{"id":35986,"date":"2026-01-10T18:40:43","date_gmt":"2026-01-10T18:40:43","guid":{"rendered":"https:\/\/toposuranos.com\/material\/?p=35986"},"modified":"2026-01-10T18:40:43","modified_gmt":"2026-01-10T18:40:43","slug":"o-algoritmo-da-divisao-euclidiana","status":"publish","type":"post","link":"https:\/\/toposuranos.com\/material\/pt\/o-algoritmo-da-divisao-euclidiana\/","title":{"rendered":"O Algoritmo da Divis\u00e3o Euclidiana"},"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>O Algoritmo da Divis\u00e3o<\/h1>\n<p style=\"text-align:center;\"><em>Nesta aula desenvolveremos o <strong>algoritmo da divis\u00e3o<\/strong> como o princ\u00edpio que formaliza, para inteiros, a decomposi\u00e7\u00e3o \u00fanica <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">b=qa+r<\/span><\/span> com <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">0\\le r&lt;|a|<\/span><\/span>. Demonstra-se primeiro a <strong>exist\u00eancia<\/strong> do quociente e do resto e, em seguida, a sua <strong>unicidade<\/strong>. Por fim, interpreta-se o significado do resto, relaciona-se a teoria com o <strong>algoritmo longo da divis\u00e3o<\/strong> como procedimento de c\u00e1lculo e antecipa-se a sua conex\u00e3o natural com a aritm\u00e9tica modular e aplica\u00e7\u00f5es computacionais.<\/em><\/p>\n<p style=\"text-align:center;\"><strong>Objetivos de Aprendizagem<\/strong><\/p>\n<p>Ao completar esta aula, o estudante ser\u00e1 capaz de:<\/p>\n<ol>\n<li><strong>Identificar<\/strong> os conceitos e pap\u00e9is b\u00e1sicos da divis\u00e3o inteira (dividendo, divisor, quociente, resto) e a no\u00e7\u00e3o de divisibilidade como caso exato.<\/li>\n<li><strong>Explicar<\/strong> o enunciado do algoritmo\/teorema da divis\u00e3o euclidiana, incluindo as condi\u00e7\u00f5es que fixam o resto (<span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">0\\le r&lt;|a|<\/span><\/span>) e o seu prop\u00f3sito para evitar ambiguidades.<\/li>\n<li><strong>Aplicar<\/strong> a decomposi\u00e7\u00e3o euclidiana <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">b=qa+r<\/span><\/span> para determinar <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">q<\/span><\/span> e <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">r<\/span><\/span> em exemplos concretos, verificando a cota do resto.<\/li>\n<li><strong>Analisar<\/strong> o tratamento de casos de acordo com o sinal do dividendo e do divisor, justificando por que a conven\u00e7\u00e3o euclidiana mant\u00e9m um resto n\u00e3o negativo.<\/li>\n<li><strong>Executar<\/strong> o algoritmo longo da divis\u00e3o como procedimento mec\u00e2nico baseado na representa\u00e7\u00e3o posicional para obter <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">q<\/span><\/span> e <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>\u00cdNDICE DE CONTE\u00daDOS<\/u>:<\/b><br \/>\n<a href=\"#1\">Divis\u00e3o e divisibilidade<\/a><br \/>\n<a href=\"#2\">Teorema da divis\u00e3o euclidiana<\/a><br \/>\n<a href=\"#3\">Demonstra\u00e7\u00e3o da divis\u00e3o euclidiana<\/a><br \/>\n<a href=\"#4\">Interpreta\u00e7\u00e3o do quociente e do resto<\/a><br \/>\n<a href=\"#5\">Algoritmo longo da divis\u00e3o<\/a><br \/>\n<a href=\"#6\">Conclus\u00e3o<\/a><br \/>\n<a href=\"#7\">Exerc\u00edcios propostos e resolvidos<\/a>\n<\/p>\n<p><a name=\"1\"><\/a><\/p>\n<h2>Divis\u00e3o e divisibilidade<\/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>Em aritm\u00e9tica, a divisibilidade descreve o caso \u00abexato\u00bb: <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">a\\mid b<\/span><\/span> significa que <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">b<\/span><\/span> se escreve exatamente como m\u00faltiplo de <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">a<\/span><\/span>. No entanto, embora nem sempre haja divisibilidade entre dois inteiros quaisquer, ainda podemos perguntar-nos \u00abquantas vezes um n\u00famero cabe dentro de outro e, se n\u00e3o o faz exatamente, o que \u00e9 que sobra\u00bb. Este \u00e9 o contexto em que surge o <strong>algoritmo da divis\u00e3o<\/strong>, que garante que um inteiro qualquer <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">b<\/span><\/span> pode ser escrito como um m\u00faltiplo de <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">a<\/span><\/span> mais um \u00abresto\u00bb.