{
  "schema_version": "ocme-reviewed-question-reader-v1",
  "question_id": "q-20261011-advance-20261015-r01-fraction-0012",
  "pilot_number": "R0012-06962",
  "number": 6962,
  "family_id": "qf-density15-fraction-sum-naive-audit",
  "title": {
    "zh-Hant": "第 6962 題：真分數和與字面錯加",
    "en": "Question 6962: True Fraction Sum versus Naive Representation Addition"
  },
  "original_question": {
    "schema_version": "ocme-question-v0.1",
    "id": "q-20261011-advance-20261015-r01-fraction-0012",
    "batch_id": "batch-20261011-advance-20261015-r01-fraction-foundations",
    "language": "zh-Hant",
    "source_mko_ids": [
      "mko-fraction"
    ],
    "title_zh": "真分數和與字面錯加",
    "stem_zh": "合法分數表示，原分母非零。a=(1)/(-2)、b=(1)/(3)。錯加候選直接以原分子相加除以原分母相加，不先約分。依序答 (真和, 錯加定義資格, 錯加值或未定義, 錯加減真和或未定義, 錯加正確資格)。",
    "answer": {
      "kind": "expression",
      "value": "(-1/6, 是, 2/1, 13/6, 否)",
      "explanation_zh": "分數真和與錯加候選稽核=(-1/6, 是, 2/1, 13/6, 否)。真和先通分；錯加另核原分母和，再比較精確差。分子分母分別加不是分數加法。表示與數值分開，有限分數核對非商集良定性證明。"
    },
    "difficulty": {
      "band": "primary_foundation",
      "level": 2,
      "skills": [
        "真分數和與字面錯加"
      ]
    },
    "variant": {
      "family_id": "qf-density15-fraction-sum-naive-audit",
      "seed": 202610151012,
      "parameters": {
        "a": [
          1,
          -2
        ],
        "b": [
          1,
          3
        ],
        "domain": "finite_fraction_representation_protocol",
        "operation": "density15_fraction_sum_naive_audit",
        "semantic_template": "qf-density15-fraction-sum-naive-audit/[\"-1/6\",\"2/1\"]"
      }
    },
    "verification": {
      "status": "candidate",
      "checks": [
        "producer_exact_integer_rational_arithmetic",
        "producer_finite_enumeration",
        "producer_source_snapshot_check",
        "producer_substantive_template_deduplication",
        "producer_json_schema"
      ],
      "warning_zh": "僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"
    },
    "provenance": {
      "generator_role": "question_producer",
      "generator_model": "model_unknown",
      "generated_at": "2026-10-11",
      "contract_version": "ocme-question-factory-v0.1"
    }
  },
  "original_english_edition": {
    "schema_version": "ocme-question-localization-v0.1",
    "question_id": "q-20261011-advance-20261015-r01-fraction-0012",
    "batch_id": "batch-20261011-advance-20261015-r01-fraction-foundations",
    "locale": "en",
    "translation_state": "ai_candidate",
    "source": {
      "question_sha256": "ec220b0c472099adb6c3a7411b0d3594f330413a3ce7ee9f0c933a9ce7c12e4b",
      "family_id": "qf-density15-fraction-sum-naive-audit",
      "seed": 202610151012,
      "semantic_template": "qf-density15-fraction-sum-naive-audit/[\"-1/6\",\"2/1\"]"
    },
    "fields": {
      "title": "True Fraction Sum versus Naive Representation Addition",
      "stem": "Use legal integer fraction representations with nonzero original denominators; negative denominators are allowed. Representation and value remain distinct. a=1/-2, b=1/3. The naive candidate adds the original numerators and original denominators separately, before reduction. Give (true sum, is naive addition defined?, naive value or undefined, naive−true defect or undefined, is naive addition correct?).",
      "answer": {
        "kind": "expression",
        "value": "(-1/6, Yes, 2/1, 13/6, No)",
        "explanation": "Compute the true sum by common denominators; separately guard the original denominator sum before evaluating the naive candidate. Adding numerators and denominators separately is not fraction addition. The exact result, retaining the requested order, is (-1/6, Yes, 2/1, 13/6, No)."
      },
      "skills": [
        "True Fraction Sum versus Naive Representation Addition"
      ],
      "verification_warning": "This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."
