{
  "schema_version": "ocme-derivation-reader-v1",
  "guide_id": "derivation-mko-bisection-method-01",
  "mko_id": "mko-bisection-method",
  "title": {
    "zh-Hant": "二分法：四步 bracket 更新、不變量與根存在的獨立責任",
    "en": "Bisection: four bracket updates, the invariant and the separate proof obligation for root existence"
  },
  "contract": {
    "schema_version": "ocme-derivation-guide-v0.1",
    "guide_id": "derivation-mko-bisection-method-01",
    "source": {
      "mko_id": "mko-bisection-method",
      "commit": "b28a5edbd46279b1db2a3f3f48bb3f447e18899f",
      "sha256": "ab834bbfb6637765ed0a0282df9021cdc9509c95f67eb41e02c6d5db1774d835",
      "english_sha256": "4a1fd3b57ecd28b4aedff341d680464833d94b18bd9445de6dea28aacb5bb7f5"
    },
    "mode": "algorithm_trace",
    "audience_prerequisites": [
      "order_basics",
      "frac_rules",
      "power_laws",
      "nat_induction",
      "archimedean",
      "case_analysis",
      "numeric_evaluation"
    ],
    "goal": {
      "zh-Hant": "對 f(x)=x²−2，從 [1,2] 與所提供的端點符號 f(1)<0、f(2)>0 出發，依所提供的中點符號執行四步二分，追蹤 bracket；證明「端點異號且 a<b」是不變量、n 步後寬度為 1/2^n、在每次所需符號可取得且更新有效的延伸執行中，寬度停止準則必在有限步內終止；再證明這些 bracket 資料本身不證明有根，並另外引用中間值定理，對連續的 f 得出最終 bracket 內有根。",
      "en": "For f(x)=x²−2, start from [1,2] with the supplied endpoint signs f(1)<0 and f(2)>0, run four bisection steps with the supplied midpoint signs and trace the brackets; prove that 'opposite endpoint signs and a<b' is an invariant, that the width after n steps is 1/2^n, and that, in an extended run with every needed sign available and effective updates, a width-based stopping rule terminates after finitely many steps; then show that the bracket data alone do not prove a root exists, and separately cite the intermediate value theorem to conclude, for the continuous f, that the final bracket contains a root."
    },
    "givens": [
      {
        "id": "G1",
        "statement": {
          "zh-Hant": "f(x)=x²−2。初始 bracket [a0,b0]=[1,2]，所提供的端點符號 s(1)=−、s(2)=+。",
          "en": "f(x)=x²−2. Initial bracket [a0,b0]=[1,2], with supplied endpoint signs s(1)=− and s(2)=+."
        }
      },
      {
        "id": "G2",
        "statement": {
          "zh-Hant": "所提供的中點符號：s(3/2)=+，s(5/4)=−，s(11/8)=−，s(23/16)=+。這些是題目提供的資料（MKO 假設 a1）；本稿以精確有理數計算 f 的值作交叉核對，不由圖形推斷。",
          "en": "Supplied midpoint signs: s(3/2)=+, s(5/4)=−, s(11/8)=− and s(23/16)=+. These are data supplied by the problem (MKO assumption a1); the guide cross-checks them by exact rational evaluation of f, not by reading a graph."
        }
      },
      {
        "id": "G3",
        "statement": {
          "zh-Hant": "演算法（MKO）：給 a<b 且端點符號相反的 [a,b]，取 m=(a+b)/2；若已觀察 f(m)=0 就停在 m；否則保留端點符號相反的半區。本題執行四步，除非提前停止。",
          "en": "Algorithm (MKO): given [a,b] with a<b and opposite endpoint signs, take m=(a+b)/2; if f(m)=0 has been observed, stop at m; otherwise keep the half whose endpoint signs are opposite. Here four steps are run unless the algorithm stops earlier."
        }
      },
      {
        "id": "G4",
        "statement": {
          "zh-Hant": "用於獨立的根存在步驟：f 是多項式，因此在 ℝ 上連續。",
          "en": "For the separate root-existence step: f is a polynomial and hence continuous on ℝ."
        }
      }
    ],
    "steps": [
      {
        "id": "S01",
        "uses": [
          "G1",
          "G2",
          "G3",
          "mko_bisect"
        ],
        "before_tex": "\\text{bisection with supplied sign data}",
        "after_tex": "[a_0,b_0]=[1,2],\\ m_{n}=\\tfrac{a_{n-1}+b_{n-1}}{2}",
        "relation": "definition",
        "rule": "mko_bisect",
        "conditions": [],
        "explanation": {
          "zh-Hant": "固定演算法與輸入資料。依 MKO，每步只依明示的端點與中點符號選半區；這只是 bracket 資料更新。本稿把兩件事分開：S02 到 S12 是演算法本身（追蹤、不變量、寬度、終止），S13、S14 才處理根是否存在。",
          "en": "Fix the algorithm and its input data. By the MKO, each step chooses a half using only the explicit endpoint and midpoint signs; this is merely a bracket-data update. The guide keeps two things apart: S02 to S12 concern the algorithm itself (trace, invariant, width, termination), and only S13 and S14 address whether a root exists."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "G1",
            "G2",
            "G3",
            "mko_bisect"
          ],
          "explanation": {
            "zh-Hant": "此步只陳述演算法與資料，不推出新事實。",
            "en": "This step only states the algorithm and the data and derives no new fact."
          }
        },
        "leaf_basis": "mko_bisect + def_intro"
      },
      {
        "id": "S02",
        "uses": [
          "G1",
          "G3"
        ],
        "before_tex": "[1,2],\\ s(1)=-,\\ s(2)=+",
        "after_tex": "\\text{valid input: } 1<2,\\ \\text{opposite endpoint signs}",
        "relation": "implies",
        "rule": "and_intro_elim",
        "conditions": [],
        "explanation": {
          "zh-Hant": "輸入合格需要兩件事：區間有序（1<2）與端點符號相反（− 與 +）。兩者都由 G1 提供。交叉核對：f(1)=1−2=−1<0，f(2)=4−2=2>0（N1、N2），與所提供的符號一致。若兩端同號，輸入不符合本演算法的端點異號契約，不能直接使用這個已證的不變量更新；這不排除取得新中點符號後找到端點異號的子區間。",
          "en": "A valid input needs two things: an ordered interval (1<2) and opposite endpoint signs (− and +). Both are supplied by G1. Cross-check: f(1)=1−2=−1<0 and f(2)=4−2=2>0 (N1, N2), agreeing with the supplied signs. Equal endpoint signs violate this algorithm's opposite-sign input contract, so its proved invariant update cannot be used directly. This does not rule out finding an opposite-sign subinterval after obtaining a new midpoint sign."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "G1",
            "and_intro_elim"
          ],
          "explanation": {
            "zh-Hant": "合格性是兩個條件的合取；精確計算只是交叉核對，演算法使用的是所提供的符號。",
            "en": "Validity is a conjunction of two conditions; the exact evaluation is only a cross-check, and the algorithm uses the supplied signs."
