## New Articles

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*Ref*]**Some formulas of partnership**by burgessAug 26[

*Rec*]**Kenosymplirostic numbers**by imaginary.iAug 14[

*Edu*]**How to find whether a given number is prime or not...**by burgessAug 12[

*Edu*]**BODMAS Rule application**by burgessAug 8[

*Edu*]**Tests of Divisibility- Simple tricks**by burgessAug 7[

*Res*]**0/0 is possible and has an answer**by imaginary.iAug 2[

*Ref*]**Sophomore's dream**by pahioJul 9[

*Res*]**examples of growth of perturbations in chemical or...**by rspuzioMay 24[

*Ref*]**proof of Stirling's approximation**by rspuzioMay 8[

*Res*]**Example of stochastic matrix of mapping**by rspuzioApr 23[

*Res*]**6. Discussion**by rspuzioApr 20[

*Res*]**5. Entanglement**by rspuzioApr 20[

*Res*]**4. Measurement**by rspuzioApr 20[

*Res*]**3. Distributed dynamical systems**by rspuzioApr 20## Latest Messages

Aug 28

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Aug 22

Aug 22

Aug 18

Aug 17

Aug 14

Aug 9

Aug 8

Aug 7

Aug 7

Aug 6

Aug 4

Aug 2

When we carry out the following operations we get a quotient
which is a Gaussian integer: a)((21+i)^20-1)/21 b)((21+i)^(21-i)-1)/21
c)((21+i)^(21-i)-i/21 and d)((341+i)^340-1)/341
Note that a) 341 is also a pseudoprime in k(1) and b) i is also one of the unities
in k(i).

Aug 25

Dear planetmath.org:
It has been several months since I have been able to perform a successful search at your site. The fact that I have not been able to find articles either by searching, nor by MSC has rendered (at least for me) planetmath.org practically unusable.
In your home page, I see that new articles are being contributed regularly. This makes me wonder: Is there a way to sidestep the search problem? Is there am alternative way to get to an article (say, "Pascal's formula", or "Pascal's rule")? Is there an alternative way of reaching articles under MSC 05A10?
Thanks in advance.>

Aug 22

In 1987, Egyptologist Gay Robins, and Charles Shute, wrote a book on the Rhind Mathematical Papyrus (RMP). Five years later Egyptologist John Legon wrote on the KP and the same class of arithmetic proportions used in the RMP. The KP and RMP report the same arithmetic proportion method to find the largest term. The method: take 1/2 of the difference, 1/2 of 5/6 (5/12 in the KP) times the number of differences (nine times 5/12 = 15/4 in the KP) plus the sum of the A.P progression (100 in the KP) divided by the number of terms (10 meant 100/10 = 10 in the KP). Finally add column 11’s result, 3 3/4, to 10, and the largest term, 13 3/4.
In unit fractions, the context of the text, add column 11: 5/12 times 9 writing 3 3/4 as 3 2/3 1/12 to 10 in column 12 beginning with the largest term 13 2/3 1/12. The scribe subtracted 5/6 nine times created remaining terms of the arithmetic progression.
Robins-Shute confused aspects of the problem by omitting the sum divided by the number of terms, a topic cited in a closely related RMP 40 problem. A scribal algebraic statement matched pairs added to 20 reporting five pairs summed to 100, a set of facts included in RMP 40.
The complete KP method found the largest term facts reported in RMP 64 and RMP 40 by John Legon in 1992. Scholars have parsed Rhind Mathematical Papyrus 40 a problem that asked that 100 loaves of bread to be shared between five men by finding the smallest term of an arithmetic progression.
C. A confirmation of the Kahun Papyrus arithmetic progression method must include discussions of RMP 40 and RMP 64. In RMP 64 Ahmes asked 10 men to share 10 hekats of barley with a differential of 1/8 defining an arithmetical progression. Robins and Shute reported: ”the scribe knew the rule that, to find the largest term of the arithmetical progression, he must add half the difference to the average number of terms as many times as there are common differences, that is, one less than the number of terms”.
1. number of terms: 10
2. arithmetical progression difference: 1/8
3. arithmetic progression sum: 10
The scribe used the following facts to find the largest term.
1. one-half of differences, 1/16, times number of terms minus one, 9,
1/16 times 9 = 9/16
2. The computed parameter(1), was found by 10, the sum, divided by 10, the number of terms. It was inserted by Robins-Shute, but had not been high-lighted, citing 1 + 1/2 + 1/16, or 1 9/16, the largest term. The remaining nine terms were found by subtracting 1/8 nine times to obtain the remaining barley shares.
That is, the KP scribe used formula 1.0:1111111
(1/2)d(n-1) + S/n = Xn (formula 1.0)
with,
d = differential, n = number of terms in the series, S = sum of the series, Xn = largest term in the series allowed three(of the four) parameters: d, n, S and Xn, to algebraically find the fourth parameter.
When n was odd, x (n/2) = S/n,
and x 1 + xn = x2 + x(n -1) = x3 + x(n -2) = … = x(n/2) = S/n,
Note that Robins-Shute omitted the sum divided by the number of terms (S/n):
A modern footnote cites Carl Friedrich Gauss implementing as a grammar school student a solution to the n = even case. Ahmes and Gauss found the sum for 1 to 100 by using d = 1 following the same rule. Ahmes and Gauss reached the sum 5050 based on 50 pairs of 101 (1 + 101 = 2 + 99 = 3 + 98 = …) by using an identical arithmetic progression rule.

