{"schema_version":"ocme-question-v0.1","id":"q-20261012-advance-20261016-r01-vector-0001","batch_id":"batch-20261012-advance-20261016-r01-vector-foundations","language":"zh-Hant","source_mko_ids":["mko-vector"],"title_zh":"兩個加權向量與總和","stem_zh":"固定標準基底的整數二分量向量。u=[0, 0]、v=[0, 0]、a=0、b=0。依序求 (a*u, b*v, a*u+b*v)。","answer":{"kind":"expression","value":"([0, 0], [0, 0], [0, 0])","explanation_zh":"有限結構核對=([0, 0], [0, 0], [0, 0])。分別把純量乘到每個分量，再逐分量相加。有限精確核對不是一般形式證明。逐分量不當內積或長度。"},"difficulty":{"band":"upper_secondary","level":1,"skills":["兩個加權向量與總和"]},"variant":{"family_id":"qf-density16-vector-weighted-combination","seed":202610161001,"parameters":{"u":[0,0],"v":[0,0],"a":0,"b":0,"domain":"Z2_fixed_basis","operation":"vector-weighted-combination","semantic_template":"qf-density16-vector-weighted-combination/{\"a\":0,\"b\":0,\"u\":[0,0],\"v\":[0,0]}"}},"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":"2026-10-12","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261012-advance-20261016-r01-vector-0002","batch_id":"batch-20261012-advance-20261016-r01-vector-foundations","language":"zh-Hant","source_mko_ids":["mko-vector"],"title_zh":"兩個加權向量與總和","stem_zh":"固定標準基底的整數二分量向量。u=[99, 1]、v=[1, 2]、a=0、b=1。依序求 (a*u, b*v, a*u+b*v)。","answer":{"kind":"expression","value":"([0, 0], [1, 2], [1, 2])","explanation_zh":"有限結構核對=([0, 0], [1, 2], [1, 2])。分別把純量乘到每個分量，再逐分量相加。有限精確核對不是一般形式證明。逐分量不當內積或長度。"},"difficulty":{"band":"upper_secondary","level":1,"skills":["兩個加權向量與總和"]},"variant":{"family_id":"qf-density16-vector-weighted-combination","seed":202610161002,"parameters":{"u":[99,1],"v":[1,2],"a":0,"b":1,"domain":"Z2_fixed_basis","operation":"vector-weighted-combination","semantic_template":"qf-density16-vector-weighted-combination/{\"a\":0,\"b\":1,\"u\":[99,1],\"v\":[1,2]}"}},"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":"2026-10-12","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261012-advance-20261016-r01-vector-0003","batch_id":"batch-20261012-advance-20261016-r01-vector-foundations","language":"zh-Hant","source_mko_ids":["mko-vector"],"title_zh":"兩個加權向量與總和","stem_zh":"固定標準基底的整數二分量向量。u=[1, 2]、v=[1, 2]、a=2、b=-2。依序求 (a*u, b*v, a*u+b*v)。","answer":{"kind":"expression","value":"([2, 4], [-2, -4], [0, 0])","explanation_zh":"有限結構核對=([2, 4], [-2, -4], [0, 0])。分別把純量乘到每個分量，再逐分量相加。有限精確核對不是一般形式證明。逐分量不當內積或長度。"},"difficulty":{"band":"upper_secondary","level":1,"skills":["兩個加權向量與總和"]},"variant":{"family_id":"qf-density16-vector-weighted-combination","seed":202610161003,"parameters":{"u":[1,2],"v":[1,2],"a":2,"b":-2,"domain":"Z2_fixed_basis","operation":"vector-weighted-combination","semantic_template":"qf-density16-vector-weighted-combination/{\"a\":2,\"b\":-2,\"u\":[1,2],\"v\":[1,2]}"}},"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":"2026-10-12","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261012-advance-20261016-r01-vector-0004","batch_id":"batch-20261012-advance-20261016-r01-vector-foundations","language":"zh-Hant","source_mko_ids":["mko-vector"],"title_zh":"兩個加權向量與總和","stem_zh":"固定標準基底的整數二分量向量。u=[2, -3]、v=[1, 4]、a=9007199254740991、b=-9007199254740991。依序求 (a*u, b*v, a*u+b*v)。","answer":{"kind":"expression","value":"([18014398509481982, -27021597764222973], [-9007199254740991, -36028797018963964], [9007199254740991, -63050394783186937])","explanation_zh":"有限結構核對=([18014398509481982, -27021597764222973], [-9007199254740991, -36028797018963964], [9007199254740991, -63050394783186937])。分別把純量乘到每個分量，再逐分量相加。有限精確核對不是一般形式證明。逐分量不當內積或長度。"},"difficulty":{"band":"upper_secondary","level":1,"skills":["兩個加權向量與總和"]},"variant":{"family_id":"qf-density16-vector-weighted-combination","seed":202610161004,"parameters":{"u":[2,-3],"v":[1,4],"a":9007199254740991,"b":-9007199254740991,"domain":"Z2_fixed_basis","operation":"vector-weighted-combination","semantic_template":"qf-density16-vector-weighted-combination/{\"a\":9007199254740991,\"b\":-9007199254740991,\"u\":[2,-3],\"v\":[1,4]}"}},"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":"2026-10-12","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261012-advance-20261016-r01-vector-0005","batch_id":"batch-20261012-advance-20261016-r01-vector-foundations","language":"zh-Hant","source_mko_ids":["mko-vector"],"title_zh":"兩個加權向量與總和","stem_zh":"固定標準基底的整數二分量向量。u=[-8, 1]、v=[1, 2]、a=2、b=-1。依序求 (a*u, b*v, a*u+b*v)。","answer":{"kind":"expression","value":"([-16, 2], [-1, -2], [-17, 0])","explanation_zh":"有限結構核對=([-16, 2], [-1, -2], [-17, 0])。分別把純量乘到每個分量，再逐分量相加。有限精確核對不是一般形式證明。逐分量不當內積或長度。"},"difficulty":{"band":"upper_secondary","level":1,"skills":["兩個加權向量與總和"]},"variant":{"family_id":"qf-density16-vector-weighted-combination","seed":202610161005,"parameters":{"u":[-8,1],"v":[1,2],"a":2,"b":-1,"domain":"Z2_fixed_basis","operation":"vector-weighted-combination","semantic_template":"qf-density16-vector-weighted-combination/{\"a\":2,\"b\":-1,\"u\":[-8,1],\"v\":[1,2]}"}},"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":"2026-10-12","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261012-advance-20261016-r01-vector-0006","batch_id":"batch-20261012-advance-20261016-r01-vector-foundations","language":"zh-Hant","source_mko_ids":["mko-vector"],"title_zh":"兩個加權向量與總和","stem_zh":"固定標準基底的整數二分量向量。u=[-7, 1]、v=[1, 3]、a=2、b=-1。依序求 (a*u, b*v, a*u+b*v)。","answer":{"kind":"expression","value":"([-14, 2], [-1, -3], [-15, -1])","explanation_zh":"有限結構核對=([-14, 2], [-1, -3], [-15, -1])。分別把純量乘到每個分量，再逐分量相加。有限精確核對不是一般形式證明。逐分量不當內積或長度。"},"difficulty":{"band":"upper_secondary","level":1,"skills":["兩個加權向量與總和"]},"variant":{"family_id":"qf-density16-vector-weighted-combination","seed":202610161006,"parameters":{"u":[-7,1],"v":[1,3],"a":2,"b":-1,"domain":"Z2_fixed_basis","operation":"vector-weighted-combination","semantic_template":"qf-density16-vector-weighted-combination/{\"a\":2,\"b\":-1,\"u\":[-7,1],\"v\":[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":"2026-10-12","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261012-advance-20261016-r01-vector-0007","batch_id":"batch-20261012-advance-20261016-r01-vector-foundations","language":"zh-Hant","source_mko_ids":["mko-vector"],"title_zh":"兩個加權向量與總和","stem_zh":"固定標準基底的整數二分量向量。u=[-6, 1]、v=[1, 4]、a=2、b=-1。依序求 (a*u, b*v, a*u+b*v)。","answer":{"kind":"expression","value":"([-12, 2], [-1, -4], [-13, -2])","explanation_zh":"有限結構核對=([-12, 2], [-1, -4], [-13, -2])。分別把純量乘到每個分量，再逐分量相加。有限精確核對不是一般形式證明。逐分量不當內積或長度。"},"difficulty":{"band":"upper_secondary","level":1,"skills":["兩個加權向量與總和"]},"variant":{"family_id":"qf-density16-vector-weighted-combination","seed":202610161007,"parameters":{"u":[-6,1],"v":[1,4],"a":2,"b":-1,"domain":"Z2_fixed_basis","operation":"vector-weighted-combination","semantic_template":"qf-density16-vector-weighted-combination/{\"a\":2,\"b\":-1,\"u\":[-6,1],\"v\":[1,4]}"}},"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":"2026-10-12","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261012-advance-20261016-r01-vector-0008","batch_id":"batch-20261012-advance-20261016-r01-vector-foundations","language":"zh-Hant","source_mko_ids":["mko-vector"],"title_zh":"兩個加權向量與總和","stem_zh":"固定標準基底的整數二分量向量。u=[-5, 1]、v=[1, 5]、a=2、b=-1。依序求 (a*u, b*v, a*u+b*v)。","answer":{"kind":"expression","value":"([-10, 2], [-1, -5], [-11, -3])","explanation_zh":"有限結構核對=([-10, 2], [-1, -5], [-11, -3])。分別把純量乘到每個分量，再逐分量相加。有限精確核對不是一般形式證明。逐分量不當內積或長度。"},"difficulty":{"band":"upper_secondary","level":1,"skills":["兩個加權向量與總和"]},"variant":{"family_id":"qf-density16-vector-weighted-combination","seed":202610161008,"parameters":{"u":[-5,1],"v":[1,5],"a":2,"b":-1,"domain":"Z2_fixed_basis","operation":"vector-weighted-combination","semantic_template":"qf-density16-vector-weighted-combination/{\"a\":2,\"b\":-1,\"u\":[-5,1],\"v\":[1,5]}"}},"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":"2026-10-12","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261012-advance-20261016-r01-vector-0009","batch_id":"batch-20261012-advance-20261016-r01-vector-foundations","language":"zh-Hant","source_mko_ids":["mko-vector"],"title_zh":"兩個加權向量與總和","stem_zh":"固定標準基底的整數二分量向量。u=[-4, 1]、v=[1, 6]、a=2、b=-1。依序求 (a*u, b*v, a*u+b*v)。","answer":{"kind":"expression","value":"([-8, 2], [-1, -6], [-9, -4])","explanation_zh":"有限結構核對=([-8, 2], [-1, -6], [-9, -4])。分別把純量乘到每個分量，再逐分量相加。有限精確核對不是一般形式證明。逐分量不當內積或長度。"},"difficulty":{"band":"upper_secondary","level":2,"skills":["兩個加權向量與總和"]},"variant":{"family_id":"qf-density16-vector-weighted-combination","seed":202610161009,"parameters":{"u":[-4,1],"v":[1,6],"a":2,"b":-1,"domain":"Z2_fixed_basis","operation":"vector-weighted-combination","semantic_template":"qf-density16-vector-weighted-combination/{\"a\":2,\"b\":-1,\"u\":[-4,1],\"v\":[1,6]}"}},"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":"2026-10-12","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261012-advance-20261016-r01-vector-0010","batch_id":"batch-20261012-advance-20261016-r01-vector-foundations","language":"zh-Hant","source_mko_ids":["mko-vector"],"title_zh":"兩個加權向量與總和","stem_zh":"固定標準基底的整數二分量向量。u=[-3, 1]、v=[1, 7]、a=2、b=-1。依序求 (a*u, b*v, a*u+b*v)。","answer":{"kind":"expression","value":"([-6, 2], [-1, -7], [-7, -5])","explanation_zh":"有限結構核對=([-6, 2], [-1, -7], [-7, -5])。分別把純量乘到每個分量，再逐分量相加。有限精確核對不是一般形式證明。逐分量不當內積或長度。"},"difficulty":{"band":"upper_secondary","level":2,"skills":["兩個加權向量與總和"]},"variant":{"family_id":"qf-density16-vector-weighted-combination","seed":202610161010,"parameters":{"u":[-3,1],"v":[1,7],"a":2,"b":-1,"domain":"Z2_fixed_basis","operation":"vector-weighted-combination","semantic_template":"qf-density16-vector-weighted-combination/{\"a\":2,\"b\":-1,\"u\":[-3,1],\"v\":[1,7]}"}},"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":"2026-10-12","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261012-advance-20261016-r01-vector-0011","batch_id":"batch-20261012-advance-20261016-r01-vector-foundations","language":"zh-Hant","source_mko_ids":["mko-vector"],"title_zh":"兩個加權向量與總和","stem_zh":"固定標準基底的整數二分量向量。u=[-2, 1]、v=[1, 8]、a=2、b=-1。依序求 (a*u, b*v, a*u+b*v)。","answer":{"kind":"expression","value":"([-4, 2], [-1, -8], [-5, -6])","explanation_zh":"有限結構核對=([-4, 2], [-1, -8], [-5, -6])。分別把純量乘到每個分量，再逐分量相加。有限精確核對不是一般形式證明。逐分量不當內積或長度。"},"difficulty":{"band":"upper_secondary","level":2,"skills":["兩個加權向量與總和"]},"variant":{"family_id":"qf-density16-vector-weighted-combination","seed":202610161011,"parameters":{"u":[-2,1],"v":[1,8],"a":2,"b":-1,"domain":"Z2_fixed_basis","operation":"vector-weighted-combination","semantic_template":"qf-density16-vector-weighted-combination/{\"a\":2,\"b\":-1,\"u\":[-2,1],\"v\":[1,8]}"}},"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":"2026-10-12","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261012-advance-20261016-r01-vector-0012","batch_id":"batch-20261012-advance-20261016-r01-vector-foundations","language":"zh-Hant","source_mko_ids":["mko-vector"],"title_zh":"兩個加權向量與總和","stem_zh":"固定標準基底的整數二分量向量。u=[-1, 1]、v=[1, 9]、a=2、b=-1。依序求 (a*u, b*v, a*u+b*v)。","answer":{"kind":"expression","value":"([-2, 2], [-1, -9], [-3, -7])","explanation_zh":"有限結構核對=([-2, 2], [-1, -9], [-3, -7])。分別把純量乘到每個分量，再逐分量相加。有限精確核對不是一般形式證明。逐分量不當內積或長度。"},"difficulty":{"band":"upper_secondary","level":2,"skills":["兩個加權向量與總和"]},"variant":{"family_id":"qf-density16-vector-weighted-combination","seed":202610161012,"parameters":{"u":[-1,1],"v":[1,9],"a":2,"b":-1,"domain":"Z2_fixed_basis","operation":"vector-weighted-combination","semantic_template":"qf-density16-vector-weighted-combination/{\"a\":2,\"b\":-1,\"u\":[-1,1],\"v\":[1,9]}"}},"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":"2026-10-12","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261012-advance-20261016-r01-vector-0013","batch_id":"batch-20261012-advance-20261016-r01-vector-foundations","language":"zh-Hant","source_mko_ids":["mko-vector"],"title_zh":"兩個加權向量與總和","stem_zh":"固定標準基底的整數二分量向量。