CHAPTER XIII — Critical Commentary & Notes¶
On Its Fourth Ineffable Effect
Historical and Philological Commentary¶
Campanus XIII.4 vs. Heiberg Euclid XIII.5 (Ordering Inversion in Book XIII): In Campanus of Novara’s Latin edition of Elements Book XIII, Heiberg’s Propositions 4 and 5 are transposed! Specifically:
Campanus XIII.4 (= Heiberg Euclid XIII.5): If a straight line $AB$ is cut in extreme and mean ratio at $C$ (with greater segment $AC$), and $AD = AC$ is added to $AB$, then the whole $BD = AB + AC$ is cut in extreme and mean ratio at $A$, with greater segment $AB$.
Campanus XIII.5 (= Heiberg Euclid XIII.4): If a straight line $AB$ is cut in extreme and mean ratio, the sum of the squares on the whole $AB$ and the lesser segment $BC$ is triple the square on the greater segment $AC$.
Pacioli follows Campanus’s numbering strictly in Chapters XIII and XIV: Chapter XIII gives Heiberg Euclid XIII.5 (citing la quarta del terzodecimo), and Chapter XIV gives Heiberg Euclid XIII.4 (citing la quinta del 13.).
Difference of Squares Identity: Notice how cleanly the product of the conjugate binomial and apotome evaluates: $(\sqrt{125} - 5)(\sqrt{125} + 5) = 125 - 25 = 100 = 10^2$.
Constantin Winterberg’s Commentary (1889 Edition)¶
Note 4 (Cap. XIII): Algebraic Proof of the Fourth Effect (Euclid XIII.5) (Sachliche Anmerkungen #4, Cap. XIII, p. 203, l. 7 from bottom)
In algebraic form, let the segment $bc$ be divided in extreme and mean ratio at $d$ ($cd > bd$), and let $ac = cd$ be adjoined to form the whole line $ad = ac + bc$:
Hypothesis: $cd^2 = bc \cdot bd$ and $ac = cd$.
Claim: $bc^2 = ad \cdot cd = ad \cdot ac$.
Proof: By hypothesis, $cd^2 = bc(bc - cd) = bc^2 - bc \cdot cd$, whence
\[bc^2 = cd^2 + bc \cdot cd = cd(cd + bc) = cd(ac + bc) = ad \cdot cd. \quad \text{Q.E.D.}\]
Anmerkung 4 (Cap. XIII, S. 203): Mathematischer Nachweis des vierten Effects (Euklid XIII.5)
In mathematischer Form (wenn $bc$ in $d$ stetig getheilt und $ac = cd$ angesetzt wird):
Voraussetzung: $cd^2 = bc \cdot bd$ und $ac = cd$.
Behauptung: $bc^2 = ad \cdot cd$.
Beweis: Nach Voraussetzung ist $cd^2 = bc(bc - cd)$, folglich:
\[ad \cdot cd = (ac + bc)\,cd = cd^2 + bc \cdot cd = bc^2. \quad \text{w. z. b. w.}\]
Philological and Folio Apparatus¶
1509 Paganini Witness Folios: 0027 (f. 6r–f. 6r)
Winterberg 1889 Italian Source:
winterberg/01_divina_proportione_italian/013_cap_XIII.txtWinterberg 1889 German Source:
winterberg/03_divina_proportione_german/013_cap_XIII.txtMarginal Woodcut Diagrams: 0 figure(s) extracted from the 1509 Venice witness.