{"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-modular-congruence"},"kind":"guides","total":1,"pages":1,"page":1,"entries":[{"id":"derivation-mko-modular-congruence-01","title":{"zh-Hant":"自然數模同餘：餘數定義、見證形式、運算保持與取消的限制","en":"Congruence modulo a positive modulus: the remainder definition, the witness form, preservation by operations and the limits of cancellation"},"summary":{"zh-Hant":"依 MKO 定義（n>0，a≡b (mod n) ⇔ a mod n=b mod n）：(1) 判定 100≡23 (mod 7)；(2) 證明它等價於見證形式「存在自然數 k、l 使 a+nk=b+nl」，並給出 100 與 23 的見證；(3) 說明同餘是等價關係；(4) 證明加法與乘法保持同餘，並由 15≡1 (mod 7) 得 115≡24、1500≡23 (mod 7)；(5) 以 2·3≡2·0 (mod 6) 說明不能無條件取消公因子；(6) 說明不同模數不能串接，並說明模數 0 被排除。","en":"From the MKO's definition (n>0, a≡b (mod n) ⇔ a mod n=b mod n): (1) decide 100≡23 (mod 7); (2) prove that this is equivalent to the witness form 'there are naturals k, l with a+nk=b+nl', and give witnesses for 100 and 23; (3) explain why congruence is an equivalence relation; (4) prove that addition and multiplication preserve congruence, and from 15≡1 (mod 7) obtain 115≡24 and 1500≡23 (mod 7); (5) use 2·3≡2·0 (mod 6) to show that a common factor cannot be cancelled unconditionally; (6) explain why different moduli cannot be chained, and why the modulus 0 is excluded."},"mko_ids":["mko-modular-congruence"]}],"title":"概念推導導讀"}