about numismatics and some problems of algebra from a byzantine manuscript of 1436 (cod. vind. phil....

8/20/2019 About numismatics and some problems of algebra from a Byzantine manuscript of 1436 (Cod. Vind. phil. gr. 65) / …

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REVUE ---------------------------------------------------NUMISMATIQUE

Dirigée parC. Morrisson, M. Amandry,

M. Bompaire, O. Picard

Secrétaires de la rédactionFr. Duyrat, A. Hostein

C. Grandjean

ISSN 0484-8942

2011(167e volume)

SOCIÉTÉ FRANÇAISE DE NUMISMATIQUE

Diffusion : Société d’édition « Les Belles Lettres »2011

----------------------------------------------------

Publiée avec le concours de l’Institut National des Sciences Humaines et Sociales

du Centre national de la recherche Scientifique


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Stefan DESCHAUER*

About Numismatics and Some Problems of Algebrafrom a Byzantine Manuscript of 1436

(Cod. Vind. phil. gr. 65)

Summary – The Cod. Vind. phil. gr. 65 is a unique mathematical document in the late Byzantineperiod. The «second book» of the manuscript contains an algebraic theory in Italian style with

many profound and hardly known problems in the history of mathematics and in the numismatics.Until today we don’t know other algebraic works in Byzantium. I present here some typical problemsinvolving 15th century money.

Résumé – Le Cod. Vind. phil. gr. 65 est un document mathématique sans équivalent de l’époquebyzantine tardive. Le « deuxième livre » du texte contient une théorie de l’algèbre à l’italienneavec beaucoup de problèmes profonds et à peine connus dans l’histoire de la mathématique et dela numismatique. Jusqu’à aujourd’hui on ne connaît pas d’autres œuvres algébriques à Byzance.J’en présente ici quelques problèmes typiques dans lesquels gurent des monnaies du XVe siècle.

Introduction

The Codex Vindobonensis phil. gr. 65 kept in the Österreichische National-bibliothek in Vienna is a mathematical manuscript written by two anonymousauthors (here called A and B). It contains 163 leaves with 36 lines each page.

The smaller text from B presents a pure collection of arithmetical problemswritten in Saloniki about the year 1430 under Turkish occupation. The posi-tional decimal system (with a point for 0) is used and also decimal fractions toemphasize. The text has been edited, translated, and commented (in German)by HUNGER, VOGEL 1963.

On the other hand 149 leaves come from the author A. He has written amethodical reckoning book – called by himself “rst book”– and with it he usesthe positional decimal system also (with a special character for 0). A problemof multiplication (the days Anno Domini) reveals us the year of writing: 1436.

But in addition the author presents methods of approximation for roots (of2nd until 4th degree), algebra, and geometry in the “second book”.1 Up to nowno other algebraic text is known from the Byzantine mathematics!

* Prof. Dr. Stefan Deschauer Technische Universität Dresden Professur für Didaktik derMathematik 01062 Dresden, Germany. Mail: [email protected].

1. Now the text of A is completely analysed, in addition it exists an unpublished transcriptionof the “rst book”.


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Without doubt his algebraic method is the Italian one at that time: It’s aquestion of an algebra by means of words, without any symbols. Obviously A has

grecianized the Italian technical terms merely: From numero (the constant), cosa (thing, our x), censo (power, our x2), cubo (cube, our x3), and quadro (square,our x4) has become j ariqmó~ (arithmós), pr` agma ( prãgma), tzv v enson (tzénson),kov ubon (kúbon), and kv adron (kádron) or tetrvagwnon (tetrágonon).2

A completely annotated edition of the “rst book” is under way. As well I plana detailed mathematical commentary of the “second book”.

In the following treatise I want to address myself to a series of demandingproblems from algebra I have found in the “second book” of A.3

Chap. 153: Boat trips4

Somebody had boat trips and took along orins (φ5).After the rst trip he had 1 –

3 of the initial amount more and 1 φ additionally.

After the second trip he had 1 –3 of the last amount more and 2 φ additionally

etc.The number of the trips was equal to the initial number of φ.At the end he had 15 φ.How many boat trips did he undertake?What was the initial amount of φ?6

Solution of the author

He estimates the number of trips: 3 or 4 (!). Thus his approach is (3 + x) trips,i. e. (3 + x) φ by start also.

After the 1st trip the boat traveller has φ = φ.

After the 1st trip he has φ = φ.

After the 3rd trip he has φ.

