{
  "schema_version": "ocme-derivation-reader-v1",
  "guide_id": "derivation-mko-dimensional-analysis-01",
  "mko_id": "mko-dimensional-analysis",
  "title": {
    "zh-Hant": "量綱分析：以指數向量核對 ½mv²=mgh 並排除 v=2gh",
    "en": "Dimensional analysis: checking ½mv²=mgh with exponent vectors and ruling out v=2gh"
  },
  "contract": {
    "schema_version": "ocme-derivation-guide-v0.1",
    "guide_id": "derivation-mko-dimensional-analysis-01",
    "source": {
      "mko_id": "mko-dimensional-analysis",
      "commit": "b28a5edbd46279b1db2a3f3f48bb3f447e18899f",
      "sha256": "05ee29db7e8f4b787e36b2e536eb249e1b92597007252752dd6347b2dffab5eb",
      "english_sha256": "cc3b3348730aa124197861366975fb0d94ed08956269687fd9bcb2c71ac58e7d"
    },
    "mode": "worked_example",
    "audience_prerequisites": [
      "cartesian_product_def",
      "numeric_evaluation",
      "eq_congruence",
      "power_laws",
      "distributive"
    ],
    "goal": {
      "zh-Hant": "以基本量綱 L、M、T 的整數指數向量 d(Q)=(a,b,c)：算出速度、加速度與兩種能量式的量綱；判定 ½mv²=mgh 量綱一致、v=2gh 不一致且無法以純數常數修正、v²=4gh 量綱一致卻在等加速度模型中錯誤；判定 m+v 不能相加；由 E=C·m^α v^β 解出 α=1、β=2；並說明量綱相同不代表單位換算因子相同。",
      "en": "With integer exponent vectors d(Q)=(a,b,c) over the base dimensions L, M, T: compute the dimensions of speed, acceleration and two energy expressions; decide that ½mv²=mgh is dimensionally consistent, that v=2gh is not and cannot be repaired by a pure-number constant, and that v²=4gh is consistent but wrong in the constant-acceleration model; decide that m+v cannot be added; solve E=C·m^α v^β for α=1 and β=2; and explain that equal dimensions do not mean equal unit conversion factors."
    },
    "givens": [
      {
        "id": "G1",
        "statement": {
          "zh-Hant": "基本量綱 L、M、T；量 Q 的量綱 [Q]=L^a M^b T^c 以整數向量 d(Q)=(a,b,c) 表示（MKO 假設 a1）。",
          "en": "Base dimensions L, M, T; the dimension [Q]=L^a M^b T^c of a quantity Q is represented by the integer vector d(Q)=(a,b,c) (MKO assumption a1)."
        }
      },
      {
        "id": "G2",
        "statement": {
          "zh-Hant": "d(m)=(0,1,0)，d(h)=(1,0,0)，d(t)=(0,0,1)；速度 v 是長度除以時間，加速度 g 是速度除以時間；½、2、4 與 C 是無因次純數（假設 a2）。",
          "en": "d(m)=(0,1,0), d(h)=(1,0,0), d(t)=(0,0,1); speed v is length divided by time and acceleration g is speed divided by time; ½, 2, 4 and C are dimensionless pure numbers (assumption a2)."
        }
      },
      {
        "id": "G3",
        "statement": {
          "zh-Hant": "等加速度模型：自靜止以加速度 g>0 運動時間 t>0 後，速度 v=gt，移動距離 h=gt²/2。",
          "en": "Constant-acceleration model: starting from rest with acceleration g>0, after time t>0 the speed is v=gt and the distance travelled is h=gt²/2."
        }
      },
      {
        "id": "G4",
        "statement": {
          "zh-Hant": "待檢查的式子：(E1) ½mv²=mgh；(E2) v=2gh；(E3) v²=4gh；(E4) 和 m+v。",
          "en": "Expressions to check: (E1) ½mv²=mgh; (E2) v=2gh; (E3) v²=4gh; (E4) the sum m+v."
        }
      }
    ],
    "steps": [
      {
        "id": "S01",
        "uses": [
          "G1",
          "mko_dim"
        ],
        "before_tex": "[Q]=L^aM^bT^c",
        "after_tex": "d(PQ)=d(P)+d(Q),\\ d(P/Q)=d(P)-d(Q),\\ d(Q^k)=k\\,d(Q)",
        "relation": "definition",
        "rule": "mko_dim",
        "conditions": [
          "k\\in\\mathbb{Z}"
        ],
        "explanation": {
          "zh-Hant": "依 MKO，把量綱寫成指數向量；乘積相加向量、商相減、整數冪倍乘。這三條就是指數律 L^a·L^a′=L^(a+a′) 等逐基本量綱套用的結果。要相等或相加的兩個量，向量必須相同。",
          "en": "By the MKO, write dimensions as exponent vectors; products add vectors, quotients subtract them and integer powers scale them. These three rules are the exponent laws such as L^a·L^a′=L^(a+a′) applied to each base dimension. Two quantities that are to be equated or added must have the same vector."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "G1",
            "mko_dim",
            "power_laws"
          ],
          "explanation": {
            "zh-Hant": "此步只寫出規則，不推出新事實；指數限於整數。",
            "en": "This step only states the rules and derives no new fact; exponents are restricted to integers."
