{"title":"欧氏算法:","authors":"S. Mneimneh","doi":"10.2307/j.ctvc774cp.28","DOIUrl":null,"url":null,"abstract":"1 The greatest common divisor Consider two positive integers a0 > a1. The greatest common divisor of a0 and a1, denoted gcd(a0, a1) is the largest positive integer g such that g|a0 and g|a1, i.e. g divides both a0 and a1. Observation 1: The gcd(a0, a1) always exists. Observation 2 (Euclid): Let a0 = q1a1 + r where 0 ≤ r < a1 (note that this representation is always possible and unique), then gcd(a0, a1) = gcd(a1, r). The proof of this fact consists of showing that d|a0 and d|a1 ⇔ d|a1 and d|r.","PeriodicalId":365299,"journal":{"name":"How to Fall Slower Than Gravity","volume":null,"pages":null},"PeriodicalIF":0.0000,"publicationDate":"2018-11-27","publicationTypes":"Journal Article","fieldsOfStudy":null,"isOpenAccess":false,"openAccessPdf":"","citationCount":"0","resultStr":"{\"title\":\"The Euclidean Algorithm:\",\"authors\":\"S. Mneimneh\",\"doi\":\"10.2307/j.ctvc774cp.28\",\"DOIUrl\":null,\"url\":null,\"abstract\":\"1 The greatest common divisor Consider two positive integers a0 > a1. The greatest common divisor of a0 and a1, denoted gcd(a0, a1) is the largest positive integer g such that g|a0 and g|a1, i.e. g divides both a0 and a1. Observation 1: The gcd(a0, a1) always exists. Observation 2 (Euclid): Let a0 = q1a1 + r where 0 ≤ r < a1 (note that this representation is always possible and unique), then gcd(a0, a1) = gcd(a1, r). The proof of this fact consists of showing that d|a0 and d|a1 ⇔ d|a1 and d|r.\",\"PeriodicalId\":365299,\"journal\":{\"name\":\"How to Fall Slower Than Gravity\",\"volume\":null,\"pages\":null},\"PeriodicalIF\":0.0000,\"publicationDate\":\"2018-11-27\",\"publicationTypes\":\"Journal Article\",\"fieldsOfStudy\":null,\"isOpenAccess\":false,\"openAccessPdf\":\"\",\"citationCount\":\"0\",\"resultStr\":null,\"platform\":\"Semanticscholar\",\"paperid\":null,\"PeriodicalName\":\"How to Fall Slower Than Gravity\",\"FirstCategoryId\":\"1085\",\"ListUrlMain\":\"https://doi.org/10.2307/j.ctvc774cp.28\",\"RegionNum\":0,\"RegionCategory\":null,\"ArticlePicture\":[],\"TitleCN\":null,\"AbstractTextCN\":null,\"PMCID\":null,\"EPubDate\":\"\",\"PubModel\":\"\",\"JCR\":\"\",\"JCRName\":\"\",\"Score\":null,\"Total\":0}","platform":"Semanticscholar","paperid":null,"PeriodicalName":"How to Fall Slower Than Gravity","FirstCategoryId":"1085","ListUrlMain":"https://doi.org/10.2307/j.ctvc774cp.28","RegionNum":0,"RegionCategory":null,"ArticlePicture":[],"TitleCN":null,"AbstractTextCN":null,"PMCID":null,"EPubDate":"","PubModel":"","JCR":"","JCRName":"","Score":null,"Total":0}
1 The greatest common divisor Consider two positive integers a0 > a1. The greatest common divisor of a0 and a1, denoted gcd(a0, a1) is the largest positive integer g such that g|a0 and g|a1, i.e. g divides both a0 and a1. Observation 1: The gcd(a0, a1) always exists. Observation 2 (Euclid): Let a0 = q1a1 + r where 0 ≤ r < a1 (note that this representation is always possible and unique), then gcd(a0, a1) = gcd(a1, r). The proof of this fact consists of showing that d|a0 and d|a1 ⇔ d|a1 and d|r.