{ "schema_version": "ocme-derivation-reader-v1", "guide_id": "derivation-mko-linear-combination-span-01", "mko_id": "mko-linear-combination-span", "title": { "zh-Hant": "線性組合與生成：以有限係數 witness 判定 span 成員", "en": "Linear combinations and span: deciding membership using finite coefficient lists as witnesses" }, "contract": { "schema_version": "ocme-derivation-guide-v0.1", "guide_id": "derivation-mko-linear-combination-span-01", "source": { "mko_id": "mko-linear-combination-span", "commit": "b28a5edbd46279b1db2a3f3f48bb3f447e18899f", "sha256": "6f4f8ff504708ab9492740d70255cb2d85c406d8eaa891eef387935b3d46dad3", "english_sha256": "3d16797b83376915e08d16c68eee0e87ae7d9b28ca5d5ba76a9e2466ce8195b3" }, "mode": "worked_example", "audience_prerequisites": [ "vector_space_axioms", "eq_congruence", "mul_inverse_nonzero", "set_extensionality", "numeric_evaluation" ], "goal": { "zh-Hant": "（一）在 ℝ² 中，v₁=(1,2)、v₂=(3,−1)，判定 b=(7,0) 是否屬於 span{v₁,v₂}：先證明兩個向量的生成集恰為 {c₁v₁+c₂v₂}，再解係數、證明唯一，並由 witness 重建 b。（二）在 ℝ³ 中，u₁=(1,0,1)、u₂=(0,1,1)，證明 t=(2,3,4) 不屬於 span{u₁,u₂}，而 t'=(2,3,5) 屬於：同為三維並不決定成員資格。", "en": "(1) In ℝ², with v₁=(1,2) and v₂=(3,−1), decide whether b=(7,0) lies in span{v₁,v₂}: first prove that the span of two vectors is exactly {c₁v₁+c₂v₂}, then solve for the coefficients, prove they are unique, and rebuild b from the witness. (2) In ℝ³, with u₁=(1,0,1) and u₂=(0,1,1), prove that t=(2,3,4) is not in span{u₁,u₂} while t'=(2,3,5) is: having three coordinates does not decide membership." }, "givens": [ { "id": "G1", "statement": { "zh-Hant": "ℝⁿ 中加法與純量乘法逐分量進行；v₁=(1,2)，v₂=(3,−1)，b=(7,0)，都在 ℝ²。", "en": "In ℝⁿ, addition and scalar multiplication act componentwise; v₁=(1,2), v₂=(3,−1) and b=(7,0) all lie in ℝ²." } }, { "id": "G2", "statement": { "zh-Hant": "u₁=(1,0,1)，u₂=(0,1,1)，t=(2,3,4)，t'=(2,3,5)，都在 ℝ³。", "en": "u₁=(1,0,1), u₂=(0,1,1), t=(2,3,4) and t'=(2,3,5) all lie in ℝ³." } }, { "id": "G3", "statement": { "zh-Hant": "每題只審核明示的有限生成族與候選 witness；結論不推及其他生成族（MKO 假設 a2）。", "en": "Each question audits only the explicitly given finite family and candidate witness; conclusions do not extend to other families (MKO assumption a2)." } } ], "steps": [ { "id": "S01", "uses": [ "mko_span_def" ], "before_tex": "\\text{family } (w_1,w_2)", "after_tex": "\\operatorname{span}\\{w_1,w_2\\}=\\{\\textstyle\\sum_{k=1}^{n}t_kx_k\\mid n\\in\\mathbb{N},\\ t_k\\in\\mathbb{R},\\ x_k\\in\\{w_1,w_2\\}\\}", "relation": "definition", "rule": "mko_span_def", "conditions": [], "explanation": { "zh-Hant": "依 MKO，有限線性組合是有限個實係數加權和；span 是所有有限長度線性組合的集合。長度 n 可以是任意自然數（n=0 時為空和，即零向量），但每一個組合都只有有限項；無限級數不是這個定義所用的有限表示式；其和若存在，仍可能有有限表示並屬於 span。", "en": "By the MKO, a finite linear combination is a finite sum weighted by real coefficients, and the span is the set of all finite-length combinations. The length n may be any natural number (n=0 gives the empty sum, the zero vector), but each combination has finitely many terms; an infinite-series expression is not the finite representation used by this definition, but its sum, when defined, may have a finite representation and lie in the span." }, "reverse_audit": { "kind": "premise_audit", "required_ids": [ "mko_span_def" ], "explanation": { "zh-Hant": "此步只寫出定義，不推出新事實；S03 會把它化為只有兩個係數的形式。", "en": "This step only states the definition and derives no new fact; S03 reduces it to a form with exactly two coefficients." } }, "leaf_basis": "mko_span_def + def_intro" }, { "id": "S02", "uses": [ "G1", "coord_ops" ], "before_tex": "v_1,v_2,b", "after_tex": "v_1,v_2,b\\in\\mathbb{R}^2;\\ \\text{witness }(c_1,c_2)\\in\\mathbb{R}^2", "relation": "implies", "rule": "coord_ops", "conditions": [], "explanation": { "zh-Hant": "三個向量都有兩個分量，屬於同一個 ℝ²，所以加權和有定義（MKO 第一段）。