2. Apart from that tzv v enson, kov ubon, and kv adron were not part of the classical Greek vocabulary.3. DESCHAUER 2005 presents a short history of Byzantine mathematics as well as a view in

content of the Codex with some (other) algebraic and geometric problems.4. See gure 1.5. Abbreviation of the monetary unit flourvi (phlurí), pl. flouriva (phluriá) in the manuscript.6. Similar problems are known from P. DELLA FRANCESCA (about 1440), L. PACIOLI (1494),

E. DE LA ROCHE (1520), and M. STIFEL (1544) – see TROPFKE 1980, p. 544.


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Author: Due to x have still a 4th trip.

His further calculation is the following:

φ =

φ

Multiplication with x: . So he linearizes the problem for the

remaining φ!Addition of (amount before the last trip) – the sum must beequal to 15:

Solution of the quadratic equation and approximation of the square root7:

Result: He had trips and just as many φ by start.

Generalization

Let a0 be the initial amount and an the amount after the nth trip.

We replace the fraction

by

for any k > 0.

Then we get the recursive formula an = 1+1

k an 1 + n, *.

From that you can also derive an explicit formula:

an = a

0 + k (1+ k )( ) 1+

1

k

n

k (1+ k + n).

The special values of the author are k = 3, a0 = n, an = 15.

7. In an earlier part of his “second book” the author has treated several approximate methodsfor working out square roots he uses now.


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Indeed he had must solve the following exponential equation:

15 = (n + 12)4

3

n

3(n + 4)

(n ≈ 3.046)

His result ≈ 3,039 achieved by linearizing (see above) leads to 14,934

approximately on the right-hand side.

Chap. 157: The gambles

3 men have hyperpyra (ǁ8). One after the other doubles the money to theother men.Finally each of them has 8 ǁ. What was the amount of everyone at thebeginning?Remark: The sum of ǁ is constant.

1. 2. 3.

8 ǁ 8 ǁ 8

ǁ

4 ǁ 4 ǁ 16

ǁ 3.

2 ǁ 14 ǁ 8 ǁ 2.

13 ǁ 7 ǁ 4

ǁ 1.

The author calculates backwards. For understanding the table in hand is useful.The constant sum of ǁ is 24. Before the nal distibution (rst line) the third manhas to double the ǁ of the others, thus they had 4 ǁ each and the third 16 ǁ(second line). A further step before the second gambler had to double the moneyof the others: The distribution was 2 ǁ, 14 ǁ, 8 ǁ – see the third line. The nextback step leads to the initial distribution: The rst man has to double the

amounts of the others.

Result: 1st 13 ǁ, 2nd 7 ǁ, 3rd 4 ǁ (see fourth line)

The author varies the problem repeatedly. His most complicated case is thefollowing:

4 men have hyperpyra (ǁ). One after the other triples the money to the othermen.Finally each of them has 9 ǁ. What was the amount of everyone at thebeginning?

8. Abbreviation of the monetary unit in the manuscript


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(The constant sum of ǁ is 36.)

His right solution (without any comment):

1st × 9 ǁ = 24 ǁ, 2nd × 9 ǁ = 8 ǁ,

3rd × 9 ǁ = 2 ǁ, 4th × 9 ǁ = 1 ǁ

For the then reader the factors probably must have been quite mysterious.

Generalization

Let’s examine the general case.n men have hyperpyra (ǁ). One after the other multiplies the money to theother men by k *. Finally each of them has 1 ǁ9. What was the amountof everyone at the beginning?

(The constant sum of ǁ is n.)

In the following steps the money of the man who has just multiplied theamounts of the others is in heavy type. The money to the right of it concernsthe men whose turn it was and the money to the left of it the men whose turn itwill be after that. The horizontal brackets are counting the number of terms.

1st step backwards:

2nd step backwards:

9. Special case for the general one.


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sth step backwards (1 ≤ s ≤ n, by induction):

Bringing the second term to the common denominator k s you get the nominator

nk s – n + s – nk s –1 + nk s –2 – nk s –2 + nk s –3 ±... – nk 2 + nk – nk + n – (s – 1) × 1= nk s – nk s –1 + 1 (telescopic sum)

Thus the sth step backwards leads to

Finally for s = n we have

The sum of the n terms is indeed.


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Examples of the author: 10;

Chap. 158: A well-known problem with a remarkable method of solution

3 men have orins (φ).

The rst says to the others: Give me1

–3 of your φ and 5 φ more.Then I will have the triple of your φ and 4 φ more.The second says to the others: Give me 1 –

2 of your φ and 3 φ more.

Then I will have the quadruple of your φ less 5 φ.The third says: I have 2 –

3 of your φ and 5 φ more.

What’s the amount of everyone?

This recreational problem is well-known from the history of mathematics11 and leads to a system of linear equations:

The author presents a remarkable method of solution. His algebra restrictsitself to a unique variable.Now his approach is that the three men have x φ all together, i. e.

