{ "schema_version": "ocme-derivation-reader-v1", "guide_id": "derivation-mko-mathematical-representation-01", "mko_id": "mko-mathematical-representation", "title": { "zh-Hant": "數學表徵：判定文字、符號、表格與序對是否表示同一個函數", "en": "Mathematical representation: deciding whether words, a formula, tables and ordered pairs represent the same function" }, "contract": { "schema_version": "ocme-derivation-guide-v0.1", "guide_id": "derivation-mko-mathematical-representation-01", "source": { "mko_id": "mko-mathematical-representation", "commit": "b28a5edbd46279b1db2a3f3f48bb3f447e18899f", "sha256": "4f70691fec7da4288ceee8ebb2ce291dfee4c28ac4a2e8aa851fa23fe5757fd5", "english_sha256": "6774f494e9469b407eedbc3e1c065e7a85ce40b95fbd32d15c7cb728f106fa88" }, "mode": "definition_walkthrough", "audience_prerequisites": [ "function_def", "cartesian_product_def", "set_extensionality", "mul_inverse_nonzero", "quantifier_negation" ], "goal": { "zh-Hant": "固定對象 f:D→ℤ，D={0,1,2,3}，f(x)=2x−1。判定：文字「輸入加倍後減 1」對應 2x−1；表格 T1={(0,−1),(1,1),(2,3),(3,5)} 表示 f；序對集合是 f 的圖形；由兩列 (1,1)、(3,5) 唯一恢復仿射規則；表格 T2 把 (3,5) 換成 (3,6) 後不表示 f，也不表示任何仿射規則；含重複輸入的 T3 不是函數表；定義在 ℤ 上的 g(x)=2x−1 與 f 不是同一個函數；以及表格不提供未列點的值。", "en": "Fix the object f:D→ℤ with D={0,1,2,3} and f(x)=2x−1. Decide: the words 'double the input, then subtract 1' correspond to 2x−1; the table T1={(0,−1),(1,1),(2,3),(3,5)} represents f; the set of ordered pairs is the graph of f; the affine rule is recovered uniquely from the two rows (1,1) and (3,5); the table T2, with (3,5) replaced by (3,6), represents neither f nor any affine rule; T3, with a repeated input, is not a function table; g(x)=2x−1 defined on ℤ is not the same function as f; and a table supplies no values at unlisted points." }, "givens": [ { "id": "G1", "statement": { "zh-Hant": "對象：f:D→ℤ，D={0,1,2,3}，f(x)=2x−1。", "en": "Object: f:D→ℤ with D={0,1,2,3} and f(x)=2x−1." } }, { "id": "G2", "statement": { "zh-Hant": "解讀規則：表格的一列 (x,y) 表示「輸入 x 的輸出是 y」。一個表格表示 f，當且僅當它的輸入恰為 D 的元素、每個各出現一次，且每一列都滿足 y=f(x)。", "en": "Interpretation rule: a table row (x,y) means 'the output at input x is y'. A table represents f exactly when its inputs are the elements of D, each appearing once, and every row satisfies y=f(x)." } }, { "id": "G3", "statement": { "zh-Hant": "文字表徵 W：「把輸入加倍，然後減 1」。", "en": "Verbal representation W: 'double the input, then subtract 1'." } }, { "id": "G4", "statement": { "zh-Hant": "T1={(0,−1),(1,1),(2,3),(3,5)}；T2={(0,−1),(1,1),(2,3),(3,6)}；T3=(0,−1),(1,1),(1,2),(2,3),(3,5)（輸入 1 出現兩次）。", "en": "T1={(0,−1),(1,1),(2,3),(3,5)}; T2={(0,−1),(1,1),(2,3),(3,6)}; T3=(0,−1),(1,1),(1,2),(2,3),(3,5) (input 1 appears twice)." } }, { "id": "G5", "statement": { "zh-Hant": "g:ℤ→ℤ，g(x)=2x−1。", "en": "g:ℤ→ℤ, g(x)=2x−1." } } ], "steps": [ { "id": "S01", "uses": [ "G1", "G2", "mko_repr" ], "before_tex": "\\text{several representations}", "after_tex": "\\text{object }f:D\\to\\mathbb{Z},\\ D=\\{0,1,2,3\\},\\ \\text{row }(x,y)\\ \\text{means}\\ f(x)=y", "relation": "definition", "rule": "mko_repr", "conditions": [], "explanation": { "zh-Hant": "依 MKO，判定對應之前要先固定對象、定義域與解讀規則。對象是 G1 的 f，定義域是四個整數，表格的解讀規則由 G2 給出。