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Number Theory in Byzantium
according to Codex
Vindobonensis phil. Graecus 65
of the 15th cent.
The numbers and the positional numeral
system
Dr. Μaria Chalkou
Ιnternational Congress MICOM 2015
Athens 22-26 of September
• Codex Vindobonensis Phil. Gr. 65, is a 15th
century Byzantine manuscript kept in the
National Library of Austria in Vienna.
• In this article we describe elements from
Number theory, as these are presented in the
Codex 65
• This Ms contains a comprehensive programme
for teaching mathematics, addressed to an
audience consisting of students probably of all
the grades of what is today’s primary and
secondary education, but also state functionaries,
merchants, craftsmen of various specialities such
as silversmiths and goldsmiths, builders, etc.
Dr. Maria Chalkou
http:// www.drchalkou.simplesite.com
Codex Vindobonensis phil. Gr. 65
• Ιt is the first Greek Ms in which there appears the
problem of the construction of a square inscribed in
an equilateral triangle, so that one of its sides
touches one side of the triangle.
Τhe theoretical part of the construction is not
recorded in the Ms; what is only asked is to calculate
the side of the square in relation to the side of the
triangle
• The Ms is of considerable importance since the
scientific research carried out indicated that it is in
essence the Mathematical Encyclopaedia of the
Byzantines, and possibly the first Mathematical
Encyclopaedia Dr. Maria Chalkou
http:// www.drchalkou.simplesite.com
Dr. Maria Chalkou
http:// www.drchalkou.simplesite.com
Dr. Maria Chalkou
http:// www.drchalkou.simplesite.com
Dr. Maria Chalkou
http:// www.drchalkou.simplesite.com
Dr. Maria Chalkou http://
www.drchalkou.simplesite.com
The initial study of the codex was published in 2004 by
NKUA as a Doctoral Thesis by the Department of
Mathematics
The improved and completed edition of the codex was
published in 2006 by the Byzantine Research Centre of
AUTΗ. This part of the codex, which had remained
unpublished until then, since studies of the content of
such Mss are rare, was called
TRACTATUS MATHEMATICUS VINDOBONENSIS
GRAECUS
Dr. Maria Chalkou
http:// www.drchalkou.simplesite.com
The units of Tractatus Mathematicus Vindobonensis Graecus
UNIT 1. Operations between real numbers.
UNIT 2. Fractions, ratios, proportions.
UNIT 3. Progressions.
UNIT 4. Problems on first-degree equations. Solution through
practical arithmetic.
UNIT 5. Interest on loans or debts.
UNIT 6. Problems of sharing in proportional parts.
UNIT 7. Problems on manufacture of gold and silver.
UNIT 8. Roots of real numbers.
UNIT 9. Solution of equations.
UNIT 10. Systems of equations.
UNIT 11. Plane geometry.
UNIT 12. Areas of plane figures.
UNIT 13. Solid geometry.
Dr. Maria Chalkou
http:// www.drchalkou.simplesite.com
Table of correspondence between Greek and Arabic numbers
(alfa) α 1 (giota) ι 10 (ro) ρ 100 ˏα 1000
(vita) β 2 (kapa) κ 20 (sigma) σ 200 ˏβ 2000
(gama) γ 3 (lamda) λ 30 (taf) τ 300 ˏγ 3000
(delta) δ 4 (mi) μ 40 (ipsilon) υ 400 ˏδ 4000
(epsilon) ε 5 (ni) ν 50 (fi) φ 500
(stigma) ς 6 (ksi) ξ 60 (chi) χ 600
(zita) ζ 7 (omikron) ο 70 (psi) ψ 700
(ita) η 8 (pi) π 80 (omega) ω 800
(thita) θ 9
(koppa) Ϟ 90
(sabi) Ϡ 900
ɥ 0
In the Transcription, and anywhere else in the Edition of the Codex 65 it was
necessary to be done, the symbol u was used instead of чto declare 0.
Dr. Maria Chalkou
http:// www.drchalkou.simplesite.com
UNIT 1
The symbols used for the numbers are the letters of
the Greek alphabet, but the calculations are done
with the then new decimal Arabic numeration.
The author appears not to have adjusted to the new
method.
Yet it must be stressed that using letters and not
numbers did not influence the result since the
positional numeral system was used, that is the
position of the letter determined its arithmetical
value.
