{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261102-r03-linear-map-image-0001","batch_id":"batch-20261014-advance-20261102-r03-linear-map-image-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"a3e3e5e85c3425e2e02a4e1108f7b3ff4b2a7057d3b76e6b5ce0cefc63245476","family_id":"qf-mod33-image-membership","seed":202611023001,"semantic_template":"qf-mod33-image-membership/{\"matrix\":[[1,1],[2,2]],\"target\":[2,4]}"},"fields":{"title":"Image Membership Uses the Full Real Column Span","stem":"Use this same exact integer matrix A=[[1, 1], [2, 2]] for the usual linear map T:R²→R², with rows and columns in the supplied order. Kernel means all real x solving A*x=0; image means the full real column span, not automatically the whole codomain. An empty basis spans the zero space. Rank2 has zero-dimensional kernel and both original columns as an image basis; rank0 has the standard kernel basis and empty image basis. At rank1 the kernel uses a primitive integer normal-equation solution, with its first nonzero coordinate positive, and the image uses the first nonzero original column. All divisions are exact rational operations. Target=[2, 4]. Give (target,rank,image_basis,in_image?,certificate). A rank2 certificate is [unique_solution,exact_coordinates]; rank0 uses zero_image or nonzero_target_zero_image; rank1 uses [column_span_cross,basis_vector,cross_value].","answer":{"kind":"expression","value":"([2, 4], 1, [[1, 2]], Yes, [column_span_cross, [1, 2], 0])","explanation":"At full rank solve A*x=target exactly, allowing noninteger rational coordinates. At rank1 test collinearity with the nonzero column; at rank0 only the zero target is in the image. These are full-real consistency checks, not an integer-only search. The exact result, retaining the requested order, is ([2, 4], 1, [[1, 2]], Yes, [column_span_cross, [1, 2], 0])."},"skills":["Image Membership Uses the Full Real Column Span"],"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-r03-linear-map-image-0002","batch_id":"batch-20261014-advance-20261102-r03-linear-map-image-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"267385bca9f87112c7f7228067a2b16efe39dcfe025df490d8f9bd619308758f","family_id":"qf-mod33-image-membership","seed":202611023002,"semantic_template":"qf-mod33-image-membership/{\"matrix\":[[1,2],[2,4]],\"target\":[3,6]}"},"fields":{"title":"Image Membership Uses the Full Real Column Span","stem":"Use this same exact integer matrix A=[[1, 2], [2, 4]] for the usual linear map T:R²→R², with rows and columns in the supplied order. Kernel means all real x solving A*x=0; image means the full real column span, not automatically the whole codomain. An empty basis spans the zero space. Rank2 has zero-dimensional kernel and both original columns as an image basis; rank0 has the standard kernel basis and empty image basis. At rank1 the kernel uses a primitive integer normal-equation solution, with its first nonzero coordinate positive, and the image uses the first nonzero original column. All divisions are exact rational operations. Target=[3, 6]. Give (target,rank,image_basis,in_image?,certificate). A rank2 certificate is [unique_solution,exact_coordinates]; rank0 uses zero_image or nonzero_target_zero_image; rank1 uses [column_span_cross,basis_vector,cross_value].","answer":{"kind":"expression","value":"([3, 6], 1, [[1, 2]], Yes, [column_span_cross, [1, 2], 0])","explanation":"At full rank solve A*x=target exactly, allowing noninteger rational coordinates. At rank1 test collinearity with the nonzero column; at rank0 only the zero target is in the image. These are full-real consistency checks, not an integer-only search. The exact result, retaining the requested order, is ([3, 6], 1, [[1, 2]], Yes, [column_span_cross, [1, 2], 0])."