之後每個判定都是對照這些固定的設定，而不是比較外觀。", "en": "By the MKO, the object, domain and interpretation rules must be fixed before deciding any correspondence. The object is f from G1, the domain is four integers, and the reading of tables is given by G2. Every later decision compares against these fixed settings, not appearances." }, "reverse_audit": { "kind": "premise_audit", "required_ids": [ "G1", "G2", "mko_repr" ], "explanation": { "zh-Hant": "此步只固定設定，不推出新事實。", "en": "This step only fixes the settings and derives no new fact." } }, "leaf_basis": "mko_repr + def_intro" }, { "id": "S02", "uses": [ "G3", "G1" ], "before_tex": "W:\\ \\text{double the input, then subtract 1}", "after_tex": "x\\mapsto 2x-1", "relation": "equivalent", "rule": "def_unfold", "conditions": [], "explanation": { "zh-Hant": "依文字的運算次序：先加倍得 2x，再減 1 得 2x−1。次序很重要：「先減 1 再加倍」是 2(x−1)=2x−2，在 x=0 給出 −2 而不是 −1。", "en": "Follow the order of operations in the words: doubling first gives 2x, then subtracting 1 gives 2x−1. The order matters: 'subtract 1, then double' is 2(x−1)=2x−2, which gives −2 at x=0 instead of −1." }, "reverse_audit": { "kind": "inverse_equivalence", "required_ids": [ "def_unfold" ], "explanation": { "zh-Hant": "把 2x−1 讀回文字：先乘 2，再減 1，正是 W；兩個方向描述同一個計算程序。", "en": "Reading 2x−1 back into words: multiply by 2, then subtract 1, which is exactly W; both directions describe the same procedure." } }, "leaf_basis": "def_unfold" }, { "id": "S03", "uses": [ "G1" ], "before_tex": "f(0),f(1),f(2),f(3)", "after_tex": "-1,\\ 1,\\ 3,\\ 5", "relation": "equivalent", "rule": "numeric_evaluation", "conditions": [], "explanation": { "zh-Hant": "在定義域的四點逐一計算：2·0−1=−1，2·1−1=1，2·2−1=3，2·3−1=5。", "en": "Compute at the four points of the domain: 2·0−1=−1, 2·1−1=1, 2·2−1=3, 2·3−1=5." }, "reverse_audit": { "kind": "inverse_equivalence", "required_ids": [ "numeric_evaluation" ], "explanation": { "zh-Hant": "精確整數計算，雙向相同。", "en": "Exact integer computation, the same both ways." } }, "leaf_basis": "numeric_evaluation" }, { "id": "S04", "uses": [ "G4", "G1", "function_def" ], "before_tex": "\\text{inputs of }T_1:\\ 0,1,2,3", "after_tex": "\\text{each }x\\in D\\ \\text{appears exactly once in }T_1", "relation": "implies", "rule": "function_def", "conditions": [], "explanation": { "zh-Hant": "T1 的輸入是 0、1、2、3，恰為 D 的四個元素，各出現一次。所以 T1 給每個 x∈D 恰好一個輸出，定義了一個 D 上的函數（記為 h₁）。", "en": "The inputs of T1 are 0, 1, 2, 3, exactly the four elements of D, each once. So T1 gives each x∈D exactly one output and defines a function on D (call it h₁)." }, "reverse_audit": { "kind": "premise_audit", "required_ids": [ "function_def", "G4" ], "explanation": { "zh-Hant": "需要「涵蓋 D」與「不重複」兩件事；只檢查其一不足以得到函數。", "en": "It needs both 'covers D' and 'no repetition'; checking only one of them does not give a function." } }, "leaf_basis": "function_def + finite_counting_basics" }, { "id": "S05", "uses": [ "S03", "S04", "G2", "mko_repr_notes" ], "before_tex": "h_1(x)\\ \\text{vs}\\ f(x)\\ \\text{for }x\\in D", "after_tex": "h_1=f\\ \\text{(}T_1\\ \\text{represents }f\\text{)}", "relation": "implies", "rule": "function_def", "conditions": [ "\\text{same domain }D" ], "explanation": { "zh-Hant": "逐列比對：(0,−1) 對 f(0)=−1，(1,1) 對 f(1)=1，(2,3) 對 f(2)=3，(3,5) 對 f(3)=5，四列全部相符。h₁ 與 f 有相同定義域 D 且每點值相同，依函數相等的定義 h₁=f。這就是 MKO 說的「共同域且每列數值相符」的明示語義橋。", "en": "Compare row by row: (0,−1) with f(0)=−1, (1,1) with f(1)=1, (2,3) with f(2)=3, (3,5) with f(3)=5; all four rows match. h₁ and f have the same domain D and the same value at every point, so by the definition of equality of functions h₁=f. This is the MKO's explicit semantic bridge: a common domain with every row matching." }, "reverse_audit": { "kind": "premise_audit", "required_ids": [ "S03", "S04", "function_def", "mko_repr_notes" ], "explanation": { "zh-Hant": "需要四列全部相符；只檢查部分列不足以推出函數相等（見 S09 的 T2）。", "en": "It needs all four rows to match; checking only some rows would not give equality of