u=[0, 1]、v=[1, 10]、a=2、b=-1。依序求 (a*u, b*v, a*u+b*v)。","answer":{"kind":"expression","value":"([0, 2], [-1, -10], [-1, -8])","explanation_zh":"有限結構核對=([0, 2], [-1, -10], [-1, -8])。分別把純量乘到每個分量，再逐分量相加。有限精確核對不是一般形式證明。逐分量不當內積或長度。"},"difficulty":{"band":"upper_secondary","level":2,"skills":["兩個加權向量與總和"]},"variant":{"family_id":"qf-density16-vector-weighted-combination","seed":202610161013,"parameters":{"u":[0,1],"v":[1,10],"a":2,"b":-1,"domain":"Z2_fixed_basis","operation":"vector-weighted-combination","semantic_template":"qf-density16-vector-weighted-combination/{\"a\":2,\"b\":-1,\"u\":[0,1],\"v\":[1,10]}"}},"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":"2026-10-12","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261012-advance-20261016-r01-vector-0014","batch_id":"batch-20261012-advance-20261016-r01-vector-foundations","language":"zh-Hant","source_mko_ids":["mko-vector"],"title_zh":"兩個加權向量與總和","stem_zh":"固定標準基底的整數二分量向量。u=[1, 1]、v=[1, 11]、a=2、b=-1。依序求 (a*u, b*v, a*u+b*v)。","answer":{"kind":"expression","value":"([2, 2], [-1, -11], [1, -9])","explanation_zh":"有限結構核對=([2, 2], [-1, -11], [1, -9])。分別把純量乘到每個分量，再逐分量相加。有限精確核對不是一般形式證明。逐分量不當內積或長度。"},"difficulty":{"band":"upper_secondary","level":2,"skills":["兩個加權向量與總和"]},"variant":{"family_id":"qf-density16-vector-weighted-combination","seed":202610161014,"parameters":{"u":[1,1],"v":[1,11],"a":2,"b":-1,"domain":"Z2_fixed_basis","operation":"vector-weighted-combination","semantic_template":"qf-density16-vector-weighted-combination/{\"a\":2,\"b\":-1,\"u\":[1,1],\"v\":[1,11]}"}},"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":"2026-10-12","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261012-advance-20261016-r01-vector-0015","batch_id":"batch-20261012-advance-20261016-r01-vector-foundations","language":"zh-Hant","source_mko_ids":["mko-vector"],"title_zh":"兩個加權向量與總和","stem_zh":"固定標準基底的整數二分量向量。u=[2, 1]、v=[1, 12]、a=2、b=-1。依序求 (a*u, b*v, a*u+b*v)。","answer":{"kind":"expression","value":"([4, 2], [-1, -12], [3, -10])","explanation_zh":"有限結構核對=([4, 2], [-1, -12], [3, -10])。分別把純量乘到每個分量，再逐分量相加。有限精確核對不是一般形式證明。逐分量不當內積或長度。"},"difficulty":{"band":"upper_secondary","level":2,"skills":["兩個加權向量與總和"]},"variant":{"family_id":"qf-density16-vector-weighted-combination","seed":202610161015,"parameters":{"u":[2,1],"v":[1,12],"a":2,"b":-1,"domain":"Z2_fixed_basis","operation":"vector-weighted-combination","semantic_template":"qf-density16-vector-weighted-combination/{\"a\":2,\"b\":-1,\"u\":[2,1],\"v\":[1,12]}"}},"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":"2026-10-12","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261012-advance-20261016-r01-vector-0016","batch_id":"batch-20261012-advance-20261016-r01-vector-foundations","language":"zh-Hant","source_mko_ids":["mko-vector"],"title_zh":"兩個加權向量與總和","stem_zh":"固定標準基底的整數二分量向量。u=[3, 1]、v=[1, 13]、a=2、b=-1。依序求 (a*u, b*v, a*u+b*v)。","answer":{"kind":"expression","value":"([6, 2], [-1, -13], [5, -11])","explanation_zh":"有限結構核對=([6, 2], [-1, -13], [5, -11])。分別把純量乘到每個分量，再逐分量相加。有限精確核對不是一般形式證明。逐分量不當內積或長度。"},"difficulty":{"band":"upper_secondary","level":2,"skills":["兩個加權向量與總和"]},"variant":{"family_id":"qf-density16-vector-weighted-combination","seed":202610161016,"parameters":{"u":[3,1],"v":[1,13],"a":2,"b":-1,"domain":"Z2_fixed_basis","operation":"vector-weighted-combination","semantic_template":"qf-density16-vector-weighted-combination/{\"a\":2,\"b\":-1,\"u\":[3,1],\"v\":[1,13]}"}},"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":"2026-10-12","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261012-advance-20261016-r01-vector-0017","batch_id":"batch-20261012-advance-20261016-r01-vector-foundations","language":"zh-Hant","source_mko_ids":["mko-vector"],"title_zh":"兩個加權向量與總和","stem_zh":"固定標準基底的整數二分量向量。u=[4, 1]、v=[1, 14]、a=2、b=-1。依序求 (a*u, b*v, a*u+b*v)。","answer":{"kind":"expression","value":"([8, 2], [-1, -14], [7, -12])","explanation_zh":"有限結構核對=([8, 2], [-1, -14], [7, -12])。分別把純量乘到每個分量，再逐分量相加。有限精確核對不是一般形式證明。逐分量不當內積或長度。"},"difficulty":{"band":"upper_secondary","level":2,"skills":["兩個加權向量與總和"]},"variant":{"family_id":"qf-density16-vector-weighted-combination","seed":202610161017,"parameters":{"u":[4,1],"v":[1,14],"a":2,"b":-1,"domain":"Z2_fixed_basis","operation":"vector-weighted-combination","semantic_template":"qf-density16-vector-weighted-combination/{\"a\":2,\"b\":-1,\"u\":[4,1],\"v\":[1,14]}"}},"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":"2026-10-12","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261012-advance-20261016-r01-vector-0018","batch_id":"batch-20261012-advance-20261016-r01-vector-foundations","language":"zh-Hant","source_mko_ids":["mko-vector"],"title_zh":"兩個加權向量與總和","stem_zh":"固定標準基底的整數二分量向量。u=[5, 1]、v=[1, 15]、a=2、b=-1。依序求 (a*u, b*v, a*u+b*v)。","answer":{"kind":"expression","value":"([10, 2], [-1, -15], [9, -13])","explanation_zh":"有限結構核對=([10, 2], [-1, -15], [9, -13])。分別把純量乘到每個分量，再逐分量相加。有限精確核對不是一般形式證明。逐分量不當內積或長度。"},"difficulty":{"band":"upper_secondary","level":3,"skills":["兩個加權向量與總和"]},"variant":{"family_id":"qf-density16-vector-weighted-combination","seed":202610161018,"parameters":{"u":[5,1],"v":[1,15],"a":2,"b":-1,"domain":"Z2_fixed_basis","operation":"vector-weighted-combination","semantic_template":"qf-density16-vector-weighted-combination/{\"a\":2,\"b\":-1,\"u\":[5,1],\"v\":[1,15]}"}},"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":"2026-10-12","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261012-advance-20261016-r01-vector-0019","batch_id":"batch-20261012-advance-20261016-r01-vector-foundations","language":"zh-Hant","source_mko_ids":["mko-vector"],"title_zh":"兩個加權向量與總和","stem_zh":"固定標準基底的整數二分量向量。u=[6, 1]、v=[1, 16]、a=2、b=-1。依序求 (a*u, b*v, a*u+b*v)。","answer":{"kind":"expression","value":"([12, 2], [-1, -16], [11, -14])","explanation_zh":"有限結構核對=([12, 2], [-1, -16], [11, -14])。分別把純量乘到每個分量，再逐分量相加。有限精確核對不是一般形式證明。逐分量不當內積或長度。"},"difficulty":{"band":"upper_secondary","level":3,"skills":["兩個加權向量與總和"]},"variant":{"family_id":"qf-density16-vector-weighted-combination","seed":202610161019,"parameters":{"u":[6,1],"v":[1,16],"a":2,"b":-1,"domain":"Z2_fixed_basis","operation":"vector-weighted-combination","semantic_template":"qf-density16-vector-weighted-combination/{\"a\":2,\"b\":-1,\"u\":[6,1],\"v\":[1,16]}"}},"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":"2026-10-12","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261012-advance-20261016-r01-vector-0020","batch_id":"batch-20261012-advance-20261016-r01-vector-foundations","language":"zh-Hant","source_mko_ids":["mko-vector"],"title_zh":"兩個加權向量與總和","stem_zh":"固定標準基底的整數二分量向量。u=[7, 1]、v=[1, 17]、a=2、b=-1。依序求 (a*u, b*v, a*u+b*v)。","answer":{"kind":"expression","value":"([14, 2], [-1, -17], [13, -15])","explanation_zh":"有限結構核對=([14, 2], [-1, -17], [13, -15])。分別把純量乘到每個分量，再逐分量相加。有限精確核對不是一般形式證明。逐分量不當內積或長度。"},"difficulty":{"band":"upper_secondary","level":3,"skills":["兩個加權向量與總和"]},"variant":{"family_id":"qf-density16-vector-weighted-combination","seed":202610161020,"parameters":{"u":[7,1],"v":[1,17],"a":2,"b":-1,"domain":"Z2_fixed_basis","operation":"vector-weighted-combination","semantic_template":"qf-density16-vector-weighted-combination/{\"a\":2,\"b\":-1,\"u\":[7,1],\"v\":[1,17]}"}},"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":"2026-10-12","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261012-advance-20261016-r01-vector-0021","batch_id":"batch-20261012-advance-20261016-r01-vector-foundations","language":"zh-Hant","source_mko_ids":["mko-vector"],"title_zh":"兩個加權向量與總和","stem_zh":"固定標準基底的整數二分量向量。u=[8, 1]、v=[1, 18]、a=2、b=-1。依序求 (a*u, b*v, a*u+b*v)。","answer":{"kind":"expression","value":"([16, 2], [-1, -18], [15, -16])","explanation_zh":"有限結構核對=([16, 2], [-1, -18], [15, -16])。分別把純量乘到每個分量，再逐分量相加。有限精確核對不是一般形式證明。逐分量不當內積或長度。"},"difficulty":{"band":"upper_secondary","level":3,"skills":["兩個加權向量與總和"]},"variant":{"family_id":"qf-density16-vector-weighted-combination","seed":202610161021,"parameters":{"u":[8,1],"v":[1,18],"a":2,"b":-1,"domain":"Z2_fixed_basis","operation":"vector-weighted-combination","semantic_template":"qf-density16-vector-weighted-combination/{\"a\":2,\"b\":-1,\"u\":[8,1],\"v\":[1,18]}"}},"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":"2026-10-12","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261012-advance-20261016-r01-vector-0022","batch_id":"batch-20261012-advance-20261016-r01-vector-foundations","language":"zh-Hant","source_mko_ids":["mko-vector"],"title_zh":"兩個加權向量與總和","stem_zh":"固定標準基底的整數二分量向量。u=[9, 1]、v=[1, 19]、a=2、b=-1。依序求 (a*u, b*v, a*u+b*v)。","answer":{"kind":"expression","value":"([18, 2], [-1, -19], [17, -17])","explanation_zh":"有限結構核對=([18, 2], [-1, -19], [17, -17])。分別把純量乘到每個分量，再逐分量相加。有限精確核對不是一般形式證明。逐分量不當內積或長度。"},"difficulty":{"band":"upper_secondary","level":3,"skills":["兩個加權向量與總和"]},"variant":{"family_id":"qf-density16-vector-weighted-combination","seed":202610161022,"parameters":{"u":[9,1],"v":[1,19],"a":2,"b":-1,"domain":"Z2_fixed_basis","operation":"vector-weighted-combination","semantic_template":"qf-density16-vector-weighted-combination/{\"a\":2,\"b\":-1,\"u\":[9,1],\"v\":[1,19]}"}},"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":"2026-10-12","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261012-advance-20261016-r01-vector-0023","batch_id":"batch-20261012-advance-20261016-r01-vector-foundations","language":"zh-Hant","source_mko_ids":["mko-vector"],"title_zh":"兩個加權向量與總和","stem_zh":"固定標準基底的整數二分量向量。u=[10, 1]、v=[1, 20]、a=2、b=-1。依序求 (a*u, b*v, a*u+b*v)。","answer":{"kind":"expression","value":"([20, 2], [-1, -20], [19, -18])","explanation_zh":"有限結構核對=([20, 2], [-1, -20], [19, -18])。分別把純量乘到每個分量，再逐分量相加。有限精確核對不是一般形式證明。逐分量不當內積或長度。"},"difficulty":{"band":"upper_secondary","level":3,"skills":["兩個加權向量與總和"]},"variant":{"family_id":"qf-density16-vector-weighted-combination","seed":202610161023,"parameters":{"u":[10,1],"v":[1,20],"a":2,"b":-1,"domain":"Z2_fixed_basis","operation":"vector-weighted-combination","semantic_template":"qf-density16-vector-weighted-combination/{\"a\":2,\"b\":-1,\"u\":[10,1],\"v\":[1,20]}"}},"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":"2026-10-12","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261012-advance-20261016-r01-vector-0024","batch_id":"batch-20261012-advance-20261016-r01-vector-foundations","language":"zh-Hant","source_mko_ids":["mko-vector"],"title_zh":"兩個加權向量與總和","stem_zh":"固定標準基底的整數二分量向量。u=[11, 1]、v=[1, 21]、a=2、b=-1。依序求 (a*u, b*v, a*u+b*v)。","answer":{"kind":"expression","value":"([22, 2], [-1, -21], [21, -19])","explanation_zh":"有限結構核對=([22, 2], [-1, -21], [21, -19])。分別把純量乘到每個分量，再逐分量相加。有限精確核對不是一般形式證明。逐分量不當內積或長度。"},"difficulty":{"band":"upper_secondary","level":3,"skills":["兩個加權向量與總和"]},"variant":{"family_id":"qf-density16-vector-weighted-combination","seed":202610161024,"parameters":{"u":[11,1],"v":[1,21],"a":2,"b":-1,"domain":"Z2_fixed_basis","operation":"vector-weighted-combination","semantic_template":"qf-density16-vector-weighted-combination/{\"a\":2,\"b\":-1,\"u\":[11,1],\"v\":[1,21]}"}},"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":"2026-10-12","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261012-advance-20261016-r01-vector-0025","batch_id":"batch-20261012-advance-20261016-r01-vector-foundations","language":"zh-Hant","source_mko_ids":["mko-vector"],"title_zh":"兩個加權向量與總和","stem_zh":"固定標準基底的整數二分量向量。u=[12, 1]、v=[1, 22]、a=2、b=-1。依序求 (a*u, b*v, a*u+b*v)。","answer":{"kind":"expression","value":"([24, 2], [-1, -22], [23, -20])","explanation_zh":"有限結構核對=([24, 2], [-1, -22], [23, -20])。分別把純量乘到每個分量，再逐分量相加。有限精確核對不是一般形式證明。逐分量不當內積或長度。"