},"skills":["Image Membership Uses the Full Real Column Span"],"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-r03-linear-map-image-0003","batch_id":"batch-20261014-advance-20261102-r03-linear-map-image-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"673cdde69ebc3aa23fa5b234ba1971d2cf4298bc9ea2ebe565c6f8ffa2cdc074","family_id":"qf-mod33-image-membership","seed":202611023003,"semantic_template":"qf-mod33-image-membership/{\"matrix\":[[1,3],[2,6]],\"target\":[4,8]}"},"fields":{"title":"Image Membership Uses the Full Real Column Span","stem":"Use this same exact integer matrix A=[[1, 3], [2, 6]] for the usual linear map T:R²→R², with rows and columns in the supplied order. Kernel means all real x solving A*x=0; image means the full real column span, not automatically the whole codomain. An empty basis spans the zero space. Rank2 has zero-dimensional kernel and both original columns as an image basis; rank0 has the standard kernel basis and empty image basis. At rank1 the kernel uses a primitive integer normal-equation solution, with its first nonzero coordinate positive, and the image uses the first nonzero original column. All divisions are exact rational operations. Target=[4, 8]. Give (target,rank,image_basis,in_image?,certificate). A rank2 certificate is [unique_solution,exact_coordinates]; rank0 uses zero_image or nonzero_target_zero_image; rank1 uses [column_span_cross,basis_vector,cross_value].","answer":{"kind":"expression","value":"([4, 8], 1, [[1, 2]], Yes, [column_span_cross, [1, 2], 0])","explanation":"At full rank solve A*x=target exactly, allowing noninteger rational coordinates. At rank1 test collinearity with the nonzero column; at rank0 only the zero target is in the image. These are full-real consistency checks, not an integer-only search. The exact result, retaining the requested order, is ([4, 8], 1, [[1, 2]], Yes, [column_span_cross, [1, 2], 0])."},"skills":["Image Membership Uses the Full Real Column Span"],"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-r03-linear-map-image-0004","batch_id":"batch-20261014-advance-20261102-r03-linear-map-image-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"8b7da2984f1406624b7c57823e06e9585811efc69bd81b96bbbc7648d96703ae","family_id":"qf-mod33-image-membership","seed":202611023004,"semantic_template":"qf-mod33-image-membership/{\"matrix\":[[1,4],[2,8]],\"target\":[5,10]}"},"fields":{"title":"Image Membership Uses the Full Real Column Span","stem":"Use this same exact integer matrix A=[[1, 4], [2, 8]] for the usual linear map T:R²→R², with rows and columns in the supplied order. Kernel means all real x solving A*x=0; image means the full real column span, not automatically the whole codomain. An empty basis spans the zero space. Rank2 has zero-dimensional kernel and both original columns as an image basis; rank0 has the standard kernel basis and empty image basis. At rank1 the kernel uses a primitive integer normal-equation solution, with its first nonzero coordinate positive, and the image uses the first nonzero original column. All divisions are exact rational operations. Target=[5, 10]. Give (target,rank,image_basis,in_image?,certificate). A rank2 certificate is [unique_solution,exact_coordinates]; rank0 uses zero_image or nonzero_target_zero_image; rank1 uses [column_span_cross,basis_vector,cross_value].","answer":{"kind":"expression","value":"([5, 10], 1, [[1, 2]], Yes, [column_span_cross, [1, 2], 0])","explanation":"At full rank solve A*x=target exactly, allowing noninteger rational coordinates. At rank1 test collinearity with the nonzero column; at rank0 only the zero target is in the image. These are full-real consistency checks, not an integer-only search. The exact result, retaining the requested order, is ([5, 10], 1, [[1, 2]], Yes, [column_span_cross, [1, 2], 0])."},"skills":["Image Membership Uses the Full Real Column Span"],"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-r03-linear-map-image-0005","batch_id":"batch-20261014-advance-20261102-r03-linear-map-image-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"872eed9bd940d4bd43fefb9ad3ef0d1abd7d81f5935de2d1095a08a9fd666fde","family_id":"qf-mod33-image-membership","seed":202611023005,"semantic_template":"qf-mod33-image-membership/{\"matrix\":[[1,5],[2,10]],\"target\":[6,12]}"},"fields":{"title":"Image Membership Uses the Full Real Column Span","stem":"Use this same exact integer matrix A=[[1, 5], [2, 10]] for the usual linear map T:R²→R², with rows and columns in the supplied order. Kernel means all real x solving A*x=0; image means the full real column span, not automatically the whole codomain. An empty basis spans the zero space. Rank2 has zero-dimensional kernel and both original columns as an image basis; rank0 has the standard kernel basis and empty image basis. At rank1 the kernel uses a primitive integer normal-equation solution, with its first nonzero coordinate positive, and the image uses the first nonzero original column. All divisions are exact rational operations. Target=[6, 12]. Give (target,rank,image_basis,in_image?,certificate). A rank2 certificate is [unique_solution,exact_coordinates]; rank0 uses zero_image or nonzero_target_zero_image; rank1 uses [column_span_cross,basis_vector,cross_value].","answer":{"kind":"expression","value":"([6, 12], 1, [[1, 2]], Yes, [column_span_cross, [1, 2], 0])","explanation":"At full rank solve A*x=target exactly, allowing noninteger rational coordinates. At rank1 test collinearity with the nonzero column; at rank0 only the zero target is in the image. These are full-real consistency checks, not an integer-only search. The exact result, retaining the requested order, is ([6, 12], 1, [[1, 2]], Yes, [column_span_cross, [1, 2], 0])."