{"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":"2"},"kind":"guides","total":299,"pages":13,"page":2,"entries":[{"id":"derivation-mko-counting-principle-01","title":{"zh-Hant":"基本計數原理：數字互異的三位偶數有幾個","en":"Basic counting principles: how many three-digit even numbers have distinct digits"},"summary":{"zh-Hant":"計算 100 到 999 之間、三個數字兩兩相異的偶數共有幾個。要求：只把互斥的類別相加；每次相乘之前，先確認每個階段的可選數對所有先前選擇都相同，不同時先分支。","en":"Count the even integers from 100 to 999 whose three digits are pairwise distinct. Requirements: add only disjoint classes, and before each multiplication confirm that the number of options at every stage is the same for all earlier choices; when it is not, branch first."},"mko_ids":["mko-counting-principle"]},{"id":"derivation-mko-factorial-01","title":{"zh-Hant":"階乘：遞迴定義的唯一性、乘積形式與 5! 的計算","en":"Factorial: uniqueness of the recursive definition, the product form and computing 5!"},"summary":{"zh-Hant":"只用 F(0)=1 與 F(n+1)=(n+1)F(n)（n∈ℕ，ℕ 含 0）：(1) 證明滿足這兩條的函數 ℕ→ℕ 只有一個；(2) 證明對所有 n 有 F(n)=∏_{k=1}^{n} k（空乘積為 1），所以 n>0 時 n! 是從 1 到 n 所有整數的乘積；(3) 證明每個 F(n)≥1，特別是 0!=1≠0；(4) 由遞迴逐步算出 5!=120。","en":"Using only F(0)=1 and F(n+1)=(n+1)F(n) (n∈ℕ, with 0∈ℕ): (1) prove that at most one function ℕ→ℕ satisfies these two equations; (2) prove F(n)=∏_{k=1}^{n} k for every n (the empty product being 1), so that for n>0, n! is the product of all integers from 1 to n; (3) prove F(n)≥1 for every n, in particular 0!=1≠0; (4) compute 5!=120 step by step from the recursion."},"mko_ids":["mko-factorial"]},{"id":"derivation-mko-absolute-value-01","title":{"zh-Hant":"絕對值：由分段定義證明非負性、零點與 |a−b|=|b−a|","en":"Absolute value: nonnegativity, the zero case and |a−b|=|b−a| from the piecewise definition"},"summary":{"zh-Hant":"只用分段定義（x≥0 時 |x|=x，x<0 時 |x|=−x）與實數的序和加法，證明對所有實數 x、a、b：(1) |x|≥0；(2) |x|=0 ⇔ x=0；(3) |−x|=|x|；(4) |a−b|=|b−a|。並計算 |−3|、|2−7|、|7−2|，說明「|x|=x 對所有 x 成立」是錯的。","en":"Using only the piecewise definition (|x|=x when x≥0, |x|=−x when x<0) together with the order and addition of the reals, prove for all reals x, a, b: (1) |x|≥0; (2) |x|=0 ⇔ x=0; (3) |−x|=|x|; (4) |a−b|=|b−a|. Then compute |−3|, |2−7| and |7−2|, and show that '|x|=x for every x' is false."},"mko_ids":["mko-absolute-value"]},{"id":"derivation-mko-fraction-01","title":{"zh-Hant":"分數：判定等值、化為最簡正分母形式與分數加法","en":"Fractions: deciding equal value, normalizing to lowest terms with a positive denominator, and adding fractions"},"summary":{"zh-Hant":"(1) 判定 (−6)/(−8) 與 9/12 是否表示同一個有理數；(2) 把 (−6)/(−8) 化成最簡、正分母的表示，並說明數值不變；(3) 計算 3/4+5/6，並說明「分子加分子、分母加分母」得到的 8/10 不是和；(4) 說明 3/0 不是合法的分數表示。","en":"(1) Decide whether (−6)/(−8) and 9/12 represent the same rational number; (2) bring (−6)/(−8) to lowest terms with a positive denominator and show that the value is unchanged; (3) compute 3/4+5/6 and show that 'numerator plus numerator over denominator plus denominator', giving 8/10, is not the sum; (4) explain why 3/0 is not a valid fraction representation."