<\/p>\n<p>Como exemplo, consideremos <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">a=3<\/span><\/span> e <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">b=16<\/span><\/span>. \u00c9 claro que 3 n\u00e3o divide 16. No entanto, \u00e9 poss\u00edvel encontrar um quociente e um resto para a opera\u00e7\u00e3o de divis\u00e3o, pois<\/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=\"Um grupo de 16 caixas foi separado em grupos de 3; como resultado, obt\u00eam-se 5 grupos de caixas e 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=\"Um grupo de 16 caixas foi separado em grupos de 3; como resultado, obt\u00eam-se 5 grupos de caixas e 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> Um grupo de 16 caixas foi separado em grupos de 3; como resultado, obt\u00eam-se 5 grupos de caixas e sobra 1<\/figcaption><\/figure>\n<p>Em consequ\u00eancia, ao dividir <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">b=16<\/span><\/span> por <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">a=3<\/span><\/span> (onde <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">b<\/span><\/span> \u00e9 o <strong>dividendo<\/strong> e <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">a<\/span><\/span> \u00e9 o <strong>divisor<\/strong>), representado por <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>, obt\u00e9m-se como <strong>quociente<\/strong> <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">q=5<\/span><\/span> e como <strong>resto<\/strong> <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">r=1<\/span><\/span>. Em geral, o <em>algoritmo da divis\u00e3o<\/em> afirma que, para quaisquer inteiros <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">a<\/span><\/span> (divisor) e <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">b<\/span><\/span> (dividendo) com <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">a\\neq 0<\/span><\/span>, existem inteiros \u00fanicos <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">q<\/span><\/span> e <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">r<\/span><\/span> tais 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>Neste exemplo, <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">16=3\\cdot 5+1<\/span><\/span>, e a condi\u00e7\u00e3o <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">0\\le r&lt;|a|<\/span><\/span> \u00e9 satisfeita porque <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">0\\le 1&lt;3<\/span><\/span>. Essa formula\u00e7\u00e3o evita ambiguidades quando o divisor (ou o dividendo) pode ser negativo e garante que o resto seja sempre n\u00e3o negativo e estritamente menor que o valor absoluto do divisor.<\/p>\n<p><a name=\"2\"><\/a><\/p>\n<h2>Teorema da divis\u00e3o euclidiana<\/h2>\n<p>O resultado de aplicar o algoritmo da divis\u00e3o \u00e9 o que se conhece como <strong>divis\u00e3o euclidiana<\/strong> e fundamenta-se no seguinte resultado.<\/p>\n<p><b>Teorema:<\/b> Sejam <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">a,b\\in\\mathbb{Z}<\/span><\/span> com <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">a\\neq 0<\/span><\/span>. Ent\u00e3o existem <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">q,r\\in\\mathbb{Z}<\/span><\/span> \u00fanicos tais 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>O inteiro <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">q<\/span><\/span> chama-se <strong>quociente<\/strong> e <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">r<\/span><\/span> chama-se <strong>resto<\/strong> da divis\u00e3o de <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">b<\/span><\/span> por <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">a<\/span><\/span>.<\/p>\n<p><a name=\"3\"><\/a><\/p>\n<h2>Demonstra\u00e7\u00e3o da divis\u00e3o euclidiana<\/h2>\n<p>Esta demonstra\u00e7\u00e3o divide-se em duas partes: primeiro demonstra-se que o quociente e o resto existem; e, em seguida, dado que existem, s\u00e3o \u00fanicos.<\/p>\n<h3>Exist\u00eancia<\/h3>\n<p>Sejam <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">a,b\\in\\mathbb{Z}<\/span><\/span> com <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">a\\neq 0<\/span><\/span> e definamos <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">d=|a|<\/span><\/span>, de modo que <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">d&gt;0<\/span><\/span>. Provaremos primeiro o caso <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">b\\ge 0<\/span><\/span>: mostraremos que existem <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">q,r\\in\\mathbb{Z}<\/span><\/span> tais que <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">b=dq+r<\/span><\/span> e <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">0\\le r&lt;d<\/span><\/span>. Em seguida, trataremos o caso <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">b&lt;0<\/span><\/span>. Ao final, substituiremos <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">d<\/span><\/span> por <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">|a|<\/span><\/span> para obter a forma <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">b=qa+r<\/span><\/span> com <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">0\\le r&lt;|a|<\/span><\/span>.