生成族有兩個向量，所以 witness 恰有兩個係數。", "en": "All three vectors have two components and lie in the same ℝ², so weighted sums are defined (MKO paragraph 1). The family has two vectors, so a witness has exactly two coefficients." }, "reverse_audit": { "kind": "premise_audit", "required_ids": [ "G1", "coord_ops" ], "explanation": { "zh-Hant": "只核維度一致與係數數量；這不推出 b 在 span 中。", "en": "Only equal dimensions and the number of coefficients are checked; this does not show that b is in the span." } }, "leaf_basis": "coord_ops" }, { "id": "S03", "uses": [ "S01", "vector_space_axioms" ], "before_tex": "\\operatorname{span}\\{w_1,w_2\\}", "after_tex": "\\{c_1w_1+c_2w_2\\mid c_1,c_2\\in\\mathbb{R}\\}", "relation": "equivalent", "rule": "L_span_two", "conditions": [], "explanation": { "zh-Hant": "定義允許任意長度、可重複使用生成向量；引理 L_span_two 說明它們都能合併成恰好兩個係數的形式。證明在 S03.1（⊆）與 S03.2（⊇），依集合外延性得等式。", "en": "The definition allows any length and repeated use of the generators; Lemma L_span_two shows they all collapse to a form with exactly two coefficients. The proof is in S03.1 (⊆) and S03.2 (⊇), and set extensionality gives the equality." }, "reverse_audit": { "kind": "inverse_equivalence", "required_ids": [ "L_span_two", "set_extensionality" ], "explanation": { "zh-Hant": "兩個包含方向都已證明，所以兩個集合相等，雙向成立。", "en": "Both inclusions are proved, so the sets are equal and the step is two-way." } }, "leaf_basis": "set_extensionality + vector_space_axioms + zero_scalar" }, { "id": "S03.1", "uses": [ "S01", "vector_space_axioms", "zero_scalar" ], "before_tex": "\\sum_{k=1}^{n}t_kx_k,\\ x_k\\in\\{w_1,w_2\\}", "after_tex": "\\Big(\\sum_{k\\in K_1}t_k\\Big)w_1+\\Big(\\sum_{k\\in K_2}t_k\\Big)w_2", "relation": "implies", "rule": "vector_space_axioms", "conditions": [], "explanation": { "zh-Hant": "（⊆）任取一個有限組合。對每一項先選一個標籤 j_k∈{1,2}，使 x_k=w_(j_k)，再以所選標籤分成 K₁、K₂，故每個下標恰屬一組。即使 w₁=w₂ 也可以全選標籤 1，令 K₂ 為空；不是用相等向量的值自動分組。依向量加法的結合與交換律重排各項，再用 (a+b)w=aw+bw 合併同一向量的係數，得到 c₁w₁+c₂w₂，其中 c₁、c₂ 是各組係數之和。若某組為空，係數和為 0，而 0·w=0，形式仍成立。", "en": "(⊆) Take any finite combination. For each term first choose a label j_k∈{1,2} with x_k=w_(j_k), then partition the indices by those chosen labels into K₁ and K₂. Every index belongs to exactly one group. Even if w₁=w₂, all labels may be 1 and K₂ empty; equal vector values do not themselves determine a partition. Reorder the terms by associativity and commutativity of vector addition, then merge coefficients of the same vector by (a+b)w=aw+bw, giving c₁w₁+c₂w₂ with c₁, c₂ the sums of the coefficients in each group. If a group is empty its coefficient sum is 0, and 0·w=0, so the form still holds." }, "reverse_audit": { "kind": "premise_audit", "required_ids": [ "vector_space_axioms", "zero_scalar", "finite_sum_rules" ], "explanation": { "zh-Hant": "只用到有限次的重排與合併，所以此引理只處理有限組合；無限級數的收斂與重排合法性須另外核對。此步只證 ⊆。", "en": "Only finitely many reorderings and mergings are used, so this lemma treats finite combinations only. Convergence and the validity of reordering an infinite series require separate justification. This step proves ⊆ only." } }, "leaf_basis": "vector_space_axioms + finite_sum_rules + zero_scalar" }, { "id": "S03.2", "uses": [ "S03.1", "S01" ], "before_tex": "c_1w_1+c_2w_2", "after_tex": "\\sum_{k=1}^{2}t_kx_k,\\ t_1=c_1,\\ x_1=w_1,\\ t_2=c_2,\\ x_2=w_2", "relation": "implies", "rule": "def_unfold", "conditions": [], "explanation": { "zh-Hant": "（⊇）c₁w₁+c₂w₂ 本身就是長度 2 的有限線性組合，所以屬於 span。兩個包含合起來，依集合外延性得 span{w₁,w₂}={c₁w₁+c₂w₂}。