functions (see T2 in S09)." } }, "leaf_basis": "function_def + numeric_evaluation" }, { "id": "S06", "uses": [ "S03", "G1", "G4" ], "before_tex": "\\Gamma_f=\\{(x,f(x))\\mid x\\in D\\}", "after_tex": "\\Gamma_f=\\{(0,-1),(1,1),(2,3),(3,5)\\}=T_1", "relation": "equivalent", "rule": "set_builder", "conditions": [], "explanation": { "zh-Hant": "序對表徵：f 的圖形是所有 (x,f(x)) 的集合；依 S03 列出四個序對，與 T1 的列完全相同，所以作為集合 Γ_f=T1。", "en": "Ordered-pair representation: the graph of f is the set of all (x,f(x)); by S03 it lists four pairs, exactly the rows of T1, so as sets Γ_f=T1." }, "reverse_audit": { "kind": "inverse_equivalence", "required_ids": [ "set_builder", "set_extensionality", "cartesian_product_def" ], "explanation": { "zh-Hant": "兩邊有相同的元素，集合相等雙向成立。", "en": "Both sides have the same elements, so the set equality holds both ways." } }, "leaf_basis": "set_builder + set_extensionality" }, { "id": "S07", "uses": [ "L_two_point" ], "before_tex": "(x_1,y_1),(x_2,y_2),\\ x_1\\neq x_2", "after_tex": "a=\\frac{y_2-y_1}{x_2-x_1},\\ b=y_1-a\\,x_1\\ \\text{(unique)}", "relation": "implies", "rule": "L_two_point", "conditions": [ "x_1\\neq x_2" ], "explanation": { "zh-Hant": "依引理 L_two_point（在 S07.1、S07.2 證明唯一性與存在性）：橫坐標不同的兩點恰好決定一個仿射規則。", "en": "By Lemma L_two_point (uniqueness and existence proved in S07.1 and S07.2): two points with different x-coordinates determine exactly one affine rule." }, "reverse_audit": { "kind": "premise_audit", "required_ids": [ "L_two_point" ], "explanation": { "zh-Hant": "依賴該引理，前提 x₁≠x₂ 不可省略。", "en": "It depends on the lemma, whose hypothesis x₁≠x₂ cannot be dropped." } }, "leaf_basis": "mul_inverse_nonzero + eq_congruence" }, { "id": "S07.1", "uses": [ "mul_inverse_nonzero", "eq_congruence" ], "before_tex": "a x_1+b=y_1,\\ a x_2+b=y_2", "after_tex": "a(x_2-x_1)=y_2-y_1\\ \\Rightarrow\\ a=\\frac{y_2-y_1}{x_2-x_1},\\ b=y_1-a\\,x_1", "relation": "implies", "rule": "mul_inverse_nonzero", "conditions": [ "x_2-x_1\\neq 0" ], "explanation": { "zh-Hant": "唯一性：若 y=ax+b 通過兩點，兩式相減消去 b 得 a(x₂−x₁)=y₂−y₁。因 x₂−x₁≠0，兩邊乘以 1/(x₂−x₁) 得 a 的唯一值；再由第一式 b=y₁−a·x₁。若 x₁=x₂，這裡就要除以 0，論證失效。", "en": "Uniqueness: if y=ax+b passes through both points, subtracting the equations eliminates b and gives a(x₂−x₁)=y₂−y₁. Since x₂−x₁≠0, multiplying by 1/(x₂−x₁) gives the only possible a, and the first equation gives b=y₁−a·x₁. If x₁=x₂ this would divide by 0 and the argument fails." }, "reverse_audit": { "kind": "premise_audit", "required_ids": [ "mul_inverse_nonzero", "eq_congruence", "distributive" ], "explanation": { "zh-Hant": "這一步只證明「若存在則唯一」；存在性在 S07.2 另外核對。", "en": "This step proves only 'if one exists, it is unique'; existence is checked separately in S07.2." } }, "leaf_basis": "mul_inverse_nonzero + eq_congruence + distributive" }, { "id": "S07.2", "uses": [ "S07.1" ], "before_tex": "a=\\frac{y_2-y_1}{x_2-x_1},\\ b=y_1-a\\,x_1", "after_tex": "a x_1+b=y_1,\\ a x_2+b=y_2", "relation": "implies", "rule": "eq_replace", "conditions": [ "x_2-x_1\\neq 0" ], "explanation": { "zh-Hant": "存在性：代回 a x₁+b=a x₁+y₁−a x₁=y₁；a x₂+b=y₁+a(x₂−x₁)=y₁+(y₂−y₁)=y₂（乘以 x₂−x₁ 後的恆等式見 N9）。所以這個規則確實通過兩點。