{"schema_version":"ocme-site-results-v1","locale":"zh-Hant","page_type":"derivations","facts":{"questions":36000,"reviewed":36000,"display_candidates":0,"knowledge":302,"guides":299,"paths":59,"batches":360,"families":1441,"atlas":302,"core":80,"evidence":9,"formal_evidence":5,"formal_subjects":4,"lean_files":6,"lean_toolchain":"leanprover/lean4:v4.30.0","question_state":"candidate","english_state":"ai_candidate","question_Lean":"NOT_RUN","canonical_truth_promotion":false},"filters":{"q":"mko-gcd-coprime"},"kind":"guides","total":1,"pages":1,"page":1,"entries":[{"id":"derivation-mko-gcd-coprime-01","title":{"zh-Hant":"最大公因數與互質：以整除刻畫驗證歐幾里得算法","en":"Greatest common divisor and coprimality: checking Euclid's algorithm against the divisibility characterization"},"summary":{"zh-Hant":"依 MKO 的整除刻畫（d∣a、d∣b，且每個公因數 c 都整除 d）：(1) 證明滿足刻畫的 d 至多一個；(2) 證明歐幾里得算法每一步保持公因數集合且必會停止，並以它算出 gcd(84,36)=12，再直接驗證 12 滿足刻畫；(3) 說明公因數 6 不是 gcd；(4) 算出 gcd(8,15)=1，說明 8 與 15 互質但都不是質數；(5) 處理含 0 的情形 gcd(0,1)=1、gcd(0,6)=6、gcd(0,0)=0。","en":"From the MKO's divisibility characterization (d∣a, d∣b, and every common divisor c divides d): (1) prove that at most one d satisfies it; (2) prove that each step of Euclid's algorithm preserves the set of common divisors and that the algorithm stops, use it to compute gcd(84,36)=12, and verify directly that 12 satisfies the characterization; (3) explain why the common divisor 6 is not the gcd; (4) compute gcd(8,15)=1 and explain that 8 and 15 are coprime although neither is prime; (5) treat the zero cases gcd(0,1)=1, gcd(0,6)=6 and gcd(0,0)=0."},"mko_ids":["mko-gcd-coprime"]}],"title":"概念推導導讀"}