Dr. Maria Chalkou
http:// www.drchalkou.simplesite.com
Examples of writing numbers
ρκγ denotes number 123. So does αβγ.
Ϡ Ϟε denotes number 995. So does θθε.
‫؍‬ετξ denotes number 5360. So does εγςu.
We observe that the numbers in the examples are symbolized
by the author in the Greek system of writing in two ways. The
second way can be used because each letter takes its value
according to its position in the number.
Dr. Maria Chalkou
http:// www.drchalkou.simplesite.com
Execution of the operations
First example η (eight) by βα (21)
η 8
βα 21
αςη 168
We observe that the operations do not differ from those we would execute today, that is we
would multiply α (one) by η (eight), and exactly below α and η we would write the result,
which is again η. Next we would multiply β (two) by η (eight) and we would write ας (16), in
which case the final result of the operation will be αςη (168).
In the above example we can observe that the numbers η, α and η are placed on the right-
hand column and denote the units. Numbers βand ςare placed on the second column and
denote the tens, and the number αon the third column and denotes one hundred.
The important element here is that the result αςη denotes the number 168 according to the
operations. This number could also be symbolized as ρξη since ρdenotes one hundred, ξ
the number 60, and ηthe number eight. That is we see that when ς(six) is in the second
place from the right, it denotes six tens, and when αis in the third place from the right, it
denotes one hundred. Therefore each letter takes an arithmetical value according to its
position in the number. Dr. Maria Chalkou
http:// www.drchalkou.simplesite.com
As a second example we mention the multiplication of αγ (13) by αε (15). This
operation appears in the manuscript as follows:
αγ α 13 1
αε 15
αθε 195
The author multiplies γ (three) by ε (five) and has ιε (15). He writes ε and
carries α.
As he says, he multiplies αγ by αε „crosswise“, that is he multiplies α by ε,
which gives a result ε, and γ by α, which gives a result γ, and adds the results
ε and γ, in which case he has a result η. He adds α, which he has carried
from the first multiplication, to η, and he places θ, which is their sum, in the
place of tens.
Finally, he multiplies α of αγ by α of αε and writes their product, which is
again α, in the place of hundreds. Thus the result of the multiplication is the
number αθε or ρϟε (195).
Dr. Maria Chalkou
http:// www.drchalkou.simplesite.com
These operations by the anonymous author are
justified by applying the distributive law of
multiplication over addition and as follows:
αγ is written as α.10+γ, and αε as α.10+ε. Therefore
the product of αγ by αε is:
αγ.αε=(α.10+γ). (α.10+ε)=
α.α.100+α.ε.10+γ.α.10+γ.ε=
α.α.100+α.ε.10+γ.α.10+α.10+ε=
α.α.100+(α.ε+γ.α+α).10+ε=
100+(5+3+1).10+5=
100+90+5=195
Dr. Maria Chalkou
http:// www.drchalkou.simplesite.com
As a third example we mention the multiplication of αε (15) by αβγ
(123).
u α ε α 0 1 5 1
α β γ α 1 2 3 1
α η δ ε 1 8 4 5
First he multiplies γ by ε and he has αε. He writes ε and carries α.
Next he multiplies α by γ, β by ε, and adds α to them. The result is
αδ. He writes δ and carries α.
He multiplies „ouden“ by γ, α by ε, β by α, and adds α to what he
finds. The result is η.
He multiplies „ouden“ by β, and α by α. The result is α. The
product of the multiplication is αηδε.
According to the distributive law of multiplication over addition we
have:
15.123= (10+5)(100+2.10+3)=
1000+2.100+3.10+5.100+10.10+3.5=
1000+8.100+4.10+5= 1845 Dr. Maria Chalkou
http:// www.drchalkou.simplesite.com
On the check of the operation of multiplication
αε 15/7= 2+1/7
ς 6/7 1.6= 6
θu 90/7= 12+6/7
The author asks for the remainder of the division of 15 by seven,
which is the unit.
Because the remainder of the division of six by seven is six, he
multiplies the unit by six and places the product, which equals six,
in a circle.
