## Wednesday, December 9, 2009

### Moscow Mathematical Papyrus (MMP)

The MMP is held in the Pushkin State Museum of Fine Arts in Moscow, Russia. Its contents cover 25 problems. The MMP has been called the Golenischev Mathematical Papyrus. Its first modern owner was Egyptologist Vladimir Golenidenov. Based on the palaeography of the hieratic text, it probably dates to the Eleventh dynasty of Egypt. Approximately 18 feet long and varying between 1 1/2 and 3 inches wide, its format was divided into 25 problems with solutions by the Soviet Orientalist Vasily Vasilievich Struve in 1930. It is one of the three well-known Mathematical Papyri along with the Rhind Mathematical Papyrus (RMP) and the Egyptian Mathematical Leather Roll (EMLR). The Moscow Mathematical Papyrus is about the same age as the EMLR, and 150 years older than the RMP. Gillings writes up the algebra correctly. Claggett reproduces Peet's 1920's and Gillings 1972 views, but adds little new beyond either.

The 25 MMP problems are written in arithmetic, algebraic, and geometric metaphors with several of the ancient definitions, and references included in Marhall Clagett's 1999 translation, and Gillings 1972 review of the text. Clagett discussed Struve's 1930 German analysis, and Peet's 1923 review, but little of the latest discoveries on the text.

MMP 6 reports the solution to a 2nd degree equation

1. (3/4)x^2 = 12

Scribal algebra solved the problem by

2. (4/3) (3/4)x^2 = (4/3)(12)

3. x^2 = 16

4. x = 4

The MMP scribe's second step actual wrote invert (3/4) and multiply by 12. That is, scribal division inverted and multiplied, a modern arithmetic technique.

More importantly, MMP 10 considered the area (A) of a circle by solving for diameter D, setting pi = 256/81 such that a formula that Gillings and earlier scholars incompletely read by stressing (8/9)(8/9) may have been involved in finding the area and circumference of a circle,

The actual MMP 10 calculations mentioned an area of an egg, a semi- circle, The circle's diameter 9 was input into an undefined formula reporting

1. A^1/2 = (8/9 x 9)  = (72/9) x (72/9)/2

writing 9 - 9/9 = 8  x 8 = 64, which meant

2. A = (72/9)(72/9) =  64/2 = 32 (cubit ^2)

meant that the MMP  scribe began with

3. A = 256/81 x (D/2)(D/2)  (area of a circle, replacing pi with 256/81 and r by D/2)

4. A = (64/81)( D)(D) was reduced to

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

RMP 41 solved, D =9, height (H) 10 using the same MMP 10 formula

6. A = (8/9)(9)(8/9)(9) = (8)(8) = 64 cubits^2

writing

7. (9 - 9/9) = (8 x 8) = 64 cubits^2

8. Volume (V) = (64) (10) = 640 cubits^3

In RMP 42  3/2 of a cubit^3 became a khar unit citing D =  10, H =10

9. A = (8/9)(10)(8/9)(10)  writing

10. A = (10 - 10/9) = (80/9)(80/9) = 6400/81 cubit^2

11. V = (6400/81)(10) = 64000/81 = 790 10/81

12 . V = (3/2)(H)(8/9)(D)(8/9)(D)  = khar unit  (algebraic geometry formula 2.0)

a third volume  formula scaled  algebraic geometry formula 2.0 by 3/2 in RMP 43 and the Kahun Papyrus created a Khar unit by

14. (3/2)V =(3/2)(3/2)(H)(8/9)(8/9)(D)D) = (4/3)(D)(4/3)(D) such that

15. V = (2/3)(H)(4/3)(D)(4/3)(D) =  Khar unit (algebraic geometry formula 2.1)

used in RMP 43. Line 5. RMP 43's raw data

4096/9 = 455 1/9 in line 3,

1/20 of 455 1/9= 22 1/2 + 1/4 + 1/180 (not 1/45) in line 4

5. 4090/180 = (22 + 1/2 + 1/4 + 1/180) hekat

and line 5 by writing the remainder 1/180 of 100 hekat unit to a hekat unit to:

100/180 = (5/9) and

(5/9)(64/64) = 320/576 =

(288 + 18 + 9)/576 + [5/288(5/5) = (25/9)ro)

(1/2 + 1/32 + 1/64)hekat + (2 + 7/9) ro, with

Ahmes recorded 7/9 ro as (1/2 + 1/4 + 1/36)ro

information that is pertinent to MMP 10 related to area and volume formulas that
created cubit, khar, 1/20 of khar = 100 hekat and hekat units related to the
Akhmim Wooden Tablet and over 40 RMP uses.

Relevant AE mathematical and metrological and other terms (per Bruce Friedman):
===================================================

"1/2 + 1/4 pesu kind of beer" is a weak strength beer made with a cheaper less potent grain.
Beer A has an exchange rate that defies the mathematical looking name.
Beer B is exchanged at half the value of beer A made with the same pesu of the stronger grain.

See problem MMP#5 [121509_see also #8 and #9] or an example of exchange of beer A and B.
The Horus-eye symbols for hekat fractions ["<" for 1/2 and "X" for 1/4] are strung as "X <" for the beer name. Consider that beer B is made with typical recipes but with poor quality ingredients per Clagett's note 7 on pages 229-230. "Weak-grained beer", "*lesser beer, or beer B" Malt-Date beer is specified a beer B most clearly in the hieratic of MMP 8. See below.

"Bank" or "idb" is the same as our rise over run calculations in modern construction. *"Bank" is not the same as "seked" see below. **See usage and an example of the "bank" in problem MMP#7 below. "Beer" is an AE staple food. See also "bread." *Typically measured in jugs or "des-jugs."

