{ "schema_version": "ocme-derivation-reader-v1", "guide_id": "derivation-mko-geometric-transformations-01", "mko_id": "mko-geometric-transformations", "title": { "zh-Hant": "幾何變換：證明基本等距並分類均勻縮放", "en": "Geometric transformations: proving the basic isometries and classifying uniform scaling" }, "contract": { "schema_version": "ocme-derivation-guide-v0.1", "guide_id": "derivation-mko-geometric-transformations-01", "source": { "mko_id": "mko-geometric-transformations", "commit": "b28a5edbd46279b1db2a3f3f48bb3f447e18899f", "sha256": "13176ddfbd967892818086ef2859a68ada71d09c120c932f688634fbec64e489", "english_sha256": "403883e15b18bf9b23ff01fe8eec209041ce5774af2f19c72ec39d5e53538c31" }, "mode": "theorem_proof", "audience_prerequisites": [ "neg_mul", "distributive", "power_laws", "abs_rules", "sqrt_def", "order_basics", "numeric_evaluation" ], "goal": { "zh-Hant": "在等單位正交實平面中證明：平移、原點逆時針 90° 旋轉 R90(x,y)=(−y,x) 以及 MKO 列出的其他旋轉與軸鏡射都是等距映射；半轉 (−x,−y) 不是所列四條軸中任何一條的鏡射；均勻縮放 Hc(x,y)=(cx,cy) 使每個距離乘 |c|。據此分類 c=0、c=−1、c=−2、c=2，並以 P=(3,1)、Q=(−1,4) 做數值核對。", "en": "On the equal-unit orthonormal real plane, prove that translations, the counterclockwise quarter-turn R90(x,y)=(−y,x) about the origin, and the other rotations and axis reflections listed in the MKO are isometries; that the half-turn (−x,−y) is not a reflection in any of the listed axes; and that uniform scaling Hc(x,y)=(cx,cy) multiplies every distance by |c|. Use this to classify c=0, c=−1, c=−2 and c=2, with a numerical check on P=(3,1), Q=(−1,4)." }, "givens": [ { "id": "G1", "statement": { "zh-Hant": "模型：等單位正交實平面 ℝ²，定義域與陪域都是整個平面；距離為 d(P,Q)=√((x2−x1)²+(y2−y1)²)。", "en": "Model: the equal-unit orthonormal real plane ℝ², with domain and codomain the whole plane; distance d(P,Q)=√((x2−x1)²+(y2−y1)²)." } }, { "id": "G2", "statement": { "zh-Hant": "映射：平移 T(x,y)=(x+tx, y+ty)；R90(x,y)=(−y,x)；R180(x,y)=(−x,−y)；R270(x,y)=(y,−x)；鏡射 x 軸 (x,−y)、y 軸 (−x,y)、y=x 為 (y,x)、y=−x 為 M(x,y)=(−y,−x)；均勻縮放 Hc(x,y)=(cx,cy)。", "en": "Maps: translation T(x,y)=(x+tx, y+ty); R90(x,y)=(−y,x); R180(x,y)=(−x,−y); R270(x,y)=(y,−x); reflections in the x-axis (x,−y), the y-axis (−x,y), y=x as (y,x), and y=−x as M(x,y)=(−y,−x); uniform scaling Hc(x,y)=(cx,cy)." } }, { "id": "G3", "statement": { "zh-Hant": "數值核對用：P=(3,1)、Q=(−1,4)；係數 c=0、−1、2、−2。", "en": "For numerical checks: P=(3,1), Q=(−1,4); coefficients c=0, −1, 2, −2." } } ], "steps": [ { "id": "S01", "uses": [ "G1", "isometry_def" ], "before_tex": "H:\\mathbb{R}^2\\to\\mathbb{R}^2", "after_tex": "H\\ \\text{isometry}\\iff\\forall P,Q\\in\\mathbb{R}^2:\\ d(H(P),H(Q))=d(P,Q)", "relation": "definition", "rule": "isometry_def", "conditions": [], "explanation": { "zh-Hant": "固定要證明的性質：等距是對「所有」點對成立的全平面性質。要證明某映射是等距，必須對任意一對點證明；要否定，只需一對點的反例。距離採 G1 的歐氏距離。", "en": "Fix the property to be proved: an isometry is a whole-plane property holding for all pairs of points. To prove a map is an isometry one must argue for an arbitrary pair; to refute it, one counterexample pair is enough. Distance is the Euclidean distance of G1." }, "reverse_audit": { "kind": "premise_audit", "required_ids": [ "G1", "isometry_def" ], "explanation": { "zh-Hant": "此步只寫出定義，不推出新事實。", "en": "This step only states the definition and derives no new fact." } }, "leaf_basis": "isometry_def + def_unfold" }, { "id": "S02", "uses": [ "G1", "dist_formula" ], "before_tex": "P=(x_1,y_1),\\ Q=(x_2,y_2)\\ \\text{arbitrary}", "after_tex": "\\Delta x=x_2-x_1,\\ \\Delta y=y_2-y_1,\\ d(P,Q)=\\sqrt{\\Delta x^2+\\Delta y^2}", "relation": "definition", "rule": "dist_formula", "conditions": [], "explanation": { "zh-Hant": "取任意兩點 P、Q，不對它們加任何條件，並為座標差取名 Δx、Δy。依距離公式 d(P,Q)=√(Δx²+Δy²)。之後各映射的論證都對這一對任意點進行，最後才能用全稱推廣。", "en": "Take two arbitrary points P and Q, with no condition on them, and name the coordinate differences Δx and Δy. By the distance formula d(P,Q)=√(Δx²+Δy²). Every later argument for a map is carried out for this arbitrary pair, which is what allows generalization at the end." }, "reverse_audit": { "kind": "premise_audit", "required_ids": [ "G1", "dist_formula", "def_intro" ], "explanation": { "zh-Hant": "只取名並寫出公式；P、Q 的任意性是 S06 使用全稱推廣的前提。", "en": "It only names and states the formula; the arbitrariness of P and Q is what S06 needs for generalization." } }, "leaf_basis": "def_intro + dist_formula" }, { "id": "S03", "uses": [ "S02", "G2" ], "before_tex": "d(T(P),T(Q))", "after_tex": "\\sqrt{((x_2+t_x)-(x_1+t_x))^2+((y_2+t_y)-(y_1+t_y))^2}=\\sqrt{\\Delta x^2+\\Delta y^2}=d(P,Q)", "relation": "equivalent", "rule": "dist_formula", "conditions": [], "explanation": { "zh-Hant": "平移：T(Q)−T(P) 的坐標是 (x2+tx)−(x1+tx)=Δx 與 (y2+ty)−(y1+ty)=Δy（加法結合、交換與反元素消去 tx、ty；N1 核對）。