{"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-prime-number"},"kind":"guides","total":1,"pages":1,"page":1,"entries":[{"id":"derivation-mko-prime-number-01","title":{"zh-Hant":"自然數質數：證明 97 是質數、91 是合數，並排除 0 與 1","en":"Prime natural numbers: proving 97 prime and 91 composite, and excluding 0 and 1"},"summary":{"zh-Hant":"依 MKO 定義（p 為質數 ⇔ 2≤p 且 p 的每個自然數因數 d 都滿足 d=1 或 d=p）：(1) 說明 0 與 1 不是質數；(2) 證明 2 是質數；(3) 以見證 7 證明 91 是合數，並說明只測 2、3、5 會誤判；(4) 證明「若 n=dk 且 d、k≥2，則 min(d,k)²≤n」，據此只需檢查 2 到 9，證明 97 是質數；(5) 證明 2 是唯一的偶質數。","en":"From the MKO's definition (p is prime ⇔ 2≤p and every natural divisor d of p satisfies d=1 or d=p): (1) explain why 0 and 1 are not prime; (2) prove that 2 is prime; (3) prove that 91 is composite with the witness 7, and show that testing only 2, 3, 5 misjudges it; (4) prove 'if n=dk with d, k≥2 then min(d,k)²≤n', and use it to prove that 97 is prime by checking only 2 to 9; (5) prove that 2 is the only even prime."},"mko_ids":["mko-prime-number"]}],"title":"概念推導導讀"}