S03.1 與 S03.2 合起來證得引理 L_span_two。", "en": "(⊇) c₁w₁+c₂w₂ is itself a finite linear combination of length 2, so it lies in the span. With both inclusions, set extensionality gives span{w₁,w₂}={c₁w₁+c₂w₂}. S03.1 and S03.2 together prove Lemma L_span_two." }, "reverse_audit": { "kind": "premise_audit", "required_ids": [ "S03.1", "def_unfold", "set_extensionality" ], "explanation": { "zh-Hant": "此方向只需要依定義認出一個長度 2 的組合。", "en": "This direction only needs to recognize a length-2 combination by the definition." } }, "leaf_basis": "def_unfold + set_extensionality" }, { "id": "S04", "uses": [ "S03", "S02" ], "before_tex": "b\\in\\operatorname{span}\\{v_1,v_2\\}", "after_tex": "\\exists c_1,c_2\\in\\mathbb{R}:\\ c_1(1,2)+c_2(3,-1)=(7,0)", "relation": "equivalent", "rule": "set_builder", "conditions": [], "explanation": { "zh-Hant": "依 S03，成員資格等價於存在兩個實係數使加權和等於 b。這樣的 (c₁,c₂) 就是 MKO 所說的 membership witness。", "en": "By S03, membership is equivalent to the existence of two real coefficients whose weighted sum is b. Such a pair (c₁,c₂) is what the MKO calls a membership witness." }, "reverse_audit": { "kind": "inverse_equivalence", "required_ids": [ "S03", "set_builder" ], "explanation": { "zh-Hant": "集合建構記號的成員條件是雙向的。", "en": "Membership in set-builder notation is two-way." } }, "leaf_basis": "set_builder + L_span_two" }, { "id": "S05", "uses": [ "G1", "coord_ops" ], "before_tex": "c_1(1,2)+c_2(3,-1)", "after_tex": "(c_1+3c_2,\\ 2c_1-c_2)", "relation": "equivalent", "rule": "coord_ops", "conditions": [], "explanation": { "zh-Hant": "逐分量：第一分量 c₁·1+c₂·3=c₁+3c₂，第二分量 c₁·2+c₂·(−1)=2c₁−c₂。", "en": "Componentwise: first component c₁·1+c₂·3=c₁+3c₂, second component c₁·2+c₂·(−1)=2c₁−c₂." }, "reverse_audit": { "kind": "inverse_equivalence", "required_ids": [ "coord_ops", "mul_comm" ], "explanation": { "zh-Hant": "逐分量的恆等改寫（N10、N11 以多項式格點核對），雙向成立。", "en": "A componentwise identity rewrite (checked on a grid in N10, N11); two-way." } }, "leaf_basis": "coord_ops + mul_comm + neg_mul" }, { "id": "S06", "uses": [ "S04", "S05", "tuple_eq" ], "before_tex": "(c_1+3c_2,\\ 2c_1-c_2)=(7,0)", "after_tex": "c_1+3c_2=7\\ \\land\\ 2c_1-c_2=0", "relation": "equivalent", "rule": "tuple_eq", "conditions": [], "explanation": { "zh-Hant": "有序對相等當且僅當兩個分量分別相等，向量方程化為兩個實數方程，兩者必須由同一組 (c₁,c₂) 同時滿足。", "en": "Ordered pairs are equal exactly when both components agree, so the vector equation becomes two real equations, which one pair (c₁,c₂) must satisfy simultaneously." }, "reverse_audit": { "kind": "inverse_equivalence", "required_ids": [ "tuple_eq" ], "explanation": { "zh-Hant": "有序對相等的條件本身是雙向的。", "en": "The equality condition for ordered pairs is itself two-way." } }, "leaf_basis": "tuple_eq" }, { "id": "S07", "uses": [ "S06" ], "before_tex": "2c_1-c_2=0", "after_tex": "c_2=2c_1", "relation": "equivalent", "rule": "eq_congruence", "conditions": [], "explanation": { "zh-Hant": "兩邊同加 c₂，得 2c₁=c₂。反向是兩邊同減 c₂。", "en": "Add c₂ to both sides to get 2c₁=c₂. The backward move subtracts c₂ from both sides." }, "reverse_audit": { "kind": "inverse_equivalence", "required_ids": [ "eq_congruence", "add_inverse" ], "explanation": { "zh-Hant": "加 c₂ 與減 c₂ 互為反運算，沒有附加條件。", "en": "Adding and subtracting c₂ undo each other, with no side condition." } }, "leaf_basis": "eq_congruence + add_inverse + add_zero" }, { "id": "S08", "uses": [ "S06", "S07" ], "before_tex": "c_1+3c_2=7,\\ c_2=2c_1", "after_tex": "(7c_1=7\\ \\land\\ c_2=2c_1)\\iff(c_1=1\\ \\land\\ c_2=2c_1)", "relation": "equivalent", "rule": "mul_inverse_nonzero", "conditions": [ "7\\neq 0" ], "explanation": { "zh-Hant": "把 c₂=2c₁ 代入第一式：c₁+6c₁=7c₁=7。