{
  "schema_version": "ocme-derivation-reader-v1",
  "guide_id": "derivation-mko-rational-number-01",
  "mko_id": "mko-rational-number",
  "title": {
    "zh-Hant": "有理數：同一個有理數的不同分數表示、交叉相乘與約分",
    "en": "Rational number: different fraction representations of one rational, cross-multiplication and reduction"
  },
  "contract": {
    "schema_version": "ocme-derivation-guide-v0.1",
    "guide_id": "derivation-mko-rational-number-01",
    "source": {
      "mko_id": "mko-rational-number",
      "commit": "b28a5edbd46279b1db2a3f3f48bb3f447e18899f",
      "sha256": "f1b52aad1867316fd94d6e07d6e94237a202d6769454898a0831d8d1bb89f23f",
      "english_sha256": "7f9330a95c6c6cb3df1c1696d1ffa62bb4442ff453291c4176febb6d0d71ff5a"
    },
    "mode": "worked_example",
    "audience_prerequisites": [
      "mul_inverse_nonzero",
      "frac_rules",
      "zero_product",
      "order_basics",
      "numeric_evaluation"
    ],
    "goal": {
      "zh-Hant": "依 MKO 的定義（存在整數 p、q，q≠0，使 x=p/q），證明 6/(−8)、−3/4、9/(−12) 表示同一個有理數，並把 6/(−8) 精確約成正規形式 −3/4；以交叉相乘證明 2/3≠3/5；說明「5/0」不表示任何數；說明有限小數 −1.25=−5/4 與整數 7=7/1 都是有理數；並在分母為正時用交叉相乘證明 −3/4<−2/3。",
      "en": "Using the MKO's definition (there are integers p, q with q≠0 and x=p/q), prove that 6/(−8), −3/4 and 9/(−12) represent the same rational number, and reduce 6/(−8) exactly to the canonical form −3/4; prove 2/3≠3/5 by cross-multiplication; show that '5/0' denotes no number; show that the finite decimal −1.25=−5/4 and the integer 7=7/1 are rational; and, with positive denominators, prove −3/4<−2/3 by cross-multiplication."
    },
    "givens": [
      {
        "id": "G1",
        "statement": {
          "zh-Hant": "在實數 ℝ 中討論，整數 ℤ⊆ℝ；對 q≠0，p/q 表示 p·(1/q)；0 沒有倒數。",
          "en": "Work inside the reals ℝ, with the integers ℤ⊆ℝ; for q≠0, p/q means p·(1/q); 0 has no reciprocal."
        }
      },
      {
        "id": "G2",
        "statement": {
          "zh-Hant": "MKO 的定義：數 x 是有理數，若存在整數 p、q，q≠0，使 x=p/q（假設 a1、a2）。",
          "en": "The MKO's definition: a number x is rational if there are integers p, q with q≠0 and x=p/q (assumptions a1, a2)."
        }
      },
      {
        "id": "G3",
        "statement": {
          "zh-Hant": "正規顯示慣例（MKO 說明）：分子與分母互質（最大公因數為 1），且分母為正。",
          "en": "Canonical display convention (MKO explanation): numerator and denominator are coprime (greatest common divisor 1), and the denominator is positive."
        }
      },
      {
        "id": "G4",
        "statement": {
          "zh-Hant": "資料：6/(−8)、−3/4、9/(−12)、2/3、3/5、字串「5/0」、有限小數 −1.25、整數 7、−2/3。",
          "en": "Data: 6/(−8), −3/4, 9/(−12), 2/3, 3/5, the string '5/0', the finite decimal −1.25, the integer 7, and −2/3."
        }
      }
    ],
    "steps": [
      {
        "id": "S01",
        "uses": [
          "G1",
          "G2",
          "mko_rat_def"
        ],
        "before_tex": "x\\in\\mathbb{Q}",
        "after_tex": "\\exists p,q\\in\\mathbb{Z},\\ q\\neq 0:\\ x=\\frac{p}{q}",
        "relation": "definition",
        "rule": "mko_rat_def",
        "conditions": [],
        "explanation": {
          "zh-Hant": "依 MKO，判定 x 是有理數，就是找出一對整數 (p,q)，q≠0，使 x=p/q。定義只要求「存在」這樣一對，並沒有說這一對唯一；不同的整數對可以表示同一個有理數，這正是以下要處理的。",
          "en": "By the MKO, deciding that x is rational means finding a pair of integers (p,q) with q≠0 and x=p/q. The definition only asks that such a pair exist and does not say it is unique; different integer pairs can represent the same rational, which is what the following steps handle."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "mko_rat_def",
            "G1"
          ],
          "explanation": {
            "zh-Hant": "q≠0 是定義的一部分（假設 a2），不能省略；q=0 時 p/q 沒有定義。",
            "en": "q≠0 is part of the definition (assumption a2) and cannot be dropped; for q=0, p/q is undefined."
          }
        },
        "leaf_basis": "mko_rat_def + mul_inverse_nonzero"
      },
      {
        "id": "S02",
        "uses": [
          "G1",
          "L_cross"
        ],
        "before_tex": "\\frac{p}{q}=\\frac{r}{s}\\quad(q\\neq0,\\ s\\neq0)",
        "after_tex": "ps=rq",
        "relation": "equivalent",
        "rule": "L_cross",
        "conditions": [
          "q\\neq 0",
          "s\\neq 0"
        ],
        "explanation": {
          "zh-Hant": "比較兩個分數是否相等的工具是引理 L_cross：分母都不為 0 時，p/q=r/s ⇔ ps=rq。引理在 S02.1、S02.2 證明，之後 S03、S04、S07、S09 都用它。",
          "en": "The tool for deciding whether two fractions are equal is Lemma L_cross: when both denominators are nonzero, p/q=r/s ⇔ ps=rq. The lemma is proved in S02.1 and S02.2 and used in S03, S04, S07 and S09."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "L_cross"
          ],
          "explanation": {
            "zh-Hant": "引理是雙條件，兩個方向都在子步證明；分母非零不可省。",
            "en": "The lemma is a biconditional with both directions proved in the sub-steps; nonzero denominators cannot be dropped."
