{ "schema_version": "ocme-derivation-reader-v1", "guide_id": "derivation-mko-modular-inverse-01", "mko_id": "mko-modular-inverse", "title": { "zh-Hant": "17模43的正規化逆元", "en": "The normalized inverse of 17 modulo 43" }, "contract": { "schema_version": "ocme-derivation-guide-v0.1", "guide_id": "derivation-mko-modular-inverse-01", "source": { "mko_id": "mko-modular-inverse", "commit": "b1b2f1dcf8c50d67a93fe52d95f8eef7b3abba9f", "sha256": "d16e0773143022c00ff50dad2cd84ef53e2d8a9078f64c2d785413c676b002ab", "english_sha256": "6d3e8307627ef6fae92d80afb7cdf3347e4ca67f4f12e28c68a96a7465d52732" }, "mode": "worked_example", "audience_prerequisites": [ "calc" ], "goal": { "zh-Hant": "17模43的唯一逆類為38；17·38=15·43+1。", "en": "The unique inverse residue class of 17 modulo 43 is represented by 38; 17*38=15*43+1." }, "givens": [ { "id": "G1", "statement": { "zh-Hant": "a=17，n=43，要求0≤x<43及17x≡1 mod43。", "en": "a=17, n=43; require 0≤x<43 and 17x congruent to 1 modulo 43." } } ], "steps": [ { "id": "S01", "uses": [ "G1" ], "before_tex": "a=17,\\quad n=43", "after_tex": "n\\ge2,\\quad43=2\\cdot17+9,\\quad0\\le9<17", "relation": "implies", "rule": "calc", "conditions": [], "explanation": { "zh-Hant": "模數域合法；第一個帶餘除法給餘數9。", "en": "The modulus domain is valid; the first exact division gives remainder 9." }, "reverse_audit": { "kind": "premise_audit", "required_ids": [ "G1" ], "explanation": { "zh-Hant": "除數17非零且餘數界明列，沒有近似除法。", "en": "The divisor 17 is nonzero and the remainder bound is explicit; no approximate division is used." } }, "leaf_basis": "calc" }, { "id": "S02", "uses": [ "G1", "S01" ], "before_tex": "17,9", "after_tex": "17=1\\cdot9+8,\\quad9=1\\cdot8+1,\\quad8=8\\cdot1+0", "relation": "implies", "rule": "calc", "conditions": [], "explanation": { "zh-Hant": "完整除法鏈的餘數依序8、1、0，均小於非零除數。", "en": "The complete division chain has remainders 8,1,0, each below its nonzero divisor." }, "reverse_audit": { "kind": "premise_audit", "required_ids": [ "G1", "S01" ], "explanation": { "zh-Hant": "到餘數0停止，不能再除以0。", "en": "Stop at remainder zero without a subsequent division by zero." } }, "leaf_basis": "calc" }, { "id": "S03", "uses": [ "G1", "S01", "S02" ], "before_tex": "u=qv+r", "after_tex": "d\\mid u,v\\iff d\\mid v,r;\\quad\\gcd(17,43)=1", "relation": "implies", "rule": "calc", "conditions": [], "explanation": { "zh-Hant": "共因數集合不變：r=u−qv與u=qv+r給雙方向；最後非零餘數1即gcd。", "en": "Common divisors are preserved in both directions using r=u-qv and u=qv+r; the last nonzero remainder 1 is the gcd." }, "reverse_audit": { "kind": "premise_audit", "required_ids": [ "G1", "S01", "S02" ], "explanation": { "zh-Hant": "gcd不只是猜1，已由整條鏈與共因數不變證明。", "en": "The gcd is not guessed; the full chain and divisor invariant prove it." } }, "leaf_basis": "calc" }, { "id": "S04", "uses": [ "G1", "S01", "S02", "S03" ], "before_tex": "1=9-8,\\quad8=17-9,\\quad9=43-2\\cdot17", "after_tex": "1=2\\cdot43-5\\cdot17", "relation": "implies", "rule": "calc", "conditions": [], "explanation": { "zh-Hant": "逐式回代給原輸入的Bezout係數x=−5、y=2。", "en": "Back-substitution gives Bezout coefficients x=-5,y=2 for the original inputs." }, "reverse_audit": { "kind": "premise_audit", "required_ids": [ "G1", "S01", "S02", "S03" ], "explanation": { "zh-Hant": "直接核86−85=1；符號錯誤會使逆元失效。", "en": "Directly check 86-85=1; a sign error would invalidate the inverse." } }, "leaf_basis": "calc" }, { "id": "S05", "uses": [ "G1", "S04" ], "before_tex": "17(-5)+43(2)=1", "after_tex": "-5\\equiv38\\pmod{43},\\quad0\\le38<43", "relation": "implies", "rule": "calc", "conditions": [], "explanation": { "zh-Hant": "把負Bezout係數加43正規化，得到候選38。", "en": "Add 43 to the negative