{"schema_version":"ocme-question-v0.1","id":"q-20261010-advance-20261014-r06-absolute-value-0001","batch_id":"batch-20261010-advance-20261014-r06-absolute-value-foundations","language":"zh-Hant","source_mko_ids":["mko-absolute-value"],"title_zh":"有理絕對值的分支","stem_zh":"有理嵌入的絕對值協議；x=-12/3。依序答 (規約x, 分支非負或負, abs(x))；0屬非負分支。","answer":{"kind":"expression","value":"(-4/1, 負, 4/1)","explanation_zh":"有理絕對值的分支與值=(-4/1, 負, 4/1)。按規約值的符號取分支，零不歸負分支。有限分段核對非全部實數性質證明。"},"difficulty":{"band":"lower_secondary","level":1,"skills":["有理絕對值的分支"]},"variant":{"family_id":"qf-density14-absolute-rational-two-branches","seed":202610146001,"parameters":{"x":[-12,3],"domain":"finite_rational_absolute_protocol","operation":"density14_absolute_rational_two_branches","semantic_template":"qf-density14-absolute-rational-two-branches/\"-4/1\""}},"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":"2026-10-10","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261010-advance-20261014-r06-absolute-value-0002","batch_id":"batch-20261010-advance-20261014-r06-absolute-value-foundations","language":"zh-Hant","source_mko_ids":["mko-absolute-value"],"title_zh":"有理絕對值的分支","stem_zh":"有理嵌入的絕對值協議；x=11/-3。依序答 (規約x, 分支非負或負, abs(x))；0屬非負分支。","answer":{"kind":"expression","value":"(-11/3, 負, 11/3)","explanation_zh":"有理絕對值的分支與值=(-11/3, 負, 11/3)。按規約值的符號取分支，零不歸負分支。有限分段核對非全部實數性質證明。"},"difficulty":{"band":"lower_secondary","level":1,"skills":["有理絕對值的分支"]},"variant":{"family_id":"qf-density14-absolute-rational-two-branches","seed":202610146002,"parameters":{"x":[11,-3],"domain":"finite_rational_absolute_protocol","operation":"density14_absolute_rational_two_branches","semantic_template":"qf-density14-absolute-rational-two-branches/\"-11/3\""}},"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":"2026-10-10","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261010-advance-20261014-r06-absolute-value-0003","batch_id":"batch-20261010-advance-20261014-r06-absolute-value-foundations","language":"zh-Hant","source_mko_ids":["mko-absolute-value"],"title_zh":"有理絕對值的分支","stem_zh":"有理嵌入的絕對值協議；x=-10/3。依序答 (規約x, 分支非負或負, abs(x))；0屬非負分支。","answer":{"kind":"expression","value":"(-10/3, 負, 10/3)","explanation_zh":"有理絕對值的分支與值=(-10/3, 負, 10/3)。按規約值的符號取分支，零不歸負分支。有限分段核對非全部實數性質證明。"},"difficulty":{"band":"lower_secondary","level":1,"skills":["有理絕對值的分支"]},"variant":{"family_id":"qf-density14-absolute-rational-two-branches","seed":202610146003,"parameters":{"x":[-10,3],"domain":"finite_rational_absolute_protocol","operation":"density14_absolute_rational_two_branches","semantic_template":"qf-density14-absolute-rational-two-branches/\"-10/3\""}},"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":"2026-10-10","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261010-advance-20261014-r06-absolute-value-0004","batch_id":"batch-20261010-advance-20261014-r06-absolute-value-foundations","language":"zh-Hant","source_mko_ids":["mko-absolute-value"],"title_zh":"有理絕對值的分支","stem_zh":"有理嵌入的絕對值協議；x=9/-3。依序答 (規約x, 分支非負或負, abs(x))；0屬非負分支。","answer":{"kind":"expression","value":"(-3/1, 負, 3/1)","explanation_zh":"有理絕對值的分支與值=(-3/1, 負, 3/1)。按規約值的符號取分支，零不歸負分支。有限分段核對非全部實數性質證明。"},"difficulty":{"band":"lower_secondary","level":1,"skills":["有理絕對值的分支"]},"variant":{"family_id":"qf-density14-absolute-rational-two-branches","seed":202610146004,"parameters":{"x":[9,-3],"domain":"finite_rational_absolute_protocol","operation":"density14_absolute_rational_two_branches","semantic_template":"qf-density14-absolute-rational-two-branches/\"-3/1\""}},"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":"2026-10-10","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261010-advance-20261014-r06-absolute-value-0005","batch_id":"batch-20261010-advance-20261014-r06-absolute-value-foundations","language":"zh-Hant","source_mko_ids":["mko-absolute-value"],"title_zh":"有理絕對值的分支","stem_zh":"有理嵌入的絕對值協議；x=-8/3。依序答 (規約x, 分支非負或負, abs(x))；0屬非負分支。","answer":{"kind":"expression","value":"(-8/3, 負, 8/3)","explanation_zh":"有理絕對值的分支與值=(-8/3, 負, 8/3)。按規約值的符號取分支，零不歸負分支。有限分段核對非全部實數性質證明。"},"difficulty":{"band":"lower_secondary","level":1,"skills":["有理絕對值的分支"]},"variant":{"family_id":"qf-density14-absolute-rational-two-branches","seed":202610146005,"parameters":{"x":[-8,3],"domain":"finite_rational_absolute_protocol","operation":"density14_absolute_rational_two_branches","semantic_template":"qf-density14-absolute-rational-two-branches/\"-8/3\""}},"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":"2026-10-10","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261010-advance-20261014-r06-absolute-value-0006","batch_id":"batch-20261010-advance-20261014-r06-absolute-value-foundations","language":"zh-Hant","source_mko_ids":["mko-absolute-value"],"title_zh":"有理絕對值的分支","stem_zh":"有理嵌入的絕對值協議；x=7/-3。依序答 (規約x, 分支非負或負, abs(x))；0屬非負分支。","answer":{"kind":"expression","value":"(-7/3, 負, 7/3)","explanation_zh":"有理絕對值的分支與值=(-7/3, 負, 7/3)。按規約值的符號取分支，零不歸負分支。有限分段核對非全部實數性質證明。"},"difficulty":{"band":"lower_secondary","level":1,"skills":["有理絕對值的分支"]},"variant":{"family_id":"qf-density14-absolute-rational-two-branches","seed":202610146006,"parameters":{"x":[7,-3],"domain":"finite_rational_absolute_protocol","operation":"density14_absolute_rational_two_branches","semantic_template":"qf-density14-absolute-rational-two-branches/\"-7/3\""}},"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":"2026-10-10","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261010-advance-20261014-r06-absolute-value-0007","batch_id":"batch-20261010-advance-20261014-r06-absolute-value-foundations","language":"zh-Hant","source_mko_ids":["mko-absolute-value"],"title_zh":"有理絕對值的分支","stem_zh":"有理嵌入的絕對值協議；x=-6/3。依序答 (規約x, 分支非負或負, abs(x))；0屬非負分支。","answer":{"kind":"expression","value":"(-2/1, 負, 2/1)","explanation_zh":"有理絕對值的分支與值=(-2/1, 負, 2/1)。按規約值的符號取分支，零不歸負分支。有限分段核對非全部實數性質證明。"},"difficulty":{"band":"lower_secondary","level":1,"skills":["有理絕對值的分支"]},"variant":{"family_id":"qf-density14-absolute-rational-two-branches","seed":202610146007,"parameters":{"x":[-6,3],"domain":"finite_rational_absolute_protocol","operation":"density14_absolute_rational_two_branches","semantic_template":"qf-density14-absolute-rational-two-branches/\"-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":"2026-10-10","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261010-advance-20261014-r06-absolute-value-0008","batch_id":"batch-20261010-advance-20261014-r06-absolute-value-foundations","language":"zh-Hant","source_mko_ids":["mko-absolute-value"],"title_zh":"有理絕對值的分支","stem_zh":"有理嵌入的絕對值協議；x=5/-3。依序答 (規約x, 分支非負或負, abs(x))；0屬非負分支。","answer":{"kind":"expression","value":"(-5/3, 負, 5/3)","explanation_zh":"有理絕對值的分支與值=(-5/3, 負, 5/3)。按規約值的符號取分支，零不歸負分支。有限分段核對非全部實數性質證明。"},"difficulty":{"band":"lower_secondary","level":1,"skills":["有理絕對值的分支"]},"variant":{"family_id":"qf-density14-absolute-rational-two-branches","seed":202610146008,"parameters":{"x":[5,-3],"domain":"finite_rational_absolute_protocol","operation":"density14_absolute_rational_two_branches","semantic_template":"qf-density14-absolute-rational-two-branches/\"-5/3\""}},"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":"2026-10-10","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261010-advance-20261014-r06-absolute-value-0009","batch_id":"batch-20261010-advance-20261014-r06-absolute-value-foundations","language":"zh-Hant","source_mko_ids":["mko-absolute-value"],"title_zh":"有理絕對值的分支","stem_zh":"有理嵌入的絕對值協議；x=-4/3。依序答 (規約x, 分支非負或負, abs(x))；0屬非負分支。","answer":{"kind":"expression","value":"(-4/3, 負, 4/3)","explanation_zh":"有理絕對值的分支與值=(-4/3, 負, 4/3)。按規約值的符號取分支，零不歸負分支。有限分段核對非全部實數性質證明。"},"difficulty":{"band":"lower_secondary","level":2,"skills":["有理絕對值的分支"]},"variant":{"family_id":"qf-density14-absolute-rational-two-branches","seed":202610146009,"parameters":{"x":[-4,3],"domain":"finite_rational_absolute_protocol","operation":"density14_absolute_rational_two_branches","semantic_template":"qf-density14-absolute-rational-two-branches/\"-4/3\""}},"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":"2026-10-10","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261010-advance-20261014-r06-absolute-value-0010","batch_id":"batch-20261010-advance-20261014-r06-absolute-value-foundations","language":"zh-Hant","source_mko_ids":["mko-absolute-value"],"title_zh":"有理絕對值的分支","stem_zh":"有理嵌入的絕對值協議；x=3/-3。依序答 (規約x, 分支非負或負, abs(x))；0屬非負分支。","answer":{"kind":"expression","value":"(-1/1, 負, 1/1)","explanation_zh":"有理絕對值的分支與值=(-1/1, 負, 1/1)。按規約值的符號取分支，零不歸負分支。有限分段核對非全部實數性質證明。"},"difficulty":{"band":"lower_secondary","level":2,"skills":["有理絕對值的分支"]},"variant":{"family_id":"qf-density14-absolute-rational-two-branches","seed":202610146010,"parameters":{"x":[3,-3],"domain":"finite_rational_absolute_protocol","operation":"density14_absolute_rational_two_branches","semantic_template":"qf-density14-absolute-rational-two-branches/\"-1/1\""}},"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":"2026-10-10","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261010-advance-20261014-r06-absolute-value-0011","batch_id":"batch-20261010-advance-20261014-r06-absolute-value-foundations","language":"zh-Hant","source_mko_ids":["mko-absolute-value"],"title_zh":"有理絕對值的分支","stem_zh":"有理嵌入的絕對值協議；x=-2/3。依序答 (規約x, 分支非負或負, abs(x))；0屬非負分支。","answer":{"kind":"expression","value":"(-2/3, 負, 2/3)","explanation_zh":"有理絕對值的分支與值=(-2/3, 負, 2/3)。按規約值的符號取分支，零不歸負分支。有限分段核對非全部實數性質證明。"},"difficulty":{"band":"lower_secondary","level":2,"skills":["有理絕對值的分支"]},"variant":{"family_id":"qf-density14-absolute-rational-two-branches","seed":202610146011,"parameters":{"x":[-2,3],"domain":"finite_rational_absolute_protocol","operation":"density14_absolute_rational_two_branches","semantic_template":"qf-density14-absolute-rational-two-branches/\"-2/3\""}},"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":"2026-10-10","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261010-advance-20261014-r06-absolute-value-0012","batch_id":"batch-20261010-advance-20261014-r06-absolute-value-foundations","language":"zh-Hant","source_mko_ids":["mko-absolute-value"],"title_zh":"有理絕對值的分支","stem_zh":"有理嵌入的絕對值協議；x=1/-3。依序答 (規約x, 分支非負或負, abs(x))；0屬非負分支。","answer":{"kind":"expression","value":"(-1/3, 負, 1/3)","explanation_zh":"有理絕對值的分支與值=(-1/3, 負, 1/3)。按規約值的符號取分支，零不歸負分支。有限分段核對非全部實數性質證明。"},"difficulty":{"band":"lower_secondary","level":2,"skills":["有理絕對值的分支"]},"variant":{"family_id":"qf-density14-absolute-rational-two-branches","seed":202610146012,"parameters":{"x":[1,-3],"domain":"finite_rational_absolute_protocol","operation":"density14_absolute_rational_two_branches","semantic_template":"qf-density14-absolute-rational-two-branches/\"-1/3\""}},"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":"2026-10-10","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261010-advance-20261014-r06-absolute-value-0013","batch_id":"batch-20261010-advance-20261014-r06-absolute-value-foundations","language":"zh-Hant","source_mko_ids":["mko-absolute-value"],"title_zh":"有理絕對值的分支","stem_zh":"有理嵌入的絕對值協議；x=0/3。依序答 (規約x, 分支非負或負, abs(x))；0屬非負分支。","answer":{"kind":"expression","value":"(0/1, 非負, 0/1)","explanation_zh":"有理絕對值的分支與值=(0/1, 非負, 0/1)。按規約值的符號取分支，零不歸負分支。有限分段核對非全部實數性質證明。"},"difficulty":{"band":"lower_secondary","level":2,"skills":["有理絕對值的分支"]},"variant":{"family_id":"qf-density14-absolute-rational-two-branches","seed":202610146013,"parameters":{"x":[0,3],"domain":"finite_rational_absolute_protocol","operation":"density14_absolute_rational_two_branches","semantic_template":"qf-density14-absolute-rational-two-branches/\"0/1\""}},"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":"2026-10-10","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261010-advance-20261014-r06-absolute-value-0014","batch_id":"batch-20261010-advance-20261014-r06-absolute-value-foundations","language":"zh-Hant","source_mko_ids":["mko-absolute-value"],"title_zh":"有理絕對值的分支","stem_zh":"有理嵌入的絕對值協議；x=-1/-3。依序答 (規約x, 分支非負或負, abs(x))；0屬非負分支。","answer":{"kind":"expression","value":"(1/3, 非負, 1/3)","explanation_zh":"有理絕對值的分支與值=(1/3, 非負, 1/3)。按規約值的符號取分支，零不歸負分支。有限分段核對非全部實數性質證明。"},"difficulty":{"band":"lower_secondary","level":2,"skills":["有理絕對值的分支"]},"variant":{"family_id":"qf-density14-absolute-rational-two-branches","seed":202610146014,"parameters":{"x":[-1,-3],"domain":"finite_rational_absolute_protocol","operation":"density14_absolute_rational_two_branches","semantic_template":"qf-density14-absolute-rational-two-branches/\"1/3\""}},"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":"2026-10-10","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261010-advance-20261014-r06-absolute-value-0015","batch_id":"batch-20261010-advance-20261014-r06-absolute-value-foundations","language":"zh-Hant","source_mko_ids":["mko-absolute-value"],"title_zh":"有理絕對值的分支","stem_zh":"有理嵌入的絕對值協議；x=2/3。