          }
        },
        "leaf_basis": "mko_dim + power_laws + vec_ops"
      },
      {
        "id": "S02",
        "uses": [
          "G2"
        ],
        "before_tex": "m,\\ h,\\ t",
        "after_tex": "d(m)=(0,1,0),\\ d(h)=(1,0,0),\\ d(t)=(0,0,1)",
        "relation": "definition",
        "rule": "def_intro",
        "conditions": [],
        "explanation": {
          "zh-Hant": "基本量的量綱由 G2 給出：質量只含 M，高度只含 L，時間只含 T。",
          "en": "The dimensions of the base quantities come from G2: mass involves only M, height only L and time only T."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "G2",
            "def_intro"
          ],
          "explanation": {
            "zh-Hant": "直接取自給定條件，不推出新事實。",
            "en": "Taken directly from the givens; no new fact is derived."
          }
        },
        "leaf_basis": "def_intro"
      },
      {
        "id": "S03",
        "uses": [
          "S01",
          "S02",
          "G2"
        ],
        "before_tex": "d(v)=d(h)-d(t)",
        "after_tex": "d(v)=(1,0,-1)",
        "relation": "equivalent",
        "rule": "vec_ops",
        "conditions": [],
        "explanation": {
          "zh-Hant": "速度是長度除以時間，商相減：(1,0,0)−(0,0,1)=(1,0,−1)，即 [v]=L T⁻¹。",
          "en": "Speed is length divided by time, and a quotient subtracts: (1,0,0)−(0,0,1)=(1,0,−1), that is [v]=L T⁻¹."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "vec_ops",
            "numeric_evaluation"
          ],
          "explanation": {
            "zh-Hant": "逐分量整數計算；反向 (1,0,−1)+(0,0,1)=(1,0,0) 還原。",
            "en": "Componentwise integer arithmetic; backwards, (1,0,−1)+(0,0,1)=(1,0,0) restores it."
          }
        },
        "leaf_basis": "vec_ops + numeric_evaluation"
      },
      {
        "id": "S04",
        "uses": [
          "S03",
          "S02",
          "G2"
        ],
        "before_tex": "d(g)=d(v)-d(t)",
        "after_tex": "d(g)=(1,0,-2)",
        "relation": "equivalent",
        "rule": "vec_ops",
        "conditions": [],
        "explanation": {
          "zh-Hant": "加速度是速度除以時間：(1,0,−1)−(0,0,1)=(1,0,−2)，即 [g]=L T⁻²。g 是具量綱的量，不是純數。",
          "en": "Acceleration is speed divided by time: (1,0,−1)−(0,0,1)=(1,0,−2), that is [g]=L T⁻². g is a dimensioned quantity, not a pure number."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "vec_ops",
            "numeric_evaluation"
          ],
          "explanation": {
            "zh-Hant": "逐分量整數計算，雙向成立。",
            "en": "Componentwise integer arithmetic, valid both ways."
          }
        },
        "leaf_basis": "vec_ops + numeric_evaluation"
      },
      {
        "id": "S05",
        "uses": [
          "S03",
          "S01"
        ],
        "before_tex": "d(v^2)=2\\,d(v)",
        "after_tex": "d(v^2)=(2,0,-2)",
        "relation": "equivalent",
        "rule": "vec_ops",
        "conditions": [
          "k=2"
        ],
        "explanation": {
          "zh-Hant": "整數冪倍乘：2·(1,0,−1)=(2,0,−2)。",
          "en": "An integer power scales: 2·(1,0,−1)=(2,0,−2)."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "vec_ops"
          ],
          "explanation": {
            "zh-Hant": "乘 2 與除以 2 在這些偶數分量上互逆，雙向成立。",
            "en": "Multiplying by 2 and halving undo each other on these even components, so both ways hold."
          }
        },
        "leaf_basis": "vec_ops + power_laws"
      },
      {
        "id": "S06",
        "uses": [
          "S02",
          "S05",
          "G2"
        ],
        "before_tex": "d(\\tfrac{1}{2}mv^2)=d(\\tfrac{1}{2})+d(m)+d(v^2)",
        "after_tex": "d(\\tfrac{1}{2}mv^2)=(2,1,-2)",
        "relation": "equivalent",
        "rule": "vec_ops",
        "conditions": [
          "d(\\tfrac{1}{2})=(0,0,0)"
        ],
        "explanation": {
          "zh-Hant": "乘積相加：(0,0,0)+(0,1,0)+(2,0,−2)=(2,1,−2)，即 [½mv²]=L² M T⁻²。½ 是純數，不影響量綱。",
          "en": "Products add: (0,0,0)+(0,1,0)+(2,0,−2)=(2,1,−2), that is [½mv²]=L² M T⁻². ½ is a pure number and does not affect the dimension."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "vec_ops",
            "numeric_evaluation"
          ],
          "explanation": {
            "zh-Hant": "逐分量整數加法，雙向是同一向量。",
            "en": "Componentwise integer addition; both sides are the same vector."
