{"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":{},"kind":"guides","total":299,"pages":13,"page":1,"entries":[{"id":"derivation-mko-point-line-plane-01","title":{"zh-Hant":"點、線、面：在座標模型中判定點是否在直線上","en":"Points, lines and planes: deciding incidence in the coordinate model"},"summary":{"zh-Hant":"在座標平面模型 ℝ² 中，設 P=(1,2)、Q=(3,6)。判定 R=(4,8) 與 S=(2,5) 是否在通過 P、Q 的直線 L 上，並據此判定 P、Q、R 是否共線。","en":"In the coordinate-plane model ℝ², let P=(1,2) and Q=(3,6). Decide whether R=(4,8) and S=(2,5) lie on the line L through P and Q, and hence whether P, Q, R are collinear."},"mko_ids":["mko-point-line-plane"]},{"id":"derivation-mko-angle-01","title":{"zh-Hant":"角：對頂角相等的逐步證明","en":"Angle: a step-by-step proof that vertical angles are equal"},"summary":{"zh-Hant":"直線 AB 與直線 CD 是兩條相異直線，相交於 O，且 O 在 A、B 之間，也在 C、D 之間。證明對頂角相等：m∠AOC=m∠BOD 且 m∠COB=m∠DOA；再以 m∠AOC=35° 為例算出四個角。","en":"Distinct lines AB and CD meet at O, with O between A and B and between C and D. Prove that vertical angles are equal: m∠AOC=m∠BOD and m∠COB=m∠DOA; then, for m∠AOC=35°, compute all four angles."},"mko_ids":["mko-angle"]},{"id":"derivation-mko-triangle-01","title":{"zh-Hant":"三角形：在座標模型中確認三點界定非退化三角形","en":"Triangle: confirming in the coordinate model that three points determine a nondegenerate triangle"},"summary":{"zh-Hant":"在座標平面模型 ℝ² 中，設 A=(0,0)、B=(4,0)、C=(1,3)。確認 A、B、C 兩兩相異且不共線，因而依 MKO 界定非退化三角形 ABC；寫出三條邊與各頂點的對邊，並說明三頂點的平均點 G=(5/3,1) 不在三角形 ABC 上。另設 D=(2,1)、E=(6,3)，說明相異三點 A、D、E 共線，不界定非退化三角形。","en":"In the coordinate-plane model ℝ², let A=(0,0), B=(4,0), C=(1,3). Confirm that A, B, C are pairwise distinct and noncollinear, so that by the MKO they determine the nondegenerate triangle ABC; write out its three sides and the side opposite each vertex, and show that the average of the vertices, G=(5/3,1), is not on triangle ABC. Then, with D=(2,1) and E=(6,3), show that the distinct points A, D, E are collinear and so do not determine a nondegenerate triangle."},"mko_ids":["mko-triangle"]},{"id":"derivation-mko-right-triangle-01","title":{"zh-Hant":"直角三角形：用座標判定哪一個內角是直角","en":"Right triangle: deciding in coordinates which interior angle is right"},"summary":{"zh-Hant":"在座標平面模型 ℝ² 中，設 A=(1,1)、B=(5,3)、C=(0,3)。證明 A、B、C 構成非退化三角形，且其在 A 的內角等於 90°（即 π/2），所以 ABC 是直角三角形；說明另外兩角不是直角、任何非退化三角形至多一個直角，並指出斜邊與兩股。再以 D=(1,4) 說明三角形 ABD 不是直角三角形。","en":"In the coordinate-plane model ℝ², let A=(1,1), B=(5,3), C=(0,3). Prove that A, B, C form a nondegenerate triangle whose interior angle at A equals 90° (that is, π/2), so ABC is a right triangle; show that the other two angles are not right and that any nondegenerate triangle has at most one right angle, and identify the hypotenuse and the legs. Then, with D=(1,4), show that triangle ABD is not a right triangle."