References For Wantzel References for pierre Laurent wantzel. Articles F Cajori, pierreLaurent wantzel, Bull. Amer. Math. Soc. 24 (1) (1917), 339347. http://intranet.woodvillehs.sa.edu.au/pages/resources/maths/History/~DZ5844.htm
Full Alphabetical Index Translate this page John (553*) Wall, C Terence (545*) Wallace, William (261*) Wallis, John (784*)Wang, Hsien Chung (649) Wangerin, Albert (46*) wantzel, pierre (1020) Waring http://intranet.woodvillehs.sa.edu.au/pages/resources/maths/History/Flllph.htm
Four Problems Of Antiquity The problem had been settled in 1837 by pierre Laurent wantzel (18141848) whohad proven that there was no way to trisect a 60 o angle in the classical http://www.cut-the-knot.org/arithmetic/antiquity.shtml
Extractions: Construct a square whose area equals that of a given circle. Often another problem is attached to the list: Construct a regular heptagon (a polygon with 7 sides.) The problems are legendary not because they did not have solutions, or the solutions they had were unusually hard. No, numerous simple solutions have been found yet by Greek mathematicians. The problem was in that all known solutions violated an important condition for this kind of problems, one condition imposed by the Greek mathematicians themselves: Valid solutions to the construction problems are assumed to consist of a finite number of steps of only two kinds: drawing a straight line with a ruler (or rather a straightedge as no marks are allowed on the ruler) and drawing a circle. You are referred to solutions of problems and as examples of existent solutions. That no solution exists subject to the self-imposed constraints have been proven only in the 19th century.
Fermat Number regular ngon is constructible. The necessity of this condition wasnot proved until 1836 by pierre wantzel. A positive integer n is http://www.fact-index.com/f/fe/fermat_number.html
Extractions: Main Page See live article Alphabetical index A Fermat number , named after Pierre de Fermat who first studied them, is a positive integer of the form where n is a nonnegative integer. The first eight Fermat numbers are If 2 n + 1 is prime , it can be shown that n must be a power of 2. In other words, every prime of the form 2 n + 1 is a Fermat number, and such primes are called Fermat primes . The only known Fermat primes are F F Table of contents 1 Basic Properties 4 Relationship to Constructible Polygons The Fermat numbers satisfy the following recurrence relations for n share a common factor i j and F i and F j have a common factor a > 1. Then a divides both and F j ; hence a divides their difference 2. Since a > 1, this means a = 2. This is a contradiction , because each Fermat number is clearly odd. As a corollary , we obtain another proof of the infinitude of the prime numbers: for each F n , choose a prime factor p n p n Here are some other basic properties of the Fermat numbers: If n F n mod 6). (See
Polygon regular polygons can be constructed with ruler and compass alone was settled by CarlFriedrich Gauss in 1796 (sufficiency)and pierre wantzel in 1836 (necessity http://www.fact-index.com/p/po/polygon.html
Extractions: Main Page See live article Alphabetical index A polygon (from the Greek poly , for "many", and gwnos , for "angle") is a closed planar path composed of a finite number of straight line segments. The term polygon sometimes also describes the interior of the polygon (the open area that this path encloses) or the union of both. The straight line segments that make up the polygon are called its sides or edges and the points where the sides meet are the polygon's vertices Table of contents 1 Names and types 4 Related links A simple non-convex hexagon A complex polygon Polygons are named according to the number of sides, combining a Greek root with the suffix -gon , e.g. pentagon dodecagon . The triangle and quadrilateral are exceptions. For larger numbers, mathematicians write the numeral itself, eg 17-gon . A variable can even be used, usually n-gon . This is useful if the number of sides is used in a formula. Polygon names Name Sides triangle quadrilateral pentagon hexagon ... nonagon or ennagon decagon hendecagon or undecagon dodecagon hectagon megagon googolgon The taxonomic classification of polygons is illustrated by the following tree: A polygon is simple if it is described by a single, non-intersecting boundary; otherwise it is called
Il Teorema Di Morley Translate this page Fu comunque solo nel 1837 che pierre wantzel (1814-1848) riuscì a dimostrare lanecessità della condizione di Gauss sui poligoni regolari e quindi anche l http://www.lorenzoroi.net/geometria/Morley.html
Extractions: Dopo un sintetico inquadramento storico del problema della trisezione di un angolo e una introduzione all'uso di alcuni strumenti della geometria dinamica, si passa a dimostrare un utile lemma sull'incentro e quindi si propone la costruzione che conduce alla dimostrazione del teorema di Morley. Di questo teorema si fornisce infine una seconda prova di carattere algebrico sfruttando i teoremi della trigonometria. trisecati duplicazione del cubo e la quadratura del cerchio poligoni regolari . Tale problema fu affrontato con successo da Gauss primi di Fermat ) della forma 2 p + 1, con p m e m = 0, 1, 2... . Ne segue che, per esempio, poligoni regolari di 7 o 9 lati non sono elementarmente costruibili. Fu comunque solo nel 1837 che Pierre Wantzel n n con l'angolo cos 3 x x 3 x ossia x - 3 x - 1 = 0. Ed essendo l'equazione di terzo grado e irriducibile, secondo il criterio di Wantzel Poiché Euclide non considerava oggetti di cui non avesse precedentemente stabilito l'esistenza con una esplicita costruzione (prima di dimostrare il teorema di Pitagora, spiega come costruire un quadrato...), solo con una certa riluttanza i matematici si sono abituati ad accettare nella geometria euclidea l'esistenza di situazioni che essi non fossero in grado di costruire. In particolare, la "storica" difficoltà di trisecare un angolo è probabilmente la ragione del perché il teorema che ci accingiamo a dimostrare non fu scoperto se non nel XX secolo.
