{"id":1787,"date":"2019-11-30T16:06:22","date_gmt":"2019-11-30T15:06:22","guid":{"rendered":"https:\/\/www.mathweb.fr\/euclide\/?p=1787"},"modified":"2020-09-29T10:41:53","modified_gmt":"2020-09-29T08:41:53","slug":"un-resultat-sur-les-racines-dun-polynome-a-coefficients-entiers","status":"publish","type":"post","link":"https:\/\/www.mathweb.fr\/euclide\/2019\/11\/30\/un-resultat-sur-les-racines-dun-polynome-a-coefficients-entiers\/","title":{"rendered":"Un r\u00e9sultat sur les racines d\u2019un polyn\u00f4me \u00e0 coefficients entiers"},"content":{"rendered":"\n<p class=\"wp-block-paragraph\">On consid\u00e8re un polyn\u00f4me \u00e0 coefficients entiers : $$P(X)=\\sum_{k=0}^n c_kX^k\\quad,\\quad \\forall\\; k\\in[0;n]\\cap\\mathbb{N},\\ c_k\\in\\mathbb{Z}.$$L&#8217;objectif ici est de d\u00e9montrer que s&#8217;il admet une racine rationnelle irr\u00e9ductible \\(\\frac{a}{b}\\) alors \\(a\\) divise \\(c_0\\) et \\(b\\) divise \\(c_ n\\).<\/p>\n\n\n\n<!--more-->\n\n\n\n<div id=\"ez-toc-container\" class=\"ez-toc-v2_0_83 counter-hierarchy ez-toc-counter ez-toc-white ez-toc-container-direction\">\n<div class=\"ez-toc-title-container\">\n<p class=\"ez-toc-title\" style=\"cursor:inherit\">Au menu sur cette page...<\/p>\n<span class=\"ez-toc-title-toggle\"><a href=\"#\" class=\"ez-toc-pull-right ez-toc-btn ez-toc-btn-xs ez-toc-btn-default ez-toc-toggle\" aria-label=\"Toggle Table of Content\"><span class=\"ez-toc-js-icon-con\"><span class=\"\"><span class=\"eztoc-hide\" style=\"display:none;\">Toggle<\/span><span class=\"ez-toc-icon-toggle-span\"><svg style=\"fill: #999;color:#999\" xmlns=\"http:\/\/www.w3.org\/2000\/svg\" class=\"list-377408\" width=\"20px\" height=\"20px\" viewBox=\"0 0 24 24\" fill=\"none\"><path d=\"M6 6H4v2h2V6zm14 0H8v2h12V6zM4 11h2v2H4v-2zm16 0H8v2h12v-2zM4 16h2v2H4v-2zm16 0H8v2h12v-2z\" fill=\"currentColor\"><\/path><\/svg><svg style=\"fill: #999;color:#999\" class=\"arrow-unsorted-368013\" xmlns=\"http:\/\/www.w3.org\/2000\/svg\" width=\"10px\" height=\"10px\" viewBox=\"0 0 24 24\" version=\"1.2\" baseProfile=\"tiny\"><path d=\"M18.2 9.3l-6.2-6.3-6.2 6.3c-.2.2-.3.4-.3.7s.1.5.3.7c.2.2.4.3.7.3h11c.3 0 .5-.1.7-.3.2-.2.3-.5.3-.7s-.1-.5-.3-.7zM5.8 14.7l6.2 6.3 6.2-6.3c.2-.2.3-.5.3-.7s-.1-.5-.3-.7c-.2-.2-.4-.3-.7-.3h-11c-.3 0-.5.1-.7.3-.2.2-.3.5-.3.7s.1.5.3.7z\"\/><\/svg><\/span><\/span><\/span><\/a><\/span><\/div>\n<nav><ul class='ez-toc-list ez-toc-list-level-1 ' ><li class='ez-toc-page-1 ez-toc-heading-level-2'><a class=\"ez-toc-link ez-toc-heading-1\" href=\"https:\/\/www.mathweb.fr\/euclide\/2019\/11\/30\/un-resultat-sur-les-racines-dun-polynome-a-coefficients-entiers\/#Demonstration\" >D\u00e9monstration<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-2'><a class=\"ez-toc-link ez-toc-heading-2\" href=\"https:\/\/www.mathweb.fr\/euclide\/2019\/11\/30\/un-resultat-sur-les-racines-dun-polynome-a-coefficients-entiers\/#Application\" >Application<\/a><\/li><\/ul><\/nav><\/div>\n<h2 class=\"wp-block-heading\"><span class=\"ez-toc-section\" id=\"Demonstration\"><\/span>D\u00e9monstration<span class=\"ez-toc-section-end\"><\/span><\/h2>\n\n\n\n<p class=\"wp-block-paragraph\">On part de l&#8217;hypoth\u00e8se selon laquelle \\(P\\left(\\frac{a}{b}\\right)=0\\) puisque \\(\\frac{a}{b}\\) est une racine du polyn\u00f4me \\(P\\) : $$ \\begin{align} &amp; P\\left(\\frac{a}{b}\\right)=0 \\\\ \\iff &amp; \\sum_{k=0}^n c_k\\left(\\frac{a}{b}\\right)^k\\\\  \\iff &amp;  c_0+\\frac{c_1a}{b}+\\frac{c_2a^2}{b^2}+\\cdots+\\frac{c_na^n}{b^n}=0\\\\   \\iff &amp;  c_0b^n+c_1ab^{n-1}+c_2a^2b^{n-2}+\\cdots+c_{n-1}a^{n-1}b+c_na^n=0.