{ "schema_version": "ocme-derivation-reader-v1", "guide_id": "derivation-mko-polynomial-01", "mko_id": "mko-polynomial", "title": { "zh-Hant": "一元有理係數多項式：合併同次項、係數向量、次數與零多項式", "en": "Univariate rational polynomials: combining like terms, coefficient vectors, degree and the zero polynomial" }, "contract": { "schema_version": "ocme-derivation-guide-v0.1", "guide_id": "derivation-mko-polynomial-01", "source": { "mko_id": "mko-polynomial", "commit": "b28a5edbd46279b1db2a3f3f48bb3f447e18899f", "sha256": "0a0d76b12363d10d9ecbe8b45d7bebb1136637db1c41e6308481c675c8db9e6c", "english_sha256": "4c5e7d9539a5612d401321e99d2e0f4667c558a65abb5f412bf6cb3518f9fbc8" }, "mode": "definition_walkthrough", "audience_prerequisites": [ "add_assoc", "add_comm", "distributive", "finite_sum_rules", "numeric_evaluation" ], "goal": { "zh-Hant": "在 ℚ[x] 中，設 p(x)=4x³+3x²−x+2−4x³+x，q 的係數向量為 [−1,1,0,0]，r(x)=x²−x²+3，z(x)=5x−5x，t(x)=5x。求 p、q、r 的係數與次數，判定 z 是零多項式並區分 z 與 r；以係數計算 p+q 與 p·q 並求次數；最後說明 p(1)=t(1) 不能推出 p=t。", "en": "In ℚ[x], let p(x)=4x³+3x²−x+2−4x³+x, let q have coefficient vector [−1,1,0,0], and let r(x)=x²−x²+3, z(x)=5x−5x and t(x)=5x. Find the coefficients and degrees of p, q and r; show that z is the zero polynomial and distinguish it from r; compute p+q and p·q from coefficients and find their degrees; finally explain why p(1)=t(1) does not give p=t." }, "givens": [ { "id": "G1", "statement": { "zh-Hant": "固定形式變數 x；ℚ[x] 的多項式由有理係數 a_k（k∈ℕ）給出，只有有限個非零，p(x)=Σ_k a_k x^k；係數向量由低次到高次列出，超出向量的係數為 0，尾端零不改變多項式（MKO 假設 a1、a2）。", "en": "Fix a formal variable x; a polynomial in ℚ[x] is given by rational coefficients a_k (k∈ℕ), only finitely many nonzero, with p(x)=Σ_k a_k x^k; coefficient vectors list coefficients from low to high power, coefficients beyond a vector are 0, and trailing zeros do not change the polynomial (MKO assumptions a1, a2)." } }, { "id": "G2", "statement": { "zh-Hant": "p(x)=4x³+3x²−x+2−4x³+x；q 的係數向量 [−1,1,0,0]；r(x)=x²−x²+3；z(x)=5x−5x；t(x)=5x。", "en": "p(x)=4x³+3x²−x+2−4x³+x; q has coefficient vector [−1,1,0,0]; r(x)=x²−x²+3; z(x)=5x−5x; t(x)=5x." } } ], "steps": [ { "id": "S01", "uses": [ "G1" ], "before_tex": "p\\in\\mathbb{Q}[x]", "after_tex": "p=\\sum_k a_k x^k,\\quad p=q\\iff a_k=b_k\\ \\text{for all}\\ k,\\quad \\deg p=\\max\\{k\\mid a_k\\neq 0\\}\\ (p\\neq 0)", "relation": "definition", "rule": "mko_polynomial", "conditions": [ "\\text{finitely many}\\ a_k\\neq 0" ], "explanation": { "zh-Hant": "依 MKO，多項式就是它的係數資料：相等逐次比較係數，次數是係數非零的最大指數，零多項式（所有係數為 0）的次數在本條目記為未定義。因為只有有限個係數非零，最大值存在。之後的判斷都由係數決定，不由字面上寫了哪些項決定。", "en": "By the MKO, a polynomial is its coefficient data: equality compares coefficients power by power, the degree is the largest index with a nonzero coefficient, and the zero polynomial (all coefficients 0) has undefined degree in this entry. Since only finitely many coefficients are nonzero, the maximum exists. Every later decision is made from the coefficients, not from which terms happen to be written." }, "reverse_audit": { "kind": "premise_audit", "required_ids": [ "G1", "mko_polynomial", "poly_eq_coeff" ], "explanation": { "zh-Hant": "此步只寫出定義；「最大值存在」用到有限支撐，零多項式被明確排除在次數定義之外。", "en": "This step only states the definitions; the existence of the maximum uses finite support, and the zero polynomial is explicitly excluded from the degree definition." } }, "leaf_basis": "mko_polynomial + poly_eq_coeff + def_intro" }, { "id": "S02", "uses": [ "G2", "add_comm", "add_assoc" ], "before_tex": "4x^3+3x^2-x+2-4x^3+x", "after_tex": "(4x^3-4x^3)+3x^2+(-x+x)+2", "relation": "equivalent", "rule": "add_comm", "conditions": [], "explanation": { "zh-Hant": "依加法交換律與結合律，把 p 的各項按 x 的次方重新排列分組：x³ 項放在一起，x 項放在一起。