因 7≠0，兩邊同乘 1/7 得 c₁=1。代入在已知 c₂=2c₁ 的條件下是等值的。", "en": "Substitute c₂=2c₁ into the first equation: c₁+6c₁=7c₁=7. Since 7≠0, multiplying both sides by 1/7 gives c₁=1. Given c₂=2c₁, the substitution is an equivalence." }, "reverse_audit": { "kind": "inverse_equivalence", "required_ids": [ "eq_replace", "mul_inverse_nonzero", "distributive" ], "explanation": { "zh-Hant": "乘 1/7 與乘 7 互為反運算，前提 7≠0；代入與代回互為反向。", "en": "Multiplying by 1/7 and by 7 undo each other, given 7≠0; substituting and substituting back are inverse moves." } }, "leaf_basis": "eq_replace + distributive + mul_inverse_nonzero + numeric_evaluation" }, { "id": "S09", "uses": [ "S07", "S08" ], "before_tex": "c_1=1,\\ c_2=2c_1", "after_tex": "(c_1,c_2)=(1,2)\\ \\text{(unique)}", "relation": "equivalent", "rule": "eq_replace", "conditions": [], "explanation": { "zh-Hant": "c₂=2·1=2。S06 到 S09 每一步都是等價變形，所以原方程組等價於 (c₁,c₂)=(1,2)：解存在而且唯一。", "en": "c₂=2·1=2. Every step from S06 to S09 is an equivalence, so the system is equivalent to (c₁,c₂)=(1,2): a solution exists and it is unique." }, "reverse_audit": { "kind": "inverse_equivalence", "required_ids": [ "eq_replace", "logic_chain" ], "explanation": { "zh-Hant": "等價鏈依 logic_chain 串接；唯一性來自整條鏈都是雙向的。", "en": "The chain of equivalences is joined by logic_chain; uniqueness comes from every link being two-way." } }, "leaf_basis": "eq_replace + logic_chain + numeric_evaluation" }, { "id": "S10", "uses": [ "S04", "S09", "G1" ], "before_tex": "1\\cdot(1,2)+2\\cdot(3,-1)", "after_tex": "(1+6,\\ 2-2)=(7,0)=b\\ \\Rightarrow\\ b\\in\\operatorname{span}\\{v_1,v_2\\}", "relation": "implies", "rule": "exists_intro", "conditions": [], "explanation": { "zh-Hant": "用 witness (1,2) 重建目標：(1,2)+(6,−2)=(7,0)=b（MKO 第二段）。依 ∃ 引入與 S04，b 屬於 span{v₁,v₂}。重建本身就是一次獨立核對。", "en": "Rebuild the target from the witness (1,2): (1,2)+(6,−2)=(7,0)=b (MKO paragraph 2). By ∃-introduction and S04, b lies in span{v₁,v₂}. The rebuild is itself an independent check." }, "reverse_audit": { "kind": "premise_audit", "required_ids": [ "S04", "exists_intro", "numeric_evaluation" ], "explanation": { "zh-Hant": "存在性只需要一個 witness；唯一性另由 S09 給出。", "en": "Existence needs one witness; uniqueness comes separately from S09." } }, "leaf_basis": "exists_intro + coord_ops + numeric_evaluation" }, { "id": "S11", "uses": [ "S06", "det2_criterion" ], "before_tex": "\\det\\begin{pmatrix}1&3\\\\2&-1\\end{pmatrix}", "after_tex": "1\\cdot(-1)-3\\cdot 2=-7\\neq 0", "relation": "implies", "rule": "det2_criterion", "conditions": [], "explanation": { "zh-Hant": "交叉核對：係數矩陣行列式為 −7≠0，依 Cramer 法則方程組恰有一解，與 S09 由代入得到的唯一解一致（MKO 第三段：ℝ² 兩個獨立生成向量以非零行列式給出唯一座標）。", "en": "Cross-check: the coefficient determinant is −7≠0, so by Cramer's rule the system has exactly one solution, matching the unique solution found by substitution in S09 (MKO paragraph 3: two independent generators of ℝ² give unique coordinates via a nonzero determinant)." }, "reverse_audit": { "kind": "premise_audit", "required_ids": [ "det2_criterion", "numeric_evaluation" ], "explanation": { "zh-Hant": "唯一性已由 S09 的等價鏈證明；此步只是一致性核對，本稿不依賴 Cramer 法則。", "en": "Uniqueness is already proved by the chain of equivalences in S09; this step is a consistency check and the guide does not depend on Cramer's rule." } }, "leaf_basis": "det2_criterion + numeric_evaluation" }, { "id": "S12", "uses": [ "G2", "S03", "coord_ops", "tuple_eq" ], "before_tex": "t\\in\\operatorname{span}\\{u_1,u_2\\}", "after_tex": "\\exists c_1,c_2:\\ c_1=2\\ \\land\\ c_2=3\\ \\land\\ c_1+c_2=4", "relation": "equivalent", "rule": "tuple_eq", "conditions": [], "explanation": { "zh-Hant": "引理 