Finally he finds the remainder of the division of 90 by seven, which
is six, and compares it with the number he has placed in the circle,
that is 6. If the two results are the same, then the multiplication is
correct. Dr. Maria Chalkou
http:// www.drchalkou.simplesite.com
His method is based on the following:
If we symbolize the multiplicand with M and the
multiplier with Ρ, then
M.Ρ ≡ M.Ρ (mod 7),
and if
M=7k+m, and Ρ=7l+n
with k, l, m, n Є ℕ and m, n < 7,
M.Ρ ≡ (7k+m).(7l+n) (mod 7),
M.Ρ ≡ m.n+(7k.l+k.n+l.m).7 (mod 7),
M.Ρ ≡ m.n (mod 7)
This is the mathematical justification of the method
with the remainders of the division by seven, which is
used by the author of the 15th century.
Dr. Maria Chalkou
http:// www.drchalkou.simplesite.com
On the check of the operation of addition
In another problem the Byzantine author deals with the addition
of final sums which result from various loan accounts. He adds
1695+1393+3454+4565 and finds 11107.
Nowadays we do not refer to a check of additions with more
than two addends. However, in Codex Vindobonensis phil. gr. 65
there is a check with the same method which the author applies
to the operation of multiplication, which is as follows:
He divides each addend by seven, in which case from the first
division he has the unit (1) as remainder, from the second one a
remainder zero (0), from the third one a remainder three (3),
and from the fourth one as remainder the unit (1).
Next he divides the sum 11107 by 7 and he finds a remainder
five (5). He adds the remainders of the four addends 1+0+3+1,
and again he finds (5). Then he writes that the addition is
correct. Dr. Maria Chalkou
http:// www.drchalkou.simplesite.com
This procedure can be justified as follows:
We symbolize the above addends with
Α1, Α2, Α3, Α4,
the respective remainders of their division by seven with
u1= 1, u2= 0, u3= 3, u4= 1,
and the respective quotients of the same divisions with
k, l, m, n .
Then we will have:
Α1+Α2+Α3+Α4 ≡ Α1+Α2+Α3+Α4 (mod 7),
In which case
7k+u1+7l+u2+7m+u3+7n+u4 ≡ Α1+Α2+Α3+Α4 (mod 7),
that is
u1+u2+u3+u4+7(k+l+m+n) ≡ Α1+Α2+Α3+Α4 (mod 7),
or
u1+u2+u3+u4 ≡ Α1+Α2+Α3+Α4 (mod 7)
Dr. Maria Chalkou
http:// www.drchalkou.simplesite.com
Thus we observe that apparently
the problems are dealt with, by
using Arithmetic, while in essence
the author seems to have a good
knowledge of Number Theory,
elements of which he uses in his
teaching
Dr. Maria Chalkou
http:// www.drchalkou.simplesite.com
Dr. Maria Chalkou http://
www.drchalkou.simplesite.com
The text after the transcription
ροα' Περὶ τοῦ πῶς ἐστὶ ἐντὸς ἡμισφαιρίου σχήματος, τεθῆναι
τετράγωνον σχῆμα ἰσόπλευρον καὶ εἰδέναι ἑκάστη πλευρὰν
πόσων σπιθαμῶν ἐστί.
Ἔστω ἡμισφαίριον σχῆμα ὅπερ ἐστὶ ἡ διάμετρος αὐτοῦ σπιθαμῶν
ιβ, τεθημένου δὲ ἐντός, σχῆμα τετράγωνον ἰσόπλευρον κατὰ τὸ
ἐγχωροῦν μεῖζον μέγεθος. Ζητεῖς εἰδέναι ἑκάστη πλευρὰν πόσων
σπιθαμῶν ἐστί. Ἔχεις δὲ τοῦτο εἰδέναι οὕτως: Πολλαπλασίασον
εἰς ἑαυτὴ τὴν διάμετρον τοῦ ἡμισφαιρίου ἥτις ἐστὶ σπιθαμῶν ιβ
ὡς εἴπομεν· ιβ-κις οὖν ιβ γίνονται ρμδ. Μέρισον ταῦτα πάντοτε
μετὰ τῶν ε. Γίνεταί δε νῦν ὁ τούτων διαμερισμὸς κη καὶ δ/ε.