此步只重排，還沒有合併。", "en": "By commutativity and associativity of addition, rearrange the terms of p into groups by power of x: the x³ terms together and the x terms together. This step only rearranges; nothing is combined yet." }, "reverse_audit": { "kind": "inverse_equivalence", "required_ids": [ "add_comm", "add_assoc" ], "explanation": { "zh-Hant": "重排與還原都只用交換律與結合律，兩式是同一個多項式。", "en": "Rearranging and restoring use only commutativity and associativity, so both forms are the same polynomial." } }, "leaf_basis": "add_comm + add_assoc" }, { "id": "S03", "uses": [ "S02", "like_terms" ], "before_tex": "(4x^3-4x^3)+3x^2+(-x+x)+2", "after_tex": "0\\cdot x^3+3x^2+0\\cdot x+2", "relation": "equivalent", "rule": "like_terms", "conditions": [], "explanation": { "zh-Hant": "合併同次項：4x³−4x³=(4−4)x³=0·x³，−x+x=(−1+1)x=0·x。這正是 MKO 要求「相同次冪的係數要先相加」。", "en": "Combine like terms: 4x³−4x³=(4−4)x³=0·x³ and −x+x=(−1+1)x=0·x. This is the MKO's requirement that coefficients of the same power be added first." }, "reverse_audit": { "kind": "inverse_equivalence", "required_ids": [ "like_terms", "distributive", "numeric_evaluation" ], "explanation": { "zh-Hant": "分配律雙向成立；4−4=0、−1+1=0 可重算（N1、N2）；N9 以次數界在格點上核對整個恆等式。", "en": "Distributivity works both ways; 4−4=0 and −1+1=0 are recomputable (N1, N2); N9 checks the whole identity on a grid with a degree bound." } }, "leaf_basis": "like_terms + distributive + numeric_evaluation" }, { "id": "S04", "uses": [ "S03", "poly_eq_coeff" ], "before_tex": "0\\cdot x^3+3x^2+0\\cdot x+2", "after_tex": "[2,0,3,0]=[2,0,3],\\quad p=3x^2+2", "relation": "equivalent", "rule": "poly_eq_coeff", "conditions": [], "explanation": { "zh-Hant": "由低次到高次讀出係數：a_0=2、a_1=0、a_2=3、a_3=0，所有 k≥4 的係數為 0。向量 [2,0,3,0] 與 [2,0,3] 在每個次方的係數都相同（索引 3 的係數在兩者中都是 0），所以表示同一個多項式 p=3x²+2。", "en": "Read the coefficients from low to high: a_0=2, a_1=0, a_2=3, a_3=0, and every coefficient with k≥4 is 0. The vectors [2,0,3,0] and [2,0,3] have the same coefficient at every power (the coefficient at index 3 is 0 in both), so they represent the same polynomial p=3x²+2." }, "reverse_audit": { "kind": "inverse_equivalence", "required_ids": [ "poly_eq_coeff" ], "explanation": { "zh-Hant": "係數逐次相同，依多項式相等的定義雙向成立；尾端零可加可去。", "en": "The coefficients agree power by power, so by the definition of polynomial equality the step is two-way; a trailing zero may be added or removed." } }, "leaf_basis": "poly_eq_coeff + def_unfold" }, { "id": "S05", "uses": [ "S01", "S04" ], "before_tex": "a_2=3\\neq 0,\\ a_k=0\\ (k\\geq 3)", "after_tex": "\\deg p=2", "relation": "implies", "rule": "mko_polynomial", "conditions": [ "p\\neq 0" ], "explanation": { "zh-Hant": "p 不是零多項式（a_2=3≠0），所以次數有定義。係數非零的指數只有 0 與 2，最大是 2，故 deg p=2。原式字面上出現 x³，但它的係數合併後是 0，不影響次數。", "en": "p is not the zero polynomial (a_2=3≠0), so its degree is defined. The indices with nonzero coefficients are 0 and 2, the largest being 2, so deg p=2. The original expression shows x³ literally, but its combined coefficient is 0 and does not affect the degree." }, "reverse_audit": { "kind": "premise_audit", "required_ids": [ "S04", "mko_polynomial" ], "explanation": { "zh-Hant": "需要先合併（S03）；若直接讀最高字面指數，會誤得 3。", "en": "The like terms must first be combined as in S03; reading the highest literal exponent would wrongly give 3." } }, "leaf_basis": "mko_polynomial + def_unfold" }, { "id": "S06", "uses": [ "G1", "G2", "S01" ], "before_tex": "q\\leftrightarrow[-1,1,0,0]", "after_tex": "q=x-1,\\quad \\deg q=1", "relation": "implies", "rule": "mko_polynomial", "conditions": [], "explanation": { "zh-Hant": "向量 [−1,1,0,0] 表示 b_0=−1、b_1=1、b_2=b_3=0，所以 q=−1+x=x−1，與 [−1,1] 是同一個多項式。