L_span_two 對任何兩個向量的族成立，所以 t∈span{u₁,u₂} 等價於存在 c₁、c₂ 使 c₁(1,0,1)+c₂(0,1,1)=(c₁, c₂, c₁+c₂) 等於 (2,3,4)，再逐分量拆成三個方程。", "en": "Lemma L_span_two holds for any family of two vectors, so t∈span{u₁,u₂} is equivalent to the existence of c₁, c₂ with c₁(1,0,1)+c₂(0,1,1)=(c₁, c₂, c₁+c₂) equal to (2,3,4), which splits componentwise into three equations." }, "reverse_audit": { "kind": "inverse_equivalence", "required_ids": [ "S03", "coord_ops", "tuple_eq" ], "explanation": { "zh-Hant": "引理、逐分量計算與有序組相等都是雙向的。", "en": "The lemma, the componentwise computation and tuple equality are all two-way." } }, "leaf_basis": "L_span_two + coord_ops + tuple_eq" }, { "id": "S13", "uses": [ "S12" ], "before_tex": "c_1=2\\ \\land\\ c_2=3\\ \\land\\ c_1+c_2=4", "after_tex": "5=4\\ \\text{false};\\ t\\notin\\operatorname{span}\\{u_1,u_2\\}", "relation": "implies", "rule": "contradiction", "conditions": [], "explanation": { "zh-Hant": "前兩個方程迫使 c₁=2、c₂=3，代入第三個得 2+3=5=4，矛盾。所以沒有任何 witness，t 不屬於 span{u₁,u₂}，儘管 t 與 u₁、u₂ 同為三維。", "en": "The first two equations force c₁=2 and c₂=3; substituting into the third gives 2+3=5=4, a contradiction. So no witness exists and t is not in span{u₁,u₂}, even though t has three coordinates like u₁ and u₂." }, "reverse_audit": { "kind": "premise_audit", "required_ids": [ "S12", "contradiction", "numeric_evaluation" ], "explanation": { "zh-Hant": "矛盾來自同一組係數要同時滿足三個方程；前兩個方程沒有選擇餘地，所以否定涵蓋所有可能的係數。", "en": "The contradiction comes from one pair of coefficients having to satisfy all three equations; the first two leave no choice, so the refutation covers every possible pair." } }, "leaf_basis": "contradiction + eq_replace + numeric_evaluation" }, { "id": "S14", "uses": [ "G2", "S03" ], "before_tex": "2u_1+3u_2", "after_tex": "(2,0,2)+(0,3,3)=(2,3,5)=t'\\ \\Rightarrow\\ t'\\in\\operatorname{span}\\{u_1,u_2\\}", "relation": "implies", "rule": "exists_intro", "conditions": [], "explanation": { "zh-Hant": "對 t'=(2,3,5)，同樣的前兩個方程給 c₁=2、c₂=3，第三個方程 2+3=5 成立。witness (2,3) 重建出 t'，所以 t'∈span{u₁,u₂}。t 與 t' 只差第三個分量，成員資格卻相反。", "en": "For t'=(2,3,5) the same first two equations give c₁=2 and c₂=3, and the third equation 2+3=5 holds. The witness (2,3) rebuilds t', so t'∈span{u₁,u₂}. t and t' differ only in the third component, yet their membership is opposite." }, "reverse_audit": { "kind": "premise_audit", "required_ids": [ "S03", "exists_intro", "numeric_evaluation" ], "explanation": { "zh-Hant": "存在性只需要一個 witness，已逐分量重建核對（N7、N8）。", "en": "Existence needs one witness, checked by rebuilding componentwise (N7, N8)." } }, "leaf_basis": "exists_intro + coord_ops + numeric_evaluation" }, { "id": "S15", "uses": [ "S09", "S10", "S13", "S14" ], "before_tex": "b,\\ t,\\ t'", "after_tex": "b\\in\\operatorname{span}\\{v_1,v_2\\}\\ \\text{via unique }(1,2);\\ t\\notin,\\ t'\\in\\operatorname{span}\\{u_1,u_2\\}", "relation": "implies", "rule": "and_intro_elim", "conditions": [], "explanation": { "zh-Hant": "合併結論：b 由唯一的 witness (1,2) 屬於 span{v₁,v₂}；t 不屬於、t' 屬於 span{u₁,u₂}。每個結論都由有限 witness 或方程組的矛盾決定，不由維度決定。", "en": "Combine the conclusions: b lies in span{v₁,v₂} via the unique witness (1,2); t is not in and t' is in span{u₁,u₂}. Each conclusion is decided by a finite witness or by a contradiction in the equations, never by dimension." }, "reverse_audit": { "kind": "sufficient_goal_reduction", "required_ids": [ "S09", "S10", "S13", "S14", "and_intro_elim" ], "explanation": { "zh-Hant": "目標 G：三個成員判定。足夠條件 H：S09、S10、S13、S14；H⇒G 由合取引入。