},"mko_ids":["mko-right-triangle"]},{"id":"derivation-mko-euclidean-length-01","title":{"zh-Hant":"歐幾里得長度：由座標精確計算兩點距離","en":"Euclidean length: computing the distance between two points exactly from coordinates"},"summary":{"zh-Hant":"在以 ℝ² 表示的歐幾里得平面中，設 P=(1,−2)、Q=(4,2)、R=(2,3)。精確算出 d(P,Q)=5；算出 d(P,R)=√26，並以平方比較證明 5.09<√26<5.1，說明 5.1 只是近似值。另對一般的實數坐標證明：距離有定義且非負、d(Q,P)=d(P,Q)、d(P,P)=0，以及 d(P,Q)=0 ⇔ P=Q。","en":"In the Euclidean plane represented by ℝ², let P=(1,−2), Q=(4,2), R=(2,3). Compute d(P,Q)=5 exactly; compute d(P,R)=√26 and prove by comparing squares that 5.09<√26<5.1, showing that 5.1 is only an approximation. For general real coordinates, also prove that the distance is defined and nonnegative, that d(Q,P)=d(P,Q), that d(P,P)=0, and that d(P,Q)=0 ⇔ P=Q."},"mko_ids":["mko-euclidean-length"]},{"id":"derivation-mko-euclid-pythagorean-theorem-01","title":{"zh-Hant":"畢達哥拉斯定理：在座標模型中以內積證明","en":"Pythagorean theorem: a proof by dot products in the coordinate model"},"summary":{"zh-Hant":"在以 ℝ² 表示的歐幾里得平面中（距離為歐幾里得長度），證明：對任何直角三角形 ABC，若直角在 C，兩股長 a=d(C,B)、b=d(C,A)，斜邊長 c=d(B,A)，則 a²+b²=c²。再以 C=(1,2)、B=(4,6)、A=(−7,8) 驗算（25+100=125），並以非直角三角形 E=(0,0)、F=(4,0)、H=(1,2) 說明沒有直角時關係一般不成立。","en":"In the Euclidean plane represented by ℝ² (with Euclidean length as distance), prove: for every right triangle ABC with the right angle at C, legs of lengths a=d(C,B) and b=d(C,A), and hypotenuse of length c=d(B,A), a²+b²=c². Then check the example C=(1,2), B=(4,6), A=(−7,8) (25+100=125), and use the non-right triangle E=(0,0), F=(4,0), H=(1,2) to show that without a right angle the relation fails in general."},"mko_ids":["mko-euclid-pythagorean-theorem"]},{"id":"derivation-mko-natural-number-01","title":{"zh-Hant":"自然數：以最小歸納集判定成員，並分辨數與數字符號","en":"Natural number: deciding membership through the smallest inductive set, and telling numbers from numerals"},"summary":{"zh-Hant":"依 OCME 慣例（0∈N），把 MKO 的「0 以及由 0 經有限次後繼所得的數」讀成「實數中包含 0 且對 x↦x+1 封閉的最小集合」。據此證明 0∈N、3∈N；說明二進位符號「11」與十進位符號「3」表示同一個自然數；證明 −1∉N 與 1/2∉N；並與「自然數從 1 開始」的慣例 N⁺=N∖{0} 比較最小元素。","en":"Under the OCME convention (0∈N), read the MKO's '0 and the numbers obtained from 0 by finitely many successor steps' as 'the smallest set of reals that contains 0 and is closed under x↦x+1'. On this basis prove 0∈N and 3∈N; show that the binary numeral '11' and the decimal numeral '3' denote the same natural number; prove −1∉N and 1/2∉N; and compare least elements with the convention N⁺=N∖{0}, in which natural numbers start at 1."},"mko_ids":["mko-natural-number"]},{"id":"derivation-mko-integer-01","title":{"zh-Hant":"整數：判定成員、負號的意義與對負號的封閉性","en":"Integer: deciding membership, the meaning of the minus sign, and closure under negation"},"summary":{"zh-Hant":"在實數中，取 Z={0}∪N⁺∪{−n ∣ n∈N⁺}（N 依 OCME 慣例含 0，N⁺=N∖{0}，−n 是 n 的加法反元素）。