          }
        },
        "leaf_basis": "vec_ops + numeric_evaluation"
      },
      {
        "id": "S07",
        "uses": [
          "S02",
          "S04"
        ],
        "before_tex": "d(mgh)=d(m)+d(g)+d(h)",
        "after_tex": "d(mgh)=(2,1,-2)",
        "relation": "equivalent",
        "rule": "vec_ops",
        "conditions": [],
        "explanation": {
          "zh-Hant": "(0,1,0)+(1,0,−2)+(1,0,0)=(2,1,−2)。",
          "en": "(0,1,0)+(1,0,−2)+(1,0,0)=(2,1,−2)."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "vec_ops",
            "numeric_evaluation"
          ],
          "explanation": {
            "zh-Hant": "逐分量整數加法，雙向成立。",
            "en": "Componentwise integer addition, valid both ways."
          }
        },
        "leaf_basis": "vec_ops + numeric_evaluation"
      },
      {
        "id": "S08",
        "uses": [
          "S06",
          "S07",
          "mko_dim",
          "mko_dim_limits"
        ],
        "before_tex": "d(\\tfrac{1}{2}mv^2)=(2,1,-2),\\ d(mgh)=(2,1,-2)",
        "after_tex": "\\text{(E1) dimensionally consistent}",
        "relation": "implies",
        "rule": "mko_dim",
        "conditions": [],
        "explanation": {
          "zh-Hant": "兩邊向量相同，所以 (E1) 通過量綱一致這個必要條件。依 MKO，這不證明 (E1) 在物理上正確；它只是沒有被量綱排除。",
          "en": "The two vectors are the same, so (E1) passes the necessary condition of dimensional consistency. By the MKO this does not prove that (E1) is physically correct; it is merely not ruled out by dimensions."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S06",
            "S07",
            "mko_dim_limits"
          ],
          "explanation": {
            "zh-Hant": "結論只是「通過必要條件」，反向（一致 ⇒ 正確）不宣稱。",
            "en": "The conclusion is only 'passes the necessary condition'; the converse (consistent ⇒ correct) is not claimed."
          }
        },
        "leaf_basis": "mko_dim + vec_ops"
      },
      {
        "id": "S09",
        "uses": [
          "S03",
          "S04",
          "S02",
          "L_const_factor"
        ],
        "before_tex": "d(v)=(1,0,-1),\\ d(2gh)=(2,0,-2)",
        "after_tex": "\\text{(E2) inconsistent};\\ d(kgh)=(2,0,-2)\\ \\forall k\\ \\text{dimensionless}",
        "relation": "implies",
        "rule": "L_const_factor",
        "conditions": [],
        "explanation": {
          "zh-Hant": "d(2gh)=(0,0,0)+(1,0,−2)+(1,0,0)=(2,0,−2)，而 d(v)=(1,0,−1)；L 分量 1≠2，向量不同，所以 (E2) 量綱不一致而被排除。依引理 L_const_factor（在 S09.1 證明），把 2 換成任何無因次常數 k 都不改變 d(kgh)，所以不可能靠調整純數修正。",
          "en": "d(2gh)=(0,0,0)+(1,0,−2)+(1,0,0)=(2,0,−2), while d(v)=(1,0,−1); the L components differ, 1≠2, so the vectors differ and (E2) is dimensionally inconsistent and ruled out. By Lemma L_const_factor (proved in S09.1), replacing 2 by any dimensionless constant k leaves d(kgh) unchanged, so no adjustment of a pure number can repair it."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "L_const_factor",
            "vec_ops",
            "mko_dim_limits"
          ],
          "explanation": {
            "zh-Hant": "向量不等只需一個分量不同。結論是 (E2) 必錯；這裡不宣稱正確的關係是什麼，那需要物理模型（S11）。",
            "en": "Unequal vectors need only one differing component. The conclusion is that (E2) is certainly wrong; what the correct relation is is not claimed here and needs a physical model (S11)."
          }
        },
        "leaf_basis": "L_const_factor + vec_ops"
      },
      {
        "id": "S09.1",
        "uses": [
          "S01",
          "G2"
        ],
        "before_tex": "d(kQ)",
        "after_tex": "d(kQ)=d(k)+d(Q)=(0,0,0)+d(Q)=d(Q)",
        "relation": "equivalent",
        "rule": "vec_ops",
        "conditions": [
          "d(k)=(0,0,0)"
        ],
        "explanation": {
          "zh-Hant": "乘積相加，且無因次常數的向量是 (0,0,0)，加零向量不變。此子步證得引理 L_const_factor。",
          "en": "Products add, and a dimensionless constant has vector (0,0,0); adding the zero vector changes nothing. This sub-step proves Lemma L_const_factor."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "vec_ops",
            "add_zero"
          ],
          "explanation": {
            "zh-Hant": "加零向量的恆等式，雙向成立；只對無因次 k 成立，具量綱常數會改變向量。",
            "en": "An identity of adding the zero vector, valid both ways; it holds only for dimensionless k, since a dimensioned constant would change the vector."