          }
        },
        "leaf_basis": "L_cross"
      },
      {
        "id": "S02.1",
        "uses": [
          "G1",
          "zero_product"
        ],
        "before_tex": "q\\neq 0,\\ s\\neq 0",
        "after_tex": "qs\\neq 0,\\ \\frac{p}{q}\\cdot qs=ps,\\ \\frac{r}{s}\\cdot qs=rq",
        "relation": "implies",
        "rule": "zero_product",
        "conditions": [
          "q\\neq 0",
          "s\\neq 0"
        ],
        "explanation": {
          "zh-Hant": "依零乘積性質，qs=0 ⇔ q=0 或 s=0；兩者都不成立，所以 qs≠0。又 (p/q)·qs=p·(1/q)·q·s=p·1·s=ps（結合律、交換律與 q·(1/q)=1）；同理 (r/s)·qs=r·(1/s)·s·q=rq。",
          "en": "By the zero-product property, qs=0 ⇔ q=0 or s=0; neither holds, so qs≠0. Also (p/q)·qs=p·(1/q)·q·s=p·1·s=ps (associativity, commutativity and q·(1/q)=1); likewise (r/s)·qs=r·(1/s)·s·q=rq."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "zero_product",
            "mul_inverse_nonzero",
            "mul_assoc",
            "mul_comm"
          ],
          "explanation": {
            "zh-Hant": "需要 q≠0 與 s≠0，才有 1/q 與 1/s。",
            "en": "q≠0 and s≠0 are needed for 1/q and 1/s to exist."
          }
        },
        "leaf_basis": "zero_product + mul_inverse_nonzero + mul_assoc + mul_comm + mul_one"
      },
      {
        "id": "S02.2",
        "uses": [
          "S02.1",
          "eq_congruence"
        ],
        "before_tex": "\\frac{p}{q}=\\frac{r}{s}",
        "after_tex": "ps=rq",
        "relation": "equivalent",
        "rule": "eq_congruence",
        "conditions": [
          "qs\\neq 0"
        ],
        "explanation": {
          "zh-Hant": "⇒：兩邊同乘 qs，依 S02.1 得 ps=rq。⇐：兩邊同乘 1/(qs)（qs≠0，所以存在），ps·(1/(qs))=p/q、rq·(1/(qs))=r/s，得 p/q=r/s。S02.1 與 S02.2 合起來證得引理 L_cross。",
          "en": "⇒: multiply both sides by qs; by S02.1 this gives ps=rq. ⇐: multiply both sides by 1/(qs), which exists since qs≠0; ps·(1/(qs))=p/q and rq·(1/(qs))=r/s, giving p/q=r/s. S02.1 and S02.2 together prove Lemma L_cross."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "eq_congruence",
            "mul_inverse_nonzero",
            "S02.1"
          ],
          "explanation": {
            "zh-Hant": "乘 qs 與乘 1/(qs) 互為反運算，前提 qs≠0。",
            "en": "Multiplying by qs and by 1/(qs) undo each other, given qs≠0."
          }
        },
        "leaf_basis": "eq_congruence + mul_inverse_nonzero + frac_rules"
      },
      {
        "id": "S03",
        "uses": [
          "S02",
          "G4"
        ],
        "before_tex": "6\\cdot 4=24=(-3)\\cdot(-8)",
        "after_tex": "\\frac{6}{-8}=\\frac{-3}{4}",
        "relation": "equivalent",
        "rule": "L_cross",
        "conditions": [
          "-8\\neq 0",
          "4\\neq 0"
        ],
        "explanation": {
          "zh-Hant": "取 p=6、q=−8、r=−3、s=4：ps=6·4=24，rq=(−3)·(−8)=24。兩者相等，依 L_cross，6/(−8)=−3/4。",
          "en": "Take p=6, q=−8, r=−3, s=4: ps=6·4=24 and rq=(−3)·(−8)=24. They are equal, so by L_cross, 6/(−8)=−3/4."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "L_cross",
            "numeric_evaluation"
          ],
          "explanation": {
            "zh-Hant": "L_cross 是雙條件；計算可重算（N1、N2、N3）。",
            "en": "L_cross is a biconditional; the computation can be redone (N1, N2, N3)."
          }
        },
        "leaf_basis": "L_cross + numeric_evaluation + neg_mul"
      },
      {
        "id": "S04",
        "uses": [
          "S02",
          "S03",
          "G4"
        ],
        "before_tex": "6\\cdot(-12)=-72=9\\cdot(-8)",
        "after_tex": "\\frac{6}{-8}=\\frac{9}{-12}=\\frac{-3}{4}",
        "relation": "equivalent",
        "rule": "L_cross",
        "conditions": [
          "-8\\neq 0",
          "-12\\neq 0"
        ],
        "explanation": {
          "zh-Hant": "取 p=6、q=−8、r=9、s=−12：ps=6·(−12)=−72，rq=9·(−8)=−72，所以 6/(−8)=9/(−12)；配合 S03，三個不同的整數對 (6,−8)、(−3,4)、(9,−12) 表示同一個有理數。這正是 MKO 所說「不同整數對可以表示同一個有理數」。",
          "en": "Take p=6, q=−8, r=9, s=−12: ps=6·(−12)=−72 and rq=9·(−8)=−72, so 6/(−8)=9/(−12); with S03, the three different integer pairs (6,−8), (−3,4) and (9,−12) represent the same rational number. This is exactly the MKO's 'different integer pairs can represent the same rational number'."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "L_cross",
            "numeric_evaluation",
            "eq_refl_symm_trans"
          ],
          "explanation": {
            "zh-Hant": "每個相等都由雙條件得到，並以等號傳遞連起來（N4 到 N6）。",
            "en": "Each equality comes from a biconditional and they are joined by transitivity of equality (N4 to N6)."
