{ "schema_version": "ocme-derivation-reader-v1", "guide_id": "derivation-mko-multivariable-real-function-01", "mko_id": "mko-multivariable-real-function", "title": { "zh-Hant": "多變數實函數：二變數二次多項式的係數次序、宣告定義域、切片與點值", "en": "Multivariable real-valued function: coefficient order, declared domain, slices and point values of a bivariate quadratic" }, "contract": { "schema_version": "ocme-derivation-guide-v0.1", "guide_id": "derivation-mko-multivariable-real-function-01", "source": { "mko_id": "mko-multivariable-real-function", "commit": "b28a5edbd46279b1db2a3f3f48bb3f447e18899f", "sha256": "f96ebe8babf93fa8de6b32cb4156ad2b710b938550218a6aff16be79ba4ef928", "english_sha256": "3bc9e3338d7c9026e1405d2719cc08c58786f3e4c4480e9803ca45cf10591239" }, "mode": "worked_example", "audience_prerequisites": [ "function_def", "cartesian_product_def", "order_basics", "distributive", "numeric_evaluation" ], "goal": { "zh-Hant": "係數向量依 [1,x,y,x²,xy,y²] 次序為 c=(1,−2,3,1,2,−1)，宣告定義域為閉矩形 R=[−1,2]×[0,3]。(1) 寫出 f 並確認 f:R→ℝ 是函數；(2) 分別核對 x、y 邊界，判定 (2,3)、(3,1)、(0,−1) 是否在 R 中；(3) 計算 f(2,1) 與 f(1,2)，並證明交換 x、y 係數得到的 g 是不同的函數；(4) 求 y=1 的 x-切片，並說明給定的 y=1 切片與指定四個點值各自都不足以決定 f。", "en": "The coefficient vector in the order [1,x,y,x²,xy,y²] is c=(1,−2,3,1,2,−1), and the declared domain is the closed rectangle R=[−1,2]×[0,3]. (1) Write f and confirm that f:R→ℝ is a function; (2) check the x and y bounds separately to decide whether (2,3), (3,1), (0,−1) lie in R; (3) compute f(2,1) and f(1,2), and prove that g, obtained by swapping the x and y coefficients, is a different function; (4) find the x-slice at y=1 and show that neither the given y=1 slice nor the four specified point values determine f." }, "givens": [ { "id": "G1", "statement": { "zh-Hant": "係數次序固定為 [1,x,y,x²,xy,y²]；f 的係數向量是 c=(1,−2,3,1,2,−1)（MKO 假設 a1：六個整數係數）。", "en": "The coefficient order is fixed as [1,x,y,x²,xy,y²]; the coefficient vector of f is c=(1,−2,3,1,2,−1) (MKO assumption a1: six integer coefficients)." } }, { "id": "G2", "statement": { "zh-Hant": "宣告定義域 R=[−1,2]×[0,3]，第一坐標為 x、第二坐標為 y；對應域 ℝ（MKO 假設 a2：只核明示閉矩形）。", "en": "Declared domain R=[−1,2]×[0,3], first coordinate x and second coordinate y; codomain ℝ (MKO assumption a2: only the explicitly declared closed rectangle is checked)." } }, { "id": "G3", "statement": { "zh-Hant": "比較函數（同一定義域 R）：g 的係數 (1,3,−2,−1,2,1)；h 的係數 (1,−2,2,1,2,0)；k 的係數 (1,−3,3,2,2,−1)。", "en": "Comparison functions (same domain R): g with coefficients (1,3,−2,−1,2,1); h with coefficients (1,−2,2,1,2,0); k with coefficients (1,−3,3,2,2,−1)." } } ], "steps": [ { "id": "S01", "uses": [ "mko_mvf_def", "G1" ], "before_tex": "c=(c_0,c_1,c_2,c_3,c_4,c_5)", "after_tex": "f(x,y)=c_0+c_1x+c_2y+c_3x^2+c_4xy+c_5y^2", "relation": "definition", "rule": "mko_mvf_def", "conditions": [], "explanation": { "zh-Hant": "MKO 固定係數次序 [1,x,y,x²,xy,y²]：第 0 個係數乘 1，第 1 個乘 x，第 2 個乘 y，第 3 個乘 x²，第 4 個乘 xy，第 5 個乘 y²。只有次序固定，係數向量才能精確重算出函數（MKO 段落 0）。", "en": "The MKO fixes the coefficient order [1,x,y,x²,xy,y²]: coefficient 0 multiplies 1, coefficient 1 multiplies x, coefficient 2 multiplies y, coefficient 3 multiplies x², coefficient 4 multiplies xy, coefficient 5 multiplies y². Only with a fixed order can the coefficient vector reproduce the function exactly (MKO paragraph 0)." }, "reverse_audit": { "kind": "premise_audit", "required_ids": [ "mko_mvf_def" ], "explanation": { "zh-Hant": "這一步只規定讀法；同一組數字換一種次序讀，會得到不同的函數（見 S09、S10）。", "en": "This step only fixes how the vector is read; reading the same numbers in another order gives a different function (see S09, S10)." } }, "leaf_basis": "mko_mvf_def + def_unfold" }, { "id": "S02", "uses": [ "S01", "G1" ], "before_tex": "c=(1,-2,3,1,2,-1)", "after_tex": "f(x,y)=1-2x+3y+x^2+2xy-y^2", "relation": "equivalent", "rule": "eq_replace", "conditions": [], "explanation": { "zh-Hant": "把 c_0=1、c_1=−2、c_2=3、c_3=1、c_4=2、c_5=−1 依次代入 S01 的通式。