},"difficulty":{"band":"upper_secondary","level":3,"skills":["兩個加權向量與總和"]},"variant":{"family_id":"qf-density16-vector-weighted-combination","seed":202610161025,"parameters":{"u":[12,1],"v":[1,22],"a":2,"b":-1,"domain":"Z2_fixed_basis","operation":"vector-weighted-combination","semantic_template":"qf-density16-vector-weighted-combination/{\"a\":2,\"b\":-1,\"u\":[12,1],\"v\":[1,22]}"}},"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":"2026-10-12","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261012-advance-20261016-r01-vector-0026","batch_id":"batch-20261012-advance-20261016-r01-vector-foundations","language":"zh-Hant","source_mko_ids":["mko-vector"],"title_zh":"向量差的兩個順序","stem_zh":"固定標準基底的整數二分量向量。u=[0, 0]、v=[0, 0]。依序答 (u, v, u−v, v−u, 兩個差的和)。","answer":{"kind":"expression","value":"([0, 0], [0, 0], [0, 0], [0, 0], [0, 0])","explanation_zh":"有限結構核對=([0, 0], [0, 0], [0, 0], [0, 0], [0, 0])。保留有序二分量，兩種減法各算一次再相加。有限精確核對不是一般形式證明。逐分量不當內積或長度。"},"difficulty":{"band":"upper_secondary","level":1,"skills":["向量差的兩個順序"]},"variant":{"family_id":"qf-density16-vector-difference-reversal","seed":202610161026,"parameters":{"u":[0,0],"v":[0,0],"domain":"Z2_fixed_basis","operation":"vector-difference-reversal","semantic_template":"qf-density16-vector-difference-reversal/{\"u\":[0,0],\"v\":[0,0]}"}},"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":"2026-10-12","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261012-advance-20261016-r01-vector-0027","batch_id":"batch-20261012-advance-20261016-r01-vector-foundations","language":"zh-Hant","source_mko_ids":["mko-vector"],"title_zh":"向量差的兩個順序","stem_zh":"固定標準基底的整數二分量向量。u=[1, -1]、v=[1, -1]。依序答 (u, v, u−v, v−u, 兩個差的和)。","answer":{"kind":"expression","value":"([1, -1], [1, -1], [0, 0], [0, 0], [0, 0])","explanation_zh":"有限結構核對=([1, -1], [1, -1], [0, 0], [0, 0], [0, 0])。保留有序二分量，兩種減法各算一次再相加。有限精確核對不是一般形式證明。逐分量不當內積或長度。"},"difficulty":{"band":"upper_secondary","level":1,"skills":["向量差的兩個順序"]},"variant":{"family_id":"qf-density16-vector-difference-reversal","seed":202610161027,"parameters":{"u":[1,-1],"v":[1,-1],"domain":"Z2_fixed_basis","operation":"vector-difference-reversal","semantic_template":"qf-density16-vector-difference-reversal/{\"u\":[1,-1],\"v\":[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":"2026-10-12","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261012-advance-20261016-r01-vector-0028","batch_id":"batch-20261012-advance-20261016-r01-vector-foundations","language":"zh-Hant","source_mko_ids":["mko-vector"],"title_zh":"向量差的兩個順序","stem_zh":"固定標準基底的整數二分量向量。u=[-10, 1]、v=[0, -1]。依序答 (u, v, u−v, v−u, 兩個差的和)。","answer":{"kind":"expression","value":"([-10, 1], [0, -1], [-10, 2], [10, -2], [0, 0])","explanation_zh":"有限結構核對=([-10, 1], [0, -1], [-10, 2], [10, -2], [0, 0])。保留有序二分量，兩種減法各算一次再相加。有限精確核對不是一般形式證明。逐分量不當內積或長度。"},"difficulty":{"band":"upper_secondary","level":1,"skills":["向量差的兩個順序"]},"variant":{"family_id":"qf-density16-vector-difference-reversal","seed":202610161028,"parameters":{"u":[-10,1],"v":[0,-1],"domain":"Z2_fixed_basis","operation":"vector-difference-reversal","semantic_template":"qf-density16-vector-difference-reversal/{\"u\":[-10,1],\"v\":[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":"2026-10-12","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261012-advance-20261016-r01-vector-0029","batch_id":"batch-20261012-advance-20261016-r01-vector-foundations","language":"zh-Hant","source_mko_ids":["mko-vector"],"title_zh":"向量差的兩個順序","stem_zh":"固定標準基底的整數二分量向量。u=[-9, 1]、v=[0, -1]。依序答 (u, v, u−v, v−u, 兩個差的和)。","answer":{"kind":"expression","value":"([-9, 1], [0, -1], [-9, 2], [9, -2], [0, 0])","explanation_zh":"有限結構核對=([-9, 1], [0, -1], [-9, 2], [9, -2], [0, 0])。保留有序二分量，兩種減法各算一次再相加。有限精確核對不是一般形式證明。逐分量不當內積或長度。"},"difficulty":{"band":"upper_secondary","level":1,"skills":["向量差的兩個順序"]},"variant":{"family_id":"qf-density16-vector-difference-reversal","seed":202610161029,"parameters":{"u":[-9,1],"v":[0,-1],"domain":"Z2_fixed_basis","operation":"vector-difference-reversal","semantic_template":"qf-density16-vector-difference-reversal/{\"u\":[-9,1],\"v\":[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":"2026-10-12","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261012-advance-20261016-r01-vector-0030","batch_id":"batch-20261012-advance-20261016-r01-vector-foundations","language":"zh-Hant","source_mko_ids":["mko-vector"],"title_zh":"向量差的兩個順序","stem_zh":"固定標準基底的整數二分量向量。u=[-8, 1]、v=[0, -1]。依序答 (u, v, u−v, v−u, 兩個差的和)。","answer":{"kind":"expression","value":"([-8, 1], [0, -1], [-8, 2], [8, -2], [0, 0])","explanation_zh":"有限結構核對=([-8, 1], [0, -1], [-8, 2], [8, -2], [0, 0])。保留有序二分量，兩種減法各算一次再相加。有限精確核對不是一般形式證明。逐分量不當內積或長度。"},"difficulty":{"band":"upper_secondary","level":1,"skills":["向量差的兩個順序"]},"variant":{"family_id":"qf-density16-vector-difference-reversal","seed":202610161030,"parameters":{"u":[-8,1],"v":[0,-1],"domain":"Z2_fixed_basis","operation":"vector-difference-reversal","semantic_template":"qf-density16-vector-difference-reversal/{\"u\":[-8,1],\"v\":[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":"2026-10-12","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261012-advance-20261016-r01-vector-0031","batch_id":"batch-20261012-advance-20261016-r01-vector-foundations","language":"zh-Hant","source_mko_ids":["mko-vector"],"title_zh":"向量差的兩個順序","stem_zh":"固定標準基底的整數二分量向量。u=[-7, 1]、v=[0, -1]。依序答 (u, v, u−v, v−u, 兩個差的和)。","answer":{"kind":"expression","value":"([-7, 1], [0, -1], [-7, 2], [7, -2], [0, 0])","explanation_zh":"有限結構核對=([-7, 1], [0, -1], [-7, 2], [7, -2], [0, 0])。保留有序二分量，兩種減法各算一次再相加。有限精確核對不是一般形式證明。逐分量不當內積或長度。"},"difficulty":{"band":"upper_secondary","level":1,"skills":["向量差的兩個順序"]},"variant":{"family_id":"qf-density16-vector-difference-reversal","seed":202610161031,"parameters":{"u":[-7,1],"v":[0,-1],"domain":"Z2_fixed_basis","operation":"vector-difference-reversal","semantic_template":"qf-density16-vector-difference-reversal/{\"u\":[-7,1],\"v\":[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":"2026-10-12","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261012-advance-20261016-r01-vector-0032","batch_id":"batch-20261012-advance-20261016-r01-vector-foundations","language":"zh-Hant","source_mko_ids":["mko-vector"],"title_zh":"向量差的兩個順序","stem_zh":"固定標準基底的整數二分量向量。u=[-6, 1]、v=[0, -1]。依序答 (u, v, u−v, v−u, 兩個差的和)。","answer":{"kind":"expression","value":"([-6, 1], [0, -1], [-6, 2], [6, -2], [0, 0])","explanation_zh":"有限結構核對=([-6, 1], [0, -1], [-6, 2], [6, -2], [0, 0])。保留有序二分量，兩種減法各算一次再相加。有限精確核對不是一般形式證明。逐分量不當內積或長度。"},"difficulty":{"band":"upper_secondary","level":1,"skills":["向量差的兩個順序"]},"variant":{"family_id":"qf-density16-vector-difference-reversal","seed":202610161032,"parameters":{"u":[-6,1],"v":[0,-1],"domain":"Z2_fixed_basis","operation":"vector-difference-reversal","semantic_template":"qf-density16-vector-difference-reversal/{\"u\":[-6,1],\"v\":[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":"2026-10-12","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261012-advance-20261016-r01-vector-0033","batch_id":"batch-20261012-advance-20261016-r01-vector-foundations","language":"zh-Hant","source_mko_ids":["mko-vector"],"title_zh":"向量差的兩個順序","stem_zh":"固定標準基底的整數二分量向量。u=[-5, 1]、v=[0, -1]。依序答 (u, v, u−v, v−u, 兩個差的和)。","answer":{"kind":"expression","value":"([-5, 1], [0, -1], [-5, 2], [5, -2], [0, 0])","explanation_zh":"有限結構核對=([-5, 1], [0, -1], [-5, 2], [5, -2], [0, 0])。保留有序二分量，兩種減法各算一次再相加。有限精確核對不是一般形式證明。逐分量不當內積或長度。"},"difficulty":{"band":"upper_secondary","level":1,"skills":["向量差的兩個順序"]},"variant":{"family_id":"qf-density16-vector-difference-reversal","seed":202610161033,"parameters":{"u":[-5,1],"v":[0,-1],"domain":"Z2_fixed_basis","operation":"vector-difference-reversal","semantic_template":"qf-density16-vector-difference-reversal/{\"u\":[-5,1],\"v\":[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":"2026-10-12","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261012-advance-20261016-r01-vector-0034","batch_id":"batch-20261012-advance-20261016-r01-vector-foundations","language":"zh-Hant","source_mko_ids":["mko-vector"],"title_zh":"向量差的兩個順序","stem_zh":"固定標準基底的整數二分量向量。u=[-4, 1]、v=[0, -1]。依序答 (u, v, u−v, v−u, 兩個差的和)。","answer":{"kind":"expression","value":"([-4, 1], [0, -1], [-4, 2], [4, -2], [0, 0])","explanation_zh":"有限結構核對=([-4, 1], [0, -1], [-4, 2], [4, -2], [0, 0])。保留有序二分量，兩種減法各算一次再相加。有限精確核對不是一般形式證明。逐分量不當內積或長度。"},"difficulty":{"band":"upper_secondary","level":2,"skills":["向量差的兩個順序"]},"variant":{"family_id":"qf-density16-vector-difference-reversal","seed":202610161034,"parameters":{"u":[-4,1],"v":[0,-1],"domain":"Z2_fixed_basis","operation":"vector-difference-reversal","semantic_template":"qf-density16-vector-difference-reversal/{\"u\":[-4,1],\"v\":[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":"2026-10-12","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261012-advance-20261016-r01-vector-0035","batch_id":"batch-20261012-advance-20261016-r01-vector-foundations","language":"zh-Hant","source_mko_ids":["mko-vector"],"title_zh":"向量差的兩個順序","stem_zh":"固定標準基底的整數二分量向量。u=[-3, 1]、v=[0, -1]。依序答 (u, v, u−v, v−u, 兩個差的和)。","answer":{"kind":"expression","value":"([-3, 1], [0, -1], [-3, 2], [3, -2], [0, 0])","explanation_zh":"有限結構核對=([-3, 1], [0, -1], [-3, 2], [3, -2], [0, 0])。保留有序二分量，兩種減法各算一次再相加。有限精確核對不是一般形式證明。逐分量不當內積或長度。"},"difficulty":{"band":"upper_secondary","level":2,"skills":["向量差的兩個順序"]},"variant":{"family_id":"qf-density16-vector-difference-reversal","seed":202610161035,"parameters":{"u":[-3,1],"v":[0,-1],"domain":"Z2_fixed_basis","operation":"vector-difference-reversal","semantic_template":"qf-density16-vector-difference-reversal/{\"u\":[-3,1],\"v\":[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":"2026-10-12","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261012-advance-20261016-r01-vector-0036","batch_id":"batch-20261012-advance-20261016-r01-vector-foundations","language":"zh-Hant","source_mko_ids":["mko-vector"],"title_zh":"向量差的兩個順序","stem_zh":"固定標準基底的整數二分量向量。u=[-2, 1]、v=[0, -1]。依序答 (u, v, u−v, v−u, 兩個差的和)。","answer":{"kind":"expression","value":"([-2, 1], [0, -1], [-2, 2], [2, -2], [0, 0])","explanation_zh":"有限結構核對=([-2, 1], [0, -1], [-2, 2], [2, -2], [0, 0])。保留有序二分量，兩種減法各算一次再相加。有限精確核對不是一般形式證明。逐分量不當內積或長度。"},"difficulty":{"band":"upper_secondary","level":2,"skills":["向量差的兩個順序"]},"variant":{"family_id":"qf-density16-vector-difference-reversal","seed":202610161036,"parameters":{"u":[-2,1],"v":[0,-1],"domain":"Z2_fixed_basis","operation":"vector-difference-reversal","semantic_template":"qf-density16-vector-difference-reversal/{\"u\":[-2,1],\"v\":[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":"2026-10-12","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261012-advance-20261016-r01-vector-0037","batch_id":"batch-20261012-advance-20261016-r01-vector-foundations","language":"zh-Hant","source_mko_ids":["mko-vector"],"title_zh":"向量差的兩個順序","stem_zh":"固定標準基底的整數二分量向量。u=[-1, 1]、v=[0, -1]。依序答 (u, v, u−v, v−u, 兩個差的和)。","answer":{"kind":"expression","value":"([-1, 1], [0, -1], [-1, 2], [1, -2], [0, 0])","explanation_zh":"有限結構核對=([-1, 1], [0, -1], [-1, 2], [1, -2], [0, 0])。保留有序二分量，兩種減法各算一次再相加。有限精確核對不是一般形式證明。逐分量不當內積或長度。"},"difficulty":{"band":"upper_secondary","level":2,"skills":["向量差的兩個順序"]},"variant":{"family_id":"qf-density16-vector-difference-reversal","seed":202610161037,"parameters":{"u":[-1,1],"v":[0,-1],"domain":"Z2_fixed_basis","operation":"vector-difference-reversal","semantic_template":"qf-density16-vector-difference-reversal/{\"u\":[-1,1],\"v\":[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":"2026-10-12","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261012-advance-20261016-r01-vector-0038","batch_id":"batch-20261012-advance-20261016-r01-vector-foundations","language":"zh-Hant","source_mko_ids":["mko-vector"],"title_zh":"向量差的兩個順序","stem_zh":"固定標準基底的整數二分量向量。u=[0, 1]、v=[0, -1]。依序答 (u, v, u−v, v−u, 兩個差的和)。","answer":{"kind":"expression","value":"([0, 1], [0, -1], [0, 2], [0, -2], [0, 0])","explanation_zh":"有限結構核對=([0, 1], [0, -1], [0, 2], [0, -2], [0, 0])。保留有序二分量，兩種減法各算一次再相加。