},"skills":["Image Membership Uses the Full Real Column Span"],"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-r03-linear-map-image-0006","batch_id":"batch-20261014-advance-20261102-r03-linear-map-image-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"9888f8739e5d6e8edc2e87fee0485fe33f89f9282add5b7fbbcf51c1847d6337","family_id":"qf-mod33-image-membership","seed":202611023006,"semantic_template":"qf-mod33-image-membership/{\"matrix\":[[1,6],[2,12]],\"target\":[7,14]}"},"fields":{"title":"Image Membership Uses the Full Real Column Span","stem":"Use this same exact integer matrix A=[[1, 6], [2, 12]] for the usual linear map T:R²→R², with rows and columns in the supplied order. Kernel means all real x solving A*x=0; image means the full real column span, not automatically the whole codomain. An empty basis spans the zero space. Rank2 has zero-dimensional kernel and both original columns as an image basis; rank0 has the standard kernel basis and empty image basis. At rank1 the kernel uses a primitive integer normal-equation solution, with its first nonzero coordinate positive, and the image uses the first nonzero original column. All divisions are exact rational operations. Target=[7, 14]. Give (target,rank,image_basis,in_image?,certificate). A rank2 certificate is [unique_solution,exact_coordinates]; rank0 uses zero_image or nonzero_target_zero_image; rank1 uses [column_span_cross,basis_vector,cross_value].","answer":{"kind":"expression","value":"([7, 14], 1, [[1, 2]], Yes, [column_span_cross, [1, 2], 0])","explanation":"At full rank solve A*x=target exactly, allowing noninteger rational coordinates. At rank1 test collinearity with the nonzero column; at rank0 only the zero target is in the image. These are full-real consistency checks, not an integer-only search. The exact result, retaining the requested order, is ([7, 14], 1, [[1, 2]], Yes, [column_span_cross, [1, 2], 0])."},"skills":["Image Membership Uses the Full Real Column Span"],"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-r03-linear-map-image-0007","batch_id":"batch-20261014-advance-20261102-r03-linear-map-image-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"fb54d2f17dabd6810dec9f9a5c061a2ef58a8abd91007b1b128be0475ffbdfe0","family_id":"qf-mod33-image-membership","seed":202611023007,"semantic_template":"qf-mod33-image-membership/{\"matrix\":[[1,7],[2,14]],\"target\":[8,16]}"},"fields":{"title":"Image Membership Uses the Full Real Column Span","stem":"Use this same exact integer matrix A=[[1, 7], [2, 14]] for the usual linear map T:R²→R², with rows and columns in the supplied order. Kernel means all real x solving A*x=0; image means the full real column span, not automatically the whole codomain. An empty basis spans the zero space. Rank2 has zero-dimensional kernel and both original columns as an image basis; rank0 has the standard kernel basis and empty image basis. At rank1 the kernel uses a primitive integer normal-equation solution, with its first nonzero coordinate positive, and the image uses the first nonzero original column. All divisions are exact rational operations. Target=[8, 16]. Give (target,rank,image_basis,in_image?,certificate). A rank2 certificate is [unique_solution,exact_coordinates]; rank0 uses zero_image or nonzero_target_zero_image; rank1 uses [column_span_cross,basis_vector,cross_value].","answer":{"kind":"expression","value":"([8, 16], 1, [[1, 2]], Yes, [column_span_cross, [1, 2], 0])","explanation":"At full rank solve A*x=target exactly, allowing noninteger rational coordinates. At rank1 test collinearity with the nonzero column; at rank0 only the zero target is in the image. These are full-real consistency checks, not an integer-only search. The exact result, retaining the requested order, is ([8, 16], 1, [[1, 2]], Yes, [column_span_cross, [1, 2], 0])."