},"mko_ids":["mko-fraction"]},{"id":"derivation-mko-ratio-proportion-01","title":{"zh-Hant":"比與比例：解 3:4=x:10、代回核對與檢查正比例資料","en":"Ratio and proportion: solving 3:4=x:10, checking back, and testing data for direct proportion"},"summary":{"zh-Hant":"在有理數中：(1) 證明對 b≠0、d≠0，比例 a:b=x:d 恰有一個有理解 x=ad/b；(2) 用它解 3:4=x:10 並代回核對；(3) 說明 4:3=x:10 的解不同，比的順序有意義；(4) 判定資料 (2,5)、(4,10)、(6,15)、(8,21) 是否符合同一個正比例 y=kx。","en":"Over the rationals: (1) prove that for b≠0 and d≠0 the proportion a:b=x:d has exactly one rational solution, x=ad/b; (2) use it to solve 3:4=x:10 and check back; (3) show that 4:3=x:10 has a different solution, so the order of a ratio matters; (4) decide whether the data (2,5), (4,10), (6,15), (8,21) fit a single direct proportion y=kx."},"mko_ids":["mko-ratio-proportion"]},{"id":"derivation-mko-domain-codomain-range-01","title":{"zh-Hant":"定義域、陪域與值域：有限表格與 h(x)=x²+1 的值域","en":"Domain, codomain and range: a finite table and the range of h(x)=x²+1"},"summary":{"zh-Hant":"(1) 設 A={−1,0,1,2}、B={0,1,2,3,4,5}、f(x)=x²+1：驗證 f:A→B 是總函數，求值域 f[A]，並說明 f[A]⊆B 但 f[A]≠B；(2) 說明輸出落在宣告陪域外或漏掉輸入的資料不是本條目的總函數；(3) 對 h:ℝ→ℝ、h(x)=x²+1，證明值域是 [1,∞)，不等於陪域 ℝ。","en":"(1) With A={−1,0,1,2}, B={0,1,2,3,4,5} and f(x)=x²+1: verify that f:A→B is a total function, find the range f[A], and show f[A]⊆B but f[A]≠B; (2) explain why data with an output outside the declared codomain, or with a missing input, is not a total function in the sense of this entry; (3) for h:ℝ→ℝ, h(x)=x²+1, prove that the range is [1,∞), which differs from the codomain ℝ."},"mko_ids":["mko-domain-codomain-range"]},{"id":"derivation-mko-composition-01","title":{"zh-Hant":"函數合成：受限合成的允許輸入、求值順序與不交換","en":"Function composition: the permitted inputs of a restricted composition, evaluation order and non-commutativity"},"summary":{"zh-Hant":"設 f:ℝ→ℝ，f(x)=x−3；g:[0,∞)→ℝ，g(u)=√u。(1) 求 g∘f 的允許輸入集合 D={x∈dom(f) ∣ f(x)∈dom(g)}，並證明 D=[3,∞)；(2) 依「先 f 後 g」計算 (g∘f)(7) 與 (g∘f)(4)，並說明 (g∘f)(1) 沒有定義；(3) 說明 f∘g 是 [0,∞) 上的總函數合成，並由 (f∘g)(4)=−1≠1=(g∘f)(4) 得到 g∘f≠f∘g；(4) 說明合成不是函數值的乘法。","en":"Let f:ℝ→ℝ, f(x)=x−3, and g:[0,∞)→ℝ, g(u)=√u. (1) Find the permitted inputs D={x∈dom(f) ∣ f(x)∈dom(g)} of g∘f and prove D=[3,∞); (2) evaluate (g∘f)(7) and (g∘f)(4) by 'f first, then g', and explain why (g∘f)(1) is undefined; (3) show that f∘g is a total composition on [0,∞), and conclude g∘f≠f∘g from (f∘g)(4)=−1≠1=(g∘f)(4); (4) show that composition is not multiplication of function values."