Ζήτει τὴν ρίζαν τοῦ κη καὶ δ/ε ἥτις ἐστὶ ε καὶ ια/λ. Τὰ γὰρ ε καὶ ια/λ
τεχνικῶς πολλαπλασιαζόμενα εἰς ἑαυτὰ πολλαπλασιάζοσιν κη καὶ
ψκα/Ϡ ἅπερ ἐστὶ κη καὶ δ/ε. Ἐστὶ δὲ ἑκάστη πλευρὰ τοῦ
τετραγώνου τοῦ σχήματος σπιθαμῶν ε καὶ ια/λ μιᾶς σπιθαμῆς
Dr. Maria Chalkou
http:// www.drchalkou.simplesite.com
The Mathematical Comments
κεφ. 171. (ροα). Μὲ πλευρὰ τμῆμα τῆς διαμέτρου ΑΒ= 12 ἑνός
κύκλου, ο συγγραφέας ζητεῖ νὰ κατασκευαστεῖ τετράγωνο μὲ
δύο διαδοχικές κορυφὲς ἐπὶ τῆς περιφέρειας καὶ τὶς ἄλλες δύο
ἐπὶ τῆς διαμέτρου τοῦ κύκλου.
Ἔστω ΓΕΖΔ τὸ ζητούμενο τετράγωνο. Τὰ ὀρθογώνια τρίγωνα ΔΖΟ
καὶ ΓΕΟ εἶναι ἴσα ἐπειδὴ ἔχουν ΔΖ=ΓΕ (ὡς πλευρὲς τετραγώνου),
καὶ ΟΖ=ΟΕ=ρ=6. Ἄρα ΟΔ=ΟΓ=χ/2, ὅπου χ ἡ πλευρὰ τοῦ ζητούμενου
τετραγώνου.
Ἐφαρμόζουμε τὸ Πυθαγόρειο θεώρημα στὸ ὀρθογώνιο τρίγωνο
ΔΖΟ:
6²= χ²+(χ/2)², ἢ
χ²= 144/5, ἢ
χ= √(144/5).
Ὁ συγγραφέας τοῦ χειρόγραφου γνωρίζει προφανῶς τή μέθοδο,
ἀλλά περιγράφει μόνο τὸ τελευταῖο βῆμα τῆς πιὸ πάνω
διαδικασίας, δηλαδὴ ὑπολογίζει κατευθείαν τὸ χ ὡς τὴν
τετραγωνικὴ ρίζα τοῦ ΑΒ²/5, χωρὶς κάποια αἰτιολόγηση.
Dr. Maria Chalkou
http:// www.drchalkou.simplesite.com
Dr. Maria Chalkou
http:// www.drchalkou.simplesite.com

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Number theory in byzantium according to codex vindobonensis

  • 1. Number Theory in Byzantium according to Codex Vindobonensis phil. Graecus 65 of the 15th cent. The numbers and the positional numeral system Dr. Μaria Chalkou Ιnternational Congress MICOM 2015 Athens 22-26 of September
  • 2. • Codex Vindobonensis Phil. Gr. 65, is a 15th century Byzantine manuscript kept in the National Library of Austria in Vienna. • In this article we describe elements from Number theory, as these are presented in the Codex 65 • This Ms contains a comprehensive programme for teaching mathematics, addressed to an audience consisting of students probably of all the grades of what is today’s primary and secondary education, but also state functionaries, merchants, craftsmen of various specialities such as silversmiths and goldsmiths, builders, etc. Dr. Maria Chalkou http:// www.drchalkou.simplesite.com
  • 3. Codex Vindobonensis phil. Gr. 65 • Ιt is the first Greek Ms in which there appears the problem of the construction of a square inscribed in an equilateral triangle, so that one of its sides touches one side of the triangle. Τhe theoretical part of the construction is not recorded in the Ms; what is only asked is to calculate the side of the square in relation to the side of the triangle • The Ms is of considerable importance since the scientific research carried out indicated that it is in essence the Mathematical Encyclopaedia of the Byzantines, and possibly the first Mathematical Encyclopaedia Dr. Maria Chalkou http:// www.drchalkou.simplesite.com
  • 4. Dr. Maria Chalkou http:// www.drchalkou.simplesite.com
  • 5. Dr. Maria Chalkou http:// www.drchalkou.simplesite.com
  • 6. Dr. Maria Chalkou http:// www.drchalkou.simplesite.com
  • 7. Dr. Maria Chalkou http:// www.drchalkou.simplesite.com
  • 8. The initial study of the codex was published in 2004 by NKUA as a Doctoral Thesis by the Department of Mathematics The improved and completed edition of the codex was published in 2006 by the Byzantine Research Centre of AUTΗ. This part of the codex, which had remained unpublished until then, since studies of the content of such Mss are rare, was called TRACTATUS MATHEMATICUS VINDOBONENSIS GRAECUS Dr. Maria Chalkou http:// www.drchalkou.simplesite.com