Aug 22

The KP http://planetmath.org/kahunpapyrusandarithmeticprogressions method was known to Ahmes in 1650 BCE

Aug 18

\begin{itemize}
\item
\end{itemize}\begin{flushleft}
\end{flushleft}
\textbf{}gg

Aug 17

561 is a pseudoprime to any base of shape (11*k+I) where k
belongs to N. This is in addition to the other bases indicated in the previous message.

Aug 14

There is an exception to "years divisible by 4 being leap years": If a
year is divisible by 100 (such years are divisible by 4), such a year is
a leap year ONLY IF it is also divisible by 400. For example, 2000 was a
leap year, but 1900 was not. Shockley's "Introduction to Number Theory"
contains a "day of the week" formula that includes the above fact.

Aug 9

Carmichael numbers constitute a sub-set of Devaraj numbers. To understand more about them
refer sequences A 104016, A 104017 and A 162290. Some interesting facts pertaining to
them will follow.

Aug 8

The reason was a dollar-sign error =o)
Corrected!
BTW, the PM search engine has long been out of order. It's quite
difficult to find entries on a wanted subject.

Aug 7

Pl. see the older entry http://planetmath.org/divisibilitytest

Aug 7

Definition: These are the impossible prime factors of 3^n - 2 (n belongs to N).
This is identical with the sequence A123239 (OEIS ).

Aug 6

Nice solution

Aug 4

Please see e.g. the entry http://planetmath.org/division
BTW, one cannot write "P(R)*0 = 0" since your P(R) is not a _number_

