiplication is effected by duplications and additions. The Egyptian mathematician will for this reason meet with "bad" and "beautiful" multipliers. 32 e.g. is a beautiful multiplier because it leads to 5 duplications; 31 on the other hand is a bad multiplier as it requires 4 duplications and 4 additions. Starting the scheme of beautiful multipliers we obtain:
2 4 8 16
3 6 12 24
20 30 40 60 80 120
In the first column we have the multipliers 2k, the second column contains the series of the sum of two adjacent numbers of the first column viz 1)
In the following we indicate lfn by ii.
85
3. 2k. By adding the two columns the same scheme in a tenfold is obtained. These multipliers are forming the first set to start with. In order to obtain as small a greatest denominator as possible in the final result the smallest term in the additional decomposition of 2N must be as great as possible. This process gives straightforward and by very simple operations, purely additive and nearly to effect by heart a series of decompositions for the prime numbers shown in Table I. The first TABLE I
I
N
N
3
2
4=
5
3
6= 5+ 1
Fractions
Sums
2 3 4 6
3+1
7
4
11
6
12 = 11
13
8
16 = 13 +
2 +
1
17
12
24 = 17 +
4 +
3 2
8=7+1
+
1
19
12
24 = 19 +
3 +
23
12
24 = 23 +
1
29
24
48
31
20
40 = 31 +
=
4 +
5 +
4
37
24
48 = 37 +
8 +
3
41
24
48 = 41 +
4 +
3
43
24
48 = 43 +
3 +
2
47
30
60 = 47 + 10 +
3
53
30
60 = 53 +
2
5 +
59
*
61
40
80 = 61 + 10
67
40
80 = 67 +
8 +
5
80 = 71 +
5 +
4
+
3
4
79
60
120 = 79 + 20 + 15 +
83
60
120 = 83 + 15 + 12 + 10 120 = 89 + 15 + 10 +
*
6 6
+
66 52+ 104 51 + 76
68
+ 114
24 +
4+ 8+ 5+ 8
+ 232
24 + 111 + 296 24 + 246 + 328 24 + 344 + 516 30 + 141 + 470 30 + 318 + 795 10
3+ 4+ 5 3 + 4 + 10
4+ 5+ 6 4 + 6 + 10
* *
58 + 174
20 + 124 + 155
8 + 10
120 = 73 + 20 + 15 + 12
*
2+ 6+ 8
*
5 +
60
* *
12
28
12 + 276
4+ 5 3+ 8 6+ 8 8 + 12 3 + 10 6 + 15
40
97
12 +
4+ 6
71
101
4+ 6+ 8+
4+ 8 3+ 4
73
89 . 60
2+ 6 3 + 15
12
29 + 12 +
*
Decompositions
40
+ 244
* + 488 + 610 '
4() + 335 + 536 40 + 568 + 710 60 + 219 + 292 + 365
+ 316 + 790 332 + 415 + 498
60 + 237 60 +
60 + 356 + 534 + 890
*
column contains the primes N, the second the multiplier N, the third the additional decomposition of 2N, the fourth the fractions of N, the last column gives the final result. So we obtained exactly the decompositions of the 2/N-table [with the
86 exception of 43 which is different ; 59, 97, 101 which cannot be found by the simplest multipliers] by one and the same direct method. We obtain a confirmation of the limitation of denominators from the decomposition of 101. It would have been very easy to take
+ 1515 !!
N
=
101
N
=
60
120
=
101 + 15 + 4
60 + 404
N
=
97
N
=
60
120
=
97 + 15 + 5 + 3
60 + 388 + 1164 + 1940.
