{"id":36006,"date":"2026-01-10T18:40:42","date_gmt":"2026-01-10T18:40:42","guid":{"rendered":"https:\/\/toposuranos.com\/material\/?p=36006"},"modified":"2026-01-10T18:40:42","modified_gmt":"2026-01-10T18:40:42","slug":"lalgorithme-de-la-division-euclidienne","status":"publish","type":"post","link":"https:\/\/toposuranos.com\/material\/fr\/lalgorithme-de-la-division-euclidienne\/","title":{"rendered":"L\u2019Algorithme de la Division Euclidienne"},"content":{"rendered":"<style>\np, ul, ol{\n  text-align: justify;\n}\nh1{\n  text-align:center;\n  text-transform: uppercase;\n}\nh2{\n  text-align:center;\n  text-transform: uppercase;\n  font-size:24pt;\n}\nh3{\n  text-align: center;\n  text-transform: uppercase;\n  font-size: 24px !important;\n}\n<\/style>\n<h1>L\u2019Algorithme de la Division<\/h1>\n<p style=\"text-align:center;\"><em>Dans ce cours, nous d\u00e9velopperons l\u2019<strong>algorithme de la division<\/strong> comme le principe qui formalise, pour les entiers, la d\u00e9composition unique <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">b=qa+r<\/span><\/span> avec <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">0\\le r&lt;|a|<\/span><\/span>. On d\u00e9montre d\u2019abord l\u2019<strong>existence<\/strong> du quotient et du reste, puis leur <strong>unicit\u00e9<\/strong>. Enfin, on interpr\u00e8te la signification du reste, on relie la th\u00e9orie \u00e0 l\u2019<strong>algorithme long de la division<\/strong> en tant que proc\u00e9dure de calcul, et on anticipe sa connexion naturelle avec l\u2019arithm\u00e9tique modulaire et les applications informatiques.<\/em><\/p>\n<p style=\"text-align:center;\"><strong>Objectifs d\u2019Apprentissage<\/strong><\/p>\n<p>\u00c0 l\u2019issue de ce cours, l\u2019\u00e9tudiant sera capable de :<\/p>\n<ol>\n<li><strong>Identifier<\/strong> les concepts et r\u00f4les de base de la division enti\u00e8re (dividende, diviseur, quotient, reste) ainsi que la notion de divisibilit\u00e9 comme cas exact.<\/li>\n<li><strong>Expliquer<\/strong> l\u2019\u00e9nonc\u00e9 de l\u2019algorithme\/th\u00e9or\u00e8me de la division euclidienne, y compris les conditions qui fixent le reste (<span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">0\\le r&lt;|a|<\/span><\/span>) et leur objectif d\u2019\u00e9viter les ambigu\u00eft\u00e9s.<\/li>\n<li><strong>Appliquer<\/strong> la d\u00e9composition euclidienne <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">b=qa+r<\/span><\/span> pour d\u00e9terminer <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">q<\/span><\/span> et <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">r<\/span><\/span> dans des exemples concrets, en v\u00e9rifiant la borne du reste.<\/li>\n<li><strong>Analyser<\/strong> le traitement des cas selon le signe du dividende et du diviseur, en justifiant pourquoi la convention euclidienne maintient un reste non n\u00e9gatif.<\/li>\n<li><strong>Ex\u00e9cuter<\/strong> l\u2019algorithme long de la division comme proc\u00e9dure m\u00e9canique fond\u00e9e sur la repr\u00e9sentation positionnelle afin d\u2019obtenir <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">q<\/span><\/span> et <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">r<\/span><\/span>.<\/li>\n<\/ol>\n<p style=\"text-align:center;\"><b><u>INDEX DES CONTENUS<\/u>:<\/b><br \/>\n<a href=\"#1\">Division et divisibilit\u00e9<\/a><br \/>\n<a href=\"#2\">Th\u00e9or\u00e8me de la division euclidienne<\/a><br \/>\n<a href=\"#3\">D\u00e9monstration de la division euclidienne<\/a><br \/>\n<a href=\"#4\">Interpr\u00e9tation du quotient et du reste<\/a><br \/>\n<a href=\"#5\">Algorithme long de la division<\/a><br \/>\n<a href=\"#6\">Conclusion<\/a><br \/>\n<a href=\"#7\">Exercices propos\u00e9s et r\u00e9solus<\/a>\n<\/p>\n<p><a name=\"1\"><\/a><\/p>\n<h2>Division et divisibilit\u00e9<\/h2>\n<p><center><iframe class=\"lazyload\" width=\"560\" height=\"315\" data-src=\"https:\/\/www.youtube.com\/embed\/nlER6VzrGwA?si=79gpsAzwc7wcWEVn\" title=\"YouTube video player\" frameborder=\"0\" allow=\"accelerometer; autoplay; clipboard-write; encrypted-media; gyroscope; picture-in-picture; web-share\" referrerpolicy=\"strict-origin-when-cross-origin\" allowfullscreen><\/iframe><\/center><\/p>\n<p>En arithm\u00e9tique, la divisibilit\u00e9 d\u00e9crit le cas \u00ab exact \u00bb : <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">a\\mid b<\/span><\/span> signifie que <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">b<\/span><\/span> s\u2019\u00e9crit exactement comme un multiple de <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">a<\/span><\/span>. Cependant, m\u00eame s\u2019il n\u2019y a pas toujours divisibilit\u00e9 entre deux entiers quelconques, nous pouvons encore nous demander \u00ab combien de fois un nombre tient dans un autre et, s\u2019il n\u2019y tient pas exactement, ce qu\u2019il en reste \u00bb. C\u2019est dans ce contexte qu\u2019appara\u00eet l\u2019<strong>algorithme de la division<\/strong>, qui garantit qu\u2019un entier quelconque <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">b<\/span><\/span> peut s\u2019\u00e9crire comme un multiple de <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">a<\/span><\/span> plus un \u00ab reste \u00bb.