{"schema_version":"ocme-site-results-v1","locale":"zh-Hant","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":"7"},"kind":"guides","total":299,"pages":13,"page":7,"entries":[{"id":"derivation-mko-finite-two-sided-ideal-01","title":{"zh-Hant":"有限雙邊理想：在上三角矩陣環中分開核對左吸收與右吸收","en":"Finite two-sided ideal: checking left and right absorption separately in an upper-triangular matrix ring"},"summary":{"zh-Hant":"父環 U 是 𝔽₂ 上 2×2 上三角矩陣環（元素記為 (a,b,c)，代表第一行 (a,b)、第二行 (0,c)），已通過有限含一環資格。（一）證明 I={(0,0,0),(0,1,0)} 是 U 的雙邊理想。（二）證明 L={(0,0,0),(1,0,0)} 滿足左吸收，但不是雙邊理想。（三）證明對角子環 D={(a,0,c)} 含 1 卻不是理想。（四）說明 {0} 與 U 本身是雙邊理想。","en":"The parent ring U is the ring of 2-by-2 upper-triangular matrices over 𝔽₂ (elements written (a,b,c) for the matrix with first row (a,b) and second row (0,c)), already qualified as a finite unital ring. (1) Prove that I={(0,0,0),(0,1,0)} is a two-sided ideal of U. (2) Prove that L={(0,0,0),(1,0,0)} satisfies left absorption but is not a two-sided ideal. (3) Prove that the diagonal subring D={(a,0,c)} contains 1 but is not an ideal. (4) Show that {0} and U itself are two-sided ideals."},"mko_ids":["mko-finite-two-sided-ideal"]},{"id":"derivation-mko-real-banach-space-01","title":{"zh-Hant":"實 Banach 空間：(ℝ², ℓ¹) 的範數、誘導度量與完備性","en":"Real Banach space: norm, induced metric and completeness of (ℝ², ℓ¹)"},"summary":{"zh-Hant":"在 V=ℝ² 上取 ℓ¹ 範數 ‖x‖₁=|x₁|+|x₂| 與同一範數誘導的度量 d₁(x,y)=‖x−y‖₁。（一）證明 ‖·‖₁ 是範數、d₁ 是度量，並算出 d₁((3,−1),(1,2))。（二）由實數的 Cauchy 完備性完整證明 (ℝ²,d₁) 完備，所以 (ℝ²,‖·‖₁) 是實 Banach 空間；並引用有限維實賦範空間完備的命名定理。（三）證明 ‖·‖₁ 不滿足平行四邊形律，所以不是由任何內積誘導。","en":"On V=ℝ² take the ℓ¹ norm ‖x‖₁=|x₁|+|x₂| and the metric induced by that same norm, d₁(x,y)=‖x−y‖₁. (1) Prove that ‖·‖₁ is a norm and d₁ a metric, and compute d₁((3,−1),(1,2)). (2) Prove completely, from the Cauchy completeness of the reals, that (ℝ²,d₁) is complete, so (ℝ²,‖·‖₁) is a real Banach space; also cite the named theorem that finite-dimensional real normed spaces are complete. (3) Prove that ‖·‖₁ violates the parallelogram law and hence is induced by no inner product."},"mko_ids":["mko-real-banach-space"]},{"id":"derivation-mko-real-inner-product-space-01","title":{"zh-Hant":"實內積空間：以 Sylvester 條件為 ℝ² 上的矩陣形式作全稱正定證書","en":"Real inner-product space: Sylvester conditions as a universal positive-definiteness certificate for a matrix form on ℝ²"},"summary":{"zh-Hant":"設 M 為第一行 (2,1)、第二行 (1,3) 的整數對稱矩陣，⟨x,y⟩_M=xᵀMy（x,y∈ℝ²）。（一）分別證明 ⟨·,·⟩_M 雙線性、對稱、正定，所以是 ℝ² 上的內積；正定性以 2×2 Sylvester 條件（完整證明其充分與必要方向）作全稱證書。（二）判定 N（行 (1,2),(2,1)）、P（行 (1,1),(1,1)）、R（行 (2,1),(0,3)）給出的形式為何不是內積。","en":"Let M be the symmetric integer matrix with first row (2,1) and second row (1,3), and ⟨x,y⟩_M=xᵀMy for x,y∈ℝ². (1) Prove separately that ⟨·,·⟩_M is bilinear, symmetric and positive definite, hence an inner product on ℝ²; positive definiteness is certified universally by the 2-by-2 Sylvester conditions, with both sufficiency and necessity proved in full. (2) Decide why the forms given by N (rows (1,2),(2,1)), P (rows (1,1),(1,1)) and R (rows (2,1),(0,3)) are not inner products."