H 由給定資料正向推出，沒有把 G 當作前提。", "en": "Goal G: the three membership decisions. Sufficient condition H: S09, S10, S13 and S14; 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": "兩個向量的 span 恰為 {c₁w₁+c₂w₂}（已證）。b=(7,0)=1·v₁+2·v₂，witness (1,2) 唯一（行列式 −7≠0 作交叉核對），所以 b∈span{v₁,v₂}。t=(2,3,4) 的方程組要求 5=4，所以 t∉span{u₁,u₂}；t'=(2,3,5)=2u₁+3u₂∈span{u₁,u₂}。同為三維不決定成員資格。", "en": "The span of two vectors is exactly {c₁w₁+c₂w₂} (proved). b=(7,0)=1·v₁+2·v₂ with the unique witness (1,2) (the determinant −7≠0 is a cross-check), so b∈span{v₁,v₂}. For t=(2,3,4) the equations demand 5=4, so t∉span{u₁,u₂}, while t'=(2,3,5)=2u₁+3u₂∈span{u₁,u₂}. Having three coordinates does not decide membership." }, "negative_controls": [ { "claim": { "zh-Hant": "t=(2,3,4) 與 u₁、u₂ 同為三維，所以 t∈span{u₁,u₂}", "en": "t=(2,3,4) has the same dimension as u₁ and u₂, so t∈span{u₁,u₂}" }, "expected": "reject", "reason": { "zh-Hant": "方程組要求 c₁=2、c₂=3 且 c₁+c₂=4，即 5=4，無解。", "en": "The equations demand c₁=2, c₂=3 and c₁+c₂=4, that is 5=4, so there is no solution." } }, { "claim": { "zh-Hant": "不把其和改寫為有限形式時，所顯示的無窮級數表示式 Σ_{k=1}^{∞} 2^{−k}v₁ 本身就是定義中允許的有限長度線性組合表示式", "en": "Without rewriting its sum finitely, the displayed infinite-series expression Σ_{k=1}^{∞} 2^{−k}v₁ itself is a finite-length linear-combination representation allowed by the definition" }, "expected": "reject", "reason": { "zh-Hant": "此表示式有無限多項，不是有限線性組合的寫法；但本例級數的和是 v₁，仍屬於 span。", "en": "This expression has infinitely many terms and is not a finite linear-combination representation; nevertheless its sum here is v₁, which lies in the span." } }, { "claim": { "zh-Hant": "(c₁,c₂)=(7,0) 是 b 的 witness", "en": "(c₁,c₂)=(7,0) is a witness for b" }, "expected": "reject", "reason": { "zh-Hant": "7(1,2)+0(3,−1)=(7,14)≠(7,0)。", "en": "7(1,2)+0(3,−1)=(7,14)≠(7,0)." } }, { "claim": { "zh-Hant": "因為 c₁+3c₂=7 有解，所以 b∈span{v₁,v₂}", "en": "Since c₁+3c₂=7 has a solution, b∈span{v₁,v₂}" }, "expected": "reject", "reason": { "zh-Hant": "必須同一組係數同時滿足兩式；結論雖然正確，理由不成立，例如 (7,0) 只滿足第一式。", "en": "One pair must satisfy both equations; the conclusion happens to be true but the reason is invalid, since for example (7,0) satisfies only the first equation." } } ], "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": "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": "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": "set_extensionality", "kind": "prerequisite", "statement": { "zh-Hant": "集合相等：A=B 當且僅當 A⊆B 且 B⊆A，即兩者有相同元素。", "en": "Set equality: A=B if and only if A⊆B and B⊆A, i.e. they have the same elements." } }, { "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_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": "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": "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": "set_builder", "kind": "prerequisite", "statement": { "zh-Hant": "集合建構式：x∈{y∈A ∣ P(y)} 當且僅當 x∈A 且 P(x)。", "en": "Set-builder notation: x∈{y∈A ∣ P(y)} if and only if x∈A and P(x)." } }, { "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": "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": "add_inverse", "kind": "prerequisite", "statement": { "zh-Hant": "加法反元素：每個 u 都有 −u，使 u+(−u)=0；u−v 定義為 u+(−v)。", "en": "Additive inverse: every u has −u with u+(−u)=0; u−v means u+(−v)." } }, { "id": "add_zero", "kind": "prerequisite", "statement": { "zh-Hant": "加法單位元：u+0=u。", "en": "Additive identity: u+0=u." } }, { "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": "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": "logic_chain", "kind": "prerequisite", "statement": { "zh-Hant": "推論的傳遞：由 P⇒Q 與 Q⇒R 得 P⇒R；⇔ 同理。", "en": "Chaining: from P⇒Q and Q⇒R infer P⇒R; likewise for ⇔." } }, { "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": "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": "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_span_def", "kind": "cited", "source": { "mko_id": "mko-linear-combination-span", "sha256": "6f4f8ff504708ab9492740d70255cb2d85c406d8eaa891eef387935b3d46dad3", "commit": "b28a5edbd46279b1db2a3f3f48bb3f447e18899f" }, "statement": { "zh-Hant": "給實向量空間中的向量族，有限線性組合為有限個實係數加權和。