係數 1 的項寫成 x²，係數 −1 的項寫成 −y²。", "en": "Substitute c_0=1, c_1=−2, c_2=3, c_3=1, c_4=2, c_5=−1 in order into the general form of S01. The term with coefficient 1 is written x², the one with coefficient −1 is written −y²." }, "reverse_audit": { "kind": "inverse_equivalence", "required_ids": [ "eq_replace", "S01" ], "explanation": { "zh-Hant": "反向：依次讀出 1、x、y、x²、xy、y² 的係數，回到 (1,−2,3,1,2,−1)。", "en": "Backward: reading off the coefficients of 1, x, y, x², xy, y² in order recovers (1,−2,3,1,2,−1)." } }, "leaf_basis": "eq_replace + mul_one" }, { "id": "S03", "uses": [ "S02", "G2", "function_def" ], "before_tex": "f(x,y)=1-2x+3y+x^2+2xy-y^2,\\ (x,y)\\in R", "after_tex": "f:R\\to\\mathbb{R}\\ \\text{is a function}", "relation": "implies", "rule": "function_def", "conditions": [], "explanation": { "zh-Hant": "函數資格需要三件事：宣告的定義域 R、對應域 ℝ、以及每個 (x,y)∈R 恰好一個值。對實數 x、y，式子只用加、減、乘，結果是唯一確定的實數，所以每點恰有一個值。坐標角色由 G2 固定：第一坐標代入 x，第二坐標代入 y。", "en": "Specifying a function requires three things: the declared domain R, the codomain ℝ, and exactly one value at each (x,y)∈R. For real x, y the expression uses only addition, subtraction and multiplication, so its value is a uniquely determined real number, and each point has exactly one value. Coordinate roles are fixed by G2: the first coordinate is substituted for x, the second for y." }, "reverse_audit": { "kind": "premise_audit", "required_ids": [ "function_def", "G2" ], "explanation": { "zh-Hant": "多項式式子在整個 ℝ² 上都能求值，但這裡的函數定義域是宣告的 R（MKO 假設 a2）；本步不宣稱連續或可微。", "en": "The polynomial expression can be evaluated on all of ℝ², but the domain of this function is the declared R (MKO assumption a2); this step asserts neither continuity nor differentiability." } }, "leaf_basis": "function_def + numeric_evaluation" }, { "id": "S04", "uses": [ "G2", "p_interval_def" ], "before_tex": "(2,3)\\in R\\ ?", "after_tex": "(2,3)\\in R", "relation": "equivalent", "rule": "p_interval_def", "conditions": [], "explanation": { "zh-Hant": "依閉矩形的定義，(2,3)∈R ⇔ 2∈[−1,2] 且 3∈[0,3]。兩個條件分別在 S04.1、S04.2 核對，都成立；注意兩個坐標都正好落在邊界上，閉區間包含端點。", "en": "By the definition of a closed rectangle, (2,3)∈R ⇔ 2∈[−1,2] and 3∈[0,3]. The two conditions are checked separately in S04.1 and S04.2 and both hold; note that both coordinates sit exactly on the boundary, and closed intervals include their endpoints." }, "reverse_audit": { "kind": "inverse_equivalence", "required_ids": [ "p_interval_def", "and_intro_elim" ], "explanation": { "zh-Hant": "矩形成員資格是兩個區間條件的合取，雙向成立；兩個條件都要核（MKO 段落 1）。", "en": "Membership in the rectangle is the conjunction of two interval conditions, two-way; both must be checked (MKO paragraph 1)." } }, "leaf_basis": "p_interval_def + and_intro_elim" }, { "id": "S04.1", "uses": [ "S04" ], "before_tex": "2\\in[-1,2]", "after_tex": "-1\\le 2\\le 2", "relation": "equivalent", "rule": "p_interval_def", "conditions": [], "explanation": { "zh-Hant": "x 邊界：−1≤2 成立，2≤2 成立（等號）。所以 x=2 在 [−1,2] 中。", "en": "x bound: −1≤2 holds and 2≤2 holds (with equality). So x=2 lies in [−1,2]." }, "reverse_audit": { "kind": "inverse_equivalence", "required_ids": [ "p_interval_def", "order_basics" ], "explanation": { "zh-Hant": "區間定義雙向成立；若誤用開區間 (−1,2)，x=2 會被錯誤排除。", "en": "The interval definition is two-way; using the open interval (−1,2) by mistake would wrongly exclude x=2." } }, "leaf_basis": "p_interval_def + order_basics" }, { "id": "S04.2", "uses": [ "S04" ], "before_tex": "3\\in[0,3]", "after_tex": "0\\le 3\\le 3", "relation": "equivalent", "rule": "p_interval_def", "conditions": [], "explanation": { "zh-Hant": "y 邊界：0≤3 成立，3≤3 成立。