{"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":"3"},"kind":"guides","total":299,"pages":13,"page":3,"entries":[{"id":"derivation-mko-surjective-01","title":{"zh-Hant":"滿射與陪域覆蓋：值域、原像與陪域變更","en":"Surjectivity and codomain coverage: range, preimages and changing the codomain"},"summary":{"zh-Hant":"設 A={1,2,3,4,5}、B={1,2,3}，函數表 f：(1,2),(2,1),(3,3),(4,2),(5,1)。確認 f 是總函數 A→B，求值域 f[A] 與各原像，判定 f 是否滿射並找出 t=2 的首個原像；再把陪域換成 B2={1,2,3,4} 與 B3={1,2} 判定合法性與滿射性；最後處理 ∅→∅ 與 ∅→{7}。","en":"Let A={1,2,3,4,5}, B={1,2,3} and the function table f: (1,2),(2,1),(3,3),(4,2),(5,1). Confirm f is a total function A→B, find the range f[A] and the preimages, decide whether f is surjective and find the first preimage of t=2; then replace the codomain by B2={1,2,3,4} and by B3={1,2}, deciding legality and surjectivity; finally treat ∅→∅ and ∅→{7}."},"mko_ids":["mko-surjective"]},{"id":"derivation-mko-bijective-01","title":{"zh-Hant":"雙射：以纖維概況判定給定函數，並與「是否存在某個雙射」分開","en":"Bijectivity: testing given functions by fiber profiles, kept apart from whether some bijection exists"},"summary":{"zh-Hant":"設 A={1,2,3}、B={4,5,6}，函數 f、g:A→B 由表給出：f(1)=5、f(2)=6、f(3)=4；g(1)=4、g(2)=4、g(3)=6。先對一般總函數證明「雙射 ⇔ 每個 b∈B 的纖維恰有一個元素」，再按陪域 4、5、6 的順序列出 f 與 g 的完整纖維概況，判定 f 是雙射、g 既非單射也非滿射。最後回答另一類問題「是否存在某個雙射」：A→B 存在（f 即是）、{1,2,3}→{4,5} 不存在、∅→∅ 恰有一個空雙射。","en":"Let A={1,2,3}, B={4,5,6}, and let f, g:A→B be given by tables: f(1)=5, f(2)=6, f(3)=4; g(1)=4, g(2)=4, g(3)=6. First prove, for any total function, the equivalence: bijective ⇔ the fiber of every b∈B has exactly one element. Then list the complete fiber profiles of f and g in the codomain order 4, 5, 6, and decide that f is bijective while g is neither injective nor surjective. Finally answer the different question of whether some bijection exists: one exists A→B (f is one), none exists {1,2,3}→{4,5}, and ∅→∅ has exactly one, the empty bijection."},"mko_ids":["mko-bijective"]},{"id":"derivation-mko-inverse-function-01","title":{"zh-Hant":"反函數：交換有限雙射表並求非零斜率仿射函數的反函數","en":"Inverse functions: exchanging a finite bijection table and inverting an affine map with nonzero slope"},"summary":{"zh-Hant":"（一）A={1,2,3}、B={4,5,6}，p(1)=5、p(2)=6、p(3)=4。先確認 p 是雙射，再交換有序對得到候選 q，驗證 q 是總函數 B→A，並逐側檢查 q(p(x))=x 與 p(q(y))=y。（二）f:ℝ→ℝ，f(x)=−4x+6（m=−4≠0）：求反函數 g(y)=uy+v，u、v 以最簡、正分母的有理數表示，並對所有實數驗證兩側恆等式。（三）說明 h(1)=4、h(2)=4、h(3)=6 的交換表不是 B→A 的函數，以及 f^{-1} 不是 1/f。","en":"(1) A={1,2,3}, B={4,5,6}, p(1)=5, p(2)=6, p(3)=4. First confirm that p is a bijection, then exchange its ordered pairs to get a candidate q, validate q as a total function B→A, and check q(p(x))=x and p(q(y))=y separately. (2) f:ℝ→ℝ, f(x)=−4x+6 (m=−4≠0): find the inverse g(y)=uy+v with u, v reduced rationals with positive denominators, and verify both identities for all reals. (3) Show that the exchanged table of h(1)=4, h(2)=4, h(3)=6 is not a function B→A, and that f^{-1} is not 1/f."