係數非零的最大指數是 1，deg q=1；向量長度 4 不代表次數 3。", "en": "The vector [−1,1,0,0] means b_0=−1, b_1=1 and b_2=b_3=0, so q=−1+x=x−1, the same polynomial as [−1,1]. The largest index with a nonzero coefficient is 1, so deg q=1; the vector length 4 does not mean degree 3." }, "reverse_audit": { "kind": "premise_audit", "required_ids": [ "G1", "mko_polynomial", "poly_eq_coeff" ], "explanation": { "zh-Hant": "次數由最大非零係數決定，與向量長度或尾端零無關。", "en": "The degree is decided by the largest nonzero coefficient, independent of the vector's length or trailing zeros." } }, "leaf_basis": "mko_polynomial + poly_eq_coeff" }, { "id": "S07", "uses": [ "G2", "like_terms", "S01" ], "before_tex": "x^2-x^2+3", "after_tex": "(1-1)x^2+3=3,\\quad [3],\\quad \\deg r=0", "relation": "implies", "rule": "like_terms", "conditions": [], "explanation": { "zh-Hant": "合併 x² 項：x²−x²=(1−1)x²=0。所以 r=3，係數向量 [3]。r 是非零常數多項式（a_0=3≠0），係數非零的最大指數是 0，deg r=0。這是 MKO para0 的例子。", "en": "Combine the x² terms: x²−x²=(1−1)x²=0. So r=3, with coefficient vector [3]. r is a nonzero constant polynomial (a_0=3≠0), and the largest index with a nonzero coefficient is 0, so deg r=0. This is the example in MKO para0." }, "reverse_audit": { "kind": "premise_audit", "required_ids": [ "like_terms", "mko_polynomial" ], "explanation": { "zh-Hant": "字面上有 x²，但合併後係數為 0；次數由合併後的資料決定（N16）。", "en": "x² appears literally, but its combined coefficient is 0; the degree is read from the combined data (N16)." } }, "leaf_basis": "like_terms + mko_polynomial + numeric_evaluation" }, { "id": "S08", "uses": [ "G2", "like_terms", "S01", "S07" ], "before_tex": "5x-5x", "after_tex": "(5-5)x=0,\\quad z=0\\ \\text{(zero polynomial)},\\quad z\\neq r", "relation": "implies", "rule": "poly_eq_coeff", "conditions": [], "explanation": { "zh-Hant": "5x−5x=(5−5)x=0·x，所以 z 的所有係數都是 0，z 是零多項式，依本條目它的次數未定義。z 與 r 不同：兩者的常數項分別為 0 與 3，0≠3，依係數相等的定義 z≠r。零多項式與非零常數多項式不可混淆。", "en": "5x−5x=(5−5)x=0·x, so every coefficient of z is 0: z is the zero polynomial, and by this entry its degree is undefined. z differs from r: their constant coefficients are 0 and 3, and 0≠3, so z≠r by the definition of equality. The zero polynomial must not be confused with a nonzero constant polynomial." }, "reverse_audit": { "kind": "premise_audit", "required_ids": [ "like_terms", "poly_eq_coeff", "mko_polynomial" ], "explanation": { "zh-Hant": "一個係數不同就足以推出不等；「未定義」是本條目的教學慣例，與 Mathlib 的 degree（⊥）或 natDegree（0）需要另外對應。", "en": "One differing coefficient is enough for inequality; 'undefined' is this entry's teaching convention and must be matched separately with Mathlib's degree (⊥) or natDegree (0)." } }, "leaf_basis": "poly_eq_coeff + like_terms + numeric_evaluation" }, { "id": "S09", "uses": [ "S04", "S06", "mko_poly_ops" ], "before_tex": "p+q=[2,0,3]+[-1,1]", "after_tex": "[2+(-1),\\,0+1,\\,3+0]=[1,1,3],\\quad p+q=3x^2+x+1", "relation": "equivalent", "rule": "mko_poly_ops", "conditions": [], "explanation": { "zh-Hant": "加法逐次相加係數；q 在 x² 的係數是 0（超出 [−1,1] 的範圍）。