},"mko_ids":["mko-composition"]},{"id":"derivation-mko-sequence-01","title":{"zh-Hant":"數列：由初始條件與遞迴確定通項，以及索引與有限前綴的陷阱","en":"Sequences: fixing the general term from an initial condition and a recurrence, and the pitfalls of indexing and finite prefixes"},"summary":{"zh-Hant":"設實數數列 a:ℕ→ℝ（ℕ 含 0）滿足 a₀=5、a_{n+1}=a_n+3。(1) 證明恰有一個這樣的數列，且 a_n=5+3n；(2) 計算 a₀ 到 a₄ 與 a₁₀，說明 a₀ 是起始項、a₁₀ 是從 1 數起的第 11 個值；(3) 寫出長度 L=4 的有限數列及其索引集合；(4) 說明前綴 5、8、11 不能唯一決定數列。","en":"Let the real sequence a:ℕ→ℝ (with 0∈ℕ) satisfy a₀=5 and a_{n+1}=a_n+3. (1) Prove that exactly one such sequence exists and that a_n=5+3n; (2) compute a₀ to a₄ and a₁₀, explaining that a₀ is the initial term and a₁₀ is the 11th value when counting from 1; (3) write down the finite sequence of length L=4 and its index set; (4) show that the prefix 5, 8, 11 does not determine the sequence."},"mko_ids":["mko-sequence"]},{"id":"derivation-mko-permutation-combination-01","title":{"zh-Hant":"排列與組合：以計數證明 P(n,k)=n!/(n−k)! 與 C(n,k)=n!/(k!(n−k)!)","en":"Permutations and combinations: proving P(n,k)=n!/(n−k)! and C(n,k)=n!/(k!(n−k)!) by counting"},"summary":{"zh-Hant":"設 X 是恰有 n 個相異物件的集合，n、k 為自然數且 0≤k≤n，選取無放回、不重複。證明：從 X 有序選取 k 個的方法數 P(n,k)=n!/(n−k)!，無序選取 k 個的方法數 C(n,k)=n!/(k!(n−k)!)，且 k=0 時兩者都等於 1。最後以 X={a,b,c,d,e}、k=3 算出 P(5,3)=60 與 C(5,3)=10。","en":"Let X be a set of exactly n distinct objects, with natural numbers n, k and 0≤k≤n, and selection without replacement and without repetition. Prove that the number of ordered selections of k objects from X is P(n,k)=n!/(n−k)!, that the number of unordered selections is C(n,k)=n!/(k!(n−k)!), and that both equal 1 when k=0. Finally compute P(5,3)=60 and C(5,3)=10 for X={a,b,c,d,e} and k=3."},"mko_ids":["mko-permutation-combination"]},{"id":"derivation-mko-coordinate-plane-01","title":{"zh-Hant":"座標平面：判定點在哪條軸或哪個象限，並比較有序對","en":"Coordinate plane: deciding axis or quadrant membership and comparing ordered pairs"},"summary":{"zh-Hant":"在固定的平面直角座標系中，給定 A=(−3,2)、B=(2,−3)、B′=(6/3,−3)、C=(0,−4)、O=(0,0)、E=(−1/2,−5)。先證明兩軸只交於原點、每個點要嘛在軸上、要嘛恰在一個象限中；再逐點判定它所在的軸或象限，並判定 A 與 B、B 與 B′ 是否為同一點。","en":"In a fixed Cartesian coordinate system, let A=(−3,2), B=(2,−3), B′=(6/3,−3), C=(0,−4), O=(0,0) and E=(−1/2,−5). First prove that the axes meet only at the origin and that every point either lies on an axis or lies in exactly one quadrant; then decide for each point its axis or quadrant, and decide whether A and B, and B and B′, are the same point."