},"mko_ids":["mko-real-inner-product-space"]},{"id":"derivation-mko-bounded-linear-operator-01","title":{"zh-Hant":"有界線性算子：矩陣在 ℓ^∞ 與 ℓ¹ 下的精確誘導算子範數","en":"Bounded linear operator: exact induced operator norms of a matrix under ℓ^∞ and ℓ¹"},"summary":{"zh-Hant":"設 A 為第一行 (1,−2)、第二行 (3,4) 的整數矩陣，T(x)=Ax:ℝ²→ℝ²。（一）在兩端都用 ℓ^∞ 範數時，證明對所有 x 有 ‖Ax‖_∞≤7‖x‖_∞，且沒有更小的常數可用，所以 ‖T‖_{∞→∞}=7（最大行絕對值和）。（二）兩端都用 ℓ¹ 範數時，證明 ‖T‖_{1→1}=6（最大列絕對值和）。（三）說明非最佳常數、有限輸入與值域無界這三個陷阱。","en":"Let A be the integer matrix with first row (1,−2) and second row (3,4), and T(x)=Ax:ℝ²→ℝ². (1) With the ℓ^∞ norm at both ends, prove that ‖Ax‖_∞≤7‖x‖_∞ for all x and that no smaller constant works, so ‖T‖_{∞→∞}=7 (the maximum absolute row sum). (2) With the ℓ¹ norm at both ends, prove ‖T‖_{1→1}=6 (the maximum absolute column sum). (3) Explain three traps: non-optimal constants, finitely many inputs, and unbounded range."},"mko_ids":["mko-bounded-linear-operator"]},{"id":"derivation-mko-random-variable-independence-01","title":{"zh-Hant":"隨機變數獨立：完整核對有限聯合表，零協方差只是後果","en":"Independence of random variables: checking the complete finite joint table, with zero covariance only a consequence"},"summary":{"zh-Hant":"（一）聯合表 J₁：X∈{0,1}、Y∈{0,1,2}，P(0,0)=1/12、P(0,1)=1/12、P(0,2)=1/6、P(1,0)=1/6、P(1,1)=1/6、P(1,2)=1/3。由聯合表聚合兩個邊際，核對全部 6 個值對的乘積式，判定 X、Y 是否獨立；並證明獨立推出任意座標事件 {X∈S} 與 {Y∈T} 獨立與 E[XY]=E[X]E[Y]。（二）聯合表 J₂：X 在 {−1,0,1} 上均勻、Y=X²。證明 Cov(X,Y)=0 但 X、Y 不獨立。（三）聯合表 J₃ 說明只核一個值對不夠。","en":"(1) Joint table J₁: X∈{0,1}, Y∈{0,1,2}, P(0,0)=1/12, P(0,1)=1/12, P(0,2)=1/6, P(1,0)=1/6, P(1,1)=1/6, P(1,2)=1/3. Aggregate both marginals from the joint table, check the product equation at all 6 value pairs, and decide whether X and Y are independent; also prove that independence implies independence of any coordinate events {X∈S} and {Y∈T} and E[XY]=E[X]E[Y]. (2) Joint table J₂: X uniform on {−1,0,1} and Y=X². Prove Cov(X,Y)=0 although X and Y are not independent. (3) Use joint table J₃ to show that checking a single value pair is not enough."},"mko_ids":["mko-random-variable-independence"]},{"id":"derivation-mko-connected-component-01","title":{"zh-Hant":"連通分量：以完整 BFS 求出有限簡單圖的全部分量","en":"Connected components: finding every component of a finite simple graph by complete BFS"},"summary":{"zh-Hant":"設 G=(V,E)，V={0,1,2,3,4,5,6}，E 恰為 {0,1}、{0,2}、{1,2}、{3,4}、{4,5} 五條邊。先證明「有 path 相連」是 V 上的等價關係，再以完整 BFS 求出每個等價類，證明 BFS 回傳的集合恰為該類，並得出 Q(G)={{0,1,2},{3,4,5},{6}}，其中孤立頂點 6 自成單點分量。","en":"Let G=(V,E) with V={0,1,2,3,4,5,6} and E consisting of exactly the five edges {0,1}, {0,2}, {1,2}, {3,4}, {4,5}. First prove that 'joined by a path' is an equivalence relation on V; then compute each equivalence class by complete BFS, prove that the set BFS returns is exactly that class, and conclude Q(G)={{0,1,2},{3,4,5},{6}}, where the isolated vertex 6 forms a singleton component."},"mko_ids":["mko-connected-component"]},{"id":"derivation-mko-finite-integral-domain-01","title":{"zh-Hant":"有限整環：以完整乘法表判定 ℤ/5ℤ 與 ℤ/6ℤ，再由有限性推出可逆","en":"Finite integral domain: deciding ℤ/5ℤ and ℤ/6ℤ from complete tables, then deriving invertibility from finiteness"},"summary":{"zh-Hant":"以完整乘法表判定 ℤ/5ℤ 是有限整環、ℤ/6ℤ 不是；再把「每個非零元可逆」作為定義之後的定理，對任何有限整環用有限性證明，並在 ℤ/5ℤ 中求出全部逆元，同時說明 ℤ/6ℤ 中 2 為何沒有逆元。","en":"Using complete multiplication tables, decide that ℤ/5ℤ is a finite integral domain and ℤ/6ℤ is not; then prove, as a theorem that comes after the definition, that every nonzero element of any finite integral domain is invertible, using finiteness; find all inverses in ℤ/5ℤ and explain why 2 has no inverse in ℤ/6ℤ."