{"schema_version":"ocme-site-results-v1","locale":"en","page_type":"derivations","facts":{"questions":36000,"reviewed":36000,"display_candidates":0,"knowledge":302,"guides":299,"paths":59,"batches":360,"families":1441,"atlas":302,"core":80,"evidence":9,"formal_evidence":5,"formal_subjects":4,"lean_files":6,"lean_toolchain":"leanprover/lean4:v4.30.0","question_state":"candidate","english_state":"ai_candidate","question_Lean":"NOT_RUN","canonical_truth_promotion":false},"filters":{"page":"6"},"kind":"guides","total":299,"pages":13,"page":6,"entries":[{"id":"derivation-mko-measurable-function-01","title":{"zh-Hant":"可測函數：以完整列舉逐一核逆像，並以合成、指示函數與像集陷阱對照","en":"Measurable function: checking every preimage by complete enumeration, with composition, indicators and the image-set trap"},"summary":{"zh-Hant":"X={0,1,2,3}、Σ_X={∅,{0,1},{2,3},X}；Y={0,1,2}、Σ_Y={∅,{0},{1,2},Y}；Z={0,1}、Σ_Z 為冪集。(1) 以完整列舉證明 f=(0↦0,1↦0,2↦1,3↦2) 可測，而 g=(0↦0,1↦1,2↦1,3↦2) 不可測；(2) 證明可測映射的合成可測，並把 φ∘f 讀回為指示函數；(3) 證明指示函數 1_E 可測若且唯若 E∈Σ_X；(4) 用 m 與 k 說明「檢查像集」既會誤拒也會誤收。","en":"X={0,1,2,3}, Σ_X={∅,{0,1},{2,3},X}; Y={0,1,2}, Σ_Y={∅,{0},{1,2},Y}; Z={0,1} with Σ_Z the power set. (1) Prove by complete enumeration that f=(0↦0,1↦0,2↦1,3↦2) is measurable and g=(0↦0,1↦1,2↦1,3↦2) is not; (2) prove that a composite of measurable maps is measurable, and read φ∘f back as an indicator function; (3) prove that an indicator 1_E is measurable exactly when E∈Σ_X; (4) use m and k to show that 'checking images' can both wrongly reject and wrongly accept."},"mko_ids":["mko-measurable-function"]},{"id":"derivation-mko-finite-markov-chain-01","title":{"zh-Hant":"有限狀態 Markov 鏈：核對行隨機矩陣，計算兩步時刻律與路徑機率","en":"Finite-state Markov chain: checking a row-stochastic matrix and computing two-step time laws and path probabilities"},"summary":{"zh-Hant":"狀態依序為 1、2、3，轉移矩陣 P 的行為 (1/2,1/2,0)、(1/4,1/2,1/4)、(0,1/3,2/3)，初始律 μ_0=(1/2,1/2,0)。(1) 核對 P 是行隨機矩陣，並說明列和不必為 1；(2) 證明一步公式 μ_{n+1}=μ_nP 且結果仍是機率向量，算出 μ_1 與 μ_2；(3) 證明路徑機率公式並計算 P(X_0=1,X_1=2,X_2=3)、P(X_0=2,X_1=3,X_2=3)；(4) 說明 P 單獨一般不能決定完整的無條件時刻律與路徑律。","en":"The states are ordered 1, 2, 3; the transition matrix P has rows (1/2,1/2,0), (1/4,1/2,1/4), (0,1/3,2/3), and the initial law is μ_0=(1/2,1/2,0). (1) Check that P is row-stochastic and explain why column sums need not be 1; (2) prove the one-step formula μ_{n+1}=μ_nP and that the result is again a probability vector, and compute μ_1 and μ_2; (3) prove the path-probability formula and compute P(X_0=1,X_1=2,X_2=3) and P(X_0=2,X_1=3,X_2=3); (4) explain why P alone generally does not determine complete unconditional marginal and path laws."},"mko_ids":["mko-finite-markov-chain"]},{"id":"derivation-mko-partial-derivative-01","title":{"zh-Hant":"偏導數：由差商極限推出二次多項式的偏導公式，並計算一點的梯度","en":"Partial derivative: deriving the partial-derivative formulas of a quadratic from the difference-quotient limit, and computing a gradient"},"summary":{"zh-Hant":"對係數依 [1,x,y,x²,xy,y²] 排列的二次多項式 f，從差商極限的定義證明 ∂f/∂x=c_1+2c_3x+c_4y、∂f/∂y=c_2+c_4x+2c_5y。