有限精確核對不是一般形式證明。逐分量不當內積或長度。"},"difficulty":{"band":"upper_secondary","level":2,"skills":["向量差的兩個順序"]},"variant":{"family_id":"qf-density16-vector-difference-reversal","seed":202610161038,"parameters":{"u":[0,1],"v":[0,-1],"domain":"Z2_fixed_basis","operation":"vector-difference-reversal","semantic_template":"qf-density16-vector-difference-reversal/{\"u\":[0,1],\"v\":[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":"2026-10-12","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261012-advance-20261016-r01-vector-0039","batch_id":"batch-20261012-advance-20261016-r01-vector-foundations","language":"zh-Hant","source_mko_ids":["mko-vector"],"title_zh":"向量差的兩個順序","stem_zh":"固定標準基底的整數二分量向量。u=[1, 1]、v=[0, -1]。依序答 (u, v, u−v, v−u, 兩個差的和)。","answer":{"kind":"expression","value":"([1, 1], [0, -1], [1, 2], [-1, -2], [0, 0])","explanation_zh":"有限結構核對=([1, 1], [0, -1], [1, 2], [-1, -2], [0, 0])。保留有序二分量，兩種減法各算一次再相加。有限精確核對不是一般形式證明。逐分量不當內積或長度。"},"difficulty":{"band":"upper_secondary","level":2,"skills":["向量差的兩個順序"]},"variant":{"family_id":"qf-density16-vector-difference-reversal","seed":202610161039,"parameters":{"u":[1,1],"v":[0,-1],"domain":"Z2_fixed_basis","operation":"vector-difference-reversal","semantic_template":"qf-density16-vector-difference-reversal/{\"u\":[1,1],\"v\":[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":"2026-10-12","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261012-advance-20261016-r01-vector-0040","batch_id":"batch-20261012-advance-20261016-r01-vector-foundations","language":"zh-Hant","source_mko_ids":["mko-vector"],"title_zh":"向量差的兩個順序","stem_zh":"固定標準基底的整數二分量向量。u=[2, 1]、v=[0, -1]。依序答 (u, v, u−v, v−u, 兩個差的和)。","answer":{"kind":"expression","value":"([2, 1], [0, -1], [2, 2], [-2, -2], [0, 0])","explanation_zh":"有限結構核對=([2, 1], [0, -1], [2, 2], [-2, -2], [0, 0])。保留有序二分量，兩種減法各算一次再相加。有限精確核對不是一般形式證明。逐分量不當內積或長度。"},"difficulty":{"band":"upper_secondary","level":2,"skills":["向量差的兩個順序"]},"variant":{"family_id":"qf-density16-vector-difference-reversal","seed":202610161040,"parameters":{"u":[2,1],"v":[0,-1],"domain":"Z2_fixed_basis","operation":"vector-difference-reversal","semantic_template":"qf-density16-vector-difference-reversal/{\"u\":[2,1],\"v\":[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":"2026-10-12","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261012-advance-20261016-r01-vector-0041","batch_id":"batch-20261012-advance-20261016-r01-vector-foundations","language":"zh-Hant","source_mko_ids":["mko-vector"],"title_zh":"向量差的兩個順序","stem_zh":"固定標準基底的整數二分量向量。u=[3, 1]、v=[0, -1]。依序答 (u, v, u−v, v−u, 兩個差的和)。","answer":{"kind":"expression","value":"([3, 1], [0, -1], [3, 2], [-3, -2], [0, 0])","explanation_zh":"有限結構核對=([3, 1], [0, -1], [3, 2], [-3, -2], [0, 0])。保留有序二分量，兩種減法各算一次再相加。有限精確核對不是一般形式證明。逐分量不當內積或長度。"},"difficulty":{"band":"upper_secondary","level":2,"skills":["向量差的兩個順序"]},"variant":{"family_id":"qf-density16-vector-difference-reversal","seed":202610161041,"parameters":{"u":[3,1],"v":[0,-1],"domain":"Z2_fixed_basis","operation":"vector-difference-reversal","semantic_template":"qf-density16-vector-difference-reversal/{\"u\":[3,1],\"v\":[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":"2026-10-12","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261012-advance-20261016-r01-vector-0042","batch_id":"batch-20261012-advance-20261016-r01-vector-foundations","language":"zh-Hant","source_mko_ids":["mko-vector"],"title_zh":"向量差的兩個順序","stem_zh":"固定標準基底的整數二分量向量。u=[4, 1]、v=[0, -1]。依序答 (u, v, u−v, v−u, 兩個差的和)。","answer":{"kind":"expression","value":"([4, 1], [0, -1], [4, 2], [-4, -2], [0, 0])","explanation_zh":"有限結構核對=([4, 1], [0, -1], [4, 2], [-4, -2], [0, 0])。保留有序二分量，兩種減法各算一次再相加。有限精確核對不是一般形式證明。逐分量不當內積或長度。"},"difficulty":{"band":"upper_secondary","level":2,"skills":["向量差的兩個順序"]},"variant":{"family_id":"qf-density16-vector-difference-reversal","seed":202610161042,"parameters":{"u":[4,1],"v":[0,-1],"domain":"Z2_fixed_basis","operation":"vector-difference-reversal","semantic_template":"qf-density16-vector-difference-reversal/{\"u\":[4,1],\"v\":[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":"2026-10-12","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261012-advance-20261016-r01-vector-0043","batch_id":"batch-20261012-advance-20261016-r01-vector-foundations","language":"zh-Hant","source_mko_ids":["mko-vector"],"title_zh":"向量差的兩個順序","stem_zh":"固定標準基底的整數二分量向量。u=[5, 1]、v=[0, -1]。依序答 (u, v, u−v, v−u, 兩個差的和)。","answer":{"kind":"expression","value":"([5, 1], [0, -1], [5, 2], [-5, -2], [0, 0])","explanation_zh":"有限結構核對=([5, 1], [0, -1], [5, 2], [-5, -2], [0, 0])。保留有序二分量，兩種減法各算一次再相加。有限精確核對不是一般形式證明。逐分量不當內積或長度。"},"difficulty":{"band":"upper_secondary","level":3,"skills":["向量差的兩個順序"]},"variant":{"family_id":"qf-density16-vector-difference-reversal","seed":202610161043,"parameters":{"u":[5,1],"v":[0,-1],"domain":"Z2_fixed_basis","operation":"vector-difference-reversal","semantic_template":"qf-density16-vector-difference-reversal/{\"u\":[5,1],\"v\":[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":"2026-10-12","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261012-advance-20261016-r01-vector-0044","batch_id":"batch-20261012-advance-20261016-r01-vector-foundations","language":"zh-Hant","source_mko_ids":["mko-vector"],"title_zh":"向量差的兩個順序","stem_zh":"固定標準基底的整數二分量向量。u=[6, 1]、v=[0, -1]。依序答 (u, v, u−v, v−u, 兩個差的和)。","answer":{"kind":"expression","value":"([6, 1], [0, -1], [6, 2], [-6, -2], [0, 0])","explanation_zh":"有限結構核對=([6, 1], [0, -1], [6, 2], [-6, -2], [0, 0])。保留有序二分量，兩種減法各算一次再相加。有限精確核對不是一般形式證明。逐分量不當內積或長度。"},"difficulty":{"band":"upper_secondary","level":3,"skills":["向量差的兩個順序"]},"variant":{"family_id":"qf-density16-vector-difference-reversal","seed":202610161044,"parameters":{"u":[6,1],"v":[0,-1],"domain":"Z2_fixed_basis","operation":"vector-difference-reversal","semantic_template":"qf-density16-vector-difference-reversal/{\"u\":[6,1],\"v\":[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":"2026-10-12","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261012-advance-20261016-r01-vector-0045","batch_id":"batch-20261012-advance-20261016-r01-vector-foundations","language":"zh-Hant","source_mko_ids":["mko-vector"],"title_zh":"向量差的兩個順序","stem_zh":"固定標準基底的整數二分量向量。u=[7, 1]、v=[0, -1]。依序答 (u, v, u−v, v−u, 兩個差的和)。","answer":{"kind":"expression","value":"([7, 1], [0, -1], [7, 2], [-7, -2], [0, 0])","explanation_zh":"有限結構核對=([7, 1], [0, -1], [7, 2], [-7, -2], [0, 0])。保留有序二分量，兩種減法各算一次再相加。有限精確核對不是一般形式證明。逐分量不當內積或長度。"},"difficulty":{"band":"upper_secondary","level":3,"skills":["向量差的兩個順序"]},"variant":{"family_id":"qf-density16-vector-difference-reversal","seed":202610161045,"parameters":{"u":[7,1],"v":[0,-1],"domain":"Z2_fixed_basis","operation":"vector-difference-reversal","semantic_template":"qf-density16-vector-difference-reversal/{\"u\":[7,1],\"v\":[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":"2026-10-12","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261012-advance-20261016-r01-vector-0046","batch_id":"batch-20261012-advance-20261016-r01-vector-foundations","language":"zh-Hant","source_mko_ids":["mko-vector"],"title_zh":"向量差的兩個順序","stem_zh":"固定標準基底的整數二分量向量。u=[8, 1]、v=[0, -1]。依序答 (u, v, u−v, v−u, 兩個差的和)。","answer":{"kind":"expression","value":"([8, 1], [0, -1], [8, 2], [-8, -2], [0, 0])","explanation_zh":"有限結構核對=([8, 1], [0, -1], [8, 2], [-8, -2], [0, 0])。保留有序二分量，兩種減法各算一次再相加。有限精確核對不是一般形式證明。逐分量不當內積或長度。"},"difficulty":{"band":"upper_secondary","level":3,"skills":["向量差的兩個順序"]},"variant":{"family_id":"qf-density16-vector-difference-reversal","seed":202610161046,"parameters":{"u":[8,1],"v":[0,-1],"domain":"Z2_fixed_basis","operation":"vector-difference-reversal","semantic_template":"qf-density16-vector-difference-reversal/{\"u\":[8,1],\"v\":[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":"2026-10-12","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261012-advance-20261016-r01-vector-0047","batch_id":"batch-20261012-advance-20261016-r01-vector-foundations","language":"zh-Hant","source_mko_ids":["mko-vector"],"title_zh":"向量差的兩個順序","stem_zh":"固定標準基底的整數二分量向量。u=[9, 1]、v=[0, -1]。依序答 (u, v, u−v, v−u, 兩個差的和)。","answer":{"kind":"expression","value":"([9, 1], [0, -1], [9, 2], [-9, -2], [0, 0])","explanation_zh":"有限結構核對=([9, 1], [0, -1], [9, 2], [-9, -2], [0, 0])。保留有序二分量，兩種減法各算一次再相加。有限精確核對不是一般形式證明。逐分量不當內積或長度。"},"difficulty":{"band":"upper_secondary","level":3,"skills":["向量差的兩個順序"]},"variant":{"family_id":"qf-density16-vector-difference-reversal","seed":202610161047,"parameters":{"u":[9,1],"v":[0,-1],"domain":"Z2_fixed_basis","operation":"vector-difference-reversal","semantic_template":"qf-density16-vector-difference-reversal/{\"u\":[9,1],\"v\":[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":"2026-10-12","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261012-advance-20261016-r01-vector-0048","batch_id":"batch-20261012-advance-20261016-r01-vector-foundations","language":"zh-Hant","source_mko_ids":["mko-vector"],"title_zh":"向量差的兩個順序","stem_zh":"固定標準基底的整數二分量向量。u=[10, 1]、v=[0, -1]。依序答 (u, v, u−v, v−u, 兩個差的和)。","answer":{"kind":"expression","value":"([10, 1], [0, -1], [10, 2], [-10, -2], [0, 0])","explanation_zh":"有限結構核對=([10, 1], [0, -1], [10, 2], [-10, -2], [0, 0])。保留有序二分量，兩種減法各算一次再相加。有限精確核對不是一般形式證明。逐分量不當內積或長度。"},"difficulty":{"band":"upper_secondary","level":3,"skills":["向量差的兩個順序"]},"variant":{"family_id":"qf-density16-vector-difference-reversal","seed":202610161048,"parameters":{"u":[10,1],"v":[0,-1],"domain":"Z2_fixed_basis","operation":"vector-difference-reversal","semantic_template":"qf-density16-vector-difference-reversal/{\"u\":[10,1],\"v\":[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":"2026-10-12","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261012-advance-20261016-r01-vector-0049","batch_id":"batch-20261012-advance-20261016-r01-vector-foundations","language":"zh-Hant","source_mko_ids":["mko-vector"],"title_zh":"向量差的兩個順序","stem_zh":"固定標準基底的整數二分量向量。u=[11, 1]、v=[0, -1]。依序答 (u, v, u−v, v−u, 兩個差的和)。","answer":{"kind":"expression","value":"([11, 1], [0, -1], [11, 2], [-11, -2], [0, 0])","explanation_zh":"有限結構核對=([11, 1], [0, -1], [11, 2], [-11, -2], [0, 0])。保留有序二分量，兩種減法各算一次再相加。有限精確核對不是一般形式證明。逐分量不當內積或長度。"},"difficulty":{"band":"upper_secondary","level":3,"skills":["向量差的兩個順序"]},"variant":{"family_id":"qf-density16-vector-difference-reversal","seed":202610161049,"parameters":{"u":[11,1],"v":[0,-1],"domain":"Z2_fixed_basis","operation":"vector-difference-reversal","semantic_template":"qf-density16-vector-difference-reversal/{\"u\":[11,1],\"v\":[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":"2026-10-12","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261012-advance-20261016-r01-vector-0050","batch_id":"batch-20261012-advance-20261016-r01-vector-foundations","language":"zh-Hant","source_mko_ids":["mko-vector"],"title_zh":"向量差的兩個順序","stem_zh":"固定標準基底的整數二分量向量。u=[12, 1]、v=[0, -1]。依序答 (u, v, u−v, v−u, 兩個差的和)。","answer":{"kind":"expression","value":"([12, 1], [0, -1], [12, 2], [-12, -2], [0, 0])","explanation_zh":"有限結構核對=([12, 1], [0, -1], [12, 2], [-12, -2], [0, 0])。保留有序二分量，兩種減法各算一次再相加。有限精確核對不是一般形式證明。逐分量不當內積或長度。"},"difficulty":{"band":"upper_secondary","level":3,"skills":["向量差的兩個順序"]},"variant":{"family_id":"qf-density16-vector-difference-reversal","seed":202610161050,"parameters":{"u":[12,1],"v":[0,-1],"domain":"Z2_fixed_basis","operation":"vector-difference-reversal","semantic_template":"qf-density16-vector-difference-reversal/{\"u\":[12,1],\"v\":[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":"2026-10-12","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261012-advance-20261016-r01-vector-0051","batch_id":"batch-20261012-advance-20261016-r01-vector-foundations","language":"zh-Hant","source_mko_ids":["mko-vector"],"title_zh":"同向量的雙純量兩路","stem_zh":"固定標準基底的整數二分量向量。u=[0, 0]、a=0、b=0。