證明 −7∈Z；說明十進位符號「3.0」表示整數 3；證明 −2.5∉Z；由加法反元素的唯一性證明 −(−a)=a，特別是 −(−4)=4；對所有 z∈Z 證明 −z∈Z；並以 5−8=5+(−8)=−3 分辨二元減法與一元負號。","en":"Inside the reals, take Z={0}∪N⁺∪{−n ∣ n∈N⁺} (N contains 0 by the OCME convention, N⁺=N∖{0}, and −n is the additive inverse of n). Prove −7∈Z; show that the decimal numeral '3.0' denotes the integer 3; prove −2.5∉Z; prove −(−a)=a from uniqueness of additive inverses, in particular −(−4)=4; prove −z∈Z for every z∈Z; and use 5−8=5+(−8)=−3 to tell binary subtraction from the unary minus sign."},"mko_ids":["mko-integer"]},{"id":"derivation-mko-rational-number-01","title":{"zh-Hant":"有理數：同一個有理數的不同分數表示、交叉相乘與約分","en":"Rational number: different fraction representations of one rational, cross-multiplication and reduction"},"summary":{"zh-Hant":"依 MKO 的定義（存在整數 p、q，q≠0，使 x=p/q），證明 6/(−8)、−3/4、9/(−12) 表示同一個有理數，並把 6/(−8) 精確約成正規形式 −3/4；以交叉相乘證明 2/3≠3/5；說明「5/0」不表示任何數；說明有限小數 −1.25=−5/4 與整數 7=7/1 都是有理數；並在分母為正時用交叉相乘證明 −3/4<−2/3。","en":"Using the MKO's definition (there are integers p, q with q≠0 and x=p/q), prove that 6/(−8), −3/4 and 9/(−12) represent the same rational number, and reduce 6/(−8) exactly to the canonical form −3/4; prove 2/3≠3/5 by cross-multiplication; show that '5/0' denotes no number; show that the finite decimal −1.25=−5/4 and the integer 7=7/1 are rational; and, with positive denominators, prove −3/4<−2/3 by cross-multiplication."},"mko_ids":["mko-rational-number"]},{"id":"derivation-mko-real-number-01","title":{"zh-Hant":"實數：在 Cauchy 完備化中證明「循環的 9」與有理數 1 是同一個實數","en":"Real number: proving in the Cauchy completion that 'repeating nines' and the rational 1 are the same real"},"summary":{"zh-Hant":"在以有理數 Cauchy 序列的等價類建構的實數中（有理數 q 嵌入為常數列的類 ι(q)），令 a_n=1−10^{−n}（n≥1），即有限小數 0.9、0.99、0.999 等。證明 (a_n) 是 Cauchy 序列，且與常數列 1 等價，所以 [(a_n)]=ι(1)；證明 ι 是單射，因此有限小數 0.999=999/1000 所表示的實數不等於 1；並說明有限小數 0.375=3/8 是有理數，在 ℝ 中就是 ι(3/8)。","en":"In the reals constructed as equivalence classes of Cauchy sequences of rationals (a rational q embedded as the class ι(q) of its constant sequence), let a_n=1−10^{−n} (n≥1), the finite decimals 0.9, 0.99, 0.999 and so on. Prove that (a_n) is a Cauchy sequence equivalent to the constant sequence 1, so [(a_n)]=ι(1); prove that ι is injective, so the real denoted by the finite decimal 0.999=999/1000 is not 1; and show that the finite decimal 0.375=3/8 is rational and is ι(3/8) in ℝ."},"mko_ids":["mko-real-number"]},{"id":"derivation-mko-arithmetic-operations-01","title":{"zh-Hant":"四則運算：同一組輸入在 ℕ、ℤ、ℚ 中是否有定義","en":"Arithmetic operations: whether the same inputs give a defined result in ℕ, ℤ and ℚ"},"summary":{"zh-Hant":"取 a=3、b=5。