Sistema de numeración babilónico

Matemáticas 1º ESO

1. SESO DEL IES LAS CUMBRES. GRAZALEMA MATEMÁTICAS 1º ESO
http://iesgrazalema.blogspot.com
SISTEMA DE NUMERACIÓN BABILÓNICO
· Babilonia – Asia: 1730 a. C. – 539 a. C.
· Sistema de numeración híbrido:
- Aditivo de base diez hasta el 60.
- Posicional para números superiores al 60.
· Dígitos:
1 10
· Ejemplos:
- Hasta el 60 → Aditivo
1 2 3
4 5 6

7 8 9
10 11 12 13
14
15 16
18 19
20 21
27 29
1

2. 30
34
38


40 42 45 46
50 53
59

1 · 60 = 60 → Se escribe como el 1, pero con un mayor espacio entre él y los dígitos
siguientes.
- Para números superiores al 60 → Posicional
1 · 60 + 1 = 61 1 · 60 + 2 = 62
1 · 60 + 9 = 69
1 · 60 + 10 = 70 1 · 60 + 12 = 72

1 · 60 + 20 =80
1 · 60 + 25 = 85
2

3. 1 · 60 + 30 = 90
1 · 60 + 38 = 98


1 · 60 + 40 = 100 1 · 60 + 43 = 143
1 · 60 + 50 = 110

1 · 60 + 59 = 119
2 · 60 = 120
2 · 60 + 1 = 121


2 · 60 + 59 = 179



3 · 60 = 180

3

4. 239 60
593
239=359
(60
=59·60
0
3·60
1
=59·13·60=3·6059


3 · 60 + 59 = 239
······························································································································································
240 60 240=˙60
004
240=40
(60
=0·60
0
4·60
1
=0·14·60=04·60=4·60

4 · 60 = 240
······························································································································································
59960
599
599=959
(60
=59·60
0
9·60
1
=59·19·60=9·6059
9 · 60 + 59 = 599
······························································································································································
600 60 600=˙60
000 10
0
600=100
(60=0·60
0
10·60
1
=0·110·60=010·60=10·60

10 · 60 = 600
4

5. 231560
51538
35
2.315=3835
(60
=35·60
0
38·60
1
=35·138·60=38·6035
38 · 60 + 35 = 2.315
······························································································································································
741460
141123 60
214 032
34
7.414=2334
(60
=34·60
0
3·60
1
2·60
2
=34·13·602·3.600=2·3.6003·6034
2 · 3.600 + 3 · 60 + 34 = 7.414
······························································································································································
3241560
24154060
015 009
15
32.415=9015
(60=15·60
0
0·60
1
9·60
2
=15·10·609·3.600=9·3.6000·6015
Para evitar confusiones, introdujeron un signo separador cuando
el valor de un orden intermedio era cero.
9 · 3.600 + 0 · 60 + 15 = 32.415
5

6. 7934660
193 132260
134 12222
146 02
26
79.346=22226
(60=26·60
0
2·60
1
22·60
2
=26·12·6022·3.600
22 · 3.600 + 2 · 60 + 26 = 79.346
······························································································································································
14831260
283247160
431 07141
112 11
52
148.312=411152
(60=52·60
0
11·60
1
44·60
2
=52·111·6041·3.600
41 · 3.600 + 11 · 60 + 52 = 148.312
······························································································································································
72564260
125 1209460
0564 0094201 60
242 34 213
02

725.642=321342
(60=2·60
0
34·60
1
21·60
2
3·60
3
=
=2·134·6021·3.6003·216.000=3·216.00021·3.60034·602
3· 216.000 + 21 · 3.600 + 21 · 3.600 + 2 = 725.642
6