},"mko_ids":["mko-inverse-function"]},{"id":"derivation-mko-random-variable-01","title":{"zh-Hant":"有限樣本空間的隨機變數：值域、原像事件與點機率","en":"Random variables on a finite sample space: range, preimage events and point probabilities"},"summary":{"zh-Hant":"S={1,2,3,4}，事件族 F=𝒫(S)，權重 w₁=1/2、w₂=1/4、w₃=1/8、w₄=1/8。隨機變數 X:S→ℝ 由表給出：X(1)=1、X(2)=0、X(3)=1、X(4)=4（即 X(s)=(s−2)²）。求 X 的值域、原像事件 {X=1}、{X≥1} 與 {X∈{2,3}}，以及每個值的點機率 P(X=t)，並證明這些點機率之和為 1。","en":"S={1,2,3,4}, event family F=𝒫(S), weights w₁=1/2, w₂=1/4, w₃=1/8, w₄=1/8. The random variable X:S→ℝ is given by the table X(1)=1, X(2)=0, X(3)=1, X(4)=4 (that is, X(s)=(s−2)²). Find the range of X, the preimage events {X=1}, {X≥1} and {X∈{2,3}}, and the point probability P(X=t) of each value, and prove that these point probabilities sum to 1."},"mko_ids":["mko-random-variable"]},{"id":"derivation-mko-expectation-01","title":{"zh-Hant":"有限模型的期望值：線性性的證明與相依變數的實例","en":"Expectation in finite models: proving linearity, with an instance of dependent variables"},"summary":{"zh-Hant":"在非空有限樣本空間 S、非負且總和為 1 的權重 w_s 下，證明 E[c]=c、E[aX+b]=aE[X]+b、E[X+Y]=E[X]+E[Y]（X、Y 定義在同一個 S 上，不假設獨立）。再在 S={1,2,3}、w₁=1/2、w₂=1/4、w₃=1/4，X=(2,−2,0)、Y=(1,3,−1) 上計算 E[X]、E[Y]、E[X+Y]、E[−3X+5]，說明 X、Y 不獨立而線性性仍成立，並以 E[XY]≠E[X]E[Y] 說明線性性不推出乘積公式。","en":"For a nonempty finite sample space S with nonnegative weights w_s summing to 1, prove E[c]=c, E[aX+b]=aE[X]+b and E[X+Y]=E[X]+E[Y] (X, Y on the same S, independence not assumed). Then on S={1,2,3} with w₁=1/2, w₂=1/4, w₃=1/4, X=(2,−2,0), Y=(1,3,−1), compute E[X], E[Y], E[X+Y] and E[−3X+5], show that X and Y are not independent while linearity still holds, and use E[XY]≠E[X]E[Y] to show that linearity does not give a product formula."},"mko_ids":["mko-expectation"]},{"id":"derivation-mko-variance-01","title":{"zh-Hant":"有限機率模型的變異數：平移、縮放與零變異數的刻畫","en":"Variance in finite probability models: translation, scaling and the zero-variance characterization"},"summary":{"zh-Hant":"在有限合法機率模型中證明 Var(X)≥0、Var(X+c)=Var(X)、Var(kX)=k²Var(X)，以及 Var(X)=0 ⇔ X 在所有正權重結果上取同一個值。再以 S={1,2,3,4}、w₁=1/2、w₂=1/4、w₃=1/4、w₄=0、X=(1,3,−1,10) 計算 μ=1、Var(X)=2、Var(X+5)=2、Var(−3X)=18，並以 Z=(2,2,2,7) 說明零權重結果上的不同值不影響零變異數。","en":"In a valid finite probability model, prove Var(X)≥0, Var(X+c)=Var(X), Var(kX)=k²Var(X), and Var(X)=0 ⇔ X takes one common value at every positive-weight outcome. Then with S={1,2,3,4}, w₁=1/2, w₂=1/4, w₃=1/4, w₄=0 and X=(1,3,−1,10), compute μ=1, Var(X)=2, Var(X+5)=2 and Var(−3X)=18, and use Z=(2,2,2,7) to show that a different value at a zero-weight outcome does not prevent zero variance."