  • 9. The units of Tractatus Mathematicus Vindobonensis Graecus UNIT 1. Operations between real numbers. UNIT 2. Fractions, ratios, proportions. UNIT 3. Progressions. UNIT 4. Problems on first-degree equations. Solution through practical arithmetic. UNIT 5. Interest on loans or debts. UNIT 6. Problems of sharing in proportional parts. UNIT 7. Problems on manufacture of gold and silver. UNIT 8. Roots of real numbers. UNIT 9. Solution of equations. UNIT 10. Systems of equations. UNIT 11. Plane geometry. UNIT 12. Areas of plane figures. UNIT 13. Solid geometry. Dr. Maria Chalkou http:// www.drchalkou.simplesite.com
  • 10. Table of correspondence between Greek and Arabic numbers (alfa) α 1 (giota) ι 10 (ro) ρ 100 ˏα 1000 (vita) β 2 (kapa) κ 20 (sigma) σ 200 ˏβ 2000 (gama) γ 3 (lamda) λ 30 (taf) τ 300 ˏγ 3000 (delta) δ 4 (mi) μ 40 (ipsilon) υ 400 ˏδ 4000 (epsilon) ε 5 (ni) ν 50 (fi) φ 500 (stigma) ς 6 (ksi) ξ 60 (chi) χ 600 (zita) ζ 7 (omikron) ο 70 (psi) ψ 700 (ita) η 8 (pi) π 80 (omega) ω 800 (thita) θ 9 (koppa) Ϟ 90 (sabi) Ϡ 900 ɥ 0 In the Transcription, and anywhere else in the Edition of the Codex 65 it was necessary to be done, the symbol u was used instead of чto declare 0. Dr. Maria Chalkou http:// www.drchalkou.simplesite.com
  • 11. UNIT 1 The symbols used for the numbers are the letters of the Greek alphabet, but the calculations are done with the then new decimal Arabic numeration. The author appears not to have adjusted to the new method. Yet it must be stressed that using letters and not numbers did not influence the result since the positional numeral system was used, that is the position of the letter determined its arithmetical value. Dr. Maria Chalkou http:// www.drchalkou.simplesite.com
  • 12. Examples of writing numbers ρκγ denotes number 123. So does αβγ. Ϡ Ϟε denotes number 995. So does θθε. ‫؍‬ετξ denotes number 5360. So does εγςu. We observe that the numbers in the examples are symbolized by the author in the Greek system of writing in two ways. The second way can be used because each letter takes its value according to its position in the number. Dr. Maria Chalkou http:// www.drchalkou.simplesite.com
  • 13. Execution of the operations First example η (eight) by βα (21) η 8 βα 21 αςη 168 We observe that the operations do not differ from those we would execute today, that is we would multiply α (one) by η (eight), and exactly below α and η we would write the result, which is again η. Next we would multiply β (two) by η (eight) and we would write ας (16), in which case the final result of the operation will be αςη (168). In the above example we can observe that the numbers η, α and η are placed on the right- hand column and denote the units. Numbers βand ςare placed on the second column and denote the tens, and the number αon the third column and denotes one hundred. The important element here is that the result αςη denotes the number 168 according to the operations. This number could also be symbolized as ρξη since ρdenotes one hundred, ξ the number 60, and ηthe number eight. That is we see that when ς(six) is in the second place from the right, it denotes six tens, and when αis in the third place from the right, it denotes one hundred. Therefore each letter takes an arithmetical value according to its position in the number. Dr. Maria Chalkou http:// www.drchalkou.simplesite.com