<\/p>\n<p>\u00c0 titre d\u2019exemple, consid\u00e9rons <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">a=3<\/span><\/span> et <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">b=16<\/span><\/span>. Il est clair que 3 ne divise pas 16. Toutefois, il est possible de trouver un quotient et un reste pour l\u2019op\u00e9ration de division, car<\/p>\n<p style=\"text-align:center;\"><span class=\"katex-eq\" data-katex-display=\"false\">16 = 5\\cdot 3 + 1<\/span>\n<figure id=\"attachment_35775\" aria-describedby=\"caption-attachment-35775\" style=\"width: 1492px\" class=\"wp-caption aligncenter\"><img decoding=\"async\" src=\"data:image\/gif;base64,R0lGODlhAQABAIAAAAAAAP\/\/\/yH5BAEAAAAALAAAAAABAAEAAAIBRAA7\" data-src=\"http:\/\/toposuranos.com\/material\/wp-content\/uploads\/2025\/12\/division16por3.jpg\" alt=\"Un grupo de 16 cajas han sido separadas en grupos de a 3, como resultado se obtienen 5 grupos de cajas y sobra 1\" width=\"1492\" height=\"230\" class=\"size-full wp-image-35775 lazyload\" \/><figcaption id=\"caption-attachment-35775\" class=\"wp-caption-text\"><noscript><img decoding=\"async\" src=\"http:\/\/toposuranos.com\/material\/wp-content\/uploads\/2025\/12\/division16por3.jpg\" alt=\"Un grupo de 16 cajas han sido separadas en grupos de a 3, como resultado se obtienen 5 grupos de cajas y sobra 1\" width=\"1492\" height=\"230\" class=\"size-full wp-image-35775 lazyload\" srcset=\"https:\/\/toposuranos.com\/material\/wp-content\/uploads\/2025\/12\/division16por3.jpg 1492w, https:\/\/toposuranos.com\/material\/wp-content\/uploads\/2025\/12\/division16por3-300x46.jpg 300w, https:\/\/toposuranos.com\/material\/wp-content\/uploads\/2025\/12\/division16por3-1024x158.jpg 1024w, https:\/\/toposuranos.com\/material\/wp-content\/uploads\/2025\/12\/division16por3-768x118.jpg 768w\" sizes=\"(max-width: 1492px) 100vw, 1492px\" \/><\/noscript> Un groupe de 16 bo\u00eetes a \u00e9t\u00e9 s\u00e9par\u00e9 en groupes de 3 ; on obtient ainsi 5 groupes de bo\u00eetes et il en reste 1<\/figcaption><\/figure>\n<p>En cons\u00e9quence, en divisant <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">b=16<\/span><\/span> par <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">a=3<\/span><\/span> (o\u00f9 <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">b<\/span><\/span> est le <strong>dividende<\/strong> et <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">a<\/span><\/span> est le <strong>diviseur<\/strong>), repr\u00e9sent\u00e9 par <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">16\/3<\/span><\/span> ou <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">16\\div 3<\/span><\/span>, on obtient comme <strong>quotient<\/strong> <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">q=5<\/span><\/span> et comme <strong>reste<\/strong> <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">r=1<\/span><\/span>. En g\u00e9n\u00e9ral, l\u2019<em>algorithme de la division<\/em> affirme que, pour tous entiers <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">a<\/span><\/span> (diviseur) et <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">b<\/span><\/span> (dividende) avec <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">a\\neq 0<\/span><\/span>, il existe des entiers uniques <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">q<\/span><\/span> et <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">r<\/span><\/span> tels que<\/p>\n<p style=\"text-align:center;\"><span class=\"katex-eq\" data-katex-display=\"false\">b=qa+r,\\qquad 0\\le r&lt;|a|<\/span>\n<p>Dans cet exemple, <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">16=3\\cdot 5+1<\/span><\/span>, et la condition <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">0\\le r&lt;|a|<\/span><\/span> est v\u00e9rifi\u00e9e car <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">0\\le 1&lt;3<\/span><\/span>. Cette formulation \u00e9vite les ambigu\u00eft\u00e9s lorsque le diviseur (ou le dividende) peut \u00eatre n\u00e9gatif et garantit que le reste est toujours non n\u00e9gatif et strictement inf\u00e9rieur \u00e0 la valeur absolue du diviseur.<\/p>\n<p><a name=\"2\"><\/a><\/p>\n<h2>Th\u00e9or\u00e8me de la division euclidienne<\/h2>\n<p>Le r\u00e9sultat de l\u2019application de l\u2019algorithme de la division est ce que l\u2019on appelle la <strong>division euclidienne<\/strong>, et il repose sur le r\u00e9sultat suivant.<\/p>\n<p><b>Th\u00e9or\u00e8me :<\/b> Soient <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">a,b\\in\\mathbb{Z}<\/span><\/span> avec <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">a\\neq 0<\/span><\/span>. Alors il existe des <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">q,r\\in\\mathbb{Z}<\/span><\/span> uniques tels que<\/p>\n<p style=\"text-align:center;\" dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\nb=qa+r,\\qquad 0\\le r&lt;|a|.\n\n<\/span>\n<p>L\u2019entier <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">q<\/span><\/span> est appel\u00e9 <strong>quotient<\/strong> et <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">r<\/span><\/span> est appel\u00e9 <strong>reste<\/strong> de la division de <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">b<\/span><\/span> par <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">a<\/span><\/span>.<\/p>\n<p><a name=\"3\"><\/a><\/p>\n<h2>D\u00e9monstration de la division euclidienne<\/h2>\n<p>Cette d\u00e9monstration se divise en deux parties : on d\u00e9montre d\u2019abord que le quotient et le reste existent ; puis, \u00e9tant donn\u00e9 qu\u2019ils existent, qu\u2019ils sont uniques.<\/p>\n<h3>Existence<\/h3>\n<p>Soient <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">a,b\\in\\mathbb{Z}<\/span><\/span> avec <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">a\\neq 0<\/span><\/span> et d\u00e9finissons <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">d=|a|<\/span><\/span>, de sorte que <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">d&gt;0<\/span><\/span>. Nous traiterons d\u2019abord le cas <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">b\\ge 0<\/span><\/span> : nous montrerons qu\u2019il existe des <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">q,r\\in\\mathbb{Z}<\/span><\/span> tels que <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">b=dq+r<\/span><\/span> et <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">0\\le r&lt;d<\/span><\/span>. Nous traiterons ensuite le cas <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">b&lt;0<\/span><\/span>. Enfin, nous remplacerons <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">d<\/span><\/span> par <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">|a|<\/span><\/span> afin d\u2019obtenir la forme <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">b=qa+r<\/span><\/span> avec <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">0\\le r&lt;|a|<\/span><\/span>.