{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261102-r01-vector-space-dimension-0001","batch_id":"batch-20261014-advance-20261102-r01-vector-space-dimension-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"3a841752e201f2ca796262810577b9f8495c985ebfc6143df26e582a4f89391f","family_id":"qf-mod33-dimension-basis-audit","seed":202611021001,"semantic_template":"qf-mod33-dimension-basis-audit/{\"dimension\":2,\"vectors\":[[0,1],[1,0]]}"},"fields":{"title":"A Basis Requires Both Independence and Spanning","stem":"Use the complete ordered real coordinate family vectors=[[1, 0], [0, 1]] in ambient R^2, limited to dimensions 2 or 3. Treat vectors as columns and calculate rank by exact rational Gaussian elimination, not floating-point samples. A basis is both independent and spanning. An independent family gives only a lower bound; a spanning family gives only an upper bound. Basis-cardinality invariance uses the finite-dimensional theorem, not a few examples. Give (ambient_dimension,vector_count,rank,independent?,spanning?,basis?).","answer":{"kind":"expression","value":"(2, 2, 2, Yes, Yes, Yes)","explanation":"Compare rank with the number of vectors for independence and with the ambient coordinate dimension for spanning. Conjoin both to qualify the supplied family as a basis; neither property alone makes its size the dimension. The exact result, retaining the requested order, is (2, 2, 2, Yes, Yes, Yes)."},"skills":["A Basis Requires Both Independence and Spanning"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261102-r01-vector-space-dimension-0002","batch_id":"batch-20261014-advance-20261102-r01-vector-space-dimension-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"c32897e46112c5eeb90a5c08b086d3c2f11318b26681838e56148b23cbe68c08","family_id":"qf-mod33-dimension-basis-audit","seed":202611021002,"semantic_template":"qf-mod33-dimension-basis-audit/{\"dimension\":2,\"vectors\":[[1,0],[1,1]]}"},"fields":{"title":"A Basis Requires Both Independence and Spanning","stem":"Use the complete ordered real coordinate family vectors=[[1, 0], [1, 1]] in ambient R^2, limited to dimensions 2 or 3. Treat vectors as columns and calculate rank by exact rational Gaussian elimination, not floating-point samples. A basis is both independent and spanning. An independent family gives only a lower bound; a spanning family gives only an upper bound. Basis-cardinality invariance uses the finite-dimensional theorem, not a few examples. Give (ambient_dimension,vector_count,rank,independent?,spanning?,basis?).","answer":{"kind":"expression","value":"(2, 2, 2, Yes, Yes, Yes)","explanation":"Compare rank with the number of vectors for independence and with the ambient coordinate dimension for spanning. Conjoin both to qualify the supplied family as a basis; neither property alone makes its size the dimension. The exact result, retaining the requested order, is (2, 2, 2, Yes, Yes, Yes)."},"skills":["A Basis Requires Both Independence and Spanning"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261102-r01-vector-space-dimension-0003","batch_id":"batch-20261014-advance-20261102-r01-vector-space-dimension-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"e9a4d6fb7ce56b749d9ec1691f46e725d3c41cbbc06df849d6b9a9f5f1947d24","family_id":"qf-mod33-dimension-basis-audit","seed":202611021003,"semantic_template":"qf-mod33-dimension-basis-audit/{\"dimension\":2,\"vectors\":[[1,0],[2,1]]}"},"fields":{"title":"A Basis Requires Both Independence and Spanning","stem":"Use the complete ordered real coordinate family vectors=[[1, 0], [2, 1]] in ambient R^2, limited to dimensions 2 or 3. Treat vectors as columns and calculate rank by exact rational Gaussian elimination, not floating-point samples. A basis is both independent and spanning. An independent family gives only a lower bound; a spanning family gives only an upper bound. Basis-cardinality invariance uses the finite-dimensional theorem, not a few examples. Give (ambient_dimension,vector_count,rank,independent?,spanning?,basis?).","answer":{"kind":"expression","value":"(2, 2, 2, Yes, Yes, Yes)","explanation":"Compare rank with the number of vectors for independence and with the ambient coordinate dimension for spanning. Conjoin both to qualify the supplied family as a basis; neither property alone makes its size the dimension. The exact result, retaining the requested order, is (2, 2, 2, Yes, Yes, Yes)."},"skills":["A Basis Requires Both Independence and Spanning"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261102-r01-vector-space-dimension-0004","batch_id":"batch-20261014-advance-20261102-r01-vector-space-dimension-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"b6eccf083c2fa1a9b674f24abd7c945b9e142bb8762ffce7c81f0d6f103485fc","family_id":"qf-mod33-dimension-basis-audit","seed":202611021004,"semantic_template":"qf-mod33-dimension-basis-audit/{\"dimension\":2,\"vectors\":[[1,0],[3,1]]}"},"fields":{"title":"A Basis Requires Both Independence and Spanning","stem":"Use the complete ordered real coordinate family vectors=[[1, 0], [3, 1]] in ambient R^2, limited to dimensions 2 or 3. Treat vectors as columns and calculate rank by exact rational Gaussian elimination, not floating-point samples. A basis is both independent and spanning. An independent family gives only a lower bound; a spanning family gives only an upper bound. Basis-cardinality invariance uses the finite-dimensional theorem, not a few examples. Give (ambient_dimension,vector_count,rank,independent?,spanning?,basis?).","answer":{"kind":"expression","value":"(2, 2, 2, Yes, Yes, Yes)","explanation":"Compare rank with the number of vectors for independence and with the ambient coordinate dimension for spanning. Conjoin both to qualify the supplied family as a basis; neither property alone makes its size the dimension. The exact result, retaining the requested order, is (2, 2, 2, Yes, Yes, Yes)."},"skills":["A Basis Requires Both Independence and Spanning"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261102-r01-vector-space-dimension-0005","batch_id":"batch-20261014-advance-20261102-r01-vector-space-dimension-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"f460d81d2d7960dbd5732f67229132860363f13b51f7fb0e593410b795b781df","family_id":"qf-mod33-dimension-basis-audit","seed":202611021005,"semantic_template":"qf-mod33-dimension-basis-audit/{\"dimension\":2,\"vectors\":[[1,0],[4,1]]}"},"fields":{"title":"A Basis Requires Both Independence and Spanning","stem":"Use the complete ordered real coordinate family vectors=[[1, 0], [4, 1]] in ambient R^2, limited to dimensions 2 or 3. Treat vectors as columns and calculate rank by exact rational Gaussian elimination, not floating-point samples. A basis is both independent and spanning. An independent family gives only a lower bound; a spanning family gives only an upper bound. Basis-cardinality invariance uses the finite-dimensional theorem, not a few examples. Give (ambient_dimension,vector_count,rank,independent?,spanning?,basis?).","answer":{"kind":"expression","value":"(2, 2, 2, Yes, Yes, Yes)","explanation":"Compare rank with the number of vectors for independence and with the ambient coordinate dimension for spanning. Conjoin both to qualify the supplied family as a basis; neither property alone makes its size the dimension. The exact result, retaining the requested order, is (2, 2, 2, Yes, Yes, Yes)."},"skills":["A Basis Requires Both Independence and Spanning"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261102-r01-vector-space-dimension-0006","batch_id":"batch-20261014-advance-20261102-r01-vector-space-dimension-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"8e937a0ac795cdcbd38d717adb63e9fa9f3efbcad9a8600d29ce0773dd701b36","family_id":"qf-mod33-dimension-basis-audit","seed":202611021006,"semantic_template":"qf-mod33-dimension-basis-audit/{\"dimension\":2,\"vectors\":[[1,0],[5,1]]}"},"fields":{"title":"A Basis Requires Both Independence and Spanning","stem":"Use the complete ordered real coordinate family vectors=[[1, 0], [5, 1]] in ambient R^2, limited to dimensions 2 or 3. Treat vectors as columns and calculate rank by exact rational Gaussian elimination, not floating-point samples. A basis is both independent and spanning. An independent family gives only a lower bound; a spanning family gives only an upper bound. Basis-cardinality invariance uses the finite-dimensional theorem, not a few examples. Give (ambient_dimension,vector_count,rank,independent?,spanning?,basis?).","answer":{"kind":"expression","value":"(2, 2, 2, Yes, Yes, Yes)","explanation":"Compare rank with the number of vectors for independence and with the ambient coordinate dimension for spanning. Conjoin both to qualify the supplied family as a basis; neither property alone makes its size the dimension. The exact result, retaining the requested order, is (2, 2, 2, Yes, Yes, Yes)."},"skills":["A Basis Requires Both Independence and Spanning"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261102-r01-vector-space-dimension-0007","batch_id":"batch-20261014-advance-20261102-r01-vector-space-dimension-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"5e5c024fb8c8bcc63643e9e0cc34da21968d8df5b4ea53ec155e6118572c1c31","family_id":"qf-mod33-dimension-basis-audit","seed":202611021007,"semantic_template":"qf-mod33-dimension-basis-audit/{\"dimension\":2,\"vectors\":[[1,0],[6,1]]}"},"fields":{"title":"A Basis Requires Both Independence and Spanning","stem":"Use the complete ordered real coordinate family vectors=[[1, 0], [6, 1]] in ambient R^2, limited to dimensions 2 or 3. Treat vectors as columns and calculate rank by exact rational Gaussian elimination, not floating-point samples. A basis is both independent and spanning. An independent family gives only a lower bound; a spanning family gives only an upper bound. Basis-cardinality invariance uses the finite-dimensional theorem, not a few examples. Give (ambient_dimension,vector_count,rank,independent?,spanning?,basis?).","answer":{"kind":"expression","value":"(2, 2, 2, Yes, Yes, Yes)","explanation":"Compare rank with the number of vectors for independence and with the ambient coordinate dimension for spanning. Conjoin both to qualify the supplied family as a basis; neither property alone makes its size the dimension. The exact result, retaining the requested order, is (2, 2, 2, Yes, Yes, Yes)."},"skills":["A Basis Requires Both Independence and Spanning"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261102-r01-vector-space-dimension-0008","batch_id":"batch-20261014-advance-20261102-r01-vector-space-dimension-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"6ce061387bc1042838c4164dff661afe3c22edef7cf8c110bca997a1b9191619","family_id":"qf-mod33-dimension-basis-audit","seed":202611021008,"semantic_template":"qf-mod33-dimension-basis-audit/{\"dimension\":2,\"vectors\":[[1,0],[7,1]]}"},"fields":{"title":"A Basis Requires Both Independence and Spanning","stem":"Use the complete ordered real coordinate family vectors=[[1, 0], [7, 1]] in ambient R^2, limited to dimensions 2 or 3. Treat vectors as columns and calculate rank by exact rational Gaussian elimination, not floating-point samples. A basis is both independent and spanning. An independent family gives only a lower bound; a spanning family gives only an upper bound. Basis-cardinality invariance uses the finite-dimensional theorem, not a few examples. Give (ambient_dimension,vector_count,rank,independent?,spanning?,basis?).","answer":{"kind":"expression","value":"(2, 2, 2, Yes, Yes, Yes)","explanation":"Compare rank with the number of vectors for independence and with the ambient coordinate dimension for spanning. Conjoin both to qualify the supplied family as a basis; neither property alone makes its size the dimension. The exact result, retaining the requested order, is (2, 2, 2, Yes, Yes, Yes)."},"skills":["A Basis Requires Both Independence and Spanning"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261102-r01-vector-space-dimension-0009","batch_id":"batch-20261014-advance-20261102-r01-vector-space-dimension-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"1a2740bf9e31b02169c877ab5ebc1aec25c352ab03a38761cf678e6ce8070c2e","family_id":"qf-mod33-dimension-basis-audit","seed":202611021009,"semantic_template":"qf-mod33-dimension-basis-audit/{\"dimension\":2,\"vectors\":[[1,0],[8,1]]}"},"fields":{"title":"A Basis Requires Both Independence and Spanning","stem":"Use the complete ordered real coordinate family vectors=[[1, 0], [8, 1]] in ambient R^2, limited to dimensions 2 or 3. Treat vectors as columns and calculate rank by exact rational Gaussian elimination, not floating-point samples. A basis is both independent and spanning. An independent family gives only a lower bound; a spanning family gives only an upper bound. Basis-cardinality invariance uses the finite-dimensional theorem, not a few examples. Give (ambient_dimension,vector_count,rank,independent?,spanning?,basis?).","answer":{"kind":"expression","value":"(2, 2, 2, Yes, Yes, Yes)","explanation":"Compare rank with the number of vectors for independence and with the ambient coordinate dimension for spanning. Conjoin both to qualify the supplied family as a basis; neither property alone makes its size the dimension. The exact result, retaining the requested order, is (2, 2, 2, Yes, Yes, Yes)."},"skills":["A Basis Requires Both Independence and Spanning"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261102-r01-vector-space-dimension-0010","batch_id":"batch-20261014-advance-20261102-r01-vector-space-dimension-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"0da782f0d5106bb9ef95609aa484766f1c291e4b1a6ed5ebbf4a31c08f5e82c9","family_id":"qf-mod33-dimension-basis-audit","seed":202611021010,"semantic_template":"qf-mod33-dimension-basis-audit/{\"dimension\":2,\"vectors\":[[1,0],[9,1]]}"},"fields":{"title":"A Basis Requires Both Independence and Spanning","stem":"Use the complete ordered real coordinate family vectors=[[1, 0], [9, 1]] in ambient R^2, limited to dimensions 2 or 3. Treat vectors as columns and calculate rank by exact rational Gaussian elimination, not floating-point samples. A basis is both independent and spanning. An independent family gives only a lower bound; a spanning family gives only an upper bound. Basis-cardinality invariance uses the finite-dimensional theorem, not a few examples. Give (ambient_dimension,vector_count,rank,independent?,spanning?,basis?).","answer":{"kind":"expression","value":"(2, 2, 2, Yes, Yes, Yes)","explanation":"Compare rank with the number of vectors for independence and with the ambient coordinate dimension for spanning. Conjoin both to qualify the supplied family as a basis; neither property alone makes its size the dimension. The exact result, retaining the requested order, is (2, 2, 2, Yes, Yes, Yes)."},"skills":["A Basis Requires Both Independence and Spanning"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261102-r01-vector-space-dimension-0011","batch_id":"batch-20261014-advance-20261102-r01-vector-space-dimension-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"3e762429e92eb692f28fc5cb08b5d44066d5f928107c829e7bcbcd34e2b8b9ae","family_id":"qf-mod33-dimension-basis-audit","seed":202611021011,"semantic_template":"qf-mod33-dimension-basis-audit/{\"dimension\":2,\"vectors\":[[1,0],[10,1]]}"},"fields":{"title":"A Basis Requires Both Independence and Spanning","stem":"Use the complete ordered real coordinate family vectors=[[1, 0], [10, 1]] in ambient R^2, limited to dimensions 2 or 3. Treat vectors as columns and calculate rank by exact rational Gaussian elimination, not floating-point samples. A basis is both independent and spanning. An independent family gives only a lower bound; a spanning family gives only an upper bound. Basis-cardinality invariance uses the finite-dimensional theorem, not a few examples. Give (ambient_dimension,vector_count,rank,independent?,spanning?,basis?).","answer":{"kind":"expression","value":"(2, 2, 2, Yes, Yes, Yes)","explanation":"Compare rank with the number of vectors for independence and with the ambient coordinate dimension for spanning. Conjoin both to qualify the supplied family as a basis; neither property alone makes its size the dimension. The exact result, retaining the requested order, is (2, 2, 2, Yes, Yes, Yes)."},"skills":["A Basis Requires Both Independence and Spanning"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261102-r01-vector-space-dimension-0012","batch_id":"batch-20261014-advance-20261102-r01-vector-space-dimension-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"935ea6f4f36f9348732bd6e998e1a10d3b0ea0b2ffafa5a4c4c0d36d05f2b26d","family_id":"qf-mod33-dimension-basis-audit","seed":202611021012,"semantic_template":"qf-mod33-dimension-basis-audit/{\"dimension\":2,\"vectors\":[[1,0],[11,1]]}"},"fields":{"title":"A Basis Requires Both Independence and Spanning","stem":"Use the complete ordered real coordinate family vectors=[[1, 0], [11, 1]] in ambient R^2, limited to dimensions 2 or 3. Treat vectors as columns and calculate rank by exact rational Gaussian elimination, not floating-point samples. A basis is both independent and spanning. An independent family gives only a lower bound; a spanning family gives only an upper bound. Basis-cardinality invariance uses the finite-dimensional theorem, not a few examples. Give (ambient_dimension,vector_count,rank,independent?,spanning?,basis?).","answer":{"kind":"expression","value":"(2, 2, 2, Yes, Yes, Yes)","explanation":"Compare rank with the number of vectors for independence and with the ambient coordinate dimension for spanning. Conjoin both to qualify the supplied family as a basis; neither property alone makes its size the dimension. The exact result, retaining the requested order, is (2, 2, 2, Yes, Yes, Yes)."},"skills":["A Basis Requires Both Independence and Spanning"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261102-r01-vector-space-dimension-0013","batch_id":"batch-20261014-advance-20261102-r01-vector-space-dimension-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"7884b93ba64c21bc9d37cbc27d5a8b37e3f131e9899e9e8e8bb2a1202b203c56","family_id":"qf-mod33-dimension-basis-audit","seed":202611021013,"semantic_template":"qf-mod33-dimension-basis-audit/{\"dimension\":2,\"vectors\":[[1,0],[12,1]]}"},"fields":{"title":"A Basis Requires Both Independence and Spanning","stem":"Use the complete ordered real coordinate family vectors=[[1, 0], [12, 1]] in ambient R^2, limited to dimensions 2 or 3. Treat vectors as columns and calculate rank by exact rational Gaussian elimination, not floating-point samples. A basis is both independent and spanning. An independent family gives only a lower bound; a spanning family gives only an upper bound. Basis-cardinality invariance uses the finite-dimensional theorem, not a few examples. Give (ambient_dimension,vector_count,rank,independent?,spanning?,basis?).","answer":{"kind":"expression","value":"(2, 2, 2, Yes, Yes, Yes)","explanation":"Compare rank with the number of vectors for independence and with the ambient coordinate dimension for spanning. Conjoin both to qualify the supplied family as a basis; neither property alone makes its size the dimension. The exact result, retaining the requested order, is (2, 2, 2, Yes, Yes, Yes)."},"skills":["A Basis Requires Both Independence and Spanning"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261102-r01-vector-space-dimension-0014","batch_id":"batch-20261014-advance-20261102-r01-vector-space-dimension-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"d5e7e3f65309b3744fc2aa250338c553ef467805eeba0dc15a693d37d3fc097b","family_id":"qf-mod33-dimension-basis-audit","seed":202611021014,"semantic_template":"qf-mod33-dimension-basis-audit/{\"dimension\":2,\"vectors\":[[1,0],[13,1]]}"},"fields":{"title":"A Basis Requires Both Independence and Spanning","stem":"Use the complete ordered real coordinate family vectors=[[1, 0], [13, 1]] in ambient R^2, limited to dimensions 2 or 3. Treat vectors as columns and calculate rank by exact rational Gaussian elimination, not floating-point samples. A basis is both independent and spanning. An independent family gives only a lower bound; a spanning family gives only an upper bound. Basis-cardinality invariance uses the finite-dimensional theorem, not a few examples. Give (ambient_dimension,vector_count,rank,independent?,spanning?,basis?).","answer":{"kind":"expression","value":"(2, 2, 2, Yes, Yes, Yes)","explanation":"Compare rank with the number of vectors for independence and with the ambient coordinate dimension for spanning. Conjoin both to qualify the supplied family as a basis; neither property alone makes its size the dimension. The exact result, retaining the requested order, is (2, 2, 2, Yes, Yes, Yes)."