依序答 (規約x, 分支非負或負, abs(x))；0屬非負分支。","answer":{"kind":"expression","value":"(2/3, 非負, 2/3)","explanation_zh":"有理絕對值的分支與值=(2/3, 非負, 2/3)。按規約值的符號取分支，零不歸負分支。有限分段核對非全部實數性質證明。"},"difficulty":{"band":"lower_secondary","level":2,"skills":["有理絕對值的分支"]},"variant":{"family_id":"qf-density14-absolute-rational-two-branches","seed":202610146015,"parameters":{"x":[2,3],"domain":"finite_rational_absolute_protocol","operation":"density14_absolute_rational_two_branches","semantic_template":"qf-density14-absolute-rational-two-branches/\"2/3\""}},"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":"2026-10-10","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261010-advance-20261014-r06-absolute-value-0016","batch_id":"batch-20261010-advance-20261014-r06-absolute-value-foundations","language":"zh-Hant","source_mko_ids":["mko-absolute-value"],"title_zh":"有理絕對值的分支","stem_zh":"有理嵌入的絕對值協議；x=-3/-3。依序答 (規約x, 分支非負或負, abs(x))；0屬非負分支。","answer":{"kind":"expression","value":"(1/1, 非負, 1/1)","explanation_zh":"有理絕對值的分支與值=(1/1, 非負, 1/1)。按規約值的符號取分支，零不歸負分支。有限分段核對非全部實數性質證明。"},"difficulty":{"band":"lower_secondary","level":2,"skills":["有理絕對值的分支"]},"variant":{"family_id":"qf-density14-absolute-rational-two-branches","seed":202610146016,"parameters":{"x":[-3,-3],"domain":"finite_rational_absolute_protocol","operation":"density14_absolute_rational_two_branches","semantic_template":"qf-density14-absolute-rational-two-branches/\"1/1\""}},"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":"2026-10-10","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261010-advance-20261014-r06-absolute-value-0017","batch_id":"batch-20261010-advance-20261014-r06-absolute-value-foundations","language":"zh-Hant","source_mko_ids":["mko-absolute-value"],"title_zh":"有理絕對值的分支","stem_zh":"有理嵌入的絕對值協議；x=4/3。依序答 (規約x, 分支非負或負, abs(x))；0屬非負分支。","answer":{"kind":"expression","value":"(4/3, 非負, 4/3)","explanation_zh":"有理絕對值的分支與值=(4/3, 非負, 4/3)。按規約值的符號取分支，零不歸負分支。有限分段核對非全部實數性質證明。"},"difficulty":{"band":"lower_secondary","level":2,"skills":["有理絕對值的分支"]},"variant":{"family_id":"qf-density14-absolute-rational-two-branches","seed":202610146017,"parameters":{"x":[4,3],"domain":"finite_rational_absolute_protocol","operation":"density14_absolute_rational_two_branches","semantic_template":"qf-density14-absolute-rational-two-branches/\"4/3\""}},"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":"2026-10-10","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261010-advance-20261014-r06-absolute-value-0018","batch_id":"batch-20261010-advance-20261014-r06-absolute-value-foundations","language":"zh-Hant","source_mko_ids":["mko-absolute-value"],"title_zh":"有理絕對值的分支","stem_zh":"有理嵌入的絕對值協議；x=-5/-3。依序答 (規約x, 分支非負或負, abs(x))；0屬非負分支。","answer":{"kind":"expression","value":"(5/3, 非負, 5/3)","explanation_zh":"有理絕對值的分支與值=(5/3, 非負, 5/3)。按規約值的符號取分支，零不歸負分支。有限分段核對非全部實數性質證明。"},"difficulty":{"band":"lower_secondary","level":3,"skills":["有理絕對值的分支"]},"variant":{"family_id":"qf-density14-absolute-rational-two-branches","seed":202610146018,"parameters":{"x":[-5,-3],"domain":"finite_rational_absolute_protocol","operation":"density14_absolute_rational_two_branches","semantic_template":"qf-density14-absolute-rational-two-branches/\"5/3\""}},"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":"2026-10-10","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261010-advance-20261014-r06-absolute-value-0019","batch_id":"batch-20261010-advance-20261014-r06-absolute-value-foundations","language":"zh-Hant","source_mko_ids":["mko-absolute-value"],"title_zh":"有理絕對值的分支","stem_zh":"有理嵌入的絕對值協議；x=6/3。依序答 (規約x, 分支非負或負, abs(x))；0屬非負分支。","answer":{"kind":"expression","value":"(2/1, 非負, 2/1)","explanation_zh":"有理絕對值的分支與值=(2/1, 非負, 2/1)。按規約值的符號取分支，零不歸負分支。有限分段核對非全部實數性質證明。"},"difficulty":{"band":"lower_secondary","level":3,"skills":["有理絕對值的分支"]},"variant":{"family_id":"qf-density14-absolute-rational-two-branches","seed":202610146019,"parameters":{"x":[6,3],"domain":"finite_rational_absolute_protocol","operation":"density14_absolute_rational_two_branches","semantic_template":"qf-density14-absolute-rational-two-branches/\"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":"2026-10-10","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261010-advance-20261014-r06-absolute-value-0020","batch_id":"batch-20261010-advance-20261014-r06-absolute-value-foundations","language":"zh-Hant","source_mko_ids":["mko-absolute-value"],"title_zh":"有理絕對值的分支","stem_zh":"有理嵌入的絕對值協議；x=-7/-3。依序答 (規約x, 分支非負或負, abs(x))；0屬非負分支。","answer":{"kind":"expression","value":"(7/3, 非負, 7/3)","explanation_zh":"有理絕對值的分支與值=(7/3, 非負, 7/3)。按規約值的符號取分支，零不歸負分支。有限分段核對非全部實數性質證明。"},"difficulty":{"band":"lower_secondary","level":3,"skills":["有理絕對值的分支"]},"variant":{"family_id":"qf-density14-absolute-rational-two-branches","seed":202610146020,"parameters":{"x":[-7,-3],"domain":"finite_rational_absolute_protocol","operation":"density14_absolute_rational_two_branches","semantic_template":"qf-density14-absolute-rational-two-branches/\"7/3\""}},"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":"2026-10-10","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261010-advance-20261014-r06-absolute-value-0021","batch_id":"batch-20261010-advance-20261014-r06-absolute-value-foundations","language":"zh-Hant","source_mko_ids":["mko-absolute-value"],"title_zh":"有理絕對值的分支","stem_zh":"有理嵌入的絕對值協議；x=8/3。依序答 (規約x, 分支非負或負, abs(x))；0屬非負分支。","answer":{"kind":"expression","value":"(8/3, 非負, 8/3)","explanation_zh":"有理絕對值的分支與值=(8/3, 非負, 8/3)。按規約值的符號取分支，零不歸負分支。有限分段核對非全部實數性質證明。"},"difficulty":{"band":"lower_secondary","level":3,"skills":["有理絕對值的分支"]},"variant":{"family_id":"qf-density14-absolute-rational-two-branches","seed":202610146021,"parameters":{"x":[8,3],"domain":"finite_rational_absolute_protocol","operation":"density14_absolute_rational_two_branches","semantic_template":"qf-density14-absolute-rational-two-branches/\"8/3\""}},"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":"2026-10-10","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261010-advance-20261014-r06-absolute-value-0022","batch_id":"batch-20261010-advance-20261014-r06-absolute-value-foundations","language":"zh-Hant","source_mko_ids":["mko-absolute-value"],"title_zh":"有理絕對值的分支","stem_zh":"有理嵌入的絕對值協議；x=-9/-3。依序答 (規約x, 分支非負或負, abs(x))；0屬非負分支。","answer":{"kind":"expression","value":"(3/1, 非負, 3/1)","explanation_zh":"有理絕對值的分支與值=(3/1, 非負, 3/1)。按規約值的符號取分支，零不歸負分支。有限分段核對非全部實數性質證明。"},"difficulty":{"band":"lower_secondary","level":3,"skills":["有理絕對值的分支"]},"variant":{"family_id":"qf-density14-absolute-rational-two-branches","seed":202610146022,"parameters":{"x":[-9,-3],"domain":"finite_rational_absolute_protocol","operation":"density14_absolute_rational_two_branches","semantic_template":"qf-density14-absolute-rational-two-branches/\"3/1\""}},"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":"2026-10-10","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261010-advance-20261014-r06-absolute-value-0023","batch_id":"batch-20261010-advance-20261014-r06-absolute-value-foundations","language":"zh-Hant","source_mko_ids":["mko-absolute-value"],"title_zh":"有理絕對值的分支","stem_zh":"有理嵌入的絕對值協議；x=10/3。依序答 (規約x, 分支非負或負, abs(x))；0屬非負分支。","answer":{"kind":"expression","value":"(10/3, 非負, 10/3)","explanation_zh":"有理絕對值的分支與值=(10/3, 非負, 10/3)。按規約值的符號取分支，零不歸負分支。有限分段核對非全部實數性質證明。"},"difficulty":{"band":"lower_secondary","level":3,"skills":["有理絕對值的分支"]},"variant":{"family_id":"qf-density14-absolute-rational-two-branches","seed":202610146023,"parameters":{"x":[10,3],"domain":"finite_rational_absolute_protocol","operation":"density14_absolute_rational_two_branches","semantic_template":"qf-density14-absolute-rational-two-branches/\"10/3\""}},"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":"2026-10-10","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261010-advance-20261014-r06-absolute-value-0024","batch_id":"batch-20261010-advance-20261014-r06-absolute-value-foundations","language":"zh-Hant","source_mko_ids":["mko-absolute-value"],"title_zh":"有理絕對值的分支","stem_zh":"有理嵌入的絕對值協議；x=-11/-3。依序答 (規約x, 分支非負或負, abs(x))；0屬非負分支。","answer":{"kind":"expression","value":"(11/3, 非負, 11/3)","explanation_zh":"有理絕對值的分支與值=(11/3, 非負, 11/3)。按規約值的符號取分支，零不歸負分支。有限分段核對非全部實數性質證明。"},"difficulty":{"band":"lower_secondary","level":3,"skills":["有理絕對值的分支"]},"variant":{"family_id":"qf-density14-absolute-rational-two-branches","seed":202610146024,"parameters":{"x":[-11,-3],"domain":"finite_rational_absolute_protocol","operation":"density14_absolute_rational_two_branches","semantic_template":"qf-density14-absolute-rational-two-branches/\"11/3\""}},"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":"2026-10-10","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261010-advance-20261014-r06-absolute-value-0025","batch_id":"batch-20261010-advance-20261014-r06-absolute-value-foundations","language":"zh-Hant","source_mko_ids":["mko-absolute-value"],"title_zh":"有理絕對值的分支","stem_zh":"有理嵌入的絕對值協議；x=12/3。依序答 (規約x, 分支非負或負, abs(x))；0屬非負分支。","answer":{"kind":"expression","value":"(4/1, 非負, 4/1)","explanation_zh":"有理絕對值的分支與值=(4/1, 非負, 4/1)。按規約值的符號取分支，零不歸負分支。有限分段核對非全部實數性質證明。"},"difficulty":{"band":"lower_secondary","level":3,"skills":["有理絕對值的分支"]},"variant":{"family_id":"qf-density14-absolute-rational-two-branches","seed":202610146025,"parameters":{"x":[12,3],"domain":"finite_rational_absolute_protocol","operation":"density14_absolute_rational_two_branches","semantic_template":"qf-density14-absolute-rational-two-branches/\"4/1\""}},"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":"2026-10-10","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261010-advance-20261014-r06-absolute-value-0026","batch_id":"batch-20261010-advance-20261014-r06-absolute-value-foundations","language":"zh-Hant","source_mko_ids":["mko-absolute-value"],"title_zh":"三點距離與路徑差","stem_zh":"有理嵌入的絕對值協議；有序points=[0/1, 0/1, 0/1]。依序求 (d01, d12, d02, d01+d12−d02) 的精確有理數。","answer":{"kind":"expression","value":"(0/1, 0/1, 0/1, 0/1)","explanation_zh":"三點距離與路徑差=(0/1, 0/1, 0/1, 0/1)。各段直接取差的絕對值；路徑量不冒充端點距離，有限差值非三角不等式普遍證明。有限分段核對非全部實數性質證明。"},"difficulty":{"band":"lower_secondary","level":1,"skills":["三點距離與路徑差"]},"variant":{"family_id":"qf-density14-absolute-three-point-distances","seed":202610146026,"parameters":{"points":[[0,1],[0,1],[0,1]],"domain":"finite_rational_absolute_protocol","operation":"density14_absolute_three_point_distances","semantic_template":"qf-density14-absolute-three-point-distances/[\"0/1\",\"0/1\",\"0/1\"]"}},"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":"2026-10-10","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261010-advance-20261014-r06-absolute-value-0027","batch_id":"batch-20261010-advance-20261014-r06-absolute-value-foundations","language":"zh-Hant","source_mko_ids":["mko-absolute-value"],"title_zh":"三點距離與路徑差","stem_zh":"有理嵌入的絕對值協議；有序points=[0/1, 1/1, 0/1]。依序求 (d01, d12, d02, d01+d12−d02) 的精確有理數。","answer":{"kind":"expression","value":"(1/1, 1/1, 0/1, 2/1)","explanation_zh":"三點距離與路徑差=(1/1, 1/1, 0/1, 2/1)。各段直接取差的絕對值；路徑量不冒充端點距離，有限差值非三角不等式普遍證明。有限分段核對非全部實數性質證明。"},"difficulty":{"band":"lower_secondary","level":1,"skills":["三點距離與路徑差"]},"variant":{"family_id":"qf-density14-absolute-three-point-distances","seed":202610146027,"parameters":{"points":[[0,1],[1,1],[0,1]],"domain":"finite_rational_absolute_protocol","operation":"density14_absolute_three_point_distances","semantic_template":"qf-density14-absolute-three-point-distances/[\"1/1\",\"1/1\",\"0/1\"]"}},"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":"2026-10-10","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261010-advance-20261014-r06-absolute-value-0028","batch_id":"batch-20261010-advance-20261014-r06-absolute-value-foundations","language":"zh-Hant","source_mko_ids":["mko-absolute-value"],"title_zh":"三點距離與路徑差","stem_zh":"有理嵌入的絕對值協議；有序points=[0/1, 2/2, 3/2]。依序求 (d01, d12, d02, d01+d12−d02) 的精確有理數。","answer":{"kind":"expression","value":"(1/1, 1/2, 3/2, 0/1)","explanation_zh":"三點距離與路徑差=(1/1, 1/2, 3/2, 0/1)。各段直接取差的絕對值；路徑量不冒充端點距離，有限差值非三角不等式普遍證明。有限分段核對非全部實數性質證明。"},"difficulty":{"band":"lower_secondary","level":1,"skills":["三點距離與路徑差"]},"variant":{"family_id":"qf-density14-absolute-three-point-distances","seed":202610146028,"parameters":{"points":[[0,1],[2,2],[3,2]],"domain":"finite_rational_absolute_protocol","operation":"density14_absolute_three_point_distances","semantic_template":"qf-density14-absolute-three-point-distances/[\"1/1\",\"1/2\",\"3/2\"]"}},"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":"2026-10-10","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261010-advance-20261014-r06-absolute-value-0029","batch_id":"batch-20261010-advance-20261014-r06-absolute-value-foundations","language":"zh-Hant","source_mko_ids":["mko-absolute-value"],"title_zh":"三點距離與路徑差","stem_zh":"有理嵌入的絕對值協議；有序points=[0/1, 3/2, 4/2]。