Date: Nov 9, 2012 6:42 PM Author: Milo Gardner Subject: Re: Why study Egyptian fraction math? B. In 1987, Egyptologist Gay Robins, and Charles Shute, wrote a book onm the Rhind Mathematical Papyrus (RMP). Five years later Egyptologist John Legon wrote on the KP and closely related arithmetic proportions used in the RMP. The KP and RMP report the same method to find the largest term in arithmetic progressions. The method: take 1/2 of the difference, 1/2 of 5/6 (5/12 in the KP) times the number of differences (nine times 5/12 = 15/4 in the KP) plus the sum of the A.P progression (100 in the KP) divided by the number of terms (10 , meaning 100/10 = 10 in the KP). Finally add column 11's result, 3 3/4, to 10, and the largest term, 13 3/4.

In unit fractions, the context of the text, add column 11: 5/12 times 9 writing 3 3/4 as 3 2/3 1/12 to 10 in column 12 beginning with the largest term 13 2/3 1/12. The scribe subtracted 5/6 nine times created remaining terms of the arithmetic progression.

Robins-Shute confused aspects of the problem by omitting the sum divided by the number of terms, a topic cited in a closely related RMP 40 problem. A scribal algebraic statement matched pairs added to 20 reporting five pairs summed to 100, a set of facts included in RMP 40.

The complete KP method found the largest term facts reported in RMP 64 and RMP 40 by John Legon in 1992. Scholars have parsed Rhind Mathematical Papyrus 40 a problem that asked that 100 loaves of bread to be shared between five men by finding the smallest term of an arithmetic progression.

C. A confirmation of the Kahun Papyrus arithmetic progression method must include discussions of RMP 40 and RMP 64. In RMP 64 Ahmes asked 10 men to share 10 hekats of barley with a differential of 1/8 defining an arithmetical progression. Robins and Shute reported: "the scribe knew the rule that, to find the largest term of the arithmetical progression, he must add half the difference to the average number of terms as many times as there are common differences, that is, one less than the number of terms".

1. number of terms: 10

2. arithmetical progression difference: 1/8

3. arithmetic progression sum: 10

The scribe used the following facts to find the largest term.

1. one-half of differences, 1/16, times number of terms minus one, 9,

1/16 times 9 = 9/16

2. The computed parameter(1), was found by 10, the sum, divided by 10, the number of terms. It was inserted by Robins-Shute, but had not been high-lighted, citing 1 + 1/2 + 1/16, or 1 9/16, the largest term. The remaining nine terms were found by subtracting 1/8 nine times to obtain the remaining barley shares.

That is, the KP scribe used formula 1.0:

(1/2)d(n-1) + S/n = Xn (formula 1.0)

with,

d = differential, n = number of terms in the series, S = sum of the series, Xn = largest term in the series allowed three(of the four) parameters: d, n, S and Xn, to algebraically find the fourth parameter.

When n was odd, x (n/2) = S/n,

and x 1 + xn = x2 + x(n -1) = x3 + x(n -2) = ... = x(n/2) = S/n,

Note that Robins-Shute omitted the sum divided by the number of terms (S/n):

A modern footnote cites Carl Friedrich Gauss implementing as a grammar school student a solution to the n = even case. Ahmes and Gauss found the sum for 1 to 100 by using d = 1 following the same rule. Ahmes and Gauss reached the sum 5050 based on 50 pairs of 101 (1 + 101 = 2 + 99 = 3 + 98 = ...) by using an identical arithmetic progression rule.

D. A four level review of a Kahun Papyrus problem that reported 1365 1/3 khar as the volume of cylinder with a diameter of 12 cubit and height of 8 cubits summarized by:

1. Level 1 shows that pi was set to 256/81, and knowing one khar equaled 3/2 of a hekat, the scribe computed 1365 1/3 khar began with the area of a circle,

A=(pi)r2

, and input pi = 256/81 and D = 2, considering:

a. A = (256/81)(D/2)(D/2) = (64/81)(D)(D)

b. A = (8/9)(8/9)(D)(D) (algebraic formula 1.0)

This formula appeared in MMP 10, RMP 41, RMP 42, RMP 43, RMP 44, RMP 45, and RMP 46. In RMP 42 Ahmes adding height (H) and created two volume formulas.

c. V = (H)(8/9)(D)(8/9)(D) cubits squared (algebraic geometry formula 1.1)

d. V = (3/2)(H)(8/9)(8/9)(D)(D) khar (algebraic geometry formula 1.2) converted cubits to khar unit

In the Kahun Papyrus and RMP 43 algebraic geometry formula 1.2 indirectly scaled by 3/2 considering

e. (3/2)V = (3/2)(H)(3/2)(8/9)(8/9)(D)(D) = (H)(4/3)(4/3)(D)(D)

f. V = (2/3)(H)[(4/3)(D)(4/3)(D)] khar (algebraic geometry formula 1.3)

g. or directly considering V = (3/2)(H)(8/9)(8/9)(D)(D) =(32/27)(D)(D)(H) =(2/3)(H)(4/3)(4/3)(D)(D) khar

published by Robins-Shute in "Rhind Mathematical Papyrus" 1987, on page 46, confirming a scribal algebraic relationship.

To numerically verify scribal algebraic geometry was used, input scribal raw data D = 12, H = 8 (from the Kahun Papyrus):

h. (4/3)(12) was reported as (16)(16) = 256 such that

i. V = (2/3)8(256)=(1365 + 1/3)khar

RMP 43 input D = 8 and H = 6 into the same formula

j. V = (2/3)(6)[(4/3)(8)(4/3)(8)] = (4)(32/3)(32/3) = 4096/9 = (455 + 1/9) khar

k. A 4-quadruple hekat (400 hekat) division required RMP 43 and the Kahun Papyrus by the volume (V) formula.

V = (2/3)(H)[(4/3)(4/3)(D)(D)] (khar)

RMP 41, 42, 44, 45, 46, and 47 also input khar times 1/20 data as 4-hekat units to scale the ancient hekat to a modern 4800 ccm. In RMP 47 400-hekat was multiplied by 1/10 and 1/20 to 10 4-hekat and 5 4-hekat, respectivelu. As important RMP 47 multiplied 100-quadruple hekat (100 1-hekat) written as (6400/64)hekat by 1/30, 1/40, 1/50, 1/60, 1/70, 1/80, 1/90 and 100 to obtain 1-hekat quotients and 1-ro remainders.

An archaeological study can test scribal feeding rates for quail, dove, duck and geese by considering RMP 83 data. This class of study can test 1/10 a hekat scaled to 480 ccm. A recent feasibility study shows1 hekat may have equaled 250-500 ccm for geese and ducks.

2. Level 2 reported relative values of 12 fowls in terms of a set-duck unit paid in the following problem by:

a. 3 re-geese unit value 8 set-ducks = 24

b. 3 terp-geese unit value 4 set-ducks = 12

c. 3 Dj. Cranes unit value 2 set-ducks = 6

d. 3 set-duck unit value 1 set-duck = 3

total value 45 set-ducks.

Not calculated, but included in the valuation of (12 - 1) = 11 with 100 - 45 = 55, cited 55/11 as the total value as 5 times the value of one set-duck, with each water fowl given a value based on the daily, 10 day, 30 day and total hekat of grain consumed at rates reported in RMP 83. RMP 83 used the same (64/64) scaling method reported in the Akhmim Wooden Tablet and in RMP 47.