根號內完全相同，所以 d(T(P),T(Q))=d(P,Q)。", "en": "Translation: the coordinates of T(Q)−T(P) are (x2+tx)−(x1+tx)=Δx and (y2+ty)−(y1+ty)=Δy (associativity, commutativity and inverses of addition cancel tx and ty; N1 checks this). The radicands are identical, so d(T(P),T(Q))=d(P,Q)." }, "reverse_audit": { "kind": "inverse_equivalence", "required_ids": [ "dist_formula", "add_inverse" ], "explanation": { "zh-Hant": "每個等號都是恆等式，整條鏈雙向成立；沒有對 tx、ty 加條件。", "en": "Each equality is an identity, so the chain holds both ways; no condition is placed on tx or ty." } }, "leaf_basis": "dist_formula + add_assoc + add_comm + add_inverse + add_zero" }, { "id": "S04", "uses": [ "S02", "G2" ], "before_tex": "R_{90}(Q)-R_{90}(P)", "after_tex": "(-y_2-(-y_1),\\ x_2-x_1)=(-\\Delta y,\\ \\Delta x)", "relation": "equivalent", "rule": "distributive", "conditions": [], "explanation": { "zh-Hant": "R90(P)=(−y1, x1)、R90(Q)=(−y2, x2)。第一坐標 −y2−(−y1)=−(y2−y1)=−Δy（負號分配，N2 核對），第二坐標 x2−x1=Δx。旋轉方向是逆時針：(1,0) 送到 (0,1)。", "en": "R90(P)=(−y1, x1) and R90(Q)=(−y2, x2). The first coordinate is −y2−(−y1)=−(y2−y1)=−Δy (distributing the sign; N2 checks this), and the second is x2−x1=Δx. The direction is counterclockwise: (1,0) goes to (0,1)." }, "reverse_audit": { "kind": "inverse_equivalence", "required_ids": [ "distributive", "neg_mul" ], "explanation": { "zh-Hant": "逐坐標的恆等變形，雙向成立。", "en": "Coordinatewise identities, valid both ways." } }, "leaf_basis": "distributive + neg_mul + eq_replace" }, { "id": "S05", "uses": [ "S04" ], "before_tex": "d(R_{90}(P),R_{90}(Q))", "after_tex": "\\sqrt{(-\\Delta y)^2+\\Delta x^2}=\\sqrt{\\Delta x^2+\\Delta y^2}=d(P,Q)", "relation": "equivalent", "rule": "L_signed_perm", "conditions": [], "explanation": { "zh-Hant": "R90(Q)−R90(P)=(−Δy, Δx) 的兩個坐標正是帶正負號、順序對調的 Δy 與 Δx，所以依引理 L_signed_perm，d(R90(P),R90(Q))=d(P,Q)。引理在 S05.1 證明。", "en": "The two coordinates of R90(Q)−R90(P)=(−Δy, Δx) are Δy and Δx with a sign and swapped order, so by Lemma L_signed_perm, d(R90(P),R90(Q))=d(P,Q). The lemma is proved in S05.1." }, "reverse_audit": { "kind": "inverse_equivalence", "required_ids": [ "L_signed_perm" ], "explanation": { "zh-Hant": "引理給出的是兩個式子的相等，雙向成立；引理在 S05.1 證明，沒有循環。", "en": "The lemma gives an equality of two expressions, valid both ways; it is proved in S05.1 without circularity." } }, "leaf_basis": "neg_mul + add_comm + eq_congruence" }, { "id": "S05.1", "uses": [ "neg_mul", "add_comm", "eq_congruence" ], "before_tex": "(\\sigma_1 a)^2+(\\sigma_2 b)^2,\\ \\sigma_i\\in\\{1,-1\\},\\ \\{a,b\\}=\\{\\Delta x,\\Delta y\\}", "after_tex": "a^2+b^2=\\Delta x^2+\\Delta y^2", "relation": "equivalent", "rule": "neg_mul", "conditions": [], "explanation": { "zh-Hant": "σ=−1 時 (−a)²=(−a)(−a)=a²（負負得正），σ=1 時顯然；所以 (σ1a)²+(σ2b)²=a²+b²（N3 核對）。a、b 是 Δx、Δy 的某種順序，依加法交換 a²+b²=Δx²+Δy²。兩個根號內相等，依等式的合同性根號值相等，即 d(H(P),H(Q))=d(P,Q)。這證得引理 L_signed_perm。", "en": "For σ=−1, (−a)²=(−a)(−a)=a² (minus times minus is plus), and for σ=1 it is clear; so (σ1a)²+(σ2b)²=a²+b² (N3 checks this). a and b are Δx and Δy in some order, so a²+b²=Δx²+Δy² by commutativity of addition. Equal radicands give equal square roots by congruence of equality, that is d(H(P),H(Q))=d(P,Q). This proves Lemma L_signed_perm." }, "reverse_audit": { "kind": "inverse_equivalence", "required_ids": [ "neg_mul", "add_comm", "eq_congruence" ], "explanation": { "zh-Hant": "全是恆等式，雙向成立；四種正負號與兩種順序都涵蓋。", "en": "All identities, valid both ways; all four sign choices and both orders are covered." } }, "leaf_basis": "neg_mul + add_comm + eq_congruence + case_analysis" }, { "id": "S06", "uses": [ "S02", "S03", "S05", "S01" ], "before_tex": "d(T(P),T(Q))=d(P,Q),\\ d(R_{90}(P),R_{90}(Q))=d(P,Q)\\ \\text{for arbitrary}\\ P,Q", "after_tex": "T\\ \\text{and}\\ R_{90}\\ \\text{are isometries}", "relation": "implies", "rule": "forall_intro", "conditions": [], "explanation": { "zh-Hant": "S03、S05 的等式是對 S02 中任意的 P、Q 證明的，沒有對它們加條件；依全稱推廣，對所有點對成立。依 S01 的定義，平移 T 與 R90 都是等距映射。", "en": "The equalities of S03 and S05 were proved for the arbitrary P and Q of S02, with no condition on them; by generalization they hold for all pairs. By the definition in S01, the translation T and R90 are isometries." }, "reverse_audit": { "kind": "sufficient_goal_reduction", "required_ids": [ "S01", "S02", "S03", "S05", "forall_intro" ], "explanation": { "zh-Hant": "目標 G：T、R90 是等距。