},"mko_ids":["mko-finite-integral-domain"]},{"id":"derivation-mko-polynomial-ring-over-finite-commutative-ring-01","title":{"zh-Hant":"有限交換環上的多項式環：在 (ℤ/4ℤ)[x] 中做加法、卷積與次數核對","en":"Polynomial ring over a finite commutative ring: addition, convolution and degree checks in (ℤ/4ℤ)[x]"},"summary":{"zh-Hant":"在 R=ℤ/4ℤ 上取 f=1+2x、g=2x。依係數序列定義計算 f+g、f·g、f·f、g·g 並正規化；證明首項係數公式 (fg)_{m+n}=f_m g_n，說明為何此處 deg(fg)=deg f+deg g 失敗；最後證明 R[x] 不是有限環。","en":"Over R=ℤ/4ℤ take f=1+2x and g=2x. Using the coefficient-sequence definition, compute and normalize f+g, f·g, f·f and g·g; prove the top-coefficient formula (fg)_{m+n}=f_m g_n and explain why deg(fg)=deg f+deg g fails here; finally prove that R[x] is not a finite ring."},"mko_ids":["mko-polynomial-ring-over-finite-commutative-ring"]},{"id":"derivation-mko-inner-product-induced-norm-01","title":{"zh-Hant":"內積誘導範數：證明 √⟨x,x⟩ 是範數並在 ℝ² 正定矩陣下精確計算","en":"Inner-product-induced norm: proving √⟨x,x⟩ is a norm and computing exactly for a positive-definite matrix on ℝ²"},"summary":{"zh-Hant":"在任一實內積空間中證明 N(v)=√⟨v,v⟩ 是範數：非負與分離、絕對齊次、三角不等式（經由完整證明的 Cauchy–Schwarz）。再在 ℝ² 上以 A=[[2,1],[1,2]] 定義的內積，對 x=(1,0)、y=(0,1) 精確算出平方範數、距離平方，並做 Cauchy–Schwarz、三角不等式與平行四邊形恆等式的實例核對。","en":"In any real inner-product space, prove that N(v)=√⟨v,v⟩ is a norm: nonnegativity and separation, absolute homogeneity, and the triangle inequality (via a fully proved Cauchy–Schwarz inequality). Then, on ℝ² with the inner product given by A=[[2,1],[1,2]], compute exactly the squared norms and the squared distance for x=(1,0) and y=(0,1), and check Cauchy–Schwarz, the triangle inequality and the parallelogram identity on this instance."},"mko_ids":["mko-inner-product-induced-norm"]},{"id":"derivation-mko-conditional-expectation-01","title":{"zh-Hant":"有限條件期望：由分割生成的 σ 代數上求 E[X|G]，含零質量區塊","en":"Finite conditional expectation: computing E[X|G] for a partition-generated σ-algebra, including a zero-mass block"},"summary":{"zh-Hant":"在 Ω={0,1,2,3,4} 上，質量依序為 1/8、3/8、1/4、1/4、0，X 的值依序為 4、0、2、6、10，G 由分割 {0,1}、{2,3}、{4} 生成。由條件期望的兩項定義（G 可測、在每個 G 事件上積分相同）求出 E[X|G] 的全部版本，證明它在兩個正質量區塊上必為 1 與 4，在零質量區塊 {4} 上可取任意值（只幾乎處處決定），並說明逐點條件化、只核總期望與不加權平均為何都錯。","en":"On Ω={0,1,2,3,4} with masses 1/8, 3/8, 1/4, 1/4, 0 and X taking the values 4, 0, 2, 6, 10 in order, let G be generated by the partition {0,1}, {2,3}, {4}. From the two defining conditions of conditional expectation (G-measurable, same integral on every event of G), find all versions of E[X|G]: prove they must equal 1 and 4 on the two positive-mass blocks and may take any value on the zero-mass block {4} (determined only almost surely), and explain why pointwise conditioning, checking only the total expectation, and the unweighted mean are all wrong."