依序求 (d01, d12, d02, d01+d12−d02) 的精確有理數。","answer":{"kind":"expression","value":"(3/2, 1/2, 2/1, 0/1)","explanation_zh":"三點距離與路徑差=(3/2, 1/2, 2/1, 0/1)。各段直接取差的絕對值；路徑量不冒充端點距離，有限差值非三角不等式普遍證明。有限分段核對非全部實數性質證明。"},"difficulty":{"band":"lower_secondary","level":1,"skills":["三點距離與路徑差"]},"variant":{"family_id":"qf-density14-absolute-three-point-distances","seed":202610146029,"parameters":{"points":[[0,1],[3,2],[4,2]],"domain":"finite_rational_absolute_protocol","operation":"density14_absolute_three_point_distances","semantic_template":"qf-density14-absolute-three-point-distances/[\"3/2\",\"1/2\",\"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":"2026-10-10","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261010-advance-20261014-r06-absolute-value-0030","batch_id":"batch-20261010-advance-20261014-r06-absolute-value-foundations","language":"zh-Hant","source_mko_ids":["mko-absolute-value"],"title_zh":"三點距離與路徑差","stem_zh":"有理嵌入的絕對值協議；有序points=[0/1, 4/2, 5/2]。依序求 (d01, d12, d02, d01+d12−d02) 的精確有理數。","answer":{"kind":"expression","value":"(2/1, 1/2, 5/2, 0/1)","explanation_zh":"三點距離與路徑差=(2/1, 1/2, 5/2, 0/1)。各段直接取差的絕對值；路徑量不冒充端點距離，有限差值非三角不等式普遍證明。有限分段核對非全部實數性質證明。"},"difficulty":{"band":"lower_secondary","level":1,"skills":["三點距離與路徑差"]},"variant":{"family_id":"qf-density14-absolute-three-point-distances","seed":202610146030,"parameters":{"points":[[0,1],[4,2],[5,2]],"domain":"finite_rational_absolute_protocol","operation":"density14_absolute_three_point_distances","semantic_template":"qf-density14-absolute-three-point-distances/[\"2/1\",\"1/2\",\"5/2\"]"}},"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":"2026-10-10","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261010-advance-20261014-r06-absolute-value-0031","batch_id":"batch-20261010-advance-20261014-r06-absolute-value-foundations","language":"zh-Hant","source_mko_ids":["mko-absolute-value"],"title_zh":"三點距離與路徑差","stem_zh":"有理嵌入的絕對值協議；有序points=[0/1, 5/2, 6/2]。依序求 (d01, d12, d02, d01+d12−d02) 的精確有理數。","answer":{"kind":"expression","value":"(5/2, 1/2, 3/1, 0/1)","explanation_zh":"三點距離與路徑差=(5/2, 1/2, 3/1, 0/1)。各段直接取差的絕對值；路徑量不冒充端點距離，有限差值非三角不等式普遍證明。有限分段核對非全部實數性質證明。"},"difficulty":{"band":"lower_secondary","level":1,"skills":["三點距離與路徑差"]},"variant":{"family_id":"qf-density14-absolute-three-point-distances","seed":202610146031,"parameters":{"points":[[0,1],[5,2],[6,2]],"domain":"finite_rational_absolute_protocol","operation":"density14_absolute_three_point_distances","semantic_template":"qf-density14-absolute-three-point-distances/[\"5/2\",\"1/2\",\"3/1\"]"}},"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":"2026-10-10","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261010-advance-20261014-r06-absolute-value-0032","batch_id":"batch-20261010-advance-20261014-r06-absolute-value-foundations","language":"zh-Hant","source_mko_ids":["mko-absolute-value"],"title_zh":"三點距離與路徑差","stem_zh":"有理嵌入的絕對值協議；有序points=[0/1, 6/2, 7/2]。依序求 (d01, d12, d02, d01+d12−d02) 的精確有理數。","answer":{"kind":"expression","value":"(3/1, 1/2, 7/2, 0/1)","explanation_zh":"三點距離與路徑差=(3/1, 1/2, 7/2, 0/1)。各段直接取差的絕對值；路徑量不冒充端點距離，有限差值非三角不等式普遍證明。有限分段核對非全部實數性質證明。"},"difficulty":{"band":"lower_secondary","level":1,"skills":["三點距離與路徑差"]},"variant":{"family_id":"qf-density14-absolute-three-point-distances","seed":202610146032,"parameters":{"points":[[0,1],[6,2],[7,2]],"domain":"finite_rational_absolute_protocol","operation":"density14_absolute_three_point_distances","semantic_template":"qf-density14-absolute-three-point-distances/[\"3/1\",\"1/2\",\"7/2\"]"}},"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":"2026-10-10","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261010-advance-20261014-r06-absolute-value-0033","batch_id":"batch-20261010-advance-20261014-r06-absolute-value-foundations","language":"zh-Hant","source_mko_ids":["mko-absolute-value"],"title_zh":"三點距離與路徑差","stem_zh":"有理嵌入的絕對值協議；有序points=[0/1, 7/2, 8/2]。依序求 (d01, d12, d02, d01+d12−d02) 的精確有理數。","answer":{"kind":"expression","value":"(7/2, 1/2, 4/1, 0/1)","explanation_zh":"三點距離與路徑差=(7/2, 1/2, 4/1, 0/1)。各段直接取差的絕對值；路徑量不冒充端點距離，有限差值非三角不等式普遍證明。有限分段核對非全部實數性質證明。"},"difficulty":{"band":"lower_secondary","level":1,"skills":["三點距離與路徑差"]},"variant":{"family_id":"qf-density14-absolute-three-point-distances","seed":202610146033,"parameters":{"points":[[0,1],[7,2],[8,2]],"domain":"finite_rational_absolute_protocol","operation":"density14_absolute_three_point_distances","semantic_template":"qf-density14-absolute-three-point-distances/[\"7/2\",\"1/2\",\"4/1\"]"}},"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":"2026-10-10","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261010-advance-20261014-r06-absolute-value-0034","batch_id":"batch-20261010-advance-20261014-r06-absolute-value-foundations","language":"zh-Hant","source_mko_ids":["mko-absolute-value"],"title_zh":"三點距離與路徑差","stem_zh":"有理嵌入的絕對值協議；有序points=[0/1, 8/2, 9/2]。依序求 (d01, d12, d02, d01+d12−d02) 的精確有理數。","answer":{"kind":"expression","value":"(4/1, 1/2, 9/2, 0/1)","explanation_zh":"三點距離與路徑差=(4/1, 1/2, 9/2, 0/1)。各段直接取差的絕對值；路徑量不冒充端點距離，有限差值非三角不等式普遍證明。有限分段核對非全部實數性質證明。"},"difficulty":{"band":"lower_secondary","level":2,"skills":["三點距離與路徑差"]},"variant":{"family_id":"qf-density14-absolute-three-point-distances","seed":202610146034,"parameters":{"points":[[0,1],[8,2],[9,2]],"domain":"finite_rational_absolute_protocol","operation":"density14_absolute_three_point_distances","semantic_template":"qf-density14-absolute-three-point-distances/[\"4/1\",\"1/2\",\"9/2\"]"}},"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":"2026-10-10","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261010-advance-20261014-r06-absolute-value-0035","batch_id":"batch-20261010-advance-20261014-r06-absolute-value-foundations","language":"zh-Hant","source_mko_ids":["mko-absolute-value"],"title_zh":"三點距離與路徑差","stem_zh":"有理嵌入的絕對值協議；有序points=[0/1, 9/2, 10/2]。依序求 (d01, d12, d02, d01+d12−d02) 的精確有理數。","answer":{"kind":"expression","value":"(9/2, 1/2, 5/1, 0/1)","explanation_zh":"三點距離與路徑差=(9/2, 1/2, 5/1, 0/1)。各段直接取差的絕對值；路徑量不冒充端點距離，有限差值非三角不等式普遍證明。有限分段核對非全部實數性質證明。"},"difficulty":{"band":"lower_secondary","level":2,"skills":["三點距離與路徑差"]},"variant":{"family_id":"qf-density14-absolute-three-point-distances","seed":202610146035,"parameters":{"points":[[0,1],[9,2],[10,2]],"domain":"finite_rational_absolute_protocol","operation":"density14_absolute_three_point_distances","semantic_template":"qf-density14-absolute-three-point-distances/[\"9/2\",\"1/2\",\"5/1\"]"}},"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":"2026-10-10","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261010-advance-20261014-r06-absolute-value-0036","batch_id":"batch-20261010-advance-20261014-r06-absolute-value-foundations","language":"zh-Hant","source_mko_ids":["mko-absolute-value"],"title_zh":"三點距離與路徑差","stem_zh":"有理嵌入的絕對值協議；有序points=[0/1, 10/2, 11/2]。依序求 (d01, d12, d02, d01+d12−d02) 的精確有理數。","answer":{"kind":"expression","value":"(5/1, 1/2, 11/2, 0/1)","explanation_zh":"三點距離與路徑差=(5/1, 1/2, 11/2, 0/1)。各段直接取差的絕對值；路徑量不冒充端點距離，有限差值非三角不等式普遍證明。有限分段核對非全部實數性質證明。"},"difficulty":{"band":"lower_secondary","level":2,"skills":["三點距離與路徑差"]},"variant":{"family_id":"qf-density14-absolute-three-point-distances","seed":202610146036,"parameters":{"points":[[0,1],[10,2],[11,2]],"domain":"finite_rational_absolute_protocol","operation":"density14_absolute_three_point_distances","semantic_template":"qf-density14-absolute-three-point-distances/[\"5/1\",\"1/2\",\"11/2\"]"}},"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":"2026-10-10","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261010-advance-20261014-r06-absolute-value-0037","batch_id":"batch-20261010-advance-20261014-r06-absolute-value-foundations","language":"zh-Hant","source_mko_ids":["mko-absolute-value"],"title_zh":"三點距離與路徑差","stem_zh":"有理嵌入的絕對值協議；有序points=[0/1, 11/2, 12/2]。依序求 (d01, d12, d02, d01+d12−d02) 的精確有理數。","answer":{"kind":"expression","value":"(11/2, 1/2, 6/1, 0/1)","explanation_zh":"三點距離與路徑差=(11/2, 1/2, 6/1, 0/1)。各段直接取差的絕對值；路徑量不冒充端點距離，有限差值非三角不等式普遍證明。有限分段核對非全部實數性質證明。"},"difficulty":{"band":"lower_secondary","level":2,"skills":["三點距離與路徑差"]},"variant":{"family_id":"qf-density14-absolute-three-point-distances","seed":202610146037,"parameters":{"points":[[0,1],[11,2],[12,2]],"domain":"finite_rational_absolute_protocol","operation":"density14_absolute_three_point_distances","semantic_template":"qf-density14-absolute-three-point-distances/[\"11/2\",\"1/2\",\"6/1\"]"}},"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":"2026-10-10","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261010-advance-20261014-r06-absolute-value-0038","batch_id":"batch-20261010-advance-20261014-r06-absolute-value-foundations","language":"zh-Hant","source_mko_ids":["mko-absolute-value"],"title_zh":"三點距離與路徑差","stem_zh":"有理嵌入的絕對值協議；有序points=[0/1, 12/2, 13/2]。依序求 (d01, d12, d02, d01+d12−d02) 的精確有理數。","answer":{"kind":"expression","value":"(6/1, 1/2, 13/2, 0/1)","explanation_zh":"三點距離與路徑差=(6/1, 1/2, 13/2, 0/1)。各段直接取差的絕對值；路徑量不冒充端點距離，有限差值非三角不等式普遍證明。有限分段核對非全部實數性質證明。"},"difficulty":{"band":"lower_secondary","level":2,"skills":["三點距離與路徑差"]},"variant":{"family_id":"qf-density14-absolute-three-point-distances","seed":202610146038,"parameters":{"points":[[0,1],[12,2],[13,2]],"domain":"finite_rational_absolute_protocol","operation":"density14_absolute_three_point_distances","semantic_template":"qf-density14-absolute-three-point-distances/[\"6/1\",\"1/2\",\"13/2\"]"}},"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":"2026-10-10","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261010-advance-20261014-r06-absolute-value-0039","batch_id":"batch-20261010-advance-20261014-r06-absolute-value-foundations","language":"zh-Hant","source_mko_ids":["mko-absolute-value"],"title_zh":"三點距離與路徑差","stem_zh":"有理嵌入的絕對值協議；有序points=[0/1, 13/2, 14/2]。依序求 (d01, d12, d02, d01+d12−d02) 的精確有理數。","answer":{"kind":"expression","value":"(13/2, 1/2, 7/1, 0/1)","explanation_zh":"三點距離與路徑差=(13/2, 1/2, 7/1, 0/1)。各段直接取差的絕對值；路徑量不冒充端點距離，有限差值非三角不等式普遍證明。有限分段核對非全部實數性質證明。"},"difficulty":{"band":"lower_secondary","level":2,"skills":["三點距離與路徑差"]},"variant":{"family_id":"qf-density14-absolute-three-point-distances","seed":202610146039,"parameters":{"points":[[0,1],[13,2],[14,2]],"domain":"finite_rational_absolute_protocol","operation":"density14_absolute_three_point_distances","semantic_template":"qf-density14-absolute-three-point-distances/[\"13/2\",\"1/2\",\"7/1\"]"}},"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":"2026-10-10","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261010-advance-20261014-r06-absolute-value-0040","batch_id":"batch-20261010-advance-20261014-r06-absolute-value-foundations","language":"zh-Hant","source_mko_ids":["mko-absolute-value"],"title_zh":"三點距離與路徑差","stem_zh":"有理嵌入的絕對值協議；有序points=[0/1, 14/2, 15/2]。依序求 (d01, d12, d02, d01+d12−d02) 的精確有理數。","answer":{"kind":"expression","value":"(7/1, 1/2, 15/2, 0/1)","explanation_zh":"三點距離與路徑差=(7/1, 1/2, 15/2, 0/1)。各段直接取差的絕對值；路徑量不冒充端點距離，有限差值非三角不等式普遍證明。有限分段核對非全部實數性質證明。"},"difficulty":{"band":"lower_secondary","level":2,"skills":["三點距離與路徑差"]},"variant":{"family_id":"qf-density14-absolute-three-point-distances","seed":202610146040,"parameters":{"points":[[0,1],[14,2],[15,2]],"domain":"finite_rational_absolute_protocol","operation":"density14_absolute_three_point_distances","semantic_template":"qf-density14-absolute-three-point-distances/[\"7/1\",\"1/2\",\"15/2\"]"}},"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":"2026-10-10","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261010-advance-20261014-r06-absolute-value-0041","batch_id":"batch-20261010-advance-20261014-r06-absolute-value-foundations","language":"zh-Hant","source_mko_ids":["mko-absolute-value"],"title_zh":"三點距離與路徑差","stem_zh":"有理嵌入的絕對值協議；有序points=[0/1, 15/2, 16/2]。依序求 (d01, d12, d02, d01+d12−d02) 的精確有理數。","answer":{"kind":"expression","value":"(15/2, 1/2, 8/1, 0/1)","explanation_zh":"三點距離與路徑差=(15/2, 1/2, 8/1, 0/1)。各段直接取差的絕對值；路徑量不冒充端點距離，有限差值非三角不等式普遍證明。有限分段核對非全部實數性質證明。"