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




Egyptian algebra, RMP 21 thru RMP 34
Posted:
Apr 29, 2011 12:50 PM


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 RobinsShute, 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, RobinsShute mentions "ab initio" related to Ahmes' steps
a. 30*(2/3 + 1/30) = 20 + 1 = 21 b. 3021 = 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)
RobinsShute 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 nonoptimal 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



