{ "schema_version": "ocme-derivation-reader-v1", "guide_id": "derivation-mko-general-linear-map-01", "mko_id": "mko-general-linear-map", "title": { "zh-Hant": "實向量空間間線性映射：矩陣映射的全稱證書與仿射、黑箱反例", "en": "Linear maps between real vector spaces: a universal certificate for a matrix map, with affine and black-box counterexamples" }, "contract": { "schema_version": "ocme-derivation-guide-v0.1", "guide_id": "derivation-mko-general-linear-map-01", "source": { "mko_id": "mko-general-linear-map", "commit": "b28a5edbd46279b1db2a3f3f48bb3f447e18899f", "sha256": "184a885b199aff8dc05ec273c0fe5b2dcdd38a9ef98c8ec790ce2668da9a07c7", "english_sha256": "efdc1e4563905094acbcc230800a69094676b62be7baef1646756189ac8456f0" }, "mode": "worked_example", "audience_prerequisites": [ "vector_space_axioms", "matrix_product_def", "distributive", "finite_sum_rules", "numeric_evaluation" ], "goal": { "zh-Hant": "設 A 為第一行 (2,−1,0)、第二行 (1,3,4) 的 2×3 整數矩陣，T:ℝ³→ℝ²，T(x)=Ax。（一）證明對所有 u,v∈ℝ³ 與所有實數 α,β，T(αu+βv)=αT(u)+βT(v)。（二）證明 S(x)=Ax+b（b=(1,0)）不是線性映射，並證明對任意 b∈ℝ²，x↦Ax+b 線性當且僅當 b=0。（三）說明黑箱 F(x)=Ax+(x₁²−x₁,0) 通過一個樣本測試，卻不是線性映射。", "en": "Let A be the 2-by-3 integer matrix with first row (2,−1,0) and second row (1,3,4), and let T:ℝ³→ℝ², T(x)=Ax. (1) Prove that T(αu+βv)=αT(u)+βT(v) for all u,v∈ℝ³ and all reals α,β. (2) Prove that S(x)=Ax+b with b=(1,0) is not linear, and that for every b∈ℝ² the map x↦Ax+b is linear if and only if b=0. (3) Show that the black box F(x)=Ax+(x₁²−x₁,0) passes one sample test and yet is not linear." }, "givens": [ { "id": "G1", "statement": { "zh-Hant": "ℝ³ 與 ℝ² 是實向量空間，加法與實數倍數逐坐標進行；兩向量相等當且僅當每個坐標相等。", "en": "ℝ³ and ℝ² are real vector spaces with coordinatewise addition and real scalar multiplication; two vectors are equal exactly when every coordinate agrees." } }, { "id": "G2", "statement": { "zh-Hant": "A 的第一行為 (2,−1,0)，第二行為 (1,3,4)；T(x)=Ax，x∈ℝ³。", "en": "A has first row (2,−1,0) and second row (1,3,4); T(x)=Ax for x∈ℝ³." } }, { "id": "G3", "statement": { "zh-Hant": "仿射映射 S_b(x)=Ax+b，b∈ℝ²；特例 S=S_b，b=(1,0)。", "en": "Affine maps S_b(x)=Ax+b with b∈ℝ²; in particular S=S_b with b=(1,0)." } }, { "id": "G4", "statement": { "zh-Hant": "黑箱 F(x)=Ax+(x₁²−x₁,0)；樣本測試只檢查 u=e₁、v=e₂、α=β=1 一組輸入。", "en": "Black box F(x)=Ax+(x₁²−x₁,0); the sample test checks only the single input u=e₁, v=e₂, α=β=1." } }, { "id": "G5", "statement": { "zh-Hant": "一致性核對用的數字：u=(1,0,2)、v=(0,1,−1)、α=3、β=−2。", "en": "Numbers for a consistency check: u=(1,0,2), v=(0,1,−1), α=3, β=−2." } } ], "steps": [ { "id": "S01", "uses": [ "G2" ], "before_tex": "T(x)=Ax,\\ A=\\begin{pmatrix}2&-1&0\\\\1&3&4\\end{pmatrix}", "after_tex": "T(x)=(2x_1-x_2+0x_3,\\ x_1+3x_2+4x_3)", "relation": "definition", "rule": "matrix_product_def", "conditions": [], "explanation": { "zh-Hant": "依矩陣乘積定義，(Ax)_i=Σ_k A_{ik}x_k。第一行 (2,−1,0) 給出 2x₁−x₂+0x₃，第二行 (1,3,4) 給出 x₁+3x₂+4x₃。之後每一步都在這個坐標公式上進行；保留 0x₃ 是為了讓每一行都寫成同一種和式。", "en": "By the definition of the matrix product, (Ax)_i=Σ_k A_{ik}x_k. Row (2,−1,0) gives 2x₁−x₂+0x₃ and row (1,3,4) gives x₁+3x₂+4x₃. Every later step works on this coordinate formula; the term 0x₃ is kept so that each row has the same form of a sum." }, "reverse_audit": { "kind": "premise_audit", "required_ids": [ "G2", "matrix_product_def" ], "explanation": { "zh-Hant": "此步只展開定義，不推出新事實；需要的只是 A 的兩行（G2）與乘積定義。尺寸相容：A 是 2×3，x 是 3 維，所以 Ax 是 2 維。", "en": "This step only unfolds a definition and derives no new fact; it needs only the two rows of A (G2) and the product definition. Sizes match: A is 2-by-3 and x has 3 coordinates, so Ax has 2." } }, "leaf_basis": "matrix_product_def" }, { "id": "S02", "uses": [ "G1", "mko_linear_map_def" ], "before_tex": "T\\ \\text{linear}", "after_tex": "\\forall u,v\\in\\mathbb{R}^3\\ \\forall\\alpha,\\beta\\in\\mathbb{R}:\\ T(\\alpha u+\\beta v)=\\alpha T(u)+\\beta T(v)", "relation": "definition", "rule": "mko_linear_map_def", "conditions": [], "explanation": { "zh-Hant": "依 MKO，線性就是這一條對所有 u、v、α、β 成立的等式。