},"difficulty":{"band":"lower_secondary","level":2,"skills":["三點距離與路徑差"]},"variant":{"family_id":"qf-density14-absolute-three-point-distances","seed":202610146041,"parameters":{"points":[[0,1],[15,2],[16,2]],"domain":"finite_rational_absolute_protocol","operation":"density14_absolute_three_point_distances","semantic_template":"qf-density14-absolute-three-point-distances/[\"15/2\",\"1/2\",\"8/1\"]"}},"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":"2026-10-10","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261010-advance-20261014-r06-absolute-value-0042","batch_id":"batch-20261010-advance-20261014-r06-absolute-value-foundations","language":"zh-Hant","source_mko_ids":["mko-absolute-value"],"title_zh":"三點距離與路徑差","stem_zh":"有理嵌入的絕對值協議；有序points=[0/1, 16/2, 17/2]。依序求 (d01, d12, d02, d01+d12−d02) 的精確有理數。","answer":{"kind":"expression","value":"(8/1, 1/2, 17/2, 0/1)","explanation_zh":"三點距離與路徑差=(8/1, 1/2, 17/2, 0/1)。各段直接取差的絕對值；路徑量不冒充端點距離，有限差值非三角不等式普遍證明。有限分段核對非全部實數性質證明。"},"difficulty":{"band":"lower_secondary","level":2,"skills":["三點距離與路徑差"]},"variant":{"family_id":"qf-density14-absolute-three-point-distances","seed":202610146042,"parameters":{"points":[[0,1],[16,2],[17,2]],"domain":"finite_rational_absolute_protocol","operation":"density14_absolute_three_point_distances","semantic_template":"qf-density14-absolute-three-point-distances/[\"8/1\",\"1/2\",\"17/2\"]"}},"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":"2026-10-10","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261010-advance-20261014-r06-absolute-value-0043","batch_id":"batch-20261010-advance-20261014-r06-absolute-value-foundations","language":"zh-Hant","source_mko_ids":["mko-absolute-value"],"title_zh":"三點距離與路徑差","stem_zh":"有理嵌入的絕對值協議；有序points=[0/1, 17/2, 18/2]。依序求 (d01, d12, d02, d01+d12−d02) 的精確有理數。","answer":{"kind":"expression","value":"(17/2, 1/2, 9/1, 0/1)","explanation_zh":"三點距離與路徑差=(17/2, 1/2, 9/1, 0/1)。各段直接取差的絕對值；路徑量不冒充端點距離，有限差值非三角不等式普遍證明。有限分段核對非全部實數性質證明。"},"difficulty":{"band":"lower_secondary","level":3,"skills":["三點距離與路徑差"]},"variant":{"family_id":"qf-density14-absolute-three-point-distances","seed":202610146043,"parameters":{"points":[[0,1],[17,2],[18,2]],"domain":"finite_rational_absolute_protocol","operation":"density14_absolute_three_point_distances","semantic_template":"qf-density14-absolute-three-point-distances/[\"17/2\",\"1/2\",\"9/1\"]"}},"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":"2026-10-10","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261010-advance-20261014-r06-absolute-value-0044","batch_id":"batch-20261010-advance-20261014-r06-absolute-value-foundations","language":"zh-Hant","source_mko_ids":["mko-absolute-value"],"title_zh":"三點距離與路徑差","stem_zh":"有理嵌入的絕對值協議；有序points=[0/1, 18/2, 19/2]。依序求 (d01, d12, d02, d01+d12−d02) 的精確有理數。","answer":{"kind":"expression","value":"(9/1, 1/2, 19/2, 0/1)","explanation_zh":"三點距離與路徑差=(9/1, 1/2, 19/2, 0/1)。各段直接取差的絕對值；路徑量不冒充端點距離，有限差值非三角不等式普遍證明。有限分段核對非全部實數性質證明。"},"difficulty":{"band":"lower_secondary","level":3,"skills":["三點距離與路徑差"]},"variant":{"family_id":"qf-density14-absolute-three-point-distances","seed":202610146044,"parameters":{"points":[[0,1],[18,2],[19,2]],"domain":"finite_rational_absolute_protocol","operation":"density14_absolute_three_point_distances","semantic_template":"qf-density14-absolute-three-point-distances/[\"9/1\",\"1/2\",\"19/2\"]"}},"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":"2026-10-10","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261010-advance-20261014-r06-absolute-value-0045","batch_id":"batch-20261010-advance-20261014-r06-absolute-value-foundations","language":"zh-Hant","source_mko_ids":["mko-absolute-value"],"title_zh":"三點距離與路徑差","stem_zh":"有理嵌入的絕對值協議；有序points=[0/1, 19/2, 20/2]。依序求 (d01, d12, d02, d01+d12−d02) 的精確有理數。","answer":{"kind":"expression","value":"(19/2, 1/2, 10/1, 0/1)","explanation_zh":"三點距離與路徑差=(19/2, 1/2, 10/1, 0/1)。各段直接取差的絕對值；路徑量不冒充端點距離，有限差值非三角不等式普遍證明。有限分段核對非全部實數性質證明。"},"difficulty":{"band":"lower_secondary","level":3,"skills":["三點距離與路徑差"]},"variant":{"family_id":"qf-density14-absolute-three-point-distances","seed":202610146045,"parameters":{"points":[[0,1],[19,2],[20,2]],"domain":"finite_rational_absolute_protocol","operation":"density14_absolute_three_point_distances","semantic_template":"qf-density14-absolute-three-point-distances/[\"19/2\",\"1/2\",\"10/1\"]"}},"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":"2026-10-10","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261010-advance-20261014-r06-absolute-value-0046","batch_id":"batch-20261010-advance-20261014-r06-absolute-value-foundations","language":"zh-Hant","source_mko_ids":["mko-absolute-value"],"title_zh":"三點距離與路徑差","stem_zh":"有理嵌入的絕對值協議；有序points=[0/1, 20/2, 21/2]。依序求 (d01, d12, d02, d01+d12−d02) 的精確有理數。","answer":{"kind":"expression","value":"(10/1, 1/2, 21/2, 0/1)","explanation_zh":"三點距離與路徑差=(10/1, 1/2, 21/2, 0/1)。各段直接取差的絕對值；路徑量不冒充端點距離，有限差值非三角不等式普遍證明。有限分段核對非全部實數性質證明。"},"difficulty":{"band":"lower_secondary","level":3,"skills":["三點距離與路徑差"]},"variant":{"family_id":"qf-density14-absolute-three-point-distances","seed":202610146046,"parameters":{"points":[[0,1],[20,2],[21,2]],"domain":"finite_rational_absolute_protocol","operation":"density14_absolute_three_point_distances","semantic_template":"qf-density14-absolute-three-point-distances/[\"10/1\",\"1/2\",\"21/2\"]"}},"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":"2026-10-10","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261010-advance-20261014-r06-absolute-value-0047","batch_id":"batch-20261010-advance-20261014-r06-absolute-value-foundations","language":"zh-Hant","source_mko_ids":["mko-absolute-value"],"title_zh":"三點距離與路徑差","stem_zh":"有理嵌入的絕對值協議；有序points=[0/1, 21/2, 22/2]。依序求 (d01, d12, d02, d01+d12−d02) 的精確有理數。","answer":{"kind":"expression","value":"(21/2, 1/2, 11/1, 0/1)","explanation_zh":"三點距離與路徑差=(21/2, 1/2, 11/1, 0/1)。各段直接取差的絕對值；路徑量不冒充端點距離，有限差值非三角不等式普遍證明。有限分段核對非全部實數性質證明。"},"difficulty":{"band":"lower_secondary","level":3,"skills":["三點距離與路徑差"]},"variant":{"family_id":"qf-density14-absolute-three-point-distances","seed":202610146047,"parameters":{"points":[[0,1],[21,2],[22,2]],"domain":"finite_rational_absolute_protocol","operation":"density14_absolute_three_point_distances","semantic_template":"qf-density14-absolute-three-point-distances/[\"21/2\",\"1/2\",\"11/1\"]"}},"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":"2026-10-10","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261010-advance-20261014-r06-absolute-value-0048","batch_id":"batch-20261010-advance-20261014-r06-absolute-value-foundations","language":"zh-Hant","source_mko_ids":["mko-absolute-value"],"title_zh":"三點距離與路徑差","stem_zh":"有理嵌入的絕對值協議；有序points=[0/1, 22/2, 23/2]。依序求 (d01, d12, d02, d01+d12−d02) 的精確有理數。","answer":{"kind":"expression","value":"(11/1, 1/2, 23/2, 0/1)","explanation_zh":"三點距離與路徑差=(11/1, 1/2, 23/2, 0/1)。各段直接取差的絕對值；路徑量不冒充端點距離，有限差值非三角不等式普遍證明。有限分段核對非全部實數性質證明。"},"difficulty":{"band":"lower_secondary","level":3,"skills":["三點距離與路徑差"]},"variant":{"family_id":"qf-density14-absolute-three-point-distances","seed":202610146048,"parameters":{"points":[[0,1],[22,2],[23,2]],"domain":"finite_rational_absolute_protocol","operation":"density14_absolute_three_point_distances","semantic_template":"qf-density14-absolute-three-point-distances/[\"11/1\",\"1/2\",\"23/2\"]"}},"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":"2026-10-10","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261010-advance-20261014-r06-absolute-value-0049","batch_id":"batch-20261010-advance-20261014-r06-absolute-value-foundations","language":"zh-Hant","source_mko_ids":["mko-absolute-value"],"title_zh":"三點距離與路徑差","stem_zh":"有理嵌入的絕對值協議；有序points=[0/1, 23/2, 24/2]。依序求 (d01, d12, d02, d01+d12−d02) 的精確有理數。","answer":{"kind":"expression","value":"(23/2, 1/2, 12/1, 0/1)","explanation_zh":"三點距離與路徑差=(23/2, 1/2, 12/1, 0/1)。各段直接取差的絕對值；路徑量不冒充端點距離，有限差值非三角不等式普遍證明。有限分段核對非全部實數性質證明。"},"difficulty":{"band":"lower_secondary","level":3,"skills":["三點距離與路徑差"]},"variant":{"family_id":"qf-density14-absolute-three-point-distances","seed":202610146049,"parameters":{"points":[[0,1],[23,2],[24,2]],"domain":"finite_rational_absolute_protocol","operation":"density14_absolute_three_point_distances","semantic_template":"qf-density14-absolute-three-point-distances/[\"23/2\",\"1/2\",\"12/1\"]"}},"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":"2026-10-10","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261010-advance-20261014-r06-absolute-value-0050","batch_id":"batch-20261010-advance-20261014-r06-absolute-value-foundations","language":"zh-Hant","source_mko_ids":["mko-absolute-value"],"title_zh":"三點距離與路徑差","stem_zh":"有理嵌入的絕對值協議；有序points=[0/1, 24/2, 25/2]。依序求 (d01, d12, d02, d01+d12−d02) 的精確有理數。","answer":{"kind":"expression","value":"(12/1, 1/2, 25/2, 0/1)","explanation_zh":"三點距離與路徑差=(12/1, 1/2, 25/2, 0/1)。各段直接取差的絕對值；路徑量不冒充端點距離，有限差值非三角不等式普遍證明。有限分段核對非全部實數性質證明。"},"difficulty":{"band":"lower_secondary","level":3,"skills":["三點距離與路徑差"]},"variant":{"family_id":"qf-density14-absolute-three-point-distances","seed":202610146050,"parameters":{"points":[[0,1],[24,2],[25,2]],"domain":"finite_rational_absolute_protocol","operation":"density14_absolute_three_point_distances","semantic_template":"qf-density14-absolute-three-point-distances/[\"12/1\",\"1/2\",\"25/2\"]"}},"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":"2026-10-10","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261010-advance-20261014-r06-absolute-value-0051","batch_id":"batch-20261010-advance-20261014-r06-absolute-value-foundations","language":"zh-Hant","source_mko_ids":["mko-absolute-value"],"title_zh":"閉半徑的四點資格","stem_zh":"有理嵌入的絕對值協議；r=0/1≥0，points=[0/1, 1/1, -1/1, 0/-2]。依序列 (r, 四個abs(point), 四個abs(point)≤r資格)，包含端點。","answer":{"kind":"expression","value":"(0/1, [0/1, 1/1, 1/1, 0/1], [是, 否, 否, 是])","explanation_zh":"閉半徑與絕對值資格=(0/1, [0/1, 1/1, 1/1, 0/1], [是, 否, 否, 是])。先取各絕對值，再用精確閉門檻比較，r=0合法；有限門檻代入不是一般不等式求解。有限分段核對非全部實數性質證明。"},"difficulty":{"band":"lower_secondary","level":1,"skills":["閉半徑的四點資格"]},"variant":{"family_id":"qf-density14-absolute-closed-radius-mask","seed":202610146051,"parameters":{"r":[0,1],"points":[[0,1],[1,1],[-1,1],[0,-2]],"domain":"finite_rational_absolute_protocol","operation":"density14_absolute_closed_radius_mask","semantic_template":"qf-density14-absolute-closed-radius-mask/[\"0/1\",[\"0/1\",\"1/1\",\"1/1\",\"0/1\"]]"}},"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":"2026-10-10","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261010-advance-20261014-r06-absolute-value-0052","batch_id":"batch-20261010-advance-20261014-r06-absolute-value-foundations","language":"zh-Hant","source_mko_ids":["mko-absolute-value"],"title_zh":"閉半徑的四點資格","stem_zh":"有理嵌入的絕對值協議；r=1/2≥0，points=[-1/2, 1/2, 3/2, 0/1]。依序列 (r, 四個abs(point), 四個abs(point)≤r資格)，包含端點。","answer":{"kind":"expression","value":"(1/2, [1/2, 1/2, 3/2, 0/1], [是, 是, 否, 是])","explanation_zh":"閉半徑與絕對值資格=(1/2, [1/2, 1/2, 3/2, 0/1], [是, 是, 否, 是])。先取各絕對值，再用精確閉門檻比較，r=0合法；有限門檻代入不是一般不等式求解。有限分段核對非全部實數性質證明。"},"difficulty":{"band":"lower_secondary","level":1,"skills":["閉半徑的四點資格"]},"variant":{"family_id":"qf-density14-absolute-closed-radius-mask","seed":202610146052,"parameters":{"r":[1,2],"points":[[-1,2],[1,2],[3,2],[0,1]],"domain":"finite_rational_absolute_protocol","operation":"density14_absolute_closed_radius_mask","semantic_template":"qf-density14-absolute-closed-radius-mask/[\"1/2\",[\"1/2\",\"1/2\",\"3/2\",\"0/1\"]]"}},"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":"2026-10-10","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261010-advance-20261014-r06-absolute-value-0053","batch_id":"batch-20261010-advance-20261014-r06-absolute-value-foundations","language":"zh-Hant","source_mko_ids":["mko-absolute-value"],"title_zh":"閉半徑的四點資格","stem_zh":"有理嵌入的絕對值協議；r=2/2≥0，points=[-2/2, 2/2, 4/2, 0/1]。依序列 (r, 四個abs(point), 四個abs(point)≤r資格)，包含端點。","answer":{"kind":"expression","value":"(1/1, [1/1, 1/1, 2/1, 0/1], [是, 是, 否, 是])","explanation_zh":"閉半徑與絕對值資格=(1/1, [1/1, 1/1, 2/1, 0/1], [是, 是, 否, 是])。先取各絕對值，再用精確閉門檻比較，r=0合法；有限門檻代入不是一般不等式求解。有限分段核對非全部實數性質證明。"},"difficulty":{"band":"lower_secondary","level":1,"skills":["閉半徑的四點資格"]},"variant":{"family_id":"qf-density14-absolute-closed-radius-mask","seed":202610146053,"parameters":{"r":[2,2],"points":[[-2,2],[2,2],[4,2],[0,1]],"domain":"finite_rational_absolute_protocol","operation":"density14_absolute_closed_radius_mask","semantic_template":"qf-density14-absolute-closed-radius-mask/[\"1/1\",[\"1/1\",\"1/1\",\"2/1\",\"0/1\"]]"}},"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":"2026-10-10","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261010-advance-20261014-r06-absolute-value-0054","batch_id":"batch-20261010-advance-20261014-r06-absolute-value-foundations","language":"zh-Hant","source_mko_ids":["mko-absolute-value"],"title_zh":"閉半徑的四點資格","stem_zh":"有理嵌入的絕對值協議；r=3/2≥0，points=[-3/2, 3/2, 5/2, 0/1]。