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Topic: Archimedes (and Ahmes) square root of 5, 6 and 7
Replies: 8   Last Post: Jan 16, 2013 6:58 PM

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Milo Gardner

Posts: 1,105
Registered: 12/3/04
Archimedes (and Ahmes) square root of 5, 6 and 7
Posted: Dec 18, 2012 5:57 PM
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A modern math three step process captures scribal notes that ended with once hard to translate unit fraction series.

Modern translations for the square roots of five (5), six (6), and seven (7) are described below (A, B, C).

Greek and Egyptian algebraic steps were finite. Decoding the algorithmic looking finite arithmetic steps have been demystified.

Scribal rational numbers n/p steps have been readable since 2006, methods that Archimedes and Ahmes considered:

A. Computed the square root of five( 5) that estimated

(Q + R)^2, R= 1/(1/2Q)

step 1: estimated

Q = 2. R,= (1//4) such that try

(2 + 1/4)^2 = 6 + *(1/4)^2; error1 = 1/16

step 2, reduced error that divided

1/16 by twice of inverted

(2 + 1/4)= 18/4 such that

1/16 x (4/18) = 1/72 = error2 = (1/72)^2

2 + 1/4 - (1/72)^2)^2 = (2 + 1285/5184)^2;

as Archimedes and Ahmes would have re-written

(2 + 1285/5184) as a unit fraction series:

a} [2 + (864 + 398 + 51 + 1/17 + 6)/5184) =

[2 + 1/6 + 1/13 + 1/39 + 1/864]^2

b} [2 + (864 + 370 + 64 + 6 +1 )/5184] =

[2 + 1/6 + 1/14 + 1/81 + 1/864 + 1/5184]^2

Note that scribal shorthand notes pondered by academics prior to 2006 suggested incomplete square root steps even though major operational aspects of the same class of arithmetic and algebraic pesu steps have been translated.

In the square root of five case a step 3 was not needed.

B. square root of six (6) ,

step 1: estimated Q = 2, R = (6 -4)/4 = 1/2 such that

(2 + 1/2)^2 = 6 + (1/2)^2, error1 = 1/4

step 2: reduced 1/4 error that divided

1/4 by twice of inverted (5/2),

1/4 x (2/10) = 1/20.

hence (2 + 1/4 - 1/400) =((2 + 99/400)*2, error2 = 1/400

Ahmes and Archimdes could have stopped at this point and recorded

(2 + 99/400) as a unit fraction series

[2 + (80 + 10 + 8 + 1)/400] =

[2 + 1/5 + 1/40 + 1/50 + 1/400 ]

step 3 (included Archimedes square root of three method).

divided 1/400 by (400/1798) = 1/1798,

hence (2 + 99/400 - (1/1798)^2 = accurate (1/1798)^2

Archimedes would have likely recorded

[2+ 1/5 + 1/40 + 1/50 + 1/400]

with a note that a longer series, with an error of (1/1798)^2, could be easily found.

C. square root of seven (7)

step 1: estimates Q = 2, R = 3/4 and (2 + 3/4)^2 =

7 + (3/4)^2 , error1 = 9/16

step 2: divides 9/16 by twice inverted (2 + 3/4) =

(9/16)(4/22) = 9/88 = (1/11 + 1/88)

(2 + 3/4 -9/88) = [2 + 1/2 + 13/88]^2 =

[2 + 1/2 + (8 + 4 + 1)/88]^2 =

[2 + 1/2 + 1/11 + 1/22 + 1/88]^2

step 3 may have been required

divide 9/88 by twice (2 + 1/2 + 13/88) =

(9/88)(88/466) =

9/466 = (1/155 + 1/155 + 1/466)

hence [2 + 1/2 + 1/11 + 1/22 + 1/88 - (2/155 +1/466)]^2 would have been recorded as a unit fraction series

Note that Archimedes and Ahmes may have paired

(1/22 - 2/155) =

and

(1/88 - 1/466) =

readers may choose the most likely final unit fraction series,

Footnote: The three step square root method may have preceded the development of the pesu. Or, equally likely, the inverse proportion pesu methology may have preceded the three step square root method.



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