},"mko_ids":["mko-coordinate-plane"]},{"id":"derivation-mko-power-root-01","title":{"zh-Hant":"自然數冪與主平方根：由遞迴算冪，並證明 √(a²)=|a|","en":"Natural-number powers and principal square roots: computing powers by recursion and proving √(a²)=|a|"},"summary":{"zh-Hant":"只用遞迴定義 a^0=1、a^(n+1)=a^n·a 算出 (−2)^n（n=0,1,2,3,4）與 0^0、0^3；證明非負平方根的唯一性，並證明對每個實數 a，√(a²)=|a|；再求 √49、√((−5)²)、方程 y²=9 的實數解集，並說明 √(−4) 在實數中沒有定義。","en":"Using only the recursive definition a^0=1, a^(n+1)=a^n·a, compute (−2)^n for n=0,1,2,3,4 and 0^0, 0^3; prove that a nonnegative square root is unique and that √(a²)=|a| for every real a; then find √49, √((−5)²) and the real solution set of y²=9, and explain why √(−4) is not defined in the reals."},"mko_ids":["mko-power-root"]},{"id":"derivation-mko-polynomial-01","title":{"zh-Hant":"一元有理係數多項式：合併同次項、係數向量、次數與零多項式","en":"Univariate rational polynomials: combining like terms, coefficient vectors, degree and the zero polynomial"},"summary":{"zh-Hant":"在 ℚ[x] 中，設 p(x)=4x³+3x²−x+2−4x³+x，q 的係數向量為 [−1,1,0,0]，r(x)=x²−x²+3，z(x)=5x−5x，t(x)=5x。求 p、q、r 的係數與次數，判定 z 是零多項式並區分 z 與 r；以係數計算 p+q 與 p·q 並求次數；最後說明 p(1)=t(1) 不能推出 p=t。","en":"In ℚ[x], let p(x)=4x³+3x²−x+2−4x³+x, let q have coefficient vector [−1,1,0,0], and let r(x)=x²−x²+3, z(x)=5x−5x and t(x)=5x. Find the coefficients and degrees of p, q and r; show that z is the zero polynomial and distinguish it from r; compute p+q and p·q from coefficients and find their degrees; finally explain why p(1)=t(1) does not give p=t."},"mko_ids":["mko-polynomial"]},{"id":"derivation-mko-factorization-01","title":{"zh-Hant":"有理係數多項式因式分解：以係數比對驗證提公因子、二項式乘積與平方差","en":"Polynomial factorization over the rationals: verifying common factors, binomial products and differences of squares by comparing coefficients"},"summary":{"zh-Hant":"在 ℚ[x] 中：(一) 把 p(x)=3x²+3x−18 分解為 3(x+3)(x−2)；(二) 把 q(x)=2x²−50 分解為 2(x−5)(x+5)；每個分解都以展開並比對每個係數驗證。(三) 證明 x²+25≠(x−5)(x+5)，且 x²+25 不能寫成兩個一次因式的乘積。(四) 說明 x³ 與 x 在 x=−1,0,1 取值相同卻不是同一個多項式。","en":"In ℚ[x]: (1) factor p(x)=3x²+3x−18 as 3(x+3)(x−2); (2) factor q(x)=2x²−50 as 2(x−5)(x+5); verify each factorization by expanding and comparing every coefficient. (3) Prove that x²+25≠(x−5)(x+5) and that x²+25 is not a product of two degree-one factors. (4) Explain why x³ and x take equal values at x=−1, 0, 1 and still are different polynomials."},"mko_ids":["mko-factorization"]},{"id":"derivation-mko-linear-function-01","title":{"zh-Hant":"仿射函數與一次函數：由兩點求 m、b，判定次數、零點與是否為線性映射","en":"Affine and degree-one functions: finding m and b from two points, and deciding degree, zero and linearity"},"summary":{"zh-Hant":"仿射函數 f:ℝ→ℝ，f(x)=mx+b 的圖形通過 (1,4) 與 (3,0)。求 m、b；判定 f 是否為次數一的函數、求它的零點，並判定 f 是否為線性代數意義下的 ℝ 線性映射。