依序列 (r, 四個abs(point), 四個abs(point)≤r資格)，包含端點。","answer":{"kind":"expression","value":"(3/2, [3/2, 3/2, 5/2, 0/1], [是, 是, 否, 是])","explanation_zh":"閉半徑與絕對值資格=(3/2, [3/2, 3/2, 5/2, 0/1], [是, 是, 否, 是])。先取各絕對值，再用精確閉門檻比較，r=0合法；有限門檻代入不是一般不等式求解。有限分段核對非全部實數性質證明。"},"difficulty":{"band":"lower_secondary","level":1,"skills":["閉半徑的四點資格"]},"variant":{"family_id":"qf-density14-absolute-closed-radius-mask","seed":202610146054,"parameters":{"r":[3,2],"points":[[-3,2],[3,2],[5,2],[0,1]],"domain":"finite_rational_absolute_protocol","operation":"density14_absolute_closed_radius_mask","semantic_template":"qf-density14-absolute-closed-radius-mask/[\"3/2\",[\"3/2\",\"3/2\",\"5/2\",\"0/1\"]]"}},"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":"2026-10-10","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261010-advance-20261014-r06-absolute-value-0055","batch_id":"batch-20261010-advance-20261014-r06-absolute-value-foundations","language":"zh-Hant","source_mko_ids":["mko-absolute-value"],"title_zh":"閉半徑的四點資格","stem_zh":"有理嵌入的絕對值協議；r=4/2≥0，points=[-4/2, 4/2, 6/2, 0/1]。依序列 (r, 四個abs(point), 四個abs(point)≤r資格)，包含端點。","answer":{"kind":"expression","value":"(2/1, [2/1, 2/1, 3/1, 0/1], [是, 是, 否, 是])","explanation_zh":"閉半徑與絕對值資格=(2/1, [2/1, 2/1, 3/1, 0/1], [是, 是, 否, 是])。先取各絕對值，再用精確閉門檻比較，r=0合法；有限門檻代入不是一般不等式求解。有限分段核對非全部實數性質證明。"},"difficulty":{"band":"lower_secondary","level":1,"skills":["閉半徑的四點資格"]},"variant":{"family_id":"qf-density14-absolute-closed-radius-mask","seed":202610146055,"parameters":{"r":[4,2],"points":[[-4,2],[4,2],[6,2],[0,1]],"domain":"finite_rational_absolute_protocol","operation":"density14_absolute_closed_radius_mask","semantic_template":"qf-density14-absolute-closed-radius-mask/[\"2/1\",[\"2/1\",\"2/1\",\"3/1\",\"0/1\"]]"}},"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":"2026-10-10","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261010-advance-20261014-r06-absolute-value-0056","batch_id":"batch-20261010-advance-20261014-r06-absolute-value-foundations","language":"zh-Hant","source_mko_ids":["mko-absolute-value"],"title_zh":"閉半徑的四點資格","stem_zh":"有理嵌入的絕對值協議；r=5/2≥0，points=[-5/2, 5/2, 7/2, 0/1]。依序列 (r, 四個abs(point), 四個abs(point)≤r資格)，包含端點。","answer":{"kind":"expression","value":"(5/2, [5/2, 5/2, 7/2, 0/1], [是, 是, 否, 是])","explanation_zh":"閉半徑與絕對值資格=(5/2, [5/2, 5/2, 7/2, 0/1], [是, 是, 否, 是])。先取各絕對值，再用精確閉門檻比較，r=0合法；有限門檻代入不是一般不等式求解。有限分段核對非全部實數性質證明。"},"difficulty":{"band":"lower_secondary","level":1,"skills":["閉半徑的四點資格"]},"variant":{"family_id":"qf-density14-absolute-closed-radius-mask","seed":202610146056,"parameters":{"r":[5,2],"points":[[-5,2],[5,2],[7,2],[0,1]],"domain":"finite_rational_absolute_protocol","operation":"density14_absolute_closed_radius_mask","semantic_template":"qf-density14-absolute-closed-radius-mask/[\"5/2\",[\"5/2\",\"5/2\",\"7/2\",\"0/1\"]]"}},"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":"2026-10-10","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261010-advance-20261014-r06-absolute-value-0057","batch_id":"batch-20261010-advance-20261014-r06-absolute-value-foundations","language":"zh-Hant","source_mko_ids":["mko-absolute-value"],"title_zh":"閉半徑的四點資格","stem_zh":"有理嵌入的絕對值協議；r=6/2≥0，points=[-6/2, 6/2, 8/2, 0/1]。依序列 (r, 四個abs(point), 四個abs(point)≤r資格)，包含端點。","answer":{"kind":"expression","value":"(3/1, [3/1, 3/1, 4/1, 0/1], [是, 是, 否, 是])","explanation_zh":"閉半徑與絕對值資格=(3/1, [3/1, 3/1, 4/1, 0/1], [是, 是, 否, 是])。先取各絕對值，再用精確閉門檻比較，r=0合法；有限門檻代入不是一般不等式求解。有限分段核對非全部實數性質證明。"},"difficulty":{"band":"lower_secondary","level":1,"skills":["閉半徑的四點資格"]},"variant":{"family_id":"qf-density14-absolute-closed-radius-mask","seed":202610146057,"parameters":{"r":[6,2],"points":[[-6,2],[6,2],[8,2],[0,1]],"domain":"finite_rational_absolute_protocol","operation":"density14_absolute_closed_radius_mask","semantic_template":"qf-density14-absolute-closed-radius-mask/[\"3/1\",[\"3/1\",\"3/1\",\"4/1\",\"0/1\"]]"}},"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":"2026-10-10","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261010-advance-20261014-r06-absolute-value-0058","batch_id":"batch-20261010-advance-20261014-r06-absolute-value-foundations","language":"zh-Hant","source_mko_ids":["mko-absolute-value"],"title_zh":"閉半徑的四點資格","stem_zh":"有理嵌入的絕對值協議；r=7/2≥0，points=[-7/2, 7/2, 9/2, 0/1]。依序列 (r, 四個abs(point), 四個abs(point)≤r資格)，包含端點。","answer":{"kind":"expression","value":"(7/2, [7/2, 7/2, 9/2, 0/1], [是, 是, 否, 是])","explanation_zh":"閉半徑與絕對值資格=(7/2, [7/2, 7/2, 9/2, 0/1], [是, 是, 否, 是])。先取各絕對值，再用精確閉門檻比較，r=0合法；有限門檻代入不是一般不等式求解。有限分段核對非全部實數性質證明。"},"difficulty":{"band":"lower_secondary","level":1,"skills":["閉半徑的四點資格"]},"variant":{"family_id":"qf-density14-absolute-closed-radius-mask","seed":202610146058,"parameters":{"r":[7,2],"points":[[-7,2],[7,2],[9,2],[0,1]],"domain":"finite_rational_absolute_protocol","operation":"density14_absolute_closed_radius_mask","semantic_template":"qf-density14-absolute-closed-radius-mask/[\"7/2\",[\"7/2\",\"7/2\",\"9/2\",\"0/1\"]]"}},"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":"2026-10-10","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261010-advance-20261014-r06-absolute-value-0059","batch_id":"batch-20261010-advance-20261014-r06-absolute-value-foundations","language":"zh-Hant","source_mko_ids":["mko-absolute-value"],"title_zh":"閉半徑的四點資格","stem_zh":"有理嵌入的絕對值協議；r=8/2≥0，points=[-8/2, 8/2, 10/2, 0/1]。依序列 (r, 四個abs(point), 四個abs(point)≤r資格)，包含端點。","answer":{"kind":"expression","value":"(4/1, [4/1, 4/1, 5/1, 0/1], [是, 是, 否, 是])","explanation_zh":"閉半徑與絕對值資格=(4/1, [4/1, 4/1, 5/1, 0/1], [是, 是, 否, 是])。先取各絕對值，再用精確閉門檻比較，r=0合法；有限門檻代入不是一般不等式求解。有限分段核對非全部實數性質證明。"},"difficulty":{"band":"lower_secondary","level":2,"skills":["閉半徑的四點資格"]},"variant":{"family_id":"qf-density14-absolute-closed-radius-mask","seed":202610146059,"parameters":{"r":[8,2],"points":[[-8,2],[8,2],[10,2],[0,1]],"domain":"finite_rational_absolute_protocol","operation":"density14_absolute_closed_radius_mask","semantic_template":"qf-density14-absolute-closed-radius-mask/[\"4/1\",[\"4/1\",\"4/1\",\"5/1\",\"0/1\"]]"}},"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":"2026-10-10","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261010-advance-20261014-r06-absolute-value-0060","batch_id":"batch-20261010-advance-20261014-r06-absolute-value-foundations","language":"zh-Hant","source_mko_ids":["mko-absolute-value"],"title_zh":"閉半徑的四點資格","stem_zh":"有理嵌入的絕對值協議；r=9/2≥0，points=[-9/2, 9/2, 11/2, 0/1]。依序列 (r, 四個abs(point), 四個abs(point)≤r資格)，包含端點。","answer":{"kind":"expression","value":"(9/2, [9/2, 9/2, 11/2, 0/1], [是, 是, 否, 是])","explanation_zh":"閉半徑與絕對值資格=(9/2, [9/2, 9/2, 11/2, 0/1], [是, 是, 否, 是])。先取各絕對值，再用精確閉門檻比較，r=0合法；有限門檻代入不是一般不等式求解。有限分段核對非全部實數性質證明。"},"difficulty":{"band":"lower_secondary","level":2,"skills":["閉半徑的四點資格"]},"variant":{"family_id":"qf-density14-absolute-closed-radius-mask","seed":202610146060,"parameters":{"r":[9,2],"points":[[-9,2],[9,2],[11,2],[0,1]],"domain":"finite_rational_absolute_protocol","operation":"density14_absolute_closed_radius_mask","semantic_template":"qf-density14-absolute-closed-radius-mask/[\"9/2\",[\"9/2\",\"9/2\",\"11/2\",\"0/1\"]]"}},"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":"2026-10-10","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261010-advance-20261014-r06-absolute-value-0061","batch_id":"batch-20261010-advance-20261014-r06-absolute-value-foundations","language":"zh-Hant","source_mko_ids":["mko-absolute-value"],"title_zh":"閉半徑的四點資格","stem_zh":"有理嵌入的絕對值協議；r=10/2≥0，points=[-10/2, 10/2, 12/2, 0/1]。依序列 (r, 四個abs(point), 四個abs(point)≤r資格)，包含端點。","answer":{"kind":"expression","value":"(5/1, [5/1, 5/1, 6/1, 0/1], [是, 是, 否, 是])","explanation_zh":"閉半徑與絕對值資格=(5/1, [5/1, 5/1, 6/1, 0/1], [是, 是, 否, 是])。先取各絕對值，再用精確閉門檻比較，r=0合法；有限門檻代入不是一般不等式求解。有限分段核對非全部實數性質證明。"},"difficulty":{"band":"lower_secondary","level":2,"skills":["閉半徑的四點資格"]},"variant":{"family_id":"qf-density14-absolute-closed-radius-mask","seed":202610146061,"parameters":{"r":[10,2],"points":[[-10,2],[10,2],[12,2],[0,1]],"domain":"finite_rational_absolute_protocol","operation":"density14_absolute_closed_radius_mask","semantic_template":"qf-density14-absolute-closed-radius-mask/[\"5/1\",[\"5/1\",\"5/1\",\"6/1\",\"0/1\"]]"}},"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":"2026-10-10","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261010-advance-20261014-r06-absolute-value-0062","batch_id":"batch-20261010-advance-20261014-r06-absolute-value-foundations","language":"zh-Hant","source_mko_ids":["mko-absolute-value"],"title_zh":"閉半徑的四點資格","stem_zh":"有理嵌入的絕對值協議；r=11/2≥0，points=[-11/2, 11/2, 13/2, 0/1]。依序列 (r, 四個abs(point), 四個abs(point)≤r資格)，包含端點。","answer":{"kind":"expression","value":"(11/2, [11/2, 11/2, 13/2, 0/1], [是, 是, 否, 是])","explanation_zh":"閉半徑與絕對值資格=(11/2, [11/2, 11/2, 13/2, 0/1], [是, 是, 否, 是])。先取各絕對值，再用精確閉門檻比較，r=0合法；有限門檻代入不是一般不等式求解。有限分段核對非全部實數性質證明。"},"difficulty":{"band":"lower_secondary","level":2,"skills":["閉半徑的四點資格"]},"variant":{"family_id":"qf-density14-absolute-closed-radius-mask","seed":202610146062,"parameters":{"r":[11,2],"points":[[-11,2],[11,2],[13,2],[0,1]],"domain":"finite_rational_absolute_protocol","operation":"density14_absolute_closed_radius_mask","semantic_template":"qf-density14-absolute-closed-radius-mask/[\"11/2\",[\"11/2\",\"11/2\",\"13/2\",\"0/1\"]]"}},"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":"2026-10-10","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261010-advance-20261014-r06-absolute-value-0063","batch_id":"batch-20261010-advance-20261014-r06-absolute-value-foundations","language":"zh-Hant","source_mko_ids":["mko-absolute-value"],"title_zh":"閉半徑的四點資格","stem_zh":"有理嵌入的絕對值協議；r=12/2≥0，points=[-12/2, 12/2, 14/2, 0/1]。依序列 (r, 四個abs(point), 四個abs(point)≤r資格)，包含端點。","answer":{"kind":"expression","value":"(6/1, [6/1, 6/1, 7/1, 0/1], [是, 是, 否, 是])","explanation_zh":"閉半徑與絕對值資格=(6/1, [6/1, 6/1, 7/1, 0/1], [是, 是, 否, 是])。先取各絕對值，再用精確閉門檻比較，r=0合法；有限門檻代入不是一般不等式求解。有限分段核對非全部實數性質證明。"},"difficulty":{"band":"lower_secondary","level":2,"skills":["閉半徑的四點資格"]},"variant":{"family_id":"qf-density14-absolute-closed-radius-mask","seed":202610146063,"parameters":{"r":[12,2],"points":[[-12,2],[12,2],[14,2],[0,1]],"domain":"finite_rational_absolute_protocol","operation":"density14_absolute_closed_radius_mask","semantic_template":"qf-density14-absolute-closed-radius-mask/[\"6/1\",[\"6/1\",\"6/1\",\"7/1\",\"0/1\"]]"}},"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":"2026-10-10","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261010-advance-20261014-r06-absolute-value-0064","batch_id":"batch-20261010-advance-20261014-r06-absolute-value-foundations","language":"zh-Hant","source_mko_ids":["mko-absolute-value"],"title_zh":"閉半徑的四點資格","stem_zh":"有理嵌入的絕對值協議；r=13/2≥0，points=[-13/2, 13/2, 15/2, 0/1]。依序列 (r, 四個abs(point), 四個abs(point)≤r資格)，包含端點。","answer":{"kind":"expression","value":"(13/2, [13/2, 13/2, 15/2, 0/1], [是, 是, 否, 是])","explanation_zh":"閉半徑與絕對值資格=(13/2, [13/2, 13/2, 15/2, 0/1], [是, 是, 否, 是])。先取各絕對值，再用精確閉門檻比較，r=0合法；有限門檻代入不是一般不等式求解。有限分段核對非全部實數性質證明。"},"difficulty":{"band":"lower_secondary","level":2,"skills":["閉半徑的四點資格"]},"variant":{"family_id":"qf-density14-absolute-closed-radius-mask","seed":202610146064,"parameters":{"r":[13,2],"points":[[-13,2],[13,2],[15,2],[0,1]],"domain":"finite_rational_absolute_protocol","operation":"density14_absolute_closed_radius_mask","semantic_template":"qf-density14-absolute-closed-radius-mask/[\"13/2\",[\"13/2\",\"13/2\",\"15/2\",\"0/1\"]]"}},"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":"2026-10-10","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261010-advance-20261014-r06-absolute-value-0065","batch_id":"batch-20261010-advance-20261014-r06-absolute-value-foundations","language":"zh-Hant","source_mko_ids":["mko-absolute-value"],"title_zh":"閉半徑的四點資格","stem_zh":"有理嵌入的絕對值協議；r=14/2≥0，points=[-14/2, 14/2, 16/2, 0/1]。依序列 (r, 四個abs(point), 四個abs(point)≤r資格)，包含端點。","answer":{"kind":"expression","value":"(7/1, [7/1, 7/1, 8/1, 0/1], [是, 是, 否, 是])","explanation_zh":"閉半徑與絕對值資格=(7/1, [7/1, 7/1, 8/1, 0/1], [是, 是, 否, 是])。先取各絕對值，再用精確閉門檻比較，r=0合法；有限門檻代入不是一般不等式求解。有限分段核對非全部實數性質證明。"},"difficulty":{"band":"lower_secondary","level":2,"skills":["閉半徑的四點資格"]},"variant":{"family_id":"qf-density14-absolute-closed-radius-mask","seed":202610146065,"parameters":{"r":[14,2],"points":[[-14,2],[14,2],[16,2],[0,1]],"domain":"finite_rational_absolute_protocol","operation":"density14_absolute_closed_radius_mask","semantic_template":"qf-density14-absolute-closed-radius-mask/[\"7/1\",[\"7/1\",\"7/1\",\"8/1\",\"0/1\"]]"}},"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":"2026-10-10","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261010-advance-20261014-r06-absolute-value-0066","batch_id":"batch-20261010-advance-20261014-r06-absolute-value-foundations","language":"zh-Hant","source_mko_ids":["mko-absolute-value"],"title_zh":"閉半徑的四點資格","stem_zh":"有理嵌入的絕對值協議；r=15/2≥0，points=[-15/2, 15/2, 17/2, 0/1]。