再對 c=(1,−2,3,1,2,−1) 算出 (1,2) 的偏導與梯度 (4,1)，並說明單一非零 h 的差商（5 或 41/10）與同時移動兩個座標（極限 5）都不是 x 偏導。","en":"For a quadratic f with coefficients in the order [1,x,y,x²,xy,y²], prove from the difference-quotient limit that ∂f/∂x=c_1+2c_3x+c_4y and ∂f/∂y=c_2+c_4x+2c_5y. Then for c=(1,−2,3,1,2,−1) compute the partials and the gradient (4,1) at (1,2), and show that a single nonzero-h quotient (5 or 41/10) and moving both coordinates at once (limit 5) are not the x-partial."},"mko_ids":["mko-partial-derivative"]},{"id":"derivation-mko-finite-convex-combination-01","title":{"zh-Hant":"有限凸組合：以精確權重核對見證，並區分負權重、近似權重與唯一性","en":"Finite convex combination: checking a witness with exact weights, and separating negative weights, approximate weights and uniqueness"},"summary":{"zh-Hant":"在 ℝ² 中令 v_1=(0,0)、v_2=(6,0)、v_3=(0,6)。(1) 核對 λ=(1/6,1/3,1/2) 是凸權重並重建 w=(2,3)；(2) 說明零權重允許、兩點凸組合是線段點；(3) 證明 u=(4,4) 不是 v_1、v_2、v_3 的凸組合（唯一的仿射權重含 −1/3），並以線性泛函上界給第二個證明；(4) 說明 0.333 三個權重不合格；(5) 加入 v_4=(3,3) 後，同一點有兩組不同的凸權重。","en":"In ℝ² let v_1=(0,0), v_2=(6,0), v_3=(0,6). (1) Check that λ=(1/6,1/3,1/2) are convex weights and reconstruct w=(2,3); (2) show that zero weights are allowed and that two-point convex combinations are segment points; (3) prove that u=(4,4) is not a convex combination of v_1, v_2, v_3 (its only affine weights include −1/3), with a second proof by a linear-functional bound; (4) show that three weights of 0.333 do not qualify; (5) after adding v_4=(3,3), exhibit two different convex weight vectors for the same point."},"mko_ids":["mko-finite-convex-combination"]},{"id":"derivation-mko-partial-differential-equation-01","title":{"zh-Hant":"偏微分方程：常係數二階算子的主部分類與二次多項式候選的符號殘差","en":"Partial differential equation: principal-part classification and symbolic residuals of quadratic polynomial candidates"},"summary":{"zh-Hant":"在 ℝ² 上給定線性常係數算子 L(u)=u_xx+2u_xy−3u_yy+u_x 與右端 g(x,y)=2x+y−2。先依明示的主部約定 a u_xx+b u_xy+c u_yy 與判別式 b²−4ac 對 L 分類；再對三個二次多項式候選 u_A=x²+xy+y²、u_B=x²+2xy、u_C=x²+xy+y²+y 精確計算符號殘差 R=L(u)−g，判定哪些候選在全 ℝ² 上滿足 L(u)=g，並說明 u_B 的點殘差在整條直線 y=−8 上為零，為何仍不構成解。","en":"On ℝ², take the linear constant-coefficient operator L(u)=u_xx+2u_xy−3u_yy+u_x and the right-hand side g(x,y)=2x+y−2. First classify L using the explicit principal-part convention a u_xx+b u_xy+c u_yy with discriminant b²−4ac; then, for the three quadratic polynomial candidates u_A=x²+xy+y², u_B=x²+2xy and u_C=x²+xy+y²+y, compute the symbolic residual R=L(u)−g exactly, decide which candidates satisfy L(u)=g on all of ℝ², and explain why u_B is still not a solution although its point residual vanishes on the whole line y=−8."