          }
        },
        "leaf_basis": "and_intro_elim + numeric_evaluation"
      },
      {
        "id": "S03",
        "uses": [
          "S02",
          "G3"
        ],
        "before_tex": "[a_0,b_0]=[1,2]",
        "after_tex": "m_1=\\tfrac{1+2}{2}=\\tfrac{3}{2},\\ 1<\\tfrac{3}{2}<2",
        "relation": "implies",
        "rule": "L_mid",
        "conditions": [
          "a_0<b_0"
        ],
        "explanation": {
          "zh-Hant": "第一步取中點 m1=(1+2)/2=3/2。依引理 L_mid，中點嚴格落在兩端之間，兩個半區 [1,3/2] 與 [3/2,2] 等寬，各為 1/2。",
          "en": "Step 1 takes the midpoint m1=(1+2)/2=3/2. By Lemma L_mid the midpoint lies strictly between the ends, and the two halves [1,3/2] and [3/2,2] have equal width 1/2."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "L_mid",
            "S02"
          ],
          "explanation": {
            "zh-Hant": "需要 a0<b0（S02）；引理在 S03.1 證明。",
            "en": "It needs a0<b0 (S02); the lemma is proved in S03.1."
          }
        },
        "leaf_basis": "L_mid + numeric_evaluation"
      },
      {
        "id": "S03.1",
        "uses": [
          "order_basics",
          "frac_rules"
        ],
        "before_tex": "a<b,\\ m=\\tfrac{a+b}{2}",
        "after_tex": "m-a=b-m=\\tfrac{b-a}{2}>0",
        "relation": "implies",
        "rule": "order_basics",
        "conditions": [
          "a<b"
        ],
        "explanation": {
          "zh-Hant": "m−a=(a+b)/2−2a/2=(b−a)/2，b−m=2b/2−(a+b)/2=(b−a)/2（N12、N13 為一次多項式恆等式）。由 a<b 得 b−a>0，乘以正數 1/2 仍為正，所以 m−a>0、b−m>0，即 a<m<b。這證得引理 L_mid。",
          "en": "m−a=(a+b)/2−2a/2=(b−a)/2 and b−m=2b/2−(a+b)/2=(b−a)/2 (N12 and N13 are degree-one polynomial identities). From a<b we get b−a>0, and multiplying by the positive number 1/2 keeps it positive, so m−a>0 and b−m>0, that is a<m<b. This proves Lemma L_mid."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "order_basics",
            "frac_rules"
          ],
          "explanation": {
            "zh-Hant": "需要 a<b；若 a=b，中點等於端點，不再有兩個非退化半區。",
            "en": "It needs a<b; if a=b the midpoint equals the endpoints and there are no longer two nondegenerate halves."
          }
        },
        "leaf_basis": "order_basics + frac_rules + distributive"
      },
      {
        "id": "S04",
        "uses": [
          "G3",
          "case_analysis"
        ],
        "before_tex": "\\{s(a),s(b)\\}=\\{-,+\\},\\ s(m)\\neq 0",
        "after_tex": "\\text{exactly one of }[a,m],\\ [m,b]\\ \\text{has opposite endpoint signs}",
        "relation": "implies",
        "rule": "L_half",
        "conditions": [
          "s(m)\\neq 0"
        ],
        "explanation": {
          "zh-Hant": "證明引理 L_half：只要中點符號不是 0，保留哪一半由符號唯一決定。這保證每一步的「保留端點異號的半區」有確定的意義。分三個子步：兩種情況，加上「情況涵蓋全部可能」的核對。",
          "en": "Prove Lemma L_half: as long as the midpoint sign is not 0, the signs determine uniquely which half to keep. This makes 'keep the half with opposite endpoint signs' well defined at every step. Three sub-steps: two cases, plus the check that the cases cover every possibility."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "L_half",
            "case_analysis"
          ],
          "explanation": {
            "zh-Hant": "引理只用符號資料，不用 f 的連續性或任何函數值；因此它也不能推出根存在。",
            "en": "The lemma uses only sign data, not continuity of f or any function value; for the same reason it cannot yield the existence of a root."
          }
        },
        "leaf_basis": "case_analysis + and_intro_elim"
      },
      {
        "id": "S04.1",
        "uses": [
          "G3"
        ],
        "before_tex": "s(m)=s(a)",
        "after_tex": "[a,m]\\ \\text{same signs},\\ [m,b]\\ \\text{opposite signs}",
        "relation": "implies",
        "rule": "def_unfold",
        "conditions": [
          "s(a)\\neq s(b)"
        ],
        "explanation": {
          "zh-Hant": "若 s(m)=s(a)，則 [a,m] 兩端同號，不能保留；[m,b] 兩端是 s(m)=s(a) 與 s(b)，而 s(a)≠s(b)，所以異號，保留 [m,b]。",
          "en": "If s(m)=s(a), then [a,m] has equal endpoint signs and cannot be kept; [m,b] has endpoint signs s(m)=s(a) and s(b), and s(a)≠s(b), so they are opposite and [m,b] is kept."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "G3",
            "def_unfold"
          ],
          "explanation": {
            "zh-Hant": "只用符號相等的傳遞性與 s(a)≠s(b)。",
            "en": "Only transitivity of equal signs and s(a)≠s(b) are used."
          }
        },
        "leaf_basis": "def_unfold + eq_refl_symm_trans"
      },
      {
        "id": "S04.2",
        "uses": [
          "G3"
        ],
        "before_tex": "s(m)=s(b)",
        "after_tex": "[a,m]\\ \\text{opposite signs},\\ [m,b]\\ \\text{same signs}",
        "relation": "implies",
        "rule": "def_unfold",
        "conditions": [
          "s(a)\\neq s(b)"
        ],
        "explanation": {
          "zh-Hant": "若 s(m)=s(b)，對稱地 [m,b] 兩端同號，[a,m] 兩端是 s(a) 與 s(m)=s(b)，異號，保留 [a,m]。",
          "en": "If s(m)=s(b), symmetrically [m,b] has equal signs, while [a,m] has endpoint signs s(a) and s(m)=s(b), which are opposite, so [a,m] is kept."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "G3",
            "def_unfold"
          ],
          "explanation": {
            "zh-Hant": "與 S04.1 對稱。",
            "en": "Symmetric to S04.1."
