8744
[Commentary:
This folio quotes some text from Girolamo Cardano, Opus novum de proportionibus
(Cardano 1570, , Proposition 170. The copy is in an unknown hand. The table below the text is exactly as given by Cardano. ]
Cardanus de proportionibus. prop.
Ut autem habeas numeros singulorum ordinum, in quavis multitudino, deducito
numerum ordinis a primo, & divide per numerum ordinis ipsius reliquum,
et illud quod proventi, ducito in numerum maximum praecedentis ordinis,
et habebis numerum quaesitum. Velut si sint undecim, volo scire breviter numeros,
qui fiunt ex variatione trium. Primum deduco pro secundo ordine 1 ex 11 fit 10,
divido per 2 numerum ordinis, exit 5, duco in 11 fit 55 numerus secundi ordinis. Inde
detraho 2, qui est numerus differentiae ordinis totij a primo ex 11, relinquitur 9,
divido 9 per 3 numerum ordinis exit 3, duco 3 in 55 numerum secundi fit 165,
numerus totij ordinis. Similiter volo numerum variatione quatuor,
dedco 3 differentiam 4 a primo ordine ab 11 relinquitur 8, divido 8 per 4
numero ordinis, exit 2, duco 2 in 195 fit 330, numeri quarti ordinis.
Similiter pro quinto detraho 4 differentiam a primo ordine, relinquitur 7,
divido per 5 numerum ordinis exit , duco in 330 numero praecedentis
ordine, fit 462 numerus quinti
[Translation: As moreover you have the numbers of a single row, of any size, you can derive the numbers of each place from the first, and divide by the number of the place, and what arises multiply by the greatest number of the preceding place,and you will have the number sought. Thus if there are eleven [objects], I want to know quickly the number that arises for three choices. First, for the second place, I take 1 from 11 which makes 10, I divide by 2, the number of the place, there comes out 5, I multiply by 11 to make 55, the number of the second place. Next I subtract 2, which is the number of the difference of all the places from the first, from 11, there remains 9, I divide 9 by 3, the number of the place, there comes out 3, I multiply 3 by 55, the second number, to make 165, the number in the third place. Similarly if I want the number for four choices, I take 3, the difference of 4 from the first place, from 11, there is left 8, I divide 8 by 4, the number of the place, there comes out 2, I multiply 2 by 165 to make 330, the number of the fourth place. Similarly for the fifth I subtract 4, the difference from the first place, there remains 7, I divide by 5, the number of the place, there comes out , I multiply by 330, the number of the previous place, to make 462, the number of the fifth place.
numerum ordinis a primo, & divide per numerum ordinis ipsius reliquum,
et illud quod proventi, ducito in numerum maximum praecedentis ordinis,
et habebis numerum quaesitum. Velut si sint undecim, volo scire breviter numeros,
qui fiunt ex variatione trium. Primum deduco pro secundo ordine 1 ex 11 fit 10,
divido per 2 numerum ordinis, exit 5, duco in 11 fit 55 numerus secundi ordinis. Inde
detraho 2, qui est numerus differentiae ordinis totij a primo ex 11, relinquitur 9,
divido 9 per 3 numerum ordinis exit 3, duco 3 in 55 numerum secundi fit 165,
numerus totij ordinis. Similiter volo numerum variatione quatuor,
dedco 3 differentiam 4 a primo ordine ab 11 relinquitur 8, divido 8 per 4
numero ordinis, exit 2, duco 2 in 195 fit 330, numeri quarti ordinis.
Similiter pro quinto detraho 4 differentiam a primo ordine, relinquitur 7,
divido per 5 numerum ordinis exit , duco in 330 numero praecedentis
ordine, fit 462 numerus quinti
[Translation: As moreover you have the numbers of a single row, of any size, you can derive the numbers of each place from the first, and divide by the number of the place, and what arises multiply by the greatest number of the preceding place,and you will have the number sought. Thus if there are eleven [objects], I want to know quickly the number that arises for three choices. First, for the second place, I take 1 from 11 which makes 10, I divide by 2, the number of the place, there comes out 5, I multiply by 11 to make 55, the number of the second place. Next I subtract 2, which is the number of the difference of all the places from the first, from 11, there remains 9, I divide 9 by 3, the number of the place, there comes out 3, I multiply 3 by 55, the second number, to make 165, the number in the third place. Similarly if I want the number for four choices, I take 3, the difference of 4 from the first place, from 11, there is left 8, I divide 8 by 4, the number of the place, there comes out 2, I multiply 2 by 165 to make 330, the number of the fourth place. Similarly for the fifth I subtract 4, the difference from the first place, there remains 7, I divide by 5, the number of the place, there comes out , I multiply by 330, the number of the previous place, to make 462, the number of the fifth place.
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