},"skills":["A Basis Requires Both Independence and Spanning"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261102-r01-vector-space-dimension-0015","batch_id":"batch-20261014-advance-20261102-r01-vector-space-dimension-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"6b183edf4710dda8afc3f951f4af7372c4b84552d7bf4073fbcce40bc1784aba","family_id":"qf-mod33-dimension-basis-audit","seed":202611021015,"semantic_template":"qf-mod33-dimension-basis-audit/{\"dimension\":2,\"vectors\":[[1,0],[14,1]]}"},"fields":{"title":"A Basis Requires Both Independence and Spanning","stem":"Use the complete ordered real coordinate family vectors=[[1, 0], [14, 1]] in ambient R^2, limited to dimensions 2 or 3. Treat vectors as columns and calculate rank by exact rational Gaussian elimination, not floating-point samples. A basis is both independent and spanning. An independent family gives only a lower bound; a spanning family gives only an upper bound. Basis-cardinality invariance uses the finite-dimensional theorem, not a few examples. Give (ambient_dimension,vector_count,rank,independent?,spanning?,basis?).","answer":{"kind":"expression","value":"(2, 2, 2, Yes, Yes, Yes)","explanation":"Compare rank with the number of vectors for independence and with the ambient coordinate dimension for spanning. Conjoin both to qualify the supplied family as a basis; neither property alone makes its size the dimension. The exact result, retaining the requested order, is (2, 2, 2, Yes, Yes, Yes)."},"skills":["A Basis Requires Both Independence and Spanning"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261102-r01-vector-space-dimension-0016","batch_id":"batch-20261014-advance-20261102-r01-vector-space-dimension-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"b28a4e0bb40bd17d78a809a060a0cc4f0c68fcc66a95f8a0e85c4d4eebc2a04c","family_id":"qf-mod33-dimension-basis-audit","seed":202611021016,"semantic_template":"qf-mod33-dimension-basis-audit/{\"dimension\":2,\"vectors\":[[1,0],[15,1]]}"},"fields":{"title":"A Basis Requires Both Independence and Spanning","stem":"Use the complete ordered real coordinate family vectors=[[1, 0], [15, 1]] in ambient R^2, limited to dimensions 2 or 3. Treat vectors as columns and calculate rank by exact rational Gaussian elimination, not floating-point samples. A basis is both independent and spanning. An independent family gives only a lower bound; a spanning family gives only an upper bound. Basis-cardinality invariance uses the finite-dimensional theorem, not a few examples. Give (ambient_dimension,vector_count,rank,independent?,spanning?,basis?).","answer":{"kind":"expression","value":"(2, 2, 2, Yes, Yes, Yes)","explanation":"Compare rank with the number of vectors for independence and with the ambient coordinate dimension for spanning. Conjoin both to qualify the supplied family as a basis; neither property alone makes its size the dimension. The exact result, retaining the requested order, is (2, 2, 2, Yes, Yes, Yes)."},"skills":["A Basis Requires Both Independence and Spanning"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261102-r01-vector-space-dimension-0017","batch_id":"batch-20261014-advance-20261102-r01-vector-space-dimension-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"0c203b32bdd43cb84fd0a24253f63d1d5ad02a5dfca4ff295d859e6181110d76","family_id":"qf-mod33-dimension-basis-audit","seed":202611021017,"semantic_template":"qf-mod33-dimension-basis-audit/{\"dimension\":2,\"vectors\":[[1,0],[16,1]]}"},"fields":{"title":"A Basis Requires Both Independence and Spanning","stem":"Use the complete ordered real coordinate family vectors=[[1, 0], [16, 1]] in ambient R^2, limited to dimensions 2 or 3. Treat vectors as columns and calculate rank by exact rational Gaussian elimination, not floating-point samples. A basis is both independent and spanning. An independent family gives only a lower bound; a spanning family gives only an upper bound. Basis-cardinality invariance uses the finite-dimensional theorem, not a few examples. Give (ambient_dimension,vector_count,rank,independent?,spanning?,basis?).","answer":{"kind":"expression","value":"(2, 2, 2, Yes, Yes, Yes)","explanation":"Compare rank with the number of vectors for independence and with the ambient coordinate dimension for spanning. Conjoin both to qualify the supplied family as a basis; neither property alone makes its size the dimension. The exact result, retaining the requested order, is (2, 2, 2, Yes, Yes, Yes)."},"skills":["A Basis Requires Both Independence and Spanning"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261102-r01-vector-space-dimension-0018","batch_id":"batch-20261014-advance-20261102-r01-vector-space-dimension-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"e463a223113476d46d7b7e474d94cab93763451afe07cb1f3f547a7dc0bdbf3b","family_id":"qf-mod33-dimension-basis-audit","seed":202611021018,"semantic_template":"qf-mod33-dimension-basis-audit/{\"dimension\":2,\"vectors\":[[1,0],[17,1]]}"},"fields":{"title":"A Basis Requires Both Independence and Spanning","stem":"Use the complete ordered real coordinate family vectors=[[1, 0], [17, 1]] in ambient R^2, limited to dimensions 2 or 3. Treat vectors as columns and calculate rank by exact rational Gaussian elimination, not floating-point samples. A basis is both independent and spanning. An independent family gives only a lower bound; a spanning family gives only an upper bound. Basis-cardinality invariance uses the finite-dimensional theorem, not a few examples. Give (ambient_dimension,vector_count,rank,independent?,spanning?,basis?).","answer":{"kind":"expression","value":"(2, 2, 2, Yes, Yes, Yes)","explanation":"Compare rank with the number of vectors for independence and with the ambient coordinate dimension for spanning. Conjoin both to qualify the supplied family as a basis; neither property alone makes its size the dimension. The exact result, retaining the requested order, is (2, 2, 2, Yes, Yes, Yes)."},"skills":["A Basis Requires Both Independence and Spanning"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261102-r01-vector-space-dimension-0019","batch_id":"batch-20261014-advance-20261102-r01-vector-space-dimension-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"6dc297871faddccc861900eaff7bb4264b3d2d472ab5d1c262f8f75f98ef3b0f","family_id":"qf-mod33-dimension-basis-audit","seed":202611021019,"semantic_template":"qf-mod33-dimension-basis-audit/{\"dimension\":2,\"vectors\":[[1,0],[18,1]]}"},"fields":{"title":"A Basis Requires Both Independence and Spanning","stem":"Use the complete ordered real coordinate family vectors=[[1, 0], [18, 1]] in ambient R^2, limited to dimensions 2 or 3. Treat vectors as columns and calculate rank by exact rational Gaussian elimination, not floating-point samples. A basis is both independent and spanning. An independent family gives only a lower bound; a spanning family gives only an upper bound. Basis-cardinality invariance uses the finite-dimensional theorem, not a few examples. Give (ambient_dimension,vector_count,rank,independent?,spanning?,basis?).","answer":{"kind":"expression","value":"(2, 2, 2, Yes, Yes, Yes)","explanation":"Compare rank with the number of vectors for independence and with the ambient coordinate dimension for spanning. Conjoin both to qualify the supplied family as a basis; neither property alone makes its size the dimension. The exact result, retaining the requested order, is (2, 2, 2, Yes, Yes, Yes)."},"skills":["A Basis Requires Both Independence and Spanning"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261102-r01-vector-space-dimension-0020","batch_id":"batch-20261014-advance-20261102-r01-vector-space-dimension-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"bc38b6aabb580b009d518f1b1f3237e6d239d4452bb42e4b4554fef750bee8e0","family_id":"qf-mod33-dimension-basis-audit","seed":202611021020,"semantic_template":"qf-mod33-dimension-basis-audit/{\"dimension\":2,\"vectors\":[[1,0],[19,1]]}"},"fields":{"title":"A Basis Requires Both Independence and Spanning","stem":"Use the complete ordered real coordinate family vectors=[[1, 0], [19, 1]] in ambient R^2, limited to dimensions 2 or 3. Treat vectors as columns and calculate rank by exact rational Gaussian elimination, not floating-point samples. A basis is both independent and spanning. An independent family gives only a lower bound; a spanning family gives only an upper bound. Basis-cardinality invariance uses the finite-dimensional theorem, not a few examples. Give (ambient_dimension,vector_count,rank,independent?,spanning?,basis?).","answer":{"kind":"expression","value":"(2, 2, 2, Yes, Yes, Yes)","explanation":"Compare rank with the number of vectors for independence and with the ambient coordinate dimension for spanning. Conjoin both to qualify the supplied family as a basis; neither property alone makes its size the dimension. The exact result, retaining the requested order, is (2, 2, 2, Yes, Yes, Yes)."},"skills":["A Basis Requires Both Independence and Spanning"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261102-r01-vector-space-dimension-0021","batch_id":"batch-20261014-advance-20261102-r01-vector-space-dimension-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"c1f187f650adbc75b1fb901adb7c85c7186af035a9f0a101e56965c2e2d5bfa9","family_id":"qf-mod33-dimension-basis-audit","seed":202611021021,"semantic_template":"qf-mod33-dimension-basis-audit/{\"dimension\":2,\"vectors\":[[1,0],[20,1]]}"},"fields":{"title":"A Basis Requires Both Independence and Spanning","stem":"Use the complete ordered real coordinate family vectors=[[1, 0], [20, 1]] in ambient R^2, limited to dimensions 2 or 3. Treat vectors as columns and calculate rank by exact rational Gaussian elimination, not floating-point samples. A basis is both independent and spanning. An independent family gives only a lower bound; a spanning family gives only an upper bound. Basis-cardinality invariance uses the finite-dimensional theorem, not a few examples. Give (ambient_dimension,vector_count,rank,independent?,spanning?,basis?).","answer":{"kind":"expression","value":"(2, 2, 2, Yes, Yes, Yes)","explanation":"Compare rank with the number of vectors for independence and with the ambient coordinate dimension for spanning. Conjoin both to qualify the supplied family as a basis; neither property alone makes its size the dimension. The exact result, retaining the requested order, is (2, 2, 2, Yes, Yes, Yes)."},"skills":["A Basis Requires Both Independence and Spanning"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261102-r01-vector-space-dimension-0022","batch_id":"batch-20261014-advance-20261102-r01-vector-space-dimension-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"a17eeedc0687d48370765627de02489eb4f6d886d2344e074b486d21e0e58206","family_id":"qf-mod33-dimension-basis-audit","seed":202611021022,"semantic_template":"qf-mod33-dimension-basis-audit/{\"dimension\":2,\"vectors\":[[1,0],[21,1]]}"},"fields":{"title":"A Basis Requires Both Independence and Spanning","stem":"Use the complete ordered real coordinate family vectors=[[1, 0], [21, 1]] in ambient R^2, limited to dimensions 2 or 3. Treat vectors as columns and calculate rank by exact rational Gaussian elimination, not floating-point samples. A basis is both independent and spanning. An independent family gives only a lower bound; a spanning family gives only an upper bound. Basis-cardinality invariance uses the finite-dimensional theorem, not a few examples. Give (ambient_dimension,vector_count,rank,independent?,spanning?,basis?).","answer":{"kind":"expression","value":"(2, 2, 2, Yes, Yes, Yes)","explanation":"Compare rank with the number of vectors for independence and with the ambient coordinate dimension for spanning. Conjoin both to qualify the supplied family as a basis; neither property alone makes its size the dimension. The exact result, retaining the requested order, is (2, 2, 2, Yes, Yes, Yes)."},"skills":["A Basis Requires Both Independence and Spanning"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261102-r01-vector-space-dimension-0023","batch_id":"batch-20261014-advance-20261102-r01-vector-space-dimension-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"e2e532badba0f92f77baa8b4260aae9de851ece27771a2c3861c07b00dddc39d","family_id":"qf-mod33-dimension-basis-audit","seed":202611021023,"semantic_template":"qf-mod33-dimension-basis-audit/{\"dimension\":2,\"vectors\":[[1,0],[22,1]]}"},"fields":{"title":"A Basis Requires Both Independence and Spanning","stem":"Use the complete ordered real coordinate family vectors=[[1, 0], [22, 1]] in ambient R^2, limited to dimensions 2 or 3. Treat vectors as columns and calculate rank by exact rational Gaussian elimination, not floating-point samples. A basis is both independent and spanning. An independent family gives only a lower bound; a spanning family gives only an upper bound. Basis-cardinality invariance uses the finite-dimensional theorem, not a few examples. Give (ambient_dimension,vector_count,rank,independent?,spanning?,basis?).","answer":{"kind":"expression","value":"(2, 2, 2, Yes, Yes, Yes)","explanation":"Compare rank with the number of vectors for independence and with the ambient coordinate dimension for spanning. Conjoin both to qualify the supplied family as a basis; neither property alone makes its size the dimension. The exact result, retaining the requested order, is (2, 2, 2, Yes, Yes, Yes)."},"skills":["A Basis Requires Both Independence and Spanning"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261102-r01-vector-space-dimension-0024","batch_id":"batch-20261014-advance-20261102-r01-vector-space-dimension-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"22d497d2c844ac27a93f1f4c469ee37a483bed4801cfb161920c12b7c3f186ef","family_id":"qf-mod33-dimension-basis-audit","seed":202611021024,"semantic_template":"qf-mod33-dimension-basis-audit/{\"dimension\":2,\"vectors\":[[1,0],[23,1]]}"},"fields":{"title":"A Basis Requires Both Independence and Spanning","stem":"Use the complete ordered real coordinate family vectors=[[1, 0], [23, 1]] in ambient R^2, limited to dimensions 2 or 3. Treat vectors as columns and calculate rank by exact rational Gaussian elimination, not floating-point samples. A basis is both independent and spanning. An independent family gives only a lower bound; a spanning family gives only an upper bound. Basis-cardinality invariance uses the finite-dimensional theorem, not a few examples. Give (ambient_dimension,vector_count,rank,independent?,spanning?,basis?).","answer":{"kind":"expression","value":"(2, 2, 2, Yes, Yes, Yes)","explanation":"Compare rank with the number of vectors for independence and with the ambient coordinate dimension for spanning. Conjoin both to qualify the supplied family as a basis; neither property alone makes its size the dimension. The exact result, retaining the requested order, is (2, 2, 2, Yes, Yes, Yes)."},"skills":["A Basis Requires Both Independence and Spanning"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261102-r01-vector-space-dimension-0025","batch_id":"batch-20261014-advance-20261102-r01-vector-space-dimension-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"cd2cf3b57b8a02e9fa5dacf73ec8c4304e6a5d2e682725203b31aaa3f5223c39","family_id":"qf-mod33-dimension-basis-audit","seed":202611021025,"semantic_template":"qf-mod33-dimension-basis-audit/{\"dimension\":2,\"vectors\":[[1,0],[24,1]]}"},"fields":{"title":"A Basis Requires Both Independence and Spanning","stem":"Use the complete ordered real coordinate family vectors=[[1, 0], [24, 1]] in ambient R^2, limited to dimensions 2 or 3. Treat vectors as columns and calculate rank by exact rational Gaussian elimination, not floating-point samples. A basis is both independent and spanning. An independent family gives only a lower bound; a spanning family gives only an upper bound. Basis-cardinality invariance uses the finite-dimensional theorem, not a few examples. Give (ambient_dimension,vector_count,rank,independent?,spanning?,basis?).","answer":{"kind":"expression","value":"(2, 2, 2, Yes, Yes, Yes)","explanation":"Compare rank with the number of vectors for independence and with the ambient coordinate dimension for spanning. Conjoin both to qualify the supplied family as a basis; neither property alone makes its size the dimension. The exact result, retaining the requested order, is (2, 2, 2, Yes, Yes, Yes)."},"skills":["A Basis Requires Both Independence and Spanning"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261102-r01-vector-space-dimension-0026","batch_id":"batch-20261014-advance-20261102-r01-vector-space-dimension-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"2e130816e1457094976338ccbba125f4e955a3cb8c20d54dca7c6703111e6fc1","family_id":"qf-mod33-dimension-exact-value","seed":202611021026,"semantic_template":"qf-mod33-dimension-exact-value/{\"dimension\":2,\"vectors\":[[0,1],[1,0]]}"},"fields":{"title":"Read Dimension from a Previously Qualified Basis","stem":"Use the complete ordered real coordinate family vectors=[[1, 0], [0, 1]] in ambient R^2, limited to dimensions 2 or 3. Treat vectors as columns and calculate rank by exact rational Gaussian elimination, not floating-point samples. A basis is both independent and spanning. An independent family gives only a lower bound; a spanning family gives only an upper bound. Basis-cardinality invariance uses the finite-dimensional theorem, not a few examples. Require this complete family to be an actual basis. Give (ambient_dimension,basis_count,rank,dimension,finite_dimensional_basis_cardinality_invariance_theorem).","answer":{"kind":"expression","value":"(2, 2, 2, 2, finite_dimensional_basis_cardinality_invariance_theorem)","explanation":"Check independence and spanning before reading the basis cardinality. A dependent or nonspanning family is rejected rather than assigned a dimension from its length. The named invariance theorem is not newly proved by this instance or recorded as Lean Evidence. The exact result, retaining the requested order, is (2, 2, 2, 2, finite_dimensional_basis_cardinality_invariance_theorem)."},"skills":["Read Dimension from a Previously Qualified Basis"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261102-r01-vector-space-dimension-0027","batch_id":"batch-20261014-advance-20261102-r01-vector-space-dimension-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"347e27cca238cabc09a374996fbba4e12fb005873d88dae5de3950bc39960a12","family_id":"qf-mod33-dimension-exact-value","seed":202611021027,"semantic_template":"qf-mod33-dimension-exact-value/{\"dimension\":2,\"vectors\":[[1,0],[1,1]]}"},"fields":{"title":"Read Dimension from a Previously Qualified Basis","stem":"Use the complete ordered real coordinate family vectors=[[1, 0], [1, 1]] in ambient R^2, limited to dimensions 2 or 3. Treat vectors as columns and calculate rank by exact rational Gaussian elimination, not floating-point samples. A basis is both independent and spanning. An independent family gives only a lower bound; a spanning family gives only an upper bound. Basis-cardinality invariance uses the finite-dimensional theorem, not a few examples. Require this complete family to be an actual basis. Give (ambient_dimension,basis_count,rank,dimension,finite_dimensional_basis_cardinality_invariance_theorem).","answer":{"kind":"expression","value":"(2, 2, 2, 2, finite_dimensional_basis_cardinality_invariance_theorem)","explanation":"Check independence and spanning before reading the basis cardinality. A dependent or nonspanning family is rejected rather than assigned a dimension from its length. The named invariance theorem is not newly proved by this instance or recorded as Lean Evidence. The exact result, retaining the requested order, is (2, 2, 2, 2, finite_dimensional_basis_cardinality_invariance_theorem)."