          }
        },
        "leaf_basis": "vec_ops + add_zero"
      },
      {
        "id": "S10",
        "uses": [
          "S05",
          "S04",
          "S02",
          "L_const_factor"
        ],
        "before_tex": "d(v^2)=(2,0,-2),\\ d(4gh)",
        "after_tex": "d(4gh)=(2,0,-2):\\ \\text{(E3) dimensionally consistent}",
        "relation": "implies",
        "rule": "mko_dim",
        "conditions": [],
        "explanation": {
          "zh-Hant": "依 L_const_factor，d(4gh)=d(gh)=(1,0,−2)+(1,0,0)=(2,0,−2)=d(v²)，所以 (E3) 量綱一致。",
          "en": "By L_const_factor, d(4gh)=d(gh)=(1,0,−2)+(1,0,0)=(2,0,−2)=d(v²), so (E3) is dimensionally consistent."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "L_const_factor",
            "vec_ops",
            "mko_dim"
          ],
          "explanation": {
            "zh-Hant": "同 S08，只是通過必要條件；S11 會顯示它仍然錯誤。",
            "en": "As in S08, it only passes the necessary condition; S11 shows that it is still wrong."
          }
        },
        "leaf_basis": "mko_dim + L_const_factor + vec_ops"
      },
      {
        "id": "S11",
        "uses": [
          "G3",
          "S10",
          "mko_dim_limits"
        ],
        "before_tex": "v=gt,\\ h=\\tfrac{1}{2}gt^2",
        "after_tex": "v^2=2gh,\\ \\text{so }v^2=4gh\\ \\text{fails for }g,t>0",
        "relation": "implies",
        "rule": "mko_dim_limits",
        "conditions": [
          "g>0,\\ t>0"
        ],
        "explanation": {
          "zh-Hant": "在 G3 的模型中 v²=2gh（S11.1），所以若 (E3) 也成立，則 2gh=4gh，即 v²=2v²，得 v²=0；但 v=gt>0，矛盾（S11.2）。(E3) 量綱一致卻在這個模型中錯誤：量綱一致是必要而非充分的條件，它查不出錯的係數 4。",
          "en": "In the model G3 we have v²=2gh (S11.1), so if (E3) also held then 2gh=4gh, that is v²=2v², giving v²=0; but v=gt>0, a contradiction (S11.2). (E3) is dimensionally consistent yet wrong in this model: consistency is necessary but not sufficient, and it cannot detect the wrong coefficient 4."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "G3",
            "mko_dim_limits"
          ],
          "explanation": {
            "zh-Hant": "需要明示的物理模型與 g、t>0；量綱分析本身無法提供這個判斷。",
            "en": "It needs the stated physical model and g, t>0; dimensional analysis alone cannot supply this verdict."
          }
        },
        "leaf_basis": "mko_dim_limits + contradiction"
      },
      {
        "id": "S11.1",
        "uses": [
          "G3"
        ],
        "before_tex": "v^2=(gt)^2,\\ 2gh=2g\\cdot\\tfrac{1}{2}gt^2",
        "after_tex": "v^2=g^2t^2=2gh",
        "relation": "equivalent",
        "rule": "power_laws",
        "conditions": [],
        "explanation": {
          "zh-Hant": "(gt)²=g²t²；2g·(½gt²)=g²t²。兩者相等，所以 v²=2gh（N5 為多項式恆等式核對）。",
          "en": "(gt)²=g²t²; 2g·(½gt²)=g²t². They are equal, so v²=2gh (N5 checks this as a polynomial identity)."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "power_laws",
            "mul_assoc",
            "mul_comm"
          ],
          "explanation": {
            "zh-Hant": "指數律與乘法交換結合律，雙向是同一個式子。",
            "en": "Exponent laws and commutativity and associativity of multiplication; both sides are the same expression."
          }
        },
        "leaf_basis": "power_laws + mul_assoc + mul_comm"
      },
      {
        "id": "S11.2",
        "uses": [
          "S11.1",
          "G3"
        ],
        "before_tex": "v^2=4gh\\ \\text{(assumed)}",
        "after_tex": "v^2=2v^2\\Rightarrow v^2=0\\Rightarrow v=0,\\ \\text{contradicting }v=gt>0",
        "relation": "implies",
        "rule": "contradiction",
        "conditions": [
          "g>0,\\ t>0"
        ],
        "explanation": {
          "zh-Hant": "假設 v²=4gh。由 S11.1，4gh=2·(2gh)=2v²，所以 v²=2v²，兩邊減 v² 得 v²=0，再依平方非負與 u²=0 ⇔ u=0 得 v=0。但 g>0、t>0 使 v=gt>0，矛盾。所以在此模型中 (E3) 不成立。",
          "en": "Assume v²=4gh. By S11.1, 4gh=2·(2gh)=2v², so v²=2v²; subtracting v² gives v²=0, and since u²=0 ⇔ u=0, v=0. But g>0 and t>0 make v=gt>0, a contradiction. So (E3) fails in this model."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S11.1",
            "square_nonneg",
            "order_basics",
            "contradiction"
          ],
          "explanation": {
            "zh-Hant": "需要 g、t>0；若 t=0（尚未運動），v=h=0，兩式都成立，這不能拿來支持 (E3)。",
            "en": "It needs g, t>0; at t=0 (before any motion) v=h=0 and both relations hold trivially, which gives no support to (E3)."