在 ℕ（含 0）、ℤ、ℚ 三個數系中，判定 a+b、a·b、a−b、a÷b 是否有定義；有定義時求出結果，並判定結果屬於哪些數系。再判定 3÷0 與 0÷0 在這三個數系中是否有定義。","en":"Take a=3 and b=5. In each of the number systems ℕ (including 0), ℤ and ℚ, decide whether a+b, a·b, a−b and a÷b are defined; when they are, compute the result and decide which systems it belongs to. Then decide whether 3÷0 and 0÷0 are defined in these three systems."},"mko_ids":["mko-arithmetic-operations"]},{"id":"derivation-mko-variable-expression-01","title":{"zh-Hant":"變數與代數式：式子的值由取值、作用域與數系共同決定","en":"Variables and expressions: a value depends on the assignment, the scope and the number system"},"summary":{"zh-Hant":"設 E=x+1、F=(x+1)÷2、H=Σ_{x=1}^{3}(x+1)。(1) 在 ℤ 中求 E 在 x=2 與 x=5 的值，並證明 E 沒有與 x 無關的固定值；(2) 求 F 在 x=4 時於 ℚ 與 ℤ 中的值；(3) 在外部指定 x=5 時求 H；(4) 判定 (x+1)+x 與 2x+1、x·x 與 2x 在 ℤ 上是否等價；(5) 區分表達式 x+1 與方程 x+1=3，並求方程在 ℤ 中的解集。","en":"Let E=x+1, F=(x+1)÷2 and H=Σ_{x=1}^{3}(x+1). (1) In ℤ, evaluate E at x=2 and at x=5 and prove that E has no fixed value independent of x; (2) evaluate F at x=4 in ℚ and in ℤ; (3) evaluate H when x=5 is assigned from outside; (4) decide whether (x+1)+x and 2x+1, and whether x·x and 2x, are equivalent on ℤ; (5) distinguish the expression x+1 from the equation x+1=3 and find the solution set of the equation in ℤ."},"mko_ids":["mko-variable-expression"]},{"id":"derivation-mko-equation-01","title":{"zh-Hant":"方程式：在定義域內解 x²/(x−1)=1/(x−1)","en":"Equation: solving x²/(x−1)=1/(x−1) inside its domain"},"summary":{"zh-Hant":"在允許範圍 ℝ 中，求方程 x²/(x−1)=1/(x−1) 的解集；並說明為什麼兩邊乘以 x−1 後得到的方程 x²=1 有兩個解，而原方程只有一個。","en":"With allowed domain ℝ, find the solution set of the equation x²/(x−1)=1/(x−1), and explain why the equation x²=1 obtained by multiplying both sides by x−1 has two solutions while the original has only one."},"mko_ids":["mko-equation"]},{"id":"derivation-mko-set-01","title":{"zh-Hant":"集合：由元素刻畫、外延相等，以及不受限制的建構為何失敗","en":"Sets: characterized by elements, equal by extensionality, and why unrestricted formation fails"},"summary":{"zh-Hant":"在論域 ℤ 中，設 A={x∈ℤ ∣ x²=4}、B={−2,2}、C={2,−2,2}、D={2}。證明 A=B 與 C=B；證明 D≠B，並區分元素 2 與集合 {2}。最後證明：在任何帶有隸屬關係的論域 U 中，都不存在 R∈U 使「y∈R ⇔ y∉y」對所有 y∈U 成立，因此不是每個條件都能形成集合。","en":"In the universe ℤ, let A={x∈ℤ ∣ x²=4}, B={−2,2}, C={2,−2,2} and D={2}. Prove A=B and C=B; prove D≠B and distinguish the element 2 from the set {2}. Finally prove that in any domain U with a membership relation there is no R∈U with 'y∈R ⇔ y∉y' for all y∈U, so not every condition forms a set."},"mko_ids":["mko-set"]},{"id":"derivation-mko-proposition-01","title":{"zh-Hant":"命題：判定候選語句是否為命題，並區分「是命題」與「已證明為真」","en":"Proposition: deciding which candidate sentences are propositions, and separating 'is a proposition' from 'is proved true'"},"summary":{"zh-Hant":"在入門古典語境中，逐一判定下列候選是否為命題；是命題者再判定真假並給出理由：s1「7 是質數」；s2「4 是質數」；s3「7 是質數嗎？」