<\/p>\n<p>Para <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">b\\in\\mathbb{Z}<\/span><\/span> com <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">b\\ge 0<\/span><\/span>, estabelecemos a proposi\u00e7\u00e3o <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">P(b)<\/span><\/span>: \u00abexistem <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">q,r\\in\\mathbb{Z}<\/span><\/span> tais que <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">b=dq+r<\/span><\/span> e <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">0\\le r&lt;d<\/span><\/span>\u00bb. Demonstraremos por indu\u00e7\u00e3o que <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">P(b)<\/span><\/span> \u00e9 v\u00e1lida para todo <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">b\\ge 0<\/span><\/span>.<\/p>\n<p><strong>Caso base:<\/strong> Para <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">b=0<\/span><\/span>, tomando <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">q=0<\/span><\/span> e <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">r=0<\/span><\/span> obt\u00e9m-se <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">0=d\\cdot 0+0<\/span><\/span> e <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">0\\le 0&lt;d<\/span><\/span>. Portanto, <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">P(0)<\/span><\/span> \u00e9 verdadeira.<\/p>\n<p><strong>Passo indutivo:<\/strong> Seja <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">k\\ge 0<\/span><\/span> e suponhamos que <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">P(k)<\/span><\/span> \u00e9 verdadeira. Ent\u00e3o existem <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">q,r\\in\\mathbb{Z}<\/span><\/span> tais 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>Somando <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">1<\/span><\/span> a ambos os lados, obt\u00e9m-se<\/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>Al\u00e9m disso, de <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">0\\le r&lt;d<\/span><\/span> segue-se que <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">r+1\\le d<\/span><\/span>. Em consequ\u00eancia, apenas podem ocorrer os casos <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>.<\/p>\n<p><b>Caso 1:<\/b> Se <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">r+1&lt;d<\/span><\/span>, definimos <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">q&#039;=q<\/span><\/span> e <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">r&#039;=r+1<\/span><\/span>. Ent\u00e3o<\/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>Caso 2:<\/b> Se <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">r+1=d<\/span><\/span>, ent\u00e3o <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">k+1=dq+d=d(q+1)+0<\/span><\/span>. Definimos <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">q&#039;=q+1<\/span><\/span> e <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">r&#039;=0<\/span><\/span>, e verifica-se que<\/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>Em ambos os casos, constru\u00edmos inteiros <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">q&#039;,r&#039;<\/span><\/span> tais que <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">k+1=dq&#039;+r&#039;<\/span><\/span> e <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">0\\le r&#039;&lt;d<\/span><\/span>. Portanto, <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">P(k+1)<\/span><\/span> \u00e9 verdadeira. Conclu\u00edmos por indu\u00e7\u00e3o que <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">P(b)<\/span><\/span> vale para todo <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">b\\ge 0<\/span><\/span>.<\/p>\n<p>Agora consideremos o caso <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">b&lt;0<\/span><\/span>. Ent\u00e3o <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">-b&gt;0<\/span><\/span>. Aplicando o resultado anterior a <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">-b<\/span><\/span>, existem <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">q,r\\in\\mathbb{Z}<\/span><\/span> tais 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>Multiplicando por <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">-1<\/span><\/span>, obt\u00e9m-se<\/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>Se <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">r=0<\/span><\/span>, basta tomar <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">q_1=-q<\/span><\/span> e <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">r_1=0<\/span><\/span>, com o que <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">b=dq_1+r_1<\/span><\/span> e <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">0\\le r_1&lt;d<\/span><\/span>.