},"mko_ids":["mko-partial-differential-equation"]},{"id":"derivation-mko-convex-set-01","title":{"zh-Hant":"凸集合：半空間與軸對齊盒的完整凸性證書，及有限點集的中點反例","en":"Convex set: complete convexity certificates for a halfspace and an axis-aligned box, and a midpoint counterexample for a finite point set"},"summary":{"zh-Hant":"在實向量空間 ℝ² 中：（一）證明閉半空間 H={(x,y) ∣ 2x−3y≤6} 是凸集合；（二）證明非空軸對齊盒 B=[−1,2]×[0,3] 是凸集合；兩個證明都必須對所有 p、q 與所有 λ∈[0,1] 成立。（三）用一個精確的中點見證否證有限點集 F={(0,0),(2,0),(0,2)} 的凸性，並說明為何抽查有限個 λ、或找不到反例，都不能證明凸性。","en":"In the real vector space ℝ²: (1) prove that the closed halfspace H={(x,y) ∣ 2x−3y≤6} is convex; (2) prove that the nonempty axis-aligned box B=[−1,2]×[0,3] is convex; both proofs must hold for all p, q and all λ∈[0,1]. (3) Refute the convexity of the finite point set F={(0,0),(2,0),(0,2)} with one exact midpoint witness, and explain why checking finitely many λ, or failing to find a counterexample, cannot prove convexity."},"mko_ids":["mko-convex-set"]},{"id":"derivation-mko-boundary-value-problem-01","title":{"zh-Hant":"邊界值問題：矩形上 Poisson–Dirichlet 問題的內部與四邊分開核對","en":"Boundary-value problem: checking the interior and all four edges separately for a Poisson–Dirichlet problem on a rectangle"},"summary":{"zh-Hant":"在矩形 Ω=(0,2)×(0,1) 上宣告 BVP：內部 Δu=u_xx+u_yy=4；Dirichlet 資料為左邊 u(0,y)=y²、右邊 u(2,y)=y²+2y+4（0≤y≤1）、下邊 u(x,0)=x²、上邊 u(x,1)=x²+x+1（0≤x≤2）。對二次多項式候選 u=x²+xy+y² 與 v=2x²+xy−2x+y，分別記錄內部 Poisson 義務（符號 Laplacian 恆等式）與四條邊的 Dirichlet 義務（精確限制到整條邊後逐係數比較），判定誰滿足此宣告的 BVP，並說明 v 為何在內部方程成立、四個角點也都吻合的情況下仍不滿足。","en":"On the rectangle Ω=(0,2)×(0,1) declare the BVP: in the interior Δu=u_xx+u_yy=4; Dirichlet data u(0,y)=y² on the left, u(2,y)=y²+2y+4 on the right (0≤y≤1), u(x,0)=x² on the bottom and u(x,1)=x²+x+1 on the top (0≤x≤2). For the quadratic polynomial candidates u=x²+xy+y² and v=2x²+xy−2x+y, record separately the interior Poisson obligation (a symbolic Laplacian identity) and the Dirichlet obligation on all four edges (exact restriction to each whole edge, then coefficient comparison); decide which candidate satisfies this declared BVP, and explain why v fails although its interior equation holds and it matches the data at all four corners."},"mko_ids":["mko-boundary-value-problem"]},{"id":"derivation-mko-convex-function-01","title":{"zh-Hant":"凸函數：二次函數 Jensen gap 的完整證書與單一 Jensen 反例","en":"Convex function: a complete certificate via the Jensen gap of a quadratic, and a single Jensen counterexample"},"summary":{"zh-Hant":"（一）證明 f(x)=2x²−3x+1 在區間 I=[−1,4] 上是凸函數：先確認 I 是凸集合，再對所有 x,y∈I 與所有 λ∈[0,1] 證明 f(λx+(1−λ)y)≤λf(x)+(1−λ)f(y)，方法是證明二次函數的 Jensen gap 等於 aλ(1−λ)(x−y)²。（二）由同一恆等式得仿射函數 ℓ(x)=3x−5 在 I 上取等號。（三）用一個精確的 Jensen 違反否證 g(x)=−x²+x 在 J=[0,2] 上的凸性，並說明幾個通過的樣本為何不能證明凸性。","en":"(1) Prove that f(x)=2x²−3x+1 is convex on the interval I=[−1,4]: first confirm that I is convex, then prove f(λx+(1−λ)y)≤λf(x)+(1−λ)f(y) for all x,y∈I and all λ∈[0,1], by showing that the Jensen gap of a quadratic equals aλ(1−λ)(x−y)². (2) From the same identity, the affine function ℓ(x)=3x−5 gives equality on I. (3) Refute the convexity of g(x)=−x²+x on J=[0,2] with one exact Jensen violation, and explain why a few passing samples cannot prove convexity."