<\/p>\n<p>Pour <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">b\\in\\mathbb{Z}<\/span><\/span> avec <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">b\\ge 0<\/span><\/span>, nous \u00e9tablissons la proposition <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">P(b)<\/span><\/span> : \u00ab il existe des <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">q,r\\in\\mathbb{Z}<\/span><\/span> tels que <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">b=dq+r<\/span><\/span> et <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">0\\le r&lt;d<\/span><\/span> \u00bb. Nous d\u00e9montrerons par r\u00e9currence que <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">P(b)<\/span><\/span> est valable pour tout <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">b\\ge 0<\/span><\/span>.<\/p>\n<p><strong>Cas de base :<\/strong> Pour <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">b=0<\/span><\/span>, en prenant <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">q=0<\/span><\/span> et <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">r=0<\/span><\/span>, on obtient <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">0=d\\cdot 0+0<\/span><\/span> et <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">0\\le 0&lt;d<\/span><\/span>. Par cons\u00e9quent, <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">P(0)<\/span><\/span> est vraie.<\/p>\n<p><strong>\u00c9tape inductive :<\/strong> Soit <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">k\\ge 0<\/span><\/span> et supposons que <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">P(k)<\/span><\/span> est vraie. Il existe alors des <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">q,r\\in\\mathbb{Z}<\/span><\/span> tels que<\/p>\n<p style=\"text-align:center;\" dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\nk=dq+r,\\qquad 0\\le r&lt;d.\n\n<\/span>\n<p>En ajoutant <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">1<\/span><\/span> aux deux membres, on obtient<\/p>\n<p style=\"text-align:center;\" dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\nk+1=dq+(r+1).\n\n<\/span>\n<p>De plus, de <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">0\\le r&lt;d<\/span><\/span>, il d\u00e9coule que <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">r+1\\le d<\/span><\/span>. En cons\u00e9quence, seuls les cas <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">r+1&lt;d<\/span><\/span> ou <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">r+1=d<\/span><\/span> peuvent se produire.<\/p>\n<p><b>Cas 1 :<\/b> Si <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">r+1&lt;d<\/span><\/span>, nous d\u00e9finissons <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">q&#039;=q<\/span><\/span> et <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">r&#039;=r+1<\/span><\/span>. Alors<\/p>\n<p style=\"text-align:center;\" dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\nk+1=dq&#039;+r&#039;,\\qquad 0\\le r&#039;&lt;d.\n\n<\/span>\n<p><b>Cas 2 :<\/b> Si <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">r+1=d<\/span><\/span>, alors <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">k+1=dq+d=d(q+1)+0<\/span><\/span>. Nous d\u00e9finissons <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">q&#039;=q+1<\/span><\/span> et <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">r&#039;=0<\/span><\/span>, et l\u2019on a<\/p>\n<p style=\"text-align:center;\" dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\nk+1=dq&#039;+r&#039;,\\qquad 0\\le r&#039;&lt;d.\n\n<\/span>\n<p>Dans les deux cas, nous avons construit des entiers <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">q&#039;,r&#039;<\/span><\/span> tels que <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">k+1=dq&#039;+r&#039;<\/span><\/span> et <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">0\\le r&#039;&lt;d<\/span><\/span>. Par cons\u00e9quent, <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">P(k+1)<\/span><\/span> est vraie. Nous concluons par r\u00e9currence que <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">P(b)<\/span><\/span> est valable pour tout <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">b\\ge 0<\/span><\/span>.<\/p>\n<p>Consid\u00e9rons maintenant le cas <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">b&lt;0<\/span><\/span>. Alors <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">-b&gt;0<\/span><\/span>. En appliquant le r\u00e9sultat pr\u00e9c\u00e9dent \u00e0 <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">-b<\/span><\/span>, il existe des <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">q,r\\in\\mathbb{Z}<\/span><\/span> tels que<\/p>\n<p style=\"text-align:center;\" dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\n-b=dq+r,\\qquad 0\\le r&lt;d.\n\n<\/span>\n<p>En multipliant par <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">-1<\/span><\/span>, on obtient<\/p>\n<p style=\"text-align:center;\" dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\nb=-dq-r.\n\n<\/span>\n<p>Si <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">r=0<\/span><\/span>, il suffit de prendre <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">q_1=-q<\/span><\/span> et <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">r_1=0<\/span><\/span>, ce qui donne <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">b=dq_1+r_1<\/span><\/span> et <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">0\\le r_1&lt;d<\/span><\/span>.<\/p>\n<p>Si <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">r&gt;0<\/span><\/span>, nous d\u00e9finissons <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">q_1=-q-1<\/span><\/span> et <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">r_1=d-r<\/span><\/span>. Comme <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">0&lt;r&lt;d<\/span><\/span>, on a <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">0&lt;d-r&lt;d<\/span><\/span>, c\u2019est-\u00e0-dire <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">0\\le r_1&lt;d<\/span><\/span>. De plus,<\/p>\n<p style=\"text-align:center;\" dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\ndq_1+r_1=d(-q-1)+(d-r)=-dq-d+d-r=-dq-r=b.