By Joran Friberg

A sequel to unforeseen hyperlinks among Egyptian and Babylonian arithmetic (World medical, 2005), this e-book is predicated at the writer s extensive and floor breaking reports of the lengthy background of Mesopotamian arithmetic, from the overdue 4th to the past due 1st millennium BC. it truly is argued within the booklet that a number of of the main well-known Greek mathematicians seem to have been conversant in numerous points of Babylonian metric algebra, a handy identify for an complex mix of geometry, metrology, and quadratic equations that's recognized from either Babylonian and pre-Babylonian mathematical clay drugs. The booklet s use of metric algebra diagrams within the Babylonian kind, the place the facet lengths and components of geometric figures are explicitly indicated, rather than completely summary lettered diagrams within the Greek type, is vital for a stronger knowing of many attention-grabbing propositions and structures in Greek mathematical works. the writer s comparisons with Babylonian arithmetic additionally bring about new solutions to a couple very important open questions within the historical past of Greek arithmetic

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Q/2) Fig. 1. A geometric explanation of the solution procedure in BM 13901 § 1 a. Note that if it is assumed here that s = ‘30’ = 30, then sq. s = ‘15’ = 15 00 and consequently A = ‘45’ = 45 00 and q = ‘1’ = 1 00! Now, consider instead BM 13901 § 1 b (Høyrup, LWS (2002), 52): BM 13901 § 1 b, literal translation explanation (floating values) My equalside inside the field I tore out, 14 30. 1, the going-out, you set. The halfpart of 1 you break. 30 and 30 you make eat each other. 15 to 14 30 you add.

5-6 19 that EF = BE. ) The square AFGH is drawn on AF, and the side GH is extended to K. With this, the construction is completed, and it remains to prove that the given line AB is cut by H in the desired way. The construction in El. 11 can be explained as follows: In Fig. 1, left, let h be the length of AB = AC, let s be the length of AF, let u = s + h be the length of CF, and let p/2 = s + h/2 be the length of EF = BE. Then, s + h/2 = EF and h/2 = EA. Now, according to El. 6, CF · AF + sq. EA = sq.

U + sq. s = S = sq. d u + s = p (p > d) B2b: sq. u + sq. s = S = sq. d u – s = q (q < d) Fig. 1. Geometric constructions of solutions in possible forerunners to El. 9, 10. A similar constructive solution to the metric algebra problem of type B2b is illustrated in Fig. 1, right. It is a likely forerunner to El. 10. In a similar way, consider the following likely forerunners to the pair El. 11* and El. 14* (Fig. 2), the proposed forerunners to El. 14. First, suppose that q is a given length and that A = sq.