Drexel dragonThe Math ForumDonate to the Math Forum



Search All of the Math Forum:

Views expressed in these public forums are not endorsed by Drexel University or The Math Forum.


Math Forum » Discussions » Inactive » math-history-list

Topic: Egyptian algebra, RMP 21 thru RMP 34
Replies: 10   Last Post: May 14, 2011 12:00 PM

Advanced Search

Back to Topic List Back to Topic List Jump to Tree View Jump to Tree View   Messages: [ Previous | Next ]
Milo Gardner

Posts: 1,105
Registered: 12/3/04
Egyptian algebra, RMP 21 thru RMP 34
Posted: Apr 29, 2011 12:50 PM
  Click to see the message monospaced in plain text Plain Text   Click to reply to this topic Reply

From my blog

http://ahmespapyrus.blogspot.com/

a discussion begins with RMP 21, and ends with RMP 34:

RMP 21. complete the series 1/5 + 1/15 + x = 1

LCM 15 was used, (yet) says Robins-Shute, gave these as Ahmes' steps

a. 15*(2/3 + 1/15) = 10 + 1 = 11
b. 15 - 11 = 4
c. 15*(1/5 + 1/15) = 3 + 1 = 4
d. 15*(1/3 + 15) + (1/5 + 1/15) = 11 + 4 = 15
e. (2/3 + 1/15) + (1/5 + 1/15) = 1 (proof)

Considering 10 + 1 = 11 as a numerator, the following rational number logic took place:

Ahmes' calculations used red, thought at times the color was omitted.

a. Find vulgar fraction: (2/3 + 1/5) = (10 + 1)/15 = 11/15

b. Find missing vulgar fraction: 15/15 - 11/15 = 4/15 = (3 + 1)/15 = 1/5 + 1/15

c. Ahmes' proof: (2/3 + 1/5) + (1/5 + 1/15) = 1

22. Complete 2/3 + 1/30 + x = 1 : find an LCM (30) ans: x = 1/5 + 1/10,
Robins-Shute mentions "ab initio" related to Ahmes' steps

a. 30*(2/3 + 1/30) = 20 + 1 = 21
b. 30-21 = 9
c. 30*(1/5 + 1/10) = 6 + 3 = 9
d. 30*(2/3 + 1/30) + (1/5 + 1/10) = 21 + 9 = 30
e. (2/3 + 1/30) + (1/5 + 1/10) = 1 (proof)

Ahmes' calculation did not use red.

a. Find vulgar fraction: (2/3 + 1/20) = (20 + 1)/30 = 21/30

b. Find missing vulgar fraction: 30/30 - 21/30 = 9/30 = (6 + 3)/30 = 1/5 + 1/10

c. Ahmes' proof: (2/3 + 1/5) + (1/5 + 1/15) = 1

Note: The aliquot parts of 30: 15, 10, 6, 5, 3, 2, 1 optimally found (6 + 3 )/30 = (1/5 + 1/10)
rejecting (6 + 2 + 1) and ( 5 + 3 + 1) alternatives, a minor rule presented in the 2/n table.

23. complete the series 1/4 + 1/8 + 1/10 + 1/30 + 1/45 + x = 2/3

Optimally find a series with LCM 45 (ans. 1/9 + 1/40 )

Ahmes' calculation:

a. Find vulgar fraction: 1/4 + 1/8 + 1/10 + 1/30 + 1/45 + x = 2/3 (multiply by 45)

b. 45/4 + 45/8 + 45/10 + 45/30 + 45/45 + 45x = 45*(2/3)

c. 11 + 1/4 + 5 + 1/2 + 1/8 + 1 + 1/2 + 45x = 30

d. 45x = 30 - (23 1/2 + 1/4 + 1/8 )= 6 + 1/8

e. 45x = 49/8,

f. x = 49/360 = (40 + 9) = 1/9 + 1/40

Ahmes omitted the last two steps, by only writing the answer 1/9 + 1/40. Gillings did not comment on the 49/360 aspect of the last step, thereby maintaining his 'ab initio' posture.

Modern rational number version:

a. 1/4 + 1/8 + 1/10 + 1/30 + 1/45 = (90 + 45 + 36 + 12 + 8 )/360= 191/360

b. Find vulgar fraction:

c. aliquot parts of 360: 180, 90 45, 40, 20, 15, 9, 5, 3, 2, 1 (find 49)

d. Find missing vulgar fraction: 49/360 = (40 + 9)/360 = 1/9 + 1/40

e. (1/4 + 1/8 + 1/10 + 1/30 + 1/45) + (1/9 + 1/40)

Robins-Shute and Gillings appropriately cite the EMLR as a related document, though neither specified the optimized LCM, red auxiliary, details used in the majority of RMP problems. The EMLR introduced non-optimal LCMs. LCM 7 converted 1/4 and 1/8 in RMP 15,

6. 9/10 = 2/3 + 1/5 + 1/30 (1.e. Ahmes' initial calculation)

9/10*(3/3) = 27/30 = (20 + 6 + 1)/30 = 2/3+ 1/5 + 1/30

as the other algebra problems collected x's and solved in a modern manner:

24. x + (1/7)x = 19

(8/7)x = 19,

x = 133/8 = 16 + 5/8 = 16 + (4 + 1)/8 = 16 + 1/2 + 1/8

25. x + (1/2)x = 16;

(3/2) x = 16,

x = 32/3 = 10 + 2/3

26. x + (1/4)x = 15

(7/4)x = 15

x = 60/7 = 8 + 4/7 = 8 + 4/7(4/4) = 8 + (16/28) = 8 + (14 + 2)/28 = 8 + 1/2 + 1/14

27. x + (1/5)x = 21

(6/5)x = 21

x = 105/6 = 17 + 1/2


28. (2/3)x - (1/3)y = 10; (2/3)y = 10

two unknowns.

29. a solution method, not a problem

1 +1/4 + 1/10 = 13 1/2

that scholars classify as a diversion.

31. x + (2/3 + 1/2 + 1/7)x = 33

(97/42)x = 33

x = 1386/97 = 14 + (28/97)

x =14 + (2/97)(56/56) + (26/97)(4/4) = 112/5432 + 104/388 =

14 + (97 + 8 + 7)/5432 + (97 + 4 + 2 + 1)/388

32. x + (2/3 + 1/4)x = 2

(23/12)x = 2

x = 24/23 = 1 + 1/23

33. x + (2/3 +1/2 + 1/7)x = 37

(97/42)x = 37

x = 1554/97 = 16 + 2/97

with 2/97 (56/56) = 112/5432 = (97 + 8 + 7)/56432

34. x + (1/2 + 1/4)x = 10

(7/4)x = 10

x= 40/7 = 5 + 5/7 (4/4) = 5 + 20/28 = 5 + (14 + 4 + 2)/28 = 5 + 1/2 + 1/7 + 1/14

Best regards to Ahmes, and scholars and amateurs alike that fairly report the 1650 BCE document as originally written,

Milo Gardner



Point your RSS reader here for a feed of the latest messages in this topic.

[Privacy Policy] [Terms of Use]

© Drexel University 1994-2014. All Rights Reserved.
The Math Forum is a research and educational enterprise of the Drexel University School of Education.