},"skills":["Read Dimension from a Previously Qualified Basis"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261102-r01-vector-space-dimension-0028","batch_id":"batch-20261014-advance-20261102-r01-vector-space-dimension-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"950a9039795ee51cac740c3ddc3356f2a5089d6ed7b5079c9a99b1fcfce52aa7","family_id":"qf-mod33-dimension-exact-value","seed":202611021028,"semantic_template":"qf-mod33-dimension-exact-value/{\"dimension\":2,\"vectors\":[[1,0],[2,1]]}"},"fields":{"title":"Read Dimension from a Previously Qualified Basis","stem":"Use the complete ordered real coordinate family vectors=[[1, 0], [2, 1]] in ambient R^2, limited to dimensions 2 or 3. Treat vectors as columns and calculate rank by exact rational Gaussian elimination, not floating-point samples. A basis is both independent and spanning. An independent family gives only a lower bound; a spanning family gives only an upper bound. Basis-cardinality invariance uses the finite-dimensional theorem, not a few examples. Require this complete family to be an actual basis. Give (ambient_dimension,basis_count,rank,dimension,finite_dimensional_basis_cardinality_invariance_theorem).","answer":{"kind":"expression","value":"(2, 2, 2, 2, finite_dimensional_basis_cardinality_invariance_theorem)","explanation":"Check independence and spanning before reading the basis cardinality. A dependent or nonspanning family is rejected rather than assigned a dimension from its length. The named invariance theorem is not newly proved by this instance or recorded as Lean Evidence. The exact result, retaining the requested order, is (2, 2, 2, 2, finite_dimensional_basis_cardinality_invariance_theorem)."},"skills":["Read Dimension from a Previously Qualified Basis"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261102-r01-vector-space-dimension-0029","batch_id":"batch-20261014-advance-20261102-r01-vector-space-dimension-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"90c855e46b90292a8ef4aff9ef3f8f84ce62cfc6266fe147ce8910ba6c6ec376","family_id":"qf-mod33-dimension-exact-value","seed":202611021029,"semantic_template":"qf-mod33-dimension-exact-value/{\"dimension\":2,\"vectors\":[[1,0],[3,1]]}"},"fields":{"title":"Read Dimension from a Previously Qualified Basis","stem":"Use the complete ordered real coordinate family vectors=[[1, 0], [3, 1]] in ambient R^2, limited to dimensions 2 or 3. Treat vectors as columns and calculate rank by exact rational Gaussian elimination, not floating-point samples. A basis is both independent and spanning. An independent family gives only a lower bound; a spanning family gives only an upper bound. Basis-cardinality invariance uses the finite-dimensional theorem, not a few examples. Require this complete family to be an actual basis. Give (ambient_dimension,basis_count,rank,dimension,finite_dimensional_basis_cardinality_invariance_theorem).","answer":{"kind":"expression","value":"(2, 2, 2, 2, finite_dimensional_basis_cardinality_invariance_theorem)","explanation":"Check independence and spanning before reading the basis cardinality. A dependent or nonspanning family is rejected rather than assigned a dimension from its length. The named invariance theorem is not newly proved by this instance or recorded as Lean Evidence. The exact result, retaining the requested order, is (2, 2, 2, 2, finite_dimensional_basis_cardinality_invariance_theorem)."},"skills":["Read Dimension from a Previously Qualified Basis"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261102-r01-vector-space-dimension-0030","batch_id":"batch-20261014-advance-20261102-r01-vector-space-dimension-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"bb0dca38d07c3b357dc62a041119cd2b1dac6f96a8fd95ee782f3fe78cfa655b","family_id":"qf-mod33-dimension-exact-value","seed":202611021030,"semantic_template":"qf-mod33-dimension-exact-value/{\"dimension\":2,\"vectors\":[[1,0],[4,1]]}"},"fields":{"title":"Read Dimension from a Previously Qualified Basis","stem":"Use the complete ordered real coordinate family vectors=[[1, 0], [4, 1]] in ambient R^2, limited to dimensions 2 or 3. Treat vectors as columns and calculate rank by exact rational Gaussian elimination, not floating-point samples. A basis is both independent and spanning. An independent family gives only a lower bound; a spanning family gives only an upper bound. Basis-cardinality invariance uses the finite-dimensional theorem, not a few examples. Require this complete family to be an actual basis. Give (ambient_dimension,basis_count,rank,dimension,finite_dimensional_basis_cardinality_invariance_theorem).","answer":{"kind":"expression","value":"(2, 2, 2, 2, finite_dimensional_basis_cardinality_invariance_theorem)","explanation":"Check independence and spanning before reading the basis cardinality. A dependent or nonspanning family is rejected rather than assigned a dimension from its length. The named invariance theorem is not newly proved by this instance or recorded as Lean Evidence. The exact result, retaining the requested order, is (2, 2, 2, 2, finite_dimensional_basis_cardinality_invariance_theorem)."},"skills":["Read Dimension from a Previously Qualified Basis"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261102-r01-vector-space-dimension-0031","batch_id":"batch-20261014-advance-20261102-r01-vector-space-dimension-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"365865c0c52dc50c9c7f913cfa2f9e18be81ba67dc16b161855852a69674f249","family_id":"qf-mod33-dimension-exact-value","seed":202611021031,"semantic_template":"qf-mod33-dimension-exact-value/{\"dimension\":2,\"vectors\":[[1,0],[5,1]]}"},"fields":{"title":"Read Dimension from a Previously Qualified Basis","stem":"Use the complete ordered real coordinate family vectors=[[1, 0], [5, 1]] in ambient R^2, limited to dimensions 2 or 3. Treat vectors as columns and calculate rank by exact rational Gaussian elimination, not floating-point samples. A basis is both independent and spanning. An independent family gives only a lower bound; a spanning family gives only an upper bound. Basis-cardinality invariance uses the finite-dimensional theorem, not a few examples. Require this complete family to be an actual basis. Give (ambient_dimension,basis_count,rank,dimension,finite_dimensional_basis_cardinality_invariance_theorem).","answer":{"kind":"expression","value":"(2, 2, 2, 2, finite_dimensional_basis_cardinality_invariance_theorem)","explanation":"Check independence and spanning before reading the basis cardinality. A dependent or nonspanning family is rejected rather than assigned a dimension from its length. The named invariance theorem is not newly proved by this instance or recorded as Lean Evidence. The exact result, retaining the requested order, is (2, 2, 2, 2, finite_dimensional_basis_cardinality_invariance_theorem)."},"skills":["Read Dimension from a Previously Qualified Basis"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261102-r01-vector-space-dimension-0032","batch_id":"batch-20261014-advance-20261102-r01-vector-space-dimension-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"076a42205efb22a726da953dc80e8191a439b0f3aaabe3ea3824681635704c87","family_id":"qf-mod33-dimension-exact-value","seed":202611021032,"semantic_template":"qf-mod33-dimension-exact-value/{\"dimension\":2,\"vectors\":[[1,0],[6,1]]}"},"fields":{"title":"Read Dimension from a Previously Qualified Basis","stem":"Use the complete ordered real coordinate family vectors=[[1, 0], [6, 1]] in ambient R^2, limited to dimensions 2 or 3. Treat vectors as columns and calculate rank by exact rational Gaussian elimination, not floating-point samples. A basis is both independent and spanning. An independent family gives only a lower bound; a spanning family gives only an upper bound. Basis-cardinality invariance uses the finite-dimensional theorem, not a few examples. Require this complete family to be an actual basis. Give (ambient_dimension,basis_count,rank,dimension,finite_dimensional_basis_cardinality_invariance_theorem).","answer":{"kind":"expression","value":"(2, 2, 2, 2, finite_dimensional_basis_cardinality_invariance_theorem)","explanation":"Check independence and spanning before reading the basis cardinality. A dependent or nonspanning family is rejected rather than assigned a dimension from its length. The named invariance theorem is not newly proved by this instance or recorded as Lean Evidence. The exact result, retaining the requested order, is (2, 2, 2, 2, finite_dimensional_basis_cardinality_invariance_theorem)."},"skills":["Read Dimension from a Previously Qualified Basis"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261102-r01-vector-space-dimension-0033","batch_id":"batch-20261014-advance-20261102-r01-vector-space-dimension-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"7457e01255624b4171c263803f36001269eeea7f6aef0275e9511fa33c357bcc","family_id":"qf-mod33-dimension-exact-value","seed":202611021033,"semantic_template":"qf-mod33-dimension-exact-value/{\"dimension\":2,\"vectors\":[[1,0],[7,1]]}"},"fields":{"title":"Read Dimension from a Previously Qualified Basis","stem":"Use the complete ordered real coordinate family vectors=[[1, 0], [7, 1]] in ambient R^2, limited to dimensions 2 or 3. Treat vectors as columns and calculate rank by exact rational Gaussian elimination, not floating-point samples. A basis is both independent and spanning. An independent family gives only a lower bound; a spanning family gives only an upper bound. Basis-cardinality invariance uses the finite-dimensional theorem, not a few examples. Require this complete family to be an actual basis. Give (ambient_dimension,basis_count,rank,dimension,finite_dimensional_basis_cardinality_invariance_theorem).","answer":{"kind":"expression","value":"(2, 2, 2, 2, finite_dimensional_basis_cardinality_invariance_theorem)","explanation":"Check independence and spanning before reading the basis cardinality. A dependent or nonspanning family is rejected rather than assigned a dimension from its length. The named invariance theorem is not newly proved by this instance or recorded as Lean Evidence. The exact result, retaining the requested order, is (2, 2, 2, 2, finite_dimensional_basis_cardinality_invariance_theorem)."},"skills":["Read Dimension from a Previously Qualified Basis"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261102-r01-vector-space-dimension-0034","batch_id":"batch-20261014-advance-20261102-r01-vector-space-dimension-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"36d5a73c030b57cbcb13f8a2a9cbf40948060dcde6c8d0b37825a7a8e6f4d3c4","family_id":"qf-mod33-dimension-exact-value","seed":202611021034,"semantic_template":"qf-mod33-dimension-exact-value/{\"dimension\":2,\"vectors\":[[1,0],[8,1]]}"},"fields":{"title":"Read Dimension from a Previously Qualified Basis","stem":"Use the complete ordered real coordinate family vectors=[[1, 0], [8, 1]] in ambient R^2, limited to dimensions 2 or 3. Treat vectors as columns and calculate rank by exact rational Gaussian elimination, not floating-point samples. A basis is both independent and spanning. An independent family gives only a lower bound; a spanning family gives only an upper bound. Basis-cardinality invariance uses the finite-dimensional theorem, not a few examples. Require this complete family to be an actual basis. Give (ambient_dimension,basis_count,rank,dimension,finite_dimensional_basis_cardinality_invariance_theorem).","answer":{"kind":"expression","value":"(2, 2, 2, 2, finite_dimensional_basis_cardinality_invariance_theorem)","explanation":"Check independence and spanning before reading the basis cardinality. A dependent or nonspanning family is rejected rather than assigned a dimension from its length. The named invariance theorem is not newly proved by this instance or recorded as Lean Evidence. The exact result, retaining the requested order, is (2, 2, 2, 2, finite_dimensional_basis_cardinality_invariance_theorem)."},"skills":["Read Dimension from a Previously Qualified Basis"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261102-r01-vector-space-dimension-0035","batch_id":"batch-20261014-advance-20261102-r01-vector-space-dimension-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"44365eaa729869407f54982c1d22f00f20c9cb022e95467ca2975f69e8cdded7","family_id":"qf-mod33-dimension-exact-value","seed":202611021035,"semantic_template":"qf-mod33-dimension-exact-value/{\"dimension\":2,\"vectors\":[[1,0],[9,1]]}"},"fields":{"title":"Read Dimension from a Previously Qualified Basis","stem":"Use the complete ordered real coordinate family vectors=[[1, 0], [9, 1]] in ambient R^2, limited to dimensions 2 or 3. Treat vectors as columns and calculate rank by exact rational Gaussian elimination, not floating-point samples. A basis is both independent and spanning. An independent family gives only a lower bound; a spanning family gives only an upper bound. Basis-cardinality invariance uses the finite-dimensional theorem, not a few examples. Require this complete family to be an actual basis. Give (ambient_dimension,basis_count,rank,dimension,finite_dimensional_basis_cardinality_invariance_theorem).","answer":{"kind":"expression","value":"(2, 2, 2, 2, finite_dimensional_basis_cardinality_invariance_theorem)","explanation":"Check independence and spanning before reading the basis cardinality. A dependent or nonspanning family is rejected rather than assigned a dimension from its length. The named invariance theorem is not newly proved by this instance or recorded as Lean Evidence. The exact result, retaining the requested order, is (2, 2, 2, 2, finite_dimensional_basis_cardinality_invariance_theorem)."},"skills":["Read Dimension from a Previously Qualified Basis"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261102-r01-vector-space-dimension-0036","batch_id":"batch-20261014-advance-20261102-r01-vector-space-dimension-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"309b3478e8e1e26ff852eaea08d142dcdae14861712cdd0e2e2b6dad202090c8","family_id":"qf-mod33-dimension-exact-value","seed":202611021036,"semantic_template":"qf-mod33-dimension-exact-value/{\"dimension\":2,\"vectors\":[[1,0],[10,1]]}"},"fields":{"title":"Read Dimension from a Previously Qualified Basis","stem":"Use the complete ordered real coordinate family vectors=[[1, 0], [10, 1]] in ambient R^2, limited to dimensions 2 or 3. Treat vectors as columns and calculate rank by exact rational Gaussian elimination, not floating-point samples. A basis is both independent and spanning. An independent family gives only a lower bound; a spanning family gives only an upper bound. Basis-cardinality invariance uses the finite-dimensional theorem, not a few examples. Require this complete family to be an actual basis. Give (ambient_dimension,basis_count,rank,dimension,finite_dimensional_basis_cardinality_invariance_theorem).","answer":{"kind":"expression","value":"(2, 2, 2, 2, finite_dimensional_basis_cardinality_invariance_theorem)","explanation":"Check independence and spanning before reading the basis cardinality. A dependent or nonspanning family is rejected rather than assigned a dimension from its length. The named invariance theorem is not newly proved by this instance or recorded as Lean Evidence. The exact result, retaining the requested order, is (2, 2, 2, 2, finite_dimensional_basis_cardinality_invariance_theorem)."},"skills":["Read Dimension from a Previously Qualified Basis"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261102-r01-vector-space-dimension-0037","batch_id":"batch-20261014-advance-20261102-r01-vector-space-dimension-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"4f11f25b3f760954274e523028a68eed1f26e6ea79a1fa0c3c2d80400fc9d9aa","family_id":"qf-mod33-dimension-exact-value","seed":202611021037,"semantic_template":"qf-mod33-dimension-exact-value/{\"dimension\":2,\"vectors\":[[1,0],[11,1]]}"},"fields":{"title":"Read Dimension from a Previously Qualified Basis","stem":"Use the complete ordered real coordinate family vectors=[[1, 0], [11, 1]] in ambient R^2, limited to dimensions 2 or 3. Treat vectors as columns and calculate rank by exact rational Gaussian elimination, not floating-point samples. A basis is both independent and spanning. An independent family gives only a lower bound; a spanning family gives only an upper bound. Basis-cardinality invariance uses the finite-dimensional theorem, not a few examples. Require this complete family to be an actual basis. Give (ambient_dimension,basis_count,rank,dimension,finite_dimensional_basis_cardinality_invariance_theorem).","answer":{"kind":"expression","value":"(2, 2, 2, 2, finite_dimensional_basis_cardinality_invariance_theorem)","explanation":"Check independence and spanning before reading the basis cardinality. A dependent or nonspanning family is rejected rather than assigned a dimension from its length. The named invariance theorem is not newly proved by this instance or recorded as Lean Evidence. The exact result, retaining the requested order, is (2, 2, 2, 2, finite_dimensional_basis_cardinality_invariance_theorem)."},"skills":["Read Dimension from a Previously Qualified Basis"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261102-r01-vector-space-dimension-0038","batch_id":"batch-20261014-advance-20261102-r01-vector-space-dimension-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"a6f6e1881e0393e5ba4d879c44c0757452760f22bdd11b7b0c10aa39e8581f67","family_id":"qf-mod33-dimension-exact-value","seed":202611021038,"semantic_template":"qf-mod33-dimension-exact-value/{\"dimension\":2,\"vectors\":[[1,0],[12,1]]}"},"fields":{"title":"Read Dimension from a Previously Qualified Basis","stem":"Use the complete ordered real coordinate family vectors=[[1, 0], [12, 1]] in ambient R^2, limited to dimensions 2 or 3. Treat vectors as columns and calculate rank by exact rational Gaussian elimination, not floating-point samples. A basis is both independent and spanning. An independent family gives only a lower bound; a spanning family gives only an upper bound. Basis-cardinality invariance uses the finite-dimensional theorem, not a few examples. Require this complete family to be an actual basis. Give (ambient_dimension,basis_count,rank,dimension,finite_dimensional_basis_cardinality_invariance_theorem).","answer":{"kind":"expression","value":"(2, 2, 2, 2, finite_dimensional_basis_cardinality_invariance_theorem)","explanation":"Check independence and spanning before reading the basis cardinality. A dependent or nonspanning family is rejected rather than assigned a dimension from its length. The named invariance theorem is not newly proved by this instance or recorded as Lean Evidence. The exact result, retaining the requested order, is (2, 2, 2, 2, finite_dimensional_basis_cardinality_invariance_theorem)."},"skills":["Read Dimension from a Previously Qualified Basis"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261102-r01-vector-space-dimension-0039","batch_id":"batch-20261014-advance-20261102-r01-vector-space-dimension-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"f2c4570c7b6a7650b073895191e606c9d4959e3ebe2424aae0bd895c37046f60","family_id":"qf-mod33-dimension-exact-value","seed":202611021039,"semantic_template":"qf-mod33-dimension-exact-value/{\"dimension\":2,\"vectors\":[[1,0],[13,1]]}"},"fields":{"title":"Read Dimension from a Previously Qualified Basis","stem":"Use the complete ordered real coordinate family vectors=[[1, 0], [13, 1]] in ambient R^2, limited to dimensions 2 or 3. Treat vectors as columns and calculate rank by exact rational Gaussian elimination, not floating-point samples. A basis is both independent and spanning. An independent family gives only a lower bound; a spanning family gives only an upper bound. Basis-cardinality invariance uses the finite-dimensional theorem, not a few examples. Require this complete family to be an actual basis. Give (ambient_dimension,basis_count,rank,dimension,finite_dimensional_basis_cardinality_invariance_theorem).","answer":{"kind":"expression","value":"(2, 2, 2, 2, finite_dimensional_basis_cardinality_invariance_theorem)","explanation":"Check independence and spanning before reading the basis cardinality. A dependent or nonspanning family is rejected rather than assigned a dimension from its length. The named invariance theorem is not newly proved by this instance or recorded as Lean Evidence. The exact result, retaining the requested order, is (2, 2, 2, 2, finite_dimensional_basis_cardinality_invariance_theorem)."