Download Amazing Traces of a Babylonian Origin in Greek Mathematics by Joran Friberg PDF

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 reviews of the lengthy historical past of Mesopotamian arithmetic, from the past due 4th to the past due 1st millennium BC. it truly is argued within the booklet that numerous of the main well-known Greek mathematicians seem to have been acquainted with a number of elements of Babylonian metric algebra, a handy identify for an complex blend of geometry, metrology, and quadratic equations that's recognized from either Babylonian and pre-Babylonian mathematical clay pills. The e-book s use of metric algebra diagrams within the Babylonian sort, the place the aspect lengths and parts of geometric figures are explicitly indicated, rather than absolutely summary lettered diagrams within the Greek sort, is key for a much better figuring out 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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Then the diameter of the circle is cut into three segments of which one is the side of the rectangle of length q. Let s be the common length of the remaining two segments, let u = s + q, and let p = u + s. Then p/2 is the length of the radius of the semicircle. Therefore, by the diagonal rule, sq. p/2 – sq. q/2 = sq. h. On the other hand, sq. p/2 – sq. q/2 = u · s. (See Fig. ) Consequently, u · s = sq. h, and it follows that u, s is a 26 Amazing Traces of a Babylonian Origin in Greek Mathematics solution to the mentioned metric algebra problem of type B1b.

Do it like this: d/2 = 10 10 = the “vertical” radius Continue like this: d/2 = 10 Write down d/2 = 10 again sq. d/2 = 1 40 d/2 – p = 10 – 2 = 8 = b/2 sq. b/2 = 1 04 sq. d/2 – sq. b/2 = 1 40 – 1 04 = 36 = sq. s/2 sqs. 36 = 6 = s/2 2·6= s = the chord Done The straightforward solution procedure is explained in Fig. 5, left Given are the diameter d of a circle and the distance p of a chord from the circumference of the circle along a radius orthogonal to the chord. The length of the chord is computed by use of the diagonal rule (see Ch.

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.

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