；s4「計算 3+4。」；s5「x>0」（x 是未指定的整數變數），以及 s5 的封閉版本：代入 x=2 與 x=−1、加上 ∀x∈ℤ、加上 ∃x∈ℤ。","en":"In an introductory classical context, decide for each candidate whether it is a proposition, and for each proposition decide whether it is true, with reasons: s1 '7 is prime'; s2 '4 is prime'; s3 'Is 7 prime?'; s4 'Compute 3+4.'; s5 'x>0' (x an unassigned integer variable), together with closed versions of s5: substituting x=2 and x=−1, prefixing ∀x∈ℤ, and prefixing ∃x∈ℤ."},"mko_ids":["mko-proposition"]},{"id":"derivation-mko-logical-connectives-01","title":{"zh-Hant":"邏輯連接詞：以整除命題檢驗 ∨、→、↔ 與括號","en":"Logical connectives: testing ∨, →, ↔ and brackets on divisibility statements"},"summary":{"zh-Hant":"對整數 n，令 P(n) 為「4∣n」，Q(n) 為「2∣n」。(1) 判定「4∣12 ∨ 6∣12」的真假，說明析取是含括或；(2) 證明 ∀n∈ℤ (P(n)→Q(n))；(3) 證明 ∀n∈ℤ (Q(n)→P(n)) 為假，因而 ∀n∈ℤ (P(n)↔Q(n)) 為假；(4) 判定 P(3)→Q(3) 與「2+2=4 → 7 是奇數」的真假，說明蘊涵不是因果；(5) 比較 ¬Q(2)∧P(2) 與 ¬(Q(2)∧P(2)) 的真假。","en":"For an integer n let P(n) be '4∣n' and Q(n) be '2∣n'. (1) Decide the truth of '4∣12 ∨ 6∣12' and explain that disjunction is inclusive; (2) prove ∀n∈ℤ (P(n)→Q(n)); (3) prove that ∀n∈ℤ (Q(n)→P(n)) is false, and hence that ∀n∈ℤ (P(n)↔Q(n)) is false; (4) decide the truth of P(3)→Q(3) and of '2+2=4 → 7 is odd', explaining that implication is not causation; (5) compare the truth of ¬Q(2)∧P(2) and ¬(Q(2)∧P(2))."},"mko_ids":["mko-logical-connectives"]},{"id":"derivation-mko-proof-01","title":{"zh-Hant":"數學證明：證明 n²+n 恆為偶數，並以反例否定「n²+n+41 恆為質數」","en":"Mathematical proof: proving n²+n is always even, and refuting 'n²+n+41 is always prime' by a counterexample"},"summary":{"zh-Hant":"(1) 在整數的指定語境中，由明示的前提與規則證明定理 T：對每個整數 n，2∣(n²+n)；再由 T 以反證法推出沒有整數 n 使 n²+n 為奇數。(2) 檢驗主張 C：「對每個自然數 n，n²+n+41 是質數」；說明 n=0 到 39 的 40 個正例為何不構成證明，並以 n=40 的反例否定 C。","en":"(1) In a stated context for the integers, prove from explicit premises and rules the theorem T: for every integer n, 2∣(n²+n); then derive from T by contradiction that no integer n makes n²+n odd. (2) Test the claim C: 'for every natural number n, n²+n+41 is prime'; explain why the 40 positive cases n=0 to 39 do not constitute a proof, and refute C by the counterexample n=40."},"mko_ids":["mko-proof"]},{"id":"derivation-mko-set-membership-01","title":{"zh-Hant":"集合隸屬：x∈A 是命題，並與 ⊆、= 及程式的成員測試區分","en":"Set membership: x∈A is a proposition, kept apart from ⊆, = and a program's membership test"},"summary":{"zh-Hant":"在型別化語境中，設 A={1,2,3}、B={1,2}、𝓕={{1,2},{3}}（元素是整數集合的集合）、E={n∈ℤ ∣ 2∣n}。判定下列命題的真假並給出理由：2∈A；4∈A；B⊆A；A⊆B；B∈𝓕；{1}∈𝓕；1000000∈E；7∈E。