},"skills":["Read Dimension from a Previously Qualified Basis"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261102-r01-vector-space-dimension-0040","batch_id":"batch-20261014-advance-20261102-r01-vector-space-dimension-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"ddcde34c427417f2cfc9378e95f12420a0bfe6f5cf744299fa92842155c8aa96","family_id":"qf-mod33-dimension-exact-value","seed":202611021040,"semantic_template":"qf-mod33-dimension-exact-value/{\"dimension\":2,\"vectors\":[[1,0],[14,1]]}"},"fields":{"title":"Read Dimension from a Previously Qualified Basis","stem":"Use the complete ordered real coordinate family vectors=[[1, 0], [14, 1]] in ambient R^2, limited to dimensions 2 or 3. Treat vectors as columns and calculate rank by exact rational Gaussian elimination, not floating-point samples. A basis is both independent and spanning. An independent family gives only a lower bound; a spanning family gives only an upper bound. Basis-cardinality invariance uses the finite-dimensional theorem, not a few examples. Require this complete family to be an actual basis. Give (ambient_dimension,basis_count,rank,dimension,finite_dimensional_basis_cardinality_invariance_theorem).","answer":{"kind":"expression","value":"(2, 2, 2, 2, finite_dimensional_basis_cardinality_invariance_theorem)","explanation":"Check independence and spanning before reading the basis cardinality. A dependent or nonspanning family is rejected rather than assigned a dimension from its length. The named invariance theorem is not newly proved by this instance or recorded as Lean Evidence. The exact result, retaining the requested order, is (2, 2, 2, 2, finite_dimensional_basis_cardinality_invariance_theorem)."},"skills":["Read Dimension from a Previously Qualified Basis"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261102-r01-vector-space-dimension-0041","batch_id":"batch-20261014-advance-20261102-r01-vector-space-dimension-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"9d1bb43657df6912f9da88832deae9274632df7abb119c6f65c42f79c6a1d51d","family_id":"qf-mod33-dimension-exact-value","seed":202611021041,"semantic_template":"qf-mod33-dimension-exact-value/{\"dimension\":2,\"vectors\":[[1,0],[15,1]]}"},"fields":{"title":"Read Dimension from a Previously Qualified Basis","stem":"Use the complete ordered real coordinate family vectors=[[1, 0], [15, 1]] in ambient R^2, limited to dimensions 2 or 3. Treat vectors as columns and calculate rank by exact rational Gaussian elimination, not floating-point samples. A basis is both independent and spanning. An independent family gives only a lower bound; a spanning family gives only an upper bound. Basis-cardinality invariance uses the finite-dimensional theorem, not a few examples. Require this complete family to be an actual basis. Give (ambient_dimension,basis_count,rank,dimension,finite_dimensional_basis_cardinality_invariance_theorem).","answer":{"kind":"expression","value":"(2, 2, 2, 2, finite_dimensional_basis_cardinality_invariance_theorem)","explanation":"Check independence and spanning before reading the basis cardinality. A dependent or nonspanning family is rejected rather than assigned a dimension from its length. The named invariance theorem is not newly proved by this instance or recorded as Lean Evidence. The exact result, retaining the requested order, is (2, 2, 2, 2, finite_dimensional_basis_cardinality_invariance_theorem)."},"skills":["Read Dimension from a Previously Qualified Basis"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261102-r01-vector-space-dimension-0042","batch_id":"batch-20261014-advance-20261102-r01-vector-space-dimension-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"9d27cea29df4ac723071a2b8d55390ca268bd329629a0b75eeb88a67edf22c42","family_id":"qf-mod33-dimension-exact-value","seed":202611021042,"semantic_template":"qf-mod33-dimension-exact-value/{\"dimension\":2,\"vectors\":[[1,0],[16,1]]}"},"fields":{"title":"Read Dimension from a Previously Qualified Basis","stem":"Use the complete ordered real coordinate family vectors=[[1, 0], [16, 1]] in ambient R^2, limited to dimensions 2 or 3. Treat vectors as columns and calculate rank by exact rational Gaussian elimination, not floating-point samples. A basis is both independent and spanning. An independent family gives only a lower bound; a spanning family gives only an upper bound. Basis-cardinality invariance uses the finite-dimensional theorem, not a few examples. Require this complete family to be an actual basis. Give (ambient_dimension,basis_count,rank,dimension,finite_dimensional_basis_cardinality_invariance_theorem).","answer":{"kind":"expression","value":"(2, 2, 2, 2, finite_dimensional_basis_cardinality_invariance_theorem)","explanation":"Check independence and spanning before reading the basis cardinality. A dependent or nonspanning family is rejected rather than assigned a dimension from its length. The named invariance theorem is not newly proved by this instance or recorded as Lean Evidence. The exact result, retaining the requested order, is (2, 2, 2, 2, finite_dimensional_basis_cardinality_invariance_theorem)."},"skills":["Read Dimension from a Previously Qualified Basis"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261102-r01-vector-space-dimension-0043","batch_id":"batch-20261014-advance-20261102-r01-vector-space-dimension-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"e2b557ed95007b7c4b71fde5bcb24c374e67627d3a2c09a2780b8ff37628c212","family_id":"qf-mod33-dimension-exact-value","seed":202611021043,"semantic_template":"qf-mod33-dimension-exact-value/{\"dimension\":2,\"vectors\":[[1,0],[17,1]]}"},"fields":{"title":"Read Dimension from a Previously Qualified Basis","stem":"Use the complete ordered real coordinate family vectors=[[1, 0], [17, 1]] in ambient R^2, limited to dimensions 2 or 3. Treat vectors as columns and calculate rank by exact rational Gaussian elimination, not floating-point samples. A basis is both independent and spanning. An independent family gives only a lower bound; a spanning family gives only an upper bound. Basis-cardinality invariance uses the finite-dimensional theorem, not a few examples. Require this complete family to be an actual basis. Give (ambient_dimension,basis_count,rank,dimension,finite_dimensional_basis_cardinality_invariance_theorem).","answer":{"kind":"expression","value":"(2, 2, 2, 2, finite_dimensional_basis_cardinality_invariance_theorem)","explanation":"Check independence and spanning before reading the basis cardinality. A dependent or nonspanning family is rejected rather than assigned a dimension from its length. The named invariance theorem is not newly proved by this instance or recorded as Lean Evidence. The exact result, retaining the requested order, is (2, 2, 2, 2, finite_dimensional_basis_cardinality_invariance_theorem)."},"skills":["Read Dimension from a Previously Qualified Basis"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261102-r01-vector-space-dimension-0044","batch_id":"batch-20261014-advance-20261102-r01-vector-space-dimension-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"c4db19d6db7fc297a66db9142215b33bd82e7a825fc9f7a745bbd9bcd1f08e7a","family_id":"qf-mod33-dimension-exact-value","seed":202611021044,"semantic_template":"qf-mod33-dimension-exact-value/{\"dimension\":2,\"vectors\":[[1,0],[18,1]]}"},"fields":{"title":"Read Dimension from a Previously Qualified Basis","stem":"Use the complete ordered real coordinate family vectors=[[1, 0], [18, 1]] in ambient R^2, limited to dimensions 2 or 3. Treat vectors as columns and calculate rank by exact rational Gaussian elimination, not floating-point samples. A basis is both independent and spanning. An independent family gives only a lower bound; a spanning family gives only an upper bound. Basis-cardinality invariance uses the finite-dimensional theorem, not a few examples. Require this complete family to be an actual basis. Give (ambient_dimension,basis_count,rank,dimension,finite_dimensional_basis_cardinality_invariance_theorem).","answer":{"kind":"expression","value":"(2, 2, 2, 2, finite_dimensional_basis_cardinality_invariance_theorem)","explanation":"Check independence and spanning before reading the basis cardinality. A dependent or nonspanning family is rejected rather than assigned a dimension from its length. The named invariance theorem is not newly proved by this instance or recorded as Lean Evidence. The exact result, retaining the requested order, is (2, 2, 2, 2, finite_dimensional_basis_cardinality_invariance_theorem)."},"skills":["Read Dimension from a Previously Qualified Basis"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261102-r01-vector-space-dimension-0045","batch_id":"batch-20261014-advance-20261102-r01-vector-space-dimension-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"730d11fb08ca36763ca406b5f015ab40db15c5552705c294feed97babd85fb28","family_id":"qf-mod33-dimension-exact-value","seed":202611021045,"semantic_template":"qf-mod33-dimension-exact-value/{\"dimension\":2,\"vectors\":[[1,0],[19,1]]}"},"fields":{"title":"Read Dimension from a Previously Qualified Basis","stem":"Use the complete ordered real coordinate family vectors=[[1, 0], [19, 1]] in ambient R^2, limited to dimensions 2 or 3. Treat vectors as columns and calculate rank by exact rational Gaussian elimination, not floating-point samples. A basis is both independent and spanning. An independent family gives only a lower bound; a spanning family gives only an upper bound. Basis-cardinality invariance uses the finite-dimensional theorem, not a few examples. Require this complete family to be an actual basis. Give (ambient_dimension,basis_count,rank,dimension,finite_dimensional_basis_cardinality_invariance_theorem).","answer":{"kind":"expression","value":"(2, 2, 2, 2, finite_dimensional_basis_cardinality_invariance_theorem)","explanation":"Check independence and spanning before reading the basis cardinality. A dependent or nonspanning family is rejected rather than assigned a dimension from its length. The named invariance theorem is not newly proved by this instance or recorded as Lean Evidence. The exact result, retaining the requested order, is (2, 2, 2, 2, finite_dimensional_basis_cardinality_invariance_theorem)."},"skills":["Read Dimension from a Previously Qualified Basis"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261102-r01-vector-space-dimension-0046","batch_id":"batch-20261014-advance-20261102-r01-vector-space-dimension-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"539d97ce385f508bf12bf2f6b112a6e9eb0aa59c710c0579c27e813038253878","family_id":"qf-mod33-dimension-exact-value","seed":202611021046,"semantic_template":"qf-mod33-dimension-exact-value/{\"dimension\":2,\"vectors\":[[1,0],[20,1]]}"},"fields":{"title":"Read Dimension from a Previously Qualified Basis","stem":"Use the complete ordered real coordinate family vectors=[[1, 0], [20, 1]] in ambient R^2, limited to dimensions 2 or 3. Treat vectors as columns and calculate rank by exact rational Gaussian elimination, not floating-point samples. A basis is both independent and spanning. An independent family gives only a lower bound; a spanning family gives only an upper bound. Basis-cardinality invariance uses the finite-dimensional theorem, not a few examples. Require this complete family to be an actual basis. Give (ambient_dimension,basis_count,rank,dimension,finite_dimensional_basis_cardinality_invariance_theorem).","answer":{"kind":"expression","value":"(2, 2, 2, 2, finite_dimensional_basis_cardinality_invariance_theorem)","explanation":"Check independence and spanning before reading the basis cardinality. A dependent or nonspanning family is rejected rather than assigned a dimension from its length. The named invariance theorem is not newly proved by this instance or recorded as Lean Evidence. The exact result, retaining the requested order, is (2, 2, 2, 2, finite_dimensional_basis_cardinality_invariance_theorem)."},"skills":["Read Dimension from a Previously Qualified Basis"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261102-r01-vector-space-dimension-0047","batch_id":"batch-20261014-advance-20261102-r01-vector-space-dimension-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"350365b9cf0e412be0c8a2fedfacaa352168543827f5cad3d0f6bdde34abd8b3","family_id":"qf-mod33-dimension-exact-value","seed":202611021047,"semantic_template":"qf-mod33-dimension-exact-value/{\"dimension\":2,\"vectors\":[[1,0],[21,1]]}"},"fields":{"title":"Read Dimension from a Previously Qualified Basis","stem":"Use the complete ordered real coordinate family vectors=[[1, 0], [21, 1]] in ambient R^2, limited to dimensions 2 or 3. Treat vectors as columns and calculate rank by exact rational Gaussian elimination, not floating-point samples. A basis is both independent and spanning. An independent family gives only a lower bound; a spanning family gives only an upper bound. Basis-cardinality invariance uses the finite-dimensional theorem, not a few examples. Require this complete family to be an actual basis. Give (ambient_dimension,basis_count,rank,dimension,finite_dimensional_basis_cardinality_invariance_theorem).","answer":{"kind":"expression","value":"(2, 2, 2, 2, finite_dimensional_basis_cardinality_invariance_theorem)","explanation":"Check independence and spanning before reading the basis cardinality. A dependent or nonspanning family is rejected rather than assigned a dimension from its length. The named invariance theorem is not newly proved by this instance or recorded as Lean Evidence. The exact result, retaining the requested order, is (2, 2, 2, 2, finite_dimensional_basis_cardinality_invariance_theorem)."},"skills":["Read Dimension from a Previously Qualified Basis"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261102-r01-vector-space-dimension-0048","batch_id":"batch-20261014-advance-20261102-r01-vector-space-dimension-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"8cbd7668e30a0bd770729e6a772d524dba322d3c6956fe8deee5742f4d3ecd32","family_id":"qf-mod33-dimension-exact-value","seed":202611021048,"semantic_template":"qf-mod33-dimension-exact-value/{\"dimension\":2,\"vectors\":[[1,0],[22,1]]}"},"fields":{"title":"Read Dimension from a Previously Qualified Basis","stem":"Use the complete ordered real coordinate family vectors=[[1, 0], [22, 1]] in ambient R^2, limited to dimensions 2 or 3. Treat vectors as columns and calculate rank by exact rational Gaussian elimination, not floating-point samples. A basis is both independent and spanning. An independent family gives only a lower bound; a spanning family gives only an upper bound. Basis-cardinality invariance uses the finite-dimensional theorem, not a few examples. Require this complete family to be an actual basis. Give (ambient_dimension,basis_count,rank,dimension,finite_dimensional_basis_cardinality_invariance_theorem).","answer":{"kind":"expression","value":"(2, 2, 2, 2, finite_dimensional_basis_cardinality_invariance_theorem)","explanation":"Check independence and spanning before reading the basis cardinality. A dependent or nonspanning family is rejected rather than assigned a dimension from its length. The named invariance theorem is not newly proved by this instance or recorded as Lean Evidence. The exact result, retaining the requested order, is (2, 2, 2, 2, finite_dimensional_basis_cardinality_invariance_theorem)."},"skills":["Read Dimension from a Previously Qualified Basis"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261102-r01-vector-space-dimension-0049","batch_id":"batch-20261014-advance-20261102-r01-vector-space-dimension-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"06760eaf2eb4c1c5eb80142130c77a297ea8b8366519da21cb2008470cd4a281","family_id":"qf-mod33-dimension-exact-value","seed":202611021049,"semantic_template":"qf-mod33-dimension-exact-value/{\"dimension\":2,\"vectors\":[[1,0],[23,1]]}"},"fields":{"title":"Read Dimension from a Previously Qualified Basis","stem":"Use the complete ordered real coordinate family vectors=[[1, 0], [23, 1]] in ambient R^2, limited to dimensions 2 or 3. Treat vectors as columns and calculate rank by exact rational Gaussian elimination, not floating-point samples. A basis is both independent and spanning. An independent family gives only a lower bound; a spanning family gives only an upper bound. Basis-cardinality invariance uses the finite-dimensional theorem, not a few examples. Require this complete family to be an actual basis. Give (ambient_dimension,basis_count,rank,dimension,finite_dimensional_basis_cardinality_invariance_theorem).","answer":{"kind":"expression","value":"(2, 2, 2, 2, finite_dimensional_basis_cardinality_invariance_theorem)","explanation":"Check independence and spanning before reading the basis cardinality. A dependent or nonspanning family is rejected rather than assigned a dimension from its length. The named invariance theorem is not newly proved by this instance or recorded as Lean Evidence. The exact result, retaining the requested order, is (2, 2, 2, 2, finite_dimensional_basis_cardinality_invariance_theorem)."},"skills":["Read Dimension from a Previously Qualified Basis"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261102-r01-vector-space-dimension-0050","batch_id":"batch-20261014-advance-20261102-r01-vector-space-dimension-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"79cbda72b5eb735e91070bbec6a5e1b6a3ea8510c5d37f6a48a395784b268fec","family_id":"qf-mod33-dimension-exact-value","seed":202611021050,"semantic_template":"qf-mod33-dimension-exact-value/{\"dimension\":2,\"vectors\":[[1,0],[24,1]]}"},"fields":{"title":"Read Dimension from a Previously Qualified Basis","stem":"Use the complete ordered real coordinate family vectors=[[1, 0], [24, 1]] in ambient R^2, limited to dimensions 2 or 3. Treat vectors as columns and calculate rank by exact rational Gaussian elimination, not floating-point samples. A basis is both independent and spanning. An independent family gives only a lower bound; a spanning family gives only an upper bound. Basis-cardinality invariance uses the finite-dimensional theorem, not a few examples. Require this complete family to be an actual basis. Give (ambient_dimension,basis_count,rank,dimension,finite_dimensional_basis_cardinality_invariance_theorem).","answer":{"kind":"expression","value":"(2, 2, 2, 2, finite_dimensional_basis_cardinality_invariance_theorem)","explanation":"Check independence and spanning before reading the basis cardinality. A dependent or nonspanning family is rejected rather than assigned a dimension from its length. The named invariance theorem is not newly proved by this instance or recorded as Lean Evidence. The exact result, retaining the requested order, is (2, 2, 2, 2, finite_dimensional_basis_cardinality_invariance_theorem)."