},"mko_ids":["mko-variance"]},{"id":"derivation-mko-order-relation-01","title":{"zh-Hant":"實數的通常順序：加法保序、正分母交叉比較、有限極值與閉區間","en":"The usual order on ℝ: translation, cross-multiplication with positive denominators, finite extrema and closed intervals"},"summary":{"zh-Hant":"證明對所有實數 a、b、c：a<b ⇔ a+c<b+c；q、s>0 時 p/q<r/s ⇔ ps<rq；非空有限實數集合有最小元素，且最小元素唯一。據此比較 M={−5/9, 1/3, −7/12, −1/2}（照此次序列出）的元素，求 M 的最小與最大元素，並判定兩端點與 1/2 是否屬於閉區間 [min M, max M]。","en":"Prove for all reals a, b, c that a<b ⇔ a+c<b+c; that p/q<r/s ⇔ ps<rq when q, s>0; and that a finite nonempty set of reals has a minimum, which is unique. Use these to compare the elements of M={−5/9, 1/3, −7/12, −1/2} (listed in this order), find the minimum and maximum of M, and decide whether the two endpoints and 1/2 belong to the closed interval [min M, max M]."},"mko_ids":["mko-order-relation"]},{"id":"derivation-mko-quadratic-function-01","title":{"zh-Hant":"頂點式二次函數：求值、對稱、全域最大值與限制區間","en":"Quadratic functions in vertex form: evaluation, symmetry, global maximum and a restricted domain"},"summary":{"zh-Hant":"q:ℝ→ℝ，q(x)=−2(x−3)²+5（a=−2、h=3、k=5）。依序求 q(1) 及其對稱點；證明對所有實數 x 有 q(6−x)=q(x)；證明 q 的全域最大值為 5、只在 x=3 取得，頂點為 (3,5)，且 q 沒有全域最小值；再說明限制在 [4,6] 時最大值是 3（在 x=4），以及 a=0 時結論不能沿用。","en":"q:ℝ→ℝ, q(x)=−2(x−3)²+5 (a=−2, h=3, k=5). Evaluate q(1) step by step and find its mirror point; prove q(6−x)=q(x) for every real x; prove that the global maximum of q is 5, attained only at x=3, that the vertex is (3,5), and that q has no global minimum; then show that on the restricted domain [4,6] the maximum is 3 (at x=4), and that the conclusions do not carry over when a=0."},"mko_ids":["mko-quadratic-function"]},{"id":"derivation-mko-rate-of-change-01","title":{"zh-Hant":"兩點間的平均變化率：資格、端點對調、仿射與二次公式","en":"Average rate of change between distinct inputs: eligibility, swapping endpoints, the affine and quadratic formulas"},"summary":{"zh-Hant":"（一）已知某指定函數 f 的 f(2)=7、f(−4)=3：判定資格並求平均變化率，化為最簡正分母分數，並說明同時對調兩端點不改變結果。（二）證明 f(x)=mx+b 的平均變化率為 m，並以 ℓ(x)=−3x+7 在 −2 與 4 之間核對。（三）證明 q(x)=a(x−h)²+k（a≠0）的平均變化率為 a(x₀+x₁−2h)，以 q(x)=−2(x−3)²+5 在 1 與 4 之間核對；並以對稱端點、改變第二端點與相同輸入三個情形說明它不是導數，也不能對相同輸入填 0。","en":"(1) For a specified function f with f(2)=7 and f(−4)=3: check eligibility, compute the average rate of change as a reduced fraction with positive denominator, and show that swapping both endpoints does not change it. (2) Prove that f(x)=mx+b has average rate m, and check it with ℓ(x)=−3x+7 between −2 and 4. (3) Prove that q(x)=a(x−h)²+k (a≠0) has average rate a(x₀+x₁−2h), check it with q(x)=−2(x−3)²+5 between 1 and 4, and use symmetric endpoints, a changed second endpoint and equal inputs to show that it is not a derivative and that 0 must not be assigned when the inputs coincide."