求 (u, (a+b)u, au+bu, 後兩路之差)。","answer":{"kind":"expression","value":"([0, 0], [0, 0], [0, 0], [0, 0])","explanation_zh":"有限結構核對=([0, 0], [0, 0], [0, 0], [0, 0])。一條路先加純量，另一條路先乘分量再加，核對實際兩路向量。有限精確核對不是一般形式證明。逐分量不當內積或長度。"},"difficulty":{"band":"upper_secondary","level":1,"skills":["同向量的雙純量兩路"]},"variant":{"family_id":"qf-density16-vector-two-scalar-additivity","seed":202610161051,"parameters":{"u":[0,0],"a":0,"b":0,"domain":"Z2_fixed_basis","operation":"vector-two-scalar-additivity","semantic_template":"qf-density16-vector-two-scalar-additivity/{\"a\":0,\"b\":0,\"u\":[0,0]}"}},"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":"2026-10-12","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261012-advance-20261016-r01-vector-0052","batch_id":"batch-20261012-advance-20261016-r01-vector-foundations","language":"zh-Hant","source_mko_ids":["mko-vector"],"title_zh":"同向量的雙純量兩路","stem_zh":"固定標準基底的整數二分量向量。u=[1, 2]、a=2、b=-2。求 (u, (a+b)u, au+bu, 後兩路之差)。","answer":{"kind":"expression","value":"([1, 2], [0, 0], [0, 0], [0, 0])","explanation_zh":"有限結構核對=([1, 2], [0, 0], [0, 0], [0, 0])。一條路先加純量，另一條路先乘分量再加，核對實際兩路向量。有限精確核對不是一般形式證明。逐分量不當內積或長度。"},"difficulty":{"band":"upper_secondary","level":1,"skills":["同向量的雙純量兩路"]},"variant":{"family_id":"qf-density16-vector-two-scalar-additivity","seed":202610161052,"parameters":{"u":[1,2],"a":2,"b":-2,"domain":"Z2_fixed_basis","operation":"vector-two-scalar-additivity","semantic_template":"qf-density16-vector-two-scalar-additivity/{\"a\":2,\"b\":-2,\"u\":[1,2]}"}},"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":"2026-10-12","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261012-advance-20261016-r01-vector-0053","batch_id":"batch-20261012-advance-20261016-r01-vector-foundations","language":"zh-Hant","source_mko_ids":["mko-vector"],"title_zh":"同向量的雙純量兩路","stem_zh":"固定標準基底的整數二分量向量。u=[1, -1]、a=-1、b=2。求 (u, (a+b)u, au+bu, 後兩路之差)。","answer":{"kind":"expression","value":"([1, -1], [1, -1], [1, -1], [0, 0])","explanation_zh":"有限結構核對=([1, -1], [1, -1], [1, -1], [0, 0])。一條路先加純量，另一條路先乘分量再加，核對實際兩路向量。有限精確核對不是一般形式證明。逐分量不當內積或長度。"},"difficulty":{"band":"upper_secondary","level":1,"skills":["同向量的雙純量兩路"]},"variant":{"family_id":"qf-density16-vector-two-scalar-additivity","seed":202610161053,"parameters":{"u":[1,-1],"a":-1,"b":2,"domain":"Z2_fixed_basis","operation":"vector-two-scalar-additivity","semantic_template":"qf-density16-vector-two-scalar-additivity/{\"a\":-1,\"b\":2,\"u\":[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":"2026-10-12","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261012-advance-20261016-r01-vector-0054","batch_id":"batch-20261012-advance-20261016-r01-vector-foundations","language":"zh-Hant","source_mko_ids":["mko-vector"],"title_zh":"同向量的雙純量兩路","stem_zh":"固定標準基底的整數二分量向量。u=[-9, 2]、a=1、b=2。求 (u, (a+b)u, au+bu, 後兩路之差)。","answer":{"kind":"expression","value":"([-9, 2], [-27, 6], [-27, 6], [0, 0])","explanation_zh":"有限結構核對=([-9, 2], [-27, 6], [-27, 6], [0, 0])。一條路先加純量，另一條路先乘分量再加，核對實際兩路向量。有限精確核對不是一般形式證明。逐分量不當內積或長度。"},"difficulty":{"band":"upper_secondary","level":1,"skills":["同向量的雙純量兩路"]},"variant":{"family_id":"qf-density16-vector-two-scalar-additivity","seed":202610161054,"parameters":{"u":[-9,2],"a":1,"b":2,"domain":"Z2_fixed_basis","operation":"vector-two-scalar-additivity","semantic_template":"qf-density16-vector-two-scalar-additivity/{\"a\":1,\"b\":2,\"u\":[-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":"2026-10-12","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261012-advance-20261016-r01-vector-0055","batch_id":"batch-20261012-advance-20261016-r01-vector-foundations","language":"zh-Hant","source_mko_ids":["mko-vector"],"title_zh":"同向量的雙純量兩路","stem_zh":"固定標準基底的整數二分量向量。u=[-8, 2]、a=1、b=2。求 (u, (a+b)u, au+bu, 後兩路之差)。","answer":{"kind":"expression","value":"([-8, 2], [-24, 6], [-24, 6], [0, 0])","explanation_zh":"有限結構核對=([-8, 2], [-24, 6], [-24, 6], [0, 0])。一條路先加純量，另一條路先乘分量再加，核對實際兩路向量。有限精確核對不是一般形式證明。逐分量不當內積或長度。"},"difficulty":{"band":"upper_secondary","level":1,"skills":["同向量的雙純量兩路"]},"variant":{"family_id":"qf-density16-vector-two-scalar-additivity","seed":202610161055,"parameters":{"u":[-8,2],"a":1,"b":2,"domain":"Z2_fixed_basis","operation":"vector-two-scalar-additivity","semantic_template":"qf-density16-vector-two-scalar-additivity/{\"a\":1,\"b\":2,\"u\":[-8,2]}"}},"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":"2026-10-12","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261012-advance-20261016-r01-vector-0056","batch_id":"batch-20261012-advance-20261016-r01-vector-foundations","language":"zh-Hant","source_mko_ids":["mko-vector"],"title_zh":"同向量的雙純量兩路","stem_zh":"固定標準基底的整數二分量向量。u=[-7, 2]、a=1、b=2。求 (u, (a+b)u, au+bu, 後兩路之差)。","answer":{"kind":"expression","value":"([-7, 2], [-21, 6], [-21, 6], [0, 0])","explanation_zh":"有限結構核對=([-7, 2], [-21, 6], [-21, 6], [0, 0])。一條路先加純量，另一條路先乘分量再加，核對實際兩路向量。有限精確核對不是一般形式證明。逐分量不當內積或長度。"},"difficulty":{"band":"upper_secondary","level":1,"skills":["同向量的雙純量兩路"]},"variant":{"family_id":"qf-density16-vector-two-scalar-additivity","seed":202610161056,"parameters":{"u":[-7,2],"a":1,"b":2,"domain":"Z2_fixed_basis","operation":"vector-two-scalar-additivity","semantic_template":"qf-density16-vector-two-scalar-additivity/{\"a\":1,\"b\":2,\"u\":[-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":"2026-10-12","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261012-advance-20261016-r01-vector-0057","batch_id":"batch-20261012-advance-20261016-r01-vector-foundations","language":"zh-Hant","source_mko_ids":["mko-vector"],"title_zh":"同向量的雙純量兩路","stem_zh":"固定標準基底的整數二分量向量。u=[-6, 2]、a=1、b=2。求 (u, (a+b)u, au+bu, 後兩路之差)。","answer":{"kind":"expression","value":"([-6, 2], [-18, 6], [-18, 6], [0, 0])","explanation_zh":"有限結構核對=([-6, 2], [-18, 6], [-18, 6], [0, 0])。一條路先加純量，另一條路先乘分量再加，核對實際兩路向量。有限精確核對不是一般形式證明。逐分量不當內積或長度。"},"difficulty":{"band":"upper_secondary","level":1,"skills":["同向量的雙純量兩路"]},"variant":{"family_id":"qf-density16-vector-two-scalar-additivity","seed":202610161057,"parameters":{"u":[-6,2],"a":1,"b":2,"domain":"Z2_fixed_basis","operation":"vector-two-scalar-additivity","semantic_template":"qf-density16-vector-two-scalar-additivity/{\"a\":1,\"b\":2,\"u\":[-6,2]}"}},"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":"2026-10-12","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261012-advance-20261016-r01-vector-0058","batch_id":"batch-20261012-advance-20261016-r01-vector-foundations","language":"zh-Hant","source_mko_ids":["mko-vector"],"title_zh":"同向量的雙純量兩路","stem_zh":"固定標準基底的整數二分量向量。u=[-5, 2]、a=1、b=2。求 (u, (a+b)u, au+bu, 後兩路之差)。","answer":{"kind":"expression","value":"([-5, 2], [-15, 6], [-15, 6], [0, 0])","explanation_zh":"有限結構核對=([-5, 2], [-15, 6], [-15, 6], [0, 0])。一條路先加純量，另一條路先乘分量再加，核對實際兩路向量。有限精確核對不是一般形式證明。逐分量不當內積或長度。"},"difficulty":{"band":"upper_secondary","level":1,"skills":["同向量的雙純量兩路"]},"variant":{"family_id":"qf-density16-vector-two-scalar-additivity","seed":202610161058,"parameters":{"u":[-5,2],"a":1,"b":2,"domain":"Z2_fixed_basis","operation":"vector-two-scalar-additivity","semantic_template":"qf-density16-vector-two-scalar-additivity/{\"a\":1,\"b\":2,\"u\":[-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":"2026-10-12","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261012-advance-20261016-r01-vector-0059","batch_id":"batch-20261012-advance-20261016-r01-vector-foundations","language":"zh-Hant","source_mko_ids":["mko-vector"],"title_zh":"同向量的雙純量兩路","stem_zh":"固定標準基底的整數二分量向量。u=[-4, 2]、a=1、b=2。求 (u, (a+b)u, au+bu, 後兩路之差)。","answer":{"kind":"expression","value":"([-4, 2], [-12, 6], [-12, 6], [0, 0])","explanation_zh":"有限結構核對=([-4, 2], [-12, 6], [-12, 6], [0, 0])。一條路先加純量，另一條路先乘分量再加，核對實際兩路向量。有限精確核對不是一般形式證明。逐分量不當內積或長度。"},"difficulty":{"band":"upper_secondary","level":2,"skills":["同向量的雙純量兩路"]},"variant":{"family_id":"qf-density16-vector-two-scalar-additivity","seed":202610161059,"parameters":{"u":[-4,2],"a":1,"b":2,"domain":"Z2_fixed_basis","operation":"vector-two-scalar-additivity","semantic_template":"qf-density16-vector-two-scalar-additivity/{\"a\":1,\"b\":2,\"u\":[-4,2]}"}},"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":"2026-10-12","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261012-advance-20261016-r01-vector-0060","batch_id":"batch-20261012-advance-20261016-r01-vector-foundations","language":"zh-Hant","source_mko_ids":["mko-vector"],"title_zh":"同向量的雙純量兩路","stem_zh":"固定標準基底的整數二分量向量。u=[-3, 2]、a=1、b=2。求 (u, (a+b)u, au+bu, 後兩路之差)。","answer":{"kind":"expression","value":"([-3, 2], [-9, 6], [-9, 6], [0, 0])","explanation_zh":"有限結構核對=([-3, 2], [-9, 6], [-9, 6], [0, 0])。一條路先加純量，另一條路先乘分量再加，核對實際兩路向量。有限精確核對不是一般形式證明。逐分量不當內積或長度。"},"difficulty":{"band":"upper_secondary","level":2,"skills":["同向量的雙純量兩路"]},"variant":{"family_id":"qf-density16-vector-two-scalar-additivity","seed":202610161060,"parameters":{"u":[-3,2],"a":1,"b":2,"domain":"Z2_fixed_basis","operation":"vector-two-scalar-additivity","semantic_template":"qf-density16-vector-two-scalar-additivity/{\"a\":1,\"b\":2,\"u\":[-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":"2026-10-12","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261012-advance-20261016-r01-vector-0061","batch_id":"batch-20261012-advance-20261016-r01-vector-foundations","language":"zh-Hant","source_mko_ids":["mko-vector"],"title_zh":"同向量的雙純量兩路","stem_zh":"固定標準基底的整數二分量向量。u=[-2, 2]、a=1、b=2。求 (u, (a+b)u, au+bu, 後兩路之差)。","answer":{"kind":"expression","value":"([-2, 2], [-6, 6], [-6, 6], [0, 0])","explanation_zh":"有限結構核對=([-2, 2], [-6, 6], [-6, 6], [0, 0])。一條路先加純量，另一條路先乘分量再加，核對實際兩路向量。有限精確核對不是一般形式證明。逐分量不當內積或長度。"},"difficulty":{"band":"upper_secondary","level":2,"skills":["同向量的雙純量兩路"]},"variant":{"family_id":"qf-density16-vector-two-scalar-additivity","seed":202610161061,"parameters":{"u":[-2,2],"a":1,"b":2,"domain":"Z2_fixed_basis","operation":"vector-two-scalar-additivity","semantic_template":"qf-density16-vector-two-scalar-additivity/{\"a\":1,\"b\":2,\"u\":[-2,2]}"}},"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":"2026-10-12","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261012-advance-20261016-r01-vector-0062","batch_id":"batch-20261012-advance-20261016-r01-vector-foundations","language":"zh-Hant","source_mko_ids":["mko-vector"],"title_zh":"同向量的雙純量兩路","stem_zh":"固定標準基底的整數二分量向量。u=[-1, 2]、a=1、b=2。求 (u, (a+b)u, au+bu, 後兩路之差)。","answer":{"kind":"expression","value":"([-1, 2], [-3, 6], [-3, 6], [0, 0])","explanation_zh":"有限結構核對=([-1, 2], [-3, 6], [-3, 6], [0, 0])。一條路先加純量，另一條路先乘分量再加，核對實際兩路向量。有限精確核對不是一般形式證明。逐分量不當內積或長度。"},"difficulty":{"band":"upper_secondary","level":2,"skills":["同向量的雙純量兩路"]},"variant":{"family_id":"qf-density16-vector-two-scalar-additivity","seed":202610161062,"parameters":{"u":[-1,2],"a":1,"b":2,"domain":"Z2_fixed_basis","operation":"vector-two-scalar-additivity","semantic_template":"qf-density16-vector-two-scalar-additivity/{\"a\":1,\"b\":2,\"u\":[-1,2]}"}},"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":"2026-10-12","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261012-advance-20261016-r01-vector-0063","batch_id":"batch-20261012-advance-20261016-r01-vector-foundations","language":"zh-Hant","source_mko_ids":["mko-vector"],"title_zh":"同向量的雙純量兩路","stem_zh":"固定標準基底的整數二分量向量。u=[0, 2]、a=1、b=2。求 (u, (a+b)u, au+bu, 後兩路之差)。","answer":{"kind":"expression","value":"([0, 2], [0, 6], [0, 6], [0, 0])","explanation_zh":"有限結構核對=([0, 2], [0, 6], [0, 6], [0, 0])。一條路先加純量，另一條路先乘分量再加，核對實際兩路向量。有限精確核對不是一般形式證明。逐分量不當內積或長度。"},"difficulty":{"band":"upper_secondary","level":2,"skills":["同向量的雙純量兩路"]},"variant":{"family_id":"qf-density16-vector-two-scalar-additivity","seed":202610161063,"parameters":{"u":[0,2],"a":1,"b":2,"domain":"Z2_fixed_basis","operation":"vector-two-scalar-additivity","semantic_template":"qf-density16-vector-two-scalar-additivity/{\"a\":1,\"b\":2,\"u\":[0,2]}"}},"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":"2026-10-12","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261012-advance-20261016-r01-vector-0064","batch_id":"batch-20261012-advance-20261016-r01-vector-foundations","language":"zh-Hant","source_mko_ids":["mko-vector"],"title_zh":"同向量的雙純量兩路","stem_zh":"固定標準基底的整數二分量向量。