},"skills":["Read Dimension from a Previously Qualified Basis"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261102-r01-vector-space-dimension-0051","batch_id":"batch-20261014-advance-20261102-r01-vector-space-dimension-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"239b86e37c24dd07af8660247807a9df07a32b7012fa1e5f081e36d69df52016","family_id":"qf-mod33-dimension-independent-lower-bound","seed":202611021051,"semantic_template":"qf-mod33-dimension-independent-lower-bound/{\"dimension\":3,\"vectors\":[[0,1,0],[1,0,0]]}"},"fields":{"title":"An Independent Family Gives Only a Lower Bound","stem":"Use the complete ordered real coordinate family vectors=[[1, 0, 0], [0, 1, 0]] in ambient R^3, limited to dimensions 2 or 3. Treat vectors as columns and calculate rank by exact rational Gaussian elimination, not floating-point samples. A basis is both independent and spanning. An independent family gives only a lower bound; a spanning family gives only an upper bound. Basis-cardinality invariance uses the finite-dimensional theorem, not a few examples. Give (vector_count,rank,independent?,lower_bound). The last field is dimension_at_least_vector_count with the numeric count, or no_lower_bound_from_dependent_family.","answer":{"kind":"expression","value":"(2, 2, Yes, dimension_at_least_2)","explanation":"Only a verified independent family supplies this size-based lower bound. If this family is dependent, the result withholds its certificate; it does not refute every possible lower bound for the ambient space. The exact result, retaining the requested order, is (2, 2, Yes, dimension_at_least_2)."},"skills":["An Independent Family Gives Only a Lower Bound"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261102-r01-vector-space-dimension-0052","batch_id":"batch-20261014-advance-20261102-r01-vector-space-dimension-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"0a50db77cc1e4d960a5f351fc08335c684a179abf2807945816ebbed5cd2558a","family_id":"qf-mod33-dimension-independent-lower-bound","seed":202611021052,"semantic_template":"qf-mod33-dimension-independent-lower-bound/{\"dimension\":3,\"vectors\":[[1,0,0],[1,1,0]]}"},"fields":{"title":"An Independent Family Gives Only a Lower Bound","stem":"Use the complete ordered real coordinate family vectors=[[1, 0, 0], [1, 1, 0]] in ambient R^3, limited to dimensions 2 or 3. Treat vectors as columns and calculate rank by exact rational Gaussian elimination, not floating-point samples. A basis is both independent and spanning. An independent family gives only a lower bound; a spanning family gives only an upper bound. Basis-cardinality invariance uses the finite-dimensional theorem, not a few examples. Give (vector_count,rank,independent?,lower_bound). The last field is dimension_at_least_vector_count with the numeric count, or no_lower_bound_from_dependent_family.","answer":{"kind":"expression","value":"(2, 2, Yes, dimension_at_least_2)","explanation":"Only a verified independent family supplies this size-based lower bound. If this family is dependent, the result withholds its certificate; it does not refute every possible lower bound for the ambient space. The exact result, retaining the requested order, is (2, 2, Yes, dimension_at_least_2)."},"skills":["An Independent Family Gives Only a Lower Bound"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261102-r01-vector-space-dimension-0053","batch_id":"batch-20261014-advance-20261102-r01-vector-space-dimension-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"965877d411df6ddb42dc74c0966a2bc6ff869e3f3a105c123f1be5054e1f6ad2","family_id":"qf-mod33-dimension-independent-lower-bound","seed":202611021053,"semantic_template":"qf-mod33-dimension-independent-lower-bound/{\"dimension\":3,\"vectors\":[[1,0,0],[2,1,0]]}"},"fields":{"title":"An Independent Family Gives Only a Lower Bound","stem":"Use the complete ordered real coordinate family vectors=[[1, 0, 0], [2, 1, 0]] in ambient R^3, limited to dimensions 2 or 3. Treat vectors as columns and calculate rank by exact rational Gaussian elimination, not floating-point samples. A basis is both independent and spanning. An independent family gives only a lower bound; a spanning family gives only an upper bound. Basis-cardinality invariance uses the finite-dimensional theorem, not a few examples. Give (vector_count,rank,independent?,lower_bound). The last field is dimension_at_least_vector_count with the numeric count, or no_lower_bound_from_dependent_family.","answer":{"kind":"expression","value":"(2, 2, Yes, dimension_at_least_2)","explanation":"Only a verified independent family supplies this size-based lower bound. If this family is dependent, the result withholds its certificate; it does not refute every possible lower bound for the ambient space. The exact result, retaining the requested order, is (2, 2, Yes, dimension_at_least_2)."},"skills":["An Independent Family Gives Only a Lower Bound"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261102-r01-vector-space-dimension-0054","batch_id":"batch-20261014-advance-20261102-r01-vector-space-dimension-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"f5e9b07ccdc461b28736d8d14f94086c450807a2d54d26d9a78886dc98084fbe","family_id":"qf-mod33-dimension-independent-lower-bound","seed":202611021054,"semantic_template":"qf-mod33-dimension-independent-lower-bound/{\"dimension\":3,\"vectors\":[[1,0,0],[3,1,0]]}"},"fields":{"title":"An Independent Family Gives Only a Lower Bound","stem":"Use the complete ordered real coordinate family vectors=[[1, 0, 0], [3, 1, 0]] in ambient R^3, limited to dimensions 2 or 3. Treat vectors as columns and calculate rank by exact rational Gaussian elimination, not floating-point samples. A basis is both independent and spanning. An independent family gives only a lower bound; a spanning family gives only an upper bound. Basis-cardinality invariance uses the finite-dimensional theorem, not a few examples. Give (vector_count,rank,independent?,lower_bound). The last field is dimension_at_least_vector_count with the numeric count, or no_lower_bound_from_dependent_family.","answer":{"kind":"expression","value":"(2, 2, Yes, dimension_at_least_2)","explanation":"Only a verified independent family supplies this size-based lower bound. If this family is dependent, the result withholds its certificate; it does not refute every possible lower bound for the ambient space. The exact result, retaining the requested order, is (2, 2, Yes, dimension_at_least_2)."},"skills":["An Independent Family Gives Only a Lower Bound"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261102-r01-vector-space-dimension-0055","batch_id":"batch-20261014-advance-20261102-r01-vector-space-dimension-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"c90b46f760306252fa8f953a05ba70f028ecf828a9d51295321505f2d061226d","family_id":"qf-mod33-dimension-independent-lower-bound","seed":202611021055,"semantic_template":"qf-mod33-dimension-independent-lower-bound/{\"dimension\":3,\"vectors\":[[1,0,0],[4,1,0]]}"},"fields":{"title":"An Independent Family Gives Only a Lower Bound","stem":"Use the complete ordered real coordinate family vectors=[[1, 0, 0], [4, 1, 0]] in ambient R^3, limited to dimensions 2 or 3. Treat vectors as columns and calculate rank by exact rational Gaussian elimination, not floating-point samples. A basis is both independent and spanning. An independent family gives only a lower bound; a spanning family gives only an upper bound. Basis-cardinality invariance uses the finite-dimensional theorem, not a few examples. Give (vector_count,rank,independent?,lower_bound). The last field is dimension_at_least_vector_count with the numeric count, or no_lower_bound_from_dependent_family.","answer":{"kind":"expression","value":"(2, 2, Yes, dimension_at_least_2)","explanation":"Only a verified independent family supplies this size-based lower bound. If this family is dependent, the result withholds its certificate; it does not refute every possible lower bound for the ambient space. The exact result, retaining the requested order, is (2, 2, Yes, dimension_at_least_2)."},"skills":["An Independent Family Gives Only a Lower Bound"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261102-r01-vector-space-dimension-0056","batch_id":"batch-20261014-advance-20261102-r01-vector-space-dimension-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"ad0b41829b0f2eb0b69b388b5ba56d0fe29957ccaca1482a731dcedaee0225fb","family_id":"qf-mod33-dimension-independent-lower-bound","seed":202611021056,"semantic_template":"qf-mod33-dimension-independent-lower-bound/{\"dimension\":3,\"vectors\":[[1,0,0],[5,1,0]]}"},"fields":{"title":"An Independent Family Gives Only a Lower Bound","stem":"Use the complete ordered real coordinate family vectors=[[1, 0, 0], [5, 1, 0]] in ambient R^3, limited to dimensions 2 or 3. Treat vectors as columns and calculate rank by exact rational Gaussian elimination, not floating-point samples. A basis is both independent and spanning. An independent family gives only a lower bound; a spanning family gives only an upper bound. Basis-cardinality invariance uses the finite-dimensional theorem, not a few examples. Give (vector_count,rank,independent?,lower_bound). The last field is dimension_at_least_vector_count with the numeric count, or no_lower_bound_from_dependent_family.","answer":{"kind":"expression","value":"(2, 2, Yes, dimension_at_least_2)","explanation":"Only a verified independent family supplies this size-based lower bound. If this family is dependent, the result withholds its certificate; it does not refute every possible lower bound for the ambient space. The exact result, retaining the requested order, is (2, 2, Yes, dimension_at_least_2)."},"skills":["An Independent Family Gives Only a Lower Bound"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261102-r01-vector-space-dimension-0057","batch_id":"batch-20261014-advance-20261102-r01-vector-space-dimension-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"a30caffe338a4be4c0825f115113b2a8c75ec295077a46442cb8682a9ef58587","family_id":"qf-mod33-dimension-independent-lower-bound","seed":202611021057,"semantic_template":"qf-mod33-dimension-independent-lower-bound/{\"dimension\":3,\"vectors\":[[1,0,0],[6,1,0]]}"},"fields":{"title":"An Independent Family Gives Only a Lower Bound","stem":"Use the complete ordered real coordinate family vectors=[[1, 0, 0], [6, 1, 0]] in ambient R^3, limited to dimensions 2 or 3. Treat vectors as columns and calculate rank by exact rational Gaussian elimination, not floating-point samples. A basis is both independent and spanning. An independent family gives only a lower bound; a spanning family gives only an upper bound. Basis-cardinality invariance uses the finite-dimensional theorem, not a few examples. Give (vector_count,rank,independent?,lower_bound). The last field is dimension_at_least_vector_count with the numeric count, or no_lower_bound_from_dependent_family.","answer":{"kind":"expression","value":"(2, 2, Yes, dimension_at_least_2)","explanation":"Only a verified independent family supplies this size-based lower bound. If this family is dependent, the result withholds its certificate; it does not refute every possible lower bound for the ambient space. The exact result, retaining the requested order, is (2, 2, Yes, dimension_at_least_2)."},"skills":["An Independent Family Gives Only a Lower Bound"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261102-r01-vector-space-dimension-0058","batch_id":"batch-20261014-advance-20261102-r01-vector-space-dimension-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"f9ee3f42a1c43d6802688b94b09239ad143eb2e6c06cf1248bbc3d33317bb1a7","family_id":"qf-mod33-dimension-independent-lower-bound","seed":202611021058,"semantic_template":"qf-mod33-dimension-independent-lower-bound/{\"dimension\":3,\"vectors\":[[1,0,0],[7,1,0]]}"},"fields":{"title":"An Independent Family Gives Only a Lower Bound","stem":"Use the complete ordered real coordinate family vectors=[[1, 0, 0], [7, 1, 0]] in ambient R^3, limited to dimensions 2 or 3. Treat vectors as columns and calculate rank by exact rational Gaussian elimination, not floating-point samples. A basis is both independent and spanning. An independent family gives only a lower bound; a spanning family gives only an upper bound. Basis-cardinality invariance uses the finite-dimensional theorem, not a few examples. Give (vector_count,rank,independent?,lower_bound). The last field is dimension_at_least_vector_count with the numeric count, or no_lower_bound_from_dependent_family.","answer":{"kind":"expression","value":"(2, 2, Yes, dimension_at_least_2)","explanation":"Only a verified independent family supplies this size-based lower bound. If this family is dependent, the result withholds its certificate; it does not refute every possible lower bound for the ambient space. The exact result, retaining the requested order, is (2, 2, Yes, dimension_at_least_2)."},"skills":["An Independent Family Gives Only a Lower Bound"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261102-r01-vector-space-dimension-0059","batch_id":"batch-20261014-advance-20261102-r01-vector-space-dimension-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"db294b6876cfcc141b4c44b7e7bd811260d6a50ecbcd09ed1201364c62976a91","family_id":"qf-mod33-dimension-independent-lower-bound","seed":202611021059,"semantic_template":"qf-mod33-dimension-independent-lower-bound/{\"dimension\":3,\"vectors\":[[1,0,0],[8,1,0]]}"},"fields":{"title":"An Independent Family Gives Only a Lower Bound","stem":"Use the complete ordered real coordinate family vectors=[[1, 0, 0], [8, 1, 0]] in ambient R^3, limited to dimensions 2 or 3. Treat vectors as columns and calculate rank by exact rational Gaussian elimination, not floating-point samples. A basis is both independent and spanning. An independent family gives only a lower bound; a spanning family gives only an upper bound. Basis-cardinality invariance uses the finite-dimensional theorem, not a few examples. Give (vector_count,rank,independent?,lower_bound). The last field is dimension_at_least_vector_count with the numeric count, or no_lower_bound_from_dependent_family.","answer":{"kind":"expression","value":"(2, 2, Yes, dimension_at_least_2)","explanation":"Only a verified independent family supplies this size-based lower bound. If this family is dependent, the result withholds its certificate; it does not refute every possible lower bound for the ambient space. The exact result, retaining the requested order, is (2, 2, Yes, dimension_at_least_2)."},"skills":["An Independent Family Gives Only a Lower Bound"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261102-r01-vector-space-dimension-0060","batch_id":"batch-20261014-advance-20261102-r01-vector-space-dimension-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"40c7d222c8f32d70822c58bc9093c807a759bd675bfdde2aac374aaed9439935","family_id":"qf-mod33-dimension-independent-lower-bound","seed":202611021060,"semantic_template":"qf-mod33-dimension-independent-lower-bound/{\"dimension\":3,\"vectors\":[[1,0,0],[9,1,0]]}"},"fields":{"title":"An Independent Family Gives Only a Lower Bound","stem":"Use the complete ordered real coordinate family vectors=[[1, 0, 0], [9, 1, 0]] in ambient R^3, limited to dimensions 2 or 3. Treat vectors as columns and calculate rank by exact rational Gaussian elimination, not floating-point samples. A basis is both independent and spanning. An independent family gives only a lower bound; a spanning family gives only an upper bound. Basis-cardinality invariance uses the finite-dimensional theorem, not a few examples. Give (vector_count,rank,independent?,lower_bound). The last field is dimension_at_least_vector_count with the numeric count, or no_lower_bound_from_dependent_family.","answer":{"kind":"expression","value":"(2, 2, Yes, dimension_at_least_2)","explanation":"Only a verified independent family supplies this size-based lower bound. If this family is dependent, the result withholds its certificate; it does not refute every possible lower bound for the ambient space. The exact result, retaining the requested order, is (2, 2, Yes, dimension_at_least_2)."},"skills":["An Independent Family Gives Only a Lower Bound"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261102-r01-vector-space-dimension-0061","batch_id":"batch-20261014-advance-20261102-r01-vector-space-dimension-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"9c1ca755345186e513ddae7ce844b3aa6f160b5fffb8f9bb15d4df8787232674","family_id":"qf-mod33-dimension-independent-lower-bound","seed":202611021061,"semantic_template":"qf-mod33-dimension-independent-lower-bound/{\"dimension\":3,\"vectors\":[[1,0,0],[10,1,0]]}"},"fields":{"title":"An Independent Family Gives Only a Lower Bound","stem":"Use the complete ordered real coordinate family vectors=[[1, 0, 0], [10, 1, 0]] in ambient R^3, limited to dimensions 2 or 3. Treat vectors as columns and calculate rank by exact rational Gaussian elimination, not floating-point samples. A basis is both independent and spanning. An independent family gives only a lower bound; a spanning family gives only an upper bound. Basis-cardinality invariance uses the finite-dimensional theorem, not a few examples. Give (vector_count,rank,independent?,lower_bound). The last field is dimension_at_least_vector_count with the numeric count, or no_lower_bound_from_dependent_family.","answer":{"kind":"expression","value":"(2, 2, Yes, dimension_at_least_2)","explanation":"Only a verified independent family supplies this size-based lower bound. If this family is dependent, the result withholds its certificate; it does not refute every possible lower bound for the ambient space. The exact result, retaining the requested order, is (2, 2, Yes, dimension_at_least_2)."},"skills":["An Independent Family Gives Only a Lower Bound"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261102-r01-vector-space-dimension-0062","batch_id":"batch-20261014-advance-20261102-r01-vector-space-dimension-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"a398682921c9d961e6bb4d5967d79b9d6919f0f13b941305f25cfc368a075e3b","family_id":"qf-mod33-dimension-independent-lower-bound","seed":202611021062,"semantic_template":"qf-mod33-dimension-independent-lower-bound/{\"dimension\":3,\"vectors\":[[1,0,0],[11,1,0]]}"},"fields":{"title":"An Independent Family Gives Only a Lower Bound","stem":"Use the complete ordered real coordinate family vectors=[[1, 0, 0], [11, 1, 0]] in ambient R^3, limited to dimensions 2 or 3. Treat vectors as columns and calculate rank by exact rational Gaussian elimination, not floating-point samples. A basis is both independent and spanning. An independent family gives only a lower bound; a spanning family gives only an upper bound. Basis-cardinality invariance uses the finite-dimensional theorem, not a few examples. Give (vector_count,rank,independent?,lower_bound). The last field is dimension_at_least_vector_count with the numeric count, or no_lower_bound_from_dependent_family.","answer":{"kind":"expression","value":"(2, 2, Yes, dimension_at_least_2)","explanation":"Only a verified independent family supplies this size-based lower bound. If this family is dependent, the result withholds its certificate; it does not refute every possible lower bound for the ambient space. The exact result, retaining the requested order, is (2, 2, Yes, dimension_at_least_2)."},"skills":["An Independent Family Gives Only a Lower Bound"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261102-r01-vector-space-dimension-0063","batch_id":"batch-20261014-advance-20261102-r01-vector-space-dimension-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"b26687ac5e2cf257143766cfc428bb1e5f6046ce8303fd2a89875fce300115d5","family_id":"qf-mod33-dimension-independent-lower-bound","seed":202611021063,"semantic_template":"qf-mod33-dimension-independent-lower-bound/{\"dimension\":3,\"vectors\":[[1,0,0],[12,1,0]]}"},"fields":{"title":"An Independent Family Gives Only a Lower Bound","stem":"Use the complete ordered real coordinate family vectors=[[1, 0, 0], [12, 1, 0]] in ambient R^3, limited to dimensions 2 or 3. Treat vectors as columns and calculate rank by exact rational Gaussian elimination, not floating-point samples. A basis is both independent and spanning. An independent family gives only a lower bound; a spanning family gives only an upper bound. Basis-cardinality invariance uses the finite-dimensional theorem, not a few examples. Give (vector_count,rank,independent?,lower_bound). The last field is dimension_at_least_vector_count with the numeric count, or no_lower_bound_from_dependent_family.","answer":{"kind":"expression","value":"(2, 2, Yes, dimension_at_least_2)","explanation":"Only a verified independent family supplies this size-based lower bound. If this family is dependent, the result withholds its certificate; it does not refute every possible lower bound for the ambient space. The exact result, retaining the requested order, is (2, 2, Yes, dimension_at_least_2)."