Wiskunde Verder zijn er stukken gewijd aan Leonhard Euler (17071783), en aan de nagenoegonbekende pierre-Laurent wantzel (1814-1848) die als eerste bewees dat een http://www.vssd.nl/hlf/wiskunde.html
Extractions: Onderwijsuitgaven van Delft University Press bestelformulier / order form download Acrobat Reader Analyse Numerieke wiskunde Statistiek en kansrekening Lineaire algebra Diversen wiskunde Regelmaat in de ruimte wiskunde dr. J.H.J. Almering e.a., geheel herzien door dr.H. Bavinck en dr.ir. R.W. Goldbach 1996 / 592 p. / ISBN 90-407-1260-3 / geb. / Euro 29,95 Aan het eind van de meeste paragrafen is een aantal oefeningen opgenomen om de lezer vertrouwd te maken met de voorafgaande leerstof. Aan het eind van elk hoofdstuk is een paragraaf met vraagstukken toegevoegd, gerangschikt overeenkomstig de behandeling van de leerstof in het betreffende hoofdstuk. Inhoud n naar IR m PDF-bestand van de inhoudsopgave (32 Kb) wiskunde dr. H. Bavinck
Haitian Math Whiz May Have Unraveled Age Over 2000 years later, in 1837, a French mathematician named pierre wantzel proclaimedthat it was impossible to trisect an angle using just a compass and a http://www.radiolakay.com/romain.htm
Extractions: Around 450 B.C., the Greek mathematician, Hippias of Ellis, began searching for a way to trisect an angle. Over 2000 years later, in 1837, a French mathematician named Pierre Wantzel proclaimed that it was impossible to trisect an angle using just a compass and a straightedge, the only tools allowed in geometric construction. But now, at the dawn of the twenty-first century, a Haitian computer program designer, Leon Romain, claims he has proven, with a "missing theorem," that it is possible to trisect an angle with those simple tools, disproving Wantzel's assertion and exploding centuries of mathematical gospel.
Ruler-and-compass Constructions - Reference Library polygons can be constructed with ruler and compass alone was settled by Carl FriedrichGauss in 1796 and (sufficiency) and pierre wantzel in 1836 (necessity http://www.campusprogram.com/reference/en/wikipedia/r/ru/ruler_and_compass_const
Extractions: Main Page See live article Alphabetical index A number of ancient problems in geometry involve the construction of lengths or angles using only an idealised ruler and compass . The ruler is indeed a straightedge , and may not be marked; the compass may only be set to already constructed distances, and used to describe circular arcs. Some famous ruler-and-compass problems have been proved impossible, in several cases by the results of Galois theory In spite of these impossibility proofs, some mathematical amateurs persist in trying to solve these problems. Many of them fail to understand that many of these problems are trivially soluble provided that other geometric transformations are allowed: for example, squaring the circle is possible using geometric constructions, but not possible using ruler and compasses alone. Mathematician Underwood Dudley has made a sideline of collecting false ruler-and-compass proofs, as well as other work by mathematical cranks , and has collected them into several books.