u=[1, 2]、a=1、b=2。求 (u, (a+b)u, au+bu, 後兩路之差)。","answer":{"kind":"expression","value":"([1, 2], [3, 6], [3, 6], [0, 0])","explanation_zh":"有限結構核對=([1, 2], [3, 6], [3, 6], [0, 0])。一條路先加純量，另一條路先乘分量再加，核對實際兩路向量。有限精確核對不是一般形式證明。逐分量不當內積或長度。"},"difficulty":{"band":"upper_secondary","level":2,"skills":["同向量的雙純量兩路"]},"variant":{"family_id":"qf-density16-vector-two-scalar-additivity","seed":202610161064,"parameters":{"u":[1,2],"a":1,"b":2,"domain":"Z2_fixed_basis","operation":"vector-two-scalar-additivity","semantic_template":"qf-density16-vector-two-scalar-additivity/{\"a\":1,\"b\":2,\"u\":[1,2]}"}},"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":"2026-10-12","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261012-advance-20261016-r01-vector-0065","batch_id":"batch-20261012-advance-20261016-r01-vector-foundations","language":"zh-Hant","source_mko_ids":["mko-vector"],"title_zh":"同向量的雙純量兩路","stem_zh":"固定標準基底的整數二分量向量。u=[2, 2]、a=1、b=2。求 (u, (a+b)u, au+bu, 後兩路之差)。","answer":{"kind":"expression","value":"([2, 2], [6, 6], [6, 6], [0, 0])","explanation_zh":"有限結構核對=([2, 2], [6, 6], [6, 6], [0, 0])。一條路先加純量，另一條路先乘分量再加，核對實際兩路向量。有限精確核對不是一般形式證明。逐分量不當內積或長度。"},"difficulty":{"band":"upper_secondary","level":2,"skills":["同向量的雙純量兩路"]},"variant":{"family_id":"qf-density16-vector-two-scalar-additivity","seed":202610161065,"parameters":{"u":[2,2],"a":1,"b":2,"domain":"Z2_fixed_basis","operation":"vector-two-scalar-additivity","semantic_template":"qf-density16-vector-two-scalar-additivity/{\"a\":1,\"b\":2,\"u\":[2,2]}"}},"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":"2026-10-12","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261012-advance-20261016-r01-vector-0066","batch_id":"batch-20261012-advance-20261016-r01-vector-foundations","language":"zh-Hant","source_mko_ids":["mko-vector"],"title_zh":"同向量的雙純量兩路","stem_zh":"固定標準基底的整數二分量向量。u=[3, 2]、a=1、b=2。求 (u, (a+b)u, au+bu, 後兩路之差)。","answer":{"kind":"expression","value":"([3, 2], [9, 6], [9, 6], [0, 0])","explanation_zh":"有限結構核對=([3, 2], [9, 6], [9, 6], [0, 0])。一條路先加純量，另一條路先乘分量再加，核對實際兩路向量。有限精確核對不是一般形式證明。逐分量不當內積或長度。"},"difficulty":{"band":"upper_secondary","level":2,"skills":["同向量的雙純量兩路"]},"variant":{"family_id":"qf-density16-vector-two-scalar-additivity","seed":202610161066,"parameters":{"u":[3,2],"a":1,"b":2,"domain":"Z2_fixed_basis","operation":"vector-two-scalar-additivity","semantic_template":"qf-density16-vector-two-scalar-additivity/{\"a\":1,\"b\":2,\"u\":[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":"2026-10-12","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261012-advance-20261016-r01-vector-0067","batch_id":"batch-20261012-advance-20261016-r01-vector-foundations","language":"zh-Hant","source_mko_ids":["mko-vector"],"title_zh":"同向量的雙純量兩路","stem_zh":"固定標準基底的整數二分量向量。u=[4, 2]、a=1、b=2。求 (u, (a+b)u, au+bu, 後兩路之差)。","answer":{"kind":"expression","value":"([4, 2], [12, 6], [12, 6], [0, 0])","explanation_zh":"有限結構核對=([4, 2], [12, 6], [12, 6], [0, 0])。一條路先加純量，另一條路先乘分量再加，核對實際兩路向量。有限精確核對不是一般形式證明。逐分量不當內積或長度。"},"difficulty":{"band":"upper_secondary","level":2,"skills":["同向量的雙純量兩路"]},"variant":{"family_id":"qf-density16-vector-two-scalar-additivity","seed":202610161067,"parameters":{"u":[4,2],"a":1,"b":2,"domain":"Z2_fixed_basis","operation":"vector-two-scalar-additivity","semantic_template":"qf-density16-vector-two-scalar-additivity/{\"a\":1,\"b\":2,\"u\":[4,2]}"}},"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":"2026-10-12","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261012-advance-20261016-r01-vector-0068","batch_id":"batch-20261012-advance-20261016-r01-vector-foundations","language":"zh-Hant","source_mko_ids":["mko-vector"],"title_zh":"同向量的雙純量兩路","stem_zh":"固定標準基底的整數二分量向量。u=[5, 2]、a=1、b=2。求 (u, (a+b)u, au+bu, 後兩路之差)。","answer":{"kind":"expression","value":"([5, 2], [15, 6], [15, 6], [0, 0])","explanation_zh":"有限結構核對=([5, 2], [15, 6], [15, 6], [0, 0])。一條路先加純量，另一條路先乘分量再加，核對實際兩路向量。有限精確核對不是一般形式證明。逐分量不當內積或長度。"},"difficulty":{"band":"upper_secondary","level":3,"skills":["同向量的雙純量兩路"]},"variant":{"family_id":"qf-density16-vector-two-scalar-additivity","seed":202610161068,"parameters":{"u":[5,2],"a":1,"b":2,"domain":"Z2_fixed_basis","operation":"vector-two-scalar-additivity","semantic_template":"qf-density16-vector-two-scalar-additivity/{\"a\":1,\"b\":2,\"u\":[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":"2026-10-12","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261012-advance-20261016-r01-vector-0069","batch_id":"batch-20261012-advance-20261016-r01-vector-foundations","language":"zh-Hant","source_mko_ids":["mko-vector"],"title_zh":"同向量的雙純量兩路","stem_zh":"固定標準基底的整數二分量向量。u=[6, 2]、a=1、b=2。求 (u, (a+b)u, au+bu, 後兩路之差)。","answer":{"kind":"expression","value":"([6, 2], [18, 6], [18, 6], [0, 0])","explanation_zh":"有限結構核對=([6, 2], [18, 6], [18, 6], [0, 0])。一條路先加純量，另一條路先乘分量再加，核對實際兩路向量。有限精確核對不是一般形式證明。逐分量不當內積或長度。"},"difficulty":{"band":"upper_secondary","level":3,"skills":["同向量的雙純量兩路"]},"variant":{"family_id":"qf-density16-vector-two-scalar-additivity","seed":202610161069,"parameters":{"u":[6,2],"a":1,"b":2,"domain":"Z2_fixed_basis","operation":"vector-two-scalar-additivity","semantic_template":"qf-density16-vector-two-scalar-additivity/{\"a\":1,\"b\":2,\"u\":[6,2]}"}},"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":"2026-10-12","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261012-advance-20261016-r01-vector-0070","batch_id":"batch-20261012-advance-20261016-r01-vector-foundations","language":"zh-Hant","source_mko_ids":["mko-vector"],"title_zh":"同向量的雙純量兩路","stem_zh":"固定標準基底的整數二分量向量。u=[7, 2]、a=1、b=2。求 (u, (a+b)u, au+bu, 後兩路之差)。","answer":{"kind":"expression","value":"([7, 2], [21, 6], [21, 6], [0, 0])","explanation_zh":"有限結構核對=([7, 2], [21, 6], [21, 6], [0, 0])。一條路先加純量，另一條路先乘分量再加，核對實際兩路向量。有限精確核對不是一般形式證明。逐分量不當內積或長度。"},"difficulty":{"band":"upper_secondary","level":3,"skills":["同向量的雙純量兩路"]},"variant":{"family_id":"qf-density16-vector-two-scalar-additivity","seed":202610161070,"parameters":{"u":[7,2],"a":1,"b":2,"domain":"Z2_fixed_basis","operation":"vector-two-scalar-additivity","semantic_template":"qf-density16-vector-two-scalar-additivity/{\"a\":1,\"b\":2,\"u\":[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":"2026-10-12","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261012-advance-20261016-r01-vector-0071","batch_id":"batch-20261012-advance-20261016-r01-vector-foundations","language":"zh-Hant","source_mko_ids":["mko-vector"],"title_zh":"同向量的雙純量兩路","stem_zh":"固定標準基底的整數二分量向量。u=[8, 2]、a=1、b=2。求 (u, (a+b)u, au+bu, 後兩路之差)。","answer":{"kind":"expression","value":"([8, 2], [24, 6], [24, 6], [0, 0])","explanation_zh":"有限結構核對=([8, 2], [24, 6], [24, 6], [0, 0])。一條路先加純量，另一條路先乘分量再加，核對實際兩路向量。有限精確核對不是一般形式證明。逐分量不當內積或長度。"},"difficulty":{"band":"upper_secondary","level":3,"skills":["同向量的雙純量兩路"]},"variant":{"family_id":"qf-density16-vector-two-scalar-additivity","seed":202610161071,"parameters":{"u":[8,2],"a":1,"b":2,"domain":"Z2_fixed_basis","operation":"vector-two-scalar-additivity","semantic_template":"qf-density16-vector-two-scalar-additivity/{\"a\":1,\"b\":2,\"u\":[8,2]}"}},"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":"2026-10-12","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261012-advance-20261016-r01-vector-0072","batch_id":"batch-20261012-advance-20261016-r01-vector-foundations","language":"zh-Hant","source_mko_ids":["mko-vector"],"title_zh":"同向量的雙純量兩路","stem_zh":"固定標準基底的整數二分量向量。u=[9, 2]、a=1、b=2。求 (u, (a+b)u, au+bu, 後兩路之差)。","answer":{"kind":"expression","value":"([9, 2], [27, 6], [27, 6], [0, 0])","explanation_zh":"有限結構核對=([9, 2], [27, 6], [27, 6], [0, 0])。一條路先加純量，另一條路先乘分量再加，核對實際兩路向量。有限精確核對不是一般形式證明。逐分量不當內積或長度。"},"difficulty":{"band":"upper_secondary","level":3,"skills":["同向量的雙純量兩路"]},"variant":{"family_id":"qf-density16-vector-two-scalar-additivity","seed":202610161072,"parameters":{"u":[9,2],"a":1,"b":2,"domain":"Z2_fixed_basis","operation":"vector-two-scalar-additivity","semantic_template":"qf-density16-vector-two-scalar-additivity/{\"a\":1,\"b\":2,\"u\":[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":"2026-10-12","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261012-advance-20261016-r01-vector-0073","batch_id":"batch-20261012-advance-20261016-r01-vector-foundations","language":"zh-Hant","source_mko_ids":["mko-vector"],"title_zh":"同向量的雙純量兩路","stem_zh":"固定標準基底的整數二分量向量。u=[10, 2]、a=1、b=2。求 (u, (a+b)u, au+bu, 後兩路之差)。","answer":{"kind":"expression","value":"([10, 2], [30, 6], [30, 6], [0, 0])","explanation_zh":"有限結構核對=([10, 2], [30, 6], [30, 6], [0, 0])。一條路先加純量，另一條路先乘分量再加，核對實際兩路向量。有限精確核對不是一般形式證明。逐分量不當內積或長度。"},"difficulty":{"band":"upper_secondary","level":3,"skills":["同向量的雙純量兩路"]},"variant":{"family_id":"qf-density16-vector-two-scalar-additivity","seed":202610161073,"parameters":{"u":[10,2],"a":1,"b":2,"domain":"Z2_fixed_basis","operation":"vector-two-scalar-additivity","semantic_template":"qf-density16-vector-two-scalar-additivity/{\"a\":1,\"b\":2,\"u\":[10,2]}"}},"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":"2026-10-12","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261012-advance-20261016-r01-vector-0074","batch_id":"batch-20261012-advance-20261016-r01-vector-foundations","language":"zh-Hant","source_mko_ids":["mko-vector"],"title_zh":"同向量的雙純量兩路","stem_zh":"固定標準基底的整數二分量向量。u=[11, 2]、a=1、b=2。求 (u, (a+b)u, au+bu, 後兩路之差)。","answer":{"kind":"expression","value":"([11, 2], [33, 6], [33, 6], [0, 0])","explanation_zh":"有限結構核對=([11, 2], [33, 6], [33, 6], [0, 0])。一條路先加純量，另一條路先乘分量再加，核對實際兩路向量。有限精確核對不是一般形式證明。逐分量不當內積或長度。"},"difficulty":{"band":"upper_secondary","level":3,"skills":["同向量的雙純量兩路"]},"variant":{"family_id":"qf-density16-vector-two-scalar-additivity","seed":202610161074,"parameters":{"u":[11,2],"a":1,"b":2,"domain":"Z2_fixed_basis","operation":"vector-two-scalar-additivity","semantic_template":"qf-density16-vector-two-scalar-additivity/{\"a\":1,\"b\":2,\"u\":[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":"2026-10-12","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261012-advance-20261016-r01-vector-0075","batch_id":"batch-20261012-advance-20261016-r01-vector-foundations","language":"zh-Hant","source_mko_ids":["mko-vector"],"title_zh":"同向量的雙純量兩路","stem_zh":"固定標準基底的整數二分量向量。u=[12, 2]、a=1、b=2。求 (u, (a+b)u, au+bu, 後兩路之差)。","answer":{"kind":"expression","value":"([12, 2], [36, 6], [36, 6], [0, 0])","explanation_zh":"有限結構核對=([12, 2], [36, 6], [36, 6], [0, 0])。一條路先加純量，另一條路先乘分量再加，核對實際兩路向量。有限精確核對不是一般形式證明。逐分量不當內積或長度。"},"difficulty":{"band":"upper_secondary","level":3,"skills":["同向量的雙純量兩路"]},"variant":{"family_id":"qf-density16-vector-two-scalar-additivity","seed":202610161075,"parameters":{"u":[12,2],"a":1,"b":2,"domain":"Z2_fixed_basis","operation":"vector-two-scalar-additivity","semantic_template":"qf-density16-vector-two-scalar-additivity/{\"a\":1,\"b\":2,\"u\":[12,2]}"}},"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":"2026-10-12","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261012-advance-20261016-r01-vector-0076","batch_id":"batch-20261012-advance-20261016-r01-vector-foundations","language":"zh-Hant","source_mko_ids":["mko-vector"],"title_zh":"加權和候選的分量殘差","stem_zh":"固定標準基底的整數二分量向量。u=[0, 0]、v=[0, 0]、a=0、b=0、候選=[0, 0]。求 (au+bv, 候選減真值, 候選相等資格)。","answer":{"kind":"expression","value":"([0, 0], [0, 0], 是)","explanation_zh":"有限結構核對=([0, 0], [0, 0], 是)。先算兩個真值分量，候選的兩個殘差都零才相等。有限精確核對不是一般形式證明。逐分量不當內積或長度。"