再證明兩個仿射函數在 ℝ 上相等當且僅當斜率與截距都相同，據此說明 f(2)=k(2)（k(x)=x）不能推出 f=k；最後處理 m=0 的兩個退化例子 g(x)=5 與 h(x)=0。","en":"The graph of the affine function f:ℝ→ℝ, f(x)=mx+b, passes through (1,4) and (3,0). Find m and b; decide whether f has degree one, find its zero, and decide whether f is an ℝ-linear map in the sense of linear algebra. Then prove that two affine functions are equal on ℝ exactly when their slopes and intercepts agree, and use it to explain why f(2)=k(2) (with k(x)=x) does not give f=k; finally treat the two degenerate examples with m=0, g(x)=5 and h(x)=0."},"mko_ids":["mko-linear-function"]},{"id":"derivation-mko-function-graph-01","title":{"zh-Hant":"實值函數圖形：成員判定與有限表格的總函數驗證","en":"Graphs of real-valued functions: membership tests and verifying a finite table as a total function"},"summary":{"zh-Hant":"(一) 設 D=[0,2]、f:D→ℝ、f(x)=x²−1。先證明 (u,v)∈G_f ⇔ u∈D 且 v=f(u)，再判定 (1,0)、(−1,0)、(2,2)、(2,3) 是否在 G_f 上。(二) 設 D_0={0,1,2}、B={0,1,4}。證明一個集合 Γ 是某個總函數 D_0→B 的圖形，當且僅當 Γ⊆D_0×B 且每個輸入恰有一個輸出；據此判定 T={(0,0),(1,1),(2,4)} 是、而 T_1={(0,0),(2,4)}、T_2={(0,0),(1,1),(1,4),(2,4)}、T_3={(0,0),(1,1),(2,4),(3,9)} 都不是。","en":"(1) Let D=[0,2], f:D→ℝ, f(x)=x²−1. First prove (u,v)∈G_f ⇔ u∈D and v=f(u), then decide whether (1,0), (−1,0), (2,2) and (2,3) lie on G_f. (2) Let D_0={0,1,2} and B={0,1,4}. Prove that a set Γ is the graph of some total function D_0→B exactly when Γ⊆D_0×B and every input has exactly one output; use it to decide that T={(0,0),(1,1),(2,4)} is such a graph, while T_1={(0,0),(2,4)}, T_2={(0,0),(1,1),(1,4),(2,4)} and T_3={(0,0),(1,1),(2,4),(3,9)} are not."},"mko_ids":["mko-function-graph"]},{"id":"derivation-mko-probability-01","title":{"zh-Hant":"有限樣本空間上的機率：驗證權重模型、計算事件機率與補事件","en":"Probability on finite sample spaces: validating a weight model, computing event probabilities and complements"},"summary":{"zh-Hant":"S={1,2,3,4}，事件族為全冪集，權重 w_1=1/2、w_2=1/3、w_3=1/6、w_4=0。驗證這是合法的有限機率模型；由定義證明 P(∅)=0、P(S)=1、互斥可加與 P(S∖E)=1−P(E)；計算 P({1,3})，說明它不等於 |E|/|S|；說明 {4} 非空但機率為 0；並判定兩組候選權重 (1/2,1/3,1/3,0) 與 (1/2,2/3,−1/6,0) 都不合法。","en":"Let S={1,2,3,4} with the full powerset as event family and weights w_1=1/2, w_2=1/3, w_3=1/6, w_4=0. Verify that this is a legal finite probability model; prove from the definition the identities P(∅)=0 and P(S)=1, additivity for disjoint events, and the complement identity P(S∖E)=1−P(E); compute P({1,3}) and explain why it differs from |E|/|S|; show that {4} is nonempty yet has probability 0; and decide that the candidate weights (1/2,1/3,1/3,0) and (1/2,2/3,−1/6,0) are both illegal."