<\/p>\n<p>Se <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">r&gt;0<\/span><\/span>, definimos <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">q_1=-q-1<\/span><\/span> e <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">r_1=d-r<\/span><\/span>. Como <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">0&lt;r&lt;d<\/span><\/span>, tem-se <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">0&lt;d-r&lt;d<\/span><\/span>, isto \u00e9, <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">0\\le r_1&lt;d<\/span><\/span>. Al\u00e9m disso,<\/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>Em consequ\u00eancia, para todo <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">b\\in\\mathbb{Z}<\/span><\/span> existem <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">q_1,r_1\\in\\mathbb{Z}<\/span><\/span> tais que <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">b=dq_1+r_1<\/span><\/span> e <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">0\\le r_1&lt;d<\/span><\/span>.<\/p>\n<p>Por fim, recordemos que <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">d=|a|<\/span><\/span>. Se <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">a&gt;0<\/span><\/span>, ent\u00e3o <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">d=a<\/span><\/span> e a igualdade <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">b=dq_1+r_1<\/span><\/span> escreve-se como <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">b=aq_1+r_1<\/span><\/span>, com <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">0\\le r_1&lt;|a|<\/span><\/span>.<\/p>\n<p>Se <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">a&lt;0<\/span><\/span>, ent\u00e3o <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">d=-a<\/span><\/span> e <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>. Definindo <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">q=-q_1<\/span><\/span> e <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">r=r_1<\/span><\/span>, obt\u00e9m-se<\/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>Com isso, fica provada a exist\u00eancia de <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">q<\/span><\/span> e <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">r<\/span><\/span> para quaisquer <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">a,b\\in\\mathbb{Z}<\/span><\/span> com <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">a\\neq 0<\/span><\/span>.<\/p>\n<h3>Unicidade<\/h3>\n<p>Suponha-se que existam <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">q,q&#039;,r,r&#039;\\in\\mathbb{Z}<\/span><\/span> tais 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>Subtraindo ambas as express\u00f5es, obt\u00e9m-se<\/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>Como <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">0\\le r,r&#039;&lt;|a|<\/span><\/span>, verifica-se <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">|r&#039;-r|&lt;|a|<\/span><\/span>. Por outro lado, o membro esquerdo \u00e9 m\u00faltiplo de <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">a<\/span><\/span> e, portanto, tamb\u00e9m \u00e9 m\u00faltiplo de <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">|a|<\/span><\/span>. O \u00fanico m\u00faltiplo inteiro de <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">|a|<\/span><\/span> com valor absoluto estritamente menor que <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">|a|<\/span><\/span> \u00e9 <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">0<\/span><\/span>; de fato, se <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">|a| m \\neq 0<\/span><\/span> com <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">m\\in\\mathbb{Z}<\/span><\/span>, ent\u00e3o <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\big| |a| m \\big| \\ge |a| <\/span><\/span>. Consequentemente, <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">r&#039;-r=0<\/span><\/span>, isto \u00e9, <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">r=r&#039;<\/span><\/span>, e ao substituir em <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">b=qa+r<\/span><\/span> deduz-se <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">q=q&#039;<\/span><\/span>. Assim, a decomposi\u00e7\u00e3o \u00e9 \u00fanica.<\/p>\n<p><a name=\"4\"><\/a><\/p>\n<h2>Interpreta\u00e7\u00e3o do quociente e do resto<\/h2>\n<ul>\n<li style=\"text-align: justify;\">\n    <strong>O que significa \u00abresto\u00bb.