},"mko_ids":["mko-convex-function"]},{"id":"derivation-mko-convex-optimization-problem-01","title":{"zh-Hant":"凸最佳化問題：資格核對、可行性、單點求值與兩候選比較，以及它們不能推出什麼","en":"Convex optimization problem: qualification, feasibility, single-point evaluation and a two-candidate comparison, and what they do not imply"},"summary":{"zh-Hant":"問題 P：在可行域 X=[0,5] 上最小化 f(x)=x²−6x+10。（一）分別核對可行域凸與目標凸，判定 P 具凸問題資格；（二）判定候選 x₁=1、x₂=4、x₃=−1 的可行性；（三）求 f(1)、f(4) 並比較兩個可行候選；（四）說明兩候選中較低者不因此是 P 的最優解，並以 P′：在 X′=(0,1] 上最小化 x² 說明凸問題不一定有最小值點。","en":"Problem P: minimize f(x)=x²−6x+10 over the feasible domain X=[0,5]. (1) Check separately that the domain and the objective are convex, so that P qualifies as a convex problem; (2) decide feasibility of the candidates x₁=1, x₂=4, x₃=−1; (3) evaluate f(1) and f(4) and compare the two feasible candidates; (4) explain why the candidate with the lower objective value is not thereby an optimal solution of P, and use P′: minimize x² over X′=(0,1], to show that a convex problem need not have a minimizer."},"mko_ids":["mko-convex-optimization-problem"]},{"id":"derivation-mko-finite-abelian-group-01","title":{"zh-Hant":"有限阿貝爾群：由完整 Cayley 表先核群資格、再核交換性","en":"Finite abelian group: from a complete Cayley table, group qualification first, then commutativity"},"summary":{"zh-Hant":"載體 G={0,1,2,3} 上給出完整的 4×4 Cayley 表（行 a、列 b 的格子是 a*b）：第 0 行 (1,2,3,0)，第 1 行 (2,3,0,1)，第 2 行 (3,0,1,2)，第 3 行 (0,1,2,3)。依序核對封閉、結合、唯一單位元與逆元，再對所有 a、b 核對 a*b=b*a，判定 (G,*) 是否為有限阿貝爾群；求元素 0 與 1 的階與它們生成的循環子群；並以 {0,1,2} 上的對稱表 a∘b=|a−b| 說明「表對稱」不能代替群資格。","en":"On the carrier G={0,1,2,3} a complete 4×4 Cayley table is given (the cell in row a, column b is a*b): row 0 is (1,2,3,0), row 1 is (2,3,0,1), row 2 is (3,0,1,2), row 3 is (0,1,2,3). Check closure, associativity, a unique identity and inverses in that order, then check a*b=b*a for all a, b, to decide whether (G,*) is a finite abelian group; find the orders of the elements 0 and 1 and the cyclic subgroups they generate; and use the symmetric table a∘b=|a−b| on {0,1,2} to show that a symmetric table cannot replace group qualification."},"mko_ids":["mko-finite-abelian-group"]},{"id":"derivation-mko-real-normed-vector-space-01","title":{"zh-Hant":"實賦範向量空間：ℓ¹ 與 ℓ^∞ 公式的全稱範數證書，以及兩個非範數公式","en":"Real normed vector space: universal norm certificates for the ℓ¹ and ℓ^∞ formulas, and two formulas that are not norms"},"summary":{"zh-Hant":"在 ℝⁿ（n≥1 固定）上，對所有向量 u、v 與所有實數 a，證明 ‖v‖₁=Σ|v_i| 與 ‖v‖_∞=max|v_i| 都滿足範數的三條公理：非負且 ‖v‖=0 ⇔ v=0、‖av‖=|a|‖v‖、‖u+v‖≤‖u‖+‖v‖。