American Mathematical Society, János Kollár, Robert's Algebraic Geometry Santa Cruz 1995, Part 2: Summer Research PDF

By American Mathematical Society, János Kollár, Robert Lazarsfeld

ISBN-10: 0821808958

ISBN-13: 9780821808955

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Extra resources for Algebraic Geometry Santa Cruz 1995, Part 2: Summer Research Institute on Algebraic Geometry, July 9-29, 1995, University of California, Santa Cruz

Example text

Eucl. 30 (left) and its proof (right) Proof. By Eucl. 27 and Eucl. 29, the lines a and b are parallel if and only if the angles α and β are equal. Eucl. 31. Drawing a parallel to a given line through a given point A. A A a a Fig. 13. Eucl. 31 (left) and the proposed construction (right) Proof. Euclid’s proof makes use of Eucl. 23 which is itself a consequence of Eucl. 22. One can also use two orthogonal lines (Eucl. 12 followed by Eucl. 11). Remark. Proclus made the following statement in his commentary: There exists at most one line through a given point A which is parallel to a given line.

1. g. Eucl. 2 and Eucl. 2 below). Eucl. 2 applies the algorithm of Eucl. 2 to real numbers. g. Hairer and Wanner, 1997, p. 67). Example. In Fig. 21 we see the Euclidean algorithm applied to a = Φ (resp. √ a = 2) and b = 1. We see that we obtain an infinite sequence of similar triangles (resp. squares) and an unending sequence of remainders c = a − b, 9 The Arabic word “algorithm” only appeared some thousand years later. The ε, though a Greek letter, came into use for this purpose only with Weierstrass many many centuries later.

48 and 50 of the Rhind papyrus, see the pictures of Fig. 23 (left): a circle in a square of 9 × 9 = 81 units is squared by cutting off corners with two sides of length 3 units. This creates a surface of 81 − 18 = 63 units. Since 63 is close to 64 = 82 = (9 − 1)2 , we obtain the “Egyptian algorithm” subtract one ninth of the diameter, then square . This is demonstrated in No. 50, where (reproductions from Peet, 1923) the area of a circle of diameter 9 is 64 . In Rhind No. 42, while computing the volume of a cylindrical container, the 1 + area for diameter 10 is given as 79 108 1 , 79 81 1 324 or , which the correct value.

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Algebraic Geometry Santa Cruz 1995, Part 2: Summer Research Institute on Algebraic Geometry, July 9-29, 1995, University of California, Santa Cruz by American Mathematical Society, János Kollár, Robert Lazarsfeld


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