並說明 1∈B 與 B∈𝓕 為何不給出 1∈𝓕，以及程式語言的成員測試為何不能取代集合隸屬的語義。","en":"In a typed context let A={1,2,3}, B={1,2}, 𝓕={{1,2},{3}} (a set whose elements are sets of integers) and E={n∈ℤ ∣ 2∣n}. Decide with reasons whether each holds: 2∈A; 4∈A; B⊆A; A⊆B; B∈𝓕; {1}∈𝓕; 1000000∈E; 7∈E. Also explain why 1∈B and B∈𝓕 do not give 1∈𝓕, and why a programming language's membership test cannot replace the semantics of set membership."},"mko_ids":["mko-set-membership"]},{"id":"derivation-mko-subset-01","title":{"zh-Hant":"子集合：以任意元素證明包含，並以反例否定包含","en":"Subset: proving inclusion from an arbitrary element and refuting it by a counterexample"},"summary":{"zh-Hant":"在論域 ℤ 中，令 A 為 6 的倍數之集、B 為偶數之集、C 為 4 的倍數之集。證明 A⊆B，且這個包含是嚴格的（A≠B）；證明 A⊄C，雖然 A 與 C 有共同元素 12；說明 B⊆B 成立（⊆ 允許相等）；並證明 {6}⊆A ⇔ 6∈A，同時分辨兩個語句的型別角色。","en":"In the universe ℤ, let A be the set of multiples of 6, B the set of even integers and C the set of multiples of 4. Prove A⊆B and that this inclusion is strict (A≠B); prove A⊄C even though A and C share the element 12; note that B⊆B holds (⊆ allows equality); and prove {6}⊆A ⇔ 6∈A while separating the type roles of the two statements."},"mko_ids":["mko-subset"]},{"id":"derivation-mko-set-operations-01","title":{"zh-Hant":"集合運算：在明示全集中計算聯集、交集、差集與補集","en":"Set operations: union, intersection, difference and complement in an explicit universe"},"summary":{"zh-Hant":"全集 U={1,2,3,4,5,6,7,8}，A={2,4,6,8}，B={3,4,5,6}。逐元素算出 A∪B、A∩B、A∖B、B∖A、U∖A、U∖B；說明 A∖B≠B∖A；說明全集換成 U′={1,2,3,4,5,6,7,8,9,10} 時 A 的補集改變；對任意集合證明 U∖(A∪B)=(U∖A)∩(U∖B) 並以本例核對；再以 A∪B=A∪(B∖A) 說明聯集為何有 6 個元素而不是 8 個。","en":"Universe U={1,2,3,4,5,6,7,8}, A={2,4,6,8}, B={3,4,5,6}. Compute A∪B, A∩B, A∖B, B∖A, U∖A and U∖B element by element; show A∖B≠B∖A; show that the complement of A changes when the universe is U′={1,2,3,4,5,6,7,8,9,10}; prove U∖(A∪B)=(U∖A)∩(U∖B) for arbitrary sets and check it on this example; and use A∪B=A∪(B∖A) to explain why the union has 6 elements, not 8."},"mko_ids":["mko-set-operations"]},{"id":"derivation-mko-sample-space-01","title":{"zh-Hant":"樣本空間：擲兩次硬幣的兩種觀察尺度與不等可能的結果","en":"Sample space: two observation scales for two coin tosses, and outcomes that are not equally likely"},"summary":{"zh-Hant":"隨機試驗：把一枚硬幣擲兩次。(1) 若依次記錄結果，寫出樣本空間 S，證明它包含所有可能基本結果且恰有 4 個元素；(2) 若只記錄正面次數，寫出樣本空間 S′={0,1,2}；(3) 在另外給定的機率模型 P（S 的每個有序結果機率 1/4）下，算出 S′ 各結果的機率 1/4、1/2、1/4，說明同一樣本空間中的結果不必等可能；(4) 以 S 上另一個合法的非均勻模型 Q 說明樣本空間本身不決定機率；(5) 分辨結果 (H,T) 與事件 {(H,T),(T,H)}。","en":"Random experiment: toss a coin twice. (1) Recording the results in order, write the sample space S, and prove that it contains every possible elementary outcome and has exactly 4 elements; (2) recording only the number of heads, write the sample space S′={0,1,2}; (3) under a separately given probability model P (each ordered outcome in S has probability 1/4), compute the probabilities 1/4, 1/2, 1/4 of the outcomes in S′, showing that outcomes in one sample space need not be equally likely; (4) use another valid, non-uniform model Q on S to show that the sample space by itself does not fix probabilities; (5) distinguish the outcome (H,T) from the event {(H,T),(T,H)}."