},"skills":["An Independent Family Gives Only a Lower Bound"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261102-r01-vector-space-dimension-0064","batch_id":"batch-20261014-advance-20261102-r01-vector-space-dimension-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"0d2ea8951e460dd706b0153dec66a9949f8a5a7789cc2ebf23e466597fe18d87","family_id":"qf-mod33-dimension-independent-lower-bound","seed":202611021064,"semantic_template":"qf-mod33-dimension-independent-lower-bound/{\"dimension\":3,\"vectors\":[[1,0,0],[13,1,0]]}"},"fields":{"title":"An Independent Family Gives Only a Lower Bound","stem":"Use the complete ordered real coordinate family vectors=[[1, 0, 0], [13, 1, 0]] in ambient R^3, limited to dimensions 2 or 3. Treat vectors as columns and calculate rank by exact rational Gaussian elimination, not floating-point samples. A basis is both independent and spanning. An independent family gives only a lower bound; a spanning family gives only an upper bound. Basis-cardinality invariance uses the finite-dimensional theorem, not a few examples. Give (vector_count,rank,independent?,lower_bound). The last field is dimension_at_least_vector_count with the numeric count, or no_lower_bound_from_dependent_family.","answer":{"kind":"expression","value":"(2, 2, Yes, dimension_at_least_2)","explanation":"Only a verified independent family supplies this size-based lower bound. If this family is dependent, the result withholds its certificate; it does not refute every possible lower bound for the ambient space. The exact result, retaining the requested order, is (2, 2, Yes, dimension_at_least_2)."},"skills":["An Independent Family Gives Only a Lower Bound"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261102-r01-vector-space-dimension-0065","batch_id":"batch-20261014-advance-20261102-r01-vector-space-dimension-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"99c845b260702ad211959a9fdf0709975a7559dbbd30cfc178b75e91f1c29f41","family_id":"qf-mod33-dimension-independent-lower-bound","seed":202611021065,"semantic_template":"qf-mod33-dimension-independent-lower-bound/{\"dimension\":3,\"vectors\":[[1,0,0],[14,1,0]]}"},"fields":{"title":"An Independent Family Gives Only a Lower Bound","stem":"Use the complete ordered real coordinate family vectors=[[1, 0, 0], [14, 1, 0]] in ambient R^3, limited to dimensions 2 or 3. Treat vectors as columns and calculate rank by exact rational Gaussian elimination, not floating-point samples. A basis is both independent and spanning. An independent family gives only a lower bound; a spanning family gives only an upper bound. Basis-cardinality invariance uses the finite-dimensional theorem, not a few examples. Give (vector_count,rank,independent?,lower_bound). The last field is dimension_at_least_vector_count with the numeric count, or no_lower_bound_from_dependent_family.","answer":{"kind":"expression","value":"(2, 2, Yes, dimension_at_least_2)","explanation":"Only a verified independent family supplies this size-based lower bound. If this family is dependent, the result withholds its certificate; it does not refute every possible lower bound for the ambient space. The exact result, retaining the requested order, is (2, 2, Yes, dimension_at_least_2)."},"skills":["An Independent Family Gives Only a Lower Bound"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261102-r01-vector-space-dimension-0066","batch_id":"batch-20261014-advance-20261102-r01-vector-space-dimension-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"fa9717c97798b98e82defcf1bcf14bdb29ad43a73cef99e4cd764d6653561dc5","family_id":"qf-mod33-dimension-independent-lower-bound","seed":202611021066,"semantic_template":"qf-mod33-dimension-independent-lower-bound/{\"dimension\":3,\"vectors\":[[1,0,0],[15,1,0]]}"},"fields":{"title":"An Independent Family Gives Only a Lower Bound","stem":"Use the complete ordered real coordinate family vectors=[[1, 0, 0], [15, 1, 0]] in ambient R^3, limited to dimensions 2 or 3. Treat vectors as columns and calculate rank by exact rational Gaussian elimination, not floating-point samples. A basis is both independent and spanning. An independent family gives only a lower bound; a spanning family gives only an upper bound. Basis-cardinality invariance uses the finite-dimensional theorem, not a few examples. Give (vector_count,rank,independent?,lower_bound). The last field is dimension_at_least_vector_count with the numeric count, or no_lower_bound_from_dependent_family.","answer":{"kind":"expression","value":"(2, 2, Yes, dimension_at_least_2)","explanation":"Only a verified independent family supplies this size-based lower bound. If this family is dependent, the result withholds its certificate; it does not refute every possible lower bound for the ambient space. The exact result, retaining the requested order, is (2, 2, Yes, dimension_at_least_2)."},"skills":["An Independent Family Gives Only a Lower Bound"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261102-r01-vector-space-dimension-0067","batch_id":"batch-20261014-advance-20261102-r01-vector-space-dimension-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"ae77c25aab7bc446ed42ee7d4db73796c15224a0ed1cb2e8d4aa6baccbbac639","family_id":"qf-mod33-dimension-independent-lower-bound","seed":202611021067,"semantic_template":"qf-mod33-dimension-independent-lower-bound/{\"dimension\":3,\"vectors\":[[1,0,0],[16,1,0]]}"},"fields":{"title":"An Independent Family Gives Only a Lower Bound","stem":"Use the complete ordered real coordinate family vectors=[[1, 0, 0], [16, 1, 0]] in ambient R^3, limited to dimensions 2 or 3. Treat vectors as columns and calculate rank by exact rational Gaussian elimination, not floating-point samples. A basis is both independent and spanning. An independent family gives only a lower bound; a spanning family gives only an upper bound. Basis-cardinality invariance uses the finite-dimensional theorem, not a few examples. Give (vector_count,rank,independent?,lower_bound). The last field is dimension_at_least_vector_count with the numeric count, or no_lower_bound_from_dependent_family.","answer":{"kind":"expression","value":"(2, 2, Yes, dimension_at_least_2)","explanation":"Only a verified independent family supplies this size-based lower bound. If this family is dependent, the result withholds its certificate; it does not refute every possible lower bound for the ambient space. The exact result, retaining the requested order, is (2, 2, Yes, dimension_at_least_2)."},"skills":["An Independent Family Gives Only a Lower Bound"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261102-r01-vector-space-dimension-0068","batch_id":"batch-20261014-advance-20261102-r01-vector-space-dimension-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"1cce499b0d0a9412cef5dd0611c04584827a5983ad45b9cd35db1d82b2238f35","family_id":"qf-mod33-dimension-independent-lower-bound","seed":202611021068,"semantic_template":"qf-mod33-dimension-independent-lower-bound/{\"dimension\":3,\"vectors\":[[1,0,0],[17,1,0]]}"},"fields":{"title":"An Independent Family Gives Only a Lower Bound","stem":"Use the complete ordered real coordinate family vectors=[[1, 0, 0], [17, 1, 0]] in ambient R^3, limited to dimensions 2 or 3. Treat vectors as columns and calculate rank by exact rational Gaussian elimination, not floating-point samples. A basis is both independent and spanning. An independent family gives only a lower bound; a spanning family gives only an upper bound. Basis-cardinality invariance uses the finite-dimensional theorem, not a few examples. Give (vector_count,rank,independent?,lower_bound). The last field is dimension_at_least_vector_count with the numeric count, or no_lower_bound_from_dependent_family.","answer":{"kind":"expression","value":"(2, 2, Yes, dimension_at_least_2)","explanation":"Only a verified independent family supplies this size-based lower bound. If this family is dependent, the result withholds its certificate; it does not refute every possible lower bound for the ambient space. The exact result, retaining the requested order, is (2, 2, Yes, dimension_at_least_2)."},"skills":["An Independent Family Gives Only a Lower Bound"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261102-r01-vector-space-dimension-0069","batch_id":"batch-20261014-advance-20261102-r01-vector-space-dimension-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"f5777158e907651cffc7cb1e96ed544546dd5218bc24effd9346a01092932703","family_id":"qf-mod33-dimension-independent-lower-bound","seed":202611021069,"semantic_template":"qf-mod33-dimension-independent-lower-bound/{\"dimension\":3,\"vectors\":[[1,0,0],[18,1,0]]}"},"fields":{"title":"An Independent Family Gives Only a Lower Bound","stem":"Use the complete ordered real coordinate family vectors=[[1, 0, 0], [18, 1, 0]] in ambient R^3, limited to dimensions 2 or 3. Treat vectors as columns and calculate rank by exact rational Gaussian elimination, not floating-point samples. A basis is both independent and spanning. An independent family gives only a lower bound; a spanning family gives only an upper bound. Basis-cardinality invariance uses the finite-dimensional theorem, not a few examples. Give (vector_count,rank,independent?,lower_bound). The last field is dimension_at_least_vector_count with the numeric count, or no_lower_bound_from_dependent_family.","answer":{"kind":"expression","value":"(2, 2, Yes, dimension_at_least_2)","explanation":"Only a verified independent family supplies this size-based lower bound. If this family is dependent, the result withholds its certificate; it does not refute every possible lower bound for the ambient space. The exact result, retaining the requested order, is (2, 2, Yes, dimension_at_least_2)."},"skills":["An Independent Family Gives Only a Lower Bound"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261102-r01-vector-space-dimension-0070","batch_id":"batch-20261014-advance-20261102-r01-vector-space-dimension-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"83ed0c11ef566b88d15704aa4dcea425973eace19e6513fa52c31a154fde0c16","family_id":"qf-mod33-dimension-independent-lower-bound","seed":202611021070,"semantic_template":"qf-mod33-dimension-independent-lower-bound/{\"dimension\":3,\"vectors\":[[1,0,0],[19,1,0]]}"},"fields":{"title":"An Independent Family Gives Only a Lower Bound","stem":"Use the complete ordered real coordinate family vectors=[[1, 0, 0], [19, 1, 0]] in ambient R^3, limited to dimensions 2 or 3. Treat vectors as columns and calculate rank by exact rational Gaussian elimination, not floating-point samples. A basis is both independent and spanning. An independent family gives only a lower bound; a spanning family gives only an upper bound. Basis-cardinality invariance uses the finite-dimensional theorem, not a few examples. Give (vector_count,rank,independent?,lower_bound). The last field is dimension_at_least_vector_count with the numeric count, or no_lower_bound_from_dependent_family.","answer":{"kind":"expression","value":"(2, 2, Yes, dimension_at_least_2)","explanation":"Only a verified independent family supplies this size-based lower bound. If this family is dependent, the result withholds its certificate; it does not refute every possible lower bound for the ambient space. The exact result, retaining the requested order, is (2, 2, Yes, dimension_at_least_2)."},"skills":["An Independent Family Gives Only a Lower Bound"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261102-r01-vector-space-dimension-0071","batch_id":"batch-20261014-advance-20261102-r01-vector-space-dimension-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"1877dac90cfed9f848153ddf919fceedf18eb41b0f34e9cb82ce0c59a520abec","family_id":"qf-mod33-dimension-independent-lower-bound","seed":202611021071,"semantic_template":"qf-mod33-dimension-independent-lower-bound/{\"dimension\":3,\"vectors\":[[1,0,0],[20,1,0]]}"},"fields":{"title":"An Independent Family Gives Only a Lower Bound","stem":"Use the complete ordered real coordinate family vectors=[[1, 0, 0], [20, 1, 0]] in ambient R^3, limited to dimensions 2 or 3. Treat vectors as columns and calculate rank by exact rational Gaussian elimination, not floating-point samples. A basis is both independent and spanning. An independent family gives only a lower bound; a spanning family gives only an upper bound. Basis-cardinality invariance uses the finite-dimensional theorem, not a few examples. Give (vector_count,rank,independent?,lower_bound). The last field is dimension_at_least_vector_count with the numeric count, or no_lower_bound_from_dependent_family.","answer":{"kind":"expression","value":"(2, 2, Yes, dimension_at_least_2)","explanation":"Only a verified independent family supplies this size-based lower bound. If this family is dependent, the result withholds its certificate; it does not refute every possible lower bound for the ambient space. The exact result, retaining the requested order, is (2, 2, Yes, dimension_at_least_2)."},"skills":["An Independent Family Gives Only a Lower Bound"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261102-r01-vector-space-dimension-0072","batch_id":"batch-20261014-advance-20261102-r01-vector-space-dimension-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"7a71a60a36f131dafeb7141651c9b364b8510c7b9308f1295b5928f3b3f3a03c","family_id":"qf-mod33-dimension-independent-lower-bound","seed":202611021072,"semantic_template":"qf-mod33-dimension-independent-lower-bound/{\"dimension\":3,\"vectors\":[[1,0,0],[21,1,0]]}"},"fields":{"title":"An Independent Family Gives Only a Lower Bound","stem":"Use the complete ordered real coordinate family vectors=[[1, 0, 0], [21, 1, 0]] in ambient R^3, limited to dimensions 2 or 3. Treat vectors as columns and calculate rank by exact rational Gaussian elimination, not floating-point samples. A basis is both independent and spanning. An independent family gives only a lower bound; a spanning family gives only an upper bound. Basis-cardinality invariance uses the finite-dimensional theorem, not a few examples. Give (vector_count,rank,independent?,lower_bound). The last field is dimension_at_least_vector_count with the numeric count, or no_lower_bound_from_dependent_family.","answer":{"kind":"expression","value":"(2, 2, Yes, dimension_at_least_2)","explanation":"Only a verified independent family supplies this size-based lower bound. If this family is dependent, the result withholds its certificate; it does not refute every possible lower bound for the ambient space. The exact result, retaining the requested order, is (2, 2, Yes, dimension_at_least_2)."},"skills":["An Independent Family Gives Only a Lower Bound"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261102-r01-vector-space-dimension-0073","batch_id":"batch-20261014-advance-20261102-r01-vector-space-dimension-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"ee4f221592cd9c83a97e8b1037baed5dc8b2f80b909d711fb224cc63161802ab","family_id":"qf-mod33-dimension-independent-lower-bound","seed":202611021073,"semantic_template":"qf-mod33-dimension-independent-lower-bound/{\"dimension\":3,\"vectors\":[[1,0,0],[22,1,0]]}"},"fields":{"title":"An Independent Family Gives Only a Lower Bound","stem":"Use the complete ordered real coordinate family vectors=[[1, 0, 0], [22, 1, 0]] in ambient R^3, limited to dimensions 2 or 3. Treat vectors as columns and calculate rank by exact rational Gaussian elimination, not floating-point samples. A basis is both independent and spanning. An independent family gives only a lower bound; a spanning family gives only an upper bound. Basis-cardinality invariance uses the finite-dimensional theorem, not a few examples. Give (vector_count,rank,independent?,lower_bound). The last field is dimension_at_least_vector_count with the numeric count, or no_lower_bound_from_dependent_family.","answer":{"kind":"expression","value":"(2, 2, Yes, dimension_at_least_2)","explanation":"Only a verified independent family supplies this size-based lower bound. If this family is dependent, the result withholds its certificate; it does not refute every possible lower bound for the ambient space. The exact result, retaining the requested order, is (2, 2, Yes, dimension_at_least_2)."},"skills":["An Independent Family Gives Only a Lower Bound"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261102-r01-vector-space-dimension-0074","batch_id":"batch-20261014-advance-20261102-r01-vector-space-dimension-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"bbc7536a2b6b5c36ef4fe7a562fd4615d20948db976770404cf48865a582f044","family_id":"qf-mod33-dimension-independent-lower-bound","seed":202611021074,"semantic_template":"qf-mod33-dimension-independent-lower-bound/{\"dimension\":3,\"vectors\":[[1,0,0],[23,1,0]]}"},"fields":{"title":"An Independent Family Gives Only a Lower Bound","stem":"Use the complete ordered real coordinate family vectors=[[1, 0, 0], [23, 1, 0]] in ambient R^3, limited to dimensions 2 or 3. Treat vectors as columns and calculate rank by exact rational Gaussian elimination, not floating-point samples. A basis is both independent and spanning. An independent family gives only a lower bound; a spanning family gives only an upper bound. Basis-cardinality invariance uses the finite-dimensional theorem, not a few examples. Give (vector_count,rank,independent?,lower_bound). The last field is dimension_at_least_vector_count with the numeric count, or no_lower_bound_from_dependent_family.","answer":{"kind":"expression","value":"(2, 2, Yes, dimension_at_least_2)","explanation":"Only a verified independent family supplies this size-based lower bound. If this family is dependent, the result withholds its certificate; it does not refute every possible lower bound for the ambient space. The exact result, retaining the requested order, is (2, 2, Yes, dimension_at_least_2)."},"skills":["An Independent Family Gives Only a Lower Bound"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261102-r01-vector-space-dimension-0075","batch_id":"batch-20261014-advance-20261102-r01-vector-space-dimension-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"98f2f42f46069106c2e76277c424cf59b99fe4c2f9547f9b8e386dc0ceacaf9d","family_id":"qf-mod33-dimension-independent-lower-bound","seed":202611021075,"semantic_template":"qf-mod33-dimension-independent-lower-bound/{\"dimension\":3,\"vectors\":[[1,0,0],[24,1,0]]}"},"fields":{"title":"An Independent Family Gives Only a Lower Bound","stem":"Use the complete ordered real coordinate family vectors=[[1, 0, 0], [24, 1, 0]] in ambient R^3, limited to dimensions 2 or 3. Treat vectors as columns and calculate rank by exact rational Gaussian elimination, not floating-point samples. A basis is both independent and spanning. An independent family gives only a lower bound; a spanning family gives only an upper bound. Basis-cardinality invariance uses the finite-dimensional theorem, not a few examples. Give (vector_count,rank,independent?,lower_bound). The last field is dimension_at_least_vector_count with the numeric count, or no_lower_bound_from_dependent_family.","answer":{"kind":"expression","value":"(2, 2, Yes, dimension_at_least_2)","explanation":"Only a verified independent family supplies this size-based lower bound. If this family is dependent, the result withholds its certificate; it does not refute every possible lower bound for the ambient space. The exact result, retaining the requested order, is (2, 2, Yes, dimension_at_least_2)."