Glimpses of Algebra and Geometry, Second Edition by Gabor Toth

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:; comprises an insert with 4-color illustrations; comprises a variety of examples and worked-out difficulties

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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.

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