足夠條件 H：對任意點對距離不變；H⇒G 由定義。H 由 S03、S05 正向證成，未把 G 當作前提；沒有用任何有限點對代替任意點對。", "en": "Goal G: T and R90 are isometries. Sufficient condition H: distance is unchanged for an arbitrary pair; H⇒G by the definition. H is proved forwards in S03 and S05, without assuming G; no finite set of pairs replaces the arbitrary pair." } }, "leaf_basis": "forall_intro + isometry_def" }, { "id": "S07", "uses": [ "S02", "G2", "S05", "S01" ], "before_tex": "R_{180},\\ R_{270},\\ \\text{reflections in }x\\text{-axis},\\ y\\text{-axis},\\ y=x,\\ y=-x", "after_tex": "(-\\Delta x,-\\Delta y),\\ (\\Delta y,-\\Delta x),\\ (\\Delta x,-\\Delta y),\\ (-\\Delta x,\\Delta y),\\ (\\Delta y,\\Delta x),\\ (-\\Delta y,-\\Delta x):\\ \\text{all isometries}", "relation": "implies", "rule": "L_signed_perm", "conditions": [], "explanation": { "zh-Hant": "逐一寫出 H(Q)−H(P)：R180 為 (−Δx,−Δy)；R270(x,y)=(y,−x) 為 (Δy,−Δx)；x 軸鏡射為 (Δx,−Δy)；y 軸鏡射為 (−Δx,Δy)；y=x 鏡射為 (Δy,Δx)；y=−x 鏡射 M 為 (−Δy,−Δx)。每一個都是帶正負號的 Δx、Δy（可能對調），所以依引理 L_signed_perm 與全稱推廣（同 S06），六個映射都是等距；R0 是恆等映射，顯然等距。", "en": "Write H(Q)−H(P) for each: R180 gives (−Δx,−Δy); R270(x,y)=(y,−x) gives (Δy,−Δx); the x-axis reflection gives (Δx,−Δy); the y-axis reflection (−Δx,Δy); the y=x reflection (Δy,Δx); the y=−x reflection M gives (−Δy,−Δx). Each is a signed Δx and Δy, possibly swapped, so by Lemma L_signed_perm and generalization (as in S06) all six maps are isometries; R0 is the identity and is clearly an isometry." }, "reverse_audit": { "kind": "premise_audit", "required_ids": [ "S02", "G2", "L_signed_perm", "forall_intro" ], "explanation": { "zh-Hant": "每個差向量都可由 G2 的公式逐坐標重算；引理的條件（帶號置換）逐一檢查成立。", "en": "Each difference vector can be recomputed coordinatewise from the formulas in G2; the lemma's hypothesis (a signed permutation) is checked case by case." } }, "leaf_basis": "L_signed_perm + forall_intro + distributive + neg_mul" }, { "id": "S08", "uses": [ "G2", "function_def" ], "before_tex": "R_{180}(x,y)=(-x,-y)", "after_tex": "R_{180}\\ \\text{differs from each of the four axis reflections}", "relation": "implies", "rule": "function_def", "conditions": [], "explanation": { "zh-Hant": "鏡射固定其軸上每一點：x 軸鏡射固定 (1,0)，y 軸鏡射固定 (0,1)，y=x 鏡射固定 (1,1)，y=−x 鏡射 M 固定 (1,−1)（M(1,−1)=(1,−1)）。半轉 (−x,−y)=(x,y) 當且僅當 2x=0 且 2y=0，即只固定原點；所以 R180 把這四點分別送到 (−1,0)、(0,−1)、(−1,−1)、(−1,1)，與四個鏡射在各自的點上取值不同。依函數相等的定義，R180 不是其中任何一個鏡射；它是中心反演，也就是半轉。", "en": "A reflection fixes every point of its axis: the x-axis reflection fixes (1,0), the y-axis reflection (0,1), the y=x reflection (1,1), and the y=−x reflection M fixes (1,−1) (M(1,−1)=(1,−1)). The half-turn has (−x,−y)=(x,y) exactly when 2x=0 and 2y=0, so it fixes only the origin; it sends those four points to (−1,0), (0,−1), (−1,−1), (−1,1), disagreeing with each reflection at that reflection's point. By the definition of equality of functions, R180 is none of these reflections; it is the point inversion, that is the half-turn." }, "reverse_audit": { "kind": "premise_audit", "required_ids": [ "G2", "function_def" ], "explanation": { "zh-Hant": "兩函數不等只需一點取值不同；本步不宣稱 R180 與鏡射在其他點都不同。", "en": "Two functions differ as soon as their values differ at one point; the step does not claim they differ at every other point." } }, "leaf_basis": "function_def + numeric_evaluation + two_nonzero + zero_product" }, { "id": "S09", "uses": [ "S02", "G2" ], "before_tex": "H_c(Q)-H_c(P)", "after_tex": "(c\\Delta x,\\ c\\Delta y),\\quad (c\\Delta x)^2+(c\\Delta y)^2=c^2(\\Delta x^2+\\Delta y^2)", "relation": "equivalent", "rule": "distributive", "conditions": [], "explanation": { "zh-Hant": "Hc(Q)−Hc(P)=(cx2−cx1, cy2−cy1)=(cΔx, cΔy)（分配律）。平方和為 c²Δx²+c²Δy²=c²(Δx²+Δy²)（(ab)²=a²b² 與分配律；N4 核對）。", "en": "Hc(Q)−Hc(P)=(cx2−cx1, cy2−cy1)=(cΔx, cΔy) (distributivity). The sum of squares is c²Δx²+c²Δy²=c²(Δx²+Δy²) ((ab)²=a²b² and distributivity; N4 checks this)." }, "reverse_audit": { "kind": "inverse_equivalence", "required_ids": [ "distributive", "power_laws" ], "explanation": { "zh-Hant": "恆等變形，雙向成立；對 c 沒有條件。", "en": "Identities, valid both ways; no condition on c." } }, "leaf_basis": "distributive + power_laws + mul_comm" }, { "id": "S10", "uses": [ "S09", "S02" ], "before_tex": "d(H_c(P),H_c(Q))=\\sqrt{c^2(\\Delta x^2+\\Delta y^2)}", "after_tex": "|c|\\sqrt{\\Delta x^2+\\Delta y^2}=|c|\\,d(P,Q)", "relation": "equivalent", "rule": "L_sqrt_scale", "conditions": [ "S=\\Delta x^2+\\Delta y^2\\ge 0" ], "explanation": { "zh-Hant": "令 S=Δx²+Δy²≥0（平方和非負）。