          }
        },
        "leaf_basis": "square_nonneg + order_basics + contradiction"
      },
      {
        "id": "S12",
        "uses": [
          "S02",
          "S03",
          "mko_dim"
        ],
        "before_tex": "m+v,\\ d(m)=(0,1,0),\\ d(v)=(1,0,-1)",
        "after_tex": "d(m)\\neq d(v):\\ m+v\\ \\text{not allowed}",
        "relation": "implies",
        "rule": "mko_dim",
        "conditions": [],
        "explanation": {
          "zh-Hant": "依 MKO，可相加的量必須量綱一致。(0,1,0) 與 (1,0,−1) 在三個分量上都不同，所以 m+v 不是有意義的量；也沒有「(1,1,−1)」這種把向量相加得到的「和的量綱」，向量相加只對應乘積。",
          "en": "By the MKO, quantities to be added must have matching dimensions. (0,1,0) and (1,0,−1) differ in all three components, so m+v is not a meaningful quantity; nor is there a 'dimension of the sum' such as (1,1,−1) obtained by adding vectors, since adding vectors corresponds to products only."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S02",
            "S03",
            "mko_dim",
            "vec_ops"
          ],
          "explanation": {
            "zh-Hant": "只需要兩個向量不同。",
            "en": "It needs only that the two vectors differ."
          }
        },
        "leaf_basis": "mko_dim + vec_ops"
      },
      {
        "id": "S13",
        "uses": [
          "S06",
          "S02",
          "S03",
          "G2"
        ],
        "before_tex": "(2,1,-2)=\\alpha(0,1,0)+\\beta(1,0,-1)",
        "after_tex": "\\alpha=1,\\ \\beta=2,\\ E=C\\,m\\,v^2",
        "relation": "equivalent",
        "rule": "vec_ops",
        "conditions": [
          "C\\ \\text{dimensionless}"
        ],
        "explanation": {
          "zh-Hant": "要求能量 E=C·m^α v^β 與 ½mv² 有相同量綱：右邊向量為 (β, α, −β)。逐分量比較：L 分量 β=2，M 分量 α=1，T 分量 −β=−2，與 β=2 一致。三個方程、兩個未知數，T 分量是真正的一致性檢查。所以 E=C m v²；量綱分析定不出 C，C=½ 要由物理論證決定。",
          "en": "Require the energy E=C·m^α v^β to have the dimension of ½mv²: the right side has vector (β, α, −β). Comparing components: L gives β=2, M gives α=1, and T gives −β=−2, consistent with β=2. Three equations in two unknowns, so the T component is a genuine consistency check. Hence E=C m v²; dimensional analysis cannot determine C, and C=½ must come from a physical argument."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "vec_ops",
            "cartesian_product_def"
          ],
          "explanation": {
            "zh-Hant": "向量相等 ⇔ 三個分量相等；代回 α=1、β=2 得 (2,1,−2)，雙向成立。",
            "en": "Vector equality ⇔ the three components are equal; substituting α=1, β=2 gives (2,1,−2), so both directions hold."
          }
        },
        "leaf_basis": "vec_ops + eq_congruence"
      },
      {
        "id": "S14",
        "uses": [
          "S06",
          "mko_dim_limits"
        ],
        "before_tex": "\\mathrm{J}=\\mathrm{kg\\,m^2\\,s^{-2}},\\ \\mathrm{erg}=\\mathrm{g\\,cm^2\\,s^{-2}}",
        "after_tex": "d=(2,1,-2)\\ \\text{for both},\\ 1\\,\\mathrm{J}=10^7\\,\\mathrm{erg}",
        "relation": "implies",
        "rule": "mko_dim_limits",
        "conditions": [],
        "explanation": {
          "zh-Hant": "焦耳與爾格都是能量單位，量綱向量同為 (2,1,−2)；但 1 g=10⁻³ kg、1 cm²=10⁻⁴ m²，所以 1 erg=10⁻⁷ J，即 1 J=10⁷ erg。量綱相同不代表數值相同；量綱分析不追蹤單位換算因子，單位名稱也不等於量綱指數。",
          "en": "The joule and the erg are both units of energy with dimension vector (2,1,−2); but 1 g=10⁻³ kg and 1 cm²=10⁻⁴ m², so 1 erg=10⁻⁷ J, that is 1 J=10⁷ erg. Equal dimensions do not mean equal numbers; dimensional analysis does not track unit conversion factors, and a unit name is not the same thing as a dimension exponent."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "mko_dim_limits",
            "power_laws",
            "numeric_evaluation"
          ],
          "explanation": {
            "zh-Hant": "需要兩組單位的定義與 10 的冪次計算；結論只是說明，與前面的量綱判定互不依賴。",
            "en": "It needs the definitions of the two unit systems and powers of 10; the conclusion is illustrative and independent of the earlier dimensional verdicts."
          }
        },
        "leaf_basis": "mko_dim_limits + power_laws + numeric_evaluation"
      },
      {
        "id": "S15",
        "uses": [
          "S08",
          "S09",
          "S10",
          "S11",
          "S12",
          "S13"
        ],
        "before_tex": "\\text{proved facts}",
        "after_tex": "\\text{(E1) consistent; (E2) inconsistent; (E3) consistent but false in G3; (E4) not allowed; }E=Cmv^2",
        "relation": "implies",
        "rule": "and_intro_elim",
        "conditions": [],
        "explanation": {
          "zh-Hant": "合併結論：(E1) 量綱一致；(E2) 不一致且純數無法修正；(E3) 一致但在等加速度模型中錯誤；m+v 不能相加；能量的量綱形式是 C m v²，C 由量綱分析定不出。",
          "en": "Combine the conclusions: (E1) is consistent; (E2) is inconsistent and cannot be repaired by a pure number; (E3) is consistent but false in the constant-acceleration model; m+v cannot be added; energy has the dimensional form C m v², with C not determined by dimensional analysis."