  • 14. As a second example we mention the multiplication of αγ (13) by αε (15). This operation appears in the manuscript as follows: αγ α 13 1 αε 15 αθε 195 The author multiplies γ (three) by ε (five) and has ιε (15). He writes ε and carries α. As he says, he multiplies αγ by αε „crosswise“, that is he multiplies α by ε, which gives a result ε, and γ by α, which gives a result γ, and adds the results ε and γ, in which case he has a result η. He adds α, which he has carried from the first multiplication, to η, and he places θ, which is their sum, in the place of tens. Finally, he multiplies α of αγ by α of αε and writes their product, which is again α, in the place of hundreds. Thus the result of the multiplication is the number αθε or ρϟε (195). Dr. Maria Chalkou http:// www.drchalkou.simplesite.com
  • 15. These operations by the anonymous author are justified by applying the distributive law of multiplication over addition and as follows: αγ is written as α.10+γ, and αε as α.10+ε. Therefore the product of αγ by αε is: αγ.αε=(α.10+γ). (α.10+ε)= α.α.100+α.ε.10+γ.α.10+γ.ε= α.α.100+α.ε.10+γ.α.10+α.10+ε= α.α.100+(α.ε+γ.α+α).10+ε= 100+(5+3+1).10+5= 100+90+5=195 Dr. Maria Chalkou http:// www.drchalkou.simplesite.com
  • 16. As a third example we mention the multiplication of αε (15) by αβγ (123). u α ε α 0 1 5 1 α β γ α 1 2 3 1 α η δ ε 1 8 4 5 First he multiplies γ by ε and he has αε. He writes ε and carries α. Next he multiplies α by γ, β by ε, and adds α to them. The result is αδ. He writes δ and carries α. He multiplies „ouden“ by γ, α by ε, β by α, and adds α to what he finds. The result is η. He multiplies „ouden“ by β, and α by α. The result is α. The product of the multiplication is αηδε. According to the distributive law of multiplication over addition we have: 15.123= (10+5)(100+2.10+3)= 1000+2.100+3.10+5.100+10.10+3.5= 1000+8.100+4.10+5= 1845 Dr. Maria Chalkou http:// www.drchalkou.simplesite.com
  • 17. On the check of the operation of multiplication αε 15/7= 2+1/7 ς 6/7 1.6= 6 θu 90/7= 12+6/7 The author asks for the remainder of the division of 15 by seven, which is the unit. Because the remainder of the division of six by seven is six, he multiplies the unit by six and places the product, which equals six, in a circle. Finally he finds the remainder of the division of 90 by seven, which is six, and compares it with the number he has placed in the circle, that is 6. If the two results are the same, then the multiplication is correct. Dr. Maria Chalkou http:// www.drchalkou.simplesite.com
  • 18. His method is based on the following: If we symbolize the multiplicand with M and the multiplier with Ρ, then M.Ρ ≡ M.Ρ (mod 7), and if M=7k+m, and Ρ=7l+n with k, l, m, n Є ℕ and m, n < 7, M.Ρ ≡ (7k+m).(7l+n) (mod 7), M.Ρ ≡ m.n+(7k.l+k.n+l.m).7 (mod 7), M.Ρ ≡ m.n (mod 7) This is the mathematical justification of the method with the remainders of the division by seven, which is used by the author of the 15th century. Dr. Maria Chalkou http:// www.drchalkou.simplesite.com