          }
        },
        "leaf_basis": "L_cross + numeric_evaluation + eq_refl_symm_trans"
      },
      {
        "id": "S05",
        "uses": [
          "G1",
          "frac_rules"
        ],
        "before_tex": "\\frac{6}{-8}",
        "after_tex": "\\frac{6\\cdot(-\\frac12)}{(-8)\\cdot(-\\frac12)}=\\frac{-3}{4}",
        "relation": "equivalent",
        "rule": "frac_rules",
        "conditions": [
          "-\\tfrac{1}{2}\\neq 0"
        ],
        "explanation": {
          "zh-Hant": "約分：gcd(6,8)=2（N20），把分子分母同乘 k=−1/2（等於同除以 −2，負號一併移到分子）。依分數規則 a/b=(ak)/(bk)（k≠0），6/(−8)=(6·(−1/2))/((−8)·(−1/2))=(−3)/4（N7、N8）。這是精確等式，不是近似。",
          "en": "Reduce: gcd(6,8)=2 (N20), so multiply numerator and denominator by k=−1/2 (that is, divide both by −2, which also moves the sign to the numerator). By the fraction rule a/b=(ak)/(bk) for k≠0, 6/(−8)=(6·(−1/2))/((−8)·(−1/2))=(−3)/4 (N7, N8). This is an exact equality, not an approximation."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "frac_rules",
            "numeric_evaluation"
          ],
          "explanation": {
            "zh-Hant": "反向：分子分母同乘 −2 回到 6/(−8)；k≠0 是規則的條件。",
            "en": "Backward: multiplying numerator and denominator by −2 returns to 6/(−8); k≠0 is the rule's condition."
          }
        },
        "leaf_basis": "frac_rules + numeric_evaluation"
      },
      {
        "id": "S06",
        "uses": [
          "S05",
          "G3",
          "gcd_def"
        ],
        "before_tex": "\\frac{-3}{4}",
        "after_tex": "\\gcd(3,4)=1,\\ 4>0:\\ \\text{canonical form}",
        "relation": "implies",
        "rule": "canonical_def",
        "conditions": [],
        "explanation": {
          "zh-Hant": "4 的正因數是 1、2、4；2 與 4 都不整除 3，所以 gcd(3,4)=1（N9）；分母 4>0。所以 −3/4 符合 G3 的正規顯示慣例。正規化改變的是表示，不是數值（S05 是等式）。",
          "en": "The positive divisors of 4 are 1, 2 and 4; neither 2 nor 4 divides 3, so gcd(3,4)=1 (N9); and the denominator 4>0. So −3/4 meets the canonical display convention of G3. Normalizing changes the representation, not the value (S05 is an equality)."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "gcd_def",
            "canonical_def",
            "divides_def"
          ],
          "explanation": {
            "zh-Hant": "只檢查 −3/4 是正規形式；正規形式的唯一性不在本稿證明（開放項目）。",
            "en": "It checks only that −3/4 is in canonical form; uniqueness of the canonical form is not proved here (open item)."
          }
        },
        "leaf_basis": "canonical_def + gcd_def + divides_def + numeric_evaluation"
      },
      {
        "id": "S07",
        "uses": [
          "S02",
          "G4"
        ],
        "before_tex": "2\\cdot5=10\\neq 9=3\\cdot3",
        "after_tex": "\\frac{2}{3}\\neq\\frac{3}{5}",
        "relation": "equivalent",
        "rule": "L_cross",
        "conditions": [
          "3\\neq 0",
          "5\\neq 0"
        ],
        "explanation": {
          "zh-Hant": "取 p=2、q=3、r=3、s=5：ps=10，rq=9，10≠9。L_cross 是雙條件，兩邊同時否定仍等價：ps≠rq ⇔ p/q≠r/s，所以 2/3≠3/5。",
          "en": "Take p=2, q=3, r=3, s=5: ps=10 and rq=9, and 10≠9. L_cross is a biconditional, so negating both sides keeps an equivalence: ps≠rq ⇔ p/q≠r/s, hence 2/3≠3/5."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "L_cross",
            "numeric_evaluation"
          ],
          "explanation": {
            "zh-Hant": "P⇔Q 與 ¬P⇔¬Q 等價（N10）。",
            "en": "P⇔Q is equivalent to ¬P⇔¬Q (N10)."