所以 p+q 的係數為 2+(−1)=1、0+1=1、3+0=3，即 3x²+x+1，次數 2。", "en": "Addition adds coefficients power by power; the x² coefficient of q is 0 (beyond [−1,1]). So the coefficients of p+q are 2+(−1)=1, 0+1=1 and 3+0=3, that is 3x²+x+1, of degree 2." }, "reverse_audit": { "kind": "inverse_equivalence", "required_ids": [ "mko_poly_ops", "poly_eq_coeff", "numeric_evaluation" ], "explanation": { "zh-Hant": "每個係數都是精確的有理數加法，可逐次核對（N3、N10），兩邊是同一個多項式。", "en": "Each coefficient is an exact rational sum and can be checked power by power (N3, N10); both sides are the same polynomial." } }, "leaf_basis": "mko_poly_ops + poly_eq_coeff + numeric_evaluation" }, { "id": "S10", "uses": [ "S04", "S06", "mko_poly_ops" ], "before_tex": "p\\cdot q,\\quad a=[2,0,3],\\ b=[-1,1]", "after_tex": "c=[-2,2,-3,3],\\quad pq=3x^3-3x^2+2x-2", "relation": "equivalent", "rule": "mko_poly_ops", "conditions": [], "explanation": { "zh-Hant": "乘積的係數是有限卷積 c_k=Σ_{i+j=k} a_i b_j，其中超出向量的 a_i、b_j 都是 0。子步 S10.1 算 c_0、c_1，S10.2 算 c_2、c_3 並說明 k≥4 時 c_k=0。", "en": "The coefficients of the product are the finite convolution c_k=Σ_{i+j=k} a_i b_j, where every a_i or b_j beyond its vector is 0. Sub-step S10.1 computes c_0 and c_1, and S10.2 computes c_2 and c_3 and shows c_k=0 for k≥4." }, "reverse_audit": { "kind": "inverse_equivalence", "required_ids": [ "mko_poly_ops", "poly_eq_coeff" ], "explanation": { "zh-Hant": "卷積公式給出乘積的每個係數，結果與 pq 逐次相同；N8 在格點上以次數界核對整個恆等式。", "en": "The convolution formula gives every coefficient of the product, and the result agrees with pq power by power; N8 checks the whole identity on a grid with a degree bound." } }, "leaf_basis": "mko_poly_ops + finite_sum_rules + poly_eq_coeff" }, { "id": "S10.1", "uses": [ "S10", "finite_sum_rules" ], "before_tex": "c_0,\\ c_1", "after_tex": "c_0=a_0b_0=-2,\\quad c_1=a_0b_1+a_1b_0=2+0=2", "relation": "equivalent", "rule": "finite_sum_rules", "conditions": [], "explanation": { "zh-Hant": "k=0 時只有 (i,j)=(0,0)：c_0=2·(−1)=−2。k=1 時有 (0,1) 與 (1,0)：c_1=2·1+0·(−1)=2。", "en": "For k=0 only (i,j)=(0,0) occurs: c_0=2·(−1)=−2. For k=1 the pairs are (0,1) and (1,0): c_1=2·1+0·(−1)=2." }, "reverse_audit": { "kind": "inverse_equivalence", "required_ids": [ "finite_sum_rules", "numeric_evaluation" ], "explanation": { "zh-Hant": "列出所有 i+j=k 的指數對，不遺漏也不重複；數值可重算（N4、N5）。", "en": "All index pairs with i+j=k are listed, none missing or repeated; the values are recomputable (N4, N5)." } }, "leaf_basis": "finite_sum_rules + numeric_evaluation" }, { "id": "S10.2", "uses": [ "S10", "finite_sum_rules" ], "before_tex": "c_2,\\ c_3,\\ c_k\\ (k\\geq 4)", "after_tex": "c_2=a_0b_2+a_1b_1+a_2b_0=0+0-3=-3,\\quad c_3=a_2b_1=3,\\quad c_k=0\\ (k\\geq 4)", "relation": "equivalent", "rule": "finite_sum_rules", "conditions": [], "explanation": { "zh-Hant": "k=2：(0,2)、(1,1)、(2,0) 給出 2·0+0·1+3·(−1)=−3。k=3：(0,3)、(1,2)、(2,1)、(3,0) 中只有 a_2b_1=3·1=3 非零（b_2=b_3=0、a_3=0）。k≥4 時 i+j=k 必有 i≥3 或 j≥2，對應的 a_i 或 b_j 為 0，所以 c_k=0。", "en": "k=2: (0,2), (1,1) and (2,0) give 2·0+0·1+3·(−1)=−3. k=3: among (0,3), (1,2), (2,1) and (3,0) only a_2b_1=3·1=3 is nonzero (b_2=b_3=0 and a_3=0). For k≥4, i+j=k forces i≥3 or j≥2, where a_i or b_j is 0, so c_k=0." }, "reverse_audit": { "kind": "inverse_equivalence", "required_ids": [ "finite_sum_rules", "numeric_evaluation", "mul_zero" ], "explanation": { "zh-Hant": "k≥4 的論證對所有 k 成立，不是逐一檢查；數值可重算（N6、N7）。", "en": "The argument for k≥4 covers every such k at once rather than checking them one by one; the values are recomputable (N6, N7)." } }, "leaf_basis": "finite_sum_rules + mul_zero + numeric_evaluation" }, { "id": "S11", "uses": [ "S10", "S10.2", "S01" ], "before_tex": "pq=3x^3-3x^2+2x-2", "after_tex": "\\deg(pq)=3=\\deg p+\\deg q", "relation": "implies", "rule": "mko_polynomial", "conditions": [ "c_3=3\\neq 0" ], "explanation": { "zh-Hant": "c_3=3≠0 且 k≥4 時 c_k=0，所以 deg(pq)=3。