    }
  },
  "givens": [
    {
      "id": "G001",
      "text": {
        "zh-Hant": "合法分數表示，原分母非零。a=(1)/(-2)、b=(1)/(3)。錯加候選直接以原分子相加除以原分母相加，不先約分。依序答 (真和, 錯加定義資格, 錯加值或未定義, 錯加減真和或未定義, 錯加正確資格)。",
        "en": "Use legal integer fraction representations with nonzero original denominators; negative denominators are allowed. Representation and value remain distinct. a=1/-2, b=1/3. The naive candidate adds the original numerators and original denominators separately, before reduction. Give (true sum, is naive addition defined?, naive value or undefined, naive−true defect or undefined, is naive addition correct?)."
      }
    }
  ],
  "steps": [
    {
      "id": "S001",
      "uses": [
        "G001"
      ],
      "before": "original integer numerator/denominator representations;original denominators nonzero",
      "after": "representation and rational value are distinct data",
      "relation": "implies",
      "rule_id": "fraction_representation_value_scope",
      "conditions": [],
      "explanation": {
        "zh-Hant": "負分母表示合法，但約分後的值不取代原始表示；涉及原分子分母的候選算法要用原資料。此有限核對不是商集良定性的一般證明。",
        "en": "Negative denominators are valid,but reduced values do not replace original representations.Candidate algorithms about raw numerators/denominators must use those raw data.This finite audit is not a general proof of quotient well-definedness."
      },
      "reverse_audit": {
        "kind": "premise_and_result_audit",
        "explanation": {
          "zh-Hant": "回到原始資料與本步條件核對，不把單向推論誤稱等價。",
          "en": "Check against the original data and this step's premises; a one-way implication is not an equivalence."
        }
      },
      "arithmetic": null
    },
    {
      "id": "S002",
      "uses": [
        "G001",
        "S001"
      ],
      "before": "true rational sum := sum(-1/2, 1/3)",
      "after": "true rational sum = -1/6",
      "relation": "calculation",
      "rule_id": "exact_rational_arithmetic",
      "conditions": [],
      "explanation": {
        "zh-Hant": "真分數加法按有理值通分相加，與字面錯加分開。",
        "en": "The true rational sum uses common-denominator addition,separate from naive representation addition."
      },
      "reverse_audit": {
        "kind": "premise_and_result_audit",
        "explanation": {
          "zh-Hant": "以精確有理數重算 true rational sum，不以小數近似替代。",
          "en": "Recompute true rational sum with exact rational arithmetic, not rounded decimals."
        }
      },
      "arithmetic": {
        "operation": "sum",
        "inputs": [
          "-1/2",
          "1/3"
        ],
        "output": "-1/6",
        "label": "true rational sum"
      }
    },
    {
      "id": "S003",
      "uses": [
        "G001",
        "S002"
      ],
      "before": "naive raw numerator sum := sum(1, 1)",
      "after": "naive raw numerator sum = 2",
      "relation": "calculation",
      "rule_id": "exact_rational_arithmetic",
      "conditions": [],
      "explanation": {
        "zh-Hant": "錯加候選直接加原始兩個分子，不使用約分後分子。",
        "en": "The naive candidate adds the two ORIGINAL numerators,not reduced numerators."
      },
      "reverse_audit": {
        "kind": "premise_and_result_audit",
        "explanation": {
          "zh-Hant": "以精確有理數重算 naive raw numerator sum，不以小數近似替代。",
          "en": "Recompute naive raw numerator sum with exact rational arithmetic, not rounded decimals."
        }
      },
      "arithmetic": {
        "operation": "sum",
        "inputs": [
          "1",
          "1"
        ],
        "output": "2",
        "label": "naive raw numerator sum"
      }
    },
    {
      "id": "S004",
      "uses": [
        "G001",
        "S003"
      ],
      "before": "naive raw denominator sum := sum(-2, 3)",
      "after": "naive raw denominator sum = 1",
      "relation": "calculation",
      "rule_id": "exact_rational_arithmetic",
      "conditions": [],
      "explanation": {
        "zh-Hant": "同樣只加原始分母；原分母各非零不保證其和非零。",
        "en": "Likewise add the ORIGINAL denominators;each being nonzero does not make their sum nonzero."
      },
      "reverse_audit": {
        "kind": "premise_and_result_audit",
        "explanation": {
          "zh-Hant": "以精確有理數重算 naive raw denominator sum，不以小數近似替代。",
          "en": "Recompute naive raw denominator sum with exact rational arithmetic, not rounded decimals."