該向量族的 span 是所有有限長度線性組合的集合，不包含未定義的無限級數。", "en": "For a vector family in a real vector space, a finite linear combination is a finite sum weighted by real coefficients. Its span is the set of all finite-length linear combinations and does not include undefined infinite series." } }, { "id": "coord_ops", "kind": "prerequisite", "statement": { "zh-Hant": "固定維度 n 的 ℝⁿ 中，向量加法與純量乘法逐分量進行；不同維度的向量不能相加。", "en": "In ℝⁿ of fixed dimension n, vector addition and scalar multiplication act componentwise; vectors of different dimensions cannot be added." } }, { "id": "tuple_eq", "kind": "prerequisite", "statement": { "zh-Hant": "兩個同維度的有序組相等，當且僅當每個對應分量相等。", "en": "Two ordered tuples of the same length are equal if and only if all corresponding components are equal." } }, { "id": "zero_scalar", "kind": "prerequisite", "statement": { "zh-Hant": "在實向量空間中 0·w=0（零向量）：因 0w=(0+0)w=0w+0w，兩邊加上 −(0w) 即得；空的有限和定義為零向量。", "en": "In a real vector space 0·w=0 (the zero vector): 0w=(0+0)w=0w+0w, and adding −(0w) to both sides gives it; the empty finite sum is defined to be the zero vector." } }, { "id": "det2_criterion", "kind": "prerequisite", "statement": { "zh-Hant": "2×2 線性方程組 px+qy=e、rx+sy=f 中，若 ps−qr≠0，則恰有一組解（Cramer 法則）。", "en": "A 2×2 linear system px+qy=e, rx+sy=f with ps−qr≠0 has exactly one solution (Cramer's rule)." } }, { "id": "L_span_two", "kind": "lemma", "proved_in": [ "S03.1", "S03.2" ], "statement": { "zh-Hant": "對實向量空間中的向量族 (w₁,w₂)：span{w₁,w₂}={c₁w₁+c₂w₂ ∣ c₁,c₂∈ℝ}。", "en": "For a family (w₁,w₂) in a real vector space: span{w₁,w₂}={c₁w₁+c₂w₂ ∣ c₁,c₂∈ℝ}." } } ], "revision_log": [], "numeric_checks": [ { "id": "N1", "step": "S08", "lhs": "1+3*2", "rhs": "7" }, { "id": "N2", "step": "S10", "lhs": "1*1+2*3", "rhs": "7" }, { "id": "N3", "step": "S10", "lhs": "1*2+2*(-1)", "rhs": "0" }, { "id": "N4", "step": "S11", "lhs": "1*(-1)-3*2", "rhs": "-7" }, { "id": "N5", "step": "S11", "lhs": "-7", "relation": "!=", "rhs": "0" }, { "id": "N6", "step": "S13", "lhs": "2+3", "relation": "!=", "rhs": "4" }, { "id": "N7", "step": "S14", "lhs": "2*1+3*0", "rhs": "2" }, { "id": "N8", "step": "S14", "lhs": "2*1+3*1", "rhs": "5" }, { "id": "N9", "lhs": "7*2+0*(-1)", "relation": "!=", "rhs": "0" }, { "id": "N10", "step": "S05", "vars": [ "c", "d" ], "degree": 1, "lhs": "c*1+d*3", "rhs": "c+3*d" }, { "id": "N11", "step": "S05", "vars": [ "c", "d" ], "degree": 1, "lhs": "c*2+d*(-1)", "rhs": "2*c-d" } ] }, "presentation": { "md_scope": { "zh-Hant": "只審核題目明示的有限生成族與候選 witness（MKO 假設 a2），具體練習採整數座標向量，所列成員 witness 恰為整數係數（MKO 假設 a1）；一般 span 定義與兩生成元引理容許所有實係數。span 由有限長度線性組合定義；無限級數不是該定義所用的表示式，其和若存在，是否在 span 中須另核有限表示。