在本例中它等於 deg p+deg q=2+1；首項係數 3 正是 p 與 q 首項係數的乘積 3·1。一般的次數乘法公式不在本稿證明。", "en": "c_3=3≠0 and c_k=0 for k≥4, so deg(pq)=3. In this example it equals deg p+deg q=2+1; the leading coefficient 3 is the product 3·1 of the leading coefficients of p and q. The general degree rule for products is not proved in this guide." }, "reverse_audit": { "kind": "premise_audit", "required_ids": [ "S10", "mko_polynomial" ], "explanation": { "zh-Hant": "只對本例宣稱；不從一個例子推出一般公式。", "en": "Claimed for this example only; no general formula is inferred from one example." } }, "leaf_basis": "mko_polynomial + numeric_evaluation" }, { "id": "S12", "uses": [ "S04", "S06", "S10", "eval_hom" ], "before_tex": "p(2),\\ q(2),\\ (pq)(2)", "after_tex": "p(2)=14,\\ q(2)=1,\\ (pq)(2)=24-12+4-2=14=p(2)\\,q(2)", "relation": "implies", "rule": "eval_hom", "conditions": [], "explanation": { "zh-Hant": "代入 x=2：p(2)=3·4+2=14，q(2)=2−1=1，(pq)(2)=3·8−3·4+2·2−2=14。依 eval_hom 應有 (pq)(2)=p(2)q(2)=14，結果一致。這只是對 S10 的一個一致性檢查；乘積本身由係數卷積確定，一個點的吻合不能證明係數正確。", "en": "Substitute x=2: p(2)=3·4+2=14, q(2)=2−1=1, and (pq)(2)=3·8−3·4+2·2−2=14. By eval_hom one must have (pq)(2)=p(2)q(2)=14, and it agrees. This is only a consistency check on S10; the product itself is determined by the coefficient convolution, and agreement at one point cannot prove the coefficients right." }, "reverse_audit": { "kind": "premise_audit", "required_ids": [ "eval_hom", "numeric_evaluation" ], "explanation": { "zh-Hant": "一致性檢查只能發現錯誤，不能確認正確；可重算（N11 至 N13）。", "en": "A consistency check can reveal an error but cannot confirm correctness; it is recomputable (N11 to N13)." } }, "leaf_basis": "eval_hom + numeric_evaluation" }, { "id": "S13", "uses": [ "S04", "G2", "poly_eq_coeff", "mko_poly_ops" ], "before_tex": "p(1)=5=t(1)", "after_tex": "p\\neq t\\quad([2,0,3]\\neq[0,5])", "relation": "implies", "rule": "poly_eq_coeff", "conditions": [], "explanation": { "zh-Hant": "p(1)=3+2=5，t(1)=5·1=5，在 x=1 的值相同。但 p 的常數項是 2，t 的常數項是 0，2≠0，依係數相等的定義 p≠t；在 x=0 也可看出 p(0)=2≠0=t(0)。MKO 指出單個代入值相等不建立多項式恆等式。", "en": "p(1)=3+2=5 and t(1)=5·1=5, so the values at x=1 agree. But the constant coefficient of p is 2 and that of t is 0, and 2≠0, so p≠t by the definition of equality; at x=0 this shows as p(0)=2≠0=t(0). The MKO notes that equality of one substituted value does not establish a polynomial identity." }, "reverse_audit": { "kind": "premise_audit", "required_ids": [ "poly_eq_coeff", "mko_poly_ops", "numeric_evaluation" ], "explanation": { "zh-Hant": "一個係數不同即足以推出不等（N14、N15）；反向「一點相等推出恆等」被明確拒絕。", "en": "One differing coefficient suffices for inequality (N14, N15); the converse inference from one equal value to an identity is explicitly rejected." } }, "leaf_basis": "poly_eq_coeff + numeric_evaluation" }, { "id": "S14", "uses": [ "S05", "S06", "S07", "S08", "S09", "S11", "S13" ], "before_tex": "p,q,r,z,p+q,pq,t", "after_tex": "\\deg p=2,\\ \\deg q=1,\\ \\deg r=0,\\ z=0\\ (\\deg\\ \\text{undefined}),\\ \\deg(p+q)=2,\\ \\deg(pq)=3,\\ p\\neq t", "relation": "implies", "rule": "and_intro_elim", "conditions": [], "explanation": { "zh-Hant": "彙整：p=3x²+2 次數 2，q=x−1 次數 1，r=3 次數 0，z 是零多項式（次數未定義）且 z≠r，p+q=3x²+x+1 次數 2，pq=3x³−3x²+2x−2 次數 3，p≠t 雖然 p(1)=t(1)。