},"mko_ids":["mko-rate-of-change"]},{"id":"derivation-mko-distance-formula-01","title":{"zh-Hant":"歐氏距離公式：計算距離、拒絕負根並以平方比較","en":"Euclidean distance formula: computing, rejecting the negative root and comparing by squares"},"summary":{"zh-Hant":"在同一個等單位正交座標系中，設 P=(1,2)、Q=(4,6)、R=(6,−1)。求 d(P,Q)，並說明候選值 −5 為何不是距離；求 d(P,R) 的精確形式並判定它不是整數；以平方比較 d(P,Q) 與 d(P,R)；最後核對重合點的距離為零，以及距離與兩點的書寫順序無關。","en":"In one equal-unit orthonormal Cartesian frame let P=(1,2), Q=(4,6) and R=(6,−1). Find d(P,Q) and explain why the candidate −5 is not a distance; find the exact form of d(P,R) and decide that it is not an integer; compare d(P,Q) with d(P,R) through their squares; finally check that coincident points have distance zero and that the distance does not depend on the order in which the two points are written."},"mko_ids":["mko-distance-formula"]},{"id":"derivation-mko-circle-01","title":{"zh-Hant":"圓：以距離平方判定點在圓內、圓上或圓外，並處理零半徑","en":"Circle: classifying points as inside, on or outside by squared distance, including radius zero"},"summary":{"zh-Hant":"在等單位正交座標系中，取中心 O=(2,−1)、半徑 r=5。判定 A=(5,3)、B=(4,1)、E=(−2,3) 各在圓內、圓上或圓外；驗證四個軸向點 O±(5,0)、O±(0,5) 都在 C(O,5) 上，並由兩個相對的軸向點算出直徑長 2r=10。再把半徑改為 r=0，證明 C(O,0)={O}、沒有圓內點，並判定 B 的位置。","en":"In an equal-unit orthonormal frame take center O=(2,−1) and radius r=5. Decide whether each of A=(5,3), B=(4,1), E=(−2,3) is strictly inside, on, or outside the circle; verify that the four axial points O±(5,0), O±(0,5) lie on C(O,5), and compute from two opposite axial points the diameter length 2r=10. Then change the radius to r=0, prove that C(O,0)={O} and that no point is strictly inside, and classify B."},"mko_ids":["mko-circle"]},{"id":"derivation-mko-area-01","title":{"zh-Hant":"面積：半開矩形的交集、聯集與單位格重算","en":"Area: intersection and union of half-open rectangles, recounted by unit cells"},"summary":{"zh-Hant":"在等單位正交座標系中，設 A=[0,4)×[0,3)、B=[2,6)×[1,5)。求 A 與 B 的面積，求出交集 A∩B 及其面積，用包含排除求 A∪B 的面積，並以「不同單位格的個數」獨立重算。再設 C=[4,7)×[0,3)（與 A 只在 x=4 處相觸）與 Z=[3,3)×[0,5)（寬為零），求 A∩C、A∪C 的面積與 Z 的面積。","en":"In an equal-unit orthonormal frame let A=[0,4)×[0,3) and B=[2,6)×[1,5). Find the areas of A and B, find the intersection A∩B and its area, find the area of A∪B by inclusion and exclusion, and recount it independently as the number of distinct unit cells. Then let C=[4,7)×[0,3) (touching A only at x=4) and Z=[3,3)×[0,5) (zero width), and find the areas of A∩C, A∪C and Z."