S07.1 與 S07.2 合起來證得引理 L_two_point。", "en": "Existence: substituting back, a x₁+b=a x₁+y₁−a x₁=y₁, and a x₂+b=y₁+a(x₂−x₁)=y₁+(y₂−y₁)=y₂ (the identity multiplied through by x₂−x₁ is N9). So the rule does pass through both points. S07.1 and S07.2 together prove Lemma L_two_point." }, "reverse_audit": { "kind": "premise_audit", "required_ids": [ "eq_replace", "distributive", "mul_inverse_nonzero" ], "explanation": { "zh-Hant": "需要 a(x₂−x₁)=y₂−y₁，這由 a 的定義與 x₂−x₁≠0 得到。", "en": "It needs a(x₂−x₁)=y₂−y₁, which follows from the definition of a and x₂−x₁≠0." } }, "leaf_basis": "eq_replace + distributive + mul_inverse_nonzero" }, { "id": "S08", "uses": [ "S07", "G4", "S03" ], "before_tex": "(1,1),(3,5)\\in T_1,\\ 1\\neq 3", "after_tex": "a=\\tfrac{5-1}{3-1}=2,\\ b=1-2=-1;\\ \\text{rows }(0,-1),(2,3)\\ \\text{also satisfy }y=2x-1", "relation": "implies", "rule": "L_two_point", "conditions": [ "1\\neq 3" ], "explanation": { "zh-Hant": "先確認兩列的橫坐標 1≠3，再依引理得 a=4/2=2、b=1−2·1=−1，恢復規則 y=2x−1。兩點只決定候選規則；其餘兩列 (0,−1)、(2,3) 仍要另外核對：2·0−1=−1、2·2−1=3，都相符。", "en": "First confirm that the x-coordinates differ, 1≠3; then the lemma gives a=4/2=2 and b=1−2·1=−1, recovering y=2x−1. Two points only determine a candidate rule; the remaining rows (0,−1) and (2,3) must still be checked: 2·0−1=−1 and 2·2−1=3, both match." }, "reverse_audit": { "kind": "premise_audit", "required_ids": [ "L_two_point", "numeric_evaluation" ], "explanation": { "zh-Hant": "需要 x₁≠x₂；核對其餘各列是必要的，因為仿射規則只由兩點決定，其他列可能不在其上（S09）。", "en": "It needs x₁≠x₂; checking the other rows is necessary, because the affine rule is determined by two points and other rows may fail to lie on it (S09)." } }, "leaf_basis": "L_two_point + numeric_evaluation" }, { "id": "S09", "uses": [ "G4", "S03", "S07" ], "before_tex": "T_2\\ \\text{row }(3,6),\\ f(3)=5", "after_tex": "T_2\\ \\text{does not represent }f\\ \\text{nor any affine rule}", "relation": "implies", "rule": "quantifier_negation", "conditions": [], "explanation": { "zh-Hant": "T2 在 x=3 的列是 6，但 f(3)=5，一列不符就否定「每一列都相符」，所以 T2 不表示 f。更進一步，若某個仿射規則表示 T2，它必通過 (0,−1) 與 (1,1)（0≠1），依 L_two_point 只能是 y=2x−1，而它在 x=3 給 5≠6，所以沒有任何仿射規則表示 T2。前三列都相符並不足夠。", "en": "The row of T2 at x=3 is 6, but f(3)=5; one mismatching row negates 'every row matches', so T2 does not represent f. Further, any affine rule representing T2 would pass through (0,−1) and (1,1) (with 0≠1), so by L_two_point it could only be y=2x−1, which gives 5≠6 at x=3; hence no affine rule represents T2. Matching the first three rows is not enough." }, "reverse_audit": { "kind": "premise_audit", "required_ids": [ "S03", "L_two_point", "quantifier_negation" ], "explanation": { "zh-Hant": "否定全稱只需一個反例；「沒有仿射規則」用到引理的唯一性。", "en": "Negating the universal statement needs one counterexample; 'no affine rule' uses the uniqueness part of the lemma." } }, "leaf_basis": "quantifier_negation + L_two_point + numeric_evaluation" }, { "id": "S10", "uses": [ "G4", "function_def", "mko_repr_notes" ], "before_tex": "T_3\\ni(1,1),(1,2)", "after_tex": "T_3\\ \\text{is not a function table}", "relation": "implies", "rule": "function_def", "conditions": [ "1\\neq 2" ], "explanation": { "zh-Hant": "函數對每個輸入指定恰好一個輸出。T3 中輸入 1 同時對應 1 與 2，而 1≠2，所以 T3 不是函數表，也就不能表示 f 或任何函數。這也是兩點橫坐標相同（垂直排列）時 L_two_point 不適用的情形：(1,1) 與 (1,2) 的斜率需要除以 0。", "en": "A function assigns exactly one output to each input. In T3 the input 1 is paired with both 1 and 2, and 1≠2, so T3 is not a function table and cannot represent f or any function. It is also the case where L_two_point does not apply because the two points share an x-coordinate (a vertical arrangement): the slope through (1,1) and (1,2) would need division by 0." }, "reverse_audit": { "kind": "premise_audit", "required_ids": [ "function_def", "mko_repr_notes" ], "explanation": { "zh-Hant": "只需要一個輸入對應兩個不同輸出；若重複的列輸出相同，那只是重複列，另行處理。", "en": "It needs only one input paired with two different outputs; if a repeated row had the same output, it would merely be a duplicate row, a separate matter." } }, "leaf_basis": "function_def" }, { "id": "S11", "uses": [ "G1", "G5", "function_def" ], "before_tex": "g:\\mathbb{Z}\\to\\mathbb{Z},\\ f:D\\to\\mathbb{Z}", "after_tex": "g\\neq f,\\ \\ g|_D=f", "relation": "implies", "rule": "function_def", "conditions": [ "D\\neq\\mathbb{Z}" ], "explanation": { "zh-Hant": "g 與 f 用同一個公式，但定義域不同（ℤ 與 D），依函數相等的定義它們不是同一個函數；例如 g(4)=7 有定義，而 4∉D。