再在 ℝ³ 中對整數座標向量 u=(3,−4,1)、v=(−1,2,5) 計算兩種範數與三角不等式、齊次性的實例；最後說明 φ(v)=|v₁| 與 ψ(v)=(‖v‖₁)² 不是範數。","en":"On ℝⁿ (n≥1 fixed), for all vectors u, v and all reals a, prove that ‖v‖₁=Σ|v_i| and ‖v‖_∞=max|v_i| both satisfy the three norm axioms: nonnegativity with ‖v‖=0 ⇔ v=0, ‖av‖=|a|‖v‖, and ‖u+v‖≤‖u‖+‖v‖. Then, in ℝ³, compute both norms and instances of the triangle inequality and homogeneity for the integer-coordinate vectors u=(3,−4,1) and v=(−1,2,5); finally show that φ(v)=|v₁| and ψ(v)=(‖v‖₁)² are not norms."},"mko_ids":["mko-real-normed-vector-space"]},{"id":"derivation-mko-cauchy-sequence-01","title":{"zh-Hant":"Cauchy 序列：倒數序列的完整證書、序列在其中無極限的空間、交替反例與最終常值模型","en":"Cauchy sequence: a complete certificate for the reciprocal sequence, a space in which it has no limit, an alternating counterexample and an ultimately constant model"},"summary":{"zh-Hant":"（一）證明 x_n=1/n（n≥1）在 (ℝ, |s−t|) 中是 Cauchy 序列：對每個 ε>0 給出 N，使所有 m,n≥N 都有 |1/m−1/n|<ε。（二）證明同一序列在 X=(0,1]（限制度量）中仍是 Cauchy，但在 X 中沒有極限。（三）證明 y_n=(−1)^n 不是 Cauchy。（四）證明最終常值序列 z（z₁=0、z₂=7、n≥3 時 z_n=5）是 Cauchy。並說明有限尾段為何不能證明 Cauchy。","en":"(1) Prove that x_n=1/n (n≥1) is a Cauchy sequence in (ℝ, |s−t|): for every ε>0 give N such that |1/m−1/n|<ε for all m,n≥N. (2) Prove that the same sequence is still Cauchy in X=(0,1] with the restricted metric, but has no limit in X. (3) Prove that y_n=(−1)^n is not Cauchy. (4) Prove that the ultimately constant sequence z (z₁=0, z₂=7, z_n=5 for n≥3) is Cauchy. Also explain why a finite tail cannot prove the Cauchy property."},"mko_ids":["mko-cauchy-sequence"]},{"id":"derivation-mko-finite-joint-distribution-01","title":{"zh-Hant":"有限聯合分布：驗證聯合質量表並由同一 joint law 聚合邊際與事件","en":"Finite joint distribution: validating a joint mass table and aggregating marginals and events from one joint law"},"summary":{"zh-Hant":"給定支撐 S={(0,0),(0,1),(1,0),(1,1),(2,1)} 與質量 p(0,0)=1/6、p(0,1)=1/3、p(1,0)=1/4、p(1,1)=1/12、p(2,1)=1/6。驗證 (S,p) 是二維有限聯合分布；由同一個 joint law 算出 X 與 Y 的邊際、矩形事件 P(X∈{1,2}, Y=1) 與單點質量 P(X=2, Y=0)；最後構造一個邊際相同但不同的 joint law，說明兩個邊際一般不能反推聯合分布。","en":"Given the support S={(0,0),(0,1),(1,0),(1,1),(2,1)} with masses p(0,0)=1/6, p(0,1)=1/3, p(1,0)=1/4, p(1,1)=1/12, p(2,1)=1/6, verify that (S,p) is a two-dimensional finite joint distribution; from this one joint law compute the marginals of X and Y, the rectangle event P(X∈{1,2}, Y=1) and the single-pair mass P(X=2, Y=0); finally construct a different joint law with the same marginals, showing that two marginals do not in general determine the joint distribution."},"mko_ids":["mko-finite-joint-distribution"]},{"id":"derivation-mko-covariance-01","title":{"zh-Hant":"協方差：對稱性、仿射規則與「零協方差不蘊含獨立」","en":"Covariance: symmetry, the affine rule, and why zero covariance does not imply independence"},"summary":{"zh-Hant":"在同一個精確有限 joint law 上：(1) 證明 Cov(Y,X)=Cov(X,Y)；(2) 證明 Cov(aX+b,cY+d)=ac·Cov(X,Y)；(3) 證明 X、Y 獨立時 Cov(X,Y)=0，因此非零協方差可否證獨立；(4) 對 X 均勻取 −1、0、1 且 Y=X² 的 joint law，算出 E[X]、E[Y]、E[XY] 與 Cov(X,Y)=0，並證明 X、Y 不獨立。","en":"On one exact finite joint law: (1) prove Cov(Y,X)=Cov(X,Y); (2) prove Cov(aX+b,cY+d)=ac·Cov(X,Y); (3) prove that independence of X and Y gives Cov(X,Y)=0, so a nonzero covariance refutes independence; (4) for the joint law with X uniform on −1, 0, 1 and Y=X², compute E[X], E[Y], E[XY] and Cov(X,Y)=0, and prove that X and Y are not independent."