不處理線性獨立、基底或維度；不改動 canonical MKO 或英文 edition。", "en": "Audits only the explicitly given finite generator families and candidate witnesses (MKO assumption a2), with integer-coordinate exercise vectors and worked membership witnesses whose coefficients happen to be integers (MKO assumption a1); the general span definition and two-generator lemma allow all real coefficients. The span is defined using finite-length linear combinations. An infinite-series expression is not such a representation; if its sum exists, membership is decided by whether that sum has a finite representation. Linear independence, bases and dimension are not treated; the canonical MKO and its English edition are not changed." }, "md_symbols": [ [ "v_1,\\ v_2", "ℝ² 中的生成族", "the generator family in ℝ²" ], [ "u_1,\\ u_2", "ℝ³ 中的生成族", "the generator family in ℝ³" ], [ "c_1,\\ c_2", "線性組合的實係數", "the real coefficients of a linear combination" ], [ "\\operatorname{span}\\{w_1,w_2\\}", "所有有限線性組合的集合", "the set of all finite linear combinations" ], [ "(c_1,c_2)", "成員資格的 witness", "a membership witness" ] ], "md_error_controls": [ { "title": "無限和不是有限線性組合 / An infinite series is not a finite linear combination", "zh": "MKO 指出的誤解之一是把無限和的寫法直接當成有限線性組合表示式。Σ_{k=1}^{∞} 2^{−k}v₁ 的寫法有無限多項，但本例級數的和是 v₁，確實屬於 span，有限表示為 1·v₁。不能從表示式有無限多項推出其和不在 span 中。S03.1 只證有限次重排與合併；無限級數須另核收斂與重排合法性。", "en": "One mistake the MKO names is treating an infinite-series expression directly as a finite linear-combination representation. The expression Σ_{k=1}^{∞} 2^{−k}v₁ has infinitely many terms, but its sum here is v₁, which lies in the span and has the finite representation 1·v₁. Infinitely many terms do not imply that the sum is outside the span. S03.1 proves only finite reorderings and mergings; infinite series need separate convergence and reordering justification." }, { "title": "只看維度不夠 / Dimension alone does not decide membership", "zh": "MKO 的另一個誤解是只看維度就宣稱某向量在 span 中。t 與 t' 都是 ℝ³ 的向量，與 u₁、u₂ 同維度，但 t 不在而 t' 在 span{u₁,u₂} 中。成員資格必須由 witness 重建或方程組的矛盾決定。", "en": "Another mistake identified in the MKO is declaring a vector in the span by looking at its dimension. t and t' are both in ℝ³, the same dimension as u₁ and u₂, but t is not in span{u₁,u₂} while t' is. Membership must be decided by rebuilding a witness or by a contradiction in the equations." }, { "title": "係數與目標分量不要混淆 / Do not confuse coefficients with target components", "zh": "witness 是係數 (c₁,c₂)，不是目標 b 的分量。把 (7,0) 當係數會得 7(1,2)+0(3,−1)=(7,14)≠b。兩個方程也必須由同一組係數同時滿足：c₁=7、c₂=0 滿足第一式，卻使第二式為 14≠0。", "en": "The witness is the coefficient pair (c₁,c₂), not the components of the target b. Using (7,0) as coefficients gives 7(1,2)+0(3,−1)=(7,14)≠b. Both equations must also hold for one pair: c₁=7, c₂=0 satisfies the first equation but makes the second 14≠0." } ], "md_open_items": [ { "zh-Hant": "引理 L_span_two 只對兩個向量的族寫出；任意有限族的情形同理（把下標依所用生成向量分組），但本稿未寫出。det2_criterion（Cramer 法則）只作交叉核對，本稿不證明它。", "en": "Lemma L_span_two is written for families of two vectors only; any finite family works the same way (group the indices by the generator used), but this is not written out here. det2_criterion (Cramer's rule) is used only as a cross-check and is not proved here." }, { "zh-Hant": "本稿不處理線性獨立、基底與維度（MKO 的 formal next）；結論只屬於題目給定的生成族（MKO 假設 a2）。", "en": "Linear independence, bases and dimension (the MKO's formal next) are not treated; the conclusions concern only the given generator families (MKO assumption a2)." } ] }, "revision": { "id": "production-20261019-adopted-96", "source_commit": "b28a5edbd46279b1db2a3f3f48bb3f447e18899f", "adopted_source_sha256": "80267e70b6712e769dacf39909b666995788a4c88937ec1d58633f8a3b9b515f", "original_source_sha256": "179ee4ab9d0ac1428d4346b01cc0e9b8f510b5eea8a68acc501365a2ddaddca5", "owner_adopted": true, "bilingual_sync": "complete_source_bound", "formal_Lean": "NOT_RUN", "independent_AI_concurrence": "NOT_MEASURED", "issue_ids": [ "BR-M053", "BR-M054" ], "scope": "Owner-adopted natural-language teaching guide; not a formal proof, canonical MKO rewrite or evidence-status promotion." } }