\n\n<\/span>\n<p>En cons\u00e9quence, pour tout <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">b\\in\\mathbb{Z}<\/span><\/span>, il existe des <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">q_1,r_1\\in\\mathbb{Z}<\/span><\/span> tels que <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">b=dq_1+r_1<\/span><\/span> et <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">0\\le r_1&lt;d<\/span><\/span>.<\/p>\n<p>Enfin, rappelons que <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">d=|a|<\/span><\/span>. Si <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">a&gt;0<\/span><\/span>, alors <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">d=a<\/span><\/span> et l\u2019\u00e9galit\u00e9 <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">b=dq_1+r_1<\/span><\/span> s\u2019\u00e9crit <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">b=aq_1+r_1<\/span><\/span>, avec <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">0\\le r_1&lt;|a|<\/span><\/span>.<\/p>\n<p>Si <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">a&lt;0<\/span><\/span>, alors <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">d=-a<\/span><\/span> et <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">b=dq_1+r_1=(-a)q_1+r_1=a(-q_1)+r_1<\/span><\/span>. En d\u00e9finissant <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">q=-q_1<\/span><\/span> et <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">r=r_1<\/span><\/span>, on obtient<\/p>\n<p style=\"text-align:center;\" dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\nb=qa+r,\\qquad 0\\le r&lt;|a|.\n\n<\/span>\n<p>Ainsi est d\u00e9montr\u00e9e l\u2019existence de <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">q<\/span><\/span> et <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">r<\/span><\/span> pour tous <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">a,b\\in\\mathbb{Z}<\/span><\/span> avec <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">a\\neq 0<\/span><\/span>.<\/p>\n<h3>Unicit\u00e9<\/h3>\n<p>Supposons qu\u2019il existe des <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">q,q&#039;,r,r&#039;\\in\\mathbb{Z}<\/span><\/span> tels que<\/p>\n<p style=\"text-align:center;\" dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\nb=qa+r=q&#039;a+r&#039;,\\qquad 0\\le r,r&#039;&lt;|a|.\n\n<\/span>\n<p>En soustrayant les deux expressions, on obtient<\/p>\n<p style=\"text-align:center;\" dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\na(q-q&#039;)=r&#039;-r.\n\n<\/span>\n<p>Comme <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">0\\le r,r&#039;&lt;|a|<\/span><\/span>, on a <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">|r&#039;-r|&lt;|a|<\/span><\/span>. En revanche, le membre de gauche est un multiple de <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">a<\/span><\/span>, et donc aussi un multiple de <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">|a|<\/span><\/span>. Le seul multiple entier de <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">|a|<\/span><\/span> dont la valeur absolue est strictement inf\u00e9rieure \u00e0 <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">|a|<\/span><\/span> est <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">0<\/span><\/span> ; en effet, si <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">|a| m \\neq 0<\/span><\/span> avec <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">m\\in\\mathbb{Z}<\/span><\/span>, alors <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\big| |a| m \\big| \\ge |a| <\/span><\/span>. Par cons\u00e9quent, <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">r&#039;-r=0<\/span><\/span>, c\u2019est-\u00e0-dire <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">r=r&#039;<\/span><\/span>, et en rempla\u00e7ant dans <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">b=qa+r<\/span><\/span>, on d\u00e9duit <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">q=q&#039;<\/span><\/span>. La d\u00e9composition est donc unique.<\/p>\n<p><a name=\"4\"><\/a><\/p>\n<h2>Interpr\u00e9tation du quotient et du reste<\/h2>\n<ul>\n<li style=\"text-align: justify;\">\n    <strong>Que signifie \u00ab reste \u00bb.<\/strong> Si <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">b=qa+r<\/span><\/span>, alors <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">r=b-qa<\/span><\/span> est ce qui \u00ab reste \u00bb lorsqu\u2019on retranche de <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">b<\/span><\/span> le multiple <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">qa<\/span><\/span>. En d\u2019autres termes, <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">r<\/span><\/span> est le r\u00e9sidu obtenu en ajustant <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">b<\/span><\/span> sur la grille des multiples de <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">a<\/span><\/span>.\n  <\/li>\n<li style=\"text-align: justify;\">\n    <strong>Pourquoi impose-t-on<\/strong> <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">0\\le r&lt;|a|<\/span><\/span>. Sans cette condition, le couple <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">(q,r)<\/span><\/span> ne serait pas unique. En effet, si <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">b=qa+r<\/span><\/span>, alors on a aussi <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">b=a(q+1)+(r-a)<\/span><\/span>. La condition <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">0\\le r&lt;|a|<\/span><\/span> force le reste \u00e0 appartenir \u00e0 une fen\u00eatre fixe <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\{0,1,2,\\dots,|a|-1\\}<\/span><\/span>, ce qui bloque les infinies \u00ab r\u00e9-\u00e9tiquetages \u00bb d\u2019un m\u00eame nombre.