},"mko_ids":["mko-covariance"]},{"id":"derivation-mko-graph-isomorphism-01","title":{"zh-Hant":"圖同構：以完整映射表逐項核對雙射與鄰接當且僅當","en":"Graph isomorphism: checking bijectivity and both directions of adjacency preservation from a complete mapping table"},"summary":{"zh-Hant":"G 的頂點為 1、2、3、4，邊為 {1,2}、{2,3}、{3,4}、{1,4}；H 的頂點為 a、b、c、d，邊為 {a,c}、{b,c}、{b,d}、{a,d}。證明映射 f：1↦a、2↦c、3↦b、4↦d 是圖同構 G→H；說明雙射 g：1↦a、2↦b、3↦c、4↦d 不是同構；並證明路徑 P（邊 {1,2}、{2,3}、{3,4}）與 H 不同構，即使 f 把 P 的每條邊都送到 H 的邊。","en":"G has vertices 1, 2, 3, 4 and edges {1,2}, {2,3}, {3,4}, {1,4}; H has vertices a, b, c, d and edges {a,c}, {b,c}, {b,d}, {a,d}. Prove that f: 1↦a, 2↦c, 3↦b, 4↦d is a graph isomorphism G→H; show that the bijection g: 1↦a, 2↦b, 3↦c, 4↦d is not an isomorphism; and prove that the path P (edges {1,2}, {2,3}, {3,4}) is not isomorphic to H, even though f sends every edge of P to an edge of H."},"mko_ids":["mko-graph-isomorphism"]},{"id":"derivation-mko-equivalence-relation-01","title":{"zh-Hant":"等價關係：在完整有限關係表上逐項核對三公理並輸出等價類","en":"Equivalence relation: checking the three axioms on a complete finite relation and emitting the classes"},"summary":{"zh-Hant":"在 X={0,1,2,3,4} 上，關係 R 完整列出為 13 個有序對：(0,0)、(1,1)、(2,2)、(3,3)、(4,4)、(0,2)、(2,0)、(1,3)、(3,1)、(1,4)、(4,1)、(3,4)、(4,3)。逐項核對反身、對稱、傳遞，證明 R 是等價關係並輸出等價類；再說明去掉 (3,4)、(4,3) 後的 R′ 雖反身且對稱，卻不是等價關係。","en":"On X={0,1,2,3,4} the relation R is listed completely as 13 ordered pairs: (0,0), (1,1), (2,2), (3,3), (4,4), (0,2), (2,0), (1,3), (3,1), (1,4), (4,1), (3,4), (4,3). Check reflexivity, symmetry and transitivity item by item, prove that R is an equivalence relation and emit its classes; then show that R′, obtained by removing (3,4) and (4,3), is reflexive and symmetric but not an equivalence relation."},"mko_ids":["mko-equivalence-relation"]},{"id":"derivation-mko-finite-unital-ring-01","title":{"zh-Hant":"有限含一環：證明模 4 的加法表與乘法表構成含一環","en":"Finite unital ring: proving that the mod-4 addition and multiplication tables form a unital ring"},"summary":{"zh-Hant":"在載體 X={0,1,2,3} 上給定完整的加法表 ⊕ 與乘法表 ⊙（即模 4 的加法與乘法）。對所有元素對與三元組證明：(X,⊕) 是阿貝爾群；⊙ 結合；⊙ 對 ⊕ 左右分配；存在唯一的雙側乘法單位元 1，且 1≠0。並指出 2⊙2=0 不妨礙它是含一環。","en":"On the carrier X={0,1,2,3}, complete addition and multiplication tables ⊕ and ⊙ (addition and multiplication mod 4) are given. Prove for all pairs and triples: (X,⊕) is an abelian group; ⊙ is associative; ⊙ distributes over ⊕ on both sides; there is a unique two-sided multiplicative identity 1, and 1≠0. Note also that 2⊙2=0 does not stop it from being a unital ring."