Trisection De L'angle en 1837 par pierre Laurent wantzel (1814-1848). Règle et compas. http://membres.lycos.fr/villemingerard/Histoire/Trisangl.htm
Extractions: NOMBRES - Curiosités, théorie et usages Accueil Dictionnaire Rubriques Index ... M'écrire Édition du: Rubrique: HISTOIRE ANTIQUITÉ Introduction Duplication du cube Trisection de l'angle Quadrature du cercle ... Heptagone Sommaire de cette page ÉQUATION pour la trisection CONSTRUCTION à l'équerre Pages voisines Règle et compas Transcendant Histoire Hilbert ... Bissection Trisection Découper un angle quelconque en deux parts égales Découper un angle quelconque en trois parts égales Bissection ÉQUATION POUR trois Idée de la démonstration avec un angle de 20° Calculons en général cos(3a) = cos(a)cos(2a) - sin(a)sin(2a) = cos(a)(cos (a) - sin (a)) - 2sin (a)cos(a) = cos(a)(2cos (a) - 1) - 2(1 - cos (a))cos(a) (a) - 3cos(a) Prenons le cas particulier de a o cos(3a) = cos(60 o L'équation, dans ce cas, devient (a) - 3cos(a) (a) - 6cos(a) - 1 En remplaçant cos(a) = x Avec v = 2x = v Voir Équation Solutions rationnelles ? Supposons que Oui, alors v = p/q fraction minimale (simplifiée) En remplaçant dans l'équation = (p/q) - 3(p/q) - 1 En multipliant par q = p - q En reformulant q = p = p (p² - 3q²) On déduit que p est divisible par q Conséquence p est divisible par q Impossible p/q est une fraction irréductible par hypothèse Et en factorisant avec p p + q = q (3p + q²) On déduit que q est divisible par p Conséquence q est divisible par p Impossible p/q est une fraction irréductible par hypothèse La supposition est fausse v n'est par rationnel
Constructions Géométriques - Constructible Translate this page o ceux pour p = 2 0 , 2 1 , 2 2 , 2 3 , 2 4. pierre Laurent wantzel (1814- 1848). Il démontre que seuls ces polygones sont constructibles. http://membres.lycos.fr/villemingerard/Geometri/Construc.htm
Extractions: Accueil Dictionnaire Rubriques Index ... M'écrire Édition du: Rubrique: CONSTRUCTIONS Constructible Bissection Sommaire de cette page CONSTRUCTIBLE: Théorème de Gauss CONSTRUCTION DES POLYGONES TYPES DE CONSTRUCTIONS Pages voisines Allumettes Centre du cercle Cycloïde Géométrie ... Triangles CONSTRUCTIBILITÉ Peut-on construire rigoureusement une figure géométrique en utilisant des outils précis: règle, compas, marque sur la règle Division de la circonférence pas possible pour CONSTRUCTIBLE: Théorème de Gauss Historique Les Grecs connaissaient de nombreuses possibilités de construction Il a fallu 2000 ans pour que Gauss (1796) démontre le théorème suivant Théorème de Gauss (formulation n°1) Il est possible de diviser la circonférence en un nombre impair de parties égales si, et seulement si, le nombre est un nombre premier de Fermat Théorème de Gauss (formulation n°2) Le polygone n'est constructible avec une règle et un compas que si : m étant entier et n premier Ou si n est formé exclusivement de combinaisons d'un nombre premier de Fermat (comme 3, 5, 17, 257, 65 537) et de puissances de deux.
Vinkelns Tredelning Och Andra Geometriska Konstruktionsproblem misstänka att problemen kanske inte kan lösas med de klassiska hjälpmedlen, mendet dröjde ända till 1837 innan fransmannen pierre wantzel bevisade att http://www.matematik.su.se/matematik/exempel/geometri/Arkimedes.html
Extractions: och cirkelns kvadratur. lika delar, konstruerar en kvadrat deliska problemet Arkimedes tredelning av en vinkel. En vinkel v (dvs AOB O . En linje genom B C och OA D CD DCO likbent (eftersom CD och CO x y DOC och DOB ger y=2x och v=x+y=3x C och D C och D B D OA (och C mellan B och D C Pierre Wantzel Ferdinand von Lindemann inte Euklides AB A och B och samma radie AB . Om C ABC AB Carl Friedrich Gauss fann 1796 en konstruktion av den regelbundna n k p ...p r, p i m m samt Arkimedes och
Chronological Indexes (Nã) Joseph Liouville 18091882) ? (Evariste Galois 1811-1832) ? (pierre Laurent wantzel 1814-1848) ? (James Joseph http://www5f.biglobe.ne.jp/~mathlife/html/mathematicians.htm
Japanese Syllabaries (Ü\¹) ? (Karl Theodor Wihelm Weierstrass 18151897) (Andrew Wiles 1953-) ? (pierre Laurent wantzel 1814-1848) http://www5f.biglobe.ne.jp/~mathlife/html/jpsyllabary.htm
Links: Henry Darcy And His Law pierre Laurent wantzel (18141848). Engineering History Sites Linksto other interesting sites. Send me your site if you have history http://biosystems.okstate.edu/darcy/Links.htm
Johns Hopkins Magazine February 1999 after generations of mathematicians had attempted in vain to solve it, that a Frenchbridge and highway engineer named pierre Louis wantzel finally cracked the http://www.jhu.edu/~jhumag/0299web/degree.html
Extractions: Deputy Director, National Foreign Language Center Japanese is without question the most daunting language for a native English speaker to tackle, according to Brecht. "I would like to learn Japanese but I don't have enough time in my lifetime. That's very depressing," says the linguist, whose center is based at Hopkins's Nitze School of Advanced International Studies (SAIS) . He notes that the State Department allows its students three times as long to learn Japanese as it does languages like Spanish or French. As Brecht explains it, the challenge with Japanese is threefold. First, there's the fact that the Japanese written code is different from the spoken code. "Therefore, you can't learn to speak the language by learning to read it," and vice versa. What's more, there are three different writing systems to master. The kanji system uses characters borrowed from Chinese. Users need to learn 10,000 to 15,000 of these characters through rote memorization; there are no mnemonic devices to help. Written Japanese also makes use of two syllabary systems: kata-kana for loan words and emphasis, and hira-gana for spelling suffixes and grammatical particles.
