43
Extractio \({1 \over 9}\) \({3 \over 5}\) de \({1 \over 7}\) \({2 \over 3}\)79
Si
|
|
|
234 |
255 |
|
\({1 \over 9}\) |
\({3 \over 5}\) |
|
\({1 \over 7}\) |
\({2 \over 3}\) |
|
extractio |
\({1~~0 \over 3~~5}\) |
|
|
80 autem \({1 \over 9}\) \({3 \over 5}\) de \({1 \over 7}\) \({2 \over 3}\) extrahere volueris, reperies prescripta 255 et 234 et
81 extrahes 234 de 255: remanebunt 21, que
82 divide suprascripta ratione per \({1~~0~~0 \over 5~~7~~9}\). Tamen
83 prius dividas per 7 et per 9 quam per 5, ideo quia 21 integraliter dividitur per 7 et per 3 que sunt de regula ipsorum 9
84; exibunt \({3~~0 \over 9~~5}\) pro residuo dicte extractionis, hoc est \({1~~0 \over 3~~5}\). De divisione autem eorum ad invicem fac ut supra.