把 g 限制在 D 上，每點值都與 f 相同，所以 g|_D=f。MKO 要求先固定定義域，原因就在此。", "en": "g and f use the same formula but have different domains (ℤ and D), so by the definition of equality of functions they are not the same function; for example g(4)=7 is defined while 4∉D. Restricting g to D gives the same value as f at every point, so g|_D=f. This is why the MKO requires fixing the domain first." }, "reverse_audit": { "kind": "premise_audit", "required_ids": [ "function_def", "set_extensionality" ], "explanation": { "zh-Hant": "定義域不同就足以使函數不等；公式相同不能補救。", "en": "Different domains suffice for the functions to differ; an identical formula cannot fix that." } }, "leaf_basis": "function_def + set_extensionality" }, { "id": "S12", "uses": [ "G1", "G4", "mko_repr_notes" ], "before_tex": "x=4,\\ x=\\tfrac{1}{2}", "after_tex": "\\text{no value supplied: }4,\\tfrac{1}{2}\\notin D", "relation": "implies", "rule": "mko_repr_notes", "conditions": [], "explanation": { "zh-Hant": "T1 只列出 D 的四點。f(4) 沒有定義，因為 4∉D；1/2 也不在 D 中。若把四個點畫出來，它們看起來在一條直線上，直線經過 (1/2,0)，但依 MKO 假設 a2，圖形外觀與未列點不提供數值，所以不能宣稱 f(1/2)=0。", "en": "T1 lists only the four points of D. f(4) is undefined because 4∉D, and 1/2 is not in D either. If the four points are plotted they look collinear, and the line passes through (1/2,0), but by MKO assumption a2 the appearance of a graph and unlisted points supply no values, so f(1/2)=0 cannot be claimed." }, "reverse_audit": { "kind": "premise_audit", "required_ids": [ "mko_repr_notes", "function_def" ], "explanation": { "zh-Hant": "只需要這些點不在定義域中；結論是「沒有資料」，不是「值為某數」。", "en": "It needs only that these points are outside the domain; the conclusion is 'no data', not 'the value is some number'." } }, "leaf_basis": "mko_repr_notes + function_def" }, { "id": "S13", "uses": [ "S02", "S05", "S06", "S08", "S09", "S10", "S11", "S12" ], "before_tex": "\\text{decisions}", "after_tex": "W,\\ 2x-1,\\ T_1,\\ \\Gamma_f\\ \\text{represent }f;\\ T_2,\\ T_3,\\ g\\ \\text{do not}", "relation": "implies", "rule": "and_intro_elim", "conditions": [], "explanation": { "zh-Hant": "合併判定：在固定的對象、定義域與解讀規則下，文字 W、公式 2x−1、表格 T1 與序對集合 Γ_f 都表示 f；兩列 (1,1)、(3,5) 唯一恢復規則；T2 不表示 f 也不表示任何仿射規則；T3 不是函數表；g 定義域不同而與 f 不等；未列點沒有值。", "en": "Combine the decisions: with the object, domain and interpretation rule fixed, the words W, the formula 2x−1, the table T1 and the pair set Γ_f all represent f; the rows (1,1) and (3,5) recover the rule uniquely; T2 represents neither f nor any affine rule; T3 is not a function table; g differs from f in its domain; unlisted points have no values." }, "reverse_audit": { "kind": "sufficient_goal_reduction", "required_ids": [ "S02", "S05", "S06", "S08", "S09", "S10", "S11", "S12", "and_intro_elim" ], "explanation": { "zh-Hant": "目標 G：本稿目標的各項判定。足夠條件 H：所列各步已證（H⇒G 由合取引入）。H 由給定條件正向推出，沒有把 G 當作前提。", "en": "Goal G: the determinations in the goal. Sufficient condition H: the listed steps, each already proved (H⇒G by conjunction introduction). H is derived forwards from the givens; G is never used as a premise." } }, "leaf_basis": "and_intro_elim" } ], "result": { "zh-Hant": "固定 f:D→ℤ（D={0,1,2,3}，f(x)=2x−1）與「列 (x,y) 表示 f(x)=y」的解讀規則後：文字「輸入加倍後減 1」、公式 2x−1、表格 T1 與序對集合 {(0,−1),(1,1),(2,3),(3,5)} 都表示 f；由 (1,1)、(3,5)（1≠3）唯一恢復 y=2x−1，其餘列另外核對相符。