所以 y=3 在 [0,3] 中，(2,3)∈R，f(2,3)=1−4+9+4+12−9=13（N11）是 f 的一個值。", "en": "y bound: 0≤3 holds and 3≤3 holds. So y=3 lies in [0,3], (2,3)∈R, and f(2,3)=1−4+9+4+12−9=13 (N11) is a value of f." }, "reverse_audit": { "kind": "inverse_equivalence", "required_ids": [ "p_interval_def", "order_basics" ], "explanation": { "zh-Hant": "與 S04.1 相同的區間定義，用在第二坐標與 y 的區間上。", "en": "The same interval definition as S04.1, applied to the second coordinate and the y-interval." } }, "leaf_basis": "p_interval_def + order_basics" }, { "id": "S05", "uses": [ "G2", "p_interval_def", "S03" ], "before_tex": "(3,1),\\ (0,-1)", "after_tex": "(3,1)\\notin R\\ (3>2),\\quad (0,-1)\\notin R\\ (-1<0)", "relation": "implies", "rule": "p_interval_def", "conditions": [], "explanation": { "zh-Hant": "(3,1)：y=1∈[0,3] 通過，但 x=3>2，x 邊界失敗。(0,−1)：x=0∈[−1,2] 通過，但 y=−1<0，y 邊界失敗。兩點都不在 R 中；每點都只有一個坐標失敗，所以只核一個坐標的檢查必定會漏掉其中之一。多項式式子在 (3,1) 仍可算出 12，但那不是宣告函數 f:R→ℝ 的值。", "en": "(3,1): y=1∈[0,3] passes, but x=3>2 fails the x bound. (0,−1): x=0∈[−1,2] passes, but y=−1<0 fails the y bound. Neither point is in R; each fails in only one coordinate, so a check of only one coordinate is bound to miss one of them. The polynomial expression still evaluates to 12 at (3,1), but that is not a value of the declared function f:R→ℝ." }, "reverse_audit": { "kind": "premise_audit", "required_ids": [ "p_interval_def", "order_basics" ], "explanation": { "zh-Hant": "否定合取只需一個條件失敗；這裡分別示範 x 失敗與 y 失敗。", "en": "A conjunction is refuted by one failing condition; here an x failure and a y failure are shown separately." } }, "leaf_basis": "p_interval_def + de_morgan_logic + order_basics" }, { "id": "S06", "uses": [ "S02", "S03" ], "before_tex": "f(2,1)", "after_tex": "1-4+3+4+4-1=7", "relation": "equivalent", "rule": "numeric_evaluation", "conditions": [ "(2,1)\\in R" ], "explanation": { "zh-Hant": "(2,1)∈R（−1≤2≤2、0≤1≤3）。代入 x=2、y=1，逐項計算見 S06.1，和為 7（N1）。", "en": "(2,1)∈R (−1≤2≤2, 0≤1≤3). Substituting x=2, y=1 and computing term by term (S06.1) gives a total of 7 (N1)." }, "reverse_audit": { "kind": "inverse_equivalence", "required_ids": [ "numeric_evaluation" ], "explanation": { "zh-Hant": "精確整數求值，兩邊是同一個數；可由 N1 重算。", "en": "Exact integer evaluation, both sides the same number; recomputed by N1." } }, "leaf_basis": "numeric_evaluation + eq_replace" }, { "id": "S06.1", "uses": [ "S02" ], "before_tex": "1,\\ -2x,\\ 3y,\\ x^2,\\ 2xy,\\ -y^2\\ \\text{at}\\ (2,1)", "after_tex": "1,\\ -4,\\ 3,\\ 4,\\ 4,\\ -1", "relation": "equivalent", "rule": "numeric_evaluation", "conditions": [], "explanation": { "zh-Hant": "依係數次序逐項：常數 1；−2x=−4；3y=3；x²=4；2xy=2·2·1=4；−y²=−1。逐項列出可以避免把 x、y 代錯位置。", "en": "Term by term in coefficient order: constant 1; −2x=−4; 3y=3; x²=4; 2xy=2·2·1=4; −y²=−1. Listing the terms one by one prevents substituting x and y into the wrong places." }, "reverse_audit": { "kind": "inverse_equivalence", "required_ids": [ "numeric_evaluation", "power_laws" ], "explanation": { "zh-Hant": "每項都是確定的整數乘積，雙向成立。", "en": "Each term is a definite integer product, so it holds both ways." } }, "leaf_basis": "numeric_evaluation + power_laws" }, { "id": "S07", "uses": [ "S02", "S03" ], "before_tex": "f(1,2)", "after_tex": "1-2+6+1+4-4=6", "relation": "equivalent", "rule": "numeric_evaluation", "conditions": [ "(1,2)\\in R" ], "explanation": { "zh-Hant": "(1,2)∈R（−1≤1≤2、0≤2≤3）。代入 x=1、y=2：1、−2、6、1、2·1·2=4、−4，和為 6（N2）。", "en": "(1,2)∈R (−1≤1≤2, 0≤2≤3). Substituting x=1, y=2: 1, −2, 6, 1, 2·1·2=4, −4, which total 6 (N2)." }, "reverse_audit": { "kind": "inverse_equivalence", "required_ids": [ "numeric_evaluation" ], "explanation": { "zh-Hant": "精確整數求值；可由 N2 重算。", "en": "Exact integer evaluation; recomputed by N2." } }, "leaf_basis": "numeric_evaluation + eq_replace" }, { "id": "S08", "uses": [ "S06", "S07" ], "before_tex": "f(2,1)=7,\\ f(1,2)=6", "after_tex": "f(2,1)\\neq f(1,2)", "relation": "implies", "rule": "numeric_evaluation", "conditions": [], "explanation": { "zh-Hant": "同樣兩個數 1 與 2，放在不同坐標位置得到不同值。