},"difficulty":{"band":"upper_secondary","level":1,"skills":["加權和候選的分量殘差"]},"variant":{"family_id":"qf-density16-vector-candidate-residual","seed":202610161076,"parameters":{"u":[0,0],"v":[0,0],"a":0,"b":0,"candidate":[0,0],"domain":"Z2_fixed_basis","operation":"vector-candidate-residual","semantic_template":"qf-density16-vector-candidate-residual/{\"a\":0,\"b\":0,\"candidate\":[0,0],\"u\":[0,0],\"v\":[0,0]}"}},"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":"2026-10-12","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261012-advance-20261016-r01-vector-0077","batch_id":"batch-20261012-advance-20261016-r01-vector-foundations","language":"zh-Hant","source_mko_ids":["mko-vector"],"title_zh":"加權和候選的分量殘差","stem_zh":"固定標準基底的整數二分量向量。u=[1, 0]、v=[0, 1]、a=1、b=1、候選=[1, 0]。求 (au+bv, 候選減真值, 候選相等資格)。","answer":{"kind":"expression","value":"([1, 1], [0, -1], 否)","explanation_zh":"有限結構核對=([1, 1], [0, -1], 否)。先算兩個真值分量，候選的兩個殘差都零才相等。有限精確核對不是一般形式證明。逐分量不當內積或長度。"},"difficulty":{"band":"upper_secondary","level":1,"skills":["加權和候選的分量殘差"]},"variant":{"family_id":"qf-density16-vector-candidate-residual","seed":202610161077,"parameters":{"u":[1,0],"v":[0,1],"a":1,"b":1,"candidate":[1,0],"domain":"Z2_fixed_basis","operation":"vector-candidate-residual","semantic_template":"qf-density16-vector-candidate-residual/{\"a\":1,\"b\":1,\"candidate\":[1,0],\"u\":[1,0],\"v\":[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":"2026-10-12","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261012-advance-20261016-r01-vector-0078","batch_id":"batch-20261012-advance-20261016-r01-vector-foundations","language":"zh-Hant","source_mko_ids":["mko-vector"],"title_zh":"加權和候選的分量殘差","stem_zh":"固定標準基底的整數二分量向量。u=[1, 2]、v=[1, 2]、a=2、b=-1、候選=[1, 2]。求 (au+bv, 候選減真值, 候選相等資格)。","answer":{"kind":"expression","value":"([1, 2], [0, 0], 是)","explanation_zh":"有限結構核對=([1, 2], [0, 0], 是)。先算兩個真值分量，候選的兩個殘差都零才相等。有限精確核對不是一般形式證明。逐分量不當內積或長度。"},"difficulty":{"band":"upper_secondary","level":1,"skills":["加權和候選的分量殘差"]},"variant":{"family_id":"qf-density16-vector-candidate-residual","seed":202610161078,"parameters":{"u":[1,2],"v":[1,2],"a":2,"b":-1,"candidate":[1,2],"domain":"Z2_fixed_basis","operation":"vector-candidate-residual","semantic_template":"qf-density16-vector-candidate-residual/{\"a\":2,\"b\":-1,\"candidate\":[1,2],\"u\":[1,2],\"v\":[1,2]}"}},"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":"2026-10-12","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261012-advance-20261016-r01-vector-0079","batch_id":"batch-20261012-advance-20261016-r01-vector-foundations","language":"zh-Hant","source_mko_ids":["mko-vector"],"title_zh":"加權和候選的分量殘差","stem_zh":"固定標準基底的整數二分量向量。u=[-9, 1]、v=[1, 2]、a=2、b=1、候選=[-17, 4]。求 (au+bv, 候選減真值, 候選相等資格)。","answer":{"kind":"expression","value":"([-17, 4], [0, 0], 是)","explanation_zh":"有限結構核對=([-17, 4], [0, 0], 是)。先算兩個真值分量，候選的兩個殘差都零才相等。有限精確核對不是一般形式證明。逐分量不當內積或長度。"},"difficulty":{"band":"upper_secondary","level":1,"skills":["加權和候選的分量殘差"]},"variant":{"family_id":"qf-density16-vector-candidate-residual","seed":202610161079,"parameters":{"u":[-9,1],"v":[1,2],"a":2,"b":1,"candidate":[-17,4],"domain":"Z2_fixed_basis","operation":"vector-candidate-residual","semantic_template":"qf-density16-vector-candidate-residual/{\"a\":2,\"b\":1,\"candidate\":[-17,4],\"u\":[-9,1],\"v\":[1,2]}"}},"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":"2026-10-12","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261012-advance-20261016-r01-vector-0080","batch_id":"batch-20261012-advance-20261016-r01-vector-foundations","language":"zh-Hant","source_mko_ids":["mko-vector"],"title_zh":"加權和候選的分量殘差","stem_zh":"固定標準基底的整數二分量向量。u=[-8, 1]、v=[1, 2]、a=2、b=1、候選=[-14, 4]。求 (au+bv, 候選減真值, 候選相等資格)。","answer":{"kind":"expression","value":"([-15, 4], [1, 0], 否)","explanation_zh":"有限結構核對=([-15, 4], [1, 0], 否)。先算兩個真值分量，候選的兩個殘差都零才相等。有限精確核對不是一般形式證明。逐分量不當內積或長度。"},"difficulty":{"band":"upper_secondary","level":1,"skills":["加權和候選的分量殘差"]},"variant":{"family_id":"qf-density16-vector-candidate-residual","seed":202610161080,"parameters":{"u":[-8,1],"v":[1,2],"a":2,"b":1,"candidate":[-14,4],"domain":"Z2_fixed_basis","operation":"vector-candidate-residual","semantic_template":"qf-density16-vector-candidate-residual/{\"a\":2,\"b\":1,\"candidate\":[-14,4],\"u\":[-8,1],\"v\":[1,2]}"}},"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":"2026-10-12","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261012-advance-20261016-r01-vector-0081","batch_id":"batch-20261012-advance-20261016-r01-vector-foundations","language":"zh-Hant","source_mko_ids":["mko-vector"],"title_zh":"加權和候選的分量殘差","stem_zh":"固定標準基底的整數二分量向量。u=[-7, 1]、v=[1, 2]、a=2、b=1、候選=[-12, 4]。求 (au+bv, 候選減真值, 候選相等資格)。","answer":{"kind":"expression","value":"([-13, 4], [1, 0], 否)","explanation_zh":"有限結構核對=([-13, 4], [1, 0], 否)。先算兩個真值分量，候選的兩個殘差都零才相等。有限精確核對不是一般形式證明。逐分量不當內積或長度。"},"difficulty":{"band":"upper_secondary","level":1,"skills":["加權和候選的分量殘差"]},"variant":{"family_id":"qf-density16-vector-candidate-residual","seed":202610161081,"parameters":{"u":[-7,1],"v":[1,2],"a":2,"b":1,"candidate":[-12,4],"domain":"Z2_fixed_basis","operation":"vector-candidate-residual","semantic_template":"qf-density16-vector-candidate-residual/{\"a\":2,\"b\":1,\"candidate\":[-12,4],\"u\":[-7,1],\"v\":[1,2]}"}},"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":"2026-10-12","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261012-advance-20261016-r01-vector-0082","batch_id":"batch-20261012-advance-20261016-r01-vector-foundations","language":"zh-Hant","source_mko_ids":["mko-vector"],"title_zh":"加權和候選的分量殘差","stem_zh":"固定標準基底的整數二分量向量。u=[-6, 1]、v=[1, 2]、a=2、b=1、候選=[-11, 4]。求 (au+bv, 候選減真值, 候選相等資格)。","answer":{"kind":"expression","value":"([-11, 4], [0, 0], 是)","explanation_zh":"有限結構核對=([-11, 4], [0, 0], 是)。先算兩個真值分量，候選的兩個殘差都零才相等。有限精確核對不是一般形式證明。逐分量不當內積或長度。"},"difficulty":{"band":"upper_secondary","level":1,"skills":["加權和候選的分量殘差"]},"variant":{"family_id":"qf-density16-vector-candidate-residual","seed":202610161082,"parameters":{"u":[-6,1],"v":[1,2],"a":2,"b":1,"candidate":[-11,4],"domain":"Z2_fixed_basis","operation":"vector-candidate-residual","semantic_template":"qf-density16-vector-candidate-residual/{\"a\":2,\"b\":1,\"candidate\":[-11,4],\"u\":[-6,1],\"v\":[1,2]}"}},"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":"2026-10-12","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261012-advance-20261016-r01-vector-0083","batch_id":"batch-20261012-advance-20261016-r01-vector-foundations","language":"zh-Hant","source_mko_ids":["mko-vector"],"title_zh":"加權和候選的分量殘差","stem_zh":"固定標準基底的整數二分量向量。u=[-5, 1]、v=[1, 2]、a=2、b=1、候選=[-8, 4]。求 (au+bv, 候選減真值, 候選相等資格)。","answer":{"kind":"expression","value":"([-9, 4], [1, 0], 否)","explanation_zh":"有限結構核對=([-9, 4], [1, 0], 否)。先算兩個真值分量，候選的兩個殘差都零才相等。有限精確核對不是一般形式證明。逐分量不當內積或長度。"},"difficulty":{"band":"upper_secondary","level":1,"skills":["加權和候選的分量殘差"]},"variant":{"family_id":"qf-density16-vector-candidate-residual","seed":202610161083,"parameters":{"u":[-5,1],"v":[1,2],"a":2,"b":1,"candidate":[-8,4],"domain":"Z2_fixed_basis","operation":"vector-candidate-residual","semantic_template":"qf-density16-vector-candidate-residual/{\"a\":2,\"b\":1,\"candidate\":[-8,4],\"u\":[-5,1],\"v\":[1,2]}"}},"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":"2026-10-12","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261012-advance-20261016-r01-vector-0084","batch_id":"batch-20261012-advance-20261016-r01-vector-foundations","language":"zh-Hant","source_mko_ids":["mko-vector"],"title_zh":"加權和候選的分量殘差","stem_zh":"固定標準基底的整數二分量向量。u=[-4, 1]、v=[1, 2]、a=2、b=1、候選=[-6, 4]。求 (au+bv, 候選減真值, 候選相等資格)。","answer":{"kind":"expression","value":"([-7, 4], [1, 0], 否)","explanation_zh":"有限結構核對=([-7, 4], [1, 0], 否)。先算兩個真值分量，候選的兩個殘差都零才相等。有限精確核對不是一般形式證明。逐分量不當內積或長度。"},"difficulty":{"band":"upper_secondary","level":2,"skills":["加權和候選的分量殘差"]},"variant":{"family_id":"qf-density16-vector-candidate-residual","seed":202610161084,"parameters":{"u":[-4,1],"v":[1,2],"a":2,"b":1,"candidate":[-6,4],"domain":"Z2_fixed_basis","operation":"vector-candidate-residual","semantic_template":"qf-density16-vector-candidate-residual/{\"a\":2,\"b\":1,\"candidate\":[-6,4],\"u\":[-4,1],\"v\":[1,2]}"}},"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":"2026-10-12","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261012-advance-20261016-r01-vector-0085","batch_id":"batch-20261012-advance-20261016-r01-vector-foundations","language":"zh-Hant","source_mko_ids":["mko-vector"],"title_zh":"加權和候選的分量殘差","stem_zh":"固定標準基底的整數二分量向量。u=[-3, 1]、v=[1, 2]、a=2、b=1、候選=[-5, 4]。求 (au+bv, 候選減真值, 候選相等資格)。","answer":{"kind":"expression","value":"([-5, 4], [0, 0], 是)","explanation_zh":"有限結構核對=([-5, 4], [0, 0], 是)。先算兩個真值分量，候選的兩個殘差都零才相等。有限精確核對不是一般形式證明。逐分量不當內積或長度。"},"difficulty":{"band":"upper_secondary","level":2,"skills":["加權和候選的分量殘差"]},"variant":{"family_id":"qf-density16-vector-candidate-residual","seed":202610161085,"parameters":{"u":[-3,1],"v":[1,2],"a":2,"b":1,"candidate":[-5,4],"domain":"Z2_fixed_basis","operation":"vector-candidate-residual","semantic_template":"qf-density16-vector-candidate-residual/{\"a\":2,\"b\":1,\"candidate\":[-5,4],\"u\":[-3,1],\"v\":[1,2]}"}},"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":"2026-10-12","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261012-advance-20261016-r01-vector-0086","batch_id":"batch-20261012-advance-20261016-r01-vector-foundations","language":"zh-Hant","source_mko_ids":["mko-vector"],"title_zh":"加權和候選的分量殘差","stem_zh":"固定標準基底的整數二分量向量。u=[-2, 1]、v=[1, 2]、a=2、b=1、候選=[-2, 4]。求 (au+bv, 候選減真值, 候選相等資格)。","answer":{"kind":"expression","value":"([-3, 4], [1, 0], 否)","explanation_zh":"有限結構核對=([-3, 4], [1, 0], 否)。先算兩個真值分量，候選的兩個殘差都零才相等。有限精確核對不是一般形式證明。逐分量不當內積或長度。"},"difficulty":{"band":"upper_secondary","level":2,"skills":["加權和候選的分量殘差"]},"variant":{"family_id":"qf-density16-vector-candidate-residual","seed":202610161086,"parameters":{"u":[-2,1],"v":[1,2],"a":2,"b":1,"candidate":[-2,4],"domain":"Z2_fixed_basis","operation":"vector-candidate-residual","semantic_template":"qf-density16-vector-candidate-residual/{\"a\":2,\"b\":1,\"candidate\":[-2,4],\"u\":[-2,1],\"v\":[1,2]}"}},"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":"2026-10-12","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261012-advance-20261016-r01-vector-0087","batch_id":"batch-20261012-advance-20261016-r01-vector-foundations","language":"zh-Hant","source_mko_ids":["mko-vector"],"title_zh":"加權和候選的分量殘差","stem_zh":"固定標準基底的整數二分量向量。u=[-1, 1]、v=[1, 2]、a=2、b=1、候選=[0, 4]。求 (au+bv, 候選減真值, 候選相等資格)。","answer":{"kind":"expression","value":"([-1, 4], [1, 0], 否)","explanation_zh":"有限結構核對=([-1, 4], [1, 0], 否)。先算兩個真值分量，候選的兩個殘差都零才相等。有限精確核對不是一般形式證明。逐分量不當內積或長度。"},"difficulty":{"band":"upper_secondary","level":2,"skills":["加權和候選的分量殘差"]},"variant":{"family_id":"qf-density16-vector-candidate-residual","seed":202610161087,"parameters":{"u":[-1,1],"v":[1,2],"a":2,"b":1,"candidate":[0,4],"domain":"Z2_fixed_basis","operation":"vector-candidate-residual","semantic_template":"qf-density16-vector-candidate-residual/{\"a\":2,\"b\":1,\"candidate\":[0,4],\"u\":[-1,1],\"v\":[1,2]}"}},"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":"2026-10-12","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261012-advance-20261016-r01-vector-0088","batch_id":"batch-20261012-advance-20261016-r01-vector-foundations","language":"zh-Hant","source_mko_ids":["mko-vector"],"title_zh":"加權和候選的分量殘差","stem_zh":"固定標準基底的整數二分量向量。u=[0, 1]、v=[1, 2]、a=2、b=1、候選=[1, 4]。求 (au+bv, 候選減真值, 候選相等資格)。","answer":{"kind":"expression","value":"([1, 4], [0, 0], 是)","explanation_zh":"有限結構核對=([1, 4], [0, 0], 是)。先算兩個真值分量，候選的兩個殘差都零才相等。有限精確核對不是一般形式證明。逐分量不當內積或長度。"