\\end{align}$$  De cette derni\u00e8re \u00e9galit\u00e9, on peut en d\u00e9duire deux r\u00e9sultats :<\/p>\n\n\n\n<ul class=\"wp-block-list\"><li>Dans un premier temps, $$\\begin{align} c_0b^n &amp; =-c_1ab^{n-1}-c_2a^2b^{n-2}-\\cdots-c_{n-1}a^{n-1}b-c_na^n\\\\ &amp; = -a\\big(c_1b^{n-1}+c_2ab^{n-2}+\\cdots+c_{n-1}a^{n-2}b+c_na^{n-1}\\big) \\end{align}$$ Or, \\(\\text{pgcd}(a,b)=1\\) donc \\(\\text{pgcd}(a,b^n)=1\\) et, d&#8217;apr\u00e8s le th\u00e9or\u00e8me (arithm\u00e9tique) de Gauss, \\(a\\) divise \\(c_0\\) ;<\/li><li>D&#8217;autre part, $$\\begin{align} c_n a^n &amp; =-c_0b^n-c_1ab^{n-1}-c_2a^2b^{n-2}-\\cdots-c_{n-1}a^{n-1}b\\\\ &amp; = -b\\big(c_0b^{n-1}+c_1ab^{n-2}+c_2a^2b^{n-3}+\\cdots+c_{n-1}a^{n-2}\\big) \\end{align}$$ Or, \\(\\text{pgcd}(a,b)=1\\) donc \\(\\text{pgcd}(a^n,b)=1\\) et, d&#8217;apr\u00e8s le th\u00e9or\u00e8me (arithm\u00e9tique) de Gauss, \\(b\\) divise \\(c_n\\).<\/li><\/ul>\n\n\n\n<h2 class=\"wp-block-heading\"><span class=\"ez-toc-section\" id=\"Application\"><\/span>Application<span class=\"ez-toc-section-end\"><\/span><\/h2>\n\n\n\n<p class=\"wp-block-paragraph\">Ce r\u00e9sultat peut nous servir par exemple \u00e0 d\u00e9montrer que \\( \\sin\\frac{\\pi}{18}\\) est irrationnel. En effet, on sait que : $$\\sin\\left(3\\times\\frac{\\pi}{18}\\right)=\\sin\\frac{\\pi}{6}=\\frac{1}{2}\\;;$$ de plus,  $$ \\sin(3x)=3\\sin x-4\\sin^3 x$$ donc, de ces deux \u00e9galit\u00e9s, on d\u00e9duit : $$ 3\\sin\\frac{\\pi}{18}-4\\sin^3 \\frac{\\pi}{18}=\\frac{1}{2}$$ soit, en posant \\( S = \\sin\\frac{\\pi}{18}\\), $$ 3S-4S^3=\\frac{1}{2}$$ ou encore : $$ 8S^3-6S+1=0.$$ <\/p>\n\n\n\n<p class=\"wp-block-paragraph\">En posant \\( X = 2S\\), l&#8217;\u00e9galit\u00e9 devient :$$X^3-3X+1=0.$$ Si \\(\\sin\\frac{\\pi}{18}\\) est rationnel alors il en est de m\u00eame pour son double \\(X\\).<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">Or, d&#8217;apr\u00e8s le r\u00e9sultat d\u00e9montr\u00e9 pr\u00e9c\u00e9demment, si l&#8217;\u00e9quation \\(X^3-3X+1=0\\) admet une solution rationnelle \\(\\frac{a}{b}\\) alors \\(a\\) divise 1 et \\(b\\) divise 1, ce qui ne laisse comme choix pour solution que \\(X=1\\) ou \\(X=-1\\).<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">Mais ni 1 ni \\(-1\\) n&#8217;est solution de l&#8217;\u00e9quation, ce qui signifie que l&#8217;\u00e9quation \\( X^3-3X+1=0\\) n&#8217;admet aucune solution rationnelle et donc que la solution connue, \u00e0 savoir \\(X\\) est irrationnelle et donc que \\(S\\) est irrationnelle.<\/p>\n","protected":false},"excerpt":{"rendered":"<p>On consid\u00e8re un polyn\u00f4me \u00e0 coefficients entiers : $$P(X)=\\sum_{k=0}^n c_kX^k\\quad,\\quad \\forall\\; k\\in[0;n]\\cap\\mathbb{N},\\ c_k\\in\\mathbb{Z}.$$L&#8217;objectif ici est de d\u00e9montrer que s&#8217;il admet une racine rationnelle irr\u00e9ductible \\(\\frac{a}{b}\\) alors \\(a\\) divise \\(c_0\\) et \\(b\\) divise \\(c_ n\\).<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[6],"tags":[25,80,81],"class_list":["post-1787","post","type-post","status-publish","format-standard","hentry","category-mathematiques","tag-demonstration","tag-polynome","tag-racines"],"yoast_head":"<!-- This site is optimized with the Yoast SEO plugin v27.6 - 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