          }
        },
        "leaf_basis": "L_cross + numeric_evaluation"
      },
      {
        "id": "S08",
        "uses": [
          "G1",
          "mul_zero"
        ],
        "before_tex": "x\\cdot 0=5\\ \\text{for some}\\ x",
        "after_tex": "0=5,\\ \\text{a contradiction}",
        "relation": "implies",
        "rule": "contradiction",
        "conditions": [],
        "explanation": {
          "zh-Hant": "若「5/0」表示某個數 x，則 x 應滿足 x·0=5。但對任何 x，x·0=0（乘以零），於是 0=5，矛盾。所以沒有任何數是 5/0；整數對 (5,0) 違反 q≠0（假設 a2），不表示任何有理數。對照：0/5=0·(1/5)=0 是合法的有理數，分子可以是 0（N11）。",
          "en": "If '5/0' denoted some number x, then x would satisfy x·0=5. But x·0=0 for every x (multiplying by zero), so 0=5, a contradiction. So no number is 5/0; the pair (5,0) violates q≠0 (assumption a2) and represents no rational. By contrast 0/5=0·(1/5)=0 is a legitimate rational; the numerator may be 0 (N11)."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "mul_zero",
            "contradiction",
            "mul_inverse_nonzero"
          ],
          "explanation": {
            "zh-Hant": "矛盾來自 x·0=0 對所有 x 成立；這正是 0 沒有倒數的原因。",
            "en": "The contradiction comes from x·0=0 for every x; this is exactly why 0 has no reciprocal."
          }
        },
        "leaf_basis": "mul_zero + contradiction + mul_inverse_nonzero"
      },
      {
        "id": "S09",
        "uses": [
          "S02",
          "G4",
          "decimal_def"
        ],
        "before_tex": "-1.25",
        "after_tex": "-\\frac{125}{100}=-\\frac{5}{4}\\in\\mathbb{Q}",
        "relation": "implies",
        "rule": "exists_intro",
        "conditions": [
          "100\\neq 0"
        ],
        "explanation": {
          "zh-Hant": "依有限小數的意義，−1.25=−(1+2/10+5/100)=−125/100（N12）。取 p=−125、q=100≠0，依定義 −1.25 是有理數。再約分：(−125)·4=−500=(−5)·100，依 L_cross，−125/100=−5/4（N13、N14）。有限小數是精確的分數，不是另一種數。",
          "en": "By the meaning of a finite decimal, −1.25=−(1+2/10+5/100)=−125/100 (N12). Taking p=−125 and q=100≠0, −1.25 is rational by definition. Reducing: (−125)·4=−500=(−5)·100, so by L_cross, −125/100=−5/4 (N13, N14). A finite decimal is an exact fraction, not a different kind of number."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "decimal_def",
            "exists_intro",
            "L_cross",
            "numeric_evaluation"
          ],
          "explanation": {
            "zh-Hant": "存在性只需一對 (p,q)；約分是另外的等式。",
            "en": "Existence needs only one pair (p,q); the reduction is a separate equality."
          }
        },
        "leaf_basis": "decimal_def + exists_intro + L_cross + numeric_evaluation"
      },
      {
        "id": "S10",
        "uses": [
          "S01",
          "mul_one"
        ],
        "before_tex": "n\\in\\mathbb{Z}",
        "after_tex": "n=\\frac{n}{1}\\in\\mathbb{Q}",
        "relation": "implies",
        "rule": "exists_intro",
        "conditions": [
          "1\\neq 0"
        ],
        "explanation": {
          "zh-Hant": "對任意整數 n：1·1=1，所以 1/1=1，n/1=n·1=n。取 p=n、q=1≠0，依定義 n∈ℚ。特別地 7=7/1 是有理數（N15）。n 是任意的，所以每個整數都是有理數。",
          "en": "For any integer n: 1·1=1, so 1/1=1 and n/1=n·1=n. Taking p=n and q=1≠0, n∈ℚ by definition. In particular 7=7/1 is rational (N15). n was arbitrary, so every integer is rational."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "exists_intro",
            "mul_one",
            "forall_intro"
          ],
          "explanation": {
            "zh-Hant": "反向（每個有理數都是整數）不成立，例如 −5/4 不是整數。",
            "en": "The converse (every rational is an integer) is false; for example −5/4 is not an integer."
          }
        },
        "leaf_basis": "exists_intro + mul_one + forall_intro"
      },
      {
        "id": "S11",
        "uses": [
          "G4",
          "L_cross_order"
        ],
        "before_tex": "(-3)\\cdot3=-9<-8=(-2)\\cdot4",
        "after_tex": "\\frac{-3}{4}<\\frac{-2}{3}",
        "relation": "equivalent",
        "rule": "L_cross_order",
        "conditions": [
          "4>0",
          "3>0"
        ],
        "explanation": {
          "zh-Hant": "兩個分母 4、3 都為正。依引理 L_cross_order（p=−3、q=4、r=−2、s=3）：ps=(−3)·3=−9，rq=(−2)·4=−8，−9<−8，所以 −3/4<−2/3（N16、N17）。引理在 S11.1 證明；分母皆為正是本引理採用的充分前提，不宣稱它是保持此比較方向的最弱條件，這也是正規形式要求分母為正的好處。",
          "en": "Both denominators 4 and 3 are positive. By Lemma L_cross_order (p=−3, q=4, r=−2, s=3): ps=(−3)·3=−9 and rq=(−2)·4=−8, and −9<−8, so −3/4<−2/3 (N16, N17). The lemma is proved in S11.1; positive denominators are sufficient hypotheses used by this lemma, not a claim of logically minimal conditions, which is one benefit of the canonical form's positive denominator."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "L_cross_order",
            "numeric_evaluation"
          ],
          "explanation": {
            "zh-Hant": "本稿在兩個分母都為正的前提下證明此雙條件；這是已證明的適用範圍，不表示其他符號情形一律不成立。",
            "en": "This guide proves the biconditional under the sufficient hypothesis that both denominators are positive; it does not claim that it fails in every other sign configuration."