(4) x1 + x2 + x3 = x

Apart from x the author describes all variables by words.

10. You nd this case at early German arithmeticians also, e. g. J. WIDMANN (1489) andChr. RUDOLFF (1525) – see TROPFKE 1980, p. 647 f.

11. See TROPFKE 1980, p. 609-611.


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Let’s reproduce roughly the solution of the author in our notation. We restrictourself to his evaluation of (1).

Let with

a) Approximation x1' = 3( x

2' + x

3' ) implies

The authors gets

b) Correction: In reality is valid. Thus you have

and

So the author gets

(6) x 1' = 34 x + 1, x2' + x 3' = 1

4 x 1.

c) Situation before handing-over of 5 φ:Let

with

according (5). With (6) follows

On the other hand you have according to (5)

With (4) follows from (8) and (9)

with the result

(10)


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Following the same method the author gets from other equations (2) and (3):

According to (4) the adding up leads to the equationthat is solved by the author:

implying

Chap. 162: Unloading of a ship12

3 men want to unload a ship.If the second and the third work 7 days, the rst nishes the job during3 days.If the rst and the third work 8 days, the second nishes the job during4 days.If the rst and the second work 6 days, the third nishes the job during2 days.

How many days does everyone need to unload the ship alone?

1. 2. 3.

386

746

782

Approach of the author:

1 day of the second = x days of the rst, i. e. 1 daily work of the second are x daily works of the rst.

With the abbreviations dw(n) for the daily work of the nth man n = 1, 2, 3,

and fw for the full work, we have the following system of linear equations:

(11) (3 + 7 x) dw(1) + 7 dw(3) = fw(12) (8 + 4 x) dw(1) + 8 dw(3) = fw(13) (6 + 6 x) dw(1) + 2 dw(3) = fw

12. See gure 2.


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If you regard 1 dw(1) as a unit, the variables x, dw(3), and fw are to determine.A comparison of (11) and (12) shows to the author

(14) 1 dw(3) = (– 5 + 3 x) dw(1)13 He multiplies this equation by 7 and substitues in (11) with the result(15) (28 x – 32) dw(1) = fw.A further substitution of (14) in (13) leads to(16) (12 x – 4) dw(1) = fw.Now the author forms the difference14 of the left sides of (15) and (16):(28 x – 32) – (12 x – 4) = 16 x – 28.

Afterwards he solves the equation3

416 28: 1 x x= =

According to his approach at the outset and to (14) he gets

(17) 1 dw(2) = 1 dw(1), 1 dw(3) = dw(1) with the consequence

(18) 7dw(1) = 4 dw(2) = 28 dw(3).

Now his comparison of (18) with the rst “service” (3 / 7 / 7) e. g. shows:

Finally it follows

The authors tests also the comparison of (18) with the other “services”,(8 / 4 / 8) respectively (6 / 6 / 2), and receives the same result.

Generally speaking the author has complete command of solving a 3 × 3system of linear equations only using one explicit variable. We nd the methodsof comparison, substraction, and substitution. With it he handles negative numbersand terms.

13. He is handling negative numbers and terms completely naturally. The term in brackets isreadable as follows: παρὰ εʹ καὶ γʹπράγματα (less 5 and 3 prágmata)

14. in the words of the author: κηʹ πράγματα παρὰ λβ ʹ παρὰ ιβ ʹ πράγματα ʹ παρὰ δʹ ἐστὶις ʹπράγματα παρὰ βηʹ. (28 prágmata less 32 less 12 prágmata less 4 is equal to 16 prágmata less 28.)


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This work problem is charming because of its combination of simultaneousworking and working in succession. The usual structure of a work problem is

the following:Let (19) be given. How many days do they need for the full job, if they work

simultaneously?

(Result: 5 days)15

An “easier method”

The author claims:

If the sums of columns in the chart are equal, then you get the result immediatly.We see that the columns in the chart above always have the sum 17, butin general the sums don’t have to be equal. This shows e. g. the following chartwith the sums 6 or 5:

1. 2. 3.

123

213

221

(Here we get fw = 9 dw(1) = 9 dw(2) = 3 dw(3).)

Let’s return to the claim of our author.One of several examples he gives for substantiating his claim is the following:

1. 2. 3.

389

749

785

Σ 20 20 20

Without any calculation he presents the correct result:

fw = 20 dw(2) = 80 dw(1) = 11 dw(3)

(From fw = 20 dw(2) the corresponding relations between fw and dw(1),dw(3) can’t be read directly, but by the solution of a system of only two linearequations.)