每個結論都由係數資料得出。", "en": "Summary: p=3x²+2 has degree 2, q=x−1 degree 1, r=3 degree 0, z is the zero polynomial (degree undefined) with z≠r, p+q=3x²+x+1 has degree 2, pq=3x³−3x²+2x−2 degree 3, and p≠t although p(1)=t(1). Every conclusion comes from the coefficient data." }, "reverse_audit": { "kind": "sufficient_goal_reduction", "required_ids": [ "S05", "S06", "S07", "S08", "S09", "S11", "S13", "and_intro_elim" ], "explanation": { "zh-Hant": "目標 G：各項結論。足夠條件 H：S05 至 S13 的各別結論；H⇒G 只是合取。H 由係數計算正向證成，沒有把 G 當作前提。", "en": "Goal G: all the stated conclusions. Sufficient condition H: the separate conclusions of S05 to S13; H⇒G is just conjunction. H is proved forwards from coefficient computations; G is never used as a premise." } }, "leaf_basis": "and_intro_elim" } ], "result": { "zh-Hant": "合併同次項後 p=3x²+2，deg p=2（不是字面上的 3）；q=[−1,1,0,0]=x−1，deg q=1（不是向量長度減一）；r=3 是零次常數；z=5x−5x 是零多項式，次數未定義且 z≠r；p+q=3x²+x+1；pq=3x³−3x²+2x−2，次數 3；p(1)=t(1)=5 但 p≠t。", "en": "After combining like terms p=3x²+2 and deg p=2 (not the literal 3); q=[−1,1,0,0]=x−1 with deg q=1 (not the vector length minus one); r=3 is a constant of degree 0; z=5x−5x is the zero polynomial, with undefined degree and z≠r; p+q=3x²+x+1; pq=3x³−3x²+2x−2 of degree 3; and p(1)=t(1)=5 while p≠t." }, "negative_controls": [ { "claim": { "zh-Hant": "p=4x³+3x²−x+2−4x³+x 的次數是 3，因為式中出現 4x³", "en": "p=4x³+3x²−x+2−4x³+x has degree 3 because 4x³ appears in it" }, "expected": "reject", "reason": { "zh-Hant": "合併後 x³ 的係數是 4−4=0，p=3x²+2，次數 2。", "en": "After combining, the x³ coefficient is 4−4=0, so p=3x²+2 of degree 2." } }, { "claim": { "zh-Hant": "係數向量 [−1,1,0,0] 長度為 4，所以 q 的次數是 3", "en": "The coefficient vector [−1,1,0,0] has length 4, so q has degree 3" }, "expected": "reject", "reason": { "zh-Hant": "最大非零係數的索引是 1，q=x−1，次數 1。", "en": "The largest index with a nonzero coefficient is 1, so q=x−1 of degree 1." } }, { "claim": { "zh-Hant": "z=5x−5x 和常數 3 一樣是零次多項式", "en": "z=5x−5x is a polynomial of degree 0, just like the constant 3" }, "expected": "reject", "reason": { "zh-Hant": "z 是零多項式，本條目中次數未定義；而且 z 的常數項 0≠3，z≠3。", "en": "z is the zero polynomial, whose degree is undefined in this entry; moreover its constant coefficient 0≠3, so z≠3." } }, { "claim": { "zh-Hant": "因為 p(1)=t(1)=5，所以 p=t", "en": "Since p(1)=t(1)=5, p=t" }, "expected": "reject", "reason": { "zh-Hant": "p(0)=2≠0=t(0)，常數項不同；單點相等不建立恆等式。", "en": "p(0)=2≠0=t(0), so the constant coefficients differ; equality at one point does not establish an identity." } } ], "status": "candidate", "gaps": [], "verification": { "structure": "checked_local_script", "mathematics": "candidate_argument", "formalization": "not_formalized" }, "rule_glossary": [ { "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": "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": "finite_sum_rules", "kind": "prerequisite", "statement": { "zh-Hant": "有限和：Σ_{i=1}^{0} a_i=0、Σ_{i=1}^{n+1} a_i=Σ_{i=1}^{n} a_i + a_{n+1}；Σ(a_i+b_i)=Σa_i+Σb_i；Σ c·a_i=c·Σa_i。", "en": "Finite sums: Σ_{i=1}^{0} a_i=0 and Σ_{i=1}^{n+1} a_i=Σ_{i=1}^{n} a_i + a_{n+1}; Σ(a_i+b_i)=Σa_i+Σb_i; Σ c·a_i=c·Σa_i." } }, { "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_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": "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": "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_polynomial", "kind": "cited", "source": { "mko_id": "mko-polynomial", "sha256": "0a0d76b12363d10d9ecbe8b45d7bebb1136637db1c41e6308481c675c8db9e6c", "commit": "b28a5edbd46279b1db2a3f3f48bb3f447e18899f" }, "statement": { "zh-Hant": "Q[x] 的一元多項式以有限個有理係數 a_k 表示，指數均為非負整數。