The fact that the Egyptian mathematician aimed at a smallest maximumdenominator is confirmed by 29, 37 for:
N = 29_ N = 20
40 = 29 + 5 + 4 + 2 20 + 116 + 145 + 290; k = 10!
leads to a quadrinomial decomposition with a greater maximum denominator; N = 37
N= 20
40 = 37 + 2
+1
20 + 370 + 740
again, has a much greater maximum denominator. As we have already shown that tl;le only possible decomposition for N = 101 is a quadrinomial trivial one the mathematician is now forced to fill up only two gaps 59, 97. Let us consider first N = 97, so N > 48 and because of k ~ 10 2N must be greater than 102. Now we have
N= 52
104 = 97 + 7
N= 53
106 = 97 + 9
N= 54
108 = 97·+ 11
N= 55
110 = 97
N= 56
112 = 97 + 8 + 7
+ 13
and we have already· obtained the decomposition 97 + 97 =56+ 679 + 776. It is however impossible to demonstrate this to be the way in which the decomposition was obtained as the result is unique. The same holds for N = 59, N ~ 30, k < 18, in which case the series
N= 30
60 =59+ 1
N= 31
62 =59+ 3
N= 32
64 =59+ 5
N= 33
66 =59+ 7
N= 34
68 =59+ 9
N= 35
70 =59+ 11
N= 36
72 =59+ 9 + 4
leads to the unique decomposition 59 +59= 36 + 236 + 531. In these two cases the mathematician is forced to use other multipliers than those from the initial scheme: 2, 3, 4, 6, 8, 12, 20, 24, 30, 40, 60. Moreover the binomial decompositions, which are possible with
87
denominators smaller than 1000 make it clear that at first no other multipliers were used:
N= 3
N=2
4=3+1
N = 19
N= 10
20 = 19 + 1
N= 5
N=3
6=5+1
N = 23
N= 12
24 = 23 + 1
N= 7
N=4
8=7+1
N = 29
N= 15
30 = 29 + 1
N=ll
N=6
12=11+1
N = 31
N= 16
32 = 31 + 1
N = 13
N=7
14 = 13 + 1
N = 37
N= 19
38 = 37 + 1
N = 17
N=9
18 = 17 + l
N = 41
N= 21
42 = 41 + 1.
The fact that N = 31 was not solved by the multiplier N = 16 might easily be explained by the greater maximum-denominator (31 + 31 = 16 + 496), but this argumentation would fail for N = 13 N = 7 13 + 13 = 7 + 91! Now one can easily check that according to the principle of the smallest maximum denominator the table contains already the "best" decompositions for 3, 5, 17, 19, 31, 37, 41, 59, 67, 73, 79, 83, 97, 101,
<
whereas for multipliers N been obtained
N= 7 N = 13
N=6
p the following ameliorations could have
12=7+3+2
6 + 14 + 21 7 + 91
-
N=7
14 = 13 + 1
N= 10
20 = 13 + 5 + 2
lO + 26 + 65
N= 12
24 = 13 + 6 + 3 + 2
12 + 26 + 52 + 78
!1
=
43
N= 42
84 = 43 + 21 + 14 + 6
42 + 86 + 129 + 301
N
=
47
N= 36
72 = 47 + 12 + 9 + 4
36 + 141 + 188 + 423
N= 40
80 = 47 + 20 + 8 + 5
40 + 94 + 235 + 376
N =53
N= 36
72 = 53 + 9 + 6 + 4
36 + 212 + 318 + 477
N = 61
N= 45
90
45 + 183 + 305 + 549
N=71
N= 42
84 = 71 + 7 + 6
N = 89
N= 63
=
61 + 15 + 9 + 5
126 = 89 + 21 + 9 + 7
42 + 426 + 497
---
63 + 267 + 623 + 801.