所以坐標角色（哪個是 x、哪個是 y）是函數資料的一部分，不能隨意交換。", "en": "The same two numbers 1 and 2 in different coordinate positions give different values. So the coordinate roles (which is x and which is y) are part of the data of the function and cannot be swapped at will." }, "reverse_audit": { "kind": "premise_audit", "required_ids": [ "S06", "S07" ], "explanation": { "zh-Hant": "只宣稱這一對點的值不同；不宣稱 f 在每一對交換點上都不同（例如 f(0,0)=f(0,0)）。", "en": "Only claims that the values at this pair of points differ; it does not claim that f differs at every swapped pair (for example f(0,0)=f(0,0))." } }, "leaf_basis": "numeric_evaluation" }, { "id": "S09", "uses": [ "S01", "G3" ], "before_tex": "g(x,y)=1+3x-2y-x^2+2xy+y^2", "after_tex": "g(x,y)=f(y,x)\\quad\\text{when }(x,y)\\in R\\ \\land\\ (y,x)\\in R", "relation": "equivalent", "rule": "eq_congruence", "conditions": [ "(x,y)\\in R", "(y,x)\\in R" ], "explanation": { "zh-Hant": "g 是交換係數 c_1↔c_2、c_3↔c_5（c_4 不變）所得、仍以 R 為定義域的函數。交換兩個多項式式子的 x、y 並重排會得到相同式子；但僅當 (x,y)∈R 且 (y,x)∈R 時，才可寫 g(x,y)=f(y,x)。本題 (2,1) 與 (1,2) 均符合；N3 核對多項式式子的恆等，不把域外求值當作 f 的值。", "en": "The function g, still with domain R, is obtained by swapping c_1↔c_2 and c_3↔c_5 while keeping c_4 fixed. Swapping the variables in the polynomial expressions and rearranging gives the same expression. However, g(x,y)=f(y,x) is valid only when both (x,y) and (y,x) lie in R. The points (2,1) and (1,2) used here satisfy this condition. N3 checks the polynomial-expression identity, not values of f outside its domain." }, "reverse_audit": { "kind": "inverse_equivalence", "required_ids": [ "add_comm", "mul_comm" ], "explanation": { "zh-Hant": "只用到加法與乘法交換律重排各項，雙向成立。", "en": "Only commutativity of addition and multiplication is used to rearrange terms, so it holds both ways." } }, "leaf_basis": "add_comm + mul_comm + eq_congruence" }, { "id": "S10", "uses": [ "S09", "S06", "S07", "function_def" ], "before_tex": "g(2,1)=f(1,2)=6,\\ f(2,1)=7", "after_tex": "g\\neq f\\ \\text{as functions on}\\ R", "relation": "implies", "rule": "function_def", "conditions": [ "(2,1)\\in R" ], "explanation": { "zh-Hant": "由 S09，g(2,1)=f(1,2)=6（N4 直接計算也得 6），而 f(2,1)=7。兩個函數相等需要在定義域每一點都同值；(2,1)∈R 上值不同，所以 g≠f。這正是 MKO 指出的誤解：交換 x 與 y 的係數會改變函數。", "en": "By S09, g(2,1)=f(1,2)=6 (N4 computes 6 directly as well), while f(2,1)=7. Equal functions must agree at every point of the domain; they differ at (2,1)∈R, so g≠f. This is the misconception named by the MKO: swapping the x and y coefficients changes the function." }, "reverse_audit": { "kind": "premise_audit", "required_ids": [ "function_def", "S09" ], "explanation": { "zh-Hant": "否定函數相等只需一個點；該點必須在宣告定義域內，(2,1) 在 R 中。", "en": "Refuting equality of functions needs one point, which must lie in the declared domain; (2,1) is in R." } }, "leaf_basis": "function_def + numeric_evaluation" }, { "id": "S11", "uses": [ "S02", "G2" ], "before_tex": "\\varphi(x)=f(x,1),\\ x\\in[-1,2]", "after_tex": "\\varphi(x)=x^2+3", "relation": "equivalent", "rule": "eq_congruence", "conditions": [ "y=1\\in[0,3]" ], "explanation": { "zh-Hant": "固定 y=1（在 [0,3] 中），x 在 [−1,2] 中變動：f(x,1)=1−2x+3+x²+2x−1=x²+3，−2x 與 2x 相消（N5 為恆等式）。這是一個單變數函數，是 f 在一條水平線上的切片。", "en": "Fix y=1 (which lies in [0,3]) and let x vary in [−1,2]: f(x,1)=1−2x+3+x²+2x−1=x²+3, with −2x and 2x cancelling (N5 is an identity). This is a single-variable function, the slice of f along one horizontal line." }, "reverse_audit": { "kind": "inverse_equivalence", "required_ids": [ "eq_congruence", "add_inverse" ], "explanation": { "zh-Hant": "化簡只用加法反元素與合併同類項，雙向成立。", "en": "The simplification uses only additive inverses and collecting like terms, so it holds both ways." } }, "leaf_basis": "eq_congruence + add_inverse + add_comm" }, { "id": "S12", "uses": [ "S11", "G3" ], "before_tex": "h(x,y)=1-2x+2y+x^2+2xy", "after_tex": "h=f+(y^2-y),\\ \\ h(x,1)=f(x,1)=x^2+3", "relation": "implies", "rule": "eq_congruence", "conditions": [], "explanation": { "zh-Hant": "h 與 f 的差是 y²−y（N6 為恆等式：3y−y=2y，−y²+y²=0）。