},"mko_ids":["mko-finite-unital-ring"]},{"id":"derivation-mko-norm-induced-metric-01","title":{"zh-Hant":"範數誘導度量：由範數公理證明 d(x,y)=‖x−y‖ 是度量，並在 ℝ³ 上以 l1 與 l∞ 精確計算","en":"Norm-induced metric: proving from the norm axioms that d(x,y)=‖x−y‖ is a metric, with exact l1 and l∞ computations in ℝ³"},"summary":{"zh-Hant":"設 (V,‖·‖) 是實賦範向量空間。證明 d(x,y)=‖x−y‖ 滿足非負、分離、對稱與三角不等式；證明 ℝⁿ 上的 l1 與 l∞ 公式都是範數，所以都誘導度量；再在 ℝ³ 上對 x=(1,−2,3)、y=(4,0,−1)、z=(2,5,1) 精確計算 d1 與 d∞。","en":"Let (V,‖·‖) be a real normed vector space. Prove that d(x,y)=‖x−y‖ is nonnegative, separates points, is symmetric and satisfies the triangle inequality; prove that the l1 and l∞ formulas on ℝⁿ are norms, so both induce metrics; then compute d1 and d∞ exactly in ℝ³ for x=(1,−2,3), y=(4,0,−1), z=(2,5,1)."},"mko_ids":["mko-norm-induced-metric"]},{"id":"derivation-mko-complete-metric-space-01","title":{"zh-Hant":"完備度量空間：以最小正距離證明有限度量空間完備，並以 (0,1] 對照","en":"Complete metric space: proving every finite metric space complete by the least positive distance, contrasted with (0,1]"},"summary":{"zh-Hant":"證明：每個至少有一點的有限度量空間 (X,d) 都是完備的（單點情形直接成立；至少兩點時用最小正距離 δ 證明每個 Cauchy 序列最終常值）。在 X0={0,1,3}、d(x,y)=|x−y| 上算出 δ=1，給一個最終常值證書與一個明示的非 Cauchy 序列；最後證明子空間 Y=(0,1] 不完備：x_n=1/n 在 Y 中是 Cauchy，但在 Y 中沒有極限。","en":"Prove that every finite metric space (X,d) with at least one point is complete (immediate for a single point; with at least two points, the least positive distance δ forces every Cauchy sequence to be ultimately constant). On X0={0,1,3} with d(x,y)=|x−y|, compute δ=1, give an ultimately constant certificate and an explicit non-Cauchy sequence; finally prove that the subspace Y=(0,1] is not complete: x_n=1/n is Cauchy in Y but has no limit in Y."},"mko_ids":["mko-complete-metric-space"]},{"id":"derivation-mko-real-bilinear-form-01","title":{"zh-Hant":"實雙線性形式：以矩陣公式 x^T A y 分槽證明雙線性並精確求值","en":"Real bilinear form: proving bilinearity slot by slot from x^T A y, with exact evaluation"},"summary":{"zh-Hant":"設 A=[[1,2],[−1,3]]，B(x,y)=x^T A y（x、y∈ℝ²）。分別證明 B 對第一槽與第二槽都是線性的（雙線性）；對 x=(2,−1)、y=(1,4) 精確求 B(x,y)=7 與 B(y,x)=−20，說明雙線性不要求對稱；並證明黑箱函數 f(x,y)=x1y1+(x2³−x2)y2 雖通過若干加法樣本卻不是雙線性。","en":"Let A=[[1,2],[−1,3]] and B(x,y)=x^T A y for x, y∈ℝ². Prove separately that B is linear in the first slot and in the second slot (bilinear); evaluate exactly B(x,y)=7 and B(y,x)=−20 for x=(2,−1), y=(1,4), showing that bilinearity does not require symmetry; and prove that the black-box function f(x,y)=x1y1+(x2³−x2)y2 passes some additivity samples but is not bilinear."},"mko_ids":["mko-real-bilinear-form"]},{"id":"derivation-mko-general-linear-map-01","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"},"summary":{"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."