<\/strong> Se <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">b=qa+r<\/span><\/span>, ent\u00e3o <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">r=b-qa<\/span><\/span> \u00e9 o que \u00absobra\u00bb ao retirar de <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">b<\/span><\/span> o m\u00faltiplo <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">qa<\/span><\/span>. Em outras palavras, <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">r<\/span><\/span> \u00e9 o res\u00edduo de ajustar <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">b<\/span><\/span> \u00e0 grelha de m\u00faltiplos 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>Por que se exige<\/strong> <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">0\\le r&lt;|a|<\/span><\/span>. Sem essa condi\u00e7\u00e3o, o par <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">(q,r)<\/span><\/span> n\u00e3o seria \u00fanico. De fato, se <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">b=qa+r<\/span><\/span>, ent\u00e3o tamb\u00e9m <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">b=a(q+1)+(r-a)<\/span><\/span>. A condi\u00e7\u00e3o <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">0\\le r&lt;|a|<\/span><\/span> for\u00e7a o resto a pertencer a uma janela fixa <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\{0,1,2,\\dots,|a|-1\\}<\/span><\/span>, e isso bloqueia as infinitas \u00abre-rotula\u00e7\u00f5es\u00bb do mesmo n\u00famero.\n  <\/li>\n<li style=\"text-align: justify;\">\n    <strong>O que acontece se<\/strong> <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">a<\/span><\/span> <strong>\u00e9 negativo.<\/strong> O teorema continua v\u00e1lido sem altera\u00e7\u00f5es: o resto mant\u00e9m-se n\u00e3o negativo e limitado por <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">|a|<\/span><\/span>. Isso \u00e9 relevante porque algumas linguagens de programa\u00e7\u00e3o utilizam truncamento em dire\u00e7\u00e3o a zero e podem produzir restos negativos, enquanto em matem\u00e1tica adota-se a conven\u00e7\u00e3o <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">0\\le r&lt;|a|<\/span><\/span> para que o resto seja um representante \u00abpadr\u00e3o\u00bb.<\/p>\n<p>    Al\u00e9m disso, quando <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">a&gt;0<\/span><\/span>, o quociente coincide com o piso:<\/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>Algoritmo longo da divis\u00e3o<\/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>A partir da divis\u00e3o euclidiana \u00e9 poss\u00edvel implementar o algoritmo longo da divis\u00e3o, aproveitando a representa\u00e7\u00e3o posicional dos n\u00fameros. Este algoritmo \u00e9 uma t\u00e9cnica de c\u00e1lculo que permite encontrar rapidamente os valores <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">q<\/span><\/span> e <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">r<\/span><\/span> quando se deseja calcular <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">b\\div a<\/span><\/span> com <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">a,b\\in\\mathbb{Z}<\/span><\/span> e <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">a\\neq 0<\/span><\/span>.<\/p>\n<p>Para descrev\u00ea-lo, revisemos primeiro alguns exemplos que ilustram o processo e as poss\u00edveis situa\u00e7\u00f5es durante a execu\u00e7\u00e3o do algoritmo.<\/p>\n<ol>\n<li>\n<p>Suponhamos que queremos calcular <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">57\\div 4<\/span><\/span>. Aqui, 57 \u00e9 o <strong>dividendo<\/strong> e 4 \u00e9 o <strong>divisor<\/strong>. Para isso, realizaremos a seguinte sequ\u00eancia de c\u00e1lculos:<\/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{Escrever a opera\u00e7\u00e3o de divis\u00e3o.} \\\\ \\\\\n\n(2) &amp; \\color{red}5&#039;\\color{black}7 \\div 4 =  \\\\ &amp; \\text{Separar o primeiro prefixo (a partir da esquerda) do dividendo que}\\\\\n\n&amp;\\text{seja maior ou igual ao divisor, neste caso 5.} \\\\ \\\\\n\n(3) &amp; 5&#039;7 \\div 4 = \\color{red}1\\color{black} \\\\ &amp; \\text{Pensar no maior n\u00famero que, multiplicado por $4$,}\\\\\n\n&amp;\\text{seja menor ou igual a $5$ e escrev\u00ea-lo \u00e0 direita da igualdade. \u00c9 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{Multiplicar o resultado pelo divisor e subtra\u00ed-lo do valor selecionado.} \\\\ \\\\\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{Selecionar e \u00abbaixar\u00bb o pr\u00f3ximo d\u00edgito.} \\\\ \\\\\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{Repetir a sequ\u00eancia com o \u00faltimo n\u00famero obtido.}\n\n\\end{array}\n\n<\/span>\n<p>O algoritmo termina quando se esgotam os d\u00edgitos a \u00abbaixar\u00bb, fornecendo como resultado o quociente <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">q=14<\/span><\/span> e o resto <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">r=1<\/span><\/span>. Em particular, o resto ser\u00e1 sempre menor que o divisor.