          }
        },
        "leaf_basis": "L_cross_order + numeric_evaluation"
      },
      {
        "id": "S11.1",
        "uses": [
          "S02.1",
          "order_basics",
          "mul_inverse_nonzero"
        ],
        "before_tex": "q>0,\\ s>0,\\ \\frac{p}{q}<\\frac{r}{s}",
        "after_tex": "ps<rq",
        "relation": "equivalent",
        "rule": "order_basics",
        "conditions": [
          "q>0",
          "s>0"
        ],
        "explanation": {
          "zh-Hant": "q>0、s>0 推出 qs>0，也推出 1/(qs)>0。⇒：不等式兩邊同乘正數 qs，方向不變，依 S02.1 的化簡得 ps<rq。⇐：兩邊同乘正數 1/(qs) 得 p/q<r/s。這證得引理 L_cross_order。若某個分母為負，乘以 qs 可能使不等號反向，引理就不適用。",
          "en": "q>0 and s>0 give qs>0 and also 1/(qs)>0. ⇒: multiplying both sides by the positive number qs keeps the direction, and the simplification in S02.1 gives ps<rq. ⇐: multiplying both sides by the positive number 1/(qs) gives p/q<r/s. This proves Lemma L_cross_order. If a denominator were negative, multiplying by qs could reverse the inequality, and the lemma would not apply."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "order_basics",
            "S02.1",
            "mul_inverse_nonzero"
          ],
          "explanation": {
            "zh-Hant": "乘以正數 qs 與乘以正數 1/(qs) 都保持不等號方向，互為反運算。",
            "en": "Multiplying by the positive qs and by the positive 1/(qs) both keep the direction of the inequality and undo each other."
          }
        },
        "leaf_basis": "order_basics + mul_inverse_nonzero + frac_rules"
      },
      {
        "id": "S12",
        "uses": [
          "S04",
          "S06",
          "S07",
          "S08",
          "S09",
          "S10",
          "S11"
        ],
        "before_tex": "\\text{S04, S06 to S11}",
        "after_tex": "\\tfrac{6}{-8}=\\tfrac{9}{-12}=\\tfrac{-3}{4};\\ \\tfrac{2}{3}\\neq\\tfrac{3}{5};\\ -1.25,\\ 7\\in\\mathbb{Q};\\ \\tfrac{-3}{4}<\\tfrac{-2}{3}",
        "relation": "implies",
        "rule": "and_intro_elim",
        "conditions": [],
        "explanation": {
          "zh-Hant": "三個整數對表示同一個有理數，正規形式是 −3/4；2/3 與 3/5 不同；「5/0」不表示任何數；−1.25=−5/4 與 7=7/1 是有理數；−3/4<−2/3。",
          "en": "Three integer pairs represent the same rational, whose canonical form is −3/4; 2/3 and 3/5 differ; '5/0' denotes no number; −1.25=−5/4 and 7=7/1 are rational; and −3/4<−2/3."
        },
        "reverse_audit": {
          "kind": "sufficient_goal_reduction",
          "required_ids": [
            "S03",
            "S04",
            "S05",
            "S06",
            "S07",
            "S08",
            "S09",
            "S10",
            "S11"
          ],
          "explanation": {
            "zh-Hant": "目標是 goal 所列各項。足夠條件 H：S03 到 S11 各自成立；H 推出目標只用合取。每一項都由定義、兩個引理與精確計算正向證成，沒有把目標當作前提。",
            "en": "The goal is the list in the goal field. Sufficient condition H: S03 to S11 each hold; H gives the goal by conjunction only. Each item is proved forwards from the definition, the two lemmas and exact computation; the goal is never used as a premise."
          }
        },
        "leaf_basis": "and_intro_elim"
      }
    ],
    "result": {
      "zh-Hant": "6/(−8)、−3/4、9/(−12) 是同一個有理數（交叉乘積 24=24、−72=−72），正規形式為 −3/4（gcd(3,4)=1，分母為正），約分是精確等式；2/3≠3/5（10≠9）；「5/0」不表示任何數；−1.25=−125/100=−5/4 與 7=7/1 都是有理數；分母為正時 −3/4<−2/3（−9<−8）。",
      "en": "6/(−8), −3/4 and 9/(−12) are the same rational (cross products 24=24 and −72=−72), with canonical form −3/4 (gcd(3,4)=1, positive denominator), and the reduction is an exact equality; 2/3≠3/5 (10≠9); '5/0' denotes no number; −1.25=−125/100=−5/4 and 7=7/1 are rational; with positive denominators, −3/4<−2/3 (−9<−8)."
    },
    "negative_controls": [
      {
        "claim": {
          "zh-Hant": "5/0 是有理數，因為 5 與 0 都是整數",
          "en": "5/0 is rational because 5 and 0 are both integers"
        },
        "expected": "reject",
        "reason": {
          "zh-Hant": "定義要求 q≠0；而且沒有任何數 x 滿足 x·0=5（S08）。",
          "en": "The definition requires q≠0, and no number x satisfies x·0=5 (S08)."
        }
      },
      {
        "claim": {
          "zh-Hant": "6/(−8) 與 −3/4 外觀不同，所以是不同的有理數",
          "en": "6/(−8) and −3/4 look different, so they are different rationals"
        },
        "expected": "reject",
        "reason": {
          "zh-Hant": "交叉乘積 6·4=24=(−3)·(−8)，依 L_cross 兩者相等（S03）。",
          "en": "The cross products 6·4=24=(−3)·(−8) agree, so by L_cross they are equal (S03)."