          }
        },
        "leaf_basis": "def_unfold + eq_refl_symm_trans"
      },
      {
        "id": "S04.3",
        "uses": [
          "S04.1",
          "S04.2",
          "case_analysis"
        ],
        "before_tex": "s(m)\\in\\{-,+\\},\\ \\{s(a),s(b)\\}=\\{-,+\\}",
        "after_tex": "s(m)=s(a)\\ \\veebar\\ s(m)=s(b)",
        "relation": "case_split",
        "rule": "case_analysis",
        "conditions": [
          "s(m)\\neq 0"
        ],
        "explanation": {
          "zh-Hant": "s(m)≠0 時只能是 − 或 +；s(a)、s(b) 恰好一個是 −、一個是 +，所以 s(m) 恰好等於其中一個。兩種情況互斥且涵蓋全部可能，配合 S04.1、S04.2，恰有一個半區被保留。S04.1 至 S04.3 證得引理 L_half。若 s(m)=0，演算法依 G3 停在 m，不進入這個引理。",
          "en": "When s(m)≠0 it is either − or +; exactly one of s(a), s(b) is − and the other +, so s(m) equals exactly one of them. The two cases are exclusive and exhaustive, and with S04.1 and S04.2 exactly one half is kept. S04.1 to S04.3 prove Lemma L_half. If s(m)=0, the algorithm stops at m by G3 and this lemma is not used."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "case_analysis",
            "S04.1",
            "S04.2"
          ],
          "explanation": {
            "zh-Hant": "分情況論證要求情況涵蓋全部可能；這裡的兩種情況正好窮盡 s(m) 的非零值。",
            "en": "Proof by cases needs exhaustive cases; here the two cases exhaust the nonzero values of s(m)."
          }
        },
        "leaf_basis": "case_analysis + and_intro_elim"
      },
      {
        "id": "S05",
        "uses": [
          "S03",
          "S04",
          "G2"
        ],
        "before_tex": "[1,2],\\ s(\\tfrac{3}{2})=+",
        "after_tex": "[a_1,b_1]=[1,\\tfrac{3}{2}]",
        "relation": "implies",
        "rule": "L_half",
        "conditions": [
          "s(\\tfrac{3}{2})\\neq 0"
        ],
        "explanation": {
          "zh-Hant": "所提供 s(3/2)=+=s(2)，依 S04.2 保留 [1,3/2]，其端點符號為 −、+。交叉核對：f(3/2)=9/4−2=1/4>0（N4）。",
          "en": "The supplied sign is s(3/2)=+=s(2), so by S04.2 keep [1,3/2], whose endpoint signs are − and +. Cross-check: f(3/2)=9/4−2=1/4>0 (N4)."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "L_half",
            "G2"
          ],
          "explanation": {
            "zh-Hant": "只用所提供的中點符號與引理；f 的精確值是交叉核對。",
            "en": "Only the supplied midpoint sign and the lemma are used; the exact value of f is a cross-check."
          }
        },
        "leaf_basis": "L_half + numeric_evaluation"
      },
      {
        "id": "S06",
        "uses": [
          "S05",
          "S04",
          "G2",
          "L_mid"
        ],
        "before_tex": "[1,\\tfrac{3}{2}],\\ m_2=\\tfrac{5}{4},\\ s(\\tfrac{5}{4})=-",
        "after_tex": "[a_2,b_2]=[\\tfrac{5}{4},\\tfrac{3}{2}]",
        "relation": "implies",
        "rule": "L_half",
        "conditions": [
          "s(\\tfrac{5}{4})\\neq 0"
        ],
        "explanation": {
          "zh-Hant": "m2=(1+3/2)/2=5/4（N5）。所提供 s(5/4)=−=s(1)，依 S04.1 保留 [5/4,3/2]。交叉核對：f(5/4)=25/16−2=−7/16<0（N6）。",
          "en": "m2=(1+3/2)/2=5/4 (N5). The supplied sign is s(5/4)=−=s(1), so by S04.1 keep [5/4,3/2]. Cross-check: f(5/4)=25/16−2=−7/16<0 (N6)."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "L_half",
            "L_mid",
            "G2"
          ],
          "explanation": {
            "zh-Hant": "同 S05：中點由 L_mid 定位，半區由 L_half 選定。",
            "en": "As in S05: the midpoint is placed by L_mid and the half chosen by L_half."
          }
        },
        "leaf_basis": "L_half + L_mid + numeric_evaluation"
      },
      {
        "id": "S07",
        "uses": [
          "S06",
          "S04",
          "G2",
          "L_mid"
        ],
        "before_tex": "[\\tfrac{5}{4},\\tfrac{3}{2}],\\ m_3=\\tfrac{11}{8},\\ s(\\tfrac{11}{8})=-",
        "after_tex": "[a_3,b_3]=[\\tfrac{11}{8},\\tfrac{3}{2}]",
        "relation": "implies",
        "rule": "L_half",
        "conditions": [
          "s(\\tfrac{11}{8})\\neq 0"
        ],
        "explanation": {
          "zh-Hant": "m3=(5/4+3/2)/2=11/8（N7）。所提供 s(11/8)=−=s(5/4)，保留 [11/8,3/2]。交叉核對：f(11/8)=121/64−2=−7/64<0（N8）。",
          "en": "m3=(5/4+3/2)/2=11/8 (N7). The supplied sign is s(11/8)=−=s(5/4), so keep [11/8,3/2]. Cross-check: f(11/8)=121/64−2=−7/64<0 (N8)."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "L_half",
            "L_mid",
            "G2"
          ],
          "explanation": {
            "zh-Hant": "同 S06。",
            "en": "As in S06."
          }
        },
        "leaf_basis": "L_half + L_mid + numeric_evaluation"
      },
      {
        "id": "S08",
        "uses": [
          "S07",
          "S04",
          "G2",
          "L_mid"
        ],
        "before_tex": "[\\tfrac{11}{8},\\tfrac{3}{2}],\\ m_4=\\tfrac{23}{16},\\ s(\\tfrac{23}{16})=+",
        "after_tex": "[a_4,b_4]=[\\tfrac{11}{8},\\tfrac{23}{16}]",
        "relation": "implies",
        "rule": "L_half",
        "conditions": [
          "s(\\tfrac{23}{16})\\neq 0"
        ],
        "explanation": {
          "zh-Hant": "m4=(11/8+3/2)/2=23/16（N9）。所提供 s(23/16)=+=s(3/2)，保留 [11/8,23/16]。交叉核對：f(23/16)=529/256−2=17/256>0（N10）。四步都沒有觀察到 f(m)=0，所以沒有提前停止；依 G3 在第四步後結束。",
          "en": "m4=(11/8+3/2)/2=23/16 (N9). The supplied sign is s(23/16)=+=s(3/2), so keep [11/8,23/16]. Cross-check: f(23/16)=529/256−2=17/256>0 (N10). No step observed f(m)=0, so there was no early stop; by G3 the run ends after step 4."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "L_half",
            "L_mid",
            "G2"
          ],
          "explanation": {
            "zh-Hant": "同 S06；結束是因為達到 G3 指定的步數。",
            "en": "As in S06; the run ends because the step count fixed in G3 is reached."