        }
      },
      "arithmetic": {
        "operation": "sum",
        "inputs": [
          "-2",
          "3"
        ],
        "output": "1",
        "label": "naive raw denominator sum"
      }
    },
    {
      "id": "S005",
      "uses": [
        "G001",
        "S004"
      ],
      "before": "defined naive fraction candidate := div(2, 1)",
      "after": "defined naive fraction candidate = 2",
      "relation": "calculation",
      "rule_id": "exact_rational_arithmetic",
      "conditions": [],
      "explanation": {
        "zh-Hant": "原分母和非零時，才給此候選一個有理值。",
        "en": "Only a nonzero original-denominator sum gives this candidate a rational value."
      },
      "reverse_audit": {
        "kind": "premise_and_result_audit",
        "explanation": {
          "zh-Hant": "以精確有理數重算 defined naive fraction candidate，不以小數近似替代。",
          "en": "Recompute defined naive fraction candidate with exact rational arithmetic, not rounded decimals."
        }
      },
      "arithmetic": {
        "operation": "div",
        "inputs": [
          "2",
          "1"
        ],
        "output": "2",
        "label": "defined naive fraction candidate"
      }
    },
    {
      "id": "S006",
      "uses": [
        "G001",
        "S005"
      ],
      "before": "naive minus true sum := sub(2, -1/6)",
      "after": "naive minus true sum = 13/6",
      "relation": "calculation",
      "rule_id": "exact_rational_arithmetic",
      "conditions": [],
      "explanation": {
        "zh-Hant": "候選減真和為有號缺陷，不能取絕對值。",
        "en": "Candidate minus true sum is the signed defect,not its absolute value."
      },
      "reverse_audit": {
        "kind": "premise_and_result_audit",
        "explanation": {
          "zh-Hant": "以精確有理數重算 naive minus true sum，不以小數近似替代。",
          "en": "Recompute naive minus true sum with exact rational arithmetic, not rounded decimals."
        }
      },
      "arithmetic": {
        "operation": "sub",
        "inputs": [
          "2",
          "-1/6"
        ],
        "output": "13/6",
        "label": "naive minus true sum"
      }
    },
    {
      "id": "S007",
      "uses": [
        "G001",
        "S006"
      ],
      "before": "raw numerator sum=2;raw denominator sum=1;true sum=-1/6",
      "after": "naive defined=是;naive=2/1;correct=否",
      "relation": "implies",
      "rule_id": "raw_fraction_naive_addition_audit",
      "conditions": [],
      "explanation": {
        "zh-Hant": "分子分母分別加不是分數加法；先約分可能改變題目指定的錯誤算法，故保留原表示。候選未定義時即使真和0，也不能判錯加正確。",
        "en": "Adding numerators and denominators separately is not fraction addition.Prior reduction may change the specified WRONG algorithm,so retain raw representations.An undefined candidate is not correct even when the true sum is0."
      },
      "reverse_audit": {
        "kind": "premise_and_result_audit",
        "explanation": {
          "zh-Hant": "回到原始資料與本步條件核對，不把單向推論誤稱等價。",
          "en": "Check against the original data and this step's premises; a one-way implication is not an equivalence."
        }
      },
      "arithmetic": null
    }
  ],
  "math_model": {
    "result": [
      "-1/6",
      true,
      "2",
      "13/6",
      false
    ],
    "raw_naive_numerator": "2",
    "raw_naive_denominator": "1"
  },
  "conclusion": {
    "canonical_value": "(-1/6, 是, 2/1, 13/6, 否)",
    "explanation": {
      "zh-Hant": "依上列完整代值及條件核對得到原題答案；原題編號、格式與狀態保持不變。",
      "en": "The full substitutions and premise checks give the original answer; source identity,format and status are unchanged."
    }
  },
  "definition_note": {
    "zh-Hant": "沿用原題明示的論域、假設與來源 MKO 定義。本題已完成第二對話數理與中英語義覆核，並由原始參數逐題獨立重算模型、答案與每個算術步驟；原題仍為 candidate。",
    "en": "Use the original question domain, assumptions and cited MKO definitions. This item has second-context mathematical and bilingual review, with the model, answer and every arithmetic step independently recomputed from the original parameters; the source question remains a candidate."