依引理 L_sqrt_scale，√(c²S)=|c|√S，所以 d(Hc(P),Hc(Q))=|c|d(P,Q)。重點是乘的是 |c|，不是 c：距離不會是負數。引理在 S10.1 至 S10.3 證明。", "en": "Let S=Δx²+Δy²≥0 (a sum of squares is nonnegative). By Lemma L_sqrt_scale, √(c²S)=|c|√S, so d(Hc(P),Hc(Q))=|c|d(P,Q). The factor is |c|, not c: a distance is never negative. The lemma is proved in S10.1 to S10.3." }, "reverse_audit": { "kind": "inverse_equivalence", "required_ids": [ "L_sqrt_scale", "square_nonneg" ], "explanation": { "zh-Hant": "引理給出兩式相等，雙向成立；前提 S≥0 由平方非負保證。", "en": "The lemma gives an equality, valid both ways; its hypothesis S≥0 follows from squares being nonnegative." } }, "leaf_basis": "sqrt_def + abs_rules + power_laws + order_basics + square_nonneg" }, { "id": "S10.1", "uses": [ "abs_rules", "sqrt_def", "order_basics" ], "before_tex": "|c|\\ge 0,\\ \\sqrt S\\ge 0", "after_tex": "|c|\\sqrt S\\ge 0", "relation": "implies", "rule": "order_basics", "conditions": [ "S\\ge 0" ], "explanation": { "zh-Hant": "|c|≥0（絕對值非負），√S≥0（主平方根非負）；兩個非負數之積非負（若其一為 0 積為 0，否則正乘正為正）。", "en": "|c|≥0 (absolute values are nonnegative) and √S≥0 (the principal root is nonnegative); a product of two nonnegative numbers is nonnegative (if one is 0 the product is 0, otherwise positive times positive is positive)." }, "reverse_audit": { "kind": "premise_audit", "required_ids": [ "abs_rules", "sqrt_def", "order_basics" ], "explanation": { "zh-Hant": "只宣稱非負，不宣稱為正（c=0 或 S=0 時為 0）。", "en": "Only nonnegativity is claimed, not positivity (it is 0 when c=0 or S=0)." } }, "leaf_basis": "abs_rules + sqrt_def + order_basics + mul_zero + case_analysis" }, { "id": "S10.2", "uses": [ "abs_rules", "neg_mul", "sqrt_def", "power_laws" ], "before_tex": "(|c|\\sqrt S)^2", "after_tex": "|c|^2(\\sqrt S)^2=c^2S", "relation": "equivalent", "rule": "power_laws", "conditions": [ "S\\ge 0" ], "explanation": { "zh-Hant": "(ab)²=a²b² 給出 |c|²(√S)²。|c|²=c²：若 c≥0，|c|=c；若 c<0，|c|=−c，(−c)²=c²（負負得正）。(√S)²=S 由主平方根定義。所以 (|c|√S)²=c²S。", "en": "(ab)²=a²b² gives |c|²(√S)². Also |c|²=c²: if c≥0 then |c|=c; if c<0 then |c|=−c and (−c)²=c² (minus times minus is plus). (√S)²=S by the definition of the principal root. So (|c|√S)²=c²S." }, "reverse_audit": { "kind": "inverse_equivalence", "required_ids": [ "power_laws", "abs_rules", "neg_mul", "sqrt_def" ], "explanation": { "zh-Hant": "每一步都是等式代換，雙向成立；|c|²=c² 的兩個情況涵蓋所有實數 c（N10 只是有限抽查）。", "en": "Each step replaces equals by equals and holds both ways; the two cases for |c|²=c² cover every real c (N10 is only a finite spot check)." } }, "leaf_basis": "power_laws + abs_rules + neg_mul + sqrt_def + case_analysis" }, { "id": "S10.3", "uses": [ "S10.1", "S10.2", "sqrt_def" ], "before_tex": "b=|c|\\sqrt S\\ge 0,\\ b^2=c^2S", "after_tex": "\\sqrt{c^2S}=|c|\\sqrt S", "relation": "implies", "rule": "sqrt_def", "conditions": [ "c^2S\\ge 0" ], "explanation": { "zh-Hant": "c²S≥0，所以 √(c²S) 是唯一滿足 b≥0 且 b²=c²S 的實數 b。由 S10.1、S10.2，b=|c|√S 滿足這兩個條件，依唯一性 √(c²S)=|c|√S。S10.1 至 S10.3 合起來證得引理 L_sqrt_scale。", "en": "c²S≥0, so √(c²S) is the unique real b with b≥0 and b²=c²S. By S10.1 and S10.2, b=|c|√S satisfies both conditions, so by uniqueness √(c²S)=|c|√S. S10.1 to S10.3 together prove Lemma L_sqrt_scale." }, "reverse_audit": { "kind": "premise_audit", "required_ids": [ "S10.1", "S10.2", "sqrt_def" ], "explanation": { "zh-Hant": "關鍵是唯一性：只驗 b²=c²S 不夠，還需 b≥0；c<0 且 S>0 時，c√S 雖平方正確卻為負，不是主平方根；S=0 時兩式都為0。", "en": "Uniqueness is the key: b²=c²S alone is not enough, b≥0 is also needed; when c<0 and S>0, c√S has the right square but is negative and is not the principal root. When S=0, both expressions are 0." } }, "leaf_basis": "sqrt_def + abs_rules + power_laws + order_basics" }, { "id": "S11", "uses": [ "S10", "G2" ], "before_tex": "c=-1:\\ d(H_{-1}(P),H_{-1}(Q))=|-1|\\,d(P,Q)", "after_tex": "H_{-1}=R_{180}\\ \\text{is an isometry}", "relation": "implies", "rule": "forall_intro", "conditions": [], "explanation": { "zh-Hant": "|−1|=1，所以對任意 P、Q 距離不變，H(−1) 是等距；而且 H(−1)(x,y)=(−x,−y) 正是半轉 R180，與 S07 一致。