By Joran Friberg

A sequel to unforeseen hyperlinks among Egyptian and Babylonian arithmetic (World clinical, 2005), this ebook relies at the writer s in depth and flooring breaking experiences of the lengthy historical past of Mesopotamian arithmetic, from the overdue 4th to the overdue 1st millennium BC. it's argued within the publication that a number of of the main well-known Greek mathematicians seem to have been accustomed to a number of facets of Babylonian metric algebra, a handy identify for an complicated blend of geometry, metrology, and quadratic equations that's identified from either Babylonian and pre-Babylonian mathematical clay pills. The ebook s use of metric algebra diagrams within the Babylonian variety, the place the part lengths and parts of geometric figures are explicitly indicated, rather than completely summary lettered diagrams within the Greek type, is key for a far better 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 few vital open questions within the historical past of Greek arithmetic

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Or q s q d d u s p u s s p (sq. p + sq. q)/2 = S/2 = sq. d = sq. u + sq. s p = u + s, q = u – s Fig. 3. Left: A geometric explanation of BM 13901 § 2 a-b. Right: El. 9. Indeed, in Fig. 3 above, left, sq. d (= the area of the oblique square) = (sq. p + sq. q)/2. This is so because sq. d plus the areas of four right triangles = sq. p, while sq. d minus the areas of four right triangles = sq. q (see Fig. 2, right). 12. Old Babylonian Solutions to Metric Algebra Problems 41 On the other hand, in view of the diagonal rule (Sec.

9 may have been to show that any quadraticlinear system of equations of type B2a: sq. u + sq. s = S, u + s = p, with S and p given, can be solved as follows: The diagram in Fig. 1, left, is constructed, with d = sqs. S. Then it can be shown, as in the proof of El. II. 9, that S = sq. u + sq. s = 2 (sq. p/2 + sq. q/2). 18 Amazing Traces of a Babylonian Origin in Greek Mathematics Consequently, u and s can be computed in the following way: (u – s)/2 = q/2 = sqs. (S/2 – sq. p/2), u = p/2 + q/2 = p/2 + sqs.

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