<\/p>\n<\/li>\n<li>\n<p>Suponhamos que queremos calcular <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">132\\div 5<\/span><\/span>. Para isso, realizaremos a seguinte sequ\u00eancia de c\u00e1lculos:<\/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{Separar o primeiro prefixo do dividendo e procurar algum natural que,}\\\\\n\n&amp; \\text{multiplicado por 5, seja menor ou igual a ele. Se n\u00e3o for poss\u00edvel,}\\\\\n\n&amp; \\text{incorporar o pr\u00f3ximo d\u00edgito at\u00e9 que seja poss\u00edvel.} \\\\ \\\\\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{Quando o passo anterior funcionar, executar o algoritmo normalmente.} \\\\ \\\\\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>Se o dividendo tiver mais d\u00edgitos, o procedimento \u00e9 executado do mesmo modo. Por exemplo, ao calcular <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">3521\\div 12<\/span><\/span> obt\u00e9m-se:<\/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>Se um dos n\u00fameros for negativo, existem duas formas aceit\u00e1veis de apresentar o resultado. No entanto, a forma padr\u00e3o em teoria dos n\u00fameros \u00e9 aquela em que o resto \u00e9 positivo e, portanto, consistente com a divis\u00e3o de Euclides. Por exemplo, se calcularmos <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">-598\\div 21<\/span><\/span>, t\u00eam-se os seguintes desenvolvimentos poss\u00edveis, ambos corretos como identidades:<\/p>\n<ul>\n<li>\n        <b>Com resto negativo:<\/b> Este \u00e9 obtido eliminando o sinal no in\u00edcio da opera\u00e7\u00e3o e restaurando-o ao final.<\/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>A partir disso, tem-se que:<\/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>Em seguida, se multiplicarmos toda a igualdade por <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">-1<\/span><\/span>, obt\u00e9m-se:<\/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 modo que, nessa representa\u00e7\u00e3o, o resultado de <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">-598\\div 21<\/span><\/span> \u00e9 quociente <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">q=-28<\/span><\/span> e resto <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">r=-10<\/span><\/span>.<\/p>\n<p>E se multiplicarmos novamente toda a igualdade por <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">-1<\/span><\/span>, obt\u00e9m-se:<\/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 modo que o resultado de <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">598\\div (-21)<\/span><\/span> \u00e9 quociente <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">q=-28<\/span><\/span> e resto <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">r=10<\/span><\/span>.<\/p>\n<\/li>\n<li>\n        <b>Com resto positivo e consistente com a divis\u00e3o de Euclides:<\/b> A partir do desenvolvimento anterior, tem-se que:<\/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>Em seguida, somando <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">0=21-21<\/span><\/span> ao lado direito dessa igualdade, obt\u00e9m-se:<\/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>Portanto, sendo consistentes com a divis\u00e3o de Euclides, o resultado de calcular <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">-598\\div 21<\/span><\/span> fornece um quociente <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">q=-29<\/span><\/span> e o resto <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">r=11<\/span><\/span>.<\/p>\n<\/li>\n<\/ul>\n<p>No que diz respeito ao que \u00e9 convencional na divis\u00e3o de Euclides, apenas o resultado com resto positivo corresponde ao quociente e ao resto euclidianos. No entanto, ambas as express\u00f5es s\u00e3o identidades corretas e podem ser \u00fateis em contextos distintos. Embora o algoritmo da divis\u00e3o longa possa produzir restos negativos, isso \u00e9 pr\u00e1tico para obter resultados de maneira mec\u00e2nica e realizar manipula\u00e7\u00f5es alg\u00e9bricas sem etapas adicionais. Por outro lado, os resultados com resto positivo, consistentes com a divis\u00e3o de Euclides, permitem fixar um representante can\u00f4nico de cada classe residual (por exemplo, em <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\{0,1,\\dots,n-1\\}<\/span><\/span>), o que facilita a rotulagem dos inteiros de forma precisa e livre de ambiguidade. Cabe salientar que tamb\u00e9m existem outras conven\u00e7\u00f5es de representantes, como os res\u00edduos sim\u00e9tricos, igualmente v\u00e1lidas uma vez fixada a regra.