By Gabor Toth
Past version offered 2000 copies in three years; Explores the sophisticated connections among quantity thought, Classical Geometry and smooth Algebra; Over a hundred and eighty illustrations, in addition to textual content and Maple records, can be found through the net facilitate knowing: http://mathsgi01.rutgers.edu/cgi-bin/wrap/gtoth/; comprises an insert with 4-color illustrations; comprises a variety of examples and worked-out difficulties
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Additional info for Glimpses of Algebra and Geometry, Second Edition
4. (b) Let pn nsn and Pn nSn be the corresponding perimeters. Use (a) to conclude that 2Pn pn and p2n pn P2n . P2n Pn + pn (c) Assume that the circle has unit radius and show that p6 6, P6 √ 4 3, p12 12 2 − √ 3, P12 24 2 − How well do p12 and P12 approximate 2π? (d) Use the inequality √ 2ab ≤ ab, a, b > 0, a+b to show10 that 2 P2n p2n < Pn pn2 and 2 1 2 1 + < + . P2n p2n Pn pn 10 This interpolation technique is due to Heinrich D¨orrie, 1940. √ 3 . 24 2. “. . There Are No Irrational Numbers at All”—Kronecker Deﬁne An Pn pn2 3 Bn and 3Pn pn .
Assuming this, if (x0 , y0 ) is on Cf , then intersecting the surface with the horizontal coordinate plane, we obtain that, near (x0 , y0 ), Cf consists of two smooth curves crossing over at (x0 , y0 ). In this case, we say that the algebraic curve Cf has a double point at (x0 , y0 ). The signiﬁcance of this concept follows from the fact that any double point on a cubic rational curve is rational! We will not prove this in general. ♣ Instead, we restrict ourselves to the most prominent class of cubic rational curves given by the so-called Weierstrass form y2 P(x), where P is a cubic polynomial with rational coefﬁcients.
9 10 2. “. . There Are No Irrational Numbers at All”—Kronecker Letting n → ∞ and assuming that |x| < 1 to assure convergence, we arrive at the geometric series formula4 1 , |x| < 1. 1−x We now turn to the proof. a1 a2 · · · ak a1 a2 · · · ak a1 a2 · · · ak · · · , where the decimal digits ai are between 0 and 9. We rewrite this as a1 a2 · · · ak (10−k + 10−2k + 10−3k + · · ·) a1 · · · ak 10−k (1 + 10−k + (10−k )2 + · · ·) a1 · · · ak 10−k a1 · · · ak 1 − 10−k 10k − 1 where we used the geometric series formula.