Search All of the Math Forum:
Views expressed in these public forums are not endorsed by
Drexel University or The Math Forum.
|
|
|
|
RMP 23, as read by RMP 66
Posted:
May 9, 2009 8:25 AM
|
|
Franz,
Your summary of RMP 23, cited below, is confusing:
In RMP 23 there is no proportional reasoning.
1. Statement of problem: complete 1/4 1/8 + 1/10 + 1/30 + 1/45 + x = 2/3 (provided by Rboins-Shute)
2. Answer: 1/4 + 1/8 + 1/9 + 1/10 + 1/30 + 1/40 + 1/45 = 1 (provided by Robins-Shute)
3." Again, a similar method is used, with 45 as the common multiplier, yielding 1/9 + 1/40 as the answer"
quote from Robins-Shute.
Translating RMP 23's intermediate calculation and answer requires the use of LCM 360, a point mentioned by Franz, but in a context that does not coincide with Ahmes' LCM 45 proof. To parse Ahmes' proof, let's follow every step as was done in RMP 66 (yesterday on the Math forum).
4. Another summary of RMP 23, taken from my Ahmes Papyrus blog says:
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) = 1
An concise analysis is required to follow Ahmes' precise proof steps. I did not do that, nor did Franz, per:
"Proportional adding up is the arithmetic method of the
Rhind Mathematical Papyrus. I explained it at length
in the case of RMP 66. Now let me demonstrate it also
in the case of RMP 23 whose working is only partly
rendered.
Complete '4 '8 '10 '30 '45 into "3
Use the proportion 1 -- 45
"3 -- 30
'4 -- 11 '2
'8 -- 5 '2 '4
'10 -- 4 '2
'30 -- 1 '2
'45 -- 1
sum 23 '2 '4 '8
30 minus 23 '2 '4 '8 equals 6 '8
'8 -- 5 '2 '8 (remains '2)
'90 -- '2
The answer is '8 '90. Alas, we can't use it, for we
have '8 already in the initial series.
'9 -- 5 (remain 1 '8)
'40 -- 1 '8
The second answer is '9 '40
'4 '8 '10 '30 '45 plus '9 '40 equals "3
'4 '8 '9 '10 '30 '40 '45 equals "3
So far RMP 23 on the level of beginners.
Advanced learners may solve another task and first
multiply the above addition by a factor of 360:
90 45 36 12 8 plus 40 9 equals 240
191 plus 49 equals 240
Regard 49 and 191 and 240 as diameters of three circles.
How long are the circumferences?
4/1 (plus 3/1) 7/2 10/3 13/4 16/5 19/6 22/7
49 times '7 of 22 equals 154
diameter 49, circumference 154
6/2 (plus 22/7) 28/9 50/16 72/23 ... 600/191
diameter 191, circumference 600
3/1 (plus 22/7) 25/8 47/15 69/22 ... 377/120
240 times '120 of 377 equals 754
diameter 240, circumference 754
In fractions:
diameter '9 '40, circumference '5 '9 '10 '60
diameter '4 '8 '10 '30 '45, circumference 1 "3
diameter '5 '10 '60 '600, circumference 1
(yielding 0.318333... for the inverse of pi)
diameter "3, circumference 2 '15 '36
diameter 1, circumference 3 '10 '24
(yielding 3.141666... for pi)"
I'll be citing Chace's transliteration of RMP 23, and its proof, to show Ahmes' use of LCM 45, rather than the required modern translation of LCM 360.
Best Regards,
Milo Gardner
|
|
|
|