\n  <\/li>\n<li style=\"text-align: justify;\">\n    <strong>Que se passe-t-il si<\/strong> <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">a<\/span><\/span> <strong>est n\u00e9gatif.<\/strong> Le th\u00e9or\u00e8me reste valable sans modification : le reste demeure non n\u00e9gatif et born\u00e9 par <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">|a|<\/span><\/span>. Cela est pertinent car certains langages de programmation utilisent la troncature vers z\u00e9ro et peuvent produire des restes n\u00e9gatifs, tandis qu\u2019en math\u00e9matiques on adopte la convention <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">0\\le r&lt;|a|<\/span><\/span> afin que le reste soit un repr\u00e9sentant \u00ab standard \u00bb.<\/p>\n<p>    De plus, lorsque <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">a&gt;0<\/span><\/span>, le quotient co\u00efncide avec la partie enti\u00e8re inf\u00e9rieure :<\/p>\n<p style=\"text-align:center;\" dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\nq=\\left\\lfloor \\frac{b}{a}\\right\\rfloor,\\qquad r=b-a\\left\\lfloor \\frac{b}{a}\\right\\rfloor.\n\n<\/span>\n<\/li>\n<\/ul>\n<p><a name=\"5\"><\/a><\/p>\n<h2>Algorithme long de la division<\/h2>\n<p><center><iframe class=\"lazyload\" width=\"560\" height=\"315\" data-src=\"https:\/\/www.youtube.com\/embed\/ABw4QYdD6Pg?si=eJiA7P6cD7RdlYYN\" title=\"YouTube video player\" frameborder=\"0\" allow=\"accelerometer; autoplay; clipboard-write; encrypted-media; gyroscope; picture-in-picture; web-share\" referrerpolicy=\"strict-origin-when-cross-origin\" allowfullscreen><\/iframe><\/center><\/p>\n<p>\u00c0 partir de la division euclidienne, il est possible de mettre en \u0153uvre l\u2019algorithme long de la division, en exploitant la repr\u00e9sentation positionnelle des nombres. Cet algorithme est une technique de calcul qui permet de trouver rapidement les valeurs <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">q<\/span><\/span> et <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">r<\/span><\/span> lorsqu\u2019on souhaite calculer <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">b\\div a<\/span><\/span> avec <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">a,b\\in\\mathbb{Z}<\/span><\/span> et <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">a\\neq 0<\/span><\/span>.<\/p>\n<p>Pour le d\u00e9crire, examinons d\u2019abord quelques exemples qui illustrent le processus et les situations possibles au cours de l\u2019ex\u00e9cution de l\u2019algorithme.<\/p>\n<ol>\n<li>\n<p>Supposons que nous souhaitions calculer <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">57\\div 4<\/span><\/span>. Ici, 57 est le <strong>dividende<\/strong> et 4 le <strong>diviseur<\/strong>. Pour y parvenir, nous effectuerons la s\u00e9quence de calculs suivante :<\/p>\n<p style=\"text-align:center;\" dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\n\\begin{array}{rll}\n\n(1) &amp; \\color{red}57 \\div 4 =\\color{black}  \\\\ &amp; \\text{\u00c9crire l\u2019op\u00e9ration de division.} \\\\ \\\\\n\n(2) &amp; \\color{red}5&#039;\\color{black}7 \\div 4 =  \\\\ &amp; \\text{S\u00e9parer le premier pr\u00e9fixe (depuis la gauche) du dividende qui}\\\\\n\n&amp;\\text{soit sup\u00e9rieur ou \u00e9gal au diviseur, en l\u2019occurrence 5.} \\\\ \\\\\n\n(3) &amp; 5&#039;7 \\div 4 = \\color{red}1\\color{black} \\\\ &amp; \\text{Chercher le plus grand nombre qui, multipli\u00e9 par $4$,}\\\\\n\n&amp;\\text{soit inf\u00e9rieur ou \u00e9gal \u00e0 $5$ et l\u2019\u00e9crire \u00e0 droite de l\u2019\u00e9galit\u00e9. C\u2019est 1.} \\\\ \\\\\n\n(4) &amp; \\begin{array}{ll}\\phantom{-|}5&#039;7 \\div 4 &amp;= 1 \\\\ \\color{red}-|\\underline{4}\\color{black} &amp; \\\\ \\color{red}\\phantom{-|}1\\color{black} &amp;  \\end{array} \\\\ &amp; \\text{Multiplier le r\u00e9sultat par le diviseur et le soustraire de la valeur s\u00e9lectionn\u00e9e.} \\\\ \\\\\n\n(5) &amp; \\begin{array}{ll}\\phantom{-|}5&#039;\\color{red}7&#039;\\color{black} \\div 4 &amp;= 1 \\\\ -|\\underline{4} &amp; \\\\ \\phantom{-|}1\\color{red}7\\color{black} &amp;  \\end{array} \\\\ &amp; \\text{S\u00e9lectionner et \u00ab abaisser \u00bb le chiffre suivant.} \\\\ \\\\\n\n(6) &amp; \\begin{array}{ll}\\phantom{-|}5&#039;7&#039; \\div 4 &amp;= 1\\color{red}4\\color{black} \\\\ -|\\underline{4} &amp; \\\\ \\phantom{-|}17 &amp; \\\\ \\color{red}-|\\underline{16}\\color{black} &amp; \\\\ \\color{red}\\phantom{-|}1\\color{black}  \\end{array} \\\\ &amp; \\text{R\u00e9p\u00e9ter la s\u00e9quence avec le dernier nombre obtenu.}\n\n\\end{array}\n\n<\/span>\n<p>L\u2019algorithme se termine lorsque tous les chiffres \u00e0 \u00ab abaisser \u00bb ont \u00e9t\u00e9 \u00e9puis\u00e9s, donnant comme r\u00e9sultat le quotient <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">q=14<\/span><\/span> et le reste <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">r=1<\/span><\/span>. En particulier, le reste sera toujours inf\u00e9rieur au diviseur.<\/p>\n<\/li>\n<li>\n<p>Supposons que nous souhaitions calculer <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">132\\div 5<\/span><\/span>. Pour y parvenir, nous effectuerons la s\u00e9quence de calculs suivante :<\/p>\n<p style=\"text-align:center;\" dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\begin{array}{rl}\n\n(1) &amp;1&#039;32 \\div 5 = \\\\\n\n&amp; \\text{S\u00e9parer le premier pr\u00e9fixe du dividende et chercher un entier naturel qui,}\\\\\n\n&amp; \\text{multipli\u00e9 par 5, soit inf\u00e9rieur ou \u00e9gal \u00e0 celui-ci. Si cela n\u2019est pas possible,}\\\\\n\n&amp; \\text{incorporer le chiffre suivant jusqu\u2019\u00e0 ce que cela le soit.} \\\\ \\\\\n\n(2) &amp;\\begin{array}{ll}\\phantom{-|}13&#039;2 \\div 5 &amp;= 2 \\\\ -|\\underline{10} &amp; \\\\ \\phantom{-|1}3 &amp; \\end{array}\\\\\n\n&amp; \\text{Lorsque l\u2019\u00e9tape pr\u00e9c\u00e9dente fonctionne, ex\u00e9cuter l\u2019algorithme normalement.