},"mko_ids":["mko-area"]},{"id":"derivation-mko-descriptive-statistics-01","title":{"zh-Hant":"描述統計：保留重數計算平均數、中位數、全距與眾數集合","en":"Descriptive statistics: mean, median, range and mode set with repetitions kept"},"summary":{"zh-Hant":"對觀測列 D=(4,10,1,3,8,1,6,3)，保留重複值，求觀測數 n、算術平均、中位數、全距與本輪約定的眾數集合；證明一般性質「最小值 ≤ 平均 ≤ 最大值」並用它核對；最後說明若先去重會得到不同（錯誤）的平均與中位數。","en":"For the observation list D=(4,10,1,3,8,1,6,3), keeping repeated values, find the number of observations n, the arithmetic mean, the median, the range and the mode set under this entry's convention; prove the general fact 'minimum ≤ mean ≤ maximum' and use it as a check; finally show that deduplicating first would give a different (wrong) mean and median."},"mko_ids":["mko-descriptive-statistics"]},{"id":"derivation-mko-matrix-01","title":{"zh-Hant":"矩陣：尺寸、一起算索引、逐項加法與轉置","en":"Matrices: dimensions, one-based indexing, entrywise addition and transpose"},"summary":{"zh-Hant":"設 M=[[2,−1,0],[5,3,7]]、N=[[1,4,−2],[0,−3,6]]（每個內層括號是一行，row）。求兩者尺寸，讀出 M(1,2)、M(2,3)，判定 M(3,1) 是否存在；求 M+N、Mᵀ、Nᵀ；對任意同尺寸實矩陣證明 (A+B)ᵀ=Aᵀ+Bᵀ 與 (Aᵀ)ᵀ=A，並在 M、N 上核對；說明 M+Mᵀ 無定義；最後以反例區分轉置與「逐行倒序」及「反矩陣」，並把程式的零起算索引對應回一起算索引。","en":"Let M=[[2,−1,0],[5,3,7]] and N=[[1,4,−2],[0,−3,6]] (each inner bracket is one row). Find their dimensions, read off M(1,2) and M(2,3), and decide whether M(3,1) exists; compute M+N, Mᵀ and Nᵀ; prove (A+B)ᵀ=Aᵀ+Bᵀ and (Aᵀ)ᵀ=A for all real matrices of equal size and check them on M and N; explain why M+Mᵀ is undefined; finally use counterexamples to separate the transpose from reversing the rows and from the inverse matrix, and translate zero-based program indices back to one-based indices."},"mko_ids":["mko-matrix"]},{"id":"derivation-mko-linear-transformation-01","title":{"zh-Hant":"線性映射：證明 v↦Mv 線性，且仿射式 Mv+b 線性當且僅當 b=0","en":"Linear maps: proving v↦Mv is linear, and Mv+b is linear exactly when b=0"},"summary":{"zh-Hant":"對任意 2×2 實矩陣 M，證明 T(v)=Mv 是 ℝ²→ℝ² 的線性映射（對所有向量 u、v 與實數 c 滿足加法相容與純量相容）；證明仿射式 F(v)=Mv+b 線性當且僅當 b=0。以 M₀=[[2,−1],[3,4]]、b=(1,−2)ᵀ、u=(1,1)ᵀ、w=(2,−3)ᵀ 為例：算出 M₀u，說明 F 不線性並給出具體失效等式；以數字示範 T 的加法相容；最後以 g(x,y)=(x²,y) 說明有限測點不能證明任意映射線性。","en":"For every 2×2 real matrix M, prove that T(v)=Mv is a linear map ℝ²→ℝ² (additivity and homogeneity for all vectors u, v and real c); prove that the affine map F(v)=Mv+b is linear exactly when b=0. For the example M₀=[[2,−1],[3,4]], b=(1,−2)ᵀ, u=(1,1)ᵀ, w=(2,−3)ᵀ: compute M₀u, show that F is not linear with an explicit failing equation, and illustrate additivity of T numerically; finally use g(x,y)=(x²,y) to show that finitely many test points cannot prove that an arbitrary map is linear."