        }
      },
      {
        "claim": {
          "zh-Hant": "把 6/(−8) 約成 −3/4 只是近似",
          "en": "Reducing 6/(−8) to −3/4 is only an approximation"
        },
        "expected": "reject",
        "reason": {
          "zh-Hant": "a/b=(ak)/(bk)（k=−1/2≠0）是精確等式（S05）。",
          "en": "a/b=(ak)/(bk) with k=−1/2≠0 is an exact equality (S05)."
        }
      },
      {
        "claim": {
          "zh-Hant": "因為 1·3>1·(−2)，所以 1/(−2)>1/3",
          "en": "Since 1·3>1·(−2), we have 1/(−2)>1/3"
        },
        "expected": "reject",
        "reason": {
          "zh-Hant": "L_cross_order 需要分母為正；事實上 −1/2<1/3（N19）。",
          "en": "L_cross_order needs positive denominators; in fact −1/2<1/3 (N19)."
        }
      }
    ],
    "status": "candidate",
    "gaps": [],
    "verification": {
      "structure": "checked_local_script",
      "mathematics": "candidate_argument",
      "formalization": "not_formalized"
    },
    "rule_glossary": [
      {
        "id": "mul_inverse_nonzero",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "非零數的倒數：在 ℚ、ℝ、ℂ 中，若 u≠0，存在 1/u 使 u·(1/u)=1；對 u≠0 定義 v/u 為 v·(1/u)。u=0 時沒有倒數，v/0 不定義。",
          "en": "Inverse of a nonzero number: in ℚ, ℝ or ℂ, if u≠0 there is 1/u with u·(1/u)=1, and v/u means v·(1/u). 0 has no inverse, so v/0 is undefined."
        }
      },
      {
        "id": "frac_rules",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "分數運算（分母皆非零）：a/c+b/c=(a+b)/c；(a/b)(c/d)=(ac)/(bd)；(a/b)/(c/d)=(ad)/(bc)（另需 c≠0）；a/b=(ak)/(bk)（k≠0）。",
          "en": "Fraction rules (all denominators nonzero): a/c+b/c=(a+b)/c; (a/b)(c/d)=(ac)/(bd); (a/b)/(c/d)=(ad)/(bc) (also needing c≠0); a/b=(ak)/(bk) for k≠0."
        }
      },
      {
        "id": "zero_product",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "零乘積性質：在 ℤ、ℚ、ℝ、ℂ（或任何整域）中，uv=0 ⇔ u=0 或 v=0。",
          "en": "Zero-product property: in ℤ, ℚ, ℝ, ℂ (or any integral domain), uv=0 ⇔ u=0 or v=0."
        }
      },
      {
        "id": "order_basics",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "實數的序：任兩數恰有 u<v、u=v、u>v 之一；u<v 且 v<w 推出 u<w；u<v 推出 u+w<v+w；u<v 且 w>0 推出 uw<vw；u<v 且 w<0 推出 uw>vw。",
          "en": "Order of the reals: exactly one of u<v, u=v, u>v holds; u<v and v<w give u<w; u<v gives u+w<v+w; u<v and w>0 give uw<vw; u<v and w<0 give uw>vw."
        }
      },
      {
        "id": "numeric_evaluation",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "明確整數或有理數算式的精確求值；可以獨立重算。",
          "en": "Exact evaluation of an explicit integer or rational expression; independently recomputable."
        }
      },
      {
        "id": "mul_assoc",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "乘法結合律：(uv)w=u(vw)。",
          "en": "Associativity of multiplication: (uv)w=u(vw)."
        }
      },
      {
        "id": "mul_comm",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "乘法交換律：uv=vu（在交換的數系中，例如 ℤ、ℚ、ℝ、ℂ）。",
          "en": "Commutativity of multiplication: uv=vu (in commutative number systems such as ℤ, ℚ, ℝ, ℂ)."
        }
      },
      {
        "id": "mul_one",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "乘法單位元：1·u=u。",
          "en": "Multiplicative identity: 1·u=u."
        }
      },
      {
        "id": "eq_congruence",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "等式同用：若 u=v，對兩邊做同一個有定義的運算 f，得 f(u)=f(v)。只對一邊做運算不屬於此規則。",
          "en": "Congruence of equality: if u=v, applying the same defined operation f to both sides gives f(u)=f(v). Operating on one side only is not an instance of this rule."
        }
      },
      {
        "id": "neg_mul",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "負號與乘法：(−u)v=−(uv)，(−u)(−v)=uv。",
          "en": "Signs and multiplication: (−u)v=−(uv) and (−u)(−v)=uv."
        }
      },
      {
        "id": "eq_refl_symm_trans",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "等號的性質：u=u；若 u=v 則 v=u；若 u=v 且 v=w 則 u=w。",
          "en": "Properties of equality: u=u; if u=v then v=u; if u=v and v=w then u=w."
        }
      },
      {
        "id": "divides_def",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "整除：對整數 a、b，a∣b 表示存在整數 k 使 b=ak。",
          "en": "Divisibility: for integers a, b, a∣b means there is an integer k with b=ak."
        }
      },
      {
        "id": "mul_zero",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "乘以零：u·0=0。理由：u·0=u·(0+0)=u·0+u·0，兩邊同減 u·0 得 0=u·0。",
          "en": "Multiplying by zero: u·0=0. Reason: u·0=u·(0+0)=u·0+u·0; subtracting u·0 from both sides gives 0=u·0."
        }
      },
      {
        "id": "contradiction",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "反證法：若假設 ¬P 推出矛盾（某命題與其否定同時成立），則 P 成立。",
          "en": "Proof by contradiction: if assuming ¬P leads to a contradiction (a statement and its negation both hold), then P holds."