} \\\\ \\\\\n\n(3) &amp;\\begin{array}{ll}\\phantom{-|}13&#039;2&#039; \\div 5 &amp;= 26 \\\\ -|\\underline{10} &amp; \\\\ \\phantom{-|1}32 &amp;  \\\\ \\phantom{\\,} -|\\underline{30} &amp; \\\\ \\phantom{-|30} 2 &amp;  \\end{array}\n\n\\end{array}<\/span>\n<\/li>\n<li>\n<p>Si le dividende comporte davantage de chiffres, la proc\u00e9dure s\u2019ex\u00e9cute de la m\u00eame mani\u00e8re. Par exemple, en calculant <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">3521\\div 12<\/span><\/span>, on obtient :<\/p>\n<p style=\"text-align:center;\" dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\begin{array}{ll}\n\n\\phantom{-|}35&#039;2&#039;1&#039; \\div 12 &amp; = 293 \\\\\n\n-|\\underline{24} &amp; \\\\\n\n\\phantom{-|}112 &amp; \\\\\n\n-|\\underline{108} &amp; \\\\\n\n\\phantom{-|10}41 &amp; \\\\\n\n\\phantom{0}-|\\underline{36} &amp; \\\\\n\n\\phantom{10-|}5 &amp; \\\\\n\n\\end{array}<\/span>\n<\/li>\n<li>\n<p>Si l\u2019un des nombres est n\u00e9gatif, il existe deux fa\u00e7ons acceptables de pr\u00e9senter le r\u00e9sultat. Toutefois, la forme standard en th\u00e9orie des nombres est celle dans laquelle le reste est positif et, par cons\u00e9quent, coh\u00e9rente avec la division d\u2019Euclide. Par exemple, si nous calculons <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">-598\\div 21<\/span><\/span>, on obtient les d\u00e9veloppements possibles suivants, tous deux corrects comme identit\u00e9s :<\/p>\n<ul>\n<li>\n        <b>Avec reste n\u00e9gatif :<\/b> Celui-ci s\u2019obtient en supprimant le signe au d\u00e9but de l\u2019op\u00e9ration et en le r\u00e9tablissant \u00e0 la fin.<\/p>\n<p style=\"text-align:center;\" dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\begin{array}{ll}\n\n\\phantom{-}59&#039;8&#039; \\div 21 &amp;= 28 \\\\\n\n-|\\underline{42} &amp; \\\\\n\n\\phantom{-|&#039;}178 &amp; \\\\\n\n\\phantom{\\,}-|\\underline{168} &amp; \\\\\n\n\\phantom{-|00}10 &amp; \\\\\n\n\\end{array}<\/span>\n<p>\u00c0 partir de cela, on a :<\/p>\n<p style=\"text-align:center;\" dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">598 = 21 \\times 28 + 10<\/span>\n<p>Puis, si l\u2019on multiplie toute l\u2019\u00e9galit\u00e9 par <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">-1<\/span><\/span>, on obtient :<\/p>\n<p style=\"text-align:center;\" dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">-598 = 21 \\times (-28) - 10<\/span>\n<p>De sorte que, dans cette repr\u00e9sentation, le r\u00e9sultat de <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">-598\\div 21<\/span><\/span> est le quotient <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">q=-28<\/span><\/span> et le reste <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">r=-10<\/span><\/span>.<\/p>\n<p>Et si l\u2019on multiplie de nouveau toute l\u2019\u00e9galit\u00e9 par <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">-1<\/span><\/span>, on obtient :<\/p>\n<p style=\"text-align:center;\" dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">598 = (-21)\\times(-28) + 10<\/span>\n<p>De sorte que le r\u00e9sultat de <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">598\\div (-21)<\/span><\/span> est le quotient <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">q=-28<\/span><\/span> et le reste <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">r=10<\/span><\/span>.<\/p>\n<\/li>\n<li>\n        <b>Avec reste positif et coh\u00e9rent avec la division d\u2019Euclide :<\/b> \u00c0 partir du d\u00e9veloppement pr\u00e9c\u00e9dent, on a :<\/p>\n<p style=\"text-align:center;\" dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">-598 = 21 \\times (-28) - 10<\/span>\n<p>Puis, en ajoutant <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">0=21-21<\/span><\/span> au membre droit de cette \u00e9galit\u00e9, on obtient :<\/p>\n<p style=\"text-align:center;\" dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">-598 = 21 \\times (-28) - 10 \\color{red}+ 21 - 21\\color{black} = 21\\times(-29) + 11<\/span>\n<p>Par cons\u00e9quent, en restant coh\u00e9rents avec la division d\u2019Euclide, le r\u00e9sultat du calcul <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">-598\\div 21<\/span><\/span> donne un quotient <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">q=-29<\/span><\/span> et un reste <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">r=11<\/span><\/span>.<\/p>\n<\/li>\n<\/ul>\n<p>En ce qui concerne la convention de la division d\u2019Euclide, seul le r\u00e9sultat avec reste positif correspond au quotient et au reste euclidiens. Toutefois, les deux expressions sont des identit\u00e9s correctes et peuvent \u00eatre utiles dans des contextes diff\u00e9rents. Bien que l\u2019algorithme de la division longue puisse produire des restes n\u00e9gatifs, cela s\u2019av\u00e8re pratique pour obtenir des r\u00e9sultats de mani\u00e8re m\u00e9canique et effectuer des manipulations alg\u00e9briques sans \u00e9tapes suppl\u00e9mentaires. En revanche, les r\u00e9sultats avec reste positif, coh\u00e9rents avec la division d\u2019Euclide, permettent de fixer un repr\u00e9sentant canonique de chaque classe r\u00e9siduelle (par exemple, dans <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\{0,1,\\dots,n-1\\}<\/span><\/span>), ce qui facilite l\u2019\u00e9tiquetage des entiers de mani\u00e8re pr\u00e9cise et sans ambigu\u00eft\u00e9. Il convient de noter qu\u2019il existe \u00e9galement d\u2019autres conventions de repr\u00e9sentants, telles que les r\u00e9sidus sym\u00e9triques, tout aussi valides une fois la r\u00e8gle fix\u00e9e.