Aug 2

under repair ... For now consider a \PMlinkexternal{generalized scribal square root method}{http://planetmath.org/squarerootof3567and29} was published in Dec. 2012.
I INTRODUCTION: An ancient square root method was decoded in 2012. Scholars for 100 years failed to fully decode Archimedes’ three steps that estimated unit fraction series answers to four to six places (modern standards), a method used by Fibonacci and Galileo.
Unresolved aspects of ancient ESTIMATION OF PI problem were reported by Kevin Brown and E.B. Davis with upper and lower limits;
(1351/780)^2 is greater than PI is greater than (265/153)^2
A. Archimedes’ actual square root of pi^2 and n^2 method, decoded in Dec 2012 and Jan 2013, calculated the higher PI limit(1351/780)^2 by:
1. step 1. guess (1 + 2/3)^2 = 1 + 4/3 + 4/9 = 2 + 3/9 + 4/9, meant 2/9 = error1
2. step 2 reduced error1 2/9
by dividing 2/9 by 2(1 + 2/3)
steps that meant
2/9 x (3/10) = 1/15
such that
(1 + 2/3 + 1/15)^2, error2 (1/15)^2 = 1/225 = error2
knowing (1 + 11/15) = 26/15
3. step 3 reduced error2 = 1/225 by dividing by 2 x (26/15) = 52/15
1/225 x (15/52) = 1/15 x (1/52) = (1/780)^2 = error 3
reached
(26/15 - 1/780)^2 = (1351/780)^2 in modern fractions
recorded a unit fraction series that began with step 2 data and subtracted 1/780
(1 + 2/3 + 1/15 - 1/780)^2
as Archimedes would have written
(1 + 2/3 + 1/30 + (13 + 6 + 4 + 2)/780)^2 = (1 + 2/3 + 1/30 + 1/60 + 1/+ 1/195 + 1/390)^2
B . The lower limit 265/153 modified step 2, used
1/17 rather than 1/15, (1+ 2/3 + 1/17) = (1 + 37/51)
such that (1 + 111/153)changed to (1 + 112/153) = 265/153
II. PROOF: Modern translations of scribal square roots of five (5), six (6), seven (7), (29) and any rational numberPlanetmathPlanetmathPlanetmath are demonstrated below (A, B, C, D).
Greek and Egyptian algebraic steps were finite. Decoding algorithmic looking finite arithmetic steps 2, 3 and 4 have been demystified.
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
(2 + 1/4)^2 = 5 + (1/4)^2; error1 = 1/16
step 2, reduced error1 that divided
1/16 by 2(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 suggested by academics prior to Dec. 2012 suggested incomplete square root steps even though major operational aspects of the same class of arithmetic and algebraic pesu steps have been found in the medieval era. In May 2013 the shorthand notes of Galileo reveal the same method was also used by Fibonacci and Archimedes.
In the square root of five 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 by (2 + 1/2)
such that
1/4 x (2/10) = 1/20.
hence (2 + 1/4 - 1/400) =((2 + 99/400)*2, error2 = 1/400
Ahmes, Archimes and Fibonacci may 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 (as included Archimedes square root of three method) was optionasl
divided 1/400 by (400/1798) = 1/1798,
hence (2 + 99/400 - (1/1798)^2 = accurate (1/1798)^2
Archimedes’ actual square root method would have recorded
[2+ 1/5 + 1/40 + 1/50 + 1/400]
with a note that a longer series, with an error of (1/1798)^2 was 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 (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 paired
(1/22 - 2/155) =
and
(1/88 - 1/466) =
readers may choose the most likely final unit fraction series,
D. ESTIMATE the square root of 29.
step 1 found R = (29-25)/2Q = 4/10 = 2/5
such that
(5 + 2/5)^2 = 29 + 4/25 = error1
STEP 2
reduce error1 4/25 by dividing by 2(5 + 2/5)
4/25 x 5/54 = 2/27
hence a final unit fraction series converted
(5 + 2/5 - 2/27)^2 by considering
(5 + 1/5 + (27-10)/135)^2 = (5 + 1/5 + 1/9 + 2/135)^2
was accurate to (2/27)^2
NOTE: the conversion of 2/135 to a unit fraction series followed Ahmes 2/n table rules
(2/5)(1/27) = (1/3 + 1/15)(1/27) = 1/51 + 1/405
MEANT THE SQUARE ROOT OF 29 WAS ESTIMATED IN TWO STEPS BY
(5 + 1/5 + 1/9 + 1/51 + 1/405)^2
footnote: Fibonacci’s square root of 17 method was appropriately cited as used by Galileo though not properly analyzed in every detail. Fibonacci guessed (4 + 1/8)^2 = (17 + 1/64) , and Fibonacci reduced the estimated 1/64 error foumd an inversePlanetmathPlanetmathPlanetmath proportion:1/64 x 8/66 = 1/528 which meant (4 + 1/8 - 1/528)^2 = (2177/528)^2 = 17.000003 is accurate to (1/528)^2 perArchimedes and not by Newton, as suggested by scholars prior to 2012.
III CONCLUSION
Unit fractionPlanetmathPlanetmath square root was formalized by 2050 BCE and used by Egyptians, Greeks, Arabs, medieval scribes and as late as Galileo. The method estimated irrational square roots of N by 1-step, 2-step, 3-step and 4-steps methods. Step 1 guessed quotients (Q) and remainders (R) = n/(2Q) with n = (N - Q^2). Step 2, 3, and 4 reduced error 1, 2 and 3 associated with the previous step by dividing by 2(Q + R).
References
1 A.B. Chace, Bull, L, Manning, H.P., Archibald, R.C., The Rhind Mathematical Papyrus, Mathematical
2 Marshall Clagett Ancient Egyptian Science, Volume III, American Philosophical Society, Philadelphia, 1999.
3 Richard Gillings, Mathematics in the Time of the Pharaohs, Dover Books, 1992, PAGE 214-217.
4 H. Schack-Schackenburg, ”Der Berliner Papyreys 6619”, Zeitscrift fur Agypyische Sprache , Vol 38 (1900), pp. 135-140 and Vol. 40 (1902), p. 65f.
5 L. E. Sigler, Fibonacci’s Liber Abaci, Leonardo’s Book of Calculation ,Springer, NY, 2002, page 491.