這說明負係數不一定破壞等距。", "en": "|−1|=1, so distance is unchanged for arbitrary P and Q and H(−1) is an isometry; indeed H(−1)(x,y)=(−x,−y) is exactly the half-turn R180, consistent with S07. This shows that a negative coefficient need not destroy isometry." }, "reverse_audit": { "kind": "premise_audit", "required_ids": [ "S10", "abs_rules", "forall_intro" ], "explanation": { "zh-Hant": "用 S10 的一般等式代入 c=−1；P、Q 任意，所以是全平面結論。", "en": "Substitute c=−1 into the general equality of S10; P and Q are arbitrary, so the conclusion is for the whole plane." } }, "leaf_basis": "abs_rules + forall_intro + numeric_evaluation" }, { "id": "S12", "uses": [ "S10", "G3" ], "before_tex": "c=-2,\\ P=(3,1),\\ Q=(-1,4)", "after_tex": "d(P,Q)=5,\\ d(H_{-2}(P),H_{-2}(Q))=10=|-2|\\cdot 5", "relation": "implies", "rule": "numeric_evaluation", "conditions": [], "explanation": { "zh-Hant": "數值核對：Q−P=(−4,3)，d(P,Q)=√(16+9)=5。H(−2)(P)=(−6,−2)、H(−2)(Q)=(2,−8)，差為 (8,−6)，距離 √(64+36)=10=2·5。倍率是 |−2|=2，與 S10 一致；不是 −2，也不是 −10。這只是一個例子，一般結論來自 S10。", "en": "Numerical check: Q−P=(−4,3), d(P,Q)=√(16+9)=5. H(−2)(P)=(−6,−2) and H(−2)(Q)=(2,−8), with difference (8,−6) and distance √(64+36)=10=2·5. The scale is |−2|=2, as S10 says; it is not −2, and the distance is not −10. This is one example; the general statement comes from S10." }, "reverse_audit": { "kind": "premise_audit", "required_ids": [ "S10", "G3", "numeric_evaluation" ], "explanation": { "zh-Hant": "可重算（N5、N6、N7）；一個例子不能取代 S10 的一般證明。", "en": "It can be recomputed (N5, N6, N7); one example cannot replace the general proof in S10." } }, "leaf_basis": "numeric_evaluation + dist_formula" }, { "id": "S13", "uses": [ "S12", "S01" ], "before_tex": "c=2:\\ H_2(0,0)=(0,0),\\ d(H_2(P),H_2(Q))=10\\neq 5=d(P,Q)", "after_tex": "H_2\\ \\text{is not an isometry}", "relation": "implies", "rule": "quantifier_negation", "conditions": [], "explanation": { "zh-Hant": "|2|=|−2|，所以同一對 P、Q 在 H2 下距離也是 10≠5。等距要求對所有點對成立，否定它只需一個反例：¬∀ 等於 ∃¬。雖然 H2 固定原點（一個不變點），這不能證明全平面等距，MKO 假設 a3 正是這個警告。", "en": "|2|=|−2|, so the same pair P, Q has image distance 10≠5 under H2 too. Isometry requires every pair, and one counterexample refutes it: ¬∀ is ∃¬. Although H2 fixes the origin (a fixed point), that cannot establish whole-plane isometry, which is exactly the warning of MKO assumption a3." }, "reverse_audit": { "kind": "premise_audit", "required_ids": [ "S12", "S01", "quantifier_negation" ], "explanation": { "zh-Hant": "一個反例足以否定全稱命題；反過來，一個不變點不足以肯定它。", "en": "One counterexample suffices to refute a universal statement; conversely, one fixed point does not suffice to establish it." } }, "leaf_basis": "quantifier_negation + isometry_def + numeric_evaluation" }, { "id": "S14", "uses": [ "S10", "G3", "S12" ], "before_tex": "c=0:\\ H_0(x,y)=(0,0)", "after_tex": "d(H_0(P),H_0(Q))=0\\neq 5:\\ \\text{degenerate constant map, not an isometry}", "relation": "implies", "rule": "quantifier_negation", "conditions": [], "explanation": { "zh-Hant": "c=0 時每一點都送到原點，是常值映射。對相異的 P、Q，d(P,Q)=5>0 但像的距離為 0。倍率 |0|=0 不是正數，所以它既不是等距，也不是 MKO 意義下的正倍率相似映射，而是退化映射。", "en": "With c=0 every point goes to the origin; the map is constant. For the distinct P and Q, d(P,Q)=5>0 but the image distance is 0. The factor |0|=0 is not positive, so the map is neither an isometry nor a positive-scale similarity in the MKO's sense; it is degenerate." }, "reverse_audit": { "kind": "premise_audit", "required_ids": [ "S10", "G3", "quantifier_negation" ], "explanation": { "zh-Hant": "一對相異點即足以否定等距；退化的判斷來自倍率 0 不是正數。", "en": "One pair of distinct points refutes isometry; degeneracy comes from the factor 0 not being positive." } }, "leaf_basis": "quantifier_negation + numeric_evaluation" }, { "id": "S15", "uses": [ "S10", "S11", "S13", "S14", "S06", "S07" ], "before_tex": "d(H_c(P),H_c(Q))=|c|\\,d(P,Q)\\ \\forall P,Q", "after_tex": "c=0:\\ \\text{degenerate};\\ |c|=1:\\ \\text{isometry};\\ \\text{other }c\\neq 0:\\ \\text{similarity of scale }|c|,\\ \\text{not an isometry}", "relation": "case_split", "rule": "case_analysis", "conditions": [], "explanation": { "zh-Hant": "分三種情況涵蓋所有實數 c。c=0：S14，退化常值映射。|c|=1（c=1 恆等、c=−1 半轉）：S10 給出距離不變，等距。其他非零 c：S10 給出每個距離乘正數 |c|，所以是倍率 |c| 的相似映射；又 |c|≠1，取 P=(0,0)、Q=(1,0)，d(P,Q)=1 而像的距離 |c|≠1，所以不是等距。這與 MKO 的分類一致。", "en": "Three cases cover every real c. c=0: S14, a degenerate constant map. |c|=1 (c=1 the identity, c=−1 the half-turn): S10 gives unchanged distances, an isometry. Any other nonzero c: S10 multiplies every distance by the positive |c|, so the map is a similarity of scale |c|; and as |c|≠1, the pair P=(0,0), Q=(1,0) has d(P,Q)=1 but image distance |c|≠1, so it is not an isometry. This agrees with the MKO's classification." }, "reverse_audit": { "kind": "sufficient_goal_reduction", "required_ids": [ "S10", "S11", "S14", "case_analysis" ], "explanation": { "zh-Hant": "目標 G：MKO 的三分類。