為了證明一個全稱命題，固定任意的 u、v∈ℝ³ 與 α、β∈ℝ，對它們不加任何額外假設；S03 到 S06 只用這些一般符號。", "en": "By the MKO, linearity is this single equation holding for all u, v, α, β. To prove a universal statement, fix arbitrary u, v∈ℝ³ and α, β∈ℝ with no further assumption on them; S03 to S06 use only these general symbols." }, "reverse_audit": { "kind": "premise_audit", "required_ids": [ "mko_linear_map_def", "G1" ], "explanation": { "zh-Hant": "此步只寫出目標並固定任意元素；之後若對 u、v、α、β 偷加條件（例如取特定數值），S07 的全稱化就不成立。", "en": "This step only states the goal and fixes arbitrary elements; if any condition were later imposed on u, v, α, β (such as particular values), the generalization in S07 would fail." } }, "leaf_basis": "mko_linear_map_def + def_unfold" }, { "id": "S03", "uses": [ "S02", "G1" ], "before_tex": "\\alpha u+\\beta v", "after_tex": "(\\alpha u_1+\\beta v_1,\\ \\alpha u_2+\\beta v_2,\\ \\alpha u_3+\\beta v_3)", "relation": "equivalent", "rule": "coord_vec_ops", "conditions": [], "explanation": { "zh-Hant": "在 ℝ³ 中純量乘法與加法都逐坐標進行，所以 αu+βv 的第 k 坐標是 αu_k+βv_k。兩邊是同一個向量的兩種寫法。", "en": "In ℝ³ scalar multiplication and addition act coordinatewise, so the k-th coordinate of αu+βv is αu_k+βv_k. Both sides are two ways of writing the same vector." }, "reverse_audit": { "kind": "inverse_equivalence", "required_ids": [ "coord_vec_ops" ], "explanation": { "zh-Hant": "反向：由坐標 (αu_k+βv_k) 依同一規則重組回 αu+βv；兩向量逐坐標相等，所以雙向成立。", "en": "Backward: the coordinates (αu_k+βv_k) reassemble into αu+βv by the same rule; the vectors agree coordinate by coordinate, so the step holds both ways." } }, "leaf_basis": "coord_vec_ops + vector_space_axioms" }, { "id": "S04", "uses": [ "S01", "S03" ], "before_tex": "[T(\\alpha u+\\beta v)]_1=2(\\alpha u_1+\\beta v_1)-(\\alpha u_2+\\beta v_2)+0(\\alpha u_3+\\beta v_3)", "after_tex": "[T(\\alpha u+\\beta v)]_1=\\alpha(2u_1-u_2+0u_3)+\\beta(2v_1-v_2+0v_3)", "relation": "equivalent", "rule": "L_row_linear", "conditions": [], "explanation": { "zh-Hant": "把 S03 的坐標代入 S01 的第一行公式，得到左式。依引理 L_row_linear（n=3，a=(2,−1,0)），和式可拆成 α 乘第一行對 u 的和，加上 β 乘第一行對 v 的和。引理在子步 S04.1、S04.2 證明，只用分配律與有限和規則，所以對每個 u、v、α、β 都成立。", "en": "Substituting the coordinates of S03 into the first-row formula of S01 gives the left side. By Lemma L_row_linear (n=3, a=(2,−1,0)) the sum splits into α times the first-row sum for u plus β times the first-row sum for v. The lemma is proved in sub-steps S04.1 and S04.2 using only distributivity and the rules for finite sums, so it holds for every u, v, α, β." }, "reverse_audit": { "kind": "inverse_equivalence", "required_ids": [ "L_row_linear" ], "explanation": { "zh-Hant": "這是對所有實數成立的恆等式；反向是把 α、β 再分配回每一項。數值核對 N1 以多項式恆等式方式重算同一式子。", "en": "This is an identity valid for all reals; the backward move distributes α and β back into each term. Numeric check N1 recomputes the same expression as a polynomial identity." } }, "leaf_basis": "distributive + mul_assoc + mul_comm + finite_sum_rules" }, { "id": "S04.1", "uses": [ "distributive" ], "before_tex": "a_k(\\alpha u_k+\\beta v_k)", "after_tex": "\\alpha(a_k u_k)+\\beta(a_k v_k)", "relation": "equivalent", "rule": "distributive", "conditions": [], "explanation": { "zh-Hant": "對單一項用分配律：a_k(αu_k+βv_k)=a_k(αu_k)+a_k(βv_k)；再用乘法結合律與交換律，a_k(αu_k)=α(a_k u_k)，a_k(βv_k)=β(a_k v_k)。", "en": "Apply distributivity to one term: a_k(αu_k+βv_k)=a_k(αu_k)+a_k(βv_k); then associativity and commutativity of multiplication give a_k(αu_k)=α(a_k u_k) and a_k(βv_k)=β(a_k v_k)." }, "reverse_audit": { "kind": "inverse_equivalence", "required_ids": [ "distributive", "mul_assoc", "mul_comm" ], "explanation": { "zh-Hant": "所用都是實數恆等式，對所有 a_k、u_k、v_k、α、β 成立，沒有附加條件，兩向都成立。", "en": "Only real-number identities are used, valid for all a_k, u_k, v_k, α, β with no side condition, so both directions hold." } }, "leaf_basis": "distributive + mul_assoc + mul_comm" }, { "id": "S04.2", "uses": [ "S04.1", "finite_sum_rules" ], "before_tex": "\\sum_{k=1}^{n}a_k(\\alpha u_k+\\beta v_k)=\\sum_{k=1}^{n}\\big(\\alpha(a_k u_k)+\\beta(a_k v_k)\\big)", "after_tex": "\\sum_{k=1}^{n}a_k(\\alpha u_k+\\beta v_k)=\\alpha\\sum_{k=1}^{n}a_k u_k+\\beta\\sum_{k=1}^{n}a_k v_k", "relation": "equivalent", "rule": "finite_sum_rules", "conditions": [], "explanation": { "zh-Hant": "左式由 S04.1 逐項改寫而來。有限和規則 Σ(x_k+y_k)=Σx_k+Σy_k 把和拆成兩個和，Σc·x_k=c·Σx_k 把常數 α、β 提出。