## Latest Messages

Aug 28

Aug 25

Aug 22

Aug 22

Aug 18

Aug 17

Aug 14

Aug 9

Aug 8

Aug 7

Aug 7

Aug 6

Aug 4

Aug 2

When we carry out the following operations we get a quotient
which is a Gaussian integer: a)((21+i)^20-1)/21 b)((21+i)^(21-i)-1)/21
c)((21+i)^(21-i)-i/21 and d)((341+i)^340-1)/341
Note that a) 341 is also a pseudoprime in k(1) and b) i is also one of the unities
in k(i).

Aug 25

Dear planetmath.org:
It has been several months since I have been able to perform a successful search at your site. The fact that I have not been able to find articles either by searching, nor by MSC has rendered (at least for me) planetmath.org practically unusable.
In your home page, I see that new articles are being contributed regularly. This makes me wonder: Is there a way to sidestep the search problem? Is there am alternative way to get to an article (say, "Pascal's formula", or "Pascal's rule")? Is there an alternative way of reaching articles under MSC 05A10?
Thanks in advance.>

Aug 22

In 1987, Egyptologist Gay Robins, and Charles Shute, wrote a book on the Rhind Mathematical Papyrus (RMP). Five years later Egyptologist John Legon wrote on the KP and the same class of arithmetic proportions used in the RMP. The KP and RMP report the same arithmetic proportion method to find the largest term. The method: take 1/2 of the difference, 1/2 of 5/6 (5/12 in the KP) times the number of differences (nine times 5/12 = 15/4 in the KP) plus the sum of the A.P progression (100 in the KP) divided by the number of terms (10 meant 100/10 = 10 in the KP). Finally add column 11’s result, 3 3/4, to 10, and the largest term, 13 3/4.
In unit fractions, the context of the text, add column 11: 5/12 times 9 writing 3 3/4 as 3 2/3 1/12 to 10 in column 12 beginning with the largest term 13 2/3 1/12. The scribe subtracted 5/6 nine times created remaining terms of the arithmetic progression.
Robins-Shute confused aspects of the problem by omitting the sum divided by the number of terms, a topic cited in a closely related RMP 40 problem. A scribal algebraic statement matched pairs added to 20 reporting five pairs summed to 100, a set of facts included in RMP 40.
The complete KP method found the largest term facts reported in RMP 64 and RMP 40 by John Legon in 1992. Scholars have parsed Rhind Mathematical Papyrus 40 a problem that asked that 100 loaves of bread to be shared between five men by finding the smallest term of an arithmetic progression.
C. A confirmation of the Kahun Papyrus arithmetic progression method must include discussions of RMP 40 and RMP 64. In RMP 64 Ahmes asked 10 men to share 10 hekats of barley with a differential of 1/8 defining an arithmetical progression. Robins and Shute reported: ”the scribe knew the rule that, to find the largest term of the arithmetical progression, he must add half the difference to the average number of terms as many times as there are common differences, that is, one less than the number of terms”.
1. number of terms: 10
2. arithmetical progression difference: 1/8
3. arithmetic progression sum: 10
The scribe used the following facts to find the largest term.
1. one-half of differences, 1/16, times number of terms minus one, 9,
1/16 times 9 = 9/16
2. The computed parameter(1), was found by 10, the sum, divided by 10, the number of terms. It was inserted by Robins-Shute, but had not been high-lighted, citing 1 + 1/2 + 1/16, or 1 9/16, the largest term. The remaining nine terms were found by subtracting 1/8 nine times to obtain the remaining barley shares.
That is, the KP scribe used formula 1.0:1111111
(1/2)d(n-1) + S/n = Xn (formula 1.0)
with,
d = differential, n = number of terms in the series, S = sum of the series, Xn = largest term in the series allowed three(of the four) parameters: d, n, S and Xn, to algebraically find the fourth parameter.
When n was odd, x (n/2) = S/n,
and x 1 + xn = x2 + x(n -1) = x3 + x(n -2) = … = x(n/2) = S/n,
Note that Robins-Shute omitted the sum divided by the number of terms (S/n):
A modern footnote cites Carl Friedrich Gauss implementing as a grammar school student a solution to the n = even case. Ahmes and Gauss found the sum for 1 to 100 by using d = 1 following the same rule. Ahmes and Gauss reached the sum 5050 based on 50 pairs of 101 (1 + 101 = 2 + 99 = 3 + 98 = …) by using an identical arithmetic progression rule.