非零多項式的次數為最大非零係數的索引；本條目把零多項式的次數記為未定義。", "en": "A univariate polynomial in Q[x] is represented by finitely many rational coefficients a_k, with nonnegative integer exponents. The degree of a nonzero polynomial is the largest index with a nonzero coefficient; this entry calls the degree of the zero polynomial undefined." } }, { "id": "mko_poly_ops", "kind": "cited", "source": { "mko_id": "mko-polynomial", "sha256": "0a0d76b12363d10d9ecbe8b45d7bebb1136637db1c41e6308481c675c8db9e6c", "commit": "b28a5edbd46279b1db2a3f3f48bb3f447e18899f" }, "statement": { "zh-Hant": "加法逐次相加係數；乘法按指數和合併，所得係數是有限卷積。代入一個數值是對同一份係數資料求值，單個代入值相等不建立多項式恆等式。", "en": "Addition adds coefficients power by power; multiplication collects by sums of exponents, and the resulting coefficients are a finite convolution. Substituting a number evaluates the same coefficient data, and equality of a single substituted value does not establish a polynomial identity." } }, { "id": "poly_eq_coeff", "kind": "prerequisite", "statement": { "zh-Hant": "ℚ[x] 中兩個多項式相等，當且僅當對每個 k，x^k 的係數都相等（未列出的係數視為 0）。", "en": "Two polynomials in ℚ[x] are equal exactly when, for every k, their coefficients of x^k are equal (coefficients not listed count as 0)." } }, { "id": "like_terms", "kind": "prerequisite", "statement": { "zh-Hant": "合併同次項：c x^k+d x^k=(c+d)x^k（分配律）；不同次的項不能合併。", "en": "Combining like terms: c x^k+d x^k=(c+d)x^k (distributivity); terms of different powers are not combined." } }, { "id": "eval_hom", "kind": "prerequisite", "statement": { "zh-Hant": "對 c∈ℚ，求值 p↦p(c)=Σ_k a_k c^k 滿足 (p+q)(c)=p(c)+q(c) 與 (pq)(c)=p(c)q(c)。", "en": "For c∈ℚ, evaluation p↦p(c)=Σ_k a_k c^k satisfies (p+q)(c)=p(c)+q(c) and (pq)(c)=p(c)q(c)." } } ], "revision_log": [], "numeric_checks": [ { "id": "N1", "step": "S03", "lhs": "4-4", "rhs": "0" }, { "id": "N2", "step": "S03", "lhs": "-1+1", "rhs": "0" }, { "id": "N3", "step": "S09", "lhs": "2+(-1)", "rhs": "1" }, { "id": "N4", "step": "S10.1", "lhs": "2*(-1)", "rhs": "-2" }, { "id": "N5", "step": "S10.1", "lhs": "2*1+0*(-1)", "rhs": "2" }, { "id": "N6", "step": "S10.2", "lhs": "2*0+0*1+3*(-1)", "rhs": "-3" }, { "id": "N7", "step": "S10.2", "lhs": "3*1", "rhs": "3" }, { "id": "N8", "step": "S10", "vars": [ "x" ], "degree": 3, "lhs": "(3*x**2+2)*(x-1)", "rhs": "3*x**3-3*x**2+2*x-2" }, { "id": "N9", "step": "S03", "vars": [ "x" ], "degree": 3, "lhs": "4*x**3+3*x**2-x+2-4*x**3+x", "rhs": "3*x**2+2" }, { "id": "N10", "step": "S09", "vars": [ "x" ], "degree": 2, "lhs": "(3*x**2+2)+(x-1)", "rhs": "3*x**2+x+1" }, { "id": "N11", "step": "S12", "lhs": "3*2**2+2", "rhs": "14" }, { "id": "N12", "step": "S12", "lhs": "3*2**3-3*2**2+2*2-2", "rhs": "14" }, { "id": "N13", "step": "S12", "lhs": "14*(2-1)", "rhs": "14" }, { "id": "N14", "step": "S13", "lhs": "3*1**2+2", "rhs": "5*1" }, { "id": "N15", "step": "S13", "lhs": "3*0**2+2", "relation": "!=", "rhs": "5*0" }, { "id": "N16", "step": "S07", "vars": [ "x" ], "degree": 2, "lhs": "x**2-x**2+3", "rhs": "3" }, { "id": "N17", "step": "S08", "vars": [ "x" ], "degree": 1, "lhs": "5*x-5*x", "rhs": "0" } ] }, "presentation": { "md_scope": { "zh-Hant": "只用 MKO 的表示：形式變數 x、有理係數、有限支撐，係數向量由低次到高次，尾端零不改變多項式，零多項式的次數在本條目記為未定義。只計算這些具體多項式；次數乘法公式 deg(pq)=deg p+deg q 只在本例核對，不作一般證明。