S04.1 與 S04.2 合起來證得引理 L_row_linear；證明對任意 n 與任意一行係數都成立。", "en": "The left side comes from rewriting each term by S04.1. The finite-sum rule Σ(x_k+y_k)=Σx_k+Σy_k splits the sum in two, and Σc·x_k=c·Σx_k pulls out the constants α and β. S04.1 and S04.2 together prove Lemma L_row_linear, for every n and every row of coefficients." }, "reverse_audit": { "kind": "inverse_equivalence", "required_ids": [ "S04.1", "finite_sum_rules" ], "explanation": { "zh-Hant": "有限和規則是等式，兩向都可用；n 是有限數，所以不涉及收斂問題。", "en": "The finite-sum rules are equations usable in both directions; n is finite, so no convergence question arises." } }, "leaf_basis": "finite_sum_rules + eq_replace" }, { "id": "S05", "uses": [ "S01", "S03", "L_row_linear" ], "before_tex": "[T(\\alpha u+\\beta v)]_2=(\\alpha u_1+\\beta v_1)+3(\\alpha u_2+\\beta v_2)+4(\\alpha u_3+\\beta v_3)", "after_tex": "[T(\\alpha u+\\beta v)]_2=\\alpha(u_1+3u_2+4u_3)+\\beta(v_1+3v_2+4v_3)", "relation": "equivalent", "rule": "L_row_linear", "conditions": [], "explanation": { "zh-Hant": "第二行同理：代入 S03 的坐標，再用引理 L_row_linear（a=(1,3,4)）把 α 與 β 提出。這裡沒有任何數值代入，u、v、α、β 仍是任意的。", "en": "The second row is handled the same way: substitute the coordinates of S03 and use Lemma L_row_linear with a=(1,3,4) to pull out α and β. No numbers are substituted; u, v, α, β remain arbitrary." }, "reverse_audit": { "kind": "inverse_equivalence", "required_ids": [ "L_row_linear" ], "explanation": { "zh-Hant": "恆等式，兩向成立；N2 以多項式恆等式核對同一式子。", "en": "An identity, valid both ways; N2 checks the same expression as a polynomial identity." } }, "leaf_basis": "distributive + finite_sum_rules" }, { "id": "S06", "uses": [ "S01", "S04", "S05", "G1" ], "before_tex": "[T(\\alpha u+\\beta v)]_i=[\\alpha T(u)+\\beta T(v)]_i\\ \\ (i=1,2)", "after_tex": "T(\\alpha u+\\beta v)=\\alpha T(u)+\\beta T(v)", "relation": "equivalent", "rule": "coord_vec_ops", "conditions": [], "explanation": { "zh-Hant": "由 S01，T(u)=(2u₁−u₂+0u₃, u₁+3u₂+4u₃)，T(v) 同理；逐坐標運算給出 αT(u)+βT(v) 的兩個坐標正是 S04、S05 的右式。兩個坐標都相等，依 G1 兩個 ℝ² 向量相等。", "en": "By S01, T(u)=(2u₁−u₂+0u₃, u₁+3u₂+4u₃) and likewise for T(v); coordinatewise operations show that the two coordinates of αT(u)+βT(v) are exactly the right sides of S04 and S05. Both coordinates agree, so by G1 the two vectors of ℝ² are equal." }, "reverse_audit": { "kind": "inverse_equivalence", "required_ids": [ "coord_vec_ops", "S04", "S05" ], "explanation": { "zh-Hant": "ℝ² 中向量相等當且僅當兩個坐標都相等，所以這一步雙向成立；只有一個坐標相等並不夠。", "en": "Vectors of ℝ² are equal exactly when both coordinates agree, so the step holds both ways; one agreeing coordinate alone would not suffice." } }, "leaf_basis": "coord_vec_ops + eq_replace" }, { "id": "S07", "uses": [ "S02", "S06" ], "before_tex": "u,v\\in\\mathbb{R}^3,\\ \\alpha,\\beta\\in\\mathbb{R}\\ \\text{arbitrary}:\\ T(\\alpha u+\\beta v)=\\alpha T(u)+\\beta T(v)", "after_tex": "\\forall u,v\\ \\forall\\alpha,\\beta:\\ T(\\alpha u+\\beta v)=\\alpha T(u)+\\beta T(v),\\ \\text{so }T\\ \\text{is linear}", "relation": "implies", "rule": "forall_intro", "conditions": [], "explanation": { "zh-Hant": "S03 到 S06 對任意固定的 u、v、α、β 成立，沒有用到它們的任何特殊性質，所以依全稱引入得到 MKO 的線性等式，T 是線性映射。這正是 MKO 所說的「矩陣分配律提供全稱證書」。同一論證把 L_row_linear 用在任一矩陣的每一行，所以每個矩陣映射 x↦Ax 都是線性的。", "en": "S03 to S06 hold for arbitrary fixed u, v, α, β and use no special property of them, so universal generalization gives the MKO's linearity equation: T is linear. This is what the MKO calls the universal certificate supplied by matrix distributivity. The same argument, applying L_row_linear to each row of any matrix, shows that every matrix map x↦Ax is linear." }, "reverse_audit": { "kind": "premise_audit", "required_ids": [ "S02", "S06", "forall_intro" ], "explanation": { "zh-Hant": "全稱引入只在元素真正任意時成立；S03 到 S06 沒有代入數值。這一步不宣稱反向，也不依賴 S08 的樣本。", "en": "Generalization is valid only because the elements were truly arbitrary; S03 to S06 substitute no numbers. No converse is claimed, and the step does not depend on the sample in S08." } }, "leaf_basis": "forall_intro + mko_linear_map_def" }, { "id": "S08", "uses": [ "S01", "G5" ], "before_tex": "u=(1,0,2),\\ v=(0,1,-1),\\ \\alpha=3,\\ \\beta=-2", "after_tex": "T(3u-2v)=T(3,-2,8)=(8,29)=3(2,9)-2(-1,-1)=3T(u)-2T(v)", "relation": "implies", "rule": "numeric_evaluation", "conditions": [], "explanation": { "zh-Hant": "一致性核對：3u−2v=(3,0,6)+(0,−2,2)=(3,−2,8)；T(3,−2,8)=(2·3−(−2), 3+3·(−2)+4·8)=(8,29)。另一邊 T(u)=(2,9)、T(v)=(−1,−1)，3T(u)−2T(v)=(6+2, 27+2)=(8,29)。兩邊相同，與 S07 一致；但這只是一組輸入，不是證明。", "en": "Consistency check: 3u−2v=(3,0,6)+(0,−2,2)=(3,−2,8); T(3,−2,8)=(2·3−(−2), 3+3·(−2)+4·8)=(8,29). On the other side T(u)=(2,9) and T(v)=(−1,−1), so 3T(u)−2T(v)=(6+2, 27+2)=(8,29). The two sides agree, consistent with S07; but this is one input and not a proof." }, "reverse_audit": { "kind": "premise_audit", "required_ids": [ "S01", "G5", "numeric_evaluation" ], "explanation": { "zh-Hant": "可由 N3 到 N6 重算。