46 Lessons in Early Geometry, part 4/10

46 Lessons in Early Geometry, part 4/10 / provisional version in my own freestyle English / a corrected version will follow in March, April or May (hopefully) / Franz Gnaedinger / February 2003 / www.seshat.ch

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Lesson 15

The Babylonians used an excellent value for the square root of 2, found on the clay tablet YBC 7289 from around 1650 or 1700 BC:

1;24,51,10 = 1 + 24/60 + 51/60x60 + 10/60x60x60

YBC 7289 1.4142129...

exact value 1.4142135...

How did the Babylonians possibly find this value?

1 1 2

2 3 4

5 7 10

12 17 24

29 41 58

70 99 140

169 239 338

408 577 816

985----1393

Divide 1393 by 985 and you obtain 1;24,51,10,3,2... Leave out the small numbers ...3,2... and keep the value 1;24,51,10.

Archimedes used the ratios 265/153 and 1351/780 as boundaries for the square root of 3.

1 1 3

2 4 6

1 2 3

3 5 9

8 14 24

4 7 12

11 19 33

30 52 90

15 26 45

41 71 123

112 194 336

56 97 168

153-----265 459

418 724 1254

209 362 627

571 989 1713

1560 2702

780----1351

Lesson 16

Victor J. Katz, A HISTORY OF MATHEMATICS, Addison-Wesley 1998, page 46, The Beginnings of Mathematics in Greece:

'A report from a visit to Egypt with Plato by Simmias of Thebes in 379 B.C.E. (from a dramatization by Plutarch of Chaeronea (first/second century C.E.)): "On our return from Egypt a party of Delians met us ... and requested Plato, as a geometer, to solve a problem set them by the god in a strange oracle. The oracle was to this effect. The present troubles of the Delians and the rest of the Greeks would be at an end when they had doubled the altar at Delos. As they not only were unable to penetrate its meaning, but failed absurdly in constructing the altar ..., they called on Plato for help in their difficulty. Plato ... replied that the god was ridiculing the Greeks for their neglect of education, deriding, as it were, our ignorance and bidding its engage in no perfunctory study of geometry; for no ordinary or near-sighted intelligence, but one well versed in the subject, was required to find two mean proportionals, that being the only way in which a body cubical in shape can be doubled with a similar increment in all dimensions. This would be done for them by Eudoxus of Cnidus ...; they were not, however to suppose that it was this the god desired, but rather that he was ordering the entire Greek nation to give up war and its miseries and cultivate the Muses, and by calming their passions through the practice of discussion and study of mathematics, so to live with one another that their relationships should be not injurious, but profitable."'

How did Eudoxus solve the problem? He is known for a ladder of numbers that allows to double the square:

1 1

2 3 a b

5 7 a+b a+b+a

12 17

29 41

70 99

.......

Domingo Gomez Morin expanded Eudoxus' ladder as follows:

1 1 1

3 4 5 a b c

12 15 19 a+b+c a+b+c+a a+b+c+a+b

46 58 73

177 223 281

............

This ladder allows to double the cube and was certainly known to Eudoxus.

The numbers of both ladders are contained in my number patterns for the approximation of the square root and cube root of 2:

1-----1 2

2-----3 4

5-----7 10

12----17 24

29----41 58

70----99 ...

1-----1-----1 2

2 2 3 4

4 5 7 8

9 12 15 18

3-----4-----5 6

7 9 11 14

16 20 25 32

36 45 57 72

12----15----19 24

27 34 43 54

61 77 87 121

138 174 219 276

46----58----73 ...

Lesson 17

Each line of a number column for the calculation of a square root provides a pair of values, one too small, one too big:

5 7 10 4 7 12 4 9 20

7/5 x 7/5 = 49/25 = 2 minus 1/25 too small

10/7 x 10/7 = 100/49 = 2 plus 2/49 too big

7/4 x 7/4 = 49/16 = 3 plus 1/16 too big

12/7 x 12/7 = 144/49 = 3 minus 3/49 too small

9/4 x 9/4 = 81/16 = 5 plus 1/16 too big

20/9 x 20/9 = 400/81 = 5 minus 5/81 too small

You can proceed more quickly by adding a pair of values and halving the sum:

(9/4 + 20/9) : 2 = 161/72 72 161 360

This leads to a powerful algorithm:

given line a b (na)

new line 2ab naa+bb (nx2ab)

Let us calculate the square root of n=2, beginning with a=b=1:

1 1 (2)

2x1x1 = 2 2x1x1 + 1x1 = 3

2 3

2x2x3 = 12 2x2x2 + 3x3 = 17

12 17

2x12x17 = 408 2x12x12 + 17x17 = 577

408 577

and so on

The numbers are contained in the lines 1, 2, 4, 8, 16, 32, 64, 128, 256, 512 ... of my first number column for the calculation of the square:

1-----1 2

2-----3 4

5 7 10

12----17 24

29 41 58

70 99 140

169 239 338

408---577 ...