y=1 時 y²−y=0，所以 h 的 y=1 切片與 f 的完全相同，都是 x²+3（N7 為恆等式）。", "en": "h differs from f by y²−y (N6 is an identity: 3y−y=2y and −y²+y²=0). At y=1, y²−y=0, so the y=1 slice of h is exactly that of f, namely x²+3 (N7 is an identity)." }, "reverse_audit": { "kind": "premise_audit", "required_ids": [ "eq_congruence", "distributive" ], "explanation": { "zh-Hant": "只宣稱在 y=1 這條線上一致；其他 y 值上的一致性不宣稱，S13 正好顯示它失敗。", "en": "Agreement is claimed only on the line y=1; agreement at other y is not claimed, and S13 shows it fails." } }, "leaf_basis": "eq_congruence + distributive + add_inverse" }, { "id": "S13", "uses": [ "S12", "function_def" ], "before_tex": "h(0,2),\\ f(0,2)", "after_tex": "h(0,2)=5\\neq 3=f(0,2)", "relation": "implies", "rule": "function_def", "conditions": [ "(0,2)\\in R" ], "explanation": { "zh-Hant": "(0,2)∈R。h(0,2)=1+4=5，f(0,2)=1+6−4=3（N8、N9），差 2 正是 y²−y 在 y=2 的值。所以 h≠f，雖然它們的 y=1 切片相同：單一切片不能決定完整函數（MKO 段落 2）。", "en": "(0,2)∈R. h(0,2)=1+4=5 and f(0,2)=1+6−4=3 (N8, N9); the difference 2 is exactly y²−y at y=2. So h≠f although their y=1 slices are identical: one slice does not determine the whole function (MKO paragraph 2)." }, "reverse_audit": { "kind": "premise_audit", "required_ids": [ "function_def", "numeric_evaluation" ], "explanation": { "zh-Hant": "一個定義域內的點就足以否定 h=f。", "en": "One point of the domain suffices to refute h=f." } }, "leaf_basis": "function_def + numeric_evaluation" }, { "id": "S14", "uses": [ "S02", "G3", "function_def" ], "before_tex": "k=f+(x^2-x)", "after_tex": "k=f\\ \\text{at}\\ (0,0),(1,0),(0,1),(1,1),\\ \\ k(2,0)=3\\neq 1=f(2,0)", "relation": "implies", "rule": "function_def", "conditions": [ "(2,0)\\in R" ], "explanation": { "zh-Hant": "k 與 f 的差是 x²−x（N10 為恆等式）。x∈{0,1} 時 x²−x=0（N12 只是這兩點的有限核對），所以 k 與 f 在四個格點上值相同（f 的值依次為 1、0、3、4）。但在 (2,0)∈R，k(2,0)=3 而 f(2,0)=1（N13、N14）。所以在本題指定四點的值相同，不足以推出兩個函數相等。", "en": "k differs from f by x²−x (N10 is an identity). For x∈{0,1}, x²−x=0 (N12 is only a finite check at these two points), so k and f agree at the four grid points (f takes the values 1, 0, 3, 4 there). But at (2,0)∈R, k(2,0)=3 while f(2,0)=1 (N13, N14). Thus agreement at these four specified points does not imply that the two functions are equal." }, "reverse_audit": { "kind": "premise_audit", "required_ids": [ "function_def", "numeric_evaluation" ], "explanation": { "zh-Hant": "這裡只否定「這四個點足以決定 f」；不討論是否有其他特殊點集能在二次多項式類中決定函數，本稿不宣稱任何這類結果。", "en": "This only refutes 'these four points determine f'; whether some other special point set determines a function within the class of quadratics is not discussed, and no such result is claimed." } }, "leaf_basis": "function_def + numeric_evaluation + distributive" }, { "id": "S15", "uses": [ "S03", "S05", "S10", "S13", "S14" ], "before_tex": "\\text{S03 to S14}", "after_tex": "f=(c,R,\\text{coordinate roles});\\ \\text{the }y=1\\text{ slice and the four listed point values each fail to determine }f", "relation": "implies", "rule": "and_intro_elim", "conditions": [], "explanation": { "zh-Hant": "總結：f 由固定次序的係數向量、宣告定義域 R 與坐標角色共同決定（S03）；成員資格要分開核 x、y（S04、S05）；交換係數得到不同函數 g（S10）；給定的 y=1 切片（S13）與指定四個點值（S14）各自不足以決定 f。", "en": "Summary: f is determined jointly by the coefficient vector in its fixed order, the declared domain R and the coordinate roles (S03); membership needs separate x and y checks (S04, S05); swapping coefficients gives a different function g (S10); neither the given y=1 slice (S13) nor the four specified point values (S14) determine f." }, "reverse_audit": { "kind": "premise_audit", "required_ids": [ "S03", "S10", "S13", "S14" ], "explanation": { "zh-Hant": "只合取已證結論，沒有新推論。", "en": "Only conjoins conclusions already proved; there is no new inference." } }, "leaf_basis": "and_intro_elim" } ], "result": { "zh-Hant": "f(x,y)=1−2x+3y+x²+2xy−y² 在宣告定義域 R=[−1,2]×[0,3] 上是函數；(2,3)∈R（f=13），(3,1) 因 x>2、(0,−1) 因 y<0 不在 R 中。f(2,1)=7、f(1,2)=6；交換係數的 g 在合法的交換點上有 g(2,1)=f(1,2)=6，所以 g≠f。