依序列 (r, 四個abs(point), 四個abs(point)≤r資格)，包含端點。","answer":{"kind":"expression","value":"(15/2, [15/2, 15/2, 17/2, 0/1], [是, 是, 否, 是])","explanation_zh":"閉半徑與絕對值資格=(15/2, [15/2, 15/2, 17/2, 0/1], [是, 是, 否, 是])。先取各絕對值，再用精確閉門檻比較，r=0合法；有限門檻代入不是一般不等式求解。有限分段核對非全部實數性質證明。"},"difficulty":{"band":"lower_secondary","level":2,"skills":["閉半徑的四點資格"]},"variant":{"family_id":"qf-density14-absolute-closed-radius-mask","seed":202610146066,"parameters":{"r":[15,2],"points":[[-15,2],[15,2],[17,2],[0,1]],"domain":"finite_rational_absolute_protocol","operation":"density14_absolute_closed_radius_mask","semantic_template":"qf-density14-absolute-closed-radius-mask/[\"15/2\",[\"15/2\",\"15/2\",\"17/2\",\"0/1\"]]"}},"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":"2026-10-10","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261010-advance-20261014-r06-absolute-value-0067","batch_id":"batch-20261010-advance-20261014-r06-absolute-value-foundations","language":"zh-Hant","source_mko_ids":["mko-absolute-value"],"title_zh":"閉半徑的四點資格","stem_zh":"有理嵌入的絕對值協議；r=16/2≥0，points=[-16/2, 16/2, 18/2, 0/1]。依序列 (r, 四個abs(point), 四個abs(point)≤r資格)，包含端點。","answer":{"kind":"expression","value":"(8/1, [8/1, 8/1, 9/1, 0/1], [是, 是, 否, 是])","explanation_zh":"閉半徑與絕對值資格=(8/1, [8/1, 8/1, 9/1, 0/1], [是, 是, 否, 是])。先取各絕對值，再用精確閉門檻比較，r=0合法；有限門檻代入不是一般不等式求解。有限分段核對非全部實數性質證明。"},"difficulty":{"band":"lower_secondary","level":2,"skills":["閉半徑的四點資格"]},"variant":{"family_id":"qf-density14-absolute-closed-radius-mask","seed":202610146067,"parameters":{"r":[16,2],"points":[[-16,2],[16,2],[18,2],[0,1]],"domain":"finite_rational_absolute_protocol","operation":"density14_absolute_closed_radius_mask","semantic_template":"qf-density14-absolute-closed-radius-mask/[\"8/1\",[\"8/1\",\"8/1\",\"9/1\",\"0/1\"]]"}},"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":"2026-10-10","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261010-advance-20261014-r06-absolute-value-0068","batch_id":"batch-20261010-advance-20261014-r06-absolute-value-foundations","language":"zh-Hant","source_mko_ids":["mko-absolute-value"],"title_zh":"閉半徑的四點資格","stem_zh":"有理嵌入的絕對值協議；r=17/2≥0，points=[-17/2, 17/2, 19/2, 0/1]。依序列 (r, 四個abs(point), 四個abs(point)≤r資格)，包含端點。","answer":{"kind":"expression","value":"(17/2, [17/2, 17/2, 19/2, 0/1], [是, 是, 否, 是])","explanation_zh":"閉半徑與絕對值資格=(17/2, [17/2, 17/2, 19/2, 0/1], [是, 是, 否, 是])。先取各絕對值，再用精確閉門檻比較，r=0合法；有限門檻代入不是一般不等式求解。有限分段核對非全部實數性質證明。"},"difficulty":{"band":"lower_secondary","level":3,"skills":["閉半徑的四點資格"]},"variant":{"family_id":"qf-density14-absolute-closed-radius-mask","seed":202610146068,"parameters":{"r":[17,2],"points":[[-17,2],[17,2],[19,2],[0,1]],"domain":"finite_rational_absolute_protocol","operation":"density14_absolute_closed_radius_mask","semantic_template":"qf-density14-absolute-closed-radius-mask/[\"17/2\",[\"17/2\",\"17/2\",\"19/2\",\"0/1\"]]"}},"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":"2026-10-10","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261010-advance-20261014-r06-absolute-value-0069","batch_id":"batch-20261010-advance-20261014-r06-absolute-value-foundations","language":"zh-Hant","source_mko_ids":["mko-absolute-value"],"title_zh":"閉半徑的四點資格","stem_zh":"有理嵌入的絕對值協議；r=18/2≥0，points=[-18/2, 18/2, 20/2, 0/1]。依序列 (r, 四個abs(point), 四個abs(point)≤r資格)，包含端點。","answer":{"kind":"expression","value":"(9/1, [9/1, 9/1, 10/1, 0/1], [是, 是, 否, 是])","explanation_zh":"閉半徑與絕對值資格=(9/1, [9/1, 9/1, 10/1, 0/1], [是, 是, 否, 是])。先取各絕對值，再用精確閉門檻比較，r=0合法；有限門檻代入不是一般不等式求解。有限分段核對非全部實數性質證明。"},"difficulty":{"band":"lower_secondary","level":3,"skills":["閉半徑的四點資格"]},"variant":{"family_id":"qf-density14-absolute-closed-radius-mask","seed":202610146069,"parameters":{"r":[18,2],"points":[[-18,2],[18,2],[20,2],[0,1]],"domain":"finite_rational_absolute_protocol","operation":"density14_absolute_closed_radius_mask","semantic_template":"qf-density14-absolute-closed-radius-mask/[\"9/1\",[\"9/1\",\"9/1\",\"10/1\",\"0/1\"]]"}},"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":"2026-10-10","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261010-advance-20261014-r06-absolute-value-0070","batch_id":"batch-20261010-advance-20261014-r06-absolute-value-foundations","language":"zh-Hant","source_mko_ids":["mko-absolute-value"],"title_zh":"閉半徑的四點資格","stem_zh":"有理嵌入的絕對值協議；r=19/2≥0，points=[-19/2, 19/2, 21/2, 0/1]。依序列 (r, 四個abs(point), 四個abs(point)≤r資格)，包含端點。","answer":{"kind":"expression","value":"(19/2, [19/2, 19/2, 21/2, 0/1], [是, 是, 否, 是])","explanation_zh":"閉半徑與絕對值資格=(19/2, [19/2, 19/2, 21/2, 0/1], [是, 是, 否, 是])。先取各絕對值，再用精確閉門檻比較，r=0合法；有限門檻代入不是一般不等式求解。有限分段核對非全部實數性質證明。"},"difficulty":{"band":"lower_secondary","level":3,"skills":["閉半徑的四點資格"]},"variant":{"family_id":"qf-density14-absolute-closed-radius-mask","seed":202610146070,"parameters":{"r":[19,2],"points":[[-19,2],[19,2],[21,2],[0,1]],"domain":"finite_rational_absolute_protocol","operation":"density14_absolute_closed_radius_mask","semantic_template":"qf-density14-absolute-closed-radius-mask/[\"19/2\",[\"19/2\",\"19/2\",\"21/2\",\"0/1\"]]"}},"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":"2026-10-10","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261010-advance-20261014-r06-absolute-value-0071","batch_id":"batch-20261010-advance-20261014-r06-absolute-value-foundations","language":"zh-Hant","source_mko_ids":["mko-absolute-value"],"title_zh":"閉半徑的四點資格","stem_zh":"有理嵌入的絕對值協議；r=20/2≥0，points=[-20/2, 20/2, 22/2, 0/1]。依序列 (r, 四個abs(point), 四個abs(point)≤r資格)，包含端點。","answer":{"kind":"expression","value":"(10/1, [10/1, 10/1, 11/1, 0/1], [是, 是, 否, 是])","explanation_zh":"閉半徑與絕對值資格=(10/1, [10/1, 10/1, 11/1, 0/1], [是, 是, 否, 是])。先取各絕對值，再用精確閉門檻比較，r=0合法；有限門檻代入不是一般不等式求解。有限分段核對非全部實數性質證明。"},"difficulty":{"band":"lower_secondary","level":3,"skills":["閉半徑的四點資格"]},"variant":{"family_id":"qf-density14-absolute-closed-radius-mask","seed":202610146071,"parameters":{"r":[20,2],"points":[[-20,2],[20,2],[22,2],[0,1]],"domain":"finite_rational_absolute_protocol","operation":"density14_absolute_closed_radius_mask","semantic_template":"qf-density14-absolute-closed-radius-mask/[\"10/1\",[\"10/1\",\"10/1\",\"11/1\",\"0/1\"]]"}},"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":"2026-10-10","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261010-advance-20261014-r06-absolute-value-0072","batch_id":"batch-20261010-advance-20261014-r06-absolute-value-foundations","language":"zh-Hant","source_mko_ids":["mko-absolute-value"],"title_zh":"閉半徑的四點資格","stem_zh":"有理嵌入的絕對值協議；r=21/2≥0，points=[-21/2, 21/2, 23/2, 0/1]。依序列 (r, 四個abs(point), 四個abs(point)≤r資格)，包含端點。","answer":{"kind":"expression","value":"(21/2, [21/2, 21/2, 23/2, 0/1], [是, 是, 否, 是])","explanation_zh":"閉半徑與絕對值資格=(21/2, [21/2, 21/2, 23/2, 0/1], [是, 是, 否, 是])。先取各絕對值，再用精確閉門檻比較，r=0合法；有限門檻代入不是一般不等式求解。有限分段核對非全部實數性質證明。"},"difficulty":{"band":"lower_secondary","level":3,"skills":["閉半徑的四點資格"]},"variant":{"family_id":"qf-density14-absolute-closed-radius-mask","seed":202610146072,"parameters":{"r":[21,2],"points":[[-21,2],[21,2],[23,2],[0,1]],"domain":"finite_rational_absolute_protocol","operation":"density14_absolute_closed_radius_mask","semantic_template":"qf-density14-absolute-closed-radius-mask/[\"21/2\",[\"21/2\",\"21/2\",\"23/2\",\"0/1\"]]"}},"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":"2026-10-10","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261010-advance-20261014-r06-absolute-value-0073","batch_id":"batch-20261010-advance-20261014-r06-absolute-value-foundations","language":"zh-Hant","source_mko_ids":["mko-absolute-value"],"title_zh":"閉半徑的四點資格","stem_zh":"有理嵌入的絕對值協議；r=22/2≥0，points=[-22/2, 22/2, 24/2, 0/1]。依序列 (r, 四個abs(point), 四個abs(point)≤r資格)，包含端點。","answer":{"kind":"expression","value":"(11/1, [11/1, 11/1, 12/1, 0/1], [是, 是, 否, 是])","explanation_zh":"閉半徑與絕對值資格=(11/1, [11/1, 11/1, 12/1, 0/1], [是, 是, 否, 是])。先取各絕對值，再用精確閉門檻比較，r=0合法；有限門檻代入不是一般不等式求解。有限分段核對非全部實數性質證明。"},"difficulty":{"band":"lower_secondary","level":3,"skills":["閉半徑的四點資格"]},"variant":{"family_id":"qf-density14-absolute-closed-radius-mask","seed":202610146073,"parameters":{"r":[22,2],"points":[[-22,2],[22,2],[24,2],[0,1]],"domain":"finite_rational_absolute_protocol","operation":"density14_absolute_closed_radius_mask","semantic_template":"qf-density14-absolute-closed-radius-mask/[\"11/1\",[\"11/1\",\"11/1\",\"12/1\",\"0/1\"]]"}},"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":"2026-10-10","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261010-advance-20261014-r06-absolute-value-0074","batch_id":"batch-20261010-advance-20261014-r06-absolute-value-foundations","language":"zh-Hant","source_mko_ids":["mko-absolute-value"],"title_zh":"閉半徑的四點資格","stem_zh":"有理嵌入的絕對值協議；r=23/2≥0，points=[-23/2, 23/2, 25/2, 0/1]。依序列 (r, 四個abs(point), 四個abs(point)≤r資格)，包含端點。","answer":{"kind":"expression","value":"(23/2, [23/2, 23/2, 25/2, 0/1], [是, 是, 否, 是])","explanation_zh":"閉半徑與絕對值資格=(23/2, [23/2, 23/2, 25/2, 0/1], [是, 是, 否, 是])。先取各絕對值，再用精確閉門檻比較，r=0合法；有限門檻代入不是一般不等式求解。有限分段核對非全部實數性質證明。"},"difficulty":{"band":"lower_secondary","level":3,"skills":["閉半徑的四點資格"]},"variant":{"family_id":"qf-density14-absolute-closed-radius-mask","seed":202610146074,"parameters":{"r":[23,2],"points":[[-23,2],[23,2],[25,2],[0,1]],"domain":"finite_rational_absolute_protocol","operation":"density14_absolute_closed_radius_mask","semantic_template":"qf-density14-absolute-closed-radius-mask/[\"23/2\",[\"23/2\",\"23/2\",\"25/2\",\"0/1\"]]"}},"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":"2026-10-10","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261010-advance-20261014-r06-absolute-value-0075","batch_id":"batch-20261010-advance-20261014-r06-absolute-value-foundations","language":"zh-Hant","source_mko_ids":["mko-absolute-value"],"title_zh":"閉半徑的四點資格","stem_zh":"有理嵌入的絕對值協議；r=24/2≥0，points=[-24/2, 24/2, 26/2, 0/1]。依序列 (r, 四個abs(point), 四個abs(point)≤r資格)，包含端點。","answer":{"kind":"expression","value":"(12/1, [12/1, 12/1, 13/1, 0/1], [是, 是, 否, 是])","explanation_zh":"閉半徑與絕對值資格=(12/1, [12/1, 12/1, 13/1, 0/1], [是, 是, 否, 是])。先取各絕對值，再用精確閉門檻比較，r=0合法；有限門檻代入不是一般不等式求解。有限分段核對非全部實數性質證明。"},"difficulty":{"band":"lower_secondary","level":3,"skills":["閉半徑的四點資格"]},"variant":{"family_id":"qf-density14-absolute-closed-radius-mask","seed":202610146075,"parameters":{"r":[24,2],"points":[[-24,2],[24,2],[26,2],[0,1]],"domain":"finite_rational_absolute_protocol","operation":"density14_absolute_closed_radius_mask","semantic_template":"qf-density14-absolute-closed-radius-mask/[\"12/1\",[\"12/1\",\"12/1\",\"13/1\",\"0/1\"]]"}},"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":"2026-10-10","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261010-advance-20261014-r06-absolute-value-0076","batch_id":"batch-20261010-advance-20261014-r06-absolute-value-foundations","language":"zh-Hant","source_mko_ids":["mko-absolute-value"],"title_zh":"絕對值目標的根與代入","stem_zh":"有理嵌入的絕對值協議；r=-1/1，queries=[0/1, 1/1, 2/1]。依序答 (r非負資格, |x|=r遞增根列表, 三個abs(query), 三個等於r資格)。","answer":{"kind":"expression","value":"(否, [], [0/1, 1/1, 2/1], [否, 否, 否])","explanation_zh":"絕對值目標資格根與代入=(否, [], [0/1, 1/1, 2/1], [否, 否, 否])。負目標無根，零只有0，正目標有正負兩根；再逐一精確代入。有限分段核對非全部實數性質證明。"},"difficulty":{"band":"lower_secondary","level":1,"skills":["絕對值目標的根與代入"]},"variant":{"family_id":"qf-density14-absolute-preimage-rational-audit","seed":202610146076,"parameters":{"r":[-1,1],"queries":[[0,1],[1,1],[2,1]],"domain":"finite_rational_absolute_protocol","operation":"density14_absolute_preimage_rational_audit","semantic_template":"qf-density14-absolute-preimage-rational-audit/[\"negative_target\",[\"0/1\",\"1/1\",\"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":"2026-10-10","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261010-advance-20261014-r06-absolute-value-0077","batch_id":"batch-20261010-advance-20261014-r06-absolute-value-foundations","language":"zh-Hant","source_mko_ids":["mko-absolute-value"],"title_zh":"絕對值目標的根與代入","stem_zh":"有理嵌入的絕對值協議；r=0/1，queries=[0/1, 1/1, -1/1]。依序答 (r非負資格, |x|=r遞增根列表, 三個abs(query), 三個等於r資格)。","answer":{"kind":"expression","value":"(是, [0/1], [0/1, 1/1, 1/1], [是, 否, 否])","explanation_zh":"絕對值目標資格根與代入=(是, [0/1], [0/1, 1/1, 1/1], [是, 否, 否])。負目標無根，零只有0，正目標有正負兩根；再逐一精確代入。有限分段核對非全部實數性質證明。"