},"difficulty":{"band":"upper_secondary","level":2,"skills":["加權和候選的分量殘差"]},"variant":{"family_id":"qf-density16-vector-candidate-residual","seed":202610161088,"parameters":{"u":[0,1],"v":[1,2],"a":2,"b":1,"candidate":[1,4],"domain":"Z2_fixed_basis","operation":"vector-candidate-residual","semantic_template":"qf-density16-vector-candidate-residual/{\"a\":2,\"b\":1,\"candidate\":[1,4],\"u\":[0,1],\"v\":[1,2]}"}},"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":"2026-10-12","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261012-advance-20261016-r01-vector-0089","batch_id":"batch-20261012-advance-20261016-r01-vector-foundations","language":"zh-Hant","source_mko_ids":["mko-vector"],"title_zh":"加權和候選的分量殘差","stem_zh":"固定標準基底的整數二分量向量。u=[1, 1]、v=[1, 2]、a=2、b=1、候選=[4, 4]。求 (au+bv, 候選減真值, 候選相等資格)。","answer":{"kind":"expression","value":"([3, 4], [1, 0], 否)","explanation_zh":"有限結構核對=([3, 4], [1, 0], 否)。先算兩個真值分量，候選的兩個殘差都零才相等。有限精確核對不是一般形式證明。逐分量不當內積或長度。"},"difficulty":{"band":"upper_secondary","level":2,"skills":["加權和候選的分量殘差"]},"variant":{"family_id":"qf-density16-vector-candidate-residual","seed":202610161089,"parameters":{"u":[1,1],"v":[1,2],"a":2,"b":1,"candidate":[4,4],"domain":"Z2_fixed_basis","operation":"vector-candidate-residual","semantic_template":"qf-density16-vector-candidate-residual/{\"a\":2,\"b\":1,\"candidate\":[4,4],\"u\":[1,1],\"v\":[1,2]}"}},"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":"2026-10-12","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261012-advance-20261016-r01-vector-0090","batch_id":"batch-20261012-advance-20261016-r01-vector-foundations","language":"zh-Hant","source_mko_ids":["mko-vector"],"title_zh":"加權和候選的分量殘差","stem_zh":"固定標準基底的整數二分量向量。u=[2, 1]、v=[1, 2]、a=2、b=1、候選=[6, 4]。求 (au+bv, 候選減真值, 候選相等資格)。","answer":{"kind":"expression","value":"([5, 4], [1, 0], 否)","explanation_zh":"有限結構核對=([5, 4], [1, 0], 否)。先算兩個真值分量，候選的兩個殘差都零才相等。有限精確核對不是一般形式證明。逐分量不當內積或長度。"},"difficulty":{"band":"upper_secondary","level":2,"skills":["加權和候選的分量殘差"]},"variant":{"family_id":"qf-density16-vector-candidate-residual","seed":202610161090,"parameters":{"u":[2,1],"v":[1,2],"a":2,"b":1,"candidate":[6,4],"domain":"Z2_fixed_basis","operation":"vector-candidate-residual","semantic_template":"qf-density16-vector-candidate-residual/{\"a\":2,\"b\":1,\"candidate\":[6,4],\"u\":[2,1],\"v\":[1,2]}"}},"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":"2026-10-12","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261012-advance-20261016-r01-vector-0091","batch_id":"batch-20261012-advance-20261016-r01-vector-foundations","language":"zh-Hant","source_mko_ids":["mko-vector"],"title_zh":"加權和候選的分量殘差","stem_zh":"固定標準基底的整數二分量向量。u=[3, 1]、v=[1, 2]、a=2、b=1、候選=[7, 4]。求 (au+bv, 候選減真值, 候選相等資格)。","answer":{"kind":"expression","value":"([7, 4], [0, 0], 是)","explanation_zh":"有限結構核對=([7, 4], [0, 0], 是)。先算兩個真值分量，候選的兩個殘差都零才相等。有限精確核對不是一般形式證明。逐分量不當內積或長度。"},"difficulty":{"band":"upper_secondary","level":2,"skills":["加權和候選的分量殘差"]},"variant":{"family_id":"qf-density16-vector-candidate-residual","seed":202610161091,"parameters":{"u":[3,1],"v":[1,2],"a":2,"b":1,"candidate":[7,4],"domain":"Z2_fixed_basis","operation":"vector-candidate-residual","semantic_template":"qf-density16-vector-candidate-residual/{\"a\":2,\"b\":1,\"candidate\":[7,4],\"u\":[3,1],\"v\":[1,2]}"}},"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":"2026-10-12","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261012-advance-20261016-r01-vector-0092","batch_id":"batch-20261012-advance-20261016-r01-vector-foundations","language":"zh-Hant","source_mko_ids":["mko-vector"],"title_zh":"加權和候選的分量殘差","stem_zh":"固定標準基底的整數二分量向量。u=[4, 1]、v=[1, 2]、a=2、b=1、候選=[10, 4]。求 (au+bv, 候選減真值, 候選相等資格)。","answer":{"kind":"expression","value":"([9, 4], [1, 0], 否)","explanation_zh":"有限結構核對=([9, 4], [1, 0], 否)。先算兩個真值分量，候選的兩個殘差都零才相等。有限精確核對不是一般形式證明。逐分量不當內積或長度。"},"difficulty":{"band":"upper_secondary","level":2,"skills":["加權和候選的分量殘差"]},"variant":{"family_id":"qf-density16-vector-candidate-residual","seed":202610161092,"parameters":{"u":[4,1],"v":[1,2],"a":2,"b":1,"candidate":[10,4],"domain":"Z2_fixed_basis","operation":"vector-candidate-residual","semantic_template":"qf-density16-vector-candidate-residual/{\"a\":2,\"b\":1,\"candidate\":[10,4],\"u\":[4,1],\"v\":[1,2]}"}},"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":"2026-10-12","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261012-advance-20261016-r01-vector-0093","batch_id":"batch-20261012-advance-20261016-r01-vector-foundations","language":"zh-Hant","source_mko_ids":["mko-vector"],"title_zh":"加權和候選的分量殘差","stem_zh":"固定標準基底的整數二分量向量。u=[5, 1]、v=[1, 2]、a=2、b=1、候選=[12, 4]。求 (au+bv, 候選減真值, 候選相等資格)。","answer":{"kind":"expression","value":"([11, 4], [1, 0], 否)","explanation_zh":"有限結構核對=([11, 4], [1, 0], 否)。先算兩個真值分量，候選的兩個殘差都零才相等。有限精確核對不是一般形式證明。逐分量不當內積或長度。"},"difficulty":{"band":"upper_secondary","level":3,"skills":["加權和候選的分量殘差"]},"variant":{"family_id":"qf-density16-vector-candidate-residual","seed":202610161093,"parameters":{"u":[5,1],"v":[1,2],"a":2,"b":1,"candidate":[12,4],"domain":"Z2_fixed_basis","operation":"vector-candidate-residual","semantic_template":"qf-density16-vector-candidate-residual/{\"a\":2,\"b\":1,\"candidate\":[12,4],\"u\":[5,1],\"v\":[1,2]}"}},"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":"2026-10-12","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261012-advance-20261016-r01-vector-0094","batch_id":"batch-20261012-advance-20261016-r01-vector-foundations","language":"zh-Hant","source_mko_ids":["mko-vector"],"title_zh":"加權和候選的分量殘差","stem_zh":"固定標準基底的整數二分量向量。u=[6, 1]、v=[1, 2]、a=2、b=1、候選=[13, 4]。求 (au+bv, 候選減真值, 候選相等資格)。","answer":{"kind":"expression","value":"([13, 4], [0, 0], 是)","explanation_zh":"有限結構核對=([13, 4], [0, 0], 是)。先算兩個真值分量，候選的兩個殘差都零才相等。有限精確核對不是一般形式證明。逐分量不當內積或長度。"},"difficulty":{"band":"upper_secondary","level":3,"skills":["加權和候選的分量殘差"]},"variant":{"family_id":"qf-density16-vector-candidate-residual","seed":202610161094,"parameters":{"u":[6,1],"v":[1,2],"a":2,"b":1,"candidate":[13,4],"domain":"Z2_fixed_basis","operation":"vector-candidate-residual","semantic_template":"qf-density16-vector-candidate-residual/{\"a\":2,\"b\":1,\"candidate\":[13,4],\"u\":[6,1],\"v\":[1,2]}"}},"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":"2026-10-12","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261012-advance-20261016-r01-vector-0095","batch_id":"batch-20261012-advance-20261016-r01-vector-foundations","language":"zh-Hant","source_mko_ids":["mko-vector"],"title_zh":"加權和候選的分量殘差","stem_zh":"固定標準基底的整數二分量向量。u=[7, 1]、v=[1, 2]、a=2、b=1、候選=[16, 4]。求 (au+bv, 候選減真值, 候選相等資格)。","answer":{"kind":"expression","value":"([15, 4], [1, 0], 否)","explanation_zh":"有限結構核對=([15, 4], [1, 0], 否)。先算兩個真值分量，候選的兩個殘差都零才相等。有限精確核對不是一般形式證明。逐分量不當內積或長度。"},"difficulty":{"band":"upper_secondary","level":3,"skills":["加權和候選的分量殘差"]},"variant":{"family_id":"qf-density16-vector-candidate-residual","seed":202610161095,"parameters":{"u":[7,1],"v":[1,2],"a":2,"b":1,"candidate":[16,4],"domain":"Z2_fixed_basis","operation":"vector-candidate-residual","semantic_template":"qf-density16-vector-candidate-residual/{\"a\":2,\"b\":1,\"candidate\":[16,4],\"u\":[7,1],\"v\":[1,2]}"}},"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":"2026-10-12","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261012-advance-20261016-r01-vector-0096","batch_id":"batch-20261012-advance-20261016-r01-vector-foundations","language":"zh-Hant","source_mko_ids":["mko-vector"],"title_zh":"加權和候選的分量殘差","stem_zh":"固定標準基底的整數二分量向量。u=[8, 1]、v=[1, 2]、a=2、b=1、候選=[18, 4]。求 (au+bv, 候選減真值, 候選相等資格)。","answer":{"kind":"expression","value":"([17, 4], [1, 0], 否)","explanation_zh":"有限結構核對=([17, 4], [1, 0], 否)。先算兩個真值分量，候選的兩個殘差都零才相等。有限精確核對不是一般形式證明。逐分量不當內積或長度。"},"difficulty":{"band":"upper_secondary","level":3,"skills":["加權和候選的分量殘差"]},"variant":{"family_id":"qf-density16-vector-candidate-residual","seed":202610161096,"parameters":{"u":[8,1],"v":[1,2],"a":2,"b":1,"candidate":[18,4],"domain":"Z2_fixed_basis","operation":"vector-candidate-residual","semantic_template":"qf-density16-vector-candidate-residual/{\"a\":2,\"b\":1,\"candidate\":[18,4],\"u\":[8,1],\"v\":[1,2]}"}},"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":"2026-10-12","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261012-advance-20261016-r01-vector-0097","batch_id":"batch-20261012-advance-20261016-r01-vector-foundations","language":"zh-Hant","source_mko_ids":["mko-vector"],"title_zh":"加權和候選的分量殘差","stem_zh":"固定標準基底的整數二分量向量。u=[9, 1]、v=[1, 2]、a=2、b=1、候選=[19, 4]。求 (au+bv, 候選減真值, 候選相等資格)。","answer":{"kind":"expression","value":"([19, 4], [0, 0], 是)","explanation_zh":"有限結構核對=([19, 4], [0, 0], 是)。先算兩個真值分量，候選的兩個殘差都零才相等。有限精確核對不是一般形式證明。逐分量不當內積或長度。"},"difficulty":{"band":"upper_secondary","level":3,"skills":["加權和候選的分量殘差"]},"variant":{"family_id":"qf-density16-vector-candidate-residual","seed":202610161097,"parameters":{"u":[9,1],"v":[1,2],"a":2,"b":1,"candidate":[19,4],"domain":"Z2_fixed_basis","operation":"vector-candidate-residual","semantic_template":"qf-density16-vector-candidate-residual/{\"a\":2,\"b\":1,\"candidate\":[19,4],\"u\":[9,1],\"v\":[1,2]}"}},"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":"2026-10-12","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261012-advance-20261016-r01-vector-0098","batch_id":"batch-20261012-advance-20261016-r01-vector-foundations","language":"zh-Hant","source_mko_ids":["mko-vector"],"title_zh":"加權和候選的分量殘差","stem_zh":"固定標準基底的整數二分量向量。u=[10, 1]、v=[1, 2]、a=2、b=1、候選=[22, 4]。求 (au+bv, 候選減真值, 候選相等資格)。","answer":{"kind":"expression","value":"([21, 4], [1, 0], 否)","explanation_zh":"有限結構核對=([21, 4], [1, 0], 否)。先算兩個真值分量，候選的兩個殘差都零才相等。有限精確核對不是一般形式證明。逐分量不當內積或長度。"},"difficulty":{"band":"upper_secondary","level":3,"skills":["加權和候選的分量殘差"]},"variant":{"family_id":"qf-density16-vector-candidate-residual","seed":202610161098,"parameters":{"u":[10,1],"v":[1,2],"a":2,"b":1,"candidate":[22,4],"domain":"Z2_fixed_basis","operation":"vector-candidate-residual","semantic_template":"qf-density16-vector-candidate-residual/{\"a\":2,\"b\":1,\"candidate\":[22,4],\"u\":[10,1],\"v\":[1,2]}"}},"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":"2026-10-12","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261012-advance-20261016-r01-vector-0099","batch_id":"batch-20261012-advance-20261016-r01-vector-foundations","language":"zh-Hant","source_mko_ids":["mko-vector"],"title_zh":"加權和候選的分量殘差","stem_zh":"固定標準基底的整數二分量向量。u=[11, 1]、v=[1, 2]、a=2、b=1、候選=[24, 4]。求 (au+bv, 候選減真值, 候選相等資格)。","answer":{"kind":"expression","value":"([23, 4], [1, 0], 否)","explanation_zh":"有限結構核對=([23, 4], [1, 0], 否)。先算兩個真值分量，候選的兩個殘差都零才相等。有限精確核對不是一般形式證明。逐分量不當內積或長度。"},"difficulty":{"band":"upper_secondary","level":3,"skills":["加權和候選的分量殘差"]},"variant":{"family_id":"qf-density16-vector-candidate-residual","seed":202610161099,"parameters":{"u":[11,1],"v":[1,2],"a":2,"b":1,"candidate":[24,4],"domain":"Z2_fixed_basis","operation":"vector-candidate-residual","semantic_template":"qf-density16-vector-candidate-residual/{\"a\":2,\"b\":1,\"candidate\":[24,4],\"u\":[11,1],\"v\":[1,2]}"}},"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":"2026-10-12","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261012-advance-20261016-r01-vector-0100","batch_id":"batch-20261012-advance-20261016-r01-vector-foundations","language":"zh-Hant","source_mko_ids":["mko-vector"],"title_zh":"加權和候選的分量殘差","stem_zh":"固定標準基底的整數二分量向量。u=[12, 1]、v=[1, 2]、a=2、b=1、候選=[25, 4]。求 (au+bv, 候選減真值, 候選相等資格)。","answer":{"kind":"expression","value":"([25, 4], [0, 0], 是)","explanation_zh":"有限結構核對=([25, 4], [0, 0], 是)。先算兩個真值分量，候選的兩個殘差都零才相等。有限精確核對不是一般形式證明。逐分量不當內積或長度。"},"difficulty":{"band":"upper_secondary","level":3,"skills":["加權和候選的分量殘差"]},"variant":{"family_id":"qf-density16-vector-candidate-residual","seed":202610161100,"parameters":{"u":[12,1],"v":[1,2],"a":2,"b":1,"candidate":[25,4],"domain":"Z2_fixed_basis","operation":"vector-candidate-residual","semantic_template":"qf-density16-vector-candidate-residual/{\"a\":2,\"b\":1,\"candidate\":[25,4],\"u\":[12,1],\"v\":[1,2]}"}},"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":"2026-10-12","contract_version":"ocme-question-factory-v0.1"}}