不處理多元多項式或其他係數環；不改動 canonical MKO 或英文 edition。", "en": "Uses only the MKO's representation: a formal variable x, rational coefficients, finite support, coefficient vectors from low to high power, trailing zeros not changing the polynomial, and the degree of the zero polynomial called undefined in this entry. Only these particular polynomials are computed; the degree rule deg(pq)=deg p+deg q is checked for this example only and not proved in general. Multivariate polynomials and other coefficient rings are not treated; the canonical MKO and its English edition are not changed." }, "md_symbols": [ [ "\\mathbb{Q}[x]", "有理係數的一元多項式全體", "all univariate polynomials with rational coefficients" ], [ "a_k", "x^k 的係數；超出列出範圍者為 0", "the coefficient of x^k; zero beyond the listed range" ], [ "[a_0,a_1,a_2]", "由低次到高次的係數向量", "a coefficient vector from low to high power" ], [ "\\deg p", "非零多項式中係數非零的最大指數", "the largest index with a nonzero coefficient, for nonzero p" ], [ "c_k=\\sum_{i+j=k}a_ib_j", "乘積的第 k 個係數（有限卷積）", "the k-th coefficient of a product (finite convolution)" ], [ "p(c)", "在 x=c 求值", "evaluation at x=c" ] ], "md_error_controls": [ { "title": "先合併同次項再讀次數 / Combine like terms before reading the degree", "zh": "MKO 指出的誤解之一是未合併同次項就取最高字面指數。p 字面上有 4x³ 與 −4x³，合併後 x³ 的係數為 0，deg p=2（S03 至 S05）。r=x²−x²+3 同理是常數 3（S07）。", "en": "One error named in the MKO is taking the highest literal exponent before combining like terms. p shows 4x³ and −4x³ literally, but the combined x³ coefficient is 0 and deg p=2 (S03 to S05). Likewise r=x²−x²+3 is the constant 3 (S07)." }, { "title": "向量長度與尾端零不決定次數 / Vector length and trailing zeros do not decide the degree", "zh": "MKO 指出的另一個誤解是用向量長度或尾端零決定次數。[−1,1,0,0] 長度為 4，但它就是 x−1，次數 1（S06）；[2,0,3,0] 與 [2,0,3] 是同一個多項式（S04）。", "en": "The other error named in the MKO is deciding the degree from the vector length or trailing zeros. [−1,1,0,0] has length 4 but is x−1, of degree 1 (S06); [2,0,3,0] and [2,0,3] are the same polynomial (S04)." }, { "title": "零多項式不是零次 / The zero polynomial is not of degree zero", "zh": "z=5x−5x 的所有係數都是 0，本條目記其次數為未定義；r=3 則是次數 0 的非零常數。兩者常數項 0 與 3 不同，所以 z≠r（S08）。Mathlib 的 degree 給零多項式 ⊥、natDegree 給 0，對應時不可混用。", "en": "Every coefficient of z=5x−5x is 0, and this entry calls its degree undefined; r=3 is a nonzero constant of degree 0. Their constant coefficients 0 and 3 differ, so z≠r (S08). Mathlib's degree gives the zero polynomial ⊥ and natDegree gives 0; the two must not be mixed when matching." }, { "title": "一點相等不是恆等 / Equal at one point is not an identity", "zh": "p(1)=t(1)=5，但 p 與 t 的係數不同，p(0)=2≠0=t(0)（S13）。S12 在 x=2 的吻合也只是一致性檢查，乘積由係數卷積確定。", "en": "p(1)=t(1)=5, but p and t have different coefficients, and p(0)=2≠0=t(0) (S13). The agreement at x=2 in S12 is likewise only a consistency check; the product is determined by the coefficient convolution." } ], "md_open_items": [ { "zh-Hant": "非零多項式乘積的次數公式 deg(pq)=deg p+deg q 只在本例核對（S11），一般證明不在本稿範圍。", "en": "The degree rule deg(pq)=deg p+deg q for nonzero polynomials is checked only for this example (S11); its general proof is outside this guide." }, { "zh-Hant": "N8 至 N10、N16、N17 是 degree 型核對：在 {0,1,至 degree} 的整數格點上比較兩邊，對次數不超過該界的一元多項式足以判定恆等；這依賴「非零多項式根的個數不超過次數」，本稿沒有證明它。", "en": "N8 to N10, N16 and N17 are degree checks: both sides are compared at the integers 0 up to the degree bound, which decides an identity of univariate polynomials within that bound; this relies on a nonzero polynomial having no more roots than its degree, which this guide does not prove." } ] }, "revision": { "id": "production-20261019-adopted-96", "source_commit": "b28a5edbd46279b1db2a3f3f48bb3f447e18899f", "adopted_source_sha256": "7aed146fa53618c3b3b6c889910ca918859436792d1f15685287e4d47d9dee02", "original_source_sha256": "7aed146fa53618c3b3b6c889910ca918859436792d1f15685287e4d47d9dee02", "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." } }