        },
        "reverse_audit": {
          "kind": "sufficient_goal_reduction",
          "required_ids": [
            "S08",
            "S09",
            "S10",
            "S11",
            "S12",
            "S13",
            "and_intro_elim"
          ],
          "explanation": {
            "zh-Hant": "目標 G：本稿目標的各項判定。足夠條件 H：S08 到 S13 各自已證（H⇒G 由合取引入）。H 由給定條件正向推出，沒有把 G 當作前提。",
            "en": "Goal G: the determinations in the goal. Sufficient condition H: S08 to S13, each already proved (H⇒G by conjunction introduction). H is derived forwards from the givens; G is never used as a premise."
          }
        },
        "leaf_basis": "and_intro_elim"
      }
    ],
    "result": {
      "zh-Hant": "d(v)=(1,0,−1)，d(g)=(1,0,−2)，d(½mv²)=d(mgh)=(2,1,−2)，所以 ½mv²=mgh 量綱一致。v=2gh 兩邊為 (1,0,−1) 與 (2,0,−2)，不一致，且任何無因次常數都無法修正。v²=4gh 量綱一致，但在等加速度模型中 v²=2gh，所以它錯誤：量綱一致只是必要條件。m+v 量綱不同，不能相加。由 (2,1,−2)=α(0,1,0)+β(1,0,−1) 得 α=1、β=2，E=C m v²，C 不由量綱決定。焦耳與爾格量綱相同但 1 J=10⁷ erg。",
      "en": "d(v)=(1,0,−1), d(g)=(1,0,−2), and d(½mv²)=d(mgh)=(2,1,−2), so ½mv²=mgh is dimensionally consistent. v=2gh has sides (1,0,−1) and (2,0,−2), which differ, and no dimensionless constant can repair it. v²=4gh is consistent, but in the constant-acceleration model v²=2gh, so it is false: consistency is only necessary. m and v have different dimensions and cannot be added. From (2,1,−2)=α(0,1,0)+β(1,0,−1), α=1 and β=2, so E=C m v² with C not fixed by dimensions. The joule and the erg have the same dimension but 1 J=10⁷ erg."
    },
    "negative_controls": [
      {
        "claim": {
          "zh-Hant": "v²=4gh 量綱一致，所以它是正確的自由落體關係",
          "en": "v²=4gh is dimensionally consistent, so it is the correct free-fall relation"
        },
        "expected": "reject",
        "reason": {
          "zh-Hant": "等加速度模型給出 v²=2gh；若 v²=4gh 也成立則 v=0，與 v=gt>0 矛盾（S11.2）。",
          "en": "The constant-acceleration model gives v²=2gh; if v²=4gh also held then v=0, contradicting v=gt>0 (S11.2)."
        }
      },
      {
        "claim": {
          "zh-Hant": "只要選對無因次常數 k，v=k·gh 就會量綱一致",
          "en": "With the right dimensionless constant k, v=k·gh becomes dimensionally consistent"
        },
        "expected": "reject",
        "reason": {
          "zh-Hant": "d(k·gh)=(2,0,−2)≠(1,0,−1)=d(v)，與 k 無關（S09、L_const_factor）。",
          "en": "d(k·gh)=(2,0,−2)≠(1,0,−1)=d(v), whatever k is (S09, L_const_factor)."
        }
      },
      {
        "claim": {
          "zh-Hant": "m+v 是有意義的量，量綱是 (0,1,0)+(1,0,−1)=(1,1,−1)",
          "en": "m+v is a meaningful quantity with dimension (0,1,0)+(1,0,−1)=(1,1,−1)"
        },
        "expected": "reject",
        "reason": {
          "zh-Hant": "只有量綱相同的量才能相加；向量相加對應的是乘積 mv，不是和（S12）。",
          "en": "Only quantities with the same dimension can be added; adding vectors corresponds to the product mv, not the sum (S12)."
        }
      },
      {
        "claim": {
          "zh-Hant": "焦耳與爾格量綱相同，所以 1 J=1 erg",
          "en": "The joule and the erg have the same dimension, so 1 J=1 erg"
        },
        "expected": "reject",
        "reason": {
          "zh-Hant": "1 J=10⁷ erg；量綱分析不追蹤換算因子（S14）。",
          "en": "1 J=10⁷ erg; dimensional analysis does not track conversion factors (S14)."
        }
      }
    ],
    "status": "candidate",
    "gaps": [],
    "verification": {
      "structure": "checked_local_script",
      "mathematics": "candidate_argument",
      "formalization": "not_formalized"
    },
    "rule_glossary": [
      {
        "id": "cartesian_product_def",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "笛卡兒積：A×B 是所有有序對 (a,b)（a∈A、b∈B）的集合；(a,b)=(c,d) ⇔ a=c 且 b=d。",
          "en": "Cartesian product: A×B is the set of ordered pairs (a,b) with a∈A and b∈B; (a,b)=(c,d) ⇔ a=c and b=d."