T2 在 x=3 的值 6≠5，不表示 f，且依兩點唯一性不表示任何仿射規則；T3 中輸入 1 對應兩個輸出，不是函數表；g:ℤ→ℤ 與 f 定義域不同而不相等；表格不提供 x=4 或 x=1/2 的值。", "en": "With f:D→ℤ (D={0,1,2,3}, f(x)=2x−1) and the reading 'a row (x,y) means f(x)=y' fixed: the words 'double the input, then subtract 1', the formula 2x−1, the table T1 and the pair set {(0,−1),(1,1),(2,3),(3,5)} all represent f; from (1,1) and (3,5) (with 1≠3) the rule y=2x−1 is recovered uniquely, and the other rows are checked separately and match. T2 has value 6≠5 at x=3, so it does not represent f, and by two-point uniqueness it represents no affine rule; in T3 the input 1 has two outputs, so it is not a function table; g:ℤ→ℤ differs from f because the domains differ; the table supplies no value at x=4 or x=1/2." }, "negative_controls": [ { "claim": { "zh-Hant": "T2 表示 f，因為四列中有三列相符，而且畫出來幾乎是一條直線", "en": "T2 represents f because three of its four rows match and the plot is almost a straight line" }, "expected": "reject", "reason": { "zh-Hant": "表示要求每一列都相符；f(3)=5≠6（S09）。", "en": "Representation needs every row to match; f(3)=5≠6 (S09)." } }, { "claim": { "zh-Hant": "g:ℤ→ℤ，g(x)=2x−1 與 f 是同一個函數", "en": "g:ℤ→ℤ with g(x)=2x−1 is the same function as f" }, "expected": "reject", "reason": { "zh-Hant": "定義域不同；例如 g(4)=7 有定義，4∉D（S11）。", "en": "The domains differ; for example g(4)=7 is defined while 4∉D (S11)." } }, { "claim": { "zh-Hant": "T3 的每一列都是序對，所以 T3 是一個函數表", "en": "Every row of T3 is an ordered pair, so T3 is a function table" }, "expected": "reject", "reason": { "zh-Hant": "輸入 1 對應 1 與 2 兩個輸出（S10）。", "en": "The input 1 is paired with two outputs, 1 and 2 (S10)." } }, { "claim": { "zh-Hant": "由 T1 可知 f(1/2)=0，因為這些點在一條直線上", "en": "From T1, f(1/2)=0, because the points lie on a line" }, "expected": "reject", "reason": { "zh-Hant": "1/2∉D，未列點與圖形外觀不提供數值（S12）。", "en": "1/2∉D, and unlisted points and the appearance of a graph supply no values (S12)." } }, { "claim": { "zh-Hant": "文字「把輸入加倍，然後減 1」描述的是 2(x−1)", "en": "The words 'double the input, then subtract 1' describe 2(x−1)" }, "expected": "reject", "reason": { "zh-Hant": "2(x−1) 是先減後乘；在 x=0 得 −2，而 W 給 −1（S02）。", "en": "2(x−1) subtracts first and then doubles; at x=0 it gives −2, while W gives −1 (S02)." } } ], "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": "set_extensionality", "kind": "prerequisite", "statement": { "zh-Hant": "集合相等：A=B 當且僅當 A⊆B 且 B⊆A，即兩者有相同元素。", "en": "Set equality: A=B if and only if A⊆B and B⊆A, i.e. they have the same elements." } }, { "id": "mul_inverse_nonzero", "kind": "prerequisite", "statement": { "zh-Hant": "非零數的倒數：在 ℚ、ℝ、ℂ 中，若 u≠0，存在 1/u 使 u·(1/u)=1；對 u≠0 定義 v/u 為 v·(1/u)。u=0 時沒有倒數，v/0 不定義。", "en": "Inverse of a nonzero number: in ℚ, ℝ or ℂ, if u≠0 there is 1/u with u·(1/u)=1, and v/u means v·(1/u). 0 has no inverse, so v/0 is undefined." } }, { "id": "quantifier_negation", "kind": "prerequisite", "statement": { "zh-Hant": "量詞否定：¬(∀x, P(x)) 等價於 ∃x, ¬P(x)；¬(∃x, P(x)) 等價於 ∀x, ¬P(x)。", "en": "Negating quantifiers: ¬(∀x, P(x)) is equivalent to ∃x, ¬P(x), and ¬(∃x, P(x)) is equivalent to ∀x, ¬P(x)." } }, { "id": "def_intro", "kind": "prerequisite", "statement": { "zh-Hant": "引入定義：可為一個命題、式子或物件取名；之後名稱與被定義者可互相替換。", "en": "Introducing a definition: a statement, expression or object may be given a name, after which the name and what it stands for are