MMP #15 - MMP #21

MMP #22 (Clag. V3; p.225 and fig. IV.6q at columns XL-XLI of page 402- largely complete and see footnote 33 on pages 236-7) called a "calculation of grain" problem by Clagett. Actually this too is a pfesu calculation. 10 hekat of upper egyptian grain are to be calculated like (or exchanged for) 100 loaves of bread of unknown pfesu.

The remainder from the 10 hekat after the 100 loaves worth of grain are subtracted is to be used for 10 des jugs of standard beer with pfesu 2 to be converted [?] to jugs of lesser beer.
Here the recipe for the lesser beer is again intimated: Deduct the 5 hekats needed to make 10 jugs of lesser beer pfesu 2 [this should read 2.5 hekats to be consistent with MMP #5 or #8].
This leaves 5 hekats for the 100 loaves. But the scribe adjusts the 5 hekats of grain to 2.5 hekats because the lesser beer recipe requires half as much grain. This leaves 7.5 hekats of grain for the 100 loaves. At this point the problem appears to be truncate

If the scribe was to complete all apparently requested work he would find the brad pfesu by dividing the 7.5 hekats of grain available by 100 loaves to find each loaf has a pfesu of 13 [+] 1/3.
i.e. 15/2 hekat = 100 loaves pfesu X

X = 100/15/2

In RMP 69-78 two-level, and higher level inventory controls of grain and other products were reported by a pesu unit. The pesu was an arithmetic inverse to the amount of grain in the product. On a macro level about 1/3 of grain was used for bread, and about 2/3 was used for several types of beers, including one with dates as an ingredient.

For example RMP 71 showed the second level use of a pesu. Beer was made in besha unit, 1/2 of a hekat of grain. 1/4 of the besha (1/8 of a hekat) was poured off, and replace by water. The pesu calculation began with a besha composed of 1/2 a hekat such that:

1/2 - 1/8 = 3/8 was returned to a hekat unit by multiplying by 8/3 or

3/8 x 8/3 = 1,

with 8/3 = 2 2/3 recorded as the pesu. This calculation is confirmed in Moscow Mathematical Papyrus (MMP) problems 12 and 16 by two strengths of beer, the first confirming the second level 2 2/3 pesu made from a besha.

The majority of RMP pesu problems began with 15 or 16 hekats, and removed 5 hekats for use with making 200 loaves of bread at 20 pesu. The remaining hekats, 10 and 11, were used for one or more beers, a methodology that was repeated several times in the MMP.

MMP #23 (Clag. V3; p.226 and fig. IV.6r at column XLII of page 403- largely complete and see footnote 34 on page 237) called a "reckoning the work of a shoemaker" problem by Clagett.
This like MMP #11 suggests work output or quota was formalized for all bureaucratically administered tasks. If the shoemaker only cuts the material for shoes he can produce 10 pairs per day. If the shoemaker only decorates (or assembles) the material for shoes he can produce 5 pairs per day.

To find how many completed cut and assembled and decorated pairs he can produce in a day:
Find that (as given) in 1 day he can cut for 10 pairs and in 2 days he can complete decorating, therefore in 3 days he can fully cut/decorate/complete 10 pairs.
Therefore 3 [+] 1/3 pairs [10/3] can be cut and completed per day.
Case closed.

MMP #24 (Clag. V3; p.226-7 and fig. IV.6s at columns XLIII-XLIV of page 404- largely complete and see footnote 35 on page 237)
Called a "calculation of grain" problem by Clagett. Actually this too is a pfesu calculation.

Given 15 hekats upper egyptian grain to be made into 200 loaves and 10 des-jugs of lesser beer with 1/10 the pfesu* of the loaves.

*Consider that the beer is 1/10th AS WEAK a the loaves. I.e. A jug of this beer is of 10 times the exchange value of one of the loaves. Find the pfesu of the bread and beer.

The scribe frames his workings to solve this alternate problem which yields the same result for the bread pfesu but has easier manipulations:

Given 15 hekats upper egyptian grain to be made into 200 loaves and 100 des-jugs of lesser beer with the same pfesu as the loaves.

300/15 = 20.

20 is the pfesu of the bread (therefore 200*loaves/20 pfesu requires 10 hekats)
*Scribe may have mistakenly shown 100 loaves instead of 200 but Struve/Clagett correct this.
and 5 hekats remain to make the 10 des-jugs of lesser beer with 1/10 the pfesu of the loaves [pfesu = 2 for beer] OR 100 des-jugs of lesser beer with the same pfesu as the loaves [irrelevant but the pfesu of 100 jugs of lesser beer made from 5 hekats grain would be found to be 20 ONLY in this case].

I believe something is bizarre about this as lesser beer pfesu 2 is otherwise shown requiring half the grain as standard beer but in MMP #24 there is no reduction!

In MMP #5 we see:
2 1/2 [hekats] x [pesu of *lesser beer*] 4 = 10 [des-jugs]

Here in MMP #24 we see:
5 [hekats] x [pesu of A STRONGER/MORE VALUABLE*lesser beer*] 2 = 10 [des-jugs]
In effect, the MMP 24 jugs should be twice or 4 times as large or twice or 4 times as valuable as those in MMP #5.
This remains unclear. See MMP #12 for related chaos.
Case closed.

MMP #25 (Clag. V3; p.227 and fig. IV.6t at column XLV of page 405- complete and see footnote 36 on page 237)
Called a "calculating a quantity" problem by Clagett.
Find X when 2x + x = 9
Solve 3X=9
Workings suggest 2+1=3; 9/3 = 3
Case closed.