},"skills":["An Independent Family Gives Only a Lower Bound"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261102-r01-vector-space-dimension-0076","batch_id":"batch-20261014-advance-20261102-r01-vector-space-dimension-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"326927e1817147c574e674fa37055e213d7a41976033df6e902f47a5163779c2","family_id":"qf-mod33-dimension-spanning-upper-bound","seed":202611021076,"semantic_template":"qf-mod33-dimension-spanning-upper-bound/{\"dimension\":2,\"vectors\":[[0,1],[1,0],[1,1]]}"},"fields":{"title":"A Spanning Family May Give a Loose Upper Bound","stem":"Use the complete ordered real coordinate family vectors=[[1, 0], [0, 1], [1, 1]] in ambient R^2, limited to dimensions 2 or 3. Treat vectors as columns and calculate rank by exact rational Gaussian elimination, not floating-point samples. A basis is both independent and spanning. An independent family gives only a lower bound; a spanning family gives only an upper bound. Basis-cardinality invariance uses the finite-dimensional theorem, not a few examples. Give (vector_count,rank,spanning?,upper_bound). The last field is dimension_at_most_vector_count with the numeric count, or no_upper_bound_from_nonspanning_family.","answer":{"kind":"expression","value":"(3, 2, Yes, dimension_at_most_3)","explanation":"Only a verified spanning family supplies this upper bound. A redundant spanning family can be larger than the exact dimension; absence of a certificate from this family does not prove that the space is infinite-dimensional. The exact result, retaining the requested order, is (3, 2, Yes, dimension_at_most_3)."},"skills":["A Spanning Family May Give a Loose Upper Bound"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261102-r01-vector-space-dimension-0077","batch_id":"batch-20261014-advance-20261102-r01-vector-space-dimension-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"4b7d568d3c9685ba67f42540af0d6ae4c5e30e2a1a176ef3c9616dad6ba92a84","family_id":"qf-mod33-dimension-spanning-upper-bound","seed":202611021077,"semantic_template":"qf-mod33-dimension-spanning-upper-bound/{\"dimension\":2,\"vectors\":[[0,1],[1,0],[2,1]]}"},"fields":{"title":"A Spanning Family May Give a Loose Upper Bound","stem":"Use the complete ordered real coordinate family vectors=[[1, 0], [0, 1], [2, 1]] in ambient R^2, limited to dimensions 2 or 3. Treat vectors as columns and calculate rank by exact rational Gaussian elimination, not floating-point samples. A basis is both independent and spanning. An independent family gives only a lower bound; a spanning family gives only an upper bound. Basis-cardinality invariance uses the finite-dimensional theorem, not a few examples. Give (vector_count,rank,spanning?,upper_bound). The last field is dimension_at_most_vector_count with the numeric count, or no_upper_bound_from_nonspanning_family.","answer":{"kind":"expression","value":"(3, 2, Yes, dimension_at_most_3)","explanation":"Only a verified spanning family supplies this upper bound. A redundant spanning family can be larger than the exact dimension; absence of a certificate from this family does not prove that the space is infinite-dimensional. The exact result, retaining the requested order, is (3, 2, Yes, dimension_at_most_3)."},"skills":["A Spanning Family May Give a Loose Upper Bound"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261102-r01-vector-space-dimension-0078","batch_id":"batch-20261014-advance-20261102-r01-vector-space-dimension-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"a95d119a1fd3d77a7d2e6e85889a829dd4381bfd740a70fd22350bbd2b290e61","family_id":"qf-mod33-dimension-spanning-upper-bound","seed":202611021078,"semantic_template":"qf-mod33-dimension-spanning-upper-bound/{\"dimension\":2,\"vectors\":[[0,1],[1,0],[3,1]]}"},"fields":{"title":"A Spanning Family May Give a Loose Upper Bound","stem":"Use the complete ordered real coordinate family vectors=[[1, 0], [0, 1], [3, 1]] in ambient R^2, limited to dimensions 2 or 3. Treat vectors as columns and calculate rank by exact rational Gaussian elimination, not floating-point samples. A basis is both independent and spanning. An independent family gives only a lower bound; a spanning family gives only an upper bound. Basis-cardinality invariance uses the finite-dimensional theorem, not a few examples. Give (vector_count,rank,spanning?,upper_bound). The last field is dimension_at_most_vector_count with the numeric count, or no_upper_bound_from_nonspanning_family.","answer":{"kind":"expression","value":"(3, 2, Yes, dimension_at_most_3)","explanation":"Only a verified spanning family supplies this upper bound. A redundant spanning family can be larger than the exact dimension; absence of a certificate from this family does not prove that the space is infinite-dimensional. The exact result, retaining the requested order, is (3, 2, Yes, dimension_at_most_3)."},"skills":["A Spanning Family May Give a Loose Upper Bound"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261102-r01-vector-space-dimension-0079","batch_id":"batch-20261014-advance-20261102-r01-vector-space-dimension-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"ce2d39d83dde57d2f9e5b66aac4f97e79a16270af0bd4bc74a6b87ec9c7c74c9","family_id":"qf-mod33-dimension-spanning-upper-bound","seed":202611021079,"semantic_template":"qf-mod33-dimension-spanning-upper-bound/{\"dimension\":2,\"vectors\":[[0,1],[1,0],[4,1]]}"},"fields":{"title":"A Spanning Family May Give a Loose Upper Bound","stem":"Use the complete ordered real coordinate family vectors=[[1, 0], [0, 1], [4, 1]] in ambient R^2, limited to dimensions 2 or 3. Treat vectors as columns and calculate rank by exact rational Gaussian elimination, not floating-point samples. A basis is both independent and spanning. An independent family gives only a lower bound; a spanning family gives only an upper bound. Basis-cardinality invariance uses the finite-dimensional theorem, not a few examples. Give (vector_count,rank,spanning?,upper_bound). The last field is dimension_at_most_vector_count with the numeric count, or no_upper_bound_from_nonspanning_family.","answer":{"kind":"expression","value":"(3, 2, Yes, dimension_at_most_3)","explanation":"Only a verified spanning family supplies this upper bound. A redundant spanning family can be larger than the exact dimension; absence of a certificate from this family does not prove that the space is infinite-dimensional. The exact result, retaining the requested order, is (3, 2, Yes, dimension_at_most_3)."},"skills":["A Spanning Family May Give a Loose Upper Bound"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261102-r01-vector-space-dimension-0080","batch_id":"batch-20261014-advance-20261102-r01-vector-space-dimension-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"75135c494c89e855f1827994fae5f0f959a51f59e60b6778b43641ee7c340832","family_id":"qf-mod33-dimension-spanning-upper-bound","seed":202611021080,"semantic_template":"qf-mod33-dimension-spanning-upper-bound/{\"dimension\":2,\"vectors\":[[0,1],[1,0],[5,1]]}"},"fields":{"title":"A Spanning Family May Give a Loose Upper Bound","stem":"Use the complete ordered real coordinate family vectors=[[1, 0], [0, 1], [5, 1]] in ambient R^2, limited to dimensions 2 or 3. Treat vectors as columns and calculate rank by exact rational Gaussian elimination, not floating-point samples. A basis is both independent and spanning. An independent family gives only a lower bound; a spanning family gives only an upper bound. Basis-cardinality invariance uses the finite-dimensional theorem, not a few examples. Give (vector_count,rank,spanning?,upper_bound). The last field is dimension_at_most_vector_count with the numeric count, or no_upper_bound_from_nonspanning_family.","answer":{"kind":"expression","value":"(3, 2, Yes, dimension_at_most_3)","explanation":"Only a verified spanning family supplies this upper bound. A redundant spanning family can be larger than the exact dimension; absence of a certificate from this family does not prove that the space is infinite-dimensional. The exact result, retaining the requested order, is (3, 2, Yes, dimension_at_most_3)."},"skills":["A Spanning Family May Give a Loose Upper Bound"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261102-r01-vector-space-dimension-0081","batch_id":"batch-20261014-advance-20261102-r01-vector-space-dimension-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"4f5f9dbcd1d233baebd635e0b7ad390d00fb6296191c63290e9baaebc1f97ab9","family_id":"qf-mod33-dimension-spanning-upper-bound","seed":202611021081,"semantic_template":"qf-mod33-dimension-spanning-upper-bound/{\"dimension\":2,\"vectors\":[[0,1],[1,0],[6,1]]}"},"fields":{"title":"A Spanning Family May Give a Loose Upper Bound","stem":"Use the complete ordered real coordinate family vectors=[[1, 0], [0, 1], [6, 1]] in ambient R^2, limited to dimensions 2 or 3. Treat vectors as columns and calculate rank by exact rational Gaussian elimination, not floating-point samples. A basis is both independent and spanning. An independent family gives only a lower bound; a spanning family gives only an upper bound. Basis-cardinality invariance uses the finite-dimensional theorem, not a few examples. Give (vector_count,rank,spanning?,upper_bound). The last field is dimension_at_most_vector_count with the numeric count, or no_upper_bound_from_nonspanning_family.","answer":{"kind":"expression","value":"(3, 2, Yes, dimension_at_most_3)","explanation":"Only a verified spanning family supplies this upper bound. A redundant spanning family can be larger than the exact dimension; absence of a certificate from this family does not prove that the space is infinite-dimensional. The exact result, retaining the requested order, is (3, 2, Yes, dimension_at_most_3)."},"skills":["A Spanning Family May Give a Loose Upper Bound"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261102-r01-vector-space-dimension-0082","batch_id":"batch-20261014-advance-20261102-r01-vector-space-dimension-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"3a92d60f09067a815e1391cc15b25fbc12c59e1c76a9104e11a05495968b2eeb","family_id":"qf-mod33-dimension-spanning-upper-bound","seed":202611021082,"semantic_template":"qf-mod33-dimension-spanning-upper-bound/{\"dimension\":2,\"vectors\":[[0,1],[1,0],[7,1]]}"},"fields":{"title":"A Spanning Family May Give a Loose Upper Bound","stem":"Use the complete ordered real coordinate family vectors=[[1, 0], [0, 1], [7, 1]] in ambient R^2, limited to dimensions 2 or 3. Treat vectors as columns and calculate rank by exact rational Gaussian elimination, not floating-point samples. A basis is both independent and spanning. An independent family gives only a lower bound; a spanning family gives only an upper bound. Basis-cardinality invariance uses the finite-dimensional theorem, not a few examples. Give (vector_count,rank,spanning?,upper_bound). The last field is dimension_at_most_vector_count with the numeric count, or no_upper_bound_from_nonspanning_family.","answer":{"kind":"expression","value":"(3, 2, Yes, dimension_at_most_3)","explanation":"Only a verified spanning family supplies this upper bound. A redundant spanning family can be larger than the exact dimension; absence of a certificate from this family does not prove that the space is infinite-dimensional. The exact result, retaining the requested order, is (3, 2, Yes, dimension_at_most_3)."},"skills":["A Spanning Family May Give a Loose Upper Bound"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261102-r01-vector-space-dimension-0083","batch_id":"batch-20261014-advance-20261102-r01-vector-space-dimension-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"9abbd79da47185648d8c0e2cb1f3eb09b02bbca9cc3c5bf912ad3a0cde153576","family_id":"qf-mod33-dimension-spanning-upper-bound","seed":202611021083,"semantic_template":"qf-mod33-dimension-spanning-upper-bound/{\"dimension\":2,\"vectors\":[[0,1],[1,0],[8,1]]}"},"fields":{"title":"A Spanning Family May Give a Loose Upper Bound","stem":"Use the complete ordered real coordinate family vectors=[[1, 0], [0, 1], [8, 1]] in ambient R^2, limited to dimensions 2 or 3. Treat vectors as columns and calculate rank by exact rational Gaussian elimination, not floating-point samples. A basis is both independent and spanning. An independent family gives only a lower bound; a spanning family gives only an upper bound. Basis-cardinality invariance uses the finite-dimensional theorem, not a few examples. Give (vector_count,rank,spanning?,upper_bound). The last field is dimension_at_most_vector_count with the numeric count, or no_upper_bound_from_nonspanning_family.","answer":{"kind":"expression","value":"(3, 2, Yes, dimension_at_most_3)","explanation":"Only a verified spanning family supplies this upper bound. A redundant spanning family can be larger than the exact dimension; absence of a certificate from this family does not prove that the space is infinite-dimensional. The exact result, retaining the requested order, is (3, 2, Yes, dimension_at_most_3)."},"skills":["A Spanning Family May Give a Loose Upper Bound"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261102-r01-vector-space-dimension-0084","batch_id":"batch-20261014-advance-20261102-r01-vector-space-dimension-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"d8ebcb02f8345aa64606c5031e059d45087f955ad2b893008ece1eb7ef75f92e","family_id":"qf-mod33-dimension-spanning-upper-bound","seed":202611021084,"semantic_template":"qf-mod33-dimension-spanning-upper-bound/{\"dimension\":2,\"vectors\":[[0,1],[1,0],[9,1]]}"},"fields":{"title":"A Spanning Family May Give a Loose Upper Bound","stem":"Use the complete ordered real coordinate family vectors=[[1, 0], [0, 1], [9, 1]] in ambient R^2, limited to dimensions 2 or 3. Treat vectors as columns and calculate rank by exact rational Gaussian elimination, not floating-point samples. A basis is both independent and spanning. An independent family gives only a lower bound; a spanning family gives only an upper bound. Basis-cardinality invariance uses the finite-dimensional theorem, not a few examples. Give (vector_count,rank,spanning?,upper_bound). The last field is dimension_at_most_vector_count with the numeric count, or no_upper_bound_from_nonspanning_family.","answer":{"kind":"expression","value":"(3, 2, Yes, dimension_at_most_3)","explanation":"Only a verified spanning family supplies this upper bound. A redundant spanning family can be larger than the exact dimension; absence of a certificate from this family does not prove that the space is infinite-dimensional. The exact result, retaining the requested order, is (3, 2, Yes, dimension_at_most_3)."},"skills":["A Spanning Family May Give a Loose Upper Bound"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261102-r01-vector-space-dimension-0085","batch_id":"batch-20261014-advance-20261102-r01-vector-space-dimension-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"fcdeb1333ea78820ff426b41a7a50ee0833ccf3280e83c9292ccdc167c1f2a8a","family_id":"qf-mod33-dimension-spanning-upper-bound","seed":202611021085,"semantic_template":"qf-mod33-dimension-spanning-upper-bound/{\"dimension\":2,\"vectors\":[[0,1],[1,0],[10,1]]}"},"fields":{"title":"A Spanning Family May Give a Loose Upper Bound","stem":"Use the complete ordered real coordinate family vectors=[[1, 0], [0, 1], [10, 1]] in ambient R^2, limited to dimensions 2 or 3. Treat vectors as columns and calculate rank by exact rational Gaussian elimination, not floating-point samples. A basis is both independent and spanning. An independent family gives only a lower bound; a spanning family gives only an upper bound. Basis-cardinality invariance uses the finite-dimensional theorem, not a few examples. Give (vector_count,rank,spanning?,upper_bound). The last field is dimension_at_most_vector_count with the numeric count, or no_upper_bound_from_nonspanning_family.","answer":{"kind":"expression","value":"(3, 2, Yes, dimension_at_most_3)","explanation":"Only a verified spanning family supplies this upper bound. A redundant spanning family can be larger than the exact dimension; absence of a certificate from this family does not prove that the space is infinite-dimensional. The exact result, retaining the requested order, is (3, 2, Yes, dimension_at_most_3)."},"skills":["A Spanning Family May Give a Loose Upper Bound"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261102-r01-vector-space-dimension-0086","batch_id":"batch-20261014-advance-20261102-r01-vector-space-dimension-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"8146856871cdd29c4ffa9697f8e187c31ee2bc5e273ff5a0fc6d73f4fe3bb8f6","family_id":"qf-mod33-dimension-spanning-upper-bound","seed":202611021086,"semantic_template":"qf-mod33-dimension-spanning-upper-bound/{\"dimension\":2,\"vectors\":[[0,1],[1,0],[11,1]]}"},"fields":{"title":"A Spanning Family May Give a Loose Upper Bound","stem":"Use the complete ordered real coordinate family vectors=[[1, 0], [0, 1], [11, 1]] in ambient R^2, limited to dimensions 2 or 3. Treat vectors as columns and calculate rank by exact rational Gaussian elimination, not floating-point samples. A basis is both independent and spanning. An independent family gives only a lower bound; a spanning family gives only an upper bound. Basis-cardinality invariance uses the finite-dimensional theorem, not a few examples. Give (vector_count,rank,spanning?,upper_bound). The last field is dimension_at_most_vector_count with the numeric count, or no_upper_bound_from_nonspanning_family.","answer":{"kind":"expression","value":"(3, 2, Yes, dimension_at_most_3)","explanation":"Only a verified spanning family supplies this upper bound. A redundant spanning family can be larger than the exact dimension; absence of a certificate from this family does not prove that the space is infinite-dimensional. The exact result, retaining the requested order, is (3, 2, Yes, dimension_at_most_3)."},"skills":["A Spanning Family May Give a Loose Upper Bound"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261102-r01-vector-space-dimension-0087","batch_id":"batch-20261014-advance-20261102-r01-vector-space-dimension-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"8a5e5192abbf16496905a767ec7c2648ec954c0a697bd9ad895e8393dec56ee8","family_id":"qf-mod33-dimension-spanning-upper-bound","seed":202611021087,"semantic_template":"qf-mod33-dimension-spanning-upper-bound/{\"dimension\":2,\"vectors\":[[0,1],[1,0],[12,1]]}"},"fields":{"title":"A Spanning Family May Give a Loose Upper Bound","stem":"Use the complete ordered real coordinate family vectors=[[1, 0], [0, 1], [12, 1]] in ambient R^2, limited to dimensions 2 or 3. Treat vectors as columns and calculate rank by exact rational Gaussian elimination, not floating-point samples. A basis is both independent and spanning. An independent family gives only a lower bound; a spanning family gives only an upper bound. Basis-cardinality invariance uses the finite-dimensional theorem, not a few examples. Give (vector_count,rank,spanning?,upper_bound). The last field is dimension_at_most_vector_count with the numeric count, or no_upper_bound_from_nonspanning_family.","answer":{"kind":"expression","value":"(3, 2, Yes, dimension_at_most_3)","explanation":"Only a verified spanning family supplies this upper bound. A redundant spanning family can be larger than the exact dimension; absence of a certificate from this family does not prove that the space is infinite-dimensional. The exact result, retaining the requested order, is (3, 2, Yes, dimension_at_most_3)."},"skills":["A Spanning Family May Give a Loose Upper Bound"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261102-r01-vector-space-dimension-0088","batch_id":"batch-20261014-advance-20261102-r01-vector-space-dimension-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"345059e25df9395cf700729dc848031fbf0e696983df903c337c32c091db156b","family_id":"qf-mod33-dimension-spanning-upper-bound","seed":202611021088,"semantic_template":"qf-mod33-dimension-spanning-upper-bound/{\"dimension\":2,\"vectors\":[[0,1],[1,0],[13,1]]}"},"fields":{"title":"A Spanning Family May Give a Loose Upper Bound","stem":"Use the complete ordered real coordinate family vectors=[[1, 0], [0, 1], [13, 1]] in ambient R^2, limited to dimensions 2 or 3. Treat vectors as columns and calculate rank by exact rational Gaussian elimination, not floating-point samples. A basis is both independent and spanning. An independent family gives only a lower bound; a spanning family gives only an upper bound. Basis-cardinality invariance uses the finite-dimensional theorem, not a few examples. Give (vector_count,rank,spanning?,upper_bound). The last field is dimension_at_most_vector_count with the numeric count, or no_upper_bound_from_nonspanning_family.","answer":{"kind":"expression","value":"(3, 2, Yes, dimension_at_most_3)","explanation":"Only a verified spanning family supplies this upper bound. A redundant spanning family can be larger than the exact dimension; absence of a certificate from this family does not prove that the space is infinite-dimensional. The exact result, retaining the requested order, is (3, 2, Yes, dimension_at_most_3)."