          }
        },
        "leaf_basis": "L_half + L_mid + numeric_evaluation"
      },
      {
        "id": "S09",
        "uses": [
          "S02",
          "S03.1",
          "S04",
          "nat_induction"
        ],
        "before_tex": "a_0<b_0,\\ s(a_0)=-,\\ s(b_0)=+",
        "after_tex": "\\forall n\\le 4:\\ a_n<b_n,\\ s(a_n)=-,\\ s(b_n)=+,\\ [a_{n}, b_{n}]\\subseteq[a_{n-1},b_{n-1}]\\ (n\\geq 1)",
        "relation": "implies",
        "rule": "nat_induction",
        "conditions": [],
        "explanation": {
          "zh-Hant": "不變量 I(n)：a_n<b_n、s(a_n)=−、s(b_n)=+。基底 I(0) 由 S02 成立。歸納步：若 I(n) 成立且 s(m)≠0，新區間是 [a_n,m] 或 [m,b_n]；由 L_mid 得 a_n<m<b_n，所以新區間仍有序且含於舊區間；由 L_half 的兩種情況，保留的半區左端符號為 −、右端為 +（S04.1 時新左端 m 與 a_n 同號 −；S04.2 時新右端 m 與 b_n 同號 +）。所以 I(n+1) 成立。S05 至 S08 的追蹤符合這個不變量。",
          "en": "Invariant I(n): a_n<b_n, s(a_n)=− and s(b_n)=+. The base I(0) holds by S02. Induction step: if I(n) holds and s(m)≠0, the new interval is [a_n,m] or [m,b_n]; L_mid gives a_n<m<b_n, so the new interval is still ordered and lies inside the old one; in both cases of L_half the kept half has sign − at its left end and + at its right end (in S04.1 the new left end m has the sign − of a_n; in S04.2 the new right end m has the sign + of b_n). So I(n+1) holds. The trace S05 to S08 agrees with this invariant."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S02",
            "L_mid",
            "L_half",
            "nat_induction"
          ],
          "explanation": {
            "zh-Hant": "歸納步需要 s(m)≠0；若觀察到 s(m)=0，演算法停止，不變量不再需要延伸。不變量只描述符號資料與區間位置，不包含「區間內有根」。",
            "en": "The induction step needs s(m)≠0; if s(m)=0 is observed, the algorithm stops and the invariant need not be extended. The invariant describes only sign data and interval positions; it does not include 'the interval contains a root'."
          }
        },
        "leaf_basis": "nat_induction + L_mid + L_half"
      },
      {
        "id": "S10",
        "uses": [
          "S03.1",
          "S09"
        ],
        "before_tex": "w_0=b_0-a_0=1",
        "after_tex": "w_n=\\tfrac{1}{2^n}",
        "relation": "implies",
        "rule": "L_width",
        "conditions": [
          "\\text{no early stop}"
        ],
        "explanation": {
          "zh-Hant": "證明引理 L_width：每次有效二分把寬度減半，n 步後寬度為 w0/2^n。這是純幾何的結果，與 f 無關。分兩個子步做歸納。",
          "en": "Prove Lemma L_width: each effective halving halves the width, so after n steps the width is w0/2^n. This is a purely geometric fact, independent of f. Two sub-steps form the induction."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "L_width",
            "S03.1"
          ],
          "explanation": {
            "zh-Hant": "只需要中點的位置；不宣稱寬度界是根的誤差界（那需要根確實在區間內，見 S14）。",
            "en": "Only the position of the midpoint is needed; it is not claimed that the width bound is an error bound for a root (that needs a root actually in the interval, see S14)."
          }
        },
        "leaf_basis": "nat_induction + L_mid + power_laws"
      },
      {
        "id": "S10.1",
        "uses": [
          "G1"
        ],
        "before_tex": "n=0",
        "after_tex": "w_0=\\tfrac{w_0}{2^0}",
        "relation": "induction_base",
        "rule": "nat_induction",
        "conditions": [],
        "explanation": {
          "zh-Hant": "基底：2^0=1，所以 w0/2^0=w0。",
          "en": "Base: 2^0=1, so w0/2^0=w0."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "power_laws"
          ],
          "explanation": {
            "zh-Hant": "只用 a^0=1。",
            "en": "Only a^0=1 is used."
          }
        },
        "leaf_basis": "nat_induction + power_laws"
      },
      {
        "id": "S10.2",
        "uses": [
          "S10.1",
          "S03.1",
          "S04"
        ],
        "before_tex": "w_n=\\tfrac{w_0}{2^n}",
        "after_tex": "w_{n+1}=\\tfrac{w_n}{2}=\\tfrac{w_0}{2^{n+1}}",
        "relation": "induction_step",
        "rule": "nat_induction",
        "conditions": [
          "\\text{no early stop}"
        ],
        "explanation": {
          "zh-Hant": "新區間是 [a_n,m] 或 [m,b_n]，由 S03.1，兩者寬度都是 (b_n−a_n)/2=w_n/2。代入歸納假設：w_(n+1)=w0/(2^n·2)=w0/2^(n+1)。依歸納原理證得引理 L_width。代入 w0=1、n=4：w4=1/16，與 23/16−11/8=1/16 一致（N11）。",
          "en": "The new interval is [a_n,m] or [m,b_n], and by S03.1 both have width (b_n−a_n)/2=w_n/2. With the induction hypothesis, w_(n+1)=w0/(2^n·2)=w0/2^(n+1). By induction this proves Lemma L_width. With w0=1 and n=4, w4=1/16, matching 23/16−11/8=1/16 (N11)."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S03.1",
            "power_laws",
            "frac_rules"
          ],
          "explanation": {
            "zh-Hant": "兩種選擇的寬度相同，所以不必知道選了哪一半。",
            "en": "Both choices have the same width, so it does not matter which half was chosen."
          }
        },
        "leaf_basis": "nat_induction + frac_rules + power_laws"
      },
      {
        "id": "S11",
        "uses": [
          "S08",
          "S10"
        ],
        "before_tex": "[a_4,b_4]=[\\tfrac{11}{8},\\tfrac{23}{16}]",
        "after_tex": "w_4=\\tfrac{1}{16},\\ m_5=\\tfrac{45}{32},\\ |x-m_5|\\le\\tfrac{1}{32}\\ \\ \\forall x\\in[a_4,b_4]",
        "relation": "implies",
        "rule": "L_mid",
        "conditions": [],
        "explanation": {
          "zh-Hant": "最終 bracket 寬度 1/16。它的中點 m5=(11/8+23/16)/2=45/32（N16）；由 L_mid，bracket 中每一點到 m5 的距離至多 (1/16)/2=1/32。這是關於區間幾何的敘述，說的是「若某點在 bracket 中」，並沒有說有這樣的根。",
          "en": "The final bracket has width 1/16. Its midpoint is m5=(11/8+23/16)/2=45/32 (N16); by L_mid every point of the bracket is within (1/16)/2=1/32 of m5. This is a statement about interval geometry, about any point that lies in the bracket; it does not say that such a root exists."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "L_mid",
            "S10"
          ],
          "explanation": {
            "zh-Hant": "只用中點與寬度；距離界套用到根之前，必須先另外證明根在區間內。",
            "en": "Only the midpoint and the width are used; before the distance bound can be applied to a root, the root must separately be shown to lie in the interval."
