23 Extractio de \({1 \over 7}\) \({1 \over 5}\) de \({1 \over 4}\) \({1 \over 3}\)

Si vero \({1 \over 7}\) \({1 \over 5}\) de \({1 \over 4}\) \({1 \over 3}\) extrahere volueris,

144 245 \({1 \over 7}\) \({1 \over 5}\) \({1 \over 4}\) \({1 \over 3}\) extractio \({5~~2~~2\phantom{0} \over 6~~7~~10}\)   \({1~~2~~6 \over 2~~8~~9}\) 1   \({4~~0~~4 \over 5~~7~~7}\)

⁵¹ reperies prescripta 245 et 144 per qualem volueris modum de duobus prescriptis modis, et extrahe 144 de 245: remanebunt 101, que suprascripta ratione divide per \({1~~0~~0\phantom{0} \over 6~~7~~10}\): exibunt \({5~~2~~2\phantom{0} \over 6~~7~~10}\) pro residuo dicte extractionis. 24 Et si \({1 \over 4}\) \({1 \over 3}\) per \({1 \over 7}\) \({1 \over 5}\) dividere vis, divide 245 per regulam de 144⁵², exibunt \({1~~2~~6 \over 2~~8~~9}\) 1. Et si 144 per regulam de 245 diviseris, habebis \({4~~0~~4 \over 5~~7~~7}\) pro eo quod⁵³ contingit integre parti ex divisione \({1 \over 7}\) \({1 \over 5}\) in \({1 \over 4}\) \({1 \over 3}\), ut in questione ostenditur.