},"mko_ids":["mko-linear-transformation"]},{"id":"derivation-mko-determinant-01","title":{"zh-Hant":"二階行列式：乘積公式、縮放公式與實數域可逆判準","en":"2×2 determinants: the product rule, the scaling rule and the invertibility criterion over ℝ"},"summary":{"zh-Hant":"對所有 2×2 實矩陣證明 det(AB)=det(A)det(B) 與 det(tM)=t²det(M)，並證明 M 在實數域可逆當且僅當 det(M)≠0。以 A=[[2,1],[−1,3]]、B=[[1,−2],[4,0]] 核對乘積公式並顯示 AB≠BA；以 t=−3 核對縮放公式；求 A⁻¹，並說明 A 雖是整數矩陣且 det(A)=7≠0，卻沒有整數反矩陣；以 C=[[1,2],[3,4]]（det=−2）說明非零行列式不必為正。","en":"For all 2×2 real matrices prove det(AB)=det(A)det(B) and det(tM)=t²det(M), and prove that M is invertible over ℝ exactly when det(M)≠0. With A=[[2,1],[−1,3]] and B=[[1,−2],[4,0]], check the product rule and show AB≠BA; check the scaling rule with t=−3; find A⁻¹ and explain why A, although an integer matrix with det(A)=7≠0, has no integer inverse; use C=[[1,2],[3,4]] (det=−2) to show that a nonzero determinant need not be positive."},"mko_ids":["mko-determinant"]},{"id":"derivation-mko-eigenvalue-01","title":{"zh-Hant":"二階實矩陣的特徵值：特徵多項式、判別式與非零特徵向量","en":"Eigenvalues of 2×2 real matrices: characteristic polynomial, discriminant and nonzero eigenvectors"},"summary":{"zh-Hant":"對 2×2 實矩陣證明：λ 是實特徵值 ⇔ det(M−λI)=0 ⇔ λ²−tr(M)λ+det(M)=0，且互異實特徵值的個數由 Δ=tr(M)²−4det(M) 的正負決定（負、零、正分別為 0、1、2）。以 M=[[4,1],[2,3]] 求出全部互異實特徵值與各自的非零特徵向量，說明候選 λ=3 的核 K₃={0} 而 3 不是特徵值，並拒絕候選特徵對 (3,0)；再以旋轉矩陣 R=[[0,−1],[1,0]]（Δ<0）與 J=[[3,1],[0,3]]（Δ=0）對照。","en":"For 2×2 real matrices prove: λ is a real eigenvalue ⇔ det(M−λI)=0 ⇔ λ²−tr(M)λ+det(M)=0, and the number of distinct real eigenvalues is decided by the sign of Δ=tr(M)²−4det(M) (negative, zero, positive give 0, 1, 2). For M=[[4,1],[2,3]] find all distinct real eigenvalues with nonzero eigenvectors, show that for the candidate λ=3 the kernel is K₃={0} and 3 is not an eigenvalue, and reject the candidate eigenpair (3,0); then contrast with the rotation R=[[0,−1],[1,0]] (Δ<0) and J=[[3,1],[0,3]] (Δ=0)."},"mko_ids":["mko-eigenvalue"]},{"id":"derivation-mko-congruence-01","title":{"zh-Hant":"全等：以 SSS 切片判定有明示對應的三角形全等","en":"Congruence: deciding congruence of labeled triangles by the SSS slice"},"summary":{"zh-Hant":"在同一個等單位正交平面中，A1=(0,0)、A2=(4,0)、A3=(0,3)，B1=(1,1)、B2=(1,5)、B3=(4,1)，對應明示為 Ai↔Bi。先以行列式驗證兩組點都是非退化三角形，再依 12、13、23 的順序精確比較平方邊長，判定 G(A,B)；說明 B 的方向與 A 相反卻仍全等；最後找出另一個對應 A1↔B1、A2↔B3、A3↔B2 的首個不等見證，並說明退化點組不合格。","en":"In one equal-unit orthonormal plane, let A1=(0,0), A2=(4,0), A3=(0,3) and B1=(1,1), B2=(1,5), B3=(4,1), with explicit correspondence Ai↔Bi. First verify by determinants that both triples are nondegenerate triangles, then compare squared side lengths exactly in the order 12, 13, 23 to decide G(A,B); explain why B has the opposite orientation and is still congruent; finally find the first unequal witness for the other correspondence A1↔B1, A2↔B3, A3↔B2, and show that a degenerate triple is not eligible."