RE: [HM] I

milo.gardner@24stex.com
Sat, 07 Nov 98 09:09:20 -0800

Val Dusek's point concerning modern proofs is
a correct one. Rigor does provide a general
swath of mathematical reason that explains
the special case under review, revealing
several hows and whys. Classical Greeks may
have been first to implement aspects of
our tradition that we now call proofs and
rigor. However, in reviewing the 'proof'
method of an Archimedes example proof, a
strong sense of Egyptian numeration, number
theory and algebraic traditions are also seen.

What are the Egyptian traditions that
Archimedes used to build his proofs?
The primary one that I would suggest
is the Middle Kingdom hieratic finite
series numeration tradition of double
checking Old Kingdom hieroglyphic
infinite series results.

As clearly known to Greeks, the Old
Kingdom Horus-Eye infinite series:

1) 1 = 1/2 + 1/4 + ... + 1/2n + ...

can be parsed into two geometric series:

2) 1/3 = 1/4 + 1/16 + 1/2n + ...

and,

3) 2/3 = 1/2 + 1/8 + 1/(2*4n) + ...

and exactly re-written as a finite series,
as the MMP, RMP, EMLR, Hibeh P. and other
Greek era documents detail.

Archimedes plausibly began with the traditional
infinite series in his famous 1/4th geometric series
and re-wrote it as a finite series as Dijksterhuis
cites as a proposition:

"Given a series of magnitudes, each of which
is equal to four times the next in order, all
the magnitudes and one-third of the least added
together will exceed the greast by one-third"

as algebraically summarized as follows:

4) 4/3 = A + B + C + D + E + E/3

One proof aspect of Archimedes' final exact series
is simply a restatement of the 1/4th geometic infinite
series to a finite series, a method that historians
may agree is present in the RMP and its 2/nth table.
I further suggest that Archimedes followed a related
'Egyptian' method to prove his In Quadrature of the
Parabola proposition by:

let A = 1, then,

1/3 = 1/4 + 1/16 + 1/64 + 1/256 + ...

(exposing equation 2))

with a last introduced at any point,
such as E/3 by:

E/3 = 1/3 - 85/255 = (256 - 255)/(3*256)

Following a finite series method very much as Hultsch
suggested in 1895 was the basis for all RMP 2/p series.

All that I add to the above is that Archimedes
may have recursively computed his initial
geometric series by beginning with 1/3 then:

1/3 - 1/4 = (4 - 3)/12 = 1/(3*4)

1/12 - 1/16 = (16 - 12)/(12*16) = 1/(3*16)

1/48 - 1/64 = (64 - 48)/(48*64) = 1/(3*64)

1/192 - 1/256 = (256 - 192)/(192*256) = 1/(3*256)

with the last exact term E/3 being a conventional
five-term stopping point, one less than Old Kingdom
Egyptians had established for 6-term (rounded off)
Horus-Eye infinite series.

Comments?

Milo Gardner
Sacramento, Calif.