But the author doesn’t explain, why just dw(2) has the sum of columns asfactor. Is the same also possible with the other dw? If so, how do you make theright choice?

15. TROPFKE 1980, p. 578-581 states only this easier type of work problems.


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Indeed it’s not difcult to give a counter-example on principle:

1. 2. 3.

296

746

791

Σ 17 17 17

The calculation leads to with factorsall different from 17.

Let’s study the problem at issue more general:1 2 3

a1b1c1

a2b2c1

a2b1c2

Σ d d d

Let 1 dw(2) = x dw(1), 1 dw(3) = y dw(1) with the unit 1 dw(1), i. e. 1 dw(1):=1.From the chart we get the system of equations

(20) a1 + a2 ( x + y) = fw, b1(1 + y) + b2 x = fw, c1(1 + x) + c2 y = fwwith the additional condition(21) a1 + b1 + c1 = a2 + b2 + c1 = a2 + b1 + c2 = d Addition of the left and the right side of (20) leads to(22) d (1 + x + y) = 3 fw.

According to (22) we have 223

(1 ) fw.a

d a x y+ + = With (20) follows

Thus we need a differentiation of cases.

α) 3a2 = d implies a2 = a1 and according to (21) b2 = b1, c2 = c1.

That leads to a1 = b1 = c1 = a2 = b2 = c2 because of (20). The system (20) isreduced to just one equation, e. g. a1(1 + x + y) = fw. With constant a1 and fwthe equation is underdetermined. The claim of the author roughly examining weset fw = d dw(i), i = 1, 2, 3, i. e. either fw = 3 a1 or fw = 3 a1 x or fw = 3 a1 y.This is only fullled for x + y = 2 resp. 2 x – y = 1 resp. 2 y – x = 1 but not truein general.


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β) 3 a2 ≠ d leads to

For calculating x, y we use the second equation of (20), and (22) again:After elimination of x we get

It’s obvious that b1 = b2 leads to the case α) again. Otherwise we have

Together with (22) you receive

Now (24) and a2 – a1 = b1 – b2 according to (21) yield to

In the same way you get from (26), (24), and a2 + b2 = d – c1 according to (21)

On account of (24), (27), and (28) we are able to formulate criterions forfw = d dw(i), i = 1, 2, 3, i. e. for fw = d or fw = dx or fw = dy.

We have:

(29) fw = d iff d = a1 + 2 a2.That’s the sum of the numbers in the rst line of the chart.

(30) fw = dx iff d = 3 b1 – a2 + a1 resp. d = 2 b1 + b2 according to (21).That’s the sum of the numbers in the second line of the chart.(31) fw = dy iff d = 3 c1 – a2 + a1 resp. d = 2 c1 + c2 according to (21).That’s the sum of the numbers in the third line of the chart.


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Result

If each of the columns in the above-mentioned chart has the same sum dwith d ≠ 3a2

(i. e. differently called numbers in the chart aren’t equal), thenis valid:fw = d dw(i) iff the i th line of the chart has the sum d also. (i = 1, 2, 3)

Remark:

In both examples from the author the sum of columns is constant (17 resp. 20).In the rst example we nd that he sum of the numbers in the frst line (only)

is 17 too, and fw = 17 dw(1) is valid.In the second example the sum of the numbers in the second line (only) is20 too, and fw = 20 dw(2) is valid.

References

DESCHAUER 2005: St. DESCHAUER, Mathematik vor der Zeitenwende – einigeGlanzlichter in einer byzantinischen Handschrift von 1436 . In: Europeanmathematics in the last centuries. Stefan Banach International MathematicalCenter / Institute of Mathematics Wrocław University, ed. W. Więsław,Wrocław, 2005, S. 7-18.

HUNGER, VOGEL 1963: H. HUNGER, K. VOGEL, Ein byzantinisches Rechenbuchdes 15. Jahrhunderts – 100 Aufgaben aus dem Codex Vindobonensis phil.Gr. 65. Text, Übersetzung und Kommentar, Wien, Denkschriften der phil.-hist. Kl. der Österreichischen Akademie der Wissenschaften, 78. Band, 2.Abh., 1963, 127 p.

TROPFKE 1980: J. TROPFKE, Geschichte der Elementarmathematik, Band 1: Arithmetik und Algebra, K. VOGEL, K. REICH, H. GERICKE (eds.), Berlin -

New York, 19804

, 742 p.


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Figure 1 - Beginning of the boat trips problem

(fol. 86r, chap. 153).


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Figure 2 - Part of the problem “Unloading of a ship”

(fol. 155v, Appendix of chap. 162).

about numismatics and some problems of algebra from a byzantine manuscript of 1436 (cod. vind. phil....

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