<\/p>\n<\/li>\n<\/ol>\n<p><a name=\"6\"><\/a><\/p>\n<h2>Conclusion<\/h2>\n<p>L\u2019algorithme de la division garantit que toute division enti\u00e8re peut \u00eatre exprim\u00e9e de mani\u00e8re unique sous la forme <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">b=qa+r<\/span><\/span> avec <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">0\\le r&lt;|a|<\/span><\/span>, ce qui fixe un crit\u00e8re standard pour interpr\u00e9ter le quotient et le reste m\u00eame lorsque des nombres n\u00e9gatifs interviennent. L\u2019algorithme long de la division n\u2019est rien d\u2019autre qu\u2019une impl\u00e9mentation pratique de cette d\u00e9composition, fond\u00e9e sur la repr\u00e9sentation positionnelle, et permet de calculer <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">q<\/span><\/span> et <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">r<\/span><\/span> de mani\u00e8re m\u00e9canique. Enfin, cette forme euclidienne \u00e9vite non seulement les ambigu\u00eft\u00e9s, mais se connecte \u00e9galement de mani\u00e8re naturelle \u00e0 l\u2019arithm\u00e9tique modulaire : le reste <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">r<\/span><\/span> agit comme un repr\u00e9sentant canonique de la classe r\u00e9siduelle de <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">b<\/span><\/span> modulo <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">a<\/span><\/span>, base de nombreuses techniques en th\u00e9orie des nombres et en applications informatiques qui pourront \u00eatre examin\u00e9es dans de prochaines livraisons.<\/p>\n<p><a name=\"7\"><\/a><\/p>\n<h2>Exercices propos\u00e9s et r\u00e9solus<\/h2>\n<h3>Division euclidienne<\/h3>\n<ol>\n<li><strong>(r\u00e9solu)<\/strong> Trouver <span class=\"katex-eq\" data-katex-display=\"false\">q<\/span> et <span class=\"katex-eq\" data-katex-display=\"false\">r<\/span> tels que\n<p style=\"text-align:center;\" dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">137 = 9q + r,\\qquad 0\\le r\\lt 9.<\/span>\n<p><strong>Solution :<\/strong> Cherchez le plus grand nombre qui, multipli\u00e9 par <span class=\"katex-eq\" data-katex-display=\"false\">9<\/span>, ne d\u00e9passe pas <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">137<\/span><\/span>. On a <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">15\\times 9 = 135<\/span><\/span> et <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">16\\times 9 = 144<\/span><\/span> ; par cons\u00e9quent, le quotient recherch\u00e9 est <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">q=15<\/span><\/span>. Calculez maintenant le reste <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">r=137-9q<\/span><\/span> et v\u00e9rifiez que <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">0\\le r\\lt 9<\/span><\/span>. En effectuant la v\u00e9rification, on obtient :<\/p>\n<p style=\"text-align:center;\" dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\nr=137-9\\cdot 15=137-135=2.\n\n<\/span>\n<p>Par cons\u00e9quent, <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">137=9\\cdot 15+2<\/span><\/span> avec <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">0\\le 2\\lt 9<\/span><\/span>.<\/p>\n<\/li>\n<li><strong>(r\u00e9solu)<\/strong> Trouver <span class=\"katex-eq\" data-katex-display=\"false\">q<\/span> et <span class=\"katex-eq\" data-katex-display=\"false\">r<\/span> tels que\n<p style=\"text-align:center;\" dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">2025 = 37q + r,\\qquad 0\\le r\\lt 37.<\/span>\n<p><strong>Solution :<\/strong> Cherchez le plus grand nombre qui, multipli\u00e9 par <span class=\"katex-eq\" data-katex-display=\"false\">37<\/span>, ne d\u00e9passe pas <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">2025<\/span><\/span>. On observe que <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">37\\times 54 = 1998<\/span><\/span> et <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">37\\times 55 = 2035<\/span><\/span> ; par cons\u00e9quent, le quotient recherch\u00e9 est <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">q=54<\/span><\/span>. Calculez maintenant le reste <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">r=2025-37q<\/span><\/span> et v\u00e9rifiez que <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">0\\le r\\lt 37<\/span><\/span>. En effectuant la v\u00e9rification, on obtient :<\/p>\n<p style=\"text-align:center;\" dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\nr=2025-37\\cdot 54=2025-1998=27.\n\n<\/span>\n<p>Par cons\u00e9quent, <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">2025=37\\cdot 54+27<\/span><\/span> avec <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">0\\le 27\\lt 37<\/span><\/span>.<\/p>\n<\/li>\n<li>Trouver <span class=\"katex-eq\" data-katex-display=\"false\">q<\/span> et <span class=\"katex-eq\" data-katex-display=\"false\">r<\/span> tels que\n<p style=\"text-align:center;\" dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">745 = 23q + r,\\qquad 0\\le r\\lt 23.<\/span>\n<\/li>\n<li>Trouver <span class=\"katex-eq\" data-katex-display=\"false\">q<\/span> et <span class=\"katex-eq\" data-katex-display=\"false\">r<\/span> tels que\n<p style=\"text-align:center;\" dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">-314 = 11q + r,\\qquad 0\\le r\\lt 11.<\/span>\n<\/li>\n<li>Trouver <span class=\"katex-eq\" data-katex-display=\"false\">q<\/span> et <span class=\"katex-eq\" data-katex-display=\"false\">r<\/span> tels que\n<p style=\"text-align:center;\" dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">598 = (-21)q + r,\\qquad 0\\le r\\lt |-21|.<\/span>\n<\/li>\n<li><strong>Reste fix\u00e9 et recherche d\u2019entiers.<\/strong> Soit <span class=\"katex-eq\" data-katex-display=\"false\">a=12<\/span>.\n<p>(a) D\u00e9crire l\u2019ensemble de tous les entiers <span class=\"katex-eq\" data-katex-display=\"false\">b<\/span> dont la division euclidienne par <span class=\"katex-eq\" data-katex-display=\"false\">12<\/span> a pour reste <span class=\"katex-eq\" data-katex-display=\"false\">r=5<\/span>.<\/p>\n<p>(b) Trouver le plus petit entier <span class=\"katex-eq\" data-katex-display=\"false\">b\\gt 1000<\/span> satisfaisant cette condition et d\u00e9terminer le quotient <span class=\"katex-eq\" data-katex-display=\"false\">q<\/span> correspondant.<\/p>\n<\/li>\n<\/ol>\n<h3>Algorithme long de la division<\/h3>\n<p>Dans chaque cas, appliquer l\u2019algorithme long de la division pour calculer le quotient <span class=\"katex-eq\" data-katex-display=\"false\">q<\/span> et le reste <span class=\"katex-eq\" data-katex-display=\"false\">r<\/span>. Transformer le r\u00e9sultat en division euclidienne lorsque l\u2019algorithme fournit un reste n\u00e9gatif.