這一步只支持這組數字；全稱結論來自 S07，S14 的黑箱說明單一樣本通過不足以推出線性。", "en": "It can be recomputed by N3 to N6. The step supports only these numbers; the universal conclusion comes from S07, and the black box of S14 shows that passing one sample does not imply linearity." } }, "leaf_basis": "numeric_evaluation + coord_vec_ops" }, { "id": "S09", "uses": [ "S07" ], "before_tex": "\\forall u,v,\\alpha,\\beta:\\ T(\\alpha u+\\beta v)=\\alpha T(u)+\\beta T(v)", "after_tex": "T(u+v)=T(u)+T(v),\\ \\ T(\\alpha u)=\\alpha T(u)", "relation": "implies", "rule": "forall_elim", "conditions": [], "explanation": { "zh-Hant": "MKO 指出單一等式同時涵蓋加法與齊次性。取 α=β=1 得可加性；取 β=0、v=0，因 0·0=0、0·T(0)=0（逐坐標 mul_zero），得 T(αu)=αT(u)。所以證明單一等式就同時證明了兩個性質。", "en": "The MKO notes that the single equation covers both additivity and homogeneity. Taking α=β=1 gives additivity; taking β=0 and v=0, since 0·0=0 and 0·T(0)=0 (coordinatewise mul_zero), gives T(αu)=αT(u). So proving the single equation proves both properties at once." }, "reverse_audit": { "kind": "premise_audit", "required_ids": [ "S07", "forall_elim", "mul_zero" ], "explanation": { "zh-Hant": "全稱命題可代入特例；反向（兩個性質合起來推出單一等式）也成立，但本步不使用也不宣稱。", "en": "A universal statement may be specialized; the converse (both properties together give the single equation) is also true, but this step neither uses nor claims it." } }, "leaf_basis": "forall_elim + mul_zero + add_zero" }, { "id": "S10", "uses": [ "mko_linear_map_def" ], "before_tex": "L:\\mathbb{R}^n\\to\\mathbb{R}^m,\\ \\forall u,v,\\alpha,\\beta:\\ L(\\alpha u+\\beta v)=\\alpha L(u)+\\beta L(v)", "after_tex": "L(0)=0", "relation": "implies", "rule": "L_zero_to_zero", "conditions": [], "explanation": { "zh-Hant": "要判斷仿射映射，先建立一個必要條件：每個線性映射都把零向量送到零向量（引理 L_zero_to_zero）。證明在子步 S10.1、S10.2，只用線性等式在 α=β=0 時的特例。", "en": "To judge affine maps, first establish a necessary condition: every linear map sends the zero vector to the zero vector (Lemma L_zero_to_zero). The proof is in sub-steps S10.1 and S10.2 and uses only the case α=β=0 of the linearity equation." }, "reverse_audit": { "kind": "premise_audit", "required_ids": [ "mko_linear_map_def", "L_zero_to_zero" ], "explanation": { "zh-Hant": "L(0)=0 是必要條件，不是充分條件：F(0)=0 但 F 不線性（S15），所以反向不宣稱。", "en": "L(0)=0 is necessary, not sufficient: F(0)=0 yet F is not linear (S15), so no converse is claimed." } }, "leaf_basis": "forall_elim + mul_zero" }, { "id": "S10.1", "uses": [ "coord_vec_ops", "mul_zero" ], "before_tex": "0u+0v", "after_tex": "0_{\\mathbb{R}^n}", "relation": "equivalent", "rule": "mul_zero", "conditions": [], "explanation": { "zh-Hant": "逐坐標：(0u+0v)_k=0·u_k+0·v_k=0+0=0，對每個 k 成立，所以 0u+0v 是零向量。這裡 u、v 可以是任意向量。", "en": "Coordinatewise: (0u+0v)_k=0·u_k+0·v_k=0+0=0 for every k, so 0u+0v is the zero vector. Here u and v may be any vectors." }, "reverse_audit": { "kind": "inverse_equivalence", "required_ids": [ "mul_zero", "add_zero", "coord_vec_ops" ], "explanation": { "zh-Hant": "兩邊是同一個向量（每個坐標都是 0），所以雙向成立。", "en": "Both sides are the same vector (every coordinate is 0), so the step holds both ways." } }, "leaf_basis": "mul_zero + add_zero + coord_vec_ops" }, { "id": "S10.2", "uses": [ "S10.1", "forall_elim" ], "before_tex": "L(0_{\\mathbb{R}^n})=L(0u+0v)", "after_tex": "L(0_{\\mathbb{R}^n})=0L(u)+0L(v)=0_{\\mathbb{R}^m}", "relation": "implies", "rule": "forall_elim", "conditions": [], "explanation": { "zh-Hant": "把線性等式代入 α=β=0：L(0u+0v)=0L(u)+0L(v)。右邊在 ℝᵐ 中逐坐標為 0·L(u)_k+0·L(v)_k=0，所以 L(0)=0。S10.1 與 S10.2 合起來證得引理 L_zero_to_zero。", "en": "Specialize the linearity equation to α=β=0: L(0u+0v)=0L(u)+0L(v). In ℝᵐ the right side has coordinates 0·L(u)_k+0·L(v)_k=0, so L(0)=0. S10.1 and S10.2 together prove Lemma L_zero_to_zero." }, "reverse_audit": { "kind": "premise_audit", "required_ids": [ "S10.1", "forall_elim", "mul_zero" ], "explanation": { "zh-Hant": "需要的只是線性等式對 α=β=0 成立；不需要 L 是矩陣映射。", "en": "Only the linearity equation at α=β=0 is needed; L need not be a matrix map." } }, "leaf_basis": "forall_elim + mul_zero + add_zero + coord_vec_ops" }, { "id": "S11", "uses": [ "G3", "S01" ], "before_tex": "S(0)=A0+b", "after_tex": "S(0)=(0,0)+(1,0)=(1,0)", "relation": "equivalent", "rule": "matrix_product_def", "conditions": [], "explanation": { "zh-Hant": "依 