},"difficulty":{"band":"lower_secondary","level":1,"skills":["絕對值目標的根與代入"]},"variant":{"family_id":"qf-density14-absolute-preimage-rational-audit","seed":202610146077,"parameters":{"r":[0,1],"queries":[[0,1],[1,1],[-1,1]],"domain":"finite_rational_absolute_protocol","operation":"density14_absolute_preimage_rational_audit","semantic_template":"qf-density14-absolute-preimage-rational-audit/[\"0/1\",[\"0/1\",\"1/1\",\"1/1\"]]"}},"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":"2026-10-10","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261010-advance-20261014-r06-absolute-value-0078","batch_id":"batch-20261010-advance-20261014-r06-absolute-value-foundations","language":"zh-Hant","source_mko_ids":["mko-absolute-value"],"title_zh":"絕對值目標的根與代入","stem_zh":"有理嵌入的絕對值協議；r=1/2，queries=[-1/2, 1/2, 3/2]。依序答 (r非負資格, |x|=r遞增根列表, 三個abs(query), 三個等於r資格)。","answer":{"kind":"expression","value":"(是, [-1/2, 1/2], [1/2, 1/2, 3/2], [是, 是, 否])","explanation_zh":"絕對值目標資格根與代入=(是, [-1/2, 1/2], [1/2, 1/2, 3/2], [是, 是, 否])。負目標無根，零只有0，正目標有正負兩根；再逐一精確代入。有限分段核對非全部實數性質證明。"},"difficulty":{"band":"lower_secondary","level":1,"skills":["絕對值目標的根與代入"]},"variant":{"family_id":"qf-density14-absolute-preimage-rational-audit","seed":202610146078,"parameters":{"r":[1,2],"queries":[[-1,2],[1,2],[3,2]],"domain":"finite_rational_absolute_protocol","operation":"density14_absolute_preimage_rational_audit","semantic_template":"qf-density14-absolute-preimage-rational-audit/[\"1/2\",[\"1/2\",\"1/2\",\"3/2\"]]"}},"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":"2026-10-10","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261010-advance-20261014-r06-absolute-value-0079","batch_id":"batch-20261010-advance-20261014-r06-absolute-value-foundations","language":"zh-Hant","source_mko_ids":["mko-absolute-value"],"title_zh":"絕對值目標的根與代入","stem_zh":"有理嵌入的絕對值協議；r=2/2，queries=[-2/2, 2/2, 4/2]。依序答 (r非負資格, |x|=r遞增根列表, 三個abs(query), 三個等於r資格)。","answer":{"kind":"expression","value":"(是, [-1/1, 1/1], [1/1, 1/1, 2/1], [是, 是, 否])","explanation_zh":"絕對值目標資格根與代入=(是, [-1/1, 1/1], [1/1, 1/1, 2/1], [是, 是, 否])。負目標無根，零只有0，正目標有正負兩根；再逐一精確代入。有限分段核對非全部實數性質證明。"},"difficulty":{"band":"lower_secondary","level":1,"skills":["絕對值目標的根與代入"]},"variant":{"family_id":"qf-density14-absolute-preimage-rational-audit","seed":202610146079,"parameters":{"r":[2,2],"queries":[[-2,2],[2,2],[4,2]],"domain":"finite_rational_absolute_protocol","operation":"density14_absolute_preimage_rational_audit","semantic_template":"qf-density14-absolute-preimage-rational-audit/[\"1/1\",[\"1/1\",\"1/1\",\"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":"2026-10-10","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261010-advance-20261014-r06-absolute-value-0080","batch_id":"batch-20261010-advance-20261014-r06-absolute-value-foundations","language":"zh-Hant","source_mko_ids":["mko-absolute-value"],"title_zh":"絕對值目標的根與代入","stem_zh":"有理嵌入的絕對值協議；r=3/2，queries=[-3/2, 3/2, 5/2]。依序答 (r非負資格, |x|=r遞增根列表, 三個abs(query), 三個等於r資格)。","answer":{"kind":"expression","value":"(是, [-3/2, 3/2], [3/2, 3/2, 5/2], [是, 是, 否])","explanation_zh":"絕對值目標資格根與代入=(是, [-3/2, 3/2], [3/2, 3/2, 5/2], [是, 是, 否])。負目標無根，零只有0，正目標有正負兩根；再逐一精確代入。有限分段核對非全部實數性質證明。"},"difficulty":{"band":"lower_secondary","level":1,"skills":["絕對值目標的根與代入"]},"variant":{"family_id":"qf-density14-absolute-preimage-rational-audit","seed":202610146080,"parameters":{"r":[3,2],"queries":[[-3,2],[3,2],[5,2]],"domain":"finite_rational_absolute_protocol","operation":"density14_absolute_preimage_rational_audit","semantic_template":"qf-density14-absolute-preimage-rational-audit/[\"3/2\",[\"3/2\",\"3/2\",\"5/2\"]]"}},"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":"2026-10-10","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261010-advance-20261014-r06-absolute-value-0081","batch_id":"batch-20261010-advance-20261014-r06-absolute-value-foundations","language":"zh-Hant","source_mko_ids":["mko-absolute-value"],"title_zh":"絕對值目標的根與代入","stem_zh":"有理嵌入的絕對值協議；r=4/2，queries=[-4/2, 4/2, 6/2]。依序答 (r非負資格, |x|=r遞增根列表, 三個abs(query), 三個等於r資格)。","answer":{"kind":"expression","value":"(是, [-2/1, 2/1], [2/1, 2/1, 3/1], [是, 是, 否])","explanation_zh":"絕對值目標資格根與代入=(是, [-2/1, 2/1], [2/1, 2/1, 3/1], [是, 是, 否])。負目標無根，零只有0，正目標有正負兩根；再逐一精確代入。有限分段核對非全部實數性質證明。"},"difficulty":{"band":"lower_secondary","level":1,"skills":["絕對值目標的根與代入"]},"variant":{"family_id":"qf-density14-absolute-preimage-rational-audit","seed":202610146081,"parameters":{"r":[4,2],"queries":[[-4,2],[4,2],[6,2]],"domain":"finite_rational_absolute_protocol","operation":"density14_absolute_preimage_rational_audit","semantic_template":"qf-density14-absolute-preimage-rational-audit/[\"2/1\",[\"2/1\",\"2/1\",\"3/1\"]]"}},"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":"2026-10-10","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261010-advance-20261014-r06-absolute-value-0082","batch_id":"batch-20261010-advance-20261014-r06-absolute-value-foundations","language":"zh-Hant","source_mko_ids":["mko-absolute-value"],"title_zh":"絕對值目標的根與代入","stem_zh":"有理嵌入的絕對值協議；r=5/2，queries=[-5/2, 5/2, 7/2]。依序答 (r非負資格, |x|=r遞增根列表, 三個abs(query), 三個等於r資格)。","answer":{"kind":"expression","value":"(是, [-5/2, 5/2], [5/2, 5/2, 7/2], [是, 是, 否])","explanation_zh":"絕對值目標資格根與代入=(是, [-5/2, 5/2], [5/2, 5/2, 7/2], [是, 是, 否])。負目標無根，零只有0，正目標有正負兩根；再逐一精確代入。有限分段核對非全部實數性質證明。"},"difficulty":{"band":"lower_secondary","level":1,"skills":["絕對值目標的根與代入"]},"variant":{"family_id":"qf-density14-absolute-preimage-rational-audit","seed":202610146082,"parameters":{"r":[5,2],"queries":[[-5,2],[5,2],[7,2]],"domain":"finite_rational_absolute_protocol","operation":"density14_absolute_preimage_rational_audit","semantic_template":"qf-density14-absolute-preimage-rational-audit/[\"5/2\",[\"5/2\",\"5/2\",\"7/2\"]]"}},"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":"2026-10-10","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261010-advance-20261014-r06-absolute-value-0083","batch_id":"batch-20261010-advance-20261014-r06-absolute-value-foundations","language":"zh-Hant","source_mko_ids":["mko-absolute-value"],"title_zh":"絕對值目標的根與代入","stem_zh":"有理嵌入的絕對值協議；r=6/2，queries=[-6/2, 6/2, 8/2]。依序答 (r非負資格, |x|=r遞增根列表, 三個abs(query), 三個等於r資格)。","answer":{"kind":"expression","value":"(是, [-3/1, 3/1], [3/1, 3/1, 4/1], [是, 是, 否])","explanation_zh":"絕對值目標資格根與代入=(是, [-3/1, 3/1], [3/1, 3/1, 4/1], [是, 是, 否])。負目標無根，零只有0，正目標有正負兩根；再逐一精確代入。有限分段核對非全部實數性質證明。"},"difficulty":{"band":"lower_secondary","level":1,"skills":["絕對值目標的根與代入"]},"variant":{"family_id":"qf-density14-absolute-preimage-rational-audit","seed":202610146083,"parameters":{"r":[6,2],"queries":[[-6,2],[6,2],[8,2]],"domain":"finite_rational_absolute_protocol","operation":"density14_absolute_preimage_rational_audit","semantic_template":"qf-density14-absolute-preimage-rational-audit/[\"3/1\",[\"3/1\",\"3/1\",\"4/1\"]]"}},"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":"2026-10-10","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261010-advance-20261014-r06-absolute-value-0084","batch_id":"batch-20261010-advance-20261014-r06-absolute-value-foundations","language":"zh-Hant","source_mko_ids":["mko-absolute-value"],"title_zh":"絕對值目標的根與代入","stem_zh":"有理嵌入的絕對值協議；r=7/2，queries=[-7/2, 7/2, 9/2]。依序答 (r非負資格, |x|=r遞增根列表, 三個abs(query), 三個等於r資格)。","answer":{"kind":"expression","value":"(是, [-7/2, 7/2], [7/2, 7/2, 9/2], [是, 是, 否])","explanation_zh":"絕對值目標資格根與代入=(是, [-7/2, 7/2], [7/2, 7/2, 9/2], [是, 是, 否])。負目標無根，零只有0，正目標有正負兩根；再逐一精確代入。有限分段核對非全部實數性質證明。"},"difficulty":{"band":"lower_secondary","level":2,"skills":["絕對值目標的根與代入"]},"variant":{"family_id":"qf-density14-absolute-preimage-rational-audit","seed":202610146084,"parameters":{"r":[7,2],"queries":[[-7,2],[7,2],[9,2]],"domain":"finite_rational_absolute_protocol","operation":"density14_absolute_preimage_rational_audit","semantic_template":"qf-density14-absolute-preimage-rational-audit/[\"7/2\",[\"7/2\",\"7/2\",\"9/2\"]]"}},"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":"2026-10-10","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261010-advance-20261014-r06-absolute-value-0085","batch_id":"batch-20261010-advance-20261014-r06-absolute-value-foundations","language":"zh-Hant","source_mko_ids":["mko-absolute-value"],"title_zh":"絕對值目標的根與代入","stem_zh":"有理嵌入的絕對值協議；r=8/2，queries=[-8/2, 8/2, 10/2]。依序答 (r非負資格, |x|=r遞增根列表, 三個abs(query), 三個等於r資格)。","answer":{"kind":"expression","value":"(是, [-4/1, 4/1], [4/1, 4/1, 5/1], [是, 是, 否])","explanation_zh":"絕對值目標資格根與代入=(是, [-4/1, 4/1], [4/1, 4/1, 5/1], [是, 是, 否])。負目標無根，零只有0，正目標有正負兩根；再逐一精確代入。有限分段核對非全部實數性質證明。"},"difficulty":{"band":"lower_secondary","level":2,"skills":["絕對值目標的根與代入"]},"variant":{"family_id":"qf-density14-absolute-preimage-rational-audit","seed":202610146085,"parameters":{"r":[8,2],"queries":[[-8,2],[8,2],[10,2]],"domain":"finite_rational_absolute_protocol","operation":"density14_absolute_preimage_rational_audit","semantic_template":"qf-density14-absolute-preimage-rational-audit/[\"4/1\",[\"4/1\",\"4/1\",\"5/1\"]]"}},"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":"2026-10-10","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261010-advance-20261014-r06-absolute-value-0086","batch_id":"batch-20261010-advance-20261014-r06-absolute-value-foundations","language":"zh-Hant","source_mko_ids":["mko-absolute-value"],"title_zh":"絕對值目標的根與代入","stem_zh":"有理嵌入的絕對值協議；r=9/2，queries=[-9/2, 9/2, 11/2]。依序答 (r非負資格, |x|=r遞增根列表, 三個abs(query), 三個等於r資格)。","answer":{"kind":"expression","value":"(是, [-9/2, 9/2], [9/2, 9/2, 11/2], [是, 是, 否])","explanation_zh":"絕對值目標資格根與代入=(是, [-9/2, 9/2], [9/2, 9/2, 11/2], [是, 是, 否])。負目標無根，零只有0，正目標有正負兩根；再逐一精確代入。有限分段核對非全部實數性質證明。"},"difficulty":{"band":"lower_secondary","level":2,"skills":["絕對值目標的根與代入"]},"variant":{"family_id":"qf-density14-absolute-preimage-rational-audit","seed":202610146086,"parameters":{"r":[9,2],"queries":[[-9,2],[9,2],[11,2]],"domain":"finite_rational_absolute_protocol","operation":"density14_absolute_preimage_rational_audit","semantic_template":"qf-density14-absolute-preimage-rational-audit/[\"9/2\",[\"9/2\",\"9/2\",\"11/2\"]]"}},"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":"2026-10-10","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261010-advance-20261014-r06-absolute-value-0087","batch_id":"batch-20261010-advance-20261014-r06-absolute-value-foundations","language":"zh-Hant","source_mko_ids":["mko-absolute-value"],"title_zh":"絕對值目標的根與代入","stem_zh":"有理嵌入的絕對值協議；r=10/2，queries=[-10/2, 10/2, 12/2]。依序答 (r非負資格, |x|=r遞增根列表, 三個abs(query), 三個等於r資格)。","answer":{"kind":"expression","value":"(是, [-5/1, 5/1], [5/1, 5/1, 6/1], [是, 是, 否])","explanation_zh":"絕對值目標資格根與代入=(是, [-5/1, 5/1], [5/1, 5/1, 6/1], [是, 是, 否])。負目標無根，零只有0，正目標有正負兩根；再逐一精確代入。有限分段核對非全部實數性質證明。"},"difficulty":{"band":"lower_secondary","level":2,"skills":["絕對值目標的根與代入"]},"variant":{"family_id":"qf-density14-absolute-preimage-rational-audit","seed":202610146087,"parameters":{"r":[10,2],"queries":[[-10,2],[10,2],[12,2]],"domain":"finite_rational_absolute_protocol","operation":"density14_absolute_preimage_rational_audit","semantic_template":"qf-density14-absolute-preimage-rational-audit/[\"5/1\",[\"5/1\",\"5/1\",\"6/1\"]]"}},"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":"2026-10-10","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261010-advance-20261014-r06-absolute-value-0088","batch_id":"batch-20261010-advance-20261014-r06-absolute-value-foundations","language":"zh-Hant","source_mko_ids":["mko-absolute-value"],"title_zh":"絕對值目標的根與代入","stem_zh":"有理嵌入的絕對值協議；r=11/2，queries=[-11/2, 11/2, 13/2]。依序答 (r非負資格, |x|=r遞增根列表, 三個abs(query), 三個等於r資格)。","answer":{"kind":"expression","value":"(是, [-11/2, 11/2], [11/2, 11/2, 13/2], [是, 是, 否])","explanation_zh":"絕對值目標資格根與代入=(是, [-11/2, 11/2], [11/2, 11/2, 13/2], [是, 是, 否])。負目標無根，零只有0，正目標有正負兩根；再逐一精確代入。有限分段核對非全部實數性質證明。"},"difficulty":{"band":"lower_secondary","level":2,"skills":["絕對值目標的根與代入"]},"variant":{"family_id":"qf-density14-absolute-preimage-rational-audit","seed":202610146088,"parameters":{"r":[11,2],"queries":[[-11,2],[11,2],[13,2]],"domain":"finite_rational_absolute_protocol","operation":"density14_absolute_preimage_rational_audit","semantic_template":"qf-density14-absolute-preimage-rational-audit/[\"11/2\",[\"11/2\",\"11/2\",\"13/2\"]]"}},"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":"2026-10-10","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261010-advance-20261014-r06-absolute-value-0089","batch_id":"batch-20261010-advance-20261014-r06-absolute-value-foundations","language":"zh-Hant","source_mko_ids":["mko-absolute-value"],"title_zh":"絕對值目標的根與代入","stem_zh":"有理嵌入的絕對值協議；r=12/2，queries=[-12/2, 12/2, 14/2]。依序答 (r非負資格, |x|=r遞增根列表, 三個abs(query), 三個等於r資格)。","answer":{"kind":"expression","value":"(是, [-6/1, 6/1], [6/1, 6/1, 7/1], [是, 是, 否])","explanation_zh":"絕對值目標資格根與代入=(是, [-6/1, 6/1], [6/1, 6/1, 7/1], [是, 是, 否])。