\n<\/p>\n<ol>\n<li><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">84\\div 6.<\/span><\/span><\/li>\n<li><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">197\\div 8.<\/span><\/span><\/li>\n<li><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">1256\\div 7.<\/span><\/span><\/li>\n<li><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">-3521\\div 12.<\/span><\/span><\/li>\n<li><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">-98765\\div 24.<\/span><\/span><\/li>\n<li><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">845\\div -13.<\/span><\/span><\/li>\n<li><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">-12345\\div -37.<\/span><\/span><\/li>\n<\/ol>\n","protected":false},"excerpt":{"rendered":"<p>L\u2019Algorithme de la Division Dans ce cours, nous d\u00e9velopperons l\u2019algorithme de la division comme le principe qui formalise, pour les entiers, la d\u00e9composition unique avec . On d\u00e9montre d\u2019abord l\u2019existence du quotient et du reste, puis leur unicit\u00e9. Enfin, on interpr\u00e8te la signification du reste, on relie la th\u00e9orie \u00e0 l\u2019algorithme long de la division [&hellip;]<\/p>\n","protected":false},"author":1,"featured_media":35980,"comment_status":"open","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"iawp_total_views":4,"footnotes":""},"categories":[569,1414],"tags":[],"class_list":["post-36006","post","type-post","status-publish","format-standard","has-post-thumbnail","hentry","category-mathematiques","category-theorie-des-nombres"],"yoast_head":"<!-- This site is optimized with the Yoast SEO plugin v27.4 - https:\/\/yoast.com\/product\/yoast-seo-wordpress\/ -->\n<title>L\u2019Algorithme de la Division Euclidienne - toposuranos.com\/material<\/title>\n<meta name=\"description\" content=\"Apprenez l\u2019algorithme de la division et la division euclidienne : existence et unicit\u00e9 de q et r, cas avec nombres n\u00e9gatifs, algorithme long, exemples et exercices.\" \/>\n<meta name=\"robots\" content=\"index, follow, max-snippet:-1, max-image-preview:large, max-video-preview:-1\" \/>\n<link rel=\"canonical\" href=\"https:\/\/toposuranos.com\/material\/fr\/lalgorithme-de-la-division-euclidienne\/\" \/>\n<meta property=\"og:locale\" content=\"es_ES\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"L\u2019Algorithme de la Division Euclidienne\" \/>\n<meta property=\"og:description\" content=\"Apprenez l\u2019algorithme de la division et la division euclidienne : existence et unicit\u00e9 de q et r, cas avec nombres n\u00e9gatifs, algorithme long, exemples et exercices.\" \/>\n<meta property=\"og:url\" content=\"https:\/\/toposuranos.com\/material\/fr\/lalgorithme-de-la-division-euclidienne\/\" \/>\n<meta property=\"og:site_name\" content=\"toposuranos.com\/material\" \/>\n<meta property=\"article:publisher\" content=\"https:\/\/www.facebook.com\/groups\/toposuranos\" \/>\n<meta property=\"article:published_time\" content=\"2026-01-10T18:40:42+00:00\" \/>\n<meta property=\"og:image\" content=\"http:\/\/toposuranos.com\/material\/wp-content\/uploads\/2026\/01\/divisioneuclidiana-1024x683.jpg\" \/>\n<meta name=\"author\" content=\"giorgio.reveco\" \/>\n<meta name=\"twitter:card\" content=\"summary_large_image\" \/>\n<meta name=\"twitter:title\" content=\"L\u2019Algorithme de la Division Euclidienne\" \/>\n<meta name=\"twitter:description\" content=\"Apprenez l\u2019algorithme de la division et la division euclidienne : existence et unicit\u00e9 de q et r, cas avec nombres n\u00e9gatifs, algorithme long, exemples et exercices.\" \/>\n<meta name=\"twitter:image\" content=\"https:\/\/toposuranos.com\/material\/wp-content\/uploads\/2026\/01\/divisioneuclidiana.jpg\" \/>\n<meta name=\"twitter:creator\" content=\"@topuranos\" \/>\n<meta name=\"twitter:site\" content=\"@topuranos\" \/>\n<meta name=\"twitter:label1\" content=\"Escrito por\" \/>\n\t<meta name=\"twitter:data1\" content=\"giorgio.reveco\" \/>\n\t<meta name=\"twitter:label2\" content=\"Tiempo de lectura\" \/>\n\t<meta name=\"twitter:data2\" content=\"12 minutos\" \/>\n<script type=\"application\/ld+json\" class=\"yoast-schema-graph\">{\"@context\":\"https:\\\/\\\/schema.org\",\"@graph\":[{\"@type\":\"Article\",\"@id\":\"https:\\\/\\\/toposuranos.com\\\/material\\\/fr\\\/lalgorithme-de-la-division-euclidienne\\\/#article\",\"isPartOf\":{\"@id\":\"https:\\\/\\\/toposuranos.com\\\/material\\\/fr\\\/lalgorithme-de-la-division-euclidienne\\\/\"},\"author\":{\"name\":\"giorgio.reveco\",\"@id\":\"https:\\\/\\\/toposuranos.com\\\/material\\\/#\\\/schema\\\/person\\\/e15164361c3f9a2a02cf6c234cf7fdc1\"},\"headline\":\"L\u2019Algorithme de la Division Euclidienne\",\"datePublished\":\"2026-01-10T18:40:42+00:00\",\"mainEntityOfPage\":{\"@id\":\"https:\\\/\\\/toposuranos.com\\\/material\\\/fr\\\/lalgorithme-de-la-division-euclidienne\\\/\"},\"wordCount\":3430,\"commentCount\":0,\"publisher\":{\"@id\":\"https:\\\/\\\/toposuranos.com\\\/material\\\/#organization\"},\"image\":{\"@id\":\"https:\\\/\\\/toposuranos.com\\\/material\\\/fr\\\/lalgorithme-de-la-division-euclidienne\\\/#primaryimage\"},\"thumbnailUrl\":\"https:\\\/\\\/toposuranos.com\\\/material\\\/wp-content\\\/uploads\\\/2026\\\/01\\\/divisioneuclidiana.jpg\",\"articleSection\":[\"Math\u00e9matiques\",\"Th\u00e9orie des nombres\"],\"inLanguage\":\"es\",\"potentialAction\":[{\"@type\":\"CommentAction\",\"name\":\"Comment\",\"target\":[\"https:\\\/\\\/toposuranos.com\\\/material\\\/fr\\\/lalgorithme-de-la-division-euclidienne\\\/#respond\"]}]},{\"@type\":\"WebPage\",\"@id\":\"https:\\\/\\\/toposuranos.com\\\/material\\\/fr\\\/lalgorithme-de-la-division-euclidienne\\\/\",\"url\":\"https:\\\/\\\/toposuranos.com\\\/material\\\/fr\\\/lalgorithme-de-la-division-euclidienne\\\/\",\"name\":\"L\u2019Algorithme de la Division Euclidienne - 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