          }
        },
        "leaf_basis": "L_mid + abs_rules + numeric_evaluation"
      },
      {
        "id": "S12",
        "uses": [
          "S10",
          "archimedean",
          "nat_induction"
        ],
        "before_tex": "\\tau>0",
        "after_tex": "\\exists N:\\ \\tfrac{1}{2^N}<\\tau",
        "relation": "implies",
        "rule": "archimedean",
        "conditions": [
          "\\tau>0"
        ],
        "explanation": {
          "zh-Hant": "寬度指標：若每次所需的中點符號都能取得、更新能有效完成，並在觀察到零時提前停止，則把停止準則改為「寬度 ≤τ 就停」後，對任何 τ>0 都在有限步內停止。這是延伸執行的條件式敘述；題目實際只提供四個中點符號。先對 n 歸納證 2^n≥n+1：n=0 時 1≥1；若 2^n≥n+1，則 2^(n+1)=2·2^n≥2n+2≥n+2。再由阿基米德性質取自然數 N>1/τ，得 2^N≥N+1>1/τ，所以 w_N=1/2^N<τ。N14 是對 n=0 到 20 的有限核對，證明本身是上面的歸納。",
          "en": "Width index: provided every requested midpoint sign is available and updates can be performed effectively, with early stopping when a zero is observed, the rule 'stop when the width is ≤τ' terminates after finitely many steps for every τ>0. This is a conditional statement about an extended run; the exercise actually supplies only four midpoint signs. First, 2^n≥n+1 by induction: for n=0, 1≥1; if 2^n≥n+1, then 2^(n+1)=2·2^n≥2n+2≥n+2. Then by the Archimedean property choose a natural number N>1/τ, so 2^N≥N+1>1/τ and w_N=1/2^N<τ. N14 is a finite check for n=0 to 20; the proof itself is the induction above."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "archimedean",
            "nat_induction",
            "S10"
          ],
          "explanation": {
            "zh-Hant": "需要 τ>0；τ=0 時寬度永遠不會 ≤0，準則不會觸發。這是演算法的終止，不是根的存在。",
            "en": "It needs τ>0; with τ=0 the width never becomes ≤0 and the rule never fires. This is termination of the algorithm, not existence of a root."
          }
        },
        "leaf_basis": "archimedean + nat_induction + order_basics"
      },
      {
        "id": "S13",
        "uses": [
          "G1",
          "G2",
          "S05",
          "S06",
          "S07",
          "S08"
        ],
        "before_tex": "g(x)=\\begin{cases}-1 & x<\\tfrac{7}{5}\\\\ 1 & x\\ge\\tfrac{7}{5}\\end{cases}",
        "after_tex": "g\\ \\text{gives the same signs and brackets, and}\\ g(x)\\neq 0\\ \\forall x",
        "relation": "implies",
        "rule": "case_analysis",
        "conditions": [],
        "explanation": {
          "zh-Hant": "符號資料本身不證明有根。取階梯函數 g：x<7/5 時 g(x)=−1，x≥7/5 時 g(x)=1。比較 1、2、3/2、5/4、11/8、23/16 與 7/5：1<7/5、5/4<7/5、11/8<7/5，其餘 ≥7/5（N17 至 N20）。所以 g 在這六點的符號與 G1、G2 完全相同，四步產生完全相同的 bracket，但 g 的值只有 −1 與 1，沒有根。演算法只看符號，無法區分 f 與 g；因此 S02 到 S12 的任何結論都不能單獨推出根存在。",
          "en": "Sign data alone do not prove a root. Take the step function g with g(x)=−1 for x<7/5 and g(x)=1 for x≥7/5. Compare 1, 2, 3/2, 5/4, 11/8 and 23/16 with 7/5: 1<7/5, 5/4<7/5 and 11/8<7/5, while the others are ≥7/5 (N17 to N20). So g has exactly the signs of G1 and G2 at these six points and the four steps produce exactly the same brackets, yet g takes only the values −1 and 1 and has no root. The algorithm sees only signs and cannot tell f from g; hence nothing in S02 to S12 by itself yields a root."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "case_analysis",
            "numeric_evaluation"
          ],
          "explanation": {
            "zh-Hant": "一個「資料相同、卻無根」的函數就足以說明資料不蘊含根存在；g 在 7/5 不連續，這正是中間值定理的前提所排除的情形。",
            "en": "One function with the same data and no root suffices to show that the data do not imply a root; g is discontinuous at 7/5, which is exactly what the hypothesis of the intermediate value theorem excludes."
          }
        },
        "leaf_basis": "case_analysis + numeric_evaluation + exists_intro"
      },
      {
        "id": "S14",
        "uses": [
          "G4",
          "poly_continuous",
          "ivt",
          "S07",
          "S08",
          "S11"
        ],
        "before_tex": "f\\ \\text{continuous on}\\ [\\tfrac{11}{8},\\tfrac{23}{16}],\\ f(\\tfrac{11}{8})=-\\tfrac{7}{64}<0<\\tfrac{17}{256}=f(\\tfrac{23}{16})",
        "after_tex": "\\exists c\\in[\\tfrac{11}{8},\\tfrac{23}{16}]:\\ f(c)=0,\\ |c-\\tfrac{45}{32}|\\le\\tfrac{1}{32}",
        "relation": "theorem_application",
        "rule": "ivt",
        "conditions": [
          "f\\ \\text{continuous on}\\ [a_4,b_4]",
          "f(a_4)<0<f(b_4)"
        ],
        "explanation": {
          "zh-Hant": "這是獨立的根存在責任。f 是多項式，在 [11/8,23/16] 上連續（G4）；端點值以精確計算得 f(11/8)=−7/64<0、f(23/16)=17/256>0（N8、N10）。中間值定理（引用，citation_verified）的兩個前提都成立，所以存在 c∈[11/8,23/16] 使 f(c)=0；再由 S11，|c−45/32|≤1/32。這個 c 就是 √2，但本稿不需要這個名稱。",
          "en": "This is the separate proof obligation for root existence. f is a polynomial and continuous on [11/8,23/16] (G4); exact evaluation gives the endpoint values f(11/8)=−7/64<0 and f(23/16)=17/256>0 (N8, N10). Both hypotheses of the intermediate value theorem (cited, citation_verified) hold, so some c∈[11/8,23/16] has f(c)=0; and by S11, |c−45/32|≤1/32. This c is √2, though the guide does not need that name."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "ivt",
            "poly_continuous",
            "G4",
            "S11"
          ],
          "explanation": {
            "zh-Hant": "定理需要連續性與端點嚴格異號的實際函數值；兩者都由本步另外提供，而不是由 bracket 更新得到。S13 的 g 缺連續性，所以同樣的論證對 g 不能做。",
            "en": "The theorem needs continuity and actual endpoint values of strictly opposite sign; both are supplied separately here rather than by the bracket updates. The g of S13 lacks continuity, so the same argument cannot be made for g."