        }
      },
      {
        "id": "exists_intro",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "∃引入：若找到 a∈D 使 P(a) 成立，則 ∃x∈D, P(x)。",
          "en": "∃-introduction: if some a∈D satisfies P(a), then ∃x∈D, P(x)."
        }
      },
      {
        "id": "forall_intro",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "∀引入（全稱推廣）：若某命題對一個「任意的」元素 x 證成，而證明沒有對 x 加任何額外假設，則它對論域中所有 x 成立。",
          "en": "∀-introduction (generalization): if a statement is proved for an arbitrary element x, with no extra assumption on x, it holds for every x in the domain."
        }
      },
      {
        "id": "and_intro_elim",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "合取：由 P 與 Q 得 P∧Q；由 P∧Q 可分別得 P 與 Q。",
          "en": "Conjunction: from P and Q infer P∧Q; from P∧Q infer P and infer Q."
        }
      },
      {
        "id": "mko_rat_def",
        "kind": "cited",
        "source": {
          "mko_id": "mko-rational-number",
          "sha256": "f1b52aad1867316fd94d6e07d6e94237a202d6769454898a0831d8d1bb89f23f",
          "commit": "b28a5edbd46279b1db2a3f3f48bb3f447e18899f"
        },
        "statement": {
          "zh-Hant": "數 x 是有理數，若存在整數 p、q 且 q≠0，使 x=p/q。不同整數對可以表示同一個有理數。",
          "en": "A number x is rational if there exist integers p and q with q≠0 such that x=p/q. Different integer pairs can represent the same rational number."
        }
      },
      {
        "id": "canonical_def",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "正規形式：p/q 中 gcd(|p|,q)=1 且 q>0。",
          "en": "Canonical form: p/q with gcd(|p|,q)=1 and q>0."
        }
      },
      {
        "id": "gcd_def",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "整數 a、b 不全為 0 時，gcd(a,b) 是同時整除 a 與 b 的最大正整數。",
          "en": "For integers a, b not both 0, gcd(a,b) is the largest positive integer dividing both a and b."
        }
      },
      {
        "id": "decimal_def",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "有限小數：整數部分加上小數點後第 j 位數字乘以 10^{−j} 之和；例如 1.25=1+2/10+5/100。",
          "en": "Finite decimal: the integer part plus the sum of the j-th digit after the point times 10^{−j}; for example 1.25=1+2/10+5/100."
        }
      },
      {
        "id": "L_cross",
        "kind": "lemma",
        "proved_in": [
          "S02.1",
          "S02.2"
        ],
        "statement": {
          "zh-Hant": "對實數 p、q、r、s，q≠0、s≠0：p/q=r/s ⇔ ps=rq。",
          "en": "For reals p, q, r, s with q≠0 and s≠0: p/q=r/s ⇔ ps=rq."
        }
      },
      {
        "id": "L_cross_order",
        "kind": "lemma",
        "proved_in": [
          "S11.1"
        ],
        "statement": {
          "zh-Hant": "對實數 p、q、r、s，q>0、s>0：p/q<r/s ⇔ ps<rq。",
          "en": "For reals p, q, r, s with q>0 and s>0: p/q<r/s ⇔ ps<rq."
        }
      }
    ],
    "revision_log": [],
    "numeric_checks": [
      {
        "id": "N1",
        "step": "S03",
        "lhs": "6*4",
        "rhs": "24"
      },
      {
        "id": "N2",
        "step": "S03",
        "lhs": "(-3)*(-8)",
        "rhs": "24"
      },
      {
        "id": "N3",
        "step": "S03",
        "lhs": "6/(-8)",
        "rhs": "-3/4"
      },
      {
        "id": "N4",
        "step": "S04",
        "lhs": "6*(-12)",
        "rhs": "-72"
      },
      {
        "id": "N5",
        "step": "S04",
        "lhs": "9*(-8)",
        "rhs": "-72"
      },
      {
        "id": "N6",
        "step": "S04",
        "lhs": "9/(-12)",
        "rhs": "-3/4"
      },
      {
        "id": "N7",
        "step": "S05",
        "lhs": "6*(-1/2)",
        "rhs": "-3"
      },
      {
        "id": "N8",
        "step": "S05",
        "lhs": "(-8)*(-1/2)",
        "rhs": "4"
      },
      {
        "id": "N9",
        "step": "S06",
        "lhs": "gcd(3,4)",
        "rhs": "1"
      },
      {
        "id": "N10",
        "step": "S07",
        "lhs": "2*5",
        "relation": "!=",
        "rhs": "3*3"
      },
      {
        "id": "N11",
        "step": "S08",
        "lhs": "0/5",
        "rhs": "0"
      },
      {
        "id": "N12",
        "step": "S09",
        "lhs": "-(1+2/10+5/100)",
        "rhs": "-125/100"
      },
      {
        "id": "N13",
        "step": "S09",
        "lhs": "(-125)*4",
        "rhs": "(-5)*100"
      },
      {
        "id": "N14",
        "step": "S09",
        "lhs": "-125/100",
        "rhs": "-5/4"
      },
      {
        "id": "N15",
        "step": "S10",
        "lhs": "7/1",
        "rhs": "7"
      },
      {
        "id": "N16",
        "step": "S11",
        "lhs": "(-3)*3",
        "relation": "<",
        "rhs": "(-2)*4"
      },
      {
        "id": "N17",
        "step": "S11",
        "lhs": "-3/4",
        "relation": "<",
        "rhs": "-2/3"
      },
      {
        "id": "N18",
        "lhs": "1*3",
        "relation": ">",
        "rhs": "1*(-2)"
      },
      {
        "id": "N19",
        "lhs": "1/(-2)",
        "relation": "<",
        "rhs": "1/3"
      },
      {
        "id": "N20",
        "step": "S05",
        "lhs": "gcd(6,8)",
        "rhs": "2"
      },
      {
        "id": "N21",
        "lhs": "-0.8",
        "relation": "!=",
        "rhs": "-3/4"
      }
    ]
  },
  "presentation": {
    "md_scope": {
      "zh-Hant": "在 ℝ 中討論分數（p/q 定義為 p·(1/q)），只用域的運算與序。證明兩個引理：分母非零時以交叉相乘判定相等，分母為正時以交叉相乘判定大小。不證明正規形式（既約且分母為正）的唯一性，不處理與 Lean Rat.divInt 的對接；不改動 canonical MKO 或英文 edition。",
      "en": "Treats fractions inside ℝ (p/q means p·(1/q)), using only field operations and order. It proves two lemmas: cross-multiplication decides equality when denominators are nonzero, and decides order when denominators are positive. It does not prove uniqueness of the canonical form (reduced, positive denominator) and does not treat the link to Lean's Rat.divInt; it does not change the canonical MKO or its English edition."