interchangeable." } }, { "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": "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": "finite_counting_basics", "kind": "prerequisite", "statement": { "zh-Hant": "有限計數：互斥有限集合 A、B 有 |A∪B|=|A|+|B|；|A×B|=|A|·|B|；有限集之間存在雙射 ⇔ 元素個數相同。", "en": "Finite counting: for disjoint finite sets A, B, |A∪B|=|A|+|B|; |A×B|=|A|·|B|; there is a bijection between finite sets ⇔ they have the same number of elements." } }, { "id": "set_builder", "kind": "prerequisite", "statement": { "zh-Hant": "集合建構式：x∈{y∈A ∣ P(y)} 當且僅當 x∈A 且 P(x)。", "en": "Set-builder notation: x∈{y∈A ∣ P(y)} if and only if x∈A and P(x)." } }, { "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": "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": "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": "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": "mko_repr", "kind": "cited", "source": { "mko_id": "mko-mathematical-representation", "sha256": "4f70691fec7da4288ceee8ebb2ce291dfee4c28ac4a2e8aa851fa23fe5757fd5", "commit": "b28a5edbd46279b1db2a3f3f48bb3f447e18899f" }, "statement": { "zh-Hant": "同一數學對象可以用文字、符號、表格、序對或圖形表示。兩個表示是否對應，必須先固定對象、定義域、標度與解讀規則；不能只依外觀宣稱等價。", "en": "The same mathematical object can be represented in words, symbols, tables, ordered pairs or diagrams. To determine whether representations correspond, first fix the object, domain, scale and interpretation rules; visual resemblance alone does not establish equivalence." } }, { "id": "mko_repr_notes", "kind": "cited", "source": { "mko_id": "mko-mathematical-representation", "sha256": "4f70691fec7da4288ceee8ebb2ce291dfee4c28ac4a2e8aa851fa23fe5757fd5", "commit": "b28a5edbd46279b1db2a3f3f48bb3f447e18899f" }, "statement": { "zh-Hant": "表格列出指定輸入與輸出，符號式則給計算規則；二者只有在共同域及每列數值相符時形成明示語義橋。兩點可決定非垂直仿射規則，但必須先確認橫坐標不同；重複輸入若對應不同輸出，就不是函數表。圖像可以支援理解，但像素位置、視覺直線感或未標示刻度不能替代明示資料。", "en": "A table lists specified inputs and outputs, while a formula gives a rule of computation; they form an explicit semantic bridge only on a common domain with every row matching. Two points determine a non-vertical affine rule, but their x-coordinates must first be checked to differ; repeated inputs with different outputs do not form a function table. A picture can support understanding, but pixel positions, a visual sense of straightness or unlabelled scales cannot replace explicit data." } }, { "id": "L_two_point", "kind": "lemma", "proved_in": [ "S07.1", "S07.2" ], "statement": { "zh-Hant": "對實數座標，若 x₁≠x₂，則恰有一個實係數仿射規則 y=ax+b 通過 (x₁,y₁) 與 (x₂,y₂)，即 a=(y₂−y₁)/(x₂−x₁)，b=y₁−a·x₁。", "en": "For real coordinates, if x₁≠x₂ then exactly one real-coefficient affine rule y=ax+b passes through (x₁,y₁) and (x₂,y₂), namely a=(y₂−y₁)/(x₂−x₁) and b=y₁−a·x₁." } } ], "revision_log": [], "numeric_checks": [ { "id": "N1", "step": "S03", "lhs": "2*0-1", "rhs": "-1" }, { "id": "N2", "step": "S03", "lhs": "2*1-1", "rhs": "1" }, { "id": "N3", "step": "S03", "lhs": "2*2-1", "rhs": "3" }, { "id": "N4", "step": "S03", "lhs": "2*3-1", "rhs": "5" }, { "id": "N5", "step": "S08", "lhs": "(5-1)/(3-1)", "rhs": "2" }, { "id": "N6", "step": "S08", "lhs": "1-2*1", "rhs": "-1" }, { "id": "N7", "step": "S09", "lhs": "2*3-1", "relation": "!=", "rhs": "6" }, { "id": "N8", "step": "S09", "lhs": "(1-(-1))/(1-0)", "rhs": "2" }, { "id": "N9", "step": "S07.2", "vars": [ "x1", "x2", "y1", "y2" ], "degree": 2, "lhs": "(y2-y1)*x2+(x2-x1)*y1-(y2-y1)*x1", "rhs": "(x2-x1)*y2" }, { "id": "N10", "step": "S02", "lhs": "2*(0-1)", "relation": "!