          }
        },
        "leaf_basis": "ivt + poly_continuous + numeric_evaluation"
      },
      {
        "id": "S15",
        "uses": [
          "S08",
          "S09",
          "S10",
          "S12",
          "S13",
          "S14"
        ],
        "before_tex": "\\text{trace, invariant, width, termination, root duty}",
        "after_tex": "[a_4,b_4]=[\\tfrac{11}{8},\\tfrac{23}{16}],\\ w_4=\\tfrac{1}{16},\\ \\text{root in}\\ [a_4,b_4]\\ \\text{only via IVT}",
        "relation": "implies",
        "rule": "and_intro_elim",
        "conditions": [],
        "explanation": {
          "zh-Hant": "彙整：四步追蹤得到 [11/8,23/16]（S08），不變量保持端點異號（S09），寬度為 1/16（S10），寬度準則必終止（S12）；這些都只是 bracket 資料的性質（S13）。區間內有根是另一個結論，來自連續性與中間值定理（S14）。",
          "en": "Summary: the four-step trace gives [11/8,23/16] (S08), the invariant keeps opposite endpoint signs (S09), the width is 1/16 (S10), and a width rule always terminates (S12); all of these are properties of bracket data only (S13). That the interval contains a root is a different conclusion, coming from continuity and the intermediate value theorem (S14)."
        },
        "reverse_audit": {
          "kind": "sufficient_goal_reduction",
          "required_ids": [
            "S08",
            "S09",
            "S10",
            "S12",
            "S13",
            "S14"
          ],
          "explanation": {
            "zh-Hant": "目標 G：追蹤結果、不變量、寬度、終止與有根的結論。足夠條件 H：S08、S09、S10、S12、S14 各別成立；S13 界定哪些結論不能只由資料得到。H 都由資料、引理與引用定理正向得出，沒有把 G 當作前提。",
            "en": "Goal G: the trace result, invariant, width, termination and the root conclusion. Sufficient condition H: S08, S09, S10, S12 and S14 each hold, with S13 marking which conclusions cannot come from the data alone. H is derived forwards from the data, the lemmas and the cited theorem; G is never used as a premise."
          }
        },
        "leaf_basis": "and_intro_elim"
      }
    ],
    "result": {
      "zh-Hant": "從 [1,2] 依所提供的符號 +、−、−、+ 執行四步二分，bracket 依序為 [1,3/2]、[5/4,3/2]、[11/8,3/2]、[11/8,23/16]，寬度 1/16=1/2⁴；端點異號是不變量，寬度準則對任何 τ>0 都會終止。這些 bracket 資料本身不證明有根（階梯函數 g 給出相同資料卻無根）；另以多項式的連續性與中間值定理，得出 f(x)=x²−2 在 [11/8,23/16] 內有根 c，且 |c−45/32|≤1/32。",
      "en": "Starting from [1,2] with the supplied signs +, −, −, +, four bisection steps give the brackets [1,3/2], [5/4,3/2], [11/8,3/2] and [11/8,23/16], of width 1/16=1/2⁴; opposite endpoint signs form an invariant, and a width rule terminates for every τ>0. These bracket data alone do not prove a root (the step function g gives the same data and has no root); separately, continuity of polynomials and the intermediate value theorem show that f(x)=x²−2 has a root c in [11/8,23/16], with |c−45/32|≤1/32."
    },
    "negative_controls": [
      {
        "claim": {
          "zh-Hant": "f(1)<0<f(2)，所以二分法本身證明了 [1,2] 中有根",
          "en": "f(1)<0<f(2), so the bisection method itself proves that [1,2] contains a root"
        },
        "expected": "reject",
        "reason": {
          "zh-Hant": "階梯函數 g 有相同的端點與中點符號，卻沒有根（S13）；有根要另用連續性與中間值定理。",
          "en": "The step function g has the same endpoint and midpoint signs and no root (S13); a root needs continuity and the intermediate value theorem separately."
        }
      },
      {
        "claim": {
          "zh-Hant": "僅由最終 bracket 寬度為 1/16，就能保證任何給出同樣符號資料的函數 h 都滿足 |h(45/32)|≤1/16。",
          "en": "The final bracket width 1/16 alone guarantees |h(45/32)|≤1/16 for every function h giving the same sign data."
        },
        "expected": "reject",
        "reason": {
          "zh-Hant": "寬度只控制 x 的距離，不一般控制函數值。取 h(x)=100(x²−2)，它給出同樣符號與 bracket，但在同一中點 |h(45/32)|=575/256>1/16。原 f 的 |f(45/32)|=23/1024≤1/16 只是本例的數值事實，並非由寬度單獨保證。",
          "en": "Width controls distances in x, not function values in general. Take h(x)=100(x²−2), which gives the same signs and brackets but at the same midpoint has |h(45/32)|=575/256>1/16. The original f happens to satisfy |f(45/32)|=23/1024≤1/16; that numerical fact is not guaranteed by width alone."
        }
      },
      {
        "claim": {
          "zh-Hant": "s(3/2)=+，所以保留 [3/2,2]",
          "en": "s(3/2)=+, so keep [3/2,2]"
        },
        "expected": "reject",
        "reason": {
          "zh-Hant": "[3/2,2] 兩端都是 +；異號的是 [1,3/2]（S05）。",
          "en": "Both ends of [3/2,2] are +; the half with opposite signs is [1,3/2] (S05)."
        }
      },
      {
        "claim": {
          "zh-Hant": "從 [1,2] 出發但所提供的符號是 s(1)=+、s(2)=+，二分法仍可正常進行",
          "en": "Starting from [1,2] with supplied signs s(1)=+ and s(2)=+, bisection can still proceed normally"
        },
        "expected": "reject",
        "reason": {
          "zh-Hant": "兩端同號不符合本演算法要求的端點異號輸入契約（S02），因此不能直接引用該 bracket 不變量；不據此宣稱不存在可由額外符號資料找出的端點異號子區間。",
          "en": "Equal endpoint signs violate this algorithm's opposite-sign input contract (S02), so the bracket invariant cannot be invoked directly. This does not claim that no opposite-sign subinterval could be found with additional sign information."
        }
      }
    ],
    "status": "candidate",
    "gaps": [],
    "verification": {
      "structure": "checked_local_script",
      "mathematics": "candidate_argument",
      "formalization": "not_formalized"
    },
    "rule_glossary": [
      {
        "id": "order_basics",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "實數的序：任兩數恰有 u<v、u=v、u>v 之一；u<v 且 v<w 推出 u<w；u<v 推出 u+w<v+w；u<v 且 w>0 推出 uw<vw；u<v 且 w<0 推出 uw>vw。",
          "en": "Order of the reals: exactly one of u<v, u=v, u>v holds; u<v and v<w give u<w; u<v gives u+w<v+w; u<v and w>0 give uw<vw; u<v and w<0 give uw>vw."
        }
      },
      {
        "id": "frac_rules",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "分數運算（分母皆非零）：a/c+b/c=(a+b)/c；(a/b)(c/d)=(ac)/(bd)；(a/b)/(c/d)=(ad)/(bc)（另需 c≠0）；a/b=(ak)/(bk)（k≠0）。",
          "en": "Fraction rules (all denominators nonzero): a/c+b/c=(a+b)/c; (a/b)(c/d)=(ac)/(bd); (a/b)/(c/d)=(ad)/(bc) (also needing c≠0); a/b=(ak)/(bk) for k≠0."