<\/p>\n<\/li>\n<\/ol>\n<p><a name=\"6\"><\/a><\/p>\n<h2>Conclus\u00e3o<\/h2>\n<p>O algoritmo da divis\u00e3o garante que toda divis\u00e3o inteira pode ser expressa de forma \u00fanica como <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">b=qa+r<\/span><\/span> com <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">0\\le r&lt;|a|<\/span><\/span>, o que estabelece um crit\u00e9rio padr\u00e3o para interpretar o quociente e o resto mesmo quando interv\u00eam n\u00fameros negativos. O algoritmo longo da divis\u00e3o n\u00e3o \u00e9 mais do que uma implementa\u00e7\u00e3o pr\u00e1tica dessa decomposi\u00e7\u00e3o, apoiada na representa\u00e7\u00e3o posicional, e permite calcular <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">q<\/span><\/span> e <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">r<\/span><\/span> de maneira mec\u00e2nica. Por fim, essa forma euclidiana n\u00e3o apenas evita ambiguidades, como tamb\u00e9m se conecta naturalmente com a aritm\u00e9tica modular: o resto <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">r<\/span><\/span> atua como um representante can\u00f4nico da classe residual de <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">b<\/span><\/span> m\u00f3dulo <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">a<\/span><\/span>, base de muitas t\u00e9cnicas em teoria dos n\u00fameros e em aplica\u00e7\u00f5es computacionais que poder\u00e3o ser revisadas em entregas futuras.<\/p>\n<p><a name=\"7\"><\/a><\/p>\n<h2>Exerc\u00edcios propostos e resolvidos<\/h2>\n<h3>Divis\u00e3o euclidiana<\/h3>\n<ol>\n<li><strong>(resolvido)<\/strong> Encontre <span class=\"katex-eq\" data-katex-display=\"false\">q<\/span> e <span class=\"katex-eq\" data-katex-display=\"false\">r<\/span> tais 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>Solu\u00e7\u00e3o:<\/strong> Procure o maior n\u00famero que, multiplicado por <span class=\"katex-eq\" data-katex-display=\"false\">9<\/span>, n\u00e3o ultrapasse <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">137<\/span><\/span>. Temos que <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">15\\times 9 = 135<\/span><\/span> e <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">16\\times 9 = 144<\/span><\/span>; portanto, o quociente procurado \u00e9 <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">q=15<\/span><\/span>. Agora calcule o resto <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">r=137-9q<\/span><\/span> e verifique que <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">0\\le r\\lt 9<\/span><\/span>. Fazendo a verifica\u00e7\u00e3o, obt\u00e9m-se:<\/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>Portanto, <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">137=9\\cdot 15+2<\/span><\/span> com <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">0\\le 2\\lt 9<\/span><\/span>.<\/p>\n<\/li>\n<li><strong>(resolvido)<\/strong> Encontre <span class=\"katex-eq\" data-katex-display=\"false\">q<\/span> e <span class=\"katex-eq\" data-katex-display=\"false\">r<\/span> tais 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>Solu\u00e7\u00e3o:<\/strong> Procure o maior n\u00famero que, multiplicado por <span class=\"katex-eq\" data-katex-display=\"false\">37<\/span>, n\u00e3o ultrapasse <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">2025<\/span><\/span>. Observe que <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">37\\times 54 = 1998<\/span><\/span> e <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">37\\times 55 = 2035<\/span><\/span>; portanto, o quociente procurado \u00e9 <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">q=54<\/span><\/span>. Agora calcule o resto <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">r=2025-37q<\/span><\/span> e verifique que <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">0\\le r\\lt 37<\/span><\/span>. Fazendo a verifica\u00e7\u00e3o, obt\u00e9m-se:<\/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>Portanto, <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">2025=37\\cdot 54+27<\/span><\/span> com <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">0\\le 27\\lt 37<\/span><\/span>.<\/p>\n<\/li>\n<li>Encontre <span class=\"katex-eq\" data-katex-display=\"false\">q<\/span> e <span class=\"katex-eq\" data-katex-display=\"false\">r<\/span> tais 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>Encontre <span class=\"katex-eq\" data-katex-display=\"false\">q<\/span> e <span class=\"katex-eq\" data-katex-display=\"false\">r<\/span> tais 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>Encontre <span class=\"katex-eq\" data-katex-display=\"false\">q<\/span> e <span class=\"katex-eq\" data-katex-display=\"false\">r<\/span> tais 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>Resto fixado e busca de inteiros.<\/strong> Seja <span class=\"katex-eq\" data-katex-display=\"false\">a=12<\/span>.