Aug 22

The KP http://planetmath.org/kahunpapyrusandarithmeticprogressions method was known to Ahmes in 1650 BCE

Aug 18

\begin{itemize}
\item
\end{itemize}\begin{flushleft}
\end{flushleft}
\textbf{}gg

Aug 17

561 is a pseudoprime to any base of shape (11*k+I) where k
belongs to N. This is in addition to the other bases indicated in the previous message.

Aug 14

There is an exception to "years divisible by 4 being leap years": If a
year is divisible by 100 (such years are divisible by 4), such a year is
a leap year ONLY IF it is also divisible by 400. For example, 2000 was a
leap year, but 1900 was not. Shockley's "Introduction to Number Theory"
contains a "day of the week" formula that includes the above fact.

Aug 9

Carmichael numbers constitute a sub-set of Devaraj numbers. To understand more about them
refer sequences A 104016, A 104017 and A 162290. Some interesting facts pertaining to
them will follow.

Aug 8

The reason was a dollar-sign error =o)
Corrected!
BTW, the PM search engine has long been out of order. It's quite
difficult to find entries on a wanted subject.

Aug 7

Pl. see the older entry http://planetmath.org/divisibilitytest

Aug 7

Definition: These are the impossible prime factors of 3^n - 2 (n belongs to N).
This is identical with the sequence A123239 (OEIS ).

Aug 6

Nice solution

Aug 4

Please see e.g. the entry http://planetmath.org/division
BTW, one cannot write "P(R)*0 = 0" since your P(R) is not a _number_