S01，A0=(0,0)（每行和式的每一項都乘 0）；加上 b=(1,0) 得 S(0)=(1,0)。", "en": "By S01, A0=(0,0) (each term of each row sum is multiplied by 0); adding b=(1,0) gives S(0)=(1,0)." }, "reverse_audit": { "kind": "inverse_equivalence", "required_ids": [ "matrix_product_def", "coord_vec_ops" ], "explanation": { "zh-Hant": "反向：(1,0)−b=(0,0)=A0，兩邊是同一個向量。", "en": "Backward: (1,0)−b=(0,0)=A0, so both sides name the same vector." } }, "leaf_basis": "matrix_product_def + mul_zero + coord_vec_ops" }, { "id": "S12", "uses": [ "S10", "S11" ], "before_tex": "S(0)=(1,0)\\neq(0,0)", "after_tex": "S\\ \\text{is not linear}", "relation": "implies", "rule": "contrapositive", "conditions": [], "explanation": { "zh-Hant": "依 L_zero_to_zero 的逆否命題：若 S(0)≠0，S 不是線性映射。具體見證：取 u=v=0、α=β=0，左邊 S(0)=(1,0)，右邊 0S(0)+0S(0)=(0,0)，等式不成立。一個反例就足以否定全稱命題。", "en": "By the contrapositive of L_zero_to_zero: if S(0)≠0, S is not linear. A concrete witness: u=v=0 and α=β=0 give left side S(0)=(1,0) and right side 0S(0)+0S(0)=(0,0), so the equation fails. One counterexample suffices to refute a universal statement." }, "reverse_audit": { "kind": "premise_audit", "required_ids": [ "S10", "S11", "contrapositive" ], "explanation": { "zh-Hant": "需要 S10 的必要條件與 S11 的計算；結論只否定線性，不否定 S 是仿射映射。", "en": "It needs the necessary condition of S10 and the computation of S11; the conclusion denies linearity only, not that S is affine." } }, "leaf_basis": "contrapositive + quantifier_negation" }, { "id": "S13", "uses": [ "S07", "S10", "G3" ], "before_tex": "S_b(x)=Ax+b\\ \\text{linear}", "after_tex": "b=0", "relation": "equivalent", "rule": "iff_intro", "conditions": [], "explanation": { "zh-Hant": "對任意 b∈ℝ²。（⇒）若 S_b 線性，依 L_zero_to_zero，b=A0+b=S_b(0)=0。（⇐）若 b=0，S_b(x)=Ax=T(x)，由 S07 線性。這正是 MKO 說的「Ax+b 只有 b=0 時才通過線性資格」。", "en": "For any b∈ℝ². (⇒) If S_b is linear, then by L_zero_to_zero, b=A0+b=S_b(0)=0. (⇐) If b=0, then S_b(x)=Ax=T(x), which is linear by S07. This is the MKO's point that Ax+b qualifies as linear only when b=0." }, "reverse_audit": { "kind": "inverse_equivalence", "required_ids": [ "S07", "S10", "iff_intro" ], "explanation": { "zh-Hant": "兩個方向各有理由：⇒ 用 S10，⇐ 用 S07；因此可以寫成等價。", "en": "Each direction has its own reason, ⇒ from S10 and ⇐ from S07, so the step is a genuine equivalence." } }, "leaf_basis": "iff_intro + L_zero_to_zero + eq_replace" }, { "id": "S14", "uses": [ "G4", "S01" ], "before_tex": "F(x)=Ax+(x_1^2-x_1,0);\\ u=e_1,\\ v=e_2,\\ \\alpha=\\beta=1", "after_tex": "F(e_1)+F(e_2)=(2,1)+(-1,3)=(1,4)=F(e_1+e_2)", "relation": "implies", "rule": "numeric_evaluation", "conditions": [], "explanation": { "zh-Hant": "Ae₁=(2,1)、Ae₂=(−1,3)、A(e₁+e₂)=(1,4)。三個輸入的第一坐標 x₁ 都是 0 或 1，所以 x₁²−x₁=0，F 在這些點與 T 相同。樣本等式 F(e₁+e₂)=F(e₁)+F(e₂) 成立，F 通過了這個測試。", "en": "Ae₁=(2,1), Ae₂=(−1,3) and A(e₁+e₂)=(1,4). For all three inputs the first coordinate x₁ is 0 or 1, so x₁²−x₁=0 and F agrees with T there. The sample equation F(e₁+e₂)=F(e₁)+F(e₂) holds, so F passes this test." }, "reverse_audit": { "kind": "premise_audit", "required_ids": [ "G4", "S01", "numeric_evaluation" ], "explanation": { "zh-Hant": "只核一組樣本，可由 N7、N8 重算；它不推出 F 線性，S15 會否定這一點。", "en": "Only one sample is checked, recomputable by N7 and N8; it does not imply that F is linear, which S15 refutes." } }, "leaf_basis": "numeric_evaluation + matrix_product_def" }, { "id": "S15", "uses": [ "G4", "S01", "S14" ], "before_tex": "F(2e_1)=A(2e_1)+(2^2-2,0)", "after_tex": "F(2e_1)=(6,2)\\neq(4,2)=2F(e_1)\\ \\Rightarrow\\ F\\ \\text{is not linear}", "relation": "implies", "rule": "quantifier_negation", "conditions": [], "explanation": { "zh-Hant": "取 u=e₁、α=2、β=0：A(2e₁)=(4,2)，x₁²−x₁=4−2=2，所以 F(2e₁)=(6,2)；而 2F(e₁)=(4,2)。線性等式在這組輸入失敗，依全稱否定，F 不是線性映射，雖然它通過了 S14 的樣本，而且 F(0)=0。", "en": "Take u=e₁, α=2, β=0: A(2e₁)=(4,2) and x₁²−x₁=4−2=2, so F(2e₁)=(6,2), whereas 2F(e₁)=(4,2). The linearity equation fails at this input, so by negating the universal statement F is not linear, even though it passed the sample of S14 and F(0)=0." }, "reverse_audit": { "kind": "premise_audit", "required_ids": [ "G4", "S01", "quantifier_negation" ], "explanation": { "zh-Hant": "否定全稱只需一個反例（N9）。