y=1 切片為 x²+3，但 h 有相同切片而 h(0,2)=5≠3=f(0,2)；k 與 f 在四個格點上相同而 k(2,0)=3≠1=f(2,0)。", "en": "f(x,y)=1−2x+3y+x²+2xy−y² is a function on the declared domain R=[−1,2]×[0,3]; (2,3)∈R (f=13), while (3,1) (x>2) and (0,−1) (y<0) are not in R. f(2,1)=7 and f(1,2)=6; the coefficient-swapped g has g(2,1)=f(1,2)=6 at these valid swapped points, so g≠f. The y=1 slice is x²+3, but h has the same slice with h(0,2)=5≠3=f(0,2); k agrees with f at four grid points yet k(2,0)=3≠1=f(2,0)." }, "negative_controls": [ { "claim": { "zh-Hant": "f(2,1)=f(1,2)，因為用的是同樣兩個數", "en": "f(2,1)=f(1,2), because the same two numbers are used" }, "expected": "reject", "reason": { "zh-Hant": "f(2,1)=7 而 f(1,2)=6（S06、S07）。", "en": "f(2,1)=7 while f(1,2)=6 (S06, S07)." } }, { "claim": { "zh-Hant": "k 與 f 在 (0,0)、(1,0)、(0,1)、(1,1) 的值相同，所以 k=f", "en": "k and f agree at (0,0), (1,0), (0,1), (1,1), so k=f" }, "expected": "reject", "reason": { "zh-Hant": "k(2,0)=3≠1=f(2,0)，而 (2,0)∈R（S14）。", "en": "k(2,0)=3≠1=f(2,0), and (2,0)∈R (S14)." } }, { "claim": { "zh-Hant": "f(3,1)=12 是宣告函數 f:R→ℝ 的一個值", "en": "f(3,1)=12 is a value of the declared function f:R→ℝ" }, "expected": "reject", "reason": { "zh-Hant": "(3,1)∉R，因為 x=3>2（S05）；式子可求值不等於該點在定義域中。", "en": "(3,1)∉R because x=3>2 (S05); being able to evaluate the expression does not put the point in the domain." } }, { "claim": { "zh-Hant": "y=1 的切片 x²+3 決定了 f", "en": "The y=1 slice x²+3 determines f" }, "expected": "reject", "reason": { "zh-Hant": "h 有相同的 y=1 切片，但 h(0,2)=5≠3=f(0,2)（S12、S13）。", "en": "h has the same y=1 slice, but h(0,2)=5≠3=f(0,2) (S12, S13)." } } ], "status": "candidate", "gaps": [], "verification": { "structure": "checked_local_script", "mathematics": "candidate_argument", "formalization": "not_formalized" }, "rule_glossary": [ { "id": "function_def", "kind": "prerequisite", "statement": { "zh-Hant": "函數：f:A→B 對每個 a∈A 指定恰好一個 f(a)∈B。兩函數相等 ⇔ 定義域、對應域相同且對每個 a 都有相同值。", "en": "Function: f:A→B assigns to each a∈A exactly one f(a)∈B. Two functions are equal ⇔ they have the same domain and codomain and agree at every a." } }, { "id": "cartesian_product_def", "kind": "prerequisite", "statement": { "zh-Hant": "笛卡兒積：A×B 是所有有序對 (a,b)（a∈A、b∈B）的集合；(a,b)=(c,d) ⇔ a=c 且 b=d。", "en": "Cartesian product: A×B is the set of ordered pairs (a,b) with a∈A and b∈B; (a,b)=(c,d) ⇔ a=c and b=d." } }, { "id": "order_basics", "kind": "prerequisite", "statement": { "zh-Hant": "實數的序：任兩數恰有 uv 之一；u0 推出 uwvw。", "en": "Order of the reals: exactly one of uv holds; u0 give uwvw." } }, { "id": "distributive", "kind": "prerequisite", "statement": { "zh-Hant": "分配律：u(v+w)=uv+uw；配合乘法交換律也有 (v+w)u=vu+wu。", "en": "Distributivity: u(v+w)=uv+uw; with commutativity also (v+w)u=vu+wu." } }, { "id": "numeric_evaluation", "kind": "prerequisite", "statement": { "zh-Hant": "明確整數或有理數算式的精確求值；可以獨立重算。", "en": "Exact evaluation of an explicit integer or rational expression; independently recomputable." } }, { "id": "def_unfold", "kind": "prerequisite", "statement": { "zh-Hant": "展開定義：把一個已定義的詞換成它的定義內容（或反向收回），命題的真假不變。", "en": "Unfolding a definition: replacing a defined term by its definition (or folding it back) does not change whether a statement is true." } }, { "id": "eq_replace", "kind": "prerequisite", "statement": { "zh-Hant": "等值代換：命題中的一個式子可換成與它相等的式子，命題的真假不變。", "en": "Replacement of equals: an expression inside a statement may be replaced by an equal expression without changing whether the statement is true." } }, { "id": "mul_one", "kind": "prerequisite", "statement": { "zh-Hant": "乘法單位元：1·u=u。", "en": "Multiplicative identity: 1·u=u." } }, { "id": "and_intro_elim", "kind": "prerequisite", "statement": { "zh-Hant": "合取：由 P 與 Q 得 P∧Q；由 P∧Q 可分別得 P 與 Q。", "en": "Conjunction: from P and Q infer P∧Q; from P∧Q infer P and infer Q." } }, { "id": "de_morgan_logic", "kind": "prerequisite", "statement": { "zh-Hant": "邏輯的德摩根律：¬(P∧Q) ⇔ ¬P∨¬Q；¬(P∨Q) ⇔ ¬P∧¬Q。", "en": "De Morgan's laws for logic: ¬(P∧Q) ⇔ ¬P∨¬Q and ¬(P∨Q) ⇔ ¬P∧¬Q." } }, { "id": "power_laws", "kind": "prerequisite", "statement": { "zh-Hant": "整數指數律（底數非零或指數為正整數）：a^m·a^n=a^{m+n}，(a^m)^n=a^{mn}，(ab)^n=a^n b^n，a^0=1。", "en": "Integer exponent laws (base nonzero, or exponents positive integers): a^m·a^n=a^{m+n}, (a^m)^n=a^{mn}, (ab)^n=a^n b^n, a^0=1." } }, { "id": "eq_congruence", "kind": "prerequisite", "statement": { "zh-Hant": "等式同用：若 u=v，對兩邊做同一個有定義的運算 f，得 f(u)=f(v)。