足夠條件 H：S10 的一般等式加上三個情況的分析；三個情況（c=0、|c|=1、其他非零 c）互斥且涵蓋全部實數。H 由 S10 正向證成，未把 G 當作前提。", "en": "Goal G: the MKO's three-way classification. Sufficient condition H: the general equality of S10 plus the analysis of three cases; the cases c=0, |c|=1 and other nonzero c are exclusive and cover all reals. H is proved forwards from S10, without assuming G." } }, "leaf_basis": "case_analysis + abs_rules + quantifier_negation + forall_intro" } ], "result": { "zh-Hant": "平移、R0、R90、R180、R270 與關於 x 軸、y 軸、y=x、y=−x 的鏡射，對任意兩點都保持距離，所以都是等距映射；半轉 (−x,−y) 只固定原點，不是所列四條軸中任何一條的鏡射。均勻縮放滿足 d(Hc(P),Hc(Q))=|c|d(P,Q)：c=0 是退化常值映射；c=±1 是等距（c=−1 即半轉）；其他非零 c（例如 c=2、c=−2）是倍率 |c| 的非等距相似映射，P=(3,1)、Q=(−1,4) 的距離 5 變成 10。H2 固定原點也不改變這個結論。", "en": "Translations, R0, R90, R180, R270 and the reflections in the x-axis, the y-axis, y=x and y=−x preserve the distance of any two points, so they are isometries; the half-turn (−x,−y) fixes only the origin and is not a reflection in any of these axes. Uniform scaling satisfies d(Hc(P),Hc(Q))=|c|d(P,Q): c=0 is a degenerate constant map; c=±1 are isometries (c=−1 is the half-turn); every other nonzero c, for example c=2 or c=−2, is a non-isometric similarity of scale |c|, turning the distance 5 of P=(3,1), Q=(−1,4) into 10. That H2 fixes the origin does not change this." }, "negative_controls": [ { "claim": { "zh-Hant": "H(−2) 的長度倍率是 −2", "en": "H(−2) has length scale −2" }, "expected": "reject", "reason": { "zh-Hant": "距離乘 |−2|=2；P、Q 的距離由 5 變 10（S10、S12）。", "en": "Distances are multiplied by |−2|=2; the distance of P and Q goes from 5 to 10 (S10, S12)." } }, { "claim": { "zh-Hant": "H2 固定原點，所以 H2 是等距映射", "en": "H2 fixes the origin, so H2 is an isometry" }, "expected": "reject", "reason": { "zh-Hant": "d(H2(P),H2(Q))=10≠5=d(P,Q)。", "en": "d(H2(P),H2(Q))=10≠5=d(P,Q)." } }, { "claim": { "zh-Hant": "半轉 (x,y)↦(−x,−y) 就是關於 x 軸的鏡射", "en": "The half-turn (x,y)↦(−x,−y) is the reflection in the x-axis" }, "expected": "reject", "reason": { "zh-Hant": "x 軸鏡射把 (1,0) 送到 (1,0)，半轉把它送到 (−1,0)。", "en": "The x-axis reflection sends (1,0) to (1,0); the half-turn sends it to (−1,0)." } }, { "claim": { "zh-Hant": "係數為負的均勻縮放都不是等距", "en": "No uniform scaling with a negative coefficient is an isometry" }, "expected": "reject", "reason": { "zh-Hant": "c=−1 時 |c|=1，H(−1) 是半轉，是等距（S11）。", "en": "For c=−1, |c|=1 and H(−1) is the half-turn, an isometry (S11)." } }, { "claim": { "zh-Hant": "繞原點逆時針 90° 把 (1,0) 送到 (0,−1)", "en": "The counterclockwise quarter-turn about the origin sends (1,0) to (0,−1)" }, "expected": "reject", "reason": { "zh-Hant": "R90(1,0)=(−0,1)=(0,1)；(0,−1) 是順時針（R270）的像。", "en": "R90(1,0)=(−0,1)=(0,1); (0,−1) is the clockwise image (R270)." } } ], "status": "candidate", "gaps": [], "verification": { "structure": "checked_local_script", "mathematics": "candidate_argument", "formalization": "not_formalized" }, "rule_glossary": [ { "id": "neg_mul", "kind": "prerequisite", "statement": { "zh-Hant": "負號與乘法：(−u)v=−(uv)，(−u)(−v)=uv。", "en": "Signs and multiplication: (−u)v=−(uv) and (−u)(−v)=uv." } }, { "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": "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": "abs_rules", "kind": "prerequisite", "statement": { "zh-Hant": "絕對值（實數）：|u|=u（u≥0）、|u|=−u（u<0）；|u|≥0；|uv|=|u||v|；三角不等式 |u+v|≤|u|+|v|；|u|v 之一；u0 推出 uwvw。", "en": "Order of the reals: exactly one of uv holds; u0 give uwvw." } }, { "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": "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": "add_assoc", "kind": "prerequisite", "statement": { "zh-Hant": "加法結合律：(u+v)+w=u+(v+w)。", "en": "Associativity of addition: (u+v)+w=u+(v+w)." } }, { "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": "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": "add_zero", "kind": "prerequisite", "statement": { "zh-Hant": "加法單位元：u+0=u。", "en": "Additive identity: u+0=u." } }, { "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": "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": "case_analysis", "kind": "prerequisite", "statement": { "zh-Hant": "分情況論證：若 P∨Q 成立，且 P⇒R、Q⇒R 都已證，則 R 成立。