Now let us approximate the square root of n=10, beginning with a=1 and b=3:

1 3 (10)

2x1x3 = 6 10x1x1 + 3x3 = 19

6 19

2x6x19 = 228 10x6x6 + 19x19 = 721

228 721

2x228x721 = 328776 10x228x228 + 721x721 = 1039681

328776 1039681

and so on

On my pocket calculator:

3/1 squared = 9

19/6 squared = 10.02777778

721/228 squared = 10.00001924

1039681/328776 squared = 10

sr6: 1 2, 4 10, 2 5, 20 49, 1960 4801, ...

sr7: 1 3, 6 16, 3 8, 48 127, 12192 32257, ...

Lesson 18

Common knowledge has the square root of 4 being 2. Did anyone ever try to prove this assumption, or should I say conjecture? Well then, let me be the one to face this challenge by drawing up a further number column of mine:

1 1 4

2 5 8

7 13 28

20 41 80

61 121 244

182 365 ...

1/1 = 2 minus 1 5/2 = 2 plus 1/2

13/7 = 2 minus 1/7 41/20 = 2 plus 1/20

121/61 = 2 minus 1/61 365/182 = 2 plus 1/182

The deviations from 2 are diminishing, but will they ever vanish? Let us apply the fast algorithm:

1 1 (4)

2x1x1 = 2 4x1x1 + 1x1 = 6*

6 / 2 = 3 / 1 = 2 plus 1

* the mistake may show that also the fast algorithm is robust (error tolerant)

2x1x3 = 6 4x1x1 + 3x3 = 13

13 / 6 = 2 plus 1/6

2x6x13 = 156 4x6x6 + 13x13 = 313

313 / 156 = 2 plus 1/156

2x156x313 = 97656 4x156x156 + 313x313 = 195313

195,313 / 97,656 = 2 plus 1/97656

The deviation is rapidly shrinking, but will it really vanish? Luckily enough we can start a number column with any pair of numbers:

1 2 4

3 6 12

1 2 4

3 6 12

1 2 4 and so on

The lines are endlessly repeating, always yielding the same ratio 2/1 = 4/2 = 2. Hence the square root of 4 is exactly 2. Glad about this result, for even the slightest deviation from 2 would have shaken the foundament of mathematics ;-)

Lesson 19

My number columns might have been the origin of the continued fractions, which, David Fowler believes, were known to the Greeks in Platon's time:

1 1 2

2 3 4

5 7 10

12 17 24

.. .. ..

1/1 2/1 3/2 4/3 7/5 10/7 17/12 24/17 ...

1/1 = 1

2/1 = 1 plus 1

3/2 = 1 plus 1/ 2

4/3 = 1 plus 1/ 2 + 1

7/5 = 1 plus 1/ 2 + 1/ 2

10/7 = 1 plus 1/ 2 + 1/ 2 + 1

17/12 = 1 plus 1/ 2 + 1/ 2 + 1/ 2

24/17 = 1 plus 1/ 2 + 1/ 2 + 1/ 2 + 1

...........................................

sr2 = 1 plus 1/ 2 + 1/ 2 + ... = (1,2,2,2,2,2,2,2,2...)

1---1 3

2 4 6

1---2 3

3---5 9

8 14 24

4---7 12

.. .. ..

1/1 3/1 2/1 3/2 5/3 9/5 7/4 12/7 ...

1 plus 2/ 2 + 2/ 2 + 2/ 2 + 2/ 2 + 2/ ...

If you consider only the marked numbers and ratios you get a simple continued fraction:

1 plus 1/ 1 + 1/ 2 + 1/ 1 + 1/ 2 + 1/ ... = (1,1,2,1,2,1...)

1 1 5

2 6 10

1 3 5

4 8 20

2 4 10

1-----2 5

3 7 15

10 22 50

5 11 25

16 36 80

8 18 40

4-----9 20

13 29 65

42 94 210

21 47 105

68 152 340

34 76 170

17----38 85

55 123 275

178 398 890

89 199 445

288 644 1440

144 322 720

72---161 ...

1 plus 4/ 2 + 4/ 2 + 4/ 2 + 4/ 2 + 4/ 2 + 4/ ...

There is again a simpler version:

2 plus 1/ 4 + 1/ 4 + 1/ 4 ... = (1,4,4,4,4,4,4...)

On the pocket calculator:

4 / + 4 = /

+ 4 = /

+ 4 = / and so on + 2 =

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