  • 19. On the check of the operation of addition In another problem the Byzantine author deals with the addition of final sums which result from various loan accounts. He adds 1695+1393+3454+4565 and finds 11107. Nowadays we do not refer to a check of additions with more than two addends. However, in Codex Vindobonensis phil. gr. 65 there is a check with the same method which the author applies to the operation of multiplication, which is as follows: He divides each addend by seven, in which case from the first division he has the unit (1) as remainder, from the second one a remainder zero (0), from the third one a remainder three (3), and from the fourth one as remainder the unit (1). Next he divides the sum 11107 by 7 and he finds a remainder five (5). He adds the remainders of the four addends 1+0+3+1, and again he finds (5). Then he writes that the addition is correct. Dr. Maria Chalkou http:// www.drchalkou.simplesite.com
  • 20. This procedure can be justified as follows: We symbolize the above addends with Α1, Α2, Α3, Α4, the respective remainders of their division by seven with u1= 1, u2= 0, u3= 3, u4= 1, and the respective quotients of the same divisions with k, l, m, n . Then we will have: Α1+Α2+Α3+Α4 ≡ Α1+Α2+Α3+Α4 (mod 7), In which case 7k+u1+7l+u2+7m+u3+7n+u4 ≡ Α1+Α2+Α3+Α4 (mod 7), that is u1+u2+u3+u4+7(k+l+m+n) ≡ Α1+Α2+Α3+Α4 (mod 7), or u1+u2+u3+u4 ≡ Α1+Α2+Α3+Α4 (mod 7) Dr. Maria Chalkou http:// www.drchalkou.simplesite.com
  • 21. Thus we observe that apparently the problems are dealt with, by using Arithmetic, while in essence the author seems to have a good knowledge of Number Theory, elements of which he uses in his teaching Dr. Maria Chalkou http:// www.drchalkou.simplesite.com
  • 22. Dr. Maria Chalkou http:// www.drchalkou.simplesite.com
  • 23. The text after the transcription ροα' Περὶ τοῦ πῶς ἐστὶ ἐντὸς ἡμισφαιρίου σχήματος, τεθῆναι τετράγωνον σχῆμα ἰσόπλευρον καὶ εἰδέναι ἑκάστη πλευρὰν πόσων σπιθαμῶν ἐστί. Ἔστω ἡμισφαίριον σχῆμα ὅπερ ἐστὶ ἡ διάμετρος αὐτοῦ σπιθαμῶν ιβ, τεθημένου δὲ ἐντός, σχῆμα τετράγωνον ἰσόπλευρον κατὰ τὸ ἐγχωροῦν μεῖζον μέγεθος. Ζητεῖς εἰδέναι ἑκάστη πλευρὰν πόσων σπιθαμῶν ἐστί. Ἔχεις δὲ τοῦτο εἰδέναι οὕτως: Πολλαπλασίασον εἰς ἑαυτὴ τὴν διάμετρον τοῦ ἡμισφαιρίου ἥτις ἐστὶ σπιθαμῶν ιβ ὡς εἴπομεν· ιβ-κις οὖν ιβ γίνονται ρμδ. Μέρισον ταῦτα πάντοτε μετὰ τῶν ε. Γίνεταί δε νῦν ὁ τούτων διαμερισμὸς κη καὶ δ/ε. Ζήτει τὴν ρίζαν τοῦ κη καὶ δ/ε ἥτις ἐστὶ ε καὶ ια/λ. Τὰ γὰρ ε καὶ ια/λ τεχνικῶς πολλαπλασιαζόμενα εἰς ἑαυτὰ πολλαπλασιάζοσιν κη καὶ ψκα/Ϡ ἅπερ ἐστὶ κη καὶ δ/ε. Ἐστὶ δὲ ἑκάστη πλευρὰ τοῦ τετραγώνου τοῦ σχήματος σπιθαμῶν ε καὶ ια/λ μιᾶς σπιθαμῆς Dr. Maria Chalkou http:// www.drchalkou.simplesite.com
  • 24. The Mathematical Comments κεφ. 171. (ροα). Μὲ πλευρὰ τμῆμα τῆς διαμέτρου ΑΒ= 12 ἑνός κύκλου, ο συγγραφέας ζητεῖ νὰ κατασκευαστεῖ τετράγωνο μὲ δύο διαδοχικές κορυφὲς ἐπὶ τῆς περιφέρειας καὶ τὶς ἄλλες δύο ἐπὶ τῆς διαμέτρου τοῦ κύκλου. Ἔστω ΓΕΖΔ τὸ ζητούμενο τετράγωνο. Τὰ ὀρθογώνια τρίγωνα ΔΖΟ καὶ ΓΕΟ εἶναι ἴσα ἐπειδὴ ἔχουν ΔΖ=ΓΕ (ὡς πλευρὲς τετραγώνου), καὶ ΟΖ=ΟΕ=ρ=6. Ἄρα ΟΔ=ΟΓ=χ/2, ὅπου χ ἡ πλευρὰ τοῦ ζητούμενου τετραγώνου. Ἐφαρμόζουμε τὸ Πυθαγόρειο θεώρημα στὸ ὀρθογώνιο τρίγωνο ΔΖΟ: 6²= χ²+(χ/2)², ἢ χ²= 144/5, ἢ χ= √(144/5). Ὁ συγγραφέας τοῦ χειρόγραφου γνωρίζει προφανῶς τή μέθοδο, ἀλλά περιγράφει μόνο τὸ τελευταῖο βῆμα τῆς πιὸ πάνω διαδικασίας, δηλαδὴ ὑπολογίζει κατευθείαν τὸ χ ὡς τὴν τετραγωνικὴ ρίζα τοῦ ΑΒ²/5, χωρὶς κάποια αἰτιολόγηση. Dr. Maria Chalkou http:// www.drchalkou.simplesite.com
  • 25. Dr. Maria Chalkou http:// www.drchalkou.simplesite.com