},"mko_ids":["mko-general-linear-map"]},{"id":"derivation-mko-event-independence-01","title":{"zh-Hant":"事件獨立：以交集乘積判定，並核對條件式、補集與零機率事件","en":"Event independence: deciding by the product rule, with the conditional form, complements and a zero-probability event"},"summary":{"zh-Hant":"公正骰子模型：Ω={1,2,3,4,5,6}，每個結果的質量為 1/6。設 A={2,4,6}、B={1,2,3,4}、D={1,3,5}、Z=∅。（一）以乘積定義判定 A 與 B 是否獨立，並在 P(B)>0 下核對 P(A|B)=P(A)。（二）證明 A 與 B 獨立推出 A 與 Bᶜ 獨立，並以數字核對。（三）判定互斥事件 A、D 是否獨立。（四）判定 A 與零機率事件 Z 是否獨立，並說明條件式為何不能使用。","en":"Fair-die model: Ω={1,2,3,4,5,6}, each outcome with mass 1/6. Let A={2,4,6}, B={1,2,3,4}, D={1,3,5} and Z=∅. (1) Decide by the product definition whether A and B are independent, and with P(B)>0 check P(A|B)=P(A). (2) Prove that independence of A and B implies independence of A and Bᶜ, and check it numerically. (3) Decide whether the mutually exclusive events A and D are independent. (4) Decide whether A and the zero-probability event Z are independent, and explain why the conditional form cannot be used."},"mko_ids":["mko-event-independence"]},{"id":"derivation-mko-graph-walk-simple-path-01","title":{"zh-Hant":"圖的 walk 與 simple path：逐對核對、長度與 trail 的區別","en":"Graph walks and simple paths: pair-by-pair checks, length and the difference from trails"},"summary":{"zh-Hant":"有限無向無 loop 簡單圖 G：V={0,1,2,3,4}，E={{0,1},{1,2},{2,3},{0,3},{1,3},{3,4}}。對頂點序列 W₁=(0,1,2,3,4)、W₂=(0,1,3,1,2)、W₃=(0,2,3)、W₄=(2,3,0,1,3,4)、W₅=(4)、W₆=(1,1) 逐一判定是否為 walk、長度為何、是否為 simple path；並證明 simple path 不重複邊，以及本圖中 simple path 的長度至多為 4。","en":"Finite undirected loop-free simple graph G: V={0,1,2,3,4}, E={{0,1},{1,2},{2,3},{0,3},{1,3},{3,4}}. For the vertex sequences W₁=(0,1,2,3,4), W₂=(0,1,3,1,2), W₃=(0,2,3), W₄=(2,3,0,1,3,4), W₅=(4) and W₆=(1,1), decide whether each is a walk, its length, and whether it is a simple path; also prove that a simple path repeats no edge and that in this graph a simple path has length at most 4."},"mko_ids":["mko-graph-walk-simple-path"]},{"id":"derivation-mko-finite-commutative-ring-01","title":{"zh-Hant":"有限交換環：逐元素對核對 ℤ₆ 的交換律，並以上三角矩陣環為非交換反例","en":"Finite commutative ring: checking commutativity of ℤ₆ over all element pairs, with an upper-triangular matrix ring as a non-commutative counterexample"},"summary":{"zh-Hant":"（一）在已通過有限含一環資格的 ℤ₆={0,1,2,3,4,5}（a⊕b=(a+b) mod 6、a⊗b=(ab) mod 6）上，核對全部 36 個有序元素對的 a⊗b=b⊗a，判定 ℤ₆ 是否為有限交換環。（二）說明 ℤ₆ 有非零零因子且 2 不可逆，所以不是整環或域。（三）在已通過資格的 ℱ₂ 上 2×2 上三角矩陣環 U 中，說明若干乘積可交換，但 U 不是交換環。","en":"(1) On ℤ₆={0,1,2,3,4,5} (a⊕b=(a+b) mod 6, a⊗b=(ab) mod 6), already qualified as a finite unital ring, check a⊗b=b⊗a for all 36 ordered element pairs and decide whether ℤ₆ is a finite commutative ring. (2) Show that ℤ₆ has nonzero zero divisors and that 2 is not invertible, so it is neither an integral domain nor a field. (3) In the qualified ring U of 2-by-2 upper-triangular matrices over 𝔽₂, show that several products commute and yet U is not commutative."},"mko_ids":["mko-finite-commutative-ring"]}],"title":"Concept derivation guides"}