這說明 MKO 假設 a2：有限樣本不能證明黑箱映射線性。", "en": "Refuting a universal statement needs one counterexample (N9). This illustrates MKO assumption a2: finitely many samples cannot prove a black-box map linear." } }, "leaf_basis": "quantifier_negation + numeric_evaluation" }, { "id": "S16", "uses": [ "S07", "S12", "S13", "S15" ], "before_tex": "\\text{S07},\\ \\text{S12},\\ \\text{S13},\\ \\text{S15}", "after_tex": "T\\ \\text{linear};\\ S\\ \\text{not linear};\\ (Ax+b\\ \\text{linear}\\iff b=0);\\ F\\ \\text{not linear}", "relation": "implies", "rule": "and_intro_elim", "conditions": [], "explanation": { "zh-Hant": "合併三個任務的結論：T 由全稱證書證得線性；S 因 S(0)≠0 不線性，且一般地 Ax+b 線性當且僅當 b=0；F 通過一個樣本卻有反例，所以不線性。", "en": "Combine the three tasks: T is linear by the universal certificate; S is not linear because S(0)≠0, and in general Ax+b is linear exactly when b=0; F passes one sample but has a counterexample, so it is not linear." }, "reverse_audit": { "kind": "sufficient_goal_reduction", "required_ids": [ "S07", "S12", "S13", "S15", "and_intro_elim" ], "explanation": { "zh-Hant": "目標 G 是三個任務的合取；足夠條件 H 是 S07、S12、S13、S15 四個已證結論，H⇒G 由合取引入。H 都由已知正向證得，沒有把 G 當作前提。", "en": "Goal G is the conjunction of the three tasks; the sufficient condition H is the four proved results S07, S12, S13 and S15, and H⇒G by conjunction introduction. Each part of H is proved forwards from the givens; G is never used as a premise." } }, "leaf_basis": "and_intro_elim" } ], "result": { "zh-Hant": "T(x)=Ax 對所有 u,v∈ℝ³ 與實數 α,β 滿足 T(αu+βv)=αT(u)+βT(v)，證書是逐行的分配律（引理 L_row_linear）。S(x)=Ax+(1,0) 不是線性，因為 S(0)=(1,0)≠0；一般地 x↦Ax+b 線性當且僅當 b=0。黑箱 F(x)=Ax+(x₁²−x₁,0) 通過樣本 F(e₁+e₂)=F(e₁)+F(e₂)，但 F(2e₁)=(6,2)≠(4,2)=2F(e₁)，所以不是線性。", "en": "T(x)=Ax satisfies T(αu+βv)=αT(u)+βT(v) for all u,v∈ℝ³ and reals α,β, certified by row-by-row distributivity (Lemma L_row_linear). S(x)=Ax+(1,0) is not linear because S(0)=(1,0)≠0; in general x↦Ax+b is linear if and only if b=0. The black box F(x)=Ax+(x₁²−x₁,0) passes the sample F(e₁+e₂)=F(e₁)+F(e₂), but F(2e₁)=(6,2)≠(4,2)=2F(e₁), so it is not linear." }, "negative_controls": [ { "claim": { "zh-Hant": "S(x)=Ax+(1,0) 的每個坐標都是 x 的一次式，所以 S 是線性映射", "en": "Each coordinate of S(x)=Ax+(1,0) is of first degree in x, so S is a linear map" }, "expected": "reject", "reason": { "zh-Hant": "S(0)=(1,0)≠0，違反每個線性映射都有的 L(0)=0（S10、S12）。", "en": "S(0)=(1,0)≠0, violating L(0)=0, which every linear map satisfies (S10, S12)." } }, { "claim": { "zh-Hant": "F 在樣本 e₁、e₂ 上滿足 F(e₁+e₂)=F(e₁)+F(e₂)，所以 F 線性", "en": "F satisfies F(e₁+e₂)=F(e₁)+F(e₂) on the samples e₁, e₂, so F is linear" }, "expected": "reject", "reason": { "zh-Hant": "F(2e₁)=(6,2)，但 2F(e₁)=(4,2)（S15）。", "en": "F(2e₁)=(6,2) while 2F(e₁)=(4,2) (S15)." } }, { "claim": { "zh-Hant": "g(x₁,x₂)=∛(x₁³+x₂³) 滿足 g(λx)=λg(x)，所以 g 線性", "en": "g(x₁,x₂)=∛(x₁³+x₂³) satisfies g(λx)=λg(x), so g is linear" }, "expected": "reject", "reason": { "zh-Hant": "可加性失敗：g(1,0)+g(0,1)=2，但 g(1,1)=∛2，而 2³=8≠2。", "en": "Additivity fails: g(1,0)+g(0,1)=2 but g(1,1)=∛2, and 2³=8≠2." } }, { "claim": { "zh-Hant": "S08 的一組數字讓兩邊相等，所以 T 線性已得證", "en": "The numbers of S08 make both sides equal, so T is proved linear" }, "expected": "reject", "reason": { "zh-Hant": "一組輸入只是一致性核對；F 也能通過一組樣本卻不線性。證明在 S04 到 S07。", "en": "One input is only a consistency check; F also passes a sample and is not linear. The proof is S04 to S07." } } ], "status": "candidate", "gaps": [], "verification": { "structure": "checked_local_script", "mathematics": "candidate_argument", "formalization": "not_formalized" }, "rule_glossary": [ { "id": "vector_space_axioms", "kind": "prerequisite", "statement": { "zh-Hant": "向量空間公理（實數或其他域 F 上）：加法成交換群；純量乘法滿足 a(u+v)=au+av、(a+b)u=au+bu、(ab)u=a(bu)、1u=u。", "en": "Vector space axioms (over ℝ or another field F): vectors form a commutative group under addition; scalar multiplication satisfies a(u+v)=au+av, (a+b)u=au+bu, (ab)u=a(bu) and 1u=u." } }, { "id": "matrix_product_def", "kind": "prerequisite", "statement": { "zh-Hant": "矩陣乘法：若 A 為 m×n、B 為 n×p，則 (AB)_{ij}=Σ_{k=1}^{n} A_{ik}B_{kj}。