\n<p>(a) Descreva o conjunto de todos os inteiros <span class=\"katex-eq\" data-katex-display=\"false\">b<\/span> cuja divis\u00e3o euclidiana por <span class=\"katex-eq\" data-katex-display=\"false\">12<\/span> tem resto <span class=\"katex-eq\" data-katex-display=\"false\">r=5<\/span>.<\/p>\n<p>(b) Encontre o menor inteiro <span class=\"katex-eq\" data-katex-display=\"false\">b\\gt 1000<\/span> que satisfa\u00e7a o item anterior e determine o quociente <span class=\"katex-eq\" data-katex-display=\"false\">q<\/span> correspondente.<\/p>\n<\/li>\n<\/ol>\n<h3>Algoritmo longo da divis\u00e3o<\/h3>\n<p>Em cada caso, aplique o algoritmo longo da divis\u00e3o para calcular o quociente <span class=\"katex-eq\" data-katex-display=\"false\">q<\/span> e o resto <span class=\"katex-eq\" data-katex-display=\"false\">r<\/span>. Transforme o resultado em divis\u00e3o euclidiana quando o algoritmo fornecer resto negativo.\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>O Algoritmo da Divis\u00e3o Nesta aula desenvolveremos o algoritmo da divis\u00e3o como o princ\u00edpio que formaliza, para inteiros, a decomposi\u00e7\u00e3o \u00fanica com . Demonstra-se primeiro a exist\u00eancia do quociente e do resto e, em seguida, a sua unicidade. Por fim, interpreta-se o significado do resto, relaciona-se a teoria com o algoritmo longo da divis\u00e3o como [&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":9,"footnotes":""},"categories":[571,1404],"tags":[],"class_list":["post-35986","post","type-post","status-publish","format-standard","has-post-thumbnail","hentry","category-matematica-pt","category-teoria-dos-numeros"],"yoast_head":"<!-- This site is optimized with the Yoast SEO plugin v26.7 - https:\/\/yoast.com\/wordpress\/plugins\/seo\/ -->\n<title>O Algoritmo da Divis\u00e3o Euclidiana - toposuranos.com\/material<\/title>\n<meta name=\"description\" content=\"Aprenda o algoritmo da divis\u00e3o e a divis\u00e3o euclidiana: exist\u00eancia e unicidade de q e r, casos com n\u00fameros negativos, algoritmo longo, exemplos e exerc\u00edcios.\" \/>\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\/pt\/o-algoritmo-da-divisao-euclidiana\/\" \/>\n<meta property=\"og:locale\" content=\"es_ES\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"O Algoritmo da Divis\u00e3o Euclidiana\" \/>\n<meta property=\"og:description\" content=\"Aprenda o algoritmo da divis\u00e3o e a divis\u00e3o euclidiana: exist\u00eancia e unicidade de q e r, casos com n\u00fameros negativos, algoritmo longo, exemplos e exerc\u00edcios.\" \/>\n<meta property=\"og:url\" content=\"https:\/\/toposuranos.com\/material\/pt\/o-algoritmo-da-divisao-euclidiana\/\" \/>\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=\"2026-01-10T18:40:43+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=\"O Algoritmo da Divis\u00e3o Euclidiana\" \/>\n<meta name=\"twitter:description\" content=\"Aprenda o algoritmo da divis\u00e3o e a divis\u00e3o euclidiana: exist\u00eancia e unicidade de q e r, casos com n\u00fameros negativos, algoritmo longo, exemplos e exerc\u00edcios.\" \/>\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=\"1 minuto\" \/>\n<script type=\"application\/ld+json\" class=\"yoast-schema-graph\">{\"@context\":\"https:\/\/schema.org\",\"@graph\":[{\"@type\":\"Article\",\"@id\":\"https:\/\/toposuranos.com\/material\/pt\/o-algoritmo-da-divisao-euclidiana\/#article\",\"isPartOf\":{\"@id\":\"https:\/\/toposuranos.com\/material\/pt\/o-algoritmo-da-divisao-euclidiana\/\"},\"author\":{\"name\":\"giorgio.reveco\",\"@id\":\"https:\/\/toposuranos.com\/material\/#\/schema\/person\/e15164361c3f9a2a02cf6c234cf7fdc1\"},\"headline\":\"O Algoritmo da Divis\u00e3o Euclidiana\",\"datePublished\":\"2026-01-10T18:40:43+00:00\",\"mainEntityOfPage\":{\"@id\":\"https:\/\/toposuranos.com\/material\/pt\/o-algoritmo-da-divisao-euclidiana\/\"},\"wordCount\":3183,\"commentCount\":0,\"publisher\":{\"@id\":\"https:\/\/toposuranos.com\/material\/#organization\"},\"image\":{\"@id\":\"https:\/\/toposuranos.com\/material\/pt\/o-algoritmo-da-divisao-euclidiana\/#primaryimage\"},\"thumbnailUrl\":\"https:\/\/toposuranos.com\/material\/wp-content\/uploads\/2026\/01\/divisioneuclidiana.jpg\",\"articleSection\":[\"Matem\u00e1tica\",\"Teoria dos N\u00fameros\"],\"inLanguage\":\"es\",\"potentialAction\":[{\"@type\":\"CommentAction\",\"name\":\"Comment\",\"target\":[\"https:\/\/toposuranos.com\/material\/pt\/o-algoritmo-da-divisao-euclidiana\/#respond\"]}]},{\"@type\":\"WebPage\",\"@id\":\"https:\/\/toposuranos.com\/material\/pt\/o-algoritmo-da-divisao-euclidiana\/\",\"url\":\"https:\/\/toposuranos.com\/material\/pt\/o-algoritmo-da-divisao-euclidiana\/\",\"name\":\"O Algoritmo da Divis\u00e3o Euclidiana - 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