        }
      },
      {
        "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": "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": "power_laws",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "整數指數律（底數非零或指數為正整數）：a^m·a^n=a^{m+n}，(a^m)^n=a^{mn}，(ab)^n=a^n b^n，a^0=1。",
          "en": "Integer exponent laws (base nonzero, or exponents positive integers): a^m·a^n=a^{m+n}, (a^m)^n=a^{mn}, (ab)^n=a^n b^n, a^0=1."
        }
      },
      {
        "id": "distributive",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "分配律：u(v+w)=uv+uw；配合乘法交換律也有 (v+w)u=vu+wu。",
          "en": "Distributivity: u(v+w)=uv+uw; with commutativity also (v+w)u=vu+wu."
        }
      },
      {
        "id": "def_intro",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "引入定義：可為一個命題、式子或物件取名；之後名稱與被定義者可互相替換。",
          "en": "Introducing a definition: a statement, expression or object may be given a name, after which the name and what it stands for are interchangeable."
        }
      },
      {
        "id": "add_zero",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "加法單位元：u+0=u。",
          "en": "Additive identity: u+0=u."
        }
      },
      {
        "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": "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": "square_nonneg",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "實數平方非負：對所有實數 u，u²≥0，且 u²=0 ⇔ u=0。",
          "en": "Squares are nonnegative: for every real u, u²≥0, and u²=0 ⇔ u=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": "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_dim",
        "kind": "cited",
        "source": {
          "mko_id": "mko-dimensional-analysis",
          "sha256": "05ee29db7e8f4b787e36b2e536eb249e1b92597007252752dd6347b2dffab5eb",
          "commit": "b28a5edbd46279b1db2a3f3f48bb3f447e18899f"
        },
        "statement": {
          "zh-Hant": "相對基本量綱 L、M、T，把物理量 Q 的量綱記為 [Q]=L^a M^b T^c，亦即指數向量 (a,b,c)。乘法相加向量、除法相減、整數冪倍乘；可相等或相加的量必須量綱一致。",
          "en": "Relative to basic dimensions L,M,T, write [Q]=L^a M^b T^c, represented by (a,b,c). Products add vectors, quotients subtract them and integer powers scale them. Quantities to be equated or added must have matching dimensions."
        }
      },
      {
        "id": "mko_dim_limits",
        "kind": "cited",
        "source": {
          "mko_id": "mko-dimensional-analysis",
          "sha256": "05ee29db7e8f4b787e36b2e536eb249e1b92597007252752dd6347b2dffab5eb",
          "commit": "b28a5edbd46279b1db2a3f3f48bb3f447e18899f"
        },
        "statement": {
          "zh-Hant": "量綱一致是候選物理方程的必要條件，不是充分條件；方程若兩側量綱不同，不能由調整純數常數修正；若一致，仍可能有錯係數。量綱分析不追蹤單位換算的數值因子。",
          "en": "Dimensional consistency is necessary but not sufficient for a physical equation; if the two sides have different dimensions, adjusting a pure-number constant cannot fix it, and if they agree there may still be a wrong coefficient. Dimensional analysis does not track the numerical factors of unit conversion."
        }
      },
      {
        "id": "vec_ops",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "ℤ³ 的逐分量運算：(a,b,c)±(a′,b′,c′)=(a±a′,b±b′,c±c′)，k(a,b,c)=(ka,kb,kc)；兩向量相等當且僅當三個分量分別相等；純數的量綱向量是 (0,0,0)。",
          "en": "Componentwise operations in ℤ³: (a,b,c)±(a′,b′,c′)=(a±a′,b±b′,c±c′) and k(a,b,c)=(ka,kb,kc); two vectors are equal exactly when the three components are equal; a pure number has dimension vector (0,0,0)."
        }
      },
      {
        "id": "L_const_factor",
        "kind": "lemma",
        "proved_in": [
          "S09.1"
        ],
        "statement": {
          "zh-Hant": "對無因次常數 k 與任意量 Q，d(kQ)=d(Q)。",
          "en": "For a dimensionless constant k and any quantity Q, d(kQ)=d(Q)."
        }
      }
    ],
    "revision_log": [],
    "numeric_checks": [
      {
        "id": "N1",
        "step": "S06",
        "lhs": "0+0+2",
        "rhs": "2"
      },
      {
        "id": "N2",
        "step": "S06",
        "lhs": "0+0-2",
        "rhs": "-2"
      },
      {
        "id": "N3",
        "step": "S07",
        "lhs": "0+1+1",
        "rhs": "2"
      },
      {
        "id": "N4",
        "step": "S09",
        "lhs": "1",
        "relation": "!=",
        "rhs": "0+1+1"
      },
      {
        "id": "N5",
        "step": "S11.1",
        "vars": [
          "g",
          "t"
        ],
        "degree": 2,
        "lhs": "(g*t)**2",
        "rhs": "2*g*(g*t**2/2)"
      },
      {
        "id": "N6",
        "step": "S11.2",
        "vars": [
          "g",
          "t"
        ],
        "degree": 2,
        "lhs": "4*g*(g*t**2/2)",
        "rhs": "2*(g*t)**2"
      },
      {
        "id": "N7",
        "step": "S13",
        "lhs": "-1*2",
        "rhs": "-2"
      },
      {
        "id": "N8",
        "step": "S14",
        "lhs": "(10**3)*(10**2)**2",
        "rhs": "10**7"
      }
    ]
  },
  "presentation": {
    "md_scope": {
      "zh-Hant": "只用 L、M、T 三個基本量綱與整數指數（MKO 假設 a1）；½、2、4 明示為無因次常數，g 是具量綱的量（假設 a2）。S11 用到的等加速度運動模型明列於 G3，不由量綱分析推出。不處理 SI 換算細節、Buckingham Pi 或無因次化；不改動 canonical MKO 或英文 edition。",
      "en": "Uses only the three base dimensions L, M, T and integer exponents (MKO assumption a1); ½, 2 and 4 are declared dimensionless, and g is a dimensioned quantity (assumption a2). The constant-acceleration model used in S11 is stated in G3 and is not derived by dimensional analysis. SI conversion details, Buckingham Pi and nondimensionalization are not treated; the canonical MKO and its English edition are not changed."