From this it is clear that only N = 13 N = 7 would give a better decomposition with a smaller N. The resulting decompositions for 7, 43, 4 7, 53 contain more terms than those of the table. Nevertheless the table contains this last decomposition of N = 43. All the others are absent! Indeed the solution given for N = 24, k = 12 gives one of the greatest values for k, which is pushed ·down by N = 42 to k = 7! In our opinion one should also consider the decompossition of 11, 23, 29
88
as the best ones because of the fact that they can only be ameliorated by exorbitant values of N, N > p! We give the results for all other N > p: N=ll
N= 12 N= 60
12 + 22 + 33 + 44 +6 +4 +3 ---55 + 33 + 20 + 12 12 + 20 + 33 + 55 * 92 + 12 + 10 + 6 15 + 115 + 138 + 230 207 + 115 + 20 + 18 20 + 36 + 207 + 230 * 29 + 15 + 10 + 6 30 + 58 + 87 + 145 ---203 + 145 + 42 + 30 30 j- 42 + 145 + 203 * 159 + 53 + 13 + 9 39 + 117 + 477 + 689 * 689 + 159 + 52 + 36 36 + 156 + 477 + 689 *.
24 = 11 120
=
N= 60 120 = N= 180 360 = N=29 N= 30 60 = N= 210 420 = N= 53 N= 117 234 = N=23
N= 468 936 =
The indirect quadrinomial decompositions, marked with *, are excluded by the "Egyptian method" as indicated above which leads immediately to direct decompositions. The remaining decompositions for N = 11 and N = 23 which would be quadrinomial instead of binomial can evidently not be considered as an improvement of the decomposition contained by the table. · Thus, finally, we are inclined to consider the decompositions of 3, 5, 7, ll, 17, 19, 23, 29, 31, 37, 41, 47, 59, 67, 73, 79, 83, 97, 101
as the best ones, given straightforward by the construction of the table; to conclude that N = 43 was replaced by a better, and the best decomposition and that N = 13, 53, 61, 71, 89 were not altered for a reason which might be found from the analysis of the compound numbers N.
§ 4. Compound numbers Finally we have to consider the decompositions of the 9 compound numbers, which are not divisible by 3: 25, 35, 49, 55, 65, 77, 85, 91, 95. We stated already the well known fact that all multiples of 3 were obtained according to the process 2)
N= 3
2
+
N=9=3x3
6
+ 18
N = 15
=
5
X
3
10
+
6
30
etc. 2) The arrangement of the fractions in BM 10250 decidedly contradicts a systematical division of the fractions of the decomposition of a prime N by s = 2, 3, 4, 5,. .. . They seem to us to have been ordered in groups belonging to the same value of N. e.g. N = 25
+ 8 + 8 + 1, +8 +1 ' + 6 + 3 + 1,
25 = 8 25 ~= 16 25 = 15
+ + +
+ + +
+
f/8 = 1/25 1/25 1/25 1f200, 1/16 = [l/25 1/50 1/400] = 1/50 1/6 = 1/10 1/25 lf50 1/150.
+
+ 1/30 + 1/150 + 1/400,
89
0. NEUGEBAUER stated in his "Vorlesungen tiber vorgriechische Mathematik" page 159: 5 + 5 = 3 + 15. Wie man sieht ist dies die kanonische Zerlegung der 2/N-Tabelle und gilt entsprechend fiir die ganze Folge der Zerlegungen fiir . ungerade durch 5 teilbare N (sofern sie noch nicht durch die von 3 = 2 + 6 ausgehenden Zerlegungen erfaszt sind)." This statement is incorrect. The numbers 25, 35, 55, 65, 85, 95 give rise to decompositions according to the Rhind papyrus: N
=
65
39 + 195
N= 35
+ 75 30 + 42
N
=
85
51+ 255
N= 55
30 + 330
N
=
95
60 + 380 + 570 '
N=
15
~5
and only the decompositions of 25, 65, 85 can be obtained by 5s + 5s
=
3s + 15s;
the other three are treated differently! Considering first N = 35 N = 6 12 = 5 + 7. 30 + 42 is evident. Secondly N = 55 gives different possibilities:
N =55
N=3
33 + 165'
the subdivision of the decomposition for N
N= 55
N = 6
12 = 11
+1
=
5, is not in the table.
1 = 330
30 + 330
is the decomposition given by the Rhind papyrus, whereas
N= 55
N= 8
16 = 11
+5
40
+ 88,
the "best" one according to the maximum-denominator is not given. The second, contained in the Rhind papyrus will, indeed, be the best for practical use, because of the greater number of divisors of 30!! But now N = 95 would result in
N= 95
N= 12
24 = 19
+5
60 + 228 !!