Full Alphabetical Index Translate this page van der (552*) Wald, Abraham (144*) Wallace, William (261*) Wallis, John (784*)Wang, Hsien Chung (649) Wangerin, Albert (46*) wantzel, pierre (1020) Waring http://www.geocities.com/Heartland/Plains/4142/matematici.html
The Hundred Greatest Theorems Karl Frederich Gauss. 1801. 8. The Impossibility of Trisecting the Angle and Doublingthe Cube. pierre wantzel. 1837. 9. The Area of a Circle. Archimedes. 225 BC. 10. http://personal.stevens.edu/~nkahl/Top100Theorems.html
Extractions: The millenium seemed to spur a lot of people to compile "Top 100" or "Best 100" lists of many things, including movies (by the American Film Institute) and books (by the Modern Library). Mathematicians were not immune, and at a mathematics conference in July, 1999, Paul and Jack Abad presented their list of "The Hundred Greatest Theorems." Their ranking is based on the following criteria: "the place the theorem holds in the literature, the quality of the proof, and the unexpectedness of the result." The list is of course as arbitrary as the movie and book list, but the theorems here are all certainly worthy results. I hope to over time include links to the proofs of them all; for now, you'll have to content yourself with the list itself and the biographies of the principals. The Irrationality of the Square Root of 2 Pythagoras and his school 500 B.C. Fundamental Theorem of Algebra Karl Frederich Gauss The Denumerability of the Rational Numbers Georg Cantor ... Pythagoras and his school 500 B.C.
Extractions: Retour à la catégorie - Mathématiques Accueil A525G Commentaires Pi est défini comme étant le rapport constant entre la circonférence et le diamètre d'un cercle. Remarque : Il a déjà fallu un certain temps à l'homme pour trouver que ce rapport est constant..., et donc pour découvrir l'existence de PI. A l'origine, ce rapport est noté P. C'est Euler qui utilisa la notation de la seizième lettre de l'alphabet grec, notation gardée par la suite vue l'importance de ses travaux. Ainsi, pour tout cercle de périmètre p, de diamètre D (de rayon R), def : p = Pi * D = 2 * Pi * R Connaître l'existence d'une constante est fort intéressant, mais connaître sa valeur l'est beaucoup plus... Elle l'est d'autant plus que Pi apparait dans de très nombreux problèmes physiques et mathématiques : Calcul de surface et de volume impliquant des cercles ou des ellipses. Par exemple, on trouve, par intégration, des formules classiques telles que :
Four Problems Of Antiquity The problem had been settled in 1837 by pierre Laurant wantzel (18141848) whohad proven that there was no way to trisect a 60 o angle in the classical http://hem.passagen.se/ceem/fourprob.htm
Extractions: Four problems of antiquity Three geometric questions raised by the early Greek mathematicians attained the status of classical problems in Mathematics. These are: Doubling of the cube Construct a cube whose volume is double that of a given one. Angle trisection Trisect an arbitrary angle. Squaring a circle Construct a square whose area equals that of a given circle. (Often another problem is attached to the list: ) Construct a regular heptagon (a polygon with 7 sides.) The problems are legendary not because they did not have solutions, or the solutions they had were unusually hard. No - numerous simple solutions have been found yet by Greek mathematicians. The problem was in that all known solutions violated by an important condition for this kind of problems, one condition imposed by the Greek mathematicians themselves: Valid solutions to the construction problems are assumed to consist of a finite number of steps of only two kinds: drawing a straight line with a ruler (or rather a straightedge as no marks are allowed on the ruler) and drawing a circle. You are referred to solutions of problems 2 and 3 as examples of existent solutions. That no solution exists subject to the self-imposed constraints have been proven only in the 19th century.