},"mko_ids":["mko-probability"]},{"id":"derivation-mko-quantifier-01","title":{"zh-Hant":"量詞與明示論域：在有限整數論域上判定全稱、存在及其否定","en":"Quantifiers and explicit domains: deciding universal and existential statements and their negations on a finite integer domain"},"summary":{"zh-Hant":"在明示有限論域 U={−2,−1,0,1,2,3} 上，判定 φ₁：∀x∈U, x²<10；φ₂：∀x∈U, x²>x；φ₃：∃x∈U, x²=2 的真假；正確寫出 φ₂ 與 φ₃ 的否定；並說明論域為空集合時全稱與存在陳述的真假。","en":"On the explicit finite domain U={−2,−1,0,1,2,3}, decide the truth of φ₁: ∀x∈U, x²<10; φ₂: ∀x∈U, x²>x; φ₃: ∃x∈U, x²=2; write the negations of φ₂ and φ₃ correctly; and settle the truth of universal and existential statements when the domain is the empty set."},"mko_ids":["mko-quantifier"]},{"id":"derivation-mko-counterexample-01","title":{"zh-Hant":"反例與論域資格：n²+n+11 的質數性、失敗集合與最小反例","en":"Counterexamples and domain eligibility: primality of n²+n+11, the failure set and the least counterexample"},"summary":{"zh-Hant":"設 U={n∈ℤ ∣ 0≤n≤12}、V={n∈ℤ ∣ 0≤n≤9}、m(n)=n²+n+11、P(n)：「m(n) 是質數」。判定 φ_U：∀n∈U P(n)；求失敗集合 F_U 與最小反例；判定候選值 −11 的資格；並判定 φ_V：∀n∈V P(n)。","en":"Let U={n∈ℤ ∣ 0≤n≤12}, V={n∈ℤ ∣ 0≤n≤9}, m(n)=n²+n+11 and P(n): 'm(n) is prime'. Decide φ_U: ∀n∈U P(n); find the failure set F_U and the least counterexample; decide whether the candidate −11 is eligible; and decide φ_V: ∀n∈V P(n)."},"mko_ids":["mko-counterexample"]},{"id":"derivation-mko-relation-01","title":{"zh-Hant":"二元關係：定義域、關係像與逆關係的逐步計算","en":"Binary relations: computing the domain, images and the converse step by step"},"summary":{"zh-Hant":"設 A={1,2,3,4}、B={2,3,4,5,6}、R={(1,2),(1,3),(3,3),(4,6)}、C={1,3}。核對 R⊆A×B；計算 Dom(R)、R[C]、R[∅]、R⁻¹ 與 Dom(R⁻¹)；判定 R 是否為函數 A→B、R⁻¹ 是否為函數 B→A。","en":"Let A={1,2,3,4}, B={2,3,4,5,6}, R={(1,2),(1,3),(3,3),(4,6)} and C={1,3}. Check R⊆A×B; compute Dom(R), R[C], R[∅], R⁻¹ and Dom(R⁻¹); decide whether R is a function A→B and whether R⁻¹ is a function B→A."},"mko_ids":["mko-relation"]},{"id":"derivation-mko-graph-01","title":{"zh-Hant":"有限簡單無向圖：鄰居、度數與握手公式","en":"Finite simple undirected graphs: neighbours, degrees and the handshake formula"},"summary":{"zh-Hant":"設 V={1,2,3,4,5}，邊以端點對清單輸入：(1,2)、(2,1)、(2,3)、(3,4)、(2,4)、(4,2)，每一項表示一條無向邊。把輸入正規化為邊集合 E，求每個頂點的鄰居集合 N(v) 與度數 deg(v)，核對度數和等於 2|E|，並對每個有限簡單無向圖證明握手公式 Σ_{v∈V} deg(v)=2|E|。","en":"Let V={1,2,3,4,5}, with edges entered as a list of endpoint pairs: (1,2), (2,1), (2,3), (3,4), (2,4), (4,2), each entry denoting an undirected edge. Normalize the input to the edge set E, find the neighbour set N(v) and degree deg(v) of every vertex, check that the degree sum equals 2|E|, and prove the handshake formula Σ_{v∈V} deg(v)=2|E| for every finite simple undirected graph."},"mko_ids":["mko-graph"]},{"id":"derivation-mko-vector-01","title":{"zh-Hant":"固定基底下的平面向量：逐分量運算、相等判定與減法方向","en":"Plane vectors in a fixed basis: componentwise operations, equality and the direction of subtraction"},"summary":{"zh-Hant":"在固定標準基底下，設 u=(3,−1)、v=(−2,4)、w=(1,3)、k=−2。