=", "rhs": "2*0-1" }, { "id": "N11", "step": "S11", "lhs": "2*4-1", "rhs": "7" } ] }, "presentation": { "md_scope": { "zh-Hant": "具體練習使用整係數仿射規則與完整明示的有限表格／序對（MKO 假設 a1）；輔助的兩點仿射引理則在實數係數與實數座標下敘述；所有核對只在列出的定義域上進行，未列點與圖形外觀不提供數值（假設 a2）。不處理一般圖形的語義或更完整的 typed semantic bridge；不改動 canonical MKO 或英文 edition。", "en": "The worked exercises use integer-coefficient affine rules and completely listed finite tables or ordered pairs (MKO assumption a1); the auxiliary two-point affine lemma is stated for real coefficients and real coordinates; every check is made on the listed domain only, and unlisted points and the appearance of a graph supply no values (assumption a2). General semantics of diagrams and a fuller typed semantic bridge are not treated; the canonical MKO and its English edition are not changed." }, "md_symbols": [ [ "D=\\{0,1,2,3\\}", "f 的定義域", "the domain of f" ], [ "f(x)=2x-1", "符號表徵", "the symbolic representation" ], [ "T_1,\\ T_2,\\ T_3", "三個待檢查的表格／序對列表", "the three tables or lists of pairs to be checked" ], [ "\\Gamma_f", "f 的圖形 {(x,f(x)) ∣ x∈D}", "the graph of f, {(x,f(x)) ∣ x∈D}" ], [ "g", "定義在 ℤ 上的同一公式", "the same formula on the domain ℤ" ] ], "md_error_controls": [ { "title": "看起來相同不等於等價 / Looking the same is not being equivalent", "zh": "MKO 指出的誤解之一：把 T2 畫出來，(3,6) 只比直線高一格，四點看起來近乎在一條線上。但對應由逐列數值決定：f(3)=5≠6（S09）。像素位置與視覺直線感不能代替明示資料。", "en": "One misconception named by the MKO: when T2 is plotted, (3,6) is only one unit above the line and the four points look almost collinear. But correspondence is decided row by row: f(3)=5≠6 (S09). Pixel positions and a visual sense of straightness cannot replace explicit data." }, { "title": "定義域與重複輸入不可忽略 / Domains and repeated inputs cannot be ignored", "zh": "MKO 指出的另一個誤解：g(x)=2x−1 在 ℤ 上與 f 公式相同，但定義域不同，所以不是同一個函數（S11）；T3 的輸入 1 對應兩個輸出，不是函數表（S10）。判定前必須先核對定義域與每個輸入只出現一次。", "en": "Another misconception named by the MKO: g(x)=2x−1 on ℤ has the same formula as f but a different domain, so it is a different function (S11); in T3 the input 1 has two outputs, so it is not a function table (S10). Before deciding, check the domain and that each input appears once." }, { "title": "兩點決定候選，其餘各列仍要核對 / Two points give a candidate; the other rows must still be checked", "zh": "由兩列恢復仿射規則需要橫坐標不同（S07.1 要除以 x₂−x₁）。恢復出的規則只是候選：T1 的其他列核對後相符（S08），T2 的第四列則不符（S09）。", "en": "Recovering an affine rule from two rows needs different x-coordinates (S07.1 divides by x₂−x₁). The recovered rule is only a candidate: the other rows of T1 are checked and match (S08), while the fourth row of T2 does not (S09)." } ], "md_open_items": [ { "zh-Hant": "圖形表徵只在錯誤控制與 S12 中討論，沒有建立座標刻度與像素的正式對應；更完整的 typed semantic bridge 屬於 MKO 的 formal next。", "en": "The diagram representation is discussed only in the error controls and S12, without a formal correspondence between coordinate scales and pixels; a fuller typed semantic bridge belongs to the MKO's formal next step." } ] }, "revision": { "id": "production-20261019-adopted-96", "source_commit": "b28a5edbd46279b1db2a3f3f48bb3f447e18899f", "adopted_source_sha256": "c57160d6a5d2e6e5cf54e5bc2ba0141c674852fb5e67bc4370040368f3ac58c2", "original_source_sha256": "9a638a2170e8b0d8bd35dbc76a4458dca4722cbd6f58352d5b1d907fd8f30ae8", "owner_adopted": true, "bilingual_sync": "complete_source_bound", "formal_Lean": "NOT_RUN", "independent_AI_concurrence": "NOT_MEASURED", "issue_ids": [ "BR-M045" ], "scope": "Owner-adopted natural-language teaching guide; not a formal proof, canonical MKO rewrite or evidence-status promotion." } }