負目標無根，零只有0，正目標有正負兩根；再逐一精確代入。有限分段核對非全部實數性質證明。"},"difficulty":{"band":"lower_secondary","level":2,"skills":["絕對值目標的根與代入"]},"variant":{"family_id":"qf-density14-absolute-preimage-rational-audit","seed":202610146089,"parameters":{"r":[12,2],"queries":[[-12,2],[12,2],[14,2]],"domain":"finite_rational_absolute_protocol","operation":"density14_absolute_preimage_rational_audit","semantic_template":"qf-density14-absolute-preimage-rational-audit/[\"6/1\",[\"6/1\",\"6/1\",\"7/1\"]]"}},"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":"2026-10-10","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261010-advance-20261014-r06-absolute-value-0090","batch_id":"batch-20261010-advance-20261014-r06-absolute-value-foundations","language":"zh-Hant","source_mko_ids":["mko-absolute-value"],"title_zh":"絕對值目標的根與代入","stem_zh":"有理嵌入的絕對值協議；r=13/2，queries=[-13/2, 13/2, 15/2]。依序答 (r非負資格, |x|=r遞增根列表, 三個abs(query), 三個等於r資格)。","answer":{"kind":"expression","value":"(是, [-13/2, 13/2], [13/2, 13/2, 15/2], [是, 是, 否])","explanation_zh":"絕對值目標資格根與代入=(是, [-13/2, 13/2], [13/2, 13/2, 15/2], [是, 是, 否])。負目標無根，零只有0，正目標有正負兩根；再逐一精確代入。有限分段核對非全部實數性質證明。"},"difficulty":{"band":"lower_secondary","level":2,"skills":["絕對值目標的根與代入"]},"variant":{"family_id":"qf-density14-absolute-preimage-rational-audit","seed":202610146090,"parameters":{"r":[13,2],"queries":[[-13,2],[13,2],[15,2]],"domain":"finite_rational_absolute_protocol","operation":"density14_absolute_preimage_rational_audit","semantic_template":"qf-density14-absolute-preimage-rational-audit/[\"13/2\",[\"13/2\",\"13/2\",\"15/2\"]]"}},"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":"2026-10-10","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261010-advance-20261014-r06-absolute-value-0091","batch_id":"batch-20261010-advance-20261014-r06-absolute-value-foundations","language":"zh-Hant","source_mko_ids":["mko-absolute-value"],"title_zh":"絕對值目標的根與代入","stem_zh":"有理嵌入的絕對值協議；r=14/2，queries=[-14/2, 14/2, 16/2]。依序答 (r非負資格, |x|=r遞增根列表, 三個abs(query), 三個等於r資格)。","answer":{"kind":"expression","value":"(是, [-7/1, 7/1], [7/1, 7/1, 8/1], [是, 是, 否])","explanation_zh":"絕對值目標資格根與代入=(是, [-7/1, 7/1], [7/1, 7/1, 8/1], [是, 是, 否])。負目標無根，零只有0，正目標有正負兩根；再逐一精確代入。有限分段核對非全部實數性質證明。"},"difficulty":{"band":"lower_secondary","level":2,"skills":["絕對值目標的根與代入"]},"variant":{"family_id":"qf-density14-absolute-preimage-rational-audit","seed":202610146091,"parameters":{"r":[14,2],"queries":[[-14,2],[14,2],[16,2]],"domain":"finite_rational_absolute_protocol","operation":"density14_absolute_preimage_rational_audit","semantic_template":"qf-density14-absolute-preimage-rational-audit/[\"7/1\",[\"7/1\",\"7/1\",\"8/1\"]]"}},"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":"2026-10-10","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261010-advance-20261014-r06-absolute-value-0092","batch_id":"batch-20261010-advance-20261014-r06-absolute-value-foundations","language":"zh-Hant","source_mko_ids":["mko-absolute-value"],"title_zh":"絕對值目標的根與代入","stem_zh":"有理嵌入的絕對值協議；r=15/2，queries=[-15/2, 15/2, 17/2]。依序答 (r非負資格, |x|=r遞增根列表, 三個abs(query), 三個等於r資格)。","answer":{"kind":"expression","value":"(是, [-15/2, 15/2], [15/2, 15/2, 17/2], [是, 是, 否])","explanation_zh":"絕對值目標資格根與代入=(是, [-15/2, 15/2], [15/2, 15/2, 17/2], [是, 是, 否])。負目標無根，零只有0，正目標有正負兩根；再逐一精確代入。有限分段核對非全部實數性質證明。"},"difficulty":{"band":"lower_secondary","level":2,"skills":["絕對值目標的根與代入"]},"variant":{"family_id":"qf-density14-absolute-preimage-rational-audit","seed":202610146092,"parameters":{"r":[15,2],"queries":[[-15,2],[15,2],[17,2]],"domain":"finite_rational_absolute_protocol","operation":"density14_absolute_preimage_rational_audit","semantic_template":"qf-density14-absolute-preimage-rational-audit/[\"15/2\",[\"15/2\",\"15/2\",\"17/2\"]]"}},"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":"2026-10-10","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261010-advance-20261014-r06-absolute-value-0093","batch_id":"batch-20261010-advance-20261014-r06-absolute-value-foundations","language":"zh-Hant","source_mko_ids":["mko-absolute-value"],"title_zh":"絕對值目標的根與代入","stem_zh":"有理嵌入的絕對值協議；r=16/2，queries=[-16/2, 16/2, 18/2]。依序答 (r非負資格, |x|=r遞增根列表, 三個abs(query), 三個等於r資格)。","answer":{"kind":"expression","value":"(是, [-8/1, 8/1], [8/1, 8/1, 9/1], [是, 是, 否])","explanation_zh":"絕對值目標資格根與代入=(是, [-8/1, 8/1], [8/1, 8/1, 9/1], [是, 是, 否])。負目標無根，零只有0，正目標有正負兩根；再逐一精確代入。有限分段核對非全部實數性質證明。"},"difficulty":{"band":"lower_secondary","level":3,"skills":["絕對值目標的根與代入"]},"variant":{"family_id":"qf-density14-absolute-preimage-rational-audit","seed":202610146093,"parameters":{"r":[16,2],"queries":[[-16,2],[16,2],[18,2]],"domain":"finite_rational_absolute_protocol","operation":"density14_absolute_preimage_rational_audit","semantic_template":"qf-density14-absolute-preimage-rational-audit/[\"8/1\",[\"8/1\",\"8/1\",\"9/1\"]]"}},"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":"2026-10-10","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261010-advance-20261014-r06-absolute-value-0094","batch_id":"batch-20261010-advance-20261014-r06-absolute-value-foundations","language":"zh-Hant","source_mko_ids":["mko-absolute-value"],"title_zh":"絕對值目標的根與代入","stem_zh":"有理嵌入的絕對值協議；r=17/2，queries=[-17/2, 17/2, 19/2]。依序答 (r非負資格, |x|=r遞增根列表, 三個abs(query), 三個等於r資格)。","answer":{"kind":"expression","value":"(是, [-17/2, 17/2], [17/2, 17/2, 19/2], [是, 是, 否])","explanation_zh":"絕對值目標資格根與代入=(是, [-17/2, 17/2], [17/2, 17/2, 19/2], [是, 是, 否])。負目標無根，零只有0，正目標有正負兩根；再逐一精確代入。有限分段核對非全部實數性質證明。"},"difficulty":{"band":"lower_secondary","level":3,"skills":["絕對值目標的根與代入"]},"variant":{"family_id":"qf-density14-absolute-preimage-rational-audit","seed":202610146094,"parameters":{"r":[17,2],"queries":[[-17,2],[17,2],[19,2]],"domain":"finite_rational_absolute_protocol","operation":"density14_absolute_preimage_rational_audit","semantic_template":"qf-density14-absolute-preimage-rational-audit/[\"17/2\",[\"17/2\",\"17/2\",\"19/2\"]]"}},"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":"2026-10-10","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261010-advance-20261014-r06-absolute-value-0095","batch_id":"batch-20261010-advance-20261014-r06-absolute-value-foundations","language":"zh-Hant","source_mko_ids":["mko-absolute-value"],"title_zh":"絕對值目標的根與代入","stem_zh":"有理嵌入的絕對值協議；r=18/2，queries=[-18/2, 18/2, 20/2]。依序答 (r非負資格, |x|=r遞增根列表, 三個abs(query), 三個等於r資格)。","answer":{"kind":"expression","value":"(是, [-9/1, 9/1], [9/1, 9/1, 10/1], [是, 是, 否])","explanation_zh":"絕對值目標資格根與代入=(是, [-9/1, 9/1], [9/1, 9/1, 10/1], [是, 是, 否])。負目標無根，零只有0，正目標有正負兩根；再逐一精確代入。有限分段核對非全部實數性質證明。"},"difficulty":{"band":"lower_secondary","level":3,"skills":["絕對值目標的根與代入"]},"variant":{"family_id":"qf-density14-absolute-preimage-rational-audit","seed":202610146095,"parameters":{"r":[18,2],"queries":[[-18,2],[18,2],[20,2]],"domain":"finite_rational_absolute_protocol","operation":"density14_absolute_preimage_rational_audit","semantic_template":"qf-density14-absolute-preimage-rational-audit/[\"9/1\",[\"9/1\",\"9/1\",\"10/1\"]]"}},"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":"2026-10-10","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261010-advance-20261014-r06-absolute-value-0096","batch_id":"batch-20261010-advance-20261014-r06-absolute-value-foundations","language":"zh-Hant","source_mko_ids":["mko-absolute-value"],"title_zh":"絕對值目標的根與代入","stem_zh":"有理嵌入的絕對值協議；r=19/2，queries=[-19/2, 19/2, 21/2]。依序答 (r非負資格, |x|=r遞增根列表, 三個abs(query), 三個等於r資格)。","answer":{"kind":"expression","value":"(是, [-19/2, 19/2], [19/2, 19/2, 21/2], [是, 是, 否])","explanation_zh":"絕對值目標資格根與代入=(是, [-19/2, 19/2], [19/2, 19/2, 21/2], [是, 是, 否])。負目標無根，零只有0，正目標有正負兩根；再逐一精確代入。有限分段核對非全部實數性質證明。"},"difficulty":{"band":"lower_secondary","level":3,"skills":["絕對值目標的根與代入"]},"variant":{"family_id":"qf-density14-absolute-preimage-rational-audit","seed":202610146096,"parameters":{"r":[19,2],"queries":[[-19,2],[19,2],[21,2]],"domain":"finite_rational_absolute_protocol","operation":"density14_absolute_preimage_rational_audit","semantic_template":"qf-density14-absolute-preimage-rational-audit/[\"19/2\",[\"19/2\",\"19/2\",\"21/2\"]]"}},"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":"2026-10-10","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261010-advance-20261014-r06-absolute-value-0097","batch_id":"batch-20261010-advance-20261014-r06-absolute-value-foundations","language":"zh-Hant","source_mko_ids":["mko-absolute-value"],"title_zh":"絕對值目標的根與代入","stem_zh":"有理嵌入的絕對值協議；r=20/2，queries=[-20/2, 20/2, 22/2]。依序答 (r非負資格, |x|=r遞增根列表, 三個abs(query), 三個等於r資格)。","answer":{"kind":"expression","value":"(是, [-10/1, 10/1], [10/1, 10/1, 11/1], [是, 是, 否])","explanation_zh":"絕對值目標資格根與代入=(是, [-10/1, 10/1], [10/1, 10/1, 11/1], [是, 是, 否])。負目標無根，零只有0，正目標有正負兩根；再逐一精確代入。有限分段核對非全部實數性質證明。"},"difficulty":{"band":"lower_secondary","level":3,"skills":["絕對值目標的根與代入"]},"variant":{"family_id":"qf-density14-absolute-preimage-rational-audit","seed":202610146097,"parameters":{"r":[20,2],"queries":[[-20,2],[20,2],[22,2]],"domain":"finite_rational_absolute_protocol","operation":"density14_absolute_preimage_rational_audit","semantic_template":"qf-density14-absolute-preimage-rational-audit/[\"10/1\",[\"10/1\",\"10/1\",\"11/1\"]]"}},"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":"2026-10-10","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261010-advance-20261014-r06-absolute-value-0098","batch_id":"batch-20261010-advance-20261014-r06-absolute-value-foundations","language":"zh-Hant","source_mko_ids":["mko-absolute-value"],"title_zh":"絕對值目標的根與代入","stem_zh":"有理嵌入的絕對值協議；r=21/2，queries=[-21/2, 21/2, 23/2]。依序答 (r非負資格, |x|=r遞增根列表, 三個abs(query), 三個等於r資格)。","answer":{"kind":"expression","value":"(是, [-21/2, 21/2], [21/2, 21/2, 23/2], [是, 是, 否])","explanation_zh":"絕對值目標資格根與代入=(是, [-21/2, 21/2], [21/2, 21/2, 23/2], [是, 是, 否])。負目標無根，零只有0，正目標有正負兩根；再逐一精確代入。有限分段核對非全部實數性質證明。"},"difficulty":{"band":"lower_secondary","level":3,"skills":["絕對值目標的根與代入"]},"variant":{"family_id":"qf-density14-absolute-preimage-rational-audit","seed":202610146098,"parameters":{"r":[21,2],"queries":[[-21,2],[21,2],[23,2]],"domain":"finite_rational_absolute_protocol","operation":"density14_absolute_preimage_rational_audit","semantic_template":"qf-density14-absolute-preimage-rational-audit/[\"21/2\",[\"21/2\",\"21/2\",\"23/2\"]]"}},"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":"2026-10-10","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261010-advance-20261014-r06-absolute-value-0099","batch_id":"batch-20261010-advance-20261014-r06-absolute-value-foundations","language":"zh-Hant","source_mko_ids":["mko-absolute-value"],"title_zh":"絕對值目標的根與代入","stem_zh":"有理嵌入的絕對值協議；r=22/2，queries=[-22/2, 22/2, 24/2]。依序答 (r非負資格, |x|=r遞增根列表, 三個abs(query), 三個等於r資格)。","answer":{"kind":"expression","value":"(是, [-11/1, 11/1], [11/1, 11/1, 12/1], [是, 是, 否])","explanation_zh":"絕對值目標資格根與代入=(是, [-11/1, 11/1], [11/1, 11/1, 12/1], [是, 是, 否])。負目標無根，零只有0，正目標有正負兩根；再逐一精確代入。有限分段核對非全部實數性質證明。"},"difficulty":{"band":"lower_secondary","level":3,"skills":["絕對值目標的根與代入"]},"variant":{"family_id":"qf-density14-absolute-preimage-rational-audit","seed":202610146099,"parameters":{"r":[22,2],"queries":[[-22,2],[22,2],[24,2]],"domain":"finite_rational_absolute_protocol","operation":"density14_absolute_preimage_rational_audit","semantic_template":"qf-density14-absolute-preimage-rational-audit/[\"11/1\",[\"11/1\",\"11/1\",\"12/1\"]]"}},"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":"2026-10-10","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261010-advance-20261014-r06-absolute-value-0100","batch_id":"batch-20261010-advance-20261014-r06-absolute-value-foundations","language":"zh-Hant","source_mko_ids":["mko-absolute-value"],"title_zh":"絕對值目標的根與代入","stem_zh":"有理嵌入的絕對值協議；r=23/2，queries=[-23/2, 23/2, 25/2]。依序答 (r非負資格, |x|=r遞增根列表, 三個abs(query), 三個等於r資格)。","answer":{"kind":"expression","value":"(是, [-23/2, 23/2], [23/2, 23/2, 25/2], [是, 是, 否])","explanation_zh":"絕對值目標資格根與代入=(是, [-23/2, 23/2], [23/2, 23/2, 25/2], [是, 是, 否])。負目標無根，零只有0，正目標有正負兩根；再逐一精確代入。有限分段核對非全部實數性質證明。"},"difficulty":{"band":"lower_secondary","level":3,"skills":["絕對值目標的根與代入"]},"variant":{"family_id":"qf-density14-absolute-preimage-rational-audit","seed":202610146100,"parameters":{"r":[23,2],"queries":[[-23,2],[23,2],[25,2]],"domain":"finite_rational_absolute_protocol","operation":"density14_absolute_preimage_rational_audit","semantic_template":"qf-density14-absolute-preimage-rational-audit/[\"23/2\",[\"23/2\",\"23/2\",\"25/2\"]]"}},"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":"2026-10-10","contract_version":"ocme-question-factory-v0.1"}}