逐分量計算 u+v、u−v、v−u、ku、0u、2u+v；求滿足 u+x=v 的向量 x；核對 k(u+v)=ku+kv；並判定 u=w 與 u=(−1,3) 是否成立。","en":"In the fixed standard basis let u=(3,−1), v=(−2,4), w=(1,3) and k=−2. Compute u+v, u−v, v−u, ku, 0u and 2u+v componentwise; find the vector x with u+x=v; check k(u+v)=ku+kv; and decide whether u=w and whether u=(−1,3)."},"mko_ids":["mko-vector"]},{"id":"derivation-mko-system-linear-equations-01","title":{"zh-Hant":"二元聯立一次方程組：唯一解公式、兩列核對與 D=0 的一致性判定","en":"Linear systems in two variables: the unique-solution formula, checking both rows, and consistency when D=0"},"summary":{"zh-Hant":"(I) 解 2x+3y=7、x−y=1：證明 D=ae−bd≠0 時的唯一解公式，求解並代入兩列核對；(II) 判定 x+2y=3、2x+4y=7；(III) 判定 x+2y=3、2x+4y=6；並以 (IV) 0x+0y=1、0x+0y=2 說明 D 與兩個 Cramer 分子都為零時不能斷定有無限多解。","en":"(I) Solve 2x+3y=7, x−y=1: prove the unique-solution formula when D=ae−bd≠0, solve, and check by substituting into both rows; (II) decide x+2y=3, 2x+4y=7; (III) decide x+2y=3, 2x+4y=6; and use (IV) 0x+0y=1, 0x+0y=2 to show that D and both Cramer numerators being zero does not imply infinitely many solutions."},"mko_ids":["mko-system-linear-equations"]},{"id":"derivation-mko-conditional-probability-01","title":{"zh-Hant":"有限模型的條件機率：正分母、兩個方向與零機率條件","en":"Conditional probability in finite models: positive denominators, the two directions and null conditioning events"},"summary":{"zh-Hant":"在 S={1,2,3,4,5,6}、權重 w=(1,2,3,2,4,0)/12 的有限模型中，設 A={2,3}、B={3,4,5,6}、Z={6}。驗證模型合法，計算 P(A|B) 與 P(B|A)，與等可能計數比 |A∩B|/|B| 及乘積 P(A)P(B) 比較，並判定 P(A|Z) 是否有定義。","en":"In the finite model on S={1,2,3,4,5,6} with weights w=(1,2,3,2,4,0)/12, let A={2,3}, B={3,4,5,6} and Z={6}. Validate the model, compute P(A|B) and P(B|A), compare them with the equiprobable count ratio |A∩B|/|B| and with the product P(A)P(B), and decide whether P(A|Z) is defined."},"mko_ids":["mko-conditional-probability"]},{"id":"derivation-mko-injective-01","title":{"zh-Hant":"單射與不同輸入的辨識：函數表的資格、碰撞見證與纖維","en":"Injectivity and distinct inputs: table qualification, collision witnesses and fibers"},"summary":{"zh-Hant":"設 A={1,2,3,4}、B={1,2,3,4,5,6}，函數表 f：(1,4),(2,6),(3,1),(4,5)；g：(1,4),(2,6),(2,6),(3,4),(4,6)；h：(1,4),(2,6),(2,5),(3,1),(4,5)。確認各表是否為總函數 A→B；判定 f、g 是否單射並列出纖維；找出 g 按字典序的首碰撞；判定 g 限制到 C={1,2} 是否單射；並說明空定義域的情形。","en":"Let A={1,2,3,4}, B={1,2,3,4,5,6}, with function tables f: (1,4),(2,6),(3,1),(4,5); g: (1,4),(2,6),(2,6),(3,4),(4,6); h: (1,4),(2,6),(2,5),(3,1),(4,5). Decide which tables are total functions A→B; decide whether f and g are injective and list their fibers; find g's first collision in lexicographic order; decide whether g restricted to C={1,2} is injective; and treat the empty domain."},"mko_ids":["mko-injective"]}],"title":"Concept derivation guides"}