},"mko_ids":["mko-conditional-expectation"]},{"id":"derivation-mko-finite-filtration-01","title":{"zh-Hant":"有限 filtration：以分割細化核對資訊嵌套，並區分 adapted 與獨立","en":"Finite filtration: checking information nesting by partition refinement, and separating it from adaptedness and independence"},"summary":{"zh-Hant":"在 Ω={0,1,2,3}（兩次擲幣：0=正正、1=正反、2=反正、3=反反）上，以分割編碼 P_0、P_1、P_2 表示時刻 0、1、2 的 σ 代數。證明「較晚分割細化較早分割 ⇔ σ 代數包含」，逐相鄰時刻完整核對，得出 (F_t) 是 filtration；給出不是 filtration 的分割序列；並證明 filtration 既不使每個過程 adapted，也不推出獨立。","en":"On Ω={0,1,2,3} (two coin tosses: 0=HH, 1=HT, 2=TH, 3=TT), represent the σ-algebras at times 0, 1, 2 by partition rows P_0, P_1, P_2. Prove that 'the later partition refines the earlier one ⇔ the σ-algebras are nested', check every adjacent pair of times completely and conclude that (F_t) is a filtration; give a sequence of partitions that is not a filtration; and show that a filtration neither makes every process adapted nor implies independence."},"mko_ids":["mko-finite-filtration"]},{"id":"derivation-mko-iid-sequence-01","title":{"zh-Hant":"獨立同分布序列：由 Bernoulli(1/3) 柱集乘積公式證明 IID","en":"IID sequence: proving IID from the Bernoulli(1/3) cylinder product formula"},"summary":{"zh-Hant":"設 p=1/3，柱集證書規定每個有限 0-1 word w（長度 n、含 k 個 1）的機率為 p^k(1−p)^{n−k}。證明：此族一致且每層總和為 1；由此（配合引用的 Kolmogorov 存在定理）得到 {0,1}^ℕ 上的機率；每個 X_i 的完整邊際都是 Bernoulli(1/3)；每個有限子族都獨立。再以三個對照說明：檢查過的 prefix、pairwise 獨立與相同期望都不足以宣稱無限 IID。","en":"Let p=1/3, and let the cylinder certificate give each finite binary word w (length n, with k ones) the probability p^k(1−p)^{n−k}. Prove that this family is consistent with total mass 1 at every level; obtain from it (with the cited Kolmogorov existence theorem) a probability on {0,1}^ℕ; show that every X_i has the complete marginal Bernoulli(1/3) and that every finite subfamily is independent. Then use three contrasts to show that a checked prefix, pairwise independence and equal expectations are each insufficient for an infinite IID claim."},"mko_ids":["mko-iid-sequence"]},{"id":"derivation-mko-equivalence-class-01","title":{"zh-Hant":"等價類：在完整有限關係上先認證、再求類，並以非等價關係作對照","en":"Equivalence class: certify a complete finite relation first, then form classes, with a non-equivalence as contrast"},"summary":{"zh-Hant":"在 X={0,1,2,3,4,5} 上，R 由 12 個明列的有序對給出。先完整認證 R 是等價關係，再依定義 [a]={x∈X ∣ aRx} 求出全部類；證明 aRb ⇒ [a]=[b]、任兩類相等或不交、各類覆蓋 X，得 K(R)={{0,3},{1,4},{2,5}}。最後以 S=R∪{(0,1),(1,0)} 說明未通過傳遞性的關係不能形成等價類。","en":"On X={0,1,2,3,4,5}, R is given by 12 explicitly listed ordered pairs. First certify completely that R is an equivalence relation, then compute every class from the definition [a]={x∈X ∣ aRx}; prove that aRb ⇒ [a]=[b], that any two classes are equal or disjoint and that the classes cover X, giving K(R)={{0,3},{1,4},{2,5}}. Finally use S=R∪{(0,1),(1,0)} to show that a relation failing transitivity cannot form equivalence classes."},"mko_ids":["mko-equivalence-class"]},{"id":"derivation-mko-finite-field-01","title":{"zh-Hant":"完整表認證 F₃ 與有限整環引理","en":"Certifying F3 and the finite-domain lemma"},"summary":{"zh-Hant":"F₃ 是域，2/2=1；有限整環的非零乘法映射給逆元。","en":"F3 is a field and 2/2=1; nonzero multiplication maps supply inverses in a finite domain."