所有情況必須涵蓋全部可能。", "en": "Proof by cases: if P∨Q holds and both P⇒R and Q⇒R are proved, then R holds. The cases must cover every possibility." } }, { "id": "forall_intro", "kind": "prerequisite", "statement": { "zh-Hant": "∀引入（全稱推廣）：若某命題對一個「任意的」元素 x 證成，而證明沒有對 x 加任何額外假設，則它對論域中所有 x 成立。", "en": "∀-introduction (generalization): if a statement is proved for an arbitrary element x, with no extra assumption on x, it holds for every x in the domain." } }, { "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": "two_nonzero", "kind": "prerequisite", "statement": { "zh-Hant": "實數中 0<1，故 0<1+1=2，所以 2≠0。", "en": "In the reals 0<1, hence 0<1+1=2, so 2≠0." } }, { "id": "zero_product", "kind": "prerequisite", "statement": { "zh-Hant": "零乘積性質：在 ℤ、ℚ、ℝ、ℂ（或任何整域）中，uv=0 ⇔ u=0 或 v=0。", "en": "Zero-product property: in ℤ, ℚ, ℝ, ℂ (or any integral domain), uv=0 ⇔ u=0 or v=0." } }, { "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": "square_nonneg", "kind": "prerequisite", "statement": { "zh-Hant": "實數平方非負：對所有實數 u，u²≥0，且 u²=0 ⇔ u=0。", "en": "Squares are nonnegative: for every real u, u²≥0, and u²=0 ⇔ u=0." } }, { "id": "mul_zero", "kind": "prerequisite", "statement": { "zh-Hant": "乘以零：u·0=0。理由：u·0=u·(0+0)=u·0+u·0，兩邊同減 u·0 得 0=u·0。", "en": "Multiplying by zero: u·0=0. Reason: u·0=u·(0+0)=u·0+u·0; subtracting u·0 from both sides gives 0=u·0." } }, { "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": "mko_transformations", "kind": "cited", "source": { "mko_id": "mko-geometric-transformations", "sha256": "13176ddfbd967892818086ef2859a68ada71d09c120c932f688634fbec64e489", "commit": "b28a5edbd46279b1db2a3f3f48bb3f447e18899f" }, "statement": { "zh-Hant": "在等單位正交實平面，平移、原點逆時針 0／90／180／270 度旋轉及關於 x 軸、y 軸、y=x、y=−x 的鏡射均為等距映射。均勻縮放 Hc(x,y)=(cx,cy)：c=0 是退化常值映射，|c|=1 是全平面等距，其他非零 c 是正倍率 |c| 的非等距相似映射。", "en": "On the equal-unit orthonormal real plane, translations, counterclockwise rotations about the origin by 0°, 90°, 180°, 270°, and reflections across the x-axis, y-axis, y=x and y=−x are isometries. For Hc(x,y)=(cx,cy), c=0 gives a degenerate constant map; |c|=1 gives a whole-plane isometry; any other nonzero c gives a non-isometric similarity with positive length scale |c|." } }, { "id": "dist_formula", "kind": "cited", "source": { "mko_id": "mko-distance-formula", "sha256": "153d0d0feabadaf33e08081677be0f81709df28ffa5bffaf0176a0f7875e5904", "commit": "b28a5edbd46279b1db2a3f3f48bb3f447e18899f" }, "statement": { "zh-Hant": "在同一等單位正交座標系，P=(x1,y1)、Q=(x2,y2) 的距離為 d(P,Q)=√((x2−x1)²+(y2−y1)²)≥0。", "en": "In one equal-unit orthonormal frame, P=(x1,y1) and Q=(x2,y2) have distance d(P,Q)=√((x2−x1)²+(y2−y1)²)≥0." } }, { "id": "isometry_def", "kind": "prerequisite", "statement": { "zh-Hant": "映射 H:ℝ²→ℝ² 是等距映射，指對所有點 P、Q 都有 d(H(P),H(Q))=d(P,Q)。這是全平面性質：要對每一對點成立。", "en": "A map H:ℝ²→ℝ² is an isometry when d(H(P),H(Q))=d(P,Q) for all points P and Q. It is a whole-plane property: it must hold for every pair of points." } }, { "id": "L_signed_perm", "kind": "lemma", "proved_in": [ "S05.1" ], "statement": { "zh-Hant": "若 H(Q)−H(P) 的兩個坐標是 ±Δx 與 ±Δy（任意正負號，順序可對調），則 d(H(P),H(Q))=d(P,Q)。", "en": "If the two coordinates of H(Q)−H(P) are ±Δx and ±Δy (any signs, in either order), then d(H(P),H(Q))=d(P,Q)." } }, { "id": "L_sqrt_scale", "kind": "lemma", "proved_in": [ "S10.1", "S10.2", "S10.3" ], "statement": { "zh-Hant": "對實數 c 與 S≥0：√(c²S)=|c|√S。", "en": "For a real c and S≥0: √(c²S)=|c|√S." } } ], "revision_log": [], "numeric_checks": [ { "id": "N1", "step": "S03", "vars": [ "x1", "x2", "t" ], "degree": 1, "lhs": "(x2+t)-(x1+t)", "rhs": "x2-x1" }, { "id": "N2", "step": "S04", "vars": [ "y1", "y2" ], "degree": 1, "lhs": "(-y2)-(-y1)", "rhs": "-(y2-y1)" }, { "id": "N3", "step": "S05.1", "vars": [ "a", "b" ], "degree": 2, "lhs": "(-a)**2+(-b)**2", "rhs": "a**2+b**2" }, { "id": "N4", "step": "S09", "vars": [ "c", "x1", "x2", "y1", "y2" ], "degree": 2, "lhs": "(c*x2-c*x1)**2+(c*y2-c*y1)**2", "rhs": "c**2*((x2-x1)**2+(y2-y1)**2)" }, { "id": "N5", "step": "S12", "lhs": "(-1-3)**2+(4-1)**2", "rhs": "25" }, { "id": "N6", "step": "S12", "lhs": "(2-(-6))**2+(-8-(-2))**2", "rhs": "100" }, { "id": "N7", "step": "S12", "lhs": "100", "rhs": "abs(-2)**2*25" }, { "id": "N8", "step": "S13", "lhs": "abs(2)*5", "relation": "!=", "rhs": "5" }, { "id": "N9", "step": "S14", "lhs": "(0*(-1)-0*3)**2+(0*4-0*1)**2", "rhs": "0" }, { "id": "N10", "step": "S10.2", "vars": [ "c" ], "range": [ -5, 5 ], "lhs": "abs(c)**2", "rhs": "c**2" }, { "id": "N11", "step": "S08", "lhs": "-1", "relation": "!