  },
  "source": {
    "input_commit": "0f7be61e2c22317a4075561dc822c87d881a50db",
    "question_binding": {
      "batch_id": "batch-20261011-advance-20261015-r01-fraction-foundations",
      "batch_path": "public/data/questions/batches/batch-20261011-advance-20261015-r01-fraction-foundations/questions.jsonl",
      "batch_sha256": "33621867facfc78fb3d0ff06bd432e33a58b1514c00596b9334e3a66c8e322fa",
      "original_line": 12,
      "original_line_sha256": "9e08e2f961aba90edebaa87f9352e2742e2da2f257a8309129559feb09f81c1f",
      "question_compact_sha256": "ec220b0c472099adb6c3a7411b0d3594f330413a3ce7ee9f0c933a9ce7c12e4b",
      "english_row_compact_sha256": "da63878ef50d533463a64ee6e6603aabc22832bdc1f2335140c0d062dec39059",
      "english_manifest_path": "public/data/i18n/editions/batch-20261011-advance-20261015-r01-fraction-foundations.en.json",
      "english_rows_path": "public/data/i18n/questions/batch-20261011-advance-20261015-r01-fraction-foundations.en.jsonl"
    },
    "mko_bindings": {
      "mko-fraction": {
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        "english_sha256": "ea8b15d28832a77ca243cb2c04a33ac67244aca632c895666d9a33fa9b4982d4",
        "original_path": "public/data/mko/fraction.json",
        "english_path": "public/data/i18n/editions/mko-fraction.en.json"
      }
    },
    "selected_manifest_sha256": "5233edba9654a6cffd8b6b2d83c932f82ded15f9c0b7c733bd049943e672acff",
    "selected_part_sha256": "c5c612ca91797c328f4489939ec6f7d885385e63ea21b334c09c421f17933d99",
    "original_candidate_sha256": "122a6d60bb9b436b3bc6936a44f85c2b79b0ba3cc521466f3e87add4e72d8ffa",
    "original_candidate_line_sha256": "122a6d60bb9b436b3bc6936a44f85c2b79b0ba3cc521466f3e87add4e72d8ffa"
  },
  "historical_source_state": {
    "review_state": {
      "derivation": "candidate_authored_and_locally_checked",
      "source_question_state": "candidate",
      "independent_AI_or_human_review": "NOT_PERFORMED",
      "formalization": "not_formalized",
      "Lean": "NOT_RUN"
    },
    "scope_note": {
      "zh-Hant": "來源綁定教學推導候選；精確核對不等於獨立AI／人工審查或一般形式證明。原題、答案、MKO、公式、AST與歷史不變。",
      "en": "Source-bound pedagogical derivation candidate.Exact checking is not independent AI/human review or general formal proof.Original questions,answers,MKOs,formulas,ASTs and history remain unchanged."
    },
    "label": "Frozen producer metadata; current per-question review is separate below."
  },
  "revision": {
    "id": "review-20261026-005-p13",
    "review_date": "2026-10-26",
    "owner_authorized_reader_publication": true,
    "scope": "One hundred sixty family prototypes and their mathematical/bilingual reasoning variants, stems, boundary statements and every distinct explanation shape were reviewed in a context independent of the producer; parameter substitutions, source bindings, full math models, both answer strings and every arithmetic step were independently reconstructed per question from the original parameters only, with negative controls. This grouped method does not claim a separate literal AI reading of every repeated sentence or an extra AI voter.",
    "review_receipts": {
      "semantic_review_sha256": "90dbc494b2d34ef51f412dcff63cfcc1a2215d038bff71f38f5dded4c64faf00",
      "question_first_witness_sha256": "b72a3b8bcf10f584396e0dd5198c657b0934a71864e5a3fa6321d7a613aef9f9"
    },
    "status": "second_context_reviewed",
    "mathematical_semantics": true,
    "bilingual_semantics": true,
    "steps_reviewed": 7,
    "method": "family_template_semantic_review_plus_independent_per_instance_reconstruction",
    "exact_question_first_check": {
      "model_sha256": "9282ccb25f57a9c2f9e2bfd6af9c3457e6606e9a21d24392d999b7dcc8662b4a",
      "arithmetic_steps": 5
    },
    "uses_answer_as_premise": false,
    "human_review": "NOT_PERFORMED",
    "formalization": "not_formalized",
    "Lean": "NOT_RUN",
    "Canonical_truth_promotion": false,
    "authority": "Owner-authorized reviewed reader publication; original question/candidate states remain historical and unchanged."
  }
}