        }
      },
      {
        "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": "nat_induction",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "數學歸納原理：對從 n₀ 起的命題族 P(n)，若 P(n₀) 成立，且對每個 n≥n₀ 都有 P(n)⇒P(n+1)，則所有 n≥n₀ 都有 P(n)。",
          "en": "Principle of induction: for a family P(n) starting at n₀, if P(n₀) holds and P(n)⇒P(n+1) for every n≥n₀, then P(n) holds for every n≥n₀."
        }
      },
      {
        "id": "archimedean",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "阿基米德性質：對任意實數 x，存在自然數 n>x；因此對任意 ε>0，存在 n 使 1/n<ε。",
          "en": "Archimedean property: for every real x there is a natural number n>x; hence for every ε>0 there is n with 1/n<ε."
        }
      },
      {
        "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": "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": "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": "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": "def_unfold",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "展開定義：把一個已定義的詞換成它的定義內容（或反向收回），命題的真假不變。",
          "en": "Unfolding a definition: replacing a defined term by its definition (or folding it back) does not change whether a statement is true."
        }
      },
      {
        "id": "eq_refl_symm_trans",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "等號的性質：u=u；若 u=v 則 v=u；若 u=v 且 v=w 則 u=w。",
          "en": "Properties of equality: u=u; if u=v then v=u; if u=v and v=w then u=w."
        }
      },
      {
        "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|<r ⇔ −r<u<r。",
          "en": "Absolute value (reals): |u|=u if u≥0 and |u|=−u if u<0; |u|≥0; |uv|=|u||v|; triangle inequality |u+v|≤|u|+|v|; |u|<r ⇔ −r<u<r."
        }
      },
      {
        "id": "exists_intro",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "∃引入：若找到 a∈D 使 P(a) 成立，則 ∃x∈D, P(x)。",
          "en": "∃-introduction: if some a∈D satisfies P(a), then ∃x∈D, P(x)."
        }
      },
      {
        "id": "mko_bisect",
        "kind": "cited",
        "source": {
          "mko_id": "mko-bisection-method",
          "sha256": "ab834bbfb6637765ed0a0282df9021cdc9509c95f67eb41e02c6d5db1774d835",
          "commit": "b28a5edbd46279b1db2a3f3f48bb3f447e18899f"
        },
        "statement": {
          "zh-Hant": "給定有序區間 [a,b] 及已提供的端點符號 bracket，每步取 m=(a+b)/2。若已觀察 f(m)=0 就停在 m；否則保留端點符號相反的半區。沒有另行使用中間值定理時，這只是 bracket 資料更新，不自行證明根存在。",
          "en": "Given an ordered interval [a,b] and supplied endpoint sign-bracket data, take m=(a+b)/2 at each step. If f(m)=0 has been observed, stop at m; otherwise retain the half whose endpoint signs are opposite. Without separately applying an intermediate value theorem, this is a bracket-data update and does not itself prove root existence."
        }
      },
      {
        "id": "ivt",
        "kind": "cited",
        "source": {
          "reference": "OpenStax, Calculus Volume 1, Section 2.4, Intermediate Value Theorem",
          "original_reference": "M. Spivak, Calculus, 4th ed., Chapter 7, Theorem 1 (intermediate value theorem, sign form)",
          "url": "https://openstax.org/books/calculus-volume-1/pages/2-4-continuity",
          "locator": "Intermediate Value Theorem and Example 2.36",
          "checked_date": "2026-10-19",
          "verification": "verified_replacement",
          "verification_note": "Continuity plus opposite endpoint signs yields a zero; sign data alone do not."
        },
        "needs_citation": false,
        "statement": {
          "zh-Hant": "中間值定理（符號形式）：若 f 在 [a,b] 上連續且 f(a)<0<f(b)，則存在 x∈[a,b] 使 f(x)=0。",
          "en": "Intermediate value theorem (sign form): if f is continuous on [a,b] and f(a)<0<f(b), then there is some x∈[a,b] with f(x)=0."
        }
      },
      {
        "id": "poly_continuous",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "多項式函數在 ℝ 上處處連續，因此在任何閉區間上連續。",
          "en": "Polynomial functions are continuous everywhere on ℝ, hence on every closed interval."
        }
      },
      {
        "id": "L_mid",
        "kind": "lemma",
        "proved_in": [
          "S03.1"
        ],
        "statement": {
          "zh-Hant": "若 a<b，m=(a+b)/2，則 a<m<b，且 m−a=b−m=(b−a)/2。",
          "en": "If a<b and m=(a+b)/2, then a<m<b and m−a=b−m=(b−a)/2."
        }
      },
      {
        "id": "L_half",
        "kind": "lemma",
        "proved_in": [
          "S04.1",
          "S04.2",
          "S04.3"
        ],
        "statement": {
          "zh-Hant": "若 s(a)、s(b) 一為 − 一為 +，且 s(m)≠0，則 [a,m] 與 [m,b] 中恰有一個端點符號相反。",
          "en": "If one of s(a), s(b) is − and the other +, and s(m)≠0, then exactly one of [a,m] and [m,b] has opposite endpoint signs."
        }
      },
      {
        "id": "L_width",
        "kind": "lemma",
        "proved_in": [
          "S10.1",
          "S10.2"
        ],
        "statement": {
          "zh-Hant": "沒有提前停止時，n 次有效二分後 b_n−a_n=(b_0−a_0)/2^n。",
          "en": "Without an early stop, after n effective halvings b_n−a_n=(b_0−a_0)/2^n."