    },
    "md_symbols": [
      [
        "d(Q)=(a,b,c)",
        "量 Q 的量綱指數向量，[Q]=L^a M^b T^c",
        "exponent vector of the quantity Q, [Q]=L^a M^b T^c"
      ],
      [
        "m,\\ h,\\ t",
        "質量、高度、時間",
        "mass, height, time"
      ],
      [
        "v,\\ g",
        "速度、加速度",
        "speed, acceleration"
      ],
      [
        "E",
        "能量（動能）",
        "energy (kinetic)"
      ],
      [
        "C,\\ \\alpha,\\ \\beta",
        "無因次常數與待求指數",
        "dimensionless constant and unknown exponents"
      ]
    ],
    "md_error_controls": [
      {
        "title": "量綱一致不等於方程正確 / Dimensional consistency does not make an equation correct",
        "zh": "MKO 指出的誤解之一：v²=4gh 兩邊都是 (2,0,−2)，但在等加速度模型中正確的是 v²=2gh（S11）。量綱分析查不出錯的無因次係數、錯的方向或錯的物理假設；它只能排除不一致的式子。",
        "en": "One misconception named by the MKO: both sides of v²=4gh are (2,0,−2), but in the constant-acceleration model the correct relation is v²=2gh (S11). Dimensional analysis cannot detect a wrong dimensionless coefficient, a wrong direction or a wrong physical assumption; it can only rule out inconsistent expressions."
      },
      {
        "title": "單位名稱不等於量綱指數 / A unit name is not a dimension exponent",
        "zh": "MKO 指出的另一個誤解：焦耳與爾格的量綱向量相同，數值卻差 10⁷ 倍（S14）。另外要注意記號：公式中的 m 是質量（量綱 M），單位「m」是公尺（量綱 L），兩者不可混淆。",
        "en": "Another misconception named by the MKO: the joule and the erg have the same dimension vector but differ by a factor of 10⁷ (S14). Notation also needs care: the m in the formulas is mass (dimension M), while the unit 'm' is the metre (dimension L); they must not be confused."
      },
      {
        "title": "純數常數修不好不一致，具量綱常數要明示 / Pure numbers cannot fix an inconsistency; dimensioned constants must be declared",
        "zh": "依 L_const_factor，乘上任何無因次 k 都不改變量綱，所以 v=k·gh 對每個 k 都不一致（S09）。反之，g 是具量綱的量 (1,0,−2)，若把它誤當純數，(E1) 的核對就會出錯；MKO 假設 a2 要求常數的量綱由題目明示。",
        "en": "By L_const_factor, multiplying by any dimensionless k leaves the dimension unchanged, so v=k·gh is inconsistent for every k (S09). Conversely, g is a dimensioned quantity (1,0,−2); treating it as a pure number would spoil the check of (E1). MKO assumption a2 requires the dimensions of constants to be stated."
      }
    ],
    "md_open_items": [
      {
        "zh-Hant": "S11 的等加速度模型（v=gt、h=gt²/2）作為給定條件使用，本稿沒有從力學原理推導它。",
        "en": "The constant-acceleration model of S11 (v=gt, h=gt²/2) is used as a given; this guide does not derive it from mechanics."
      },
      {
        "zh-Hant": "SI 換算、Buckingham Pi 定理與無因次化屬於 MKO 的 formal next。",
        "en": "SI conversion, the Buckingham Pi theorem and nondimensionalization belong to the MKO's formal next step."
      }
    ]
  },
  "revision": {
    "id": "production-20261019-adopted-96",
    "source_commit": "b28a5edbd46279b1db2a3f3f48bb3f447e18899f",
    "adopted_source_sha256": "d37c1bae6edd943432231306417829058ff4d2192474aeccdb559f8b130beb94",
    "original_source_sha256": "d37c1bae6edd943432231306417829058ff4d2192474aeccdb559f8b130beb94",
    "owner_adopted": true,
    "bilingual_sync": "complete_source_bound",
    "formal_Lean": "NOT_RUN",
    "independent_AI_concurrence": "NOT_MEASURED",
    "issue_ids": [],
    "scope": "Owner-adopted natural-language teaching guide; not a formal proof, canonical MKO rewrite or evidence-status promotion."
  }
}