{"schema_version":"ocme-site-results-v1","locale":"en","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-factorial"},"kind":"guides","total":1,"pages":1,"page":1,"entries":[{"id":"derivation-mko-factorial-01","title":{"zh-Hant":"階乘：遞迴定義的唯一性、乘積形式與 5! 的計算","en":"Factorial: uniqueness of the recursive definition, the product form and computing 5!"},"summary":{"zh-Hant":"只用 F(0)=1 與 F(n+1)=(n+1)F(n)（n∈ℕ，ℕ 含 0）：(1) 證明滿足這兩條的函數 ℕ→ℕ 只有一個；(2) 證明對所有 n 有 F(n)=∏_{k=1}^{n} k（空乘積為 1），所以 n>0 時 n! 是從 1 到 n 所有整數的乘積；(3) 證明每個 F(n)≥1，特別是 0!=1≠0；(4) 由遞迴逐步算出 5!=120。","en":"Using only F(0)=1 and F(n+1)=(n+1)F(n) (n∈ℕ, with 0∈ℕ): (1) prove that at most one function ℕ→ℕ satisfies these two equations; (2) prove F(n)=∏_{k=1}^{n} k for every n (the empty product being 1), so that for n>0, n! is the product of all integers from 1 to n; (3) prove F(n)≥1 for every n, in particular 0!=1≠0; (4) compute 5!=120 step by step from the recursion."},"mko_ids":["mko-factorial"]}],"title":"Concept derivation guides"}