},"mko_ids":["mko-congruence"]},{"id":"derivation-mko-similarity-01","title":{"zh-Hant":"相似：以指定正倍率判定標記三角形相似並求面積","en":"Similarity: deciding labeled triangle similarity at a specified positive scale and finding the area"},"summary":{"zh-Hant":"在同一等單位正交平面中，A1=(0,0)、A2=(4,0)、A3=(0,2)，B1=(1,1)、B2=(1,7)、B3=(4,1)，對應 Ai↔Bi，倍率方向為 |BiBj|=k|AiAj|，指定 k=3/2。判定兩者在 k=3/2 下相似，並由已知面積 Area(A)=4 求 Area(B)；再說明 k=2 只被否定為見證、k=−3/2 不合資格，以及方向相反不改變正倍率。","en":"In one equal-unit orthonormal plane, let A1=(0,0), A2=(4,0), A3=(0,2) and B1=(1,1), B2=(1,7), B3=(4,1), with correspondence Ai↔Bi and scale direction |BiBj|=k|AiAj|, at the specified k=3/2. Decide that they are similar at k=3/2, and from the given area Area(A)=4 find Area(B); then explain why k=2 is rejected only as a witness, why k=−3/2 is not eligible, and why opposite orientation does not change the positive scale."},"mko_ids":["mko-similarity"]},{"id":"derivation-mko-geometric-transformations-01","title":{"zh-Hant":"幾何變換：證明基本等距並分類均勻縮放","en":"Geometric transformations: proving the basic isometries and classifying uniform scaling"},"summary":{"zh-Hant":"在等單位正交實平面中證明：平移、原點逆時針 90° 旋轉 R90(x,y)=(−y,x) 以及 MKO 列出的其他旋轉與軸鏡射都是等距映射；半轉 (−x,−y) 不是所列四條軸中任何一條的鏡射；均勻縮放 Hc(x,y)=(cx,cy) 使每個距離乘 |c|。據此分類 c=0、c=−1、c=−2、c=2，並以 P=(3,1)、Q=(−1,4) 做數值核對。","en":"On the equal-unit orthonormal real plane, prove that translations, the counterclockwise quarter-turn R90(x,y)=(−y,x) about the origin, and the other rotations and axis reflections listed in the MKO are isometries; that the half-turn (−x,−y) is not a reflection in any of the listed axes; and that uniform scaling Hc(x,y)=(cx,cy) multiplies every distance by |c|. Use this to classify c=0, c=−1, c=−2 and c=2, with a numerical check on P=(3,1), Q=(−1,4)."},"mko_ids":["mko-geometric-transformations"]},{"id":"derivation-mko-sequence-limit-01","title":{"zh-Hant":"數列極限：倒數模型的精確尾界與 ε-N 證明","en":"Sequence limits: the exact tail cutoff and an ε-N proof for the reciprocal model"},"summary":{"zh-Hant":"對 u(n)=2−3/(n+1)（n∈ℕ0，即 MKO 模型 L=2、c=−3、d=1）：(1) 證明候選尾界 N 對 ε 合法當且僅當 3/(N+1)<ε；(2) 求 ε=1/2 的最小合法 N，並說明 N=5 為何不合法；(3) 以 ε-N 定義證明 u(n)→2；(4) 說明 N 必須依 ε 選取。","en":"For u(n)=2−3/(n+1) with n∈ℕ0 (the MKO model with L=2, c=−3, d=1): (1) prove that a candidate cutoff N is valid for ε exactly when 3/(N+1)<ε; (2) find the least valid N for ε=1/2 and explain why N=5 fails; (3) prove u(n)→2 from the ε-N definition; (4) show that N must be chosen after ε."