},"mko_ids":["mko-sample-space"]},{"id":"derivation-mko-event-01","title":{"zh-Hant":"事件：在完整事件族與較粗事件族中判定事件","en":"Event: deciding what counts as an event under the full event family and under a coarser one"},"summary":{"zh-Hant":"擲一顆骰子一次，S={1,2,3,4,5,6}。(1) 在有限初等模型 F=𝒫(S) 中，說明 E={2,4,6} 是事件，判定觀察結果為 4 與 3 時 E 是否發生；分辨結果 4 與事件 {4}；說明空事件與必然事件。(2) 在一般模型 (S,F₀) 中，F₀={∅,O,E,S}，O={1,3,5}：證明 F₀ 是 S 上的事件族（σ-代數），並證明 {6}∉F₀，所以雖然 S 有限且 {6}⊆S，{6} 在這個模型中不是事件。","en":"Roll a die once, S={1,2,3,4,5,6}. (1) In the finite elementary model F=𝒫(S), show that E={2,4,6} is an event, and decide whether E occurs when the observed outcome is 4 and when it is 3; distinguish the outcome 4 from the event {4}; describe the empty and the sure event. (2) In the general model (S,F₀) with F₀={∅,O,E,S} and O={1,3,5}: prove that F₀ is an event family (σ-algebra) on S, and prove {6}∉F₀, so although S is finite and {6}⊆S, {6} is not an event in this model."},"mko_ids":["mko-event"]},{"id":"derivation-mko-function-mapping-01","title":{"zh-Hant":"函數映射：判定對應是否為函數，並分辨陪域、值域與圖形","en":"Function mapping: deciding whether a rule is a function, and separating codomain, range and graph"},"summary":{"zh-Hant":"設 X={−2,−1,0,1,2}、Y={0,1,2,3,4,5}，f:X→Y，f(x)=x²。證明 f 是函數；算出值域 f(X)={0,1,4} 並說明 f(X)≠Y；寫出圖形 Γ_f；說明把陪域改成 {0,1,4} 會得到不同的函數。再判定兩個非例：反向規則 s(y)=「X 中滿足 x²=y 的 x」不是 Y→X 的函數；g(x)=x+4 不是 X→Y 的函數。","en":"Let X={−2,−1,0,1,2}, Y={0,1,2,3,4,5} and f:X→Y, f(x)=x². Prove that f is a function; compute the range f(X)={0,1,4} and show f(X)≠Y; write down the graph Γ_f; and show that changing the codomain to {0,1,4} gives a different function. Then decide two non-examples: the reverse rule s(y)='the x in X with x²=y' is not a function Y→X, and g(x)=x+4 is not a function X→Y."},"mko_ids":["mko-function-mapping"]},{"id":"derivation-mko-tends-to-relation-01","title":{"zh-Hant":"趨近關係：x→a 不是 x=a，也不保證函數值收斂","en":"Tends-to relation: x→a is not x=a, and does not guarantee that function values converge"},"summary":{"zh-Hant":"在 (ℝ, d(x,y)=|x−y|) 中：(1) 證明數列 x_n=1+1/n 趨近 1，而每一項都有 x_n≠1；(2) 對 g(x)=1（x>1）、g(x)=0（x≤1），證明沿 x_n 有 g(x_n)→1，但 lim_{x→1} g(x) 不存在。由此說明 x→1 本身既不要求 x 等於 1，也不決定函數值是否收斂。","en":"In (ℝ, d(x,y)=|x−y|): (1) prove that the sequence x_n=1+1/n tends to 1 while every term satisfies x_n≠1; (2) for g(x)=1 (x>1) and g(x)=0 (x≤1), prove that g(x_n)→1 along x_n, but lim_{x→1} g(x) does not exist. This shows that x→1 by itself neither requires x to equal 1 nor decides whether function values converge."},"mko_ids":["mko-tends-to-relation"]}],"title":"概念推導導讀"}