},"skills":["A Spanning Family May Give a Loose Upper Bound"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261102-r01-vector-space-dimension-0089","batch_id":"batch-20261014-advance-20261102-r01-vector-space-dimension-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"d7275c6c5ad8f774de1faa8cc1993afd0a54253b9ce44da2d32f5304da4ac901","family_id":"qf-mod33-dimension-spanning-upper-bound","seed":202611021089,"semantic_template":"qf-mod33-dimension-spanning-upper-bound/{\"dimension\":2,\"vectors\":[[0,1],[1,0],[14,1]]}"},"fields":{"title":"A Spanning Family May Give a Loose Upper Bound","stem":"Use the complete ordered real coordinate family vectors=[[1, 0], [0, 1], [14, 1]] in ambient R^2, limited to dimensions 2 or 3. Treat vectors as columns and calculate rank by exact rational Gaussian elimination, not floating-point samples. A basis is both independent and spanning. An independent family gives only a lower bound; a spanning family gives only an upper bound. Basis-cardinality invariance uses the finite-dimensional theorem, not a few examples. Give (vector_count,rank,spanning?,upper_bound). The last field is dimension_at_most_vector_count with the numeric count, or no_upper_bound_from_nonspanning_family.","answer":{"kind":"expression","value":"(3, 2, Yes, dimension_at_most_3)","explanation":"Only a verified spanning family supplies this upper bound. A redundant spanning family can be larger than the exact dimension; absence of a certificate from this family does not prove that the space is infinite-dimensional. The exact result, retaining the requested order, is (3, 2, Yes, dimension_at_most_3)."},"skills":["A Spanning Family May Give a Loose Upper Bound"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261102-r01-vector-space-dimension-0090","batch_id":"batch-20261014-advance-20261102-r01-vector-space-dimension-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"eeea62bf66934cc68625c2be4546e013035feb026880ae9c786a60a89f1d9097","family_id":"qf-mod33-dimension-spanning-upper-bound","seed":202611021090,"semantic_template":"qf-mod33-dimension-spanning-upper-bound/{\"dimension\":2,\"vectors\":[[0,1],[1,0],[15,1]]}"},"fields":{"title":"A Spanning Family May Give a Loose Upper Bound","stem":"Use the complete ordered real coordinate family vectors=[[1, 0], [0, 1], [15, 1]] in ambient R^2, limited to dimensions 2 or 3. Treat vectors as columns and calculate rank by exact rational Gaussian elimination, not floating-point samples. A basis is both independent and spanning. An independent family gives only a lower bound; a spanning family gives only an upper bound. Basis-cardinality invariance uses the finite-dimensional theorem, not a few examples. Give (vector_count,rank,spanning?,upper_bound). The last field is dimension_at_most_vector_count with the numeric count, or no_upper_bound_from_nonspanning_family.","answer":{"kind":"expression","value":"(3, 2, Yes, dimension_at_most_3)","explanation":"Only a verified spanning family supplies this upper bound. A redundant spanning family can be larger than the exact dimension; absence of a certificate from this family does not prove that the space is infinite-dimensional. The exact result, retaining the requested order, is (3, 2, Yes, dimension_at_most_3)."},"skills":["A Spanning Family May Give a Loose Upper Bound"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261102-r01-vector-space-dimension-0091","batch_id":"batch-20261014-advance-20261102-r01-vector-space-dimension-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"3a68ef3f114e2f3eea7c79437918cd63bfe586fb15534ad86ea1be299c4026cf","family_id":"qf-mod33-dimension-spanning-upper-bound","seed":202611021091,"semantic_template":"qf-mod33-dimension-spanning-upper-bound/{\"dimension\":2,\"vectors\":[[0,1],[1,0],[16,1]]}"},"fields":{"title":"A Spanning Family May Give a Loose Upper Bound","stem":"Use the complete ordered real coordinate family vectors=[[1, 0], [0, 1], [16, 1]] in ambient R^2, limited to dimensions 2 or 3. Treat vectors as columns and calculate rank by exact rational Gaussian elimination, not floating-point samples. A basis is both independent and spanning. An independent family gives only a lower bound; a spanning family gives only an upper bound. Basis-cardinality invariance uses the finite-dimensional theorem, not a few examples. Give (vector_count,rank,spanning?,upper_bound). The last field is dimension_at_most_vector_count with the numeric count, or no_upper_bound_from_nonspanning_family.","answer":{"kind":"expression","value":"(3, 2, Yes, dimension_at_most_3)","explanation":"Only a verified spanning family supplies this upper bound. A redundant spanning family can be larger than the exact dimension; absence of a certificate from this family does not prove that the space is infinite-dimensional. The exact result, retaining the requested order, is (3, 2, Yes, dimension_at_most_3)."},"skills":["A Spanning Family May Give a Loose Upper Bound"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261102-r01-vector-space-dimension-0092","batch_id":"batch-20261014-advance-20261102-r01-vector-space-dimension-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"6a3bc545939963ae97e76e45236bfe06e5ef9647c2b5ea1594ce3c63faa9c0d2","family_id":"qf-mod33-dimension-spanning-upper-bound","seed":202611021092,"semantic_template":"qf-mod33-dimension-spanning-upper-bound/{\"dimension\":2,\"vectors\":[[0,1],[1,0],[17,1]]}"},"fields":{"title":"A Spanning Family May Give a Loose Upper Bound","stem":"Use the complete ordered real coordinate family vectors=[[1, 0], [0, 1], [17, 1]] in ambient R^2, limited to dimensions 2 or 3. Treat vectors as columns and calculate rank by exact rational Gaussian elimination, not floating-point samples. A basis is both independent and spanning. An independent family gives only a lower bound; a spanning family gives only an upper bound. Basis-cardinality invariance uses the finite-dimensional theorem, not a few examples. Give (vector_count,rank,spanning?,upper_bound). The last field is dimension_at_most_vector_count with the numeric count, or no_upper_bound_from_nonspanning_family.","answer":{"kind":"expression","value":"(3, 2, Yes, dimension_at_most_3)","explanation":"Only a verified spanning family supplies this upper bound. A redundant spanning family can be larger than the exact dimension; absence of a certificate from this family does not prove that the space is infinite-dimensional. The exact result, retaining the requested order, is (3, 2, Yes, dimension_at_most_3)."},"skills":["A Spanning Family May Give a Loose Upper Bound"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261102-r01-vector-space-dimension-0093","batch_id":"batch-20261014-advance-20261102-r01-vector-space-dimension-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"14e3591554d9ef355d9bd420728e860858625b1daff3bb0241b612e42968da11","family_id":"qf-mod33-dimension-spanning-upper-bound","seed":202611021093,"semantic_template":"qf-mod33-dimension-spanning-upper-bound/{\"dimension\":2,\"vectors\":[[0,1],[1,0],[18,1]]}"},"fields":{"title":"A Spanning Family May Give a Loose Upper Bound","stem":"Use the complete ordered real coordinate family vectors=[[1, 0], [0, 1], [18, 1]] in ambient R^2, limited to dimensions 2 or 3. Treat vectors as columns and calculate rank by exact rational Gaussian elimination, not floating-point samples. A basis is both independent and spanning. An independent family gives only a lower bound; a spanning family gives only an upper bound. Basis-cardinality invariance uses the finite-dimensional theorem, not a few examples. Give (vector_count,rank,spanning?,upper_bound). The last field is dimension_at_most_vector_count with the numeric count, or no_upper_bound_from_nonspanning_family.","answer":{"kind":"expression","value":"(3, 2, Yes, dimension_at_most_3)","explanation":"Only a verified spanning family supplies this upper bound. A redundant spanning family can be larger than the exact dimension; absence of a certificate from this family does not prove that the space is infinite-dimensional. The exact result, retaining the requested order, is (3, 2, Yes, dimension_at_most_3)."},"skills":["A Spanning Family May Give a Loose Upper Bound"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261102-r01-vector-space-dimension-0094","batch_id":"batch-20261014-advance-20261102-r01-vector-space-dimension-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"73a6f52c3e0c1c54c3b4c6c048011edacfe1f52ade2c35f1ffed6ca9e9aed879","family_id":"qf-mod33-dimension-spanning-upper-bound","seed":202611021094,"semantic_template":"qf-mod33-dimension-spanning-upper-bound/{\"dimension\":2,\"vectors\":[[0,1],[1,0],[19,1]]}"},"fields":{"title":"A Spanning Family May Give a Loose Upper Bound","stem":"Use the complete ordered real coordinate family vectors=[[1, 0], [0, 1], [19, 1]] in ambient R^2, limited to dimensions 2 or 3. Treat vectors as columns and calculate rank by exact rational Gaussian elimination, not floating-point samples. A basis is both independent and spanning. An independent family gives only a lower bound; a spanning family gives only an upper bound. Basis-cardinality invariance uses the finite-dimensional theorem, not a few examples. Give (vector_count,rank,spanning?,upper_bound). The last field is dimension_at_most_vector_count with the numeric count, or no_upper_bound_from_nonspanning_family.","answer":{"kind":"expression","value":"(3, 2, Yes, dimension_at_most_3)","explanation":"Only a verified spanning family supplies this upper bound. A redundant spanning family can be larger than the exact dimension; absence of a certificate from this family does not prove that the space is infinite-dimensional. The exact result, retaining the requested order, is (3, 2, Yes, dimension_at_most_3)."},"skills":["A Spanning Family May Give a Loose Upper Bound"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261102-r01-vector-space-dimension-0095","batch_id":"batch-20261014-advance-20261102-r01-vector-space-dimension-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"10f4ec6cd98453ce4c55a3379b184c1d179fac1a6bac661f2e1ad48109e3606c","family_id":"qf-mod33-dimension-spanning-upper-bound","seed":202611021095,"semantic_template":"qf-mod33-dimension-spanning-upper-bound/{\"dimension\":2,\"vectors\":[[0,1],[1,0],[20,1]]}"},"fields":{"title":"A Spanning Family May Give a Loose Upper Bound","stem":"Use the complete ordered real coordinate family vectors=[[1, 0], [0, 1], [20, 1]] in ambient R^2, limited to dimensions 2 or 3. Treat vectors as columns and calculate rank by exact rational Gaussian elimination, not floating-point samples. A basis is both independent and spanning. An independent family gives only a lower bound; a spanning family gives only an upper bound. Basis-cardinality invariance uses the finite-dimensional theorem, not a few examples. Give (vector_count,rank,spanning?,upper_bound). The last field is dimension_at_most_vector_count with the numeric count, or no_upper_bound_from_nonspanning_family.","answer":{"kind":"expression","value":"(3, 2, Yes, dimension_at_most_3)","explanation":"Only a verified spanning family supplies this upper bound. A redundant spanning family can be larger than the exact dimension; absence of a certificate from this family does not prove that the space is infinite-dimensional. The exact result, retaining the requested order, is (3, 2, Yes, dimension_at_most_3)."},"skills":["A Spanning Family May Give a Loose Upper Bound"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261102-r01-vector-space-dimension-0096","batch_id":"batch-20261014-advance-20261102-r01-vector-space-dimension-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"5a1c029a063bd23aee1d0b1726ddec2122e9dc9261ee1028fd23372d2c98b54e","family_id":"qf-mod33-dimension-spanning-upper-bound","seed":202611021096,"semantic_template":"qf-mod33-dimension-spanning-upper-bound/{\"dimension\":2,\"vectors\":[[0,1],[1,0],[21,1]]}"},"fields":{"title":"A Spanning Family May Give a Loose Upper Bound","stem":"Use the complete ordered real coordinate family vectors=[[1, 0], [0, 1], [21, 1]] in ambient R^2, limited to dimensions 2 or 3. Treat vectors as columns and calculate rank by exact rational Gaussian elimination, not floating-point samples. A basis is both independent and spanning. An independent family gives only a lower bound; a spanning family gives only an upper bound. Basis-cardinality invariance uses the finite-dimensional theorem, not a few examples. Give (vector_count,rank,spanning?,upper_bound). The last field is dimension_at_most_vector_count with the numeric count, or no_upper_bound_from_nonspanning_family.","answer":{"kind":"expression","value":"(3, 2, Yes, dimension_at_most_3)","explanation":"Only a verified spanning family supplies this upper bound. A redundant spanning family can be larger than the exact dimension; absence of a certificate from this family does not prove that the space is infinite-dimensional. The exact result, retaining the requested order, is (3, 2, Yes, dimension_at_most_3)."},"skills":["A Spanning Family May Give a Loose Upper Bound"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261102-r01-vector-space-dimension-0097","batch_id":"batch-20261014-advance-20261102-r01-vector-space-dimension-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"9647f25bb720f5268d4897411cca7b349d1d72700f4a700cc2e40e6cf2fc58ca","family_id":"qf-mod33-dimension-spanning-upper-bound","seed":202611021097,"semantic_template":"qf-mod33-dimension-spanning-upper-bound/{\"dimension\":2,\"vectors\":[[0,1],[1,0],[22,1]]}"},"fields":{"title":"A Spanning Family May Give a Loose Upper Bound","stem":"Use the complete ordered real coordinate family vectors=[[1, 0], [0, 1], [22, 1]] in ambient R^2, limited to dimensions 2 or 3. Treat vectors as columns and calculate rank by exact rational Gaussian elimination, not floating-point samples. A basis is both independent and spanning. An independent family gives only a lower bound; a spanning family gives only an upper bound. Basis-cardinality invariance uses the finite-dimensional theorem, not a few examples. Give (vector_count,rank,spanning?,upper_bound). The last field is dimension_at_most_vector_count with the numeric count, or no_upper_bound_from_nonspanning_family.","answer":{"kind":"expression","value":"(3, 2, Yes, dimension_at_most_3)","explanation":"Only a verified spanning family supplies this upper bound. A redundant spanning family can be larger than the exact dimension; absence of a certificate from this family does not prove that the space is infinite-dimensional. The exact result, retaining the requested order, is (3, 2, Yes, dimension_at_most_3)."},"skills":["A Spanning Family May Give a Loose Upper Bound"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261102-r01-vector-space-dimension-0098","batch_id":"batch-20261014-advance-20261102-r01-vector-space-dimension-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"8b3c6e89da08206c67e78fc28611ebba5030b46070333456491403d0ee0345c5","family_id":"qf-mod33-dimension-spanning-upper-bound","seed":202611021098,"semantic_template":"qf-mod33-dimension-spanning-upper-bound/{\"dimension\":2,\"vectors\":[[0,1],[1,0],[23,1]]}"},"fields":{"title":"A Spanning Family May Give a Loose Upper Bound","stem":"Use the complete ordered real coordinate family vectors=[[1, 0], [0, 1], [23, 1]] in ambient R^2, limited to dimensions 2 or 3. Treat vectors as columns and calculate rank by exact rational Gaussian elimination, not floating-point samples. A basis is both independent and spanning. An independent family gives only a lower bound; a spanning family gives only an upper bound. Basis-cardinality invariance uses the finite-dimensional theorem, not a few examples. Give (vector_count,rank,spanning?,upper_bound). The last field is dimension_at_most_vector_count with the numeric count, or no_upper_bound_from_nonspanning_family.","answer":{"kind":"expression","value":"(3, 2, Yes, dimension_at_most_3)","explanation":"Only a verified spanning family supplies this upper bound. A redundant spanning family can be larger than the exact dimension; absence of a certificate from this family does not prove that the space is infinite-dimensional. The exact result, retaining the requested order, is (3, 2, Yes, dimension_at_most_3)."},"skills":["A Spanning Family May Give a Loose Upper Bound"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261102-r01-vector-space-dimension-0099","batch_id":"batch-20261014-advance-20261102-r01-vector-space-dimension-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"d992b813de63315b513cb06617754e4b9da7a74921aeadc02ea36411fb88fbba","family_id":"qf-mod33-dimension-spanning-upper-bound","seed":202611021099,"semantic_template":"qf-mod33-dimension-spanning-upper-bound/{\"dimension\":2,\"vectors\":[[0,1],[1,0],[24,1]]}"},"fields":{"title":"A Spanning Family May Give a Loose Upper Bound","stem":"Use the complete ordered real coordinate family vectors=[[1, 0], [0, 1], [24, 1]] in ambient R^2, limited to dimensions 2 or 3. Treat vectors as columns and calculate rank by exact rational Gaussian elimination, not floating-point samples. A basis is both independent and spanning. An independent family gives only a lower bound; a spanning family gives only an upper bound. Basis-cardinality invariance uses the finite-dimensional theorem, not a few examples. Give (vector_count,rank,spanning?,upper_bound). The last field is dimension_at_most_vector_count with the numeric count, or no_upper_bound_from_nonspanning_family.","answer":{"kind":"expression","value":"(3, 2, Yes, dimension_at_most_3)","explanation":"Only a verified spanning family supplies this upper bound. A redundant spanning family can be larger than the exact dimension; absence of a certificate from this family does not prove that the space is infinite-dimensional. The exact result, retaining the requested order, is (3, 2, Yes, dimension_at_most_3)."},"skills":["A Spanning Family May Give a Loose Upper Bound"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261102-r01-vector-space-dimension-0100","batch_id":"batch-20261014-advance-20261102-r01-vector-space-dimension-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"7f2b9d7f0000ac1b05fde14818ee722b49a8884b873e1be89e9b448b45b3d429","family_id":"qf-mod33-dimension-spanning-upper-bound","seed":202611021100,"semantic_template":"qf-mod33-dimension-spanning-upper-bound/{\"dimension\":2,\"vectors\":[[0,1],[1,0],[25,1]]}"},"fields":{"title":"A Spanning Family May Give a Loose Upper Bound","stem":"Use the complete ordered real coordinate family vectors=[[1, 0], [0, 1], [25, 1]] in ambient R^2, limited to dimensions 2 or 3. Treat vectors as columns and calculate rank by exact rational Gaussian elimination, not floating-point samples. A basis is both independent and spanning. An independent family gives only a lower bound; a spanning family gives only an upper bound. Basis-cardinality invariance uses the finite-dimensional theorem, not a few examples. Give (vector_count,rank,spanning?,upper_bound). The last field is dimension_at_most_vector_count with the numeric count, or no_upper_bound_from_nonspanning_family.","answer":{"kind":"expression","value":"(3, 2, Yes, dimension_at_most_3)","explanation":"Only a verified spanning family supplies this upper bound. A redundant spanning family can be larger than the exact dimension; absence of a certificate from this family does not prove that the space is infinite-dimensional. The exact result, retaining the requested order, is (3, 2, Yes, dimension_at_most_3)."},"skills":["A Spanning Family May Give a Loose Upper Bound"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}