矩陣乘法一般不可交換。", "en": "Matrix product: for A of size m×n and B of size n×p, (AB)_{ij}=Σ_{k=1}^{n} A_{ik}B_{kj}. Matrix multiplication is in general not commutative." } }, { "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": "finite_sum_rules", "kind": "prerequisite", "statement": { "zh-Hant": "有限和：Σ_{i=1}^{0} a_i=0、Σ_{i=1}^{n+1} a_i=Σ_{i=1}^{n} a_i + a_{n+1}；Σ(a_i+b_i)=Σa_i+Σb_i；Σ c·a_i=c·Σa_i。", "en": "Finite sums: Σ_{i=1}^{0} a_i=0 and Σ_{i=1}^{n+1} a_i=Σ_{i=1}^{n} a_i + a_{n+1}; Σ(a_i+b_i)=Σa_i+Σb_i; Σ c·a_i=c·Σa_i." } }, { "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": "def_unfold", "kind": "prerequisite", "statement": { "zh-Hant": "展開定義：把一個已定義的詞換成它的定義內容（或反向收回），命題的真假不變。", "en": "Unfolding a definition: replacing a defined term by its definition (or folding it back) does not change whether a statement is true." } }, { "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": "eq_replace", "kind": "prerequisite", "statement": { "zh-Hant": "等值代換：命題中的一個式子可換成與它相等的式子，命題的真假不變。", "en": "Replacement of equals: an expression inside a statement may be replaced by an equal expression without changing whether the statement is true." } }, { "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": "forall_elim", "kind": "prerequisite", "statement": { "zh-Hant": "∀消去（代入特例）：若 ∀x∈D, P(x) 成立且 a∈D，則 P(a) 成立。", "en": "∀-elimination (instantiation): if ∀x∈D, P(x) holds and a∈D, then P(a) holds." } }, { "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": "add_zero", "kind": "prerequisite", "statement": { "zh-Hant": "加法單位元：u+0=u。", "en": "Additive identity: u+0=u." } }, { "id": "contrapositive", "kind": "prerequisite", "statement": { "zh-Hant": "逆否命題：P⇒Q 與 ¬Q⇒¬P 等價。逆命題 Q⇒P 一般不與 P⇒Q 等價。", "en": "Contrapositive: P⇒Q is equivalent to ¬Q⇒¬P. The converse Q⇒P is in general not equivalent to P⇒Q." } }, { "id": "quantifier_negation", "kind": "prerequisite", "statement": { "zh-Hant": "量詞否定：¬(∀x, P(x)) 等價於 ∃x, ¬P(x)；¬(∃x, P(x)) 等價於 ∀x, ¬P(x)。", "en": "Negating quantifiers: ¬(∀x, P(x)) is equivalent to ∃x, ¬P(x), and ¬(∃x, P(x)) is equivalent to ∀x, ¬P(x)." } }, { "id": "iff_intro", "kind": "prerequisite", "statement": { "zh-Hant": "雙條件：P⇒Q 與 Q⇒P 都成立時，P⇔Q 成立；反之亦然。", "en": "Biconditional: P⇔Q holds exactly when both P⇒Q and Q⇒P hold." } }, { "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_linear_map_def", "kind": "cited", "source": { "mko_id": "mko-general-linear-map", "sha256": "184a885b199aff8dc05ec273c0fe5b2dcdd38a9ef98c8ec790ce2668da9a07c7", "commit": "b28a5edbd46279b1db2a3f3f48bb3f447e18899f" }, "statement": { "zh-Hant": "獨立來源追加式勘誤（使用者已採用於本地修訂；未合併或發布）：實線性映射 T:V→W 對所有 u,v 與實數 α,β 滿足 T(αu+βv)=αT(u)+βT(v)。有限維矩陣公式提供全稱座標證書。對 S(x)=Ax+b，線性當且僅當 b=0；任何非零平移都因 S(0)=b≠0 而不線性，不只是「通常」不線性。原始引文與 canonical MKO 保留。", "en": "Independent append-only source erratum (owner-adopted in this local revision; not merged or published): a real linear map T:V→W satisfies T(αu+βv)=αT(u)+βT(v) for all u,v and real α,β. A finite-dimensional matrix formula supplies a universal coordinate certificate. For S(x)=Ax+b, linearity holds if and only if b=0; every nonzero translation fails because S(0)=b≠0, not merely in typical cases. The original citation and canonical MKO are retained." } }, { "id": "coord_vec_ops", "kind": "prerequisite", "statement": { "zh-Hant": "在 ℝⁿ 中 (u+v)_k=u_k+v_k、(αu)_k=αu_k；零向量各坐標為 0；兩向量相等當且僅當所有坐標相等。", "en": "In ℝⁿ, (u+v)_k=u_k+v_k and (αu)_k=αu_k; the zero vector has all coordinates 0; two vectors are equal exactly when all coordinates agree." } }, { "id": "L_row_linear", "kind": "lemma", "proved_in": [ "S04.1", "S04.2" ], "statement": { "zh-Hant": "對實數 a_k、u_k、v_k（k=1 到 n）與實數 α、β：Σ_{k=1}^{n} a_k(αu_k+βv_k)=αΣ_{k=1}^{n} a_k u_k+βΣ_{k=1}^{n} a_k v_k。", "en": "For reals a_k, u_k, v_k (k=1 to n) and reals α, β: Σ_{k=1}^{n} a_k(αu_k+βv_k)=αΣ_{k=1}^{n} a_k u_k+βΣ_{k=1}^{n} a_k v_k." } }, { "id": "L_zero_to_zero", "kind": "lemma", "proved_in": [ "S10.1", "S10.2" ], "statement": { "zh-Hant": "若 L:ℝⁿ→ℝᵐ 對所有 u,v 與實數 α,β 滿足 L(αu+βv)=αL(u)+βL(v)，則 L(0)=0。", "en": "If L:ℝⁿ→ℝᵐ satisfies L(αu+βv)=αL(u)+βL(v) for all u,v and reals α,β, then L(0)=0." } } ], "revision_log": [], "numeric_checks": [ { "id": "N1", "step": "S04", "vars": [ "al", "be", "u1", "u2", "v1", "v2" ], "degree": 2, "lhs": "2*(al*u1+be*v1)-(al*u2+be*v2)", "rhs": "al*(2*u1-u2)+be*(2*v1-v2)" }, { "id": "N2", "step": "S05", "vars": [ "al", "be", "u1", "u2", "u3", "v1", "v2", "v3" ], "degree": 2, "lhs": "(al*u1+be*v1)+3*(al*u2+be*v2)+4*(al*u3+be*v3)", "rhs": "al*(u1+3*u2+4*u3)+be*(v1+3*v2+4*v3)" }, { "id": "N3", "step": "S08", "lhs": "2*3-(-2)", "rhs": "8" }, { "id": "N4", "step": "S08", "lhs": "3+3*(-2)+4*8", "rhs": "29" }, { "id": "N5", "step": "S08", "lhs": "3*(2*1-0)-2*(2*0-1)", "rhs": "8" }, { "id": "N6", "step": "S08", "lhs": "3*(1+3*0+4*2)-2*(0+3*1+4*(-1))", "rhs": "29" }, { "id": "N7", "step": "S14", "lhs": "(2*1-1+0)+(1**2-1)", "rhs": "2+(-1)" }, { "id": "N8", "step": "S14", "lhs": "1+3*1+4*0", "rhs": "1+3" }, { "id": "N9", "step": "S15", "lhs": "2*2+(2**2-2)", "relation": "!