Aug 2

under repair ... For now consider a \PMlinkexternal{generalized scribal square root method}{http://planetmath.org/squarerootof3567and29} was published in Dec. 2012.
I INTRODUCTION: An ancient square root method was decoded in 2012. Scholars for 100 years failed to fully decode Archimedes’ three steps that estimated unit fraction series answers to four to six places (modern standards), a method used by Fibonacci and Galileo.
Unresolved aspects of ancient ESTIMATION OF PI problem were reported by Kevin Brown and E.B. Davis with upper and lower limits;
(1351/780)^2 is greater than PI is greater than (265/153)^2
A. Archimedes’ actual square root of pi^2 and n^2 method, decoded in Dec 2012 and Jan 2013, calculated the higher PI limit(1351/780)^2 by:
1. step 1. guess (1 + 2/3)^2 = 1 + 4/3 + 4/9 = 2 + 3/9 + 4/9, meant 2/9 = error1
2. step 2 reduced error1 2/9
by dividing 2/9 by 2(1 + 2/3)
steps that meant
2/9 x (3/10) = 1/15
such that
(1 + 2/3 + 1/15)^2, error2 (1/15)^2 = 1/225 = error2
knowing (1 + 11/15) = 26/15
3. step 3 reduced error2 = 1/225 by dividing by 2 x (26/15) = 52/15
1/225 x (15/52) = 1/15 x (1/52) = (1/780)^2 = error 3
reached
(26/15 - 1/780)^2 = (1351/780)^2 in modern fractions
recorded a unit fraction series that began with step 2 data and subtracted 1/780
(1 + 2/3 + 1/15 - 1/780)^2
as Archimedes would have written
(1 + 2/3 + 1/30 + (13 + 6 + 4 + 2)/780)^2 = (1 + 2/3 + 1/30 + 1/60 + 1/+ 1/195 + 1/390)^2
B . The lower limit 265/153 modified step 2, used
1/17 rather than 1/15, (1+ 2/3 + 1/17) = (1 + 37/51)
such that (1 + 111/153)changed to (1 + 112/153) = 265/153
II. PROOF: Modern translations of scribal square roots of five (5), six (6), seven (7), (29) and any rational numberPlanetmathPlanetmathPlanetmath are demonstrated below (A, B, C, D).
Greek and Egyptian algebraic steps were finite. Decoding algorithmic looking finite arithmetic steps 2, 3 and 4 have been demystified.
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
(2 + 1/4)^2 = 5 + (1/4)^2; error1 = 1/16
step 2, reduced error1 that divided
1/16 by 2(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 suggested by academics prior to Dec. 2012 suggested incomplete square root steps even though major operational aspects of the same class of arithmetic and algebraic pesu steps have been found in the medieval era. In May 2013 the shorthand notes of Galileo reveal the same method was also used by Fibonacci and Archimedes.
In the square root of five 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 by (2 + 1/2)
such that
1/4 x (2/10) = 1/20.
hence (2 + 1/4 - 1/400) =((2 + 99/400)*2, error2 = 1/400
Ahmes, Archimes and Fibonacci may 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 (as included Archimedes square root of three method) was optionasl
divided 1/400 by (400/1798) = 1/1798,
hence (2 + 99/400 - (1/1798)^2 = accurate (1/1798)^2
Archimedes’ actual square root method would have recorded
[2+ 1/5 + 1/40 + 1/50 + 1/400]
with a note that a longer series, with an error of (1/1798)^2 was 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 (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 paired
(1/22 - 2/155) =
and
(1/88 - 1/466) =
readers may choose the most likely final unit fraction series,
D. ESTIMATE the square root of 29.
step 1 found R = (29-25)/2Q = 4/10 = 2/5
such that
(5 + 2/5)^2 = 29 + 4/25 = error1
STEP 2
reduce error1 4/25 by dividing by 2(5 + 2/5)
4/25 x 5/54 = 2/27
hence a final unit fraction series converted
(5 + 2/5 - 2/27)^2 by considering
(5 + 1/5 + (27-10)/135)^2 = (5 + 1/5 + 1/9 + 2/135)^2
was accurate to (2/27)^2
NOTE: the conversion of 2/135 to a unit fraction series followed Ahmes 2/n table rules
(2/5)(1/27) = (1/3 + 1/15)(1/27) = 1/51 + 1/405
MEANT THE SQUARE ROOT OF 29 WAS ESTIMATED IN TWO STEPS BY
(5 + 1/5 + 1/9 + 1/51 + 1/405)^2
footnote: Fibonacci’s square root of 17 method was appropriately cited as used by Galileo though not properly analyzed in every detail. Fibonacci guessed (4 + 1/8)^2 = (17 + 1/64) , and Fibonacci reduced the estimated 1/64 error foumd an inversePlanetmathPlanetmathPlanetmath proportion:1/64 x 8/66 = 1/528 which meant (4 + 1/8 - 1/528)^2 = (2177/528)^2 = 17.000003 is accurate to (1/528)^2 perArchimedes and not by Newton, as suggested by scholars prior to 2012.
III CONCLUSION
Unit fractionPlanetmathPlanetmath square root was formalized by 2050 BCE and used by Egyptians, Greeks, Arabs, medieval scribes and as late as Galileo. The method estimated irrational square roots of N by 1-step, 2-step, 3-step and 4-steps methods. Step 1 guessed quotients (Q) and remainders (R) = n/(2Q) with n = (N - Q^2). Step 2, 3, and 4 reduced error 1, 2 and 3 associated with the previous step by dividing by 2(Q + R).
References
1 A.B. Chace, Bull, L, Manning, H.P., Archibald, R.C., The Rhind Mathematical Papyrus, Mathematical
2 Marshall Clagett Ancient Egyptian Science, Volume III, American Philosophical Society, Philadelphia, 1999.
3 Richard Gillings, Mathematics in the Time of the Pharaohs, Dover Books, 1992, PAGE 214-217.
4 H. Schack-Schackenburg, ”Der Berliner Papyreys 6619”, Zeitscrift fur Agypyische Sprache , Vol 38 (1900), pp. 135-140 and Vol. 40 (1902), p. 65f.
5 L. E. Sigler, Fibonacci’s Liber Abaci, Leonardo’s Book of Calculation ,Springer, NY, 2002, page 491.