By Joran Friberg

ISBN-10: 9812704523

ISBN-13: 9789812704528

A sequel to unforeseen hyperlinks among Egyptian and Babylonian arithmetic (World medical, 2005), this ebook is predicated at the writer s in depth and flooring breaking reports of the lengthy historical past of Mesopotamian arithmetic, from the past due 4th to the overdue 1st millennium BC. it's argued within the e-book that a number of of the main well-known Greek mathematicians seem to have been conversant in quite a few facets of Babylonian metric algebra, a handy identify for an tricky mixture of geometry, metrology, and quadratic equations that's recognized from either Babylonian and pre-Babylonian mathematical clay capsules. The e-book s use of metric algebra diagrams within the Babylonian type, 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 much better realizing of many attention-grabbing propositions and structures in Greek mathematical works. the writer s comparisons with Babylonian arithmetic additionally result in new solutions to a few very important open questions within the background of Greek arithmetic

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**Extra resources for Amazing Traces of a Babylonian Origin in Greek Mathematics**

**Sample text**

Q and p + m + q sq. m – sq. q and m – q given given given sq. m – sq. q = ? p, m, q = ? p, m, q = ? 1 problem 1 problem 2 values for c 1 problem 1 problem 1 problem Note: In 9b-c it is tacitly assumed that p – q = q – r. In §§ 1-4 of TMS 5, the given values of the coefficient c are allowed to vary in the same way as the given values of the coefficient c in § 5 of BM 80209 (Sec. 10 above), but much more extensively. 11. A Large Old Babylonian Catalog Text of a Similar Kind c s c s c s 1 2 3 4 3" 2' 3' 4 3' 4 1 3" 1 2' 1 3' 14 1 3' 4 2 2' 3 3' 44 30 1 7 27 17 127 7 1/7 7 2 1/7 1 77 277 177 1277 1 11 2 11 35 1 11 11 2 11 11 1 11 7 2 11 7 1 3" 2' 3' 11 7 2 3" 2' 3' 11 7 10 05 4 05 33 35 = 5 · 7 4 05 = 5 · 7 · 7 6 25 55 = 5 · 11 12 50 10 05 = 5 · 11 · 11 6 25 = 5 · 7 · 11 12 50 = 2 · 5 · 7 · 11 55 Probably in order to save space, the values given for c in this text make use of some otherwise undocumented notations for fractions.

P + sq. q = 21 40, p + q = 50 sq. p + sq. q = 21 40, p – q = 10 sq. p + sq. q = 21 15, q = p – 1/7 p sq. p + sq. q = 28 15, p = q + 1/7 q sq. p + sq. q = 21 40, p · q = 10 sq. p + sq. q = 28 20, q = 1/4 p sq. p + sq. q = 25 25, q = 3" p + 5 sq. p + sq. q + sq. r + sq. s = 27 05, q = 3" p, r = 2' q, s = 3' r 1 h sq. s – 3' s = 5 3 a. sq. p + sq. q + sq. r = 10 12 45, q = 1/7 p, r =1/7 q 3 b sq. p + sq. q + sq. r = 23 20, p – q = q – r = 10 2 h sq. p + sq. q + sq. (p – q) = 23 20, p + q = 50 ······················································································· s = 30 s = 30 s = 30 s = 20 s = 30 s = 30 s = 30 p = 30, q = 20 p = 30, q = 20 p = 3 30, q = 3 p = 4, q = 3 3 0 p = 30, q = 20 p = 40, q = 10 p = 30, q = 25 1i 3c s = 10 4 s + sq.

14 is that he starts with any rectangle with the given area A, say one with the sides u, s = A, 1. He then constructs the diagram in Fig. 2, right, in the case when p = u + s. In this way, he manages to construct the side h of a square with the given area A, as the upright side of a right triangle. Therefore, proposition El. 14 in the inverted form that Euclid chose to give to it (with u and s, hence also p and q, given from the beginning rather than A and p) may very well have replaced an original proposition El.

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