=", "rhs": "1" }, { "id": "N12", "step": "S11", "lhs": "abs(-1)", "rhs": "1" }, { "id": "N13", "step": "S15", "lhs": "abs(3)", "relation": "!=", "rhs": "1" }, { "id": "N14", "step": "S07", "vars": [ "x1", "x2", "y1", "y2" ], "degree": 2, "lhs": "(y2-y1)**2+(-(x2-x1))**2", "rhs": "(x2-x1)**2+(y2-y1)**2" } ] }, "presentation": { "md_scope": { "zh-Hant": "只處理 MKO 列出的映射：平移、繞原點逆時針 0°、90°、180°、270° 旋轉、四條指定軸的鏡射、均勻縮放；距離用等單位正交座標的歐氏距離。不處理一般旋轉角、一般直線鏡射或等距映射的完整分類。不改動 canonical MKO 或英文 edition。", "en": "Treats only the maps listed in the MKO: translations, counterclockwise rotations about the origin by 0°, 90°, 180°, 270°, reflections in the four specified lines, and uniform scaling, with Euclidean distance in an equal-unit orthonormal frame. It does not treat general rotation angles, reflections in general lines, or the full classification of isometries. It does not change the canonical MKO or its English edition." }, "md_symbols": [ [ "d(P,Q)", "P、Q 的歐氏距離", "the Euclidean distance between P and Q" ], [ "\\Delta x,\\ \\Delta y", "Q 與 P 的座標差 x2−x1、y2−y1", "the coordinate differences x2−x1 and y2−y1 of Q and P" ], [ "R_{90}", "繞原點逆時針 90° 旋轉 (x,y)↦(−y,x)", "counterclockwise quarter-turn about the origin, (x,y)↦(−y,x)" ], [ "M", "關於 y=−x 的鏡射 (x,y)↦(−y,−x)", "reflection in y=−x, (x,y)↦(−y,−x)" ], [ "H_c", "均勻縮放 (x,y)↦(cx,cy)", "uniform scaling (x,y)↦(cx,cy)" ] ], "md_error_controls": [ { "title": "負係數不是負長度倍率 / A negative coefficient is not a negative length scale", "zh": "MKO 的誤解之一是把負係數當負長度倍率。S10 證明距離乘的是 |c|：c=−2 的倍率是 2（距離 5 變 10，S12），c=−1 甚至是等距（S11）。在固定 |c| 下，係數的負號額外帶來半轉效果，不改變相對於 +|c| 的長度倍率；完整係數 c=−2 仍把長度放大 2 倍。距離不會因負號變成負數。", "en": "One misconception the MKO lists is reading a negative coefficient as a negative length scale. S10 proves that distances are multiplied by |c|: c=−2 has scale 2 (distance 5 becomes 10, S12), and c=−1 is even an isometry (S11). At fixed |c|, the negative sign adds a half-turn without changing the length factor relative to +|c|. The complete coefficient c=−2 still doubles lengths; its sign never makes a distance negative." }, { "title": "一個不變點不推出全域等距 / One fixed point does not give global isometry", "zh": "H2 固定原點，但 d(H2(P),H2(Q))=10≠5。等距是全平面的全稱性質（S01）：肯定它要對任意點對證明（S06 的作法），否定它只需一個反例（S13）。從一個重合點或不變點推論等距是 MKO 假設 a3 明言禁止的。", "en": "H2 fixes the origin, yet d(H2(P),H2(Q))=10≠5. Isometry is a universal whole-plane property (S01): establishing it needs an argument for an arbitrary pair (as in S06), and refuting it needs one counterexample (S13). Inferring isometry from one coincident or fixed point is exactly what MKO assumption a3 forbids." }, { "title": "旋轉方向、中心與鏡射軸要明示 / State the rotation direction, centre and mirror axis", "zh": "逆時針 90° 是 (x,y)↦(−y,x)，把 (1,0) 送到 (0,1)；順時針 90° 則是 (y,−x)，即本稿的 R270，把 (1,0) 送到 (0,−1)。半轉 (−x,−y) 不是 x 軸鏡射：x 軸鏡射固定 (1,0)，半轉把它送到 (−1,0)（S08）。混用方向或軸會得到錯誤的像點，即使它仍是等距。", "en": "The counterclockwise quarter-turn is (x,y)↦(−y,x), sending (1,0) to (0,1); the clockwise one is (y,−x), which is R270 here and sends (1,0) to (0,−1). The half-turn (−x,−y) is not the x-axis reflection: that reflection fixes (1,0) while the half-turn sends it to (−1,0) (S08). Mixing up directions or axes gives wrong image points even though the map is still an isometry." } ], "md_open_items": [ { "zh-Hant": "本稿只證明 MKO 列出的有限種公式或映射族（平移參數仍可為任意實數）是等距；不處理一般角度旋轉、一般直線鏡射或平面等距的完整分類。", "en": "This guide proves isometry only for the finitely many formula types or map families listed in the MKO (translation parameters still range over arbitrary reals); it does not treat rotations by general angles, reflections in general lines, or the complete classification of plane isometries." }, { "zh-Hant": "N10 是 |c|²=c² 在 −5 到 5 的整數上的有限抽查；一般證明在 S10.2 用兩個情況完成。", "en": "N10 is a finite spot check of |c|²=c² at the integers from −5 to 5; the general proof is the two-case argument of S10.2." } ] }, "revision": { "id": "production-20261019-adopted-96", "source_commit": "b28a5edbd46279b1db2a3f3f48bb3f447e18899f", "adopted_source_sha256": "b2381974fe84d166a9e1c4c15b5317ce2f5a357d19a292ebdd8111cd97ee96cf", "original_source_sha256": "e7f90378f82483f1ef33bf0f1ac86bbee7369e81b121701b1725f67cb43d1d8d", "owner_adopted": true, "bilingual_sync": "complete_source_bound", "formal_Lean": "NOT_RUN", "independent_AI_concurrence": "NOT_MEASURED", "issue_ids": [ "BR-M022", "BR-M023", "BR-M024" ], "scope": "Owner-adopted natural-language teaching guide; not a formal proof, canonical MKO rewrite or evidence-status promotion." } }