只對一邊做運算不屬於此規則。", "en": "Congruence of equality: if u=v, applying the same defined operation f to both sides gives f(u)=f(v). Operating on one side only is not an instance of this rule." } }, { "id": "add_comm", "kind": "prerequisite", "statement": { "zh-Hant": "加法交換律：u+v=v+u。", "en": "Commutativity of addition: u+v=v+u." } }, { "id": "mul_comm", "kind": "prerequisite", "statement": { "zh-Hant": "乘法交換律：uv=vu（在交換的數系中，例如 ℤ、ℚ、ℝ、ℂ）。", "en": "Commutativity of multiplication: uv=vu (in commutative number systems such as ℤ, ℚ, ℝ, ℂ)." } }, { "id": "add_inverse", "kind": "prerequisite", "statement": { "zh-Hant": "加法反元素：每個 u 都有 −u，使 u+(−u)=0；u−v 定義為 u+(−v)。", "en": "Additive inverse: every u has −u with u+(−u)=0; u−v means u+(−v)." } }, { "id": "mko_mvf_def", "kind": "cited", "source": { "mko_id": "mko-multivariable-real-function", "sha256": "f96ebe8babf93fa8de6b32cb4156ad2b710b938550218a6aff16be79ba4ef928", "commit": "b28a5edbd46279b1db2a3f3f48bb3f447e18899f" }, "statement": { "zh-Hant": "本條目先限二變數實多項式 f:ℝ²→ℝ 的明示切片，係數依 [1,x,y,x²,xy,y²] 固定次序。函數資格需明示定義域與座標角色；本輪不自動宣稱連續、可微或向量值。", "en": "This entry restricts attention to explicit slices of bivariate real polynomials f:ℝ²→ℝ, with coefficients ordered as [1,x,y,x²,x*y,y²]. Function qualification requires a declared domain and coordinate roles; continuity, differentiability and vector-valued codomains are not automatically asserted." } }, { "id": "p_interval_def", "kind": "prerequisite", "statement": { "zh-Hant": "閉區間與閉矩形：u∈[α,β] ⇔ α≤u≤β；(x,y)∈[α,β]×[γ,δ] ⇔ x∈[α,β] 且 y∈[γ,δ]。", "en": "Closed intervals and rectangles: u∈[α,β] ⇔ α≤u≤β; (x,y)∈[α,β]×[γ,δ] ⇔ x∈[α,β] and y∈[γ,δ]." } } ], "revision_log": [], "numeric_checks": [ { "id": "N1", "step": "S06", "lhs": "1-2*2+3*1+2**2+2*2*1-1**2", "rhs": "7" }, { "id": "N2", "step": "S07", "lhs": "1-2*1+3*2+1**2+2*1*2-2**2", "rhs": "6" }, { "id": "N3", "step": "S09", "vars": [ "x", "y" ], "degree": 2, "lhs": "1+3*x-2*y-x**2+2*x*y+y**2", "rhs": "1-2*y+3*x+y**2+2*y*x-x**2" }, { "id": "N4", "step": "S10", "lhs": "1+3*2-2*1-2**2+2*2*1+1**2", "rhs": "6" }, { "id": "N5", "step": "S11", "vars": [ "x" ], "degree": 2, "lhs": "1-2*x+3*1+x**2+2*x*1-1**2", "rhs": "x**2+3" }, { "id": "N6", "step": "S12", "vars": [ "x", "y" ], "degree": 2, "lhs": "1-2*x+2*y+x**2+2*x*y", "rhs": "(1-2*x+3*y+x**2+2*x*y-y**2)+(y**2-y)" }, { "id": "N7", "step": "S12", "vars": [ "x" ], "degree": 2, "lhs": "1-2*x+2*1+x**2+2*x*1", "rhs": "x**2+3" }, { "id": "N8", "step": "S13", "lhs": "1-2*0+2*2+0**2+2*0*2", "rhs": "5" }, { "id": "N9", "step": "S13", "lhs": "1-2*0+3*2+0**2+2*0*2-2**2", "rhs": "3" }, { "id": "N10", "step": "S14", "vars": [ "x", "y" ], "degree": 2, "lhs": "1-3*x+3*y+2*x**2+2*x*y-y**2", "rhs": "(1-2*x+3*y+x**2+2*x*y-y**2)+(x**2-x)" }, { "id": "N11", "step": "S04.2", "lhs": "1-2*2+3*3+2**2+2*2*3-3**2", "rhs": "13" }, { "id": "N12", "step": "S14", "vars": [ "x" ], "range": [ 0, 1 ], "lhs": "x**2-x", "rhs": "0" }, { "id": "N13", "step": "S14", "lhs": "1-3*2+3*0+2*2**2+2*2*0-0**2", "rhs": "3" }, { "id": "N14", "step": "S14", "lhs": "1-2*2+3*0+2**2+2*2*0-0**2", "rhs": "1" }, { "id": "N15", "step": "S05", "lhs": "1-2*3+3*1+3**2+2*3*1-1**2", "rhs": "12" } ] }, "presentation": { "md_scope": { "zh-Hant": "只處理 G1、G2 指定的二變數整數係數二次多項式切片與閉矩形定義域（MKO 假設 a1、a2）。不宣稱連續、可微或向量值；不證明多項式恆等定理或任何點集對二次多項式的唯一決定性。