},"mko_ids":["mko-sequence-limit"]},{"id":"derivation-mko-function-limit-01","title":{"zh-Hant":"函數極限：可去因子商、仿射 δ 界與不等的單側極限","en":"Function limits: a removable-factor quotient, affine δ bounds and unequal one-sided limits"},"summary":{"zh-Hant":"在去心實數論域上：(1) 以 ε-δ 證明 f(x)=(x²−9)/(x−3) 滿足 lim_(x→3) f(x)=6，而 f(3) 仍無定義；(2) 對仿射 g(x)=4x−1 在 a=2、ε=1/2，證明 δ=1/8 合法（等號 |m|δ=ε 允許），δ=1/7 不合法；(3) 對分支函數 h(x)=x+1（x<1）、h(x)=2x+3（x>1），求左右極限 2 與 5，並證明兩側極限不存在。","en":"On punctured real domains: (1) prove by ε-δ that f(x)=(x²−9)/(x−3) has lim_(x→3) f(x)=6 while f(3) stays undefined; (2) for the affine g(x)=4x−1 at a=2 with ε=1/2, prove that δ=1/8 is valid (the equality |m|δ=ε is allowed) and δ=1/7 is not; (3) for the branch function h(x)=x+1 (x<1), h(x)=2x+3 (x>1), find the one-sided limits 2 and 5 and prove that the two-sided limit does not exist."},"mko_ids":["mko-function-limit"]},{"id":"derivation-mko-continuity-01","title":{"zh-Hant":"連續：分支函數的中心值與可去商的唯一補點","en":"Continuity: centre values of branch functions and the unique continuous extension across a removable singularity"},"summary":{"zh-Hant":"(1) 對 f_V(x)=2x+1（x<1）、f_V(1)=V、f_V(x)=5x−2（x>1），證明 V=3 時 f_V 在 1 連續，V=4 時不連續；(2) 對 g(x)=2x+1（x<1）、g(1)=3、g(x)=5x−1（x>1），證明 g 在 1 不連續；(3) 對 q(x)=(x²−4)/(x−2)（論域 ℝ∖{2}），證明全實數連續延拓在 2 的值必為 4，且 4 確實可行。","en":"(1) For f_V(x)=2x+1 (x<1), f_V(1)=V, f_V(x)=5x−2 (x>1), prove that f_V is continuous at 1 when V=3 and not when V=4; (2) for g(x)=2x+1 (x<1), g(1)=3, g(x)=5x−1 (x>1), prove that g is not continuous at 1; (3) for q(x)=(x²−4)/(x−2) on ℝ∖{2}, prove that a continuous extension to all real numbers must take the value 4 at 2, and that 4 does work."},"mko_ids":["mko-continuity"]},{"id":"derivation-mko-derivative-01","title":{"zh-Hant":"導數：由差商極限求二次式導數並判定尖點無導數","en":"The derivative: a quadratic's derivative from the difference-quotient limit, and no derivative at a corner"},"summary":{"zh-Hant":"對 f(t)=3t²−2t+5：先對 h≠0 化簡差商 (f(1+h)−f(1))/h=4+3h，再以 ε-δ 證明 f'(1)=4；推導一般公式：差商為 2a2x+a1+a2h、導數為 2a2x+a1；說明有限步 h=1/10 的差商 4.3 不是導數、a2=0 的仿射邊界；最後證明 g(t)=|t−2| 在 t=2 沒有導數（不是 0），而在 t>2 處導數為 1。","en":"For f(t)=3t²−2t+5: first simplify the difference quotient for h≠0, (f(1+h)−f(1))/h=4+3h, then prove f'(1)=4 by ε-δ; derive the general formulas: the quotient is 2a2x+a1+a2h and the derivative 2a2x+a1; explain why the finite-step quotient 4.3 at h=1/10 is not the derivative, and the affine boundary a2=0; finally prove that g(t)=|t−2| has no derivative at t=2 (not 0), while its derivative is 1 for t>2."},"mko_ids":["mko-derivative"]}],"title":"Concept derivation guides"}