Bezout coefficient to normalize the candidate to 38." }, "reverse_audit": { "kind": "premise_audit", "required_ids": [ "G1", "S04" ], "explanation": { "zh-Hant": "−5與38是同一逆類，不是兩個不同逆元類。", "en": "-5 and 38 represent the same inverse class, not distinct inverse classes." } }, "leaf_basis": "calc" }, { "id": "S06", "uses": [ "G1", "S05" ], "before_tex": "x=38", "after_tex": "17\\cdot38=646=15\\cdot43+1\\Rightarrow17\\cdot38\\equiv1\\pmod{43}", "relation": "implies", "rule": "mko_definition", "conditions": [], "explanation": { "zh-Hant": "整數乘積再核，證候選真是逆元。", "en": "An independent exact product check confirms the candidate is an inverse." }, "reverse_audit": { "kind": "premise_audit", "required_ids": [ "G1", "S05" ], "explanation": { "zh-Hant": "等式先在Z成立，再推出模43同餘，不混淆兩層。", "en": "The equality is in Z before passing to congruence modulo 43." } }, "leaf_basis": "mko_definition" }, { "id": "S07", "uses": [ "G1", "S04", "S06" ], "before_tex": "17x\\equiv17x'\\equiv1\\pmod{43}", "after_tex": "43\\mid17(x-x');\\quad x-x'=(2\\cdot43-5\\cdot17)(x-x')\\Rightarrow43\\mid x-x'", "relation": "implies", "rule": "calc", "conditions": [], "explanation": { "zh-Hant": "以已證Bezout式消去17，證逆元只在模43意義下唯一。", "en": "Use the established Bezout identity to cancel 17 and prove uniqueness modulo 43." }, "reverse_audit": { "kind": "premise_audit", "required_ids": [ "G1", "S04", "S06" ], "explanation": { "zh-Hant": "若x與x'均在0到42，差為43倍且絕對值小43，才得整數相等。", "en": "Only if both are normalized to 0 through 42 does the multiple-of-43 difference force integer equality." } }, "leaf_basis": "calc" }, { "id": "S08", "uses": [ "G1", "S03" ], "before_tex": "6x+15y=1\\text{ proposed}", "after_tex": "3\\mid6x+15y\\text{ but }3\\nmid1\\Rightarrow6\\text{ has no inverse mod }15", "relation": "implies", "rule": "calc", "conditions": [], "explanation": { "zh-Hant": "nonunit示例gcd(6,15)=3，不可能給Bezout1，因此不能用有理1/6代替模逆元。", "en": "For nonunit 6 modulo 15, gcd 3 makes Bezout 1 impossible; rational 1/6 is not a modular inverse." }, "reverse_audit": { "kind": "premise_audit", "required_ids": [ "G1", "S03" ], "explanation": { "zh-Hant": "此無解證書對所有整數x、y成立，不僅有限搜尋。", "en": "The obstruction applies to all integer x,y, not just a bounded search." } }, "leaf_basis": "calc" } ], "result": { "zh-Hant": "17模43的唯一逆類為38；17·38=15·43+1。", "en": "The unique inverse residue class of 17 modulo 43 is represented by 38; 17*38=15*43+1." }, "negative_controls": [ { "claim": { "zh-Hant": "逆元為5。", "en": "The inverse is 5." }, "expected": "reject", "reason": { "zh-Hant": "17·5=85≡42而非1 mod43。", "en": "17*5=85 is 42, not 1, modulo 43." } }, { "claim": { "zh-Hant": "6模15的逆元可寫1/6。", "en": "The inverse of 6 modulo 15 is 1/6." }, "expected": "reject", "reason": { "zh-Hant": "3整除任何6x+15y，不可能等1。", "en": "Every 6x+15y is divisible by 3 and cannot equal 1." } }, { "claim": { "zh-Hant": "逆元作整數唯一。", "en": "An inverse is unique as an integer." }, "expected": "reject", "reason": { "zh-Hant": "38與−5均滿足同餘，只在正規化範圍選38。", "en": "38 and -5 both satisfy the congruence; normalization selects 38." } } ], "status": "candidate", "gaps": [], "verification": { "structure": "checked_local_script", "mathematics": "candidate_argument", "formalization": "not_formalized" }, "rule_glossary": [ { "id": "calc", "kind": "prerequisite", "statement": { "zh-Hant": "假設精確整數算術、帶餘除法、整除與gcd的定義；同餘等式始終綁定明示模數。", "en": "Assume exact integer arithmetic, division with remainder, and definitions of divisibility and gcd; every congruence has an explicit modulus." } }, { "id": "mko_definition", "kind": "cited", "statement": { "zh-Hant": "使用extended Euclid給exact Bezout certificate ax+ny=1，輸出0≤x mod n