},"mko_ids":["mko-finite-field"]},{"id":"derivation-mko-real-hilbert-space-01","title":{"zh-Hant":"同一矩陣誘導的有限維 Hilbert 證書","en":"A finite-dimensional Hilbert certificate from one matrix"},"summary":{"zh-Hant":"驗 R^2 配 A=[[2,1],[1,2]] 的內積、範數、度量與完備性一致。","en":"Verify the consistent inner product, norm, metric and completeness of R^2 with A=[[2,1],[1,2]]."},"mko_ids":["mko-real-hilbert-space"]},{"id":"derivation-mko-martingale-01","title":{"zh-Hant":"兩次公平擲幣的完整鞅檢查","en":"A complete two-coin martingale audit"},"summary":{"zh-Hant":"在同一空間逐時核鞅的全部責任。","en":"Audit every martingale obligation on one space."},"mko_ids":["mko-martingale"]},{"id":"derivation-mko-quotient-set-01","title":{"zh-Hant":"模2關係在四元素載體上的商集","en":"The quotient of a four-element carrier by parity"},"summary":{"zh-Hant":"X/R={ {0,2},{1,3} }，且投影值是整個類。","en":"X/R={{0,2},{1,3}}, and projection values are whole classes."},"mko_ids":["mko-quotient-set"]},{"id":"derivation-mko-cardinality-comparison-01","title":{"zh-Hant":"完整映射表的基數證書","en":"Cardinality certificates from full maps"},"summary":{"zh-Hant":"有單射A→B且|A|=2<3=|B|；反向滿射非單射。","en":"There is an injection A to B with |A|=2<3=|B|; a reverse surjection is not injective."},"mko_ids":["mko-cardinality-comparison"]},{"id":"derivation-mko-cumulative-distribution-function-01","title":{"zh-Hant":"三點分布的完整CDF","en":"The complete CDF of a three-point law"},"summary":{"zh-Hant":"算出全實軸CDF、右連續值與跳躍。","en":"Compute the CDF on the whole real line, including its right-continuous values and its jumps."},"mko_ids":["mko-cumulative-distribution-function"]},{"id":"derivation-mko-convergence-in-probability-01","title":{"zh-Hant":"eventual equality的全稱尾端證書","en":"A universal tail certificate from eventual equality"},"summary":{"zh-Hant":"以共同有限空間明證依機率收斂。","en":"Prove convergence in probability on a common finite space."},"mko_ids":["mko-convergence-in-probability"]},{"id":"derivation-mko-at-most-countable-set-01","title":{"zh-Hant":"三條自然數列的全稱編碼","en":"A universal code for three natural-number tracks"},"summary":{"zh-Hant":"N₀×{0,1,2}以c(n,j)=3n+j單射到N₀，故至多可數。","en":"N0 times {0,1,2} injects into N0 by c(n,j)=3n+j and is at most countable."},"mko_ids":["mko-at-most-countable-set"]},{"id":"derivation-mko-bipartite-graph-01","title":{"zh-Hant":"含孤立點的二分證書","en":"A bipartition certificate with an isolated vertex"},"summary":{"zh-Hant":"核兩部的全部責任，區分失敗partition與非二分。","en":"Audit every bipartition obligation and distinguish one failed partition from nonbipartiteness."},"mko_ids":["mko-bipartite-graph"]},{"id":"derivation-mko-convergence-in-distribution-01","title":{"zh-Hant":"Bernoulli參數尾端的CDF證書","en":"A Bernoulli parameter-tail CDF certificate"},"summary":{"zh-Hant":"在極限CDF連續點證依分布收斂。","en":"Prove distributional convergence at continuity points of the limiting CDF."},"mko_ids":["mko-convergence-in-distribution"]},{"id":"derivation-mko-finite-quotient-ring-01","title":{"zh-Hant":"Z/4Z 對偶數理想取商","en":"Quotienting Z/4Z by its even ideal"},"summary":{"zh-Hant":"I={0,2}為proper雙邊理想，R/I有兩個cosets及良定雙表。","en":"I={0,2} is a proper two-sided ideal; R/I has two cosets and well-defined operation tables."},"mko_ids":["mko-finite-quotient-ring"]}],"title":"概念推導導讀"}