第 27415 題：整數p進賦值

原題與已知條件

prime p、absolute-value factor extraction與v_p(0)=infinity責任。操作=pval-boundary；資料={"mode":"negative_normalization","p":7,"a":-735}。請依 exact_ordered_tuple_declared_by_interface 欄序回答。

• G001 — prime p、absolute-value factor extraction與v_p(0)=infinity責任。操作=pval-boundary；資料={"mode":"negative_normalization","p":7,"a":-735}。請依 exact_ordered_tuple_declared_by_interface 欄序回答。

定義與適用範圍

沿用原題明示的論域、假設與來源 MKO 定義。本題已完成第二對話數理與中英語義覆核，並由原始參數逐題獨立重算模型、答案與每個算術步驟；原題仍為 candidate。

• mko-p-adic-valuation-integers

逐步推導

source roles = [pval-boundary]

derived = [[mode, p, integer_a, absolute_a, valuation, negative_integer_valid, scope]]

Valuation boundary ORIGINAL negative ORIGINAL division step1 := div(735, 7)

Valuation boundary ORIGINAL negative ORIGINAL division step1 = 105

Valuation boundary ORIGINAL negative dividing prime power1 := mul(1, 7)

Valuation boundary ORIGINAL negative dividing prime power1 = 7

source roles = [Valuation boundary ORIGINAL negative, 7, 735, 735, 1]

derived = [105, 7, 0, 是]

Valuation boundary ORIGINAL negative ORIGINAL division step2 := div(105, 7)

Valuation boundary ORIGINAL negative ORIGINAL division step2 = 15

Valuation boundary ORIGINAL negative dividing prime power2 := mul(7, 7)

Valuation boundary ORIGINAL negative dividing prime power2 = 49

source roles = [Valuation boundary ORIGINAL negative, 7, 735, 105, 2]

derived = [15, 49, 0, 是]

Valuation boundary ORIGINAL negative terminal factorization product := mul(49, 15)

Valuation boundary ORIGINAL negative terminal factorization product = 735

source roles = [Valuation boundary ORIGINAL negative, 7, -735, 735, 2, 15, [7, 49]]

derived = [735, 1, 是, 是]

逐步反向核對

• S009 · premise_and_result_audit — 回查原始完整資料、步驟操作數與本步前提；有限核對、條件式定理引用與一般形式證明分開。
• S008 · premise_and_result_audit — 以精確有理數重算 Valuation boundary ORIGINAL negative terminal factorization product，不以小數近似替代。
• S007 · premise_and_result_audit — 回查原始完整資料、步驟操作數與本步前提；有限核對、條件式定理引用與一般形式證明分開。
• S006 · premise_and_result_audit — 以精確有理數重算 Valuation boundary ORIGINAL negative dividing prime power2，不以小數近似替代。
• S005 · premise_and_result_audit — 以精確有理數重算 Valuation boundary ORIGINAL negative ORIGINAL division step2，不以小數近似替代。
• S004 · premise_and_result_audit — 回查原始完整資料、步驟操作數與本步前提；有限核對、條件式定理引用與一般形式證明分開。
• S003 · premise_and_result_audit — 以精確有理數重算 Valuation boundary ORIGINAL negative dividing prime power1，不以小數近似替代。
• S002 · premise_and_result_audit — 以精確有理數重算 Valuation boundary ORIGINAL negative ORIGINAL division step1，不以小數近似替代。
• S001 · premise_and_result_audit — 回查原始完整資料、步驟操作數與本步前提；有限核對、條件式定理引用與一般形式證明分開。

結論

(negative_normalization, 7, -735, 735, 2, 是, negative_integer_is_valid)

依上列完整代值及條件核對得到原題答案；原題編號、格式與狀態保持不變。

本題覆核與來源綁定

覆核日期：2026-10-26

原題、答案、步驟表達式、數值 AST 與既有歷史保持原樣；原來的候選狀態屬於來源歷史，不覆寫為形式化證明。