CHAPTER XII — Critical Commentary & Notes¶
On Its Third Singular Effect
Historical and Philological Commentary¶
Radical Arithmetic in Chapter XII (Euclid XIII.3): Pacioli's step-by-step calculation with apotomes is a tour de force of Renaissance abacus algebra:
Halving $\sqrt{125}$ requires dividing the radicand by $2^2 = 4$: $\frac{1}{2}\sqrt{125} = \sqrt{\frac{125}{4}} = \sqrt{31\frac{1}{4}}$.
Subtracting $\sqrt{125} - \sqrt{31\frac{1}{4}} = 2\sqrt{31\frac{1}{4}} - \sqrt{31\frac{1}{4}} = \sqrt{31\frac{1}{4}}$ gives $b + \frac{a}{2} = (15 - 2\frac{1}{2}) - \sqrt{31\frac{1}{4}} = 12\frac{1}{2} - \sqrt{31\frac{1}{4}}$.
Squaring $12\frac{1}{2} - \sqrt{31\frac{1}{4}}$ gives $(12\frac{1}{2})^2 + 31\frac{1}{4} - 2(12\frac{1}{2})\sqrt{31\frac{1}{4}} = 156\frac{1}{4} + 31\frac{1}{4} - \sqrt{625 \times 31\frac{1}{4}} = 187\frac{1}{2} - \sqrt{19531\frac{1}{4}}$.
Squaring $\frac{a}{2} = \sqrt{31\frac{1}{4}} - 2\frac{1}{2}$ gives $31\frac{1}{4} + 6\frac{1}{4} - \sqrt{25 \times 31\frac{1}{4}} = 37\frac{1}{2} - \sqrt{781\frac{1}{4}}$.
Multiplying by $5$ scales the rational part by $5$ ($5 \times 37\frac{1}{2} = 187\frac{1}{2}$) and the radicand by $5^2 = 25$ ($25 \times 781\frac{1}{4} = 19531\frac{1}{4}$).
Constantin Winterberg’s Commentary (1889 Edition)¶
Note 3 (Cap. XII): Algebraic Proof of the Third Effect (Euclid XIII.3) (Sachliche Anmerkungen #3, Cap. XII, p. 203, l. 3 from top)
Expressed algebraically for the line $ac$ divided in extreme and mean ratio at $b$ ($ab > bc$):
Hypothesis: $ab^2 = ac \cdot bc = (ab + bc)\,bc$.
Claim: $\left(bc + \frac{ab}{2}\right)^2 = 5\left(\frac{ab}{2}\right)^2$.
Proof: Since $ab^2 = bc^2 + ab \cdot bc$ by hypothesis, adding $\left(\frac{ab}{2}\right)^2$ to both sides yields
\[\left(bc + \frac{ab}{2}\right)^2 = bc^2 + ab \cdot bc + \left(\frac{ab}{2}\right)^2 = ab^2 + \frac{ab^2}{4} = 5\left(\frac{ab}{2}\right)^2.\]Substituting the numerical values of Chapter XII ($ac = 10$, $ab = \sqrt{125} - 5$, $bc = 15 - \sqrt{125}$) confirms the identity of both sides immediately.
Anmerkung 3 (Cap. XII, S. 203): Mathematischer Nachweis des dritten Effects (Euklid XIII.3)
Mathematisch ausgedrückt heisst dies mit Bezug auf die in $b$ stetig getheilte Linie $ac$:
Voraussetzung: $ab^2 = ac \cdot bc$.
Behauptung: $\left(bc + \frac{ab}{2}\right)^2 = 5\left(\frac{ab}{2}\right)^2$.
Beweis: $ab^2 = bc^2 + ab \cdot bc$ (nach Voraussetzung), daher analog wie in Anmerkung 1 durch Hinzufügung von $\left(\frac{ab}{2}\right)^2$ rechts und links:
\[\left(bc + \frac{ab}{2}\right)^2 = 5\left(\frac{ab}{2}\right)^2.\]Setzt man die Werthe des in Cap. XII behandelten Beispiels ein ($ab = \sqrt{125} - 5$, $bc = 15 - \sqrt{125}$), so ergibt sich die Richtigkeit durch Uebereinstimmung der beiden Seiten der Gleichung ohne weiteres.
Philological and Folio Apparatus¶
1509 Paganini Witness Folios: 0027 (f. 6r–f. 6r)
Winterberg 1889 Italian Source:
winterberg/01_divina_proportione_italian/012_cap_XII.txtWinterberg 1889 German Source:
winterberg/03_divina_proportione_german/012_cap_XII.txtMarginal Woodcut Diagrams: 0 figure(s) extracted from the 1509 Venice witness.