不改動 canonical MKO 或其英文 edition。", "en": "Treats only the bivariate integer-coefficient quadratic slice and the closed rectangular domain given by G1 and G2 (MKO assumptions a1, a2). Continuity, differentiability and vector values are not asserted; neither the polynomial identity theorem nor the unisolvence of any point set for quadratics is proved. The canonical MKO and its English edition are not changed." }, "md_symbols": [ [ "c=(c_0,c_1,c_2,c_3,c_4,c_5)", "依 [1,x,y,x²,xy,y²] 次序的係數", "coefficients in the order [1,x,y,x²,xy,y²]" ], [ "f(x,y)", "由 c=(1,−2,3,1,2,−1) 給出的函數", "the function given by c=(1,−2,3,1,2,−1)" ], [ "R=[-1,2]\\times[0,3]", "宣告的閉矩形定義域", "the declared closed rectangular domain" ], [ "g", "交換 x、y 係數後的函數", "the function with x and y coefficients swapped" ], [ "\\varphi(x)=f(x,1)", "y=1 的 x-切片", "the x-slice at y=1" ], [ "h,\\ k", "與 f 在切片或有限點上一致的比較函數", "comparison functions agreeing with f on a slice or at finitely many points" ] ], "md_error_controls": [ { "title": "交換 x 與 y 係數 / Swapping the x and y coefficients", "zh": "MKO 指出的第一個誤解是交換 x 與 y 係數。係數向量必須依 [1,x,y,x²,xy,y²] 讀；若誤讀成 [1,y,x,y²,xy,x²]，同一組數字給出的是 g；當 (x,y)、(y,x) 均在 R 時有 g(x,y)=f(y,x)，而 g(2,1)=6≠7=f(2,1)。代入時也要依坐標角色：第一坐標代 x，第二坐標代 y（S06.1 逐項列出就是為此）。", "en": "The first misconception named by the MKO is swapping the x and y coefficients. The coefficient vector must be read in the order [1,x,y,x²,xy,y²]; read as [1,y,x,y²,xy,x²], the same numbers give g; the identity g(x,y)=f(y,x) holds when both (x,y) and (y,x) lie in R, and g(2,1)=6≠7=f(2,1). Substitution must also respect the coordinate roles: the first coordinate goes to x, the second to y (which is why S06.1 lists the terms one by one)." }, { "title": "幾個點值不能宣稱全域等價 / A few point values cannot establish global equivalence", "zh": "MKO 的第二個誤解是以幾個點值宣稱全域函數等價、連續或可微。S14 的 k 與 f 在 (0,0)、(1,0)、(0,1)、(1,1) 完全一致，卻在 (2,0) 不同；S13 的 h 與 f 的整條 y=1 切片一致，卻在 (0,2) 不同。本題指定四點的核對不足以保證全域相等；若另外知道多項式的次數上界，足夠且適當的有限點可判定恆等，本稿不使用該判準。", "en": "The second misconception named by the MKO is using a few point values to claim global equivalence, continuity or differentiability. The function k of S14 agrees with f at (0,0), (1,0), (0,1), (1,1) but differs at (2,0); the function h of S13 agrees with f on the whole slice y=1 but differs at (0,2). Checking these four points is insufficient for global equality. With a known polynomial-degree bound, a suitable sufficiently large finite point set can decide an identity; that criterion is not used here." }, { "title": "矩形定義域要分開核兩個坐標 / Check both coordinates of a rectangular domain", "zh": "(0,−1) 的 x=0 在 [−1,2] 中，若只核 x 就會誤判它在 R 中；(3,1) 的 y=1 在 [0,3] 中，若只核 y 也會誤判。另外 R 是閉矩形，邊界點如 (2,3) 屬於 R。", "en": "For (0,−1), x=0 lies in [−1,2], so checking only x would wrongly put it in R; for (3,1), y=1 lies in [0,3], so checking only y would also err. Also R is a closed rectangle, so boundary points such as (2,3) belong to R." } ], "md_open_items": [ { "zh-Hant": "本稿不宣稱 f 連續或可微；偏導數與梯度屬於後續條目（MKO 的 formal next）。", "en": "This guide does not assert that f is continuous or differentiable; partial derivatives and the gradient belong to a later entry (the MKO's formal next)." }, { "zh-Hant": "N12 只是 x∈{0,1} 兩點的有限核對，不是恆等式；本稿正是用它說明在指定四個格點上一致不足以推出全域相等。", "en": "N12 is only a finite check at the two points x∈{0,1}, not an identity; the guide uses it precisely to show that agreement at these four grid points is insufficient for global equality." } ] }, "revision": { "id": "production-20261019-adopted-96", "source_commit": "b28a5edbd46279b1db2a3f3f48bb3f447e18899f", "adopted_source_sha256": "896db668ee991856b4a5f03263420b664073e8a89c748e47bac90da89b7ed88b", "original_source_sha256": "896db668ee991856b4a5f03263420b664073e8a89c748e47bac90da89b7ed88b", "owner_adopted": true, "bilingual_sync": "complete_source_bound", "formal_Lean": "NOT_RUN", "independent_AI_concurrence": "NOT_MEASURED", "issue_ids": [], "scope": "Owner-adopted natural-language teaching guide; not a formal proof, canonical MKO rewrite or evidence-status promotion." } }