    },
    "md_symbols": [
      [
        "\\mathbb{Q}",
        "有理數集合",
        "the set of rational numbers"
      ],
      [
        "\\frac{p}{q}",
        "p·(1/q)，q≠0",
        "p·(1/q), with q≠0"
      ],
      [
        "(p,q)",
        "表示有理數的一對整數",
        "a pair of integers representing a rational"
      ],
      [
        "\\gcd(a,b)",
        "最大公因數",
        "greatest common divisor"
      ],
      [
        "k",
        "約分時分子分母同乘的非零數",
        "the nonzero factor applied to numerator and denominator when reducing"
      ]
    ],
    "md_error_controls": [
      {
        "title": "分母不可為零 / The denominator may not be zero",
        "zh": "MKO 指出的誤解之一是允許零分母。「5/0」看起來像分數，但若它是某個數 x，就要有 x·0=5，而 x·0 永遠是 0（S08）。所以 (5,0) 不表示任何有理數；相對地，0/5=0 是合法的有理數。",
        "en": "One misconception the MKO names is allowing a zero denominator. '5/0' looks like a fraction, but if it were a number x we would need x·0=5, while x·0 is always 0 (S08). So (5,0) represents no rational; by contrast 0/5=0 is a legitimate rational."
      },
      {
        "title": "外觀不同的分數可以相等 / Different-looking fractions can be equal",
        "zh": "MKO 的另一誤解是把不同分數外觀當成不同數值。6/(−8)、−3/4、9/(−12) 看起來不同，卻是同一個有理數（S03、S04）；判定要用交叉相乘 L_cross，而不是比較分子或分母。比較大小時還要先讓分母為正：1/(−2) 與 1/3 若直接交叉相乘，1·3=3>−2=1·(−2) 會誤判 1/(−2)>1/3，事實上 −1/2<1/3（N18、N19）。",
        "en": "Another misconception the MKO names is treating different-looking fractions as different values. 6/(−8), −3/4 and 9/(−12) look different yet are one rational (S03, S04); equality is decided by cross-multiplication L_cross, not by comparing numerators or denominators. For order, denominators must first be made positive: cross-multiplying 1/(−2) and 1/3 directly gives 1·3=3>−2=1·(−2) and would wrongly suggest 1/(−2)>1/3, while in fact −1/2<1/3 (N18, N19)."
      },
      {
        "title": "約分是精確等式，不是近似 / Reduction is an exact equality, not an approximation",
        "zh": "MKO 提醒不要把約分視為近似。6/(−8)=−3/4 由 a/b=(ak)/(bk)（k≠0）得到，是精確等式（S05）。真正改變數值的是四捨五入：把 −0.75 寫成 −0.8，而 −0.8≠−3/4（N21）。",
        "en": "The MKO warns against treating reduction as an approximation. 6/(−8)=−3/4 follows from a/b=(ak)/(bk) with k≠0 and is an exact equality (S05). What does change the value is rounding: writing −0.75 as −0.8, and −0.8≠−3/4 (N21)."
      }
    ],
    "md_open_items": [
      {
        "zh-Hant": "本稿只檢查 −3/4 是正規形式，不證明每個有理數恰有一個正規形式（唯一性需要例如歐幾里得引理）。",
        "en": "This guide only checks that −3/4 is in canonical form; it does not prove that each rational has exactly one canonical form (uniqueness needs, for example, Euclid's lemma)."
      },
      {
        "zh-Hant": "與 Lean Rat.divInt、非零分母與正規化分子分母聲明的對接（MKO formal next）未在此處理。",
        "en": "The link to Lean's Rat.divInt and the statements about nonzero denominators and normalized numerator and denominator (MKO formal next) is not treated here."
      }
    ]
  },
  "revision": {
    "id": "production-20261019-adopted-96",
    "source_commit": "b28a5edbd46279b1db2a3f3f48bb3f447e18899f",
    "adopted_source_sha256": "7082af66df1018fc25b71890f73e46bebc2e117da1918db747a20e274bd8bc19",
    "original_source_sha256": "b6cd86f150bfd74850fefee9d1f793b0b5d8bf486f024406abf8d1a96e33ec1f",
    "owner_adopted": true,
    "bilingual_sync": "complete_source_bound",
    "formal_Lean": "NOT_RUN",
    "independent_AI_concurrence": "NOT_MEASURED",
    "issue_ids": [
      "BR-M004"
    ],
    "scope": "Owner-adopted natural-language teaching guide; not a formal proof, canonical MKO rewrite or evidence-status promotion."
  }
}