        }
      }
    ],
    "revision_log": [],
    "numeric_checks": [
      {
        "id": "N1",
        "step": "S02",
        "lhs": "1**2-2",
        "relation": "<",
        "rhs": "0"
      },
      {
        "id": "N2",
        "step": "S02",
        "lhs": "2**2-2",
        "relation": ">",
        "rhs": "0"
      },
      {
        "id": "N3",
        "step": "S03",
        "lhs": "(1+2)/2",
        "rhs": "3/2"
      },
      {
        "id": "N4",
        "step": "S05",
        "lhs": "(3/2)**2-2",
        "rhs": "1/4"
      },
      {
        "id": "N5",
        "step": "S06",
        "lhs": "(1+3/2)/2",
        "rhs": "5/4"
      },
      {
        "id": "N6",
        "step": "S06",
        "lhs": "(5/4)**2-2",
        "rhs": "-7/16"
      },
      {
        "id": "N7",
        "step": "S07",
        "lhs": "(5/4+3/2)/2",
        "rhs": "11/8"
      },
      {
        "id": "N8",
        "step": "S07",
        "lhs": "(11/8)**2-2",
        "rhs": "-7/64"
      },
      {
        "id": "N9",
        "step": "S08",
        "lhs": "(11/8+3/2)/2",
        "rhs": "23/16"
      },
      {
        "id": "N10",
        "step": "S08",
        "lhs": "(23/16)**2-2",
        "rhs": "17/256"
      },
      {
        "id": "N11",
        "step": "S10.2",
        "lhs": "23/16-11/8",
        "rhs": "1/2**4"
      },
      {
        "id": "N12",
        "step": "S03.1",
        "vars": [
          "a",
          "b"
        ],
        "degree": 1,
        "lhs": "(a+b)/2-a",
        "rhs": "(b-a)/2"
      },
      {
        "id": "N13",
        "step": "S03.1",
        "vars": [
          "a",
          "b"
        ],
        "degree": 1,
        "lhs": "b-(a+b)/2",
        "rhs": "(b-a)/2"
      },
      {
        "id": "N14",
        "step": "S12",
        "vars": [
          "n"
        ],
        "range": [
          0,
          20
        ],
        "lhs": "2**n",
        "relation": ">=",
        "rhs": "n+1"
      },
      {
        "id": "N15",
        "lhs": "100*((11/8)**2-2)",
        "rhs": "-175/16"
      },
      {
        "id": "N16",
        "step": "S11",
        "lhs": "(11/8+23/16)/2",
        "rhs": "45/32"
      },
      {
        "id": "N17",
        "step": "S13",
        "lhs": "11/8-7/5",
        "relation": "<",
        "rhs": "0"
      },
      {
        "id": "N18",
        "step": "S13",
        "lhs": "5/4-7/5",
        "relation": "<",
        "rhs": "0"
      },
      {
        "id": "N19",
        "step": "S13",
        "lhs": "23/16-7/5",
        "relation": ">",
        "rhs": "0"
      },
      {
        "id": "N20",
        "step": "S13",
        "lhs": "3/2-7/5",
        "relation": ">",
        "rhs": "0"
      },
      {
        "id": "N21",
        "step": "S11",
        "lhs": "23/16-45/32",
        "rhs": "1/32"
      }
    ]
  },
  "presentation": {
    "md_scope": {
      "zh-Hant": "在 MKO 的範圍內工作：端點與中點符號由題目直接提供（假設 a1），演算法只做 bracket 資料更新；寬度界 w/2^n 是幾何減半結果，不是根存在證明（假設 a2）。根存在作為獨立責任，以引用的中間值定理（citation_verified）與多項式的連續性處理。不處理浮點誤差、收斂速度比較或其他求根法；不改動 canonical MKO 或英文 edition。",
      "en": "Works inside the MKO's scope: endpoint and midpoint signs are supplied directly by the problem (assumption a1), and the algorithm only updates bracket data; the width bound w/2^n is a geometric halving result, not a proof that a root exists (assumption a2). Root existence is handled as a separate proof obligation with the cited intermediate value theorem (citation_verified) and continuity of polynomials. Floating-point error, comparisons of convergence speed and other root-finding methods are not treated; the canonical MKO and its English edition are not changed."
    },
    "md_symbols": [
      [
        "[a_n,b_n]",
        "第 n 步後的 bracket",
        "the bracket after step n"
      ],
      [
        "m_n",
        "第 n 步的中點 (a_{n-1}+b_{n-1})/2",
        "the midpoint (a_{n-1}+b_{n-1})/2 of step n"
      ],
      [
        "s(x)",
        "所提供的 f(x) 的符號，取 −、0、+",
        "the supplied sign of f(x), one of −, 0, +"
      ],
      [
        "w_n=b_n-a_n",
        "第 n 步後的寬度",
        "the width after step n"
      ],
      [
        "\\tau",
        "寬度停止準則的容許值",
        "the tolerance of a width-based stopping rule"
      ]
    ],
    "md_error_controls": [
      {
        "title": "異號不等於有根 / Opposite signs alone do not prove a root exists",
        "zh": "MKO 指出的誤解之一是看到異號就省略中間值定理而直接宣稱有根。S13 的階梯函數 g 與 f 在所有用到的點上符號相同，產生相同的 bracket，卻沒有根。有根需要連續性加上中間值定理（S14），這是 bracket 更新之外的獨立責任。",
        "en": "One mistake named by the MKO is to skip the intermediate value theorem and declare a root on seeing opposite signs. The step function g of S13 has the same signs as f at every point used and produces the same brackets, yet has no root. A root needs continuity plus the intermediate value theorem (S14), a proof obligation separate from the bracket updates."
      },
      {
        "title": "區間寬度不是函數值誤差 / Width is not function-value error",
        "zh": "MKO 指出的另一誤解。最終寬度 1/16 只界定 x 方向的距離（且要先知道根在區間內）。函數 h(x)=100(x²−2) 有完全相同的符號與 bracket，但 h(11/8)=−175/16，絕對值約 10.9（N15），遠大於 1/16。寬度界與 |f(x)| 的大小是兩回事。",
        "en": "The other mistake named by the MKO. The final width 1/16 bounds only distances in the x-direction (and only once the root is known to lie in the interval). The function h(x)=100(x²−2) has exactly the same signs and brackets, but h(11/8)=−175/16, about 10.9 in absolute value (N15), far larger than 1/16. The width bound and the size of |f(x)| are different things."
      },
      {
        "title": "選半區時看端點符號 / Choose the half by endpoint signs",
        "zh": "最容易做錯的是保留錯的一半：s(m) 與 s(a) 同號時要保留 [m,b]，不是 [a,m]（S04.1）。檢查方法是看保留的區間兩端是否一負一正；S09 的不變量要求左端為 −、右端為 +。",
        "en": "The step most easily done wrong is keeping the wrong half: when s(m) equals s(a), keep [m,b], not [a,m] (S04.1). The check is whether the kept interval has one negative and one positive end; the invariant of S09 requires − at the left end and + at the right."
      }
    ],
    "md_open_items": [
      {
        "zh-Hant": "中間值定理以引用方式使用（citation_verified），本稿沒有證明它；多項式的連續性作為前置知識。",
        "en": "The intermediate value theorem is used by citation (citation_verified) and not proved here; continuity of polynomials is a prerequisite."
      },
      {
        "zh-Hant": "N14 是 n=0 到 20 的有限核對，2^n≥n+1 對所有 n 的證明是 S12 中的歸納；N12、N13 是一次多項式恆等式的網格核對。",
        "en": "N14 is a finite check for n=0 to 20; the proof that 2^n≥n+1 for all n is the induction in S12; N12 and N13 are grid checks of degree-one polynomial identities."
      }
    ]
  },
  "revision": {
    "id": "production-20261019-adopted-96",
    "source_commit": "b28a5edbd46279b1db2a3f3f48bb3f447e18899f",
    "adopted_source_sha256": "fc6815b2e34c1a498109d7c25fa634c56e668faddafc9d05c16b74cc9206e49d",
    "original_source_sha256": "6785e2b4757cf8bcdb04bf8c2e24e02150a9f5d955df275d9ba4228ece867aaa",
    "owner_adopted": true,
    "bilingual_sync": "complete_source_bound",
    "formal_Lean": "NOT_RUN",
    "independent_AI_concurrence": "NOT_MEASURED",
    "issue_ids": [
      "BR-M047",
      "BR-M048"
    ],
    "scope": "Owner-adopted natural-language teaching guide; not a formal proof, canonical MKO rewrite or evidence-status promotion."
  }
}