=", "rhs": "2*(2*1+(1**2-1))" }, { "id": "N10", "lhs": "2**3", "relation": "!=", "rhs": "2" } ] }, "presentation": { "md_scope": { "zh-Hant": "在 MKO 的範圍內工作：題庫明示 ℝⁿ 到 ℝᵐ 的矩陣或矩陣加平移公式（假設 a1）；線性以 MKO 的單一等式定義；全稱證書來自矩陣乘法對加法與純量的分配律（假設 a2）。有限個樣本只作一致性核對，不作證明。不處理無限維空間、連續性或有界性（後者屬有界線性算子節點），不改動 canonical MKO 或英文 edition。", "en": "Works inside the MKO's scope: exercises state a matrix or matrix-plus-bias formula from ℝⁿ to ℝᵐ (assumption a1); linearity is the MKO's single equation; the universal certificate comes from distributivity of the matrix product over addition and scalars (assumption a2). Finitely many samples serve only as consistency checks, never as proofs. Infinite-dimensional spaces, continuity and boundedness (the bounded-linear-operator node) are not treated, and the canonical MKO and its English edition are not changed." }, "md_symbols": [ [ "A=\\begin{pmatrix}2&-1&0\\\\1&3&4\\end{pmatrix}", "給定的 2×3 整數矩陣", "the given 2-by-3 integer matrix" ], [ "T(x)=Ax", "由 A 定義的映射 ℝ³→ℝ²", "the map ℝ³→ℝ² defined by A" ], [ "S_b(x)=Ax+b", "帶平移 b∈ℝ² 的仿射映射；S=S_{(1,0)}", "the affine map with translation b∈ℝ²; S=S_{(1,0)}" ], [ "F", "只在樣本上被測試的黑箱映射", "a black-box map tested only on samples" ], [ "u,v", "ℝ³ 中任意向量，坐標 u_k、v_k", "arbitrary vectors of ℝ³ with coordinates u_k, v_k" ], [ "\\alpha,\\beta", "任意實數純量", "arbitrary real scalars" ], [ "e_1,e_2,e_3", "ℝ³ 的標準基向量", "the standard basis vectors of ℝ³" ] ], "md_error_controls": [ { "title": "仿射映射不一定是線性映射 / An affine map need not be linear", "zh": "MKO 指出的常見誤解是把任意仿射映射稱為線性。中學裡的「一次函數」y=mx+c 在線性代數中是仿射映射；它線性當且僅當 c=0。本稿的 S13 給出 Ax+b 的完整判準：線性必須把 0 送到 0，S(0)=(1,0) 就足以否定。", "en": "A common misconception identified in the MKO is to call every affine map linear. The school 'linear function' y=mx+c is, in linear algebra, an affine map, and it is linear exactly when c=0. S13 gives the full criterion for Ax+b: a linear map must send 0 to 0, and S(0)=(1,0) already refutes linearity." }, { "title": "樣本通過不是證明 / Passing samples is not a proof", "zh": "S14 的黑箱 F 在 e₁、e₂、e₁+e₂ 上與 T 一致，樣本等式成立，而且 F(0)=0；但 F(2e₁)≠2F(e₁)。有限個輸入無法排除在其他點出現的非線性項（這裡是 x₁²−x₁，它在 x₁=0、1 時恰好為 0）。全稱結論只能來自 S04 到 S07 的符號證明。", "en": "The black box F of S14 agrees with T on e₁, e₂ and e₁+e₂, the sample equation holds and F(0)=0; yet F(2e₁)≠2F(e₁). Finitely many inputs cannot exclude a nonlinear term that shows up elsewhere (here x₁²−x₁, which happens to vanish at x₁=0 and 1). The universal conclusion can come only from the symbolic proof in S04 to S07." }, { "title": "只核一半性質 / Checking only half of the property", "zh": "單一等式包含可加性（α=β=1）與齊次性（β=0），見 S09。只核其中一個性質不夠：例如 g(x₁,x₂)=∛(x₁³+x₂³) 對所有實數 λ 滿足 g(λx)=λg(x)，但 g(1,0)+g(0,1)=2，而 g(1,1)=∛2≠2，所以不可加，也不線性。", "en": "The single equation contains additivity (α=β=1) and homogeneity (β=0); see S09. Checking only one of them is not enough: for example g(x₁,x₂)=∛(x₁³+x₂³) satisfies g(λx)=λg(x) for every real λ, but g(1,0)+g(0,1)=2 while g(1,1)=∛2≠2, so g is not additive and not linear." } ], "md_open_items": [ { "zh-Hant": "本稿只處理有限維實坐標空間 ℝⁿ→ℝᵐ；一般實向量空間之間的線性映射（例如函數空間上的微分）不在範圍內。", "en": "This guide treats only finite-dimensional real coordinate spaces ℝⁿ→ℝᵐ; linear maps between general real vector spaces (such as differentiation on function spaces) are outside its scope." }, { "zh-Hant": "錯誤控制中 g 的齊次性只用到實數立方根是奇函數且 ∛(λ³t)=λ∛t；本稿未另立步驟證明。", "en": "The homogeneity of g in the error controls uses only that the real cube root is odd and ∛(λ³t)=λ∛t; the guide does not prove this in a separate step." } ] }, "revision": { "id": "production-20261019-adopted-96", "source_commit": "b28a5edbd46279b1db2a3f3f48bb3f447e18899f", "adopted_source_sha256": "9dace050059ee6d62ad50b29fac28662fb293d07b70ac44f1e242468648fd39c", "original_source_sha256": "a43b3c0cd76ed5cdfcda4c4d4db12aa689aa56d4c5cc554015137ae8813dc921", "owner_adopted": true, "bilingual_sync": "complete_source_bound", "formal_Lean": "NOT_RUN", "independent_AI_concurrence": "NOT_MEASURED", "issue_ids": [ "BR-M075" ], "scope": "Owner-adopted natural-language teaching guide; not a formal proof, canonical MKO rewrite or evidence-status promotion." } }