This shows, that this more complicated factorising of N = 95 = 5 x 19 has escaped the attention of the constructors of the 2/N-table! The Rhind papyrus shows the reducible trinomial decomposition 95 + 95
=
60 + 380 + 570 .
This fact allows us to conclude that the ·greater N were automatically decomposed according to the general scheme with gradually increasing multipliers! Indeed:
N = 95
N = 60
120 = 95 + 25 = 95 + 15 + 10
90 leads to a trinomial decomposition as 25 is not a divisor of 60! And where even such a simple compound number N was not an object of further research in order to diminish the number of components and the greatest denominator we can safely assume that for the same reason N = (47), 53, 61, 71, 89 were not changed too. The two remaining compound numbers N = 49, N = 77 can be treated by N = 4 8 = 7 + 1 resulting in a subdivision of the decomposition of
N= 7. § 5.
The operations of the Rhind-papyrus
Having shown how the decompositions from the Rhind papyrus can be obtained straightforward from one and the same principle we wish to make a few remarks which make clear the reason for the more complicated, in our opinion impossible interpretations of B. L. VAN DER WAERDEN and 0. NEUGEBAUER. The first operations on fractions, which lead VAN DER W AERDEN to his hypothesis that the table was obtained by simple dividing out 2 : N, the operations being given only in a highly abbreviated form, are nothing but the check of the table given in full. Indeed, here are the results given before the "division", according to VAN DER WAERDEN, could have begun. In a scheme as e.g. for 1
N = 31 l
31
20
1
124
4
155
5
2
2
20
5
4
20
we have nothing of a division resulting in 2 : 31 = 20 + 124 merely a complete check! 2 + 4 + 5 + 20 = 1 being evident [10 + 1 = 20]. Moreover a division 2 : 7 would be easier by 2:7 = 6
+ 155 but +5+4+
+ 14 + 21;
again a completion of 1 + 2 + 12, having been obtained as 1/60 of 95 would be more easily effected by 3 + 12 than by 4 + 6. The choice of this last one is only justified by the "denominator < 1000"-condition! As to the theory of 0. NEUGEBAUER: he started with the number 1 +;, found in the calculations of the Rhind papyrus and supposed, that by gradually complicating the numbers 1 + ~ and calculating 7f = 1 - ~ the table was constructed with 25 different combinations ;;, {j, such that (1 + ~) N =iii, which is a rather complicated problem.
91 In our opinion 0. NEUGEBAUER also started from the wrong side. Our Table I shows immediately, that the fractions 1f are those of the fourth column; in our scheme, however, the 7f are given by one and the same procedure, the fractions 1 + ~ the completions of {i to 2, need not to be calculated, the decomposition being obtained straightforward. Conclusions The operations in the papyrus Rhind are merely the check by the pupil of the 2/N-table. 2. The 2/N-table is a three symbol table; all denominators are less than 1000. 3. The decompositions for prime numbers are direct, at most quadrinomial. 4. The trivial decomposition of 2/101 is the only possible one according to the property stated in 2. 5. The greatest denominator should be· as small as possible. 6. According to the introduction of a smaller unit by simple multipliers, the table can be constructed in a very short time. 7. The decomposition of 2/95 is reducible to a binomial one. This proves that for greater values of N the table obtained by simple multipliers was not improved any more. 8. The value N = 43 has been replaced by the "best" one according to the principle of smallest maximum denominator at the cost of one . more term in the decomposition. 9. The immediately obtained decompositions of § 6. 1.
3, 5, 7, 11, 17, 19, 23, 29, 31, 37, 41, 47, 59, 67, 73, 79, 83, 97, 101 are unique or the best ones. Only 13, 53, 61, 71, 89couldhave been improved. 10. For higher prime numbers only 103, 107, 109, 139 could give rise to decompositions in a non trivial way.