},"skills":["Image Membership Uses the Full Real Column Span"],"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-r03-linear-map-image-0008","batch_id":"batch-20261014-advance-20261102-r03-linear-map-image-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"dd4c2028ba72c474f38aa615b32a9831dfc8a69fd0794d0db6102877ff4bc06c","family_id":"qf-mod33-image-membership","seed":202611023008,"semantic_template":"qf-mod33-image-membership/{\"matrix\":[[1,8],[2,16]],\"target\":[9,18]}"},"fields":{"title":"Image Membership Uses the Full Real Column Span","stem":"Use this same exact integer matrix A=[[1, 8], [2, 16]] for the usual linear map T:R²→R², with rows and columns in the supplied order. Kernel means all real x solving A*x=0; image means the full real column span, not automatically the whole codomain. An empty basis spans the zero space. Rank2 has zero-dimensional kernel and both original columns as an image basis; rank0 has the standard kernel basis and empty image basis. At rank1 the kernel uses a primitive integer normal-equation solution, with its first nonzero coordinate positive, and the image uses the first nonzero original column. All divisions are exact rational operations. Target=[9, 18]. Give (target,rank,image_basis,in_image?,certificate). A rank2 certificate is [unique_solution,exact_coordinates]; rank0 uses zero_image or nonzero_target_zero_image; rank1 uses [column_span_cross,basis_vector,cross_value].","answer":{"kind":"expression","value":"([9, 18], 1, [[1, 2]], Yes, [column_span_cross, [1, 2], 0])","explanation":"At full rank solve A*x=target exactly, allowing noninteger rational coordinates. At rank1 test collinearity with the nonzero column; at rank0 only the zero target is in the image. These are full-real consistency checks, not an integer-only search. The exact result, retaining the requested order, is ([9, 18], 1, [[1, 2]], Yes, [column_span_cross, [1, 2], 0])."},"skills":["Image Membership Uses the Full Real Column Span"],"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-r03-linear-map-image-0009","batch_id":"batch-20261014-advance-20261102-r03-linear-map-image-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"de58b521a79141613831a95b39878e55ba316298acf6fbf36e16a2e4638597f2","family_id":"qf-mod33-image-membership","seed":202611023009,"semantic_template":"qf-mod33-image-membership/{\"matrix\":[[1,9],[2,18]],\"target\":[10,20]}"},"fields":{"title":"Image Membership Uses the Full Real Column Span","stem":"Use this same exact integer matrix A=[[1, 9], [2, 18]] for the usual linear map T:R²→R², with rows and columns in the supplied order. Kernel means all real x solving A*x=0; image means the full real column span, not automatically the whole codomain. An empty basis spans the zero space. Rank2 has zero-dimensional kernel and both original columns as an image basis; rank0 has the standard kernel basis and empty image basis. At rank1 the kernel uses a primitive integer normal-equation solution, with its first nonzero coordinate positive, and the image uses the first nonzero original column. All divisions are exact rational operations. Target=[10, 20]. Give (target,rank,image_basis,in_image?,certificate). A rank2 certificate is [unique_solution,exact_coordinates]; rank0 uses zero_image or nonzero_target_zero_image; rank1 uses [column_span_cross,basis_vector,cross_value].","answer":{"kind":"expression","value":"([10, 20], 1, [[1, 2]], Yes, [column_span_cross, [1, 2], 0])","explanation":"At full rank solve A*x=target exactly, allowing noninteger rational coordinates. At rank1 test collinearity with the nonzero column; at rank0 only the zero target is in the image. These are full-real consistency checks, not an integer-only search. The exact result, retaining the requested order, is ([10, 20], 1, [[1, 2]], Yes, [column_span_cross, [1, 2], 0])."},"skills":["Image Membership Uses the Full Real Column Span"],"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-r03-linear-map-image-0010","batch_id":"batch-20261014-advance-20261102-r03-linear-map-image-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"8e61523f94942388e43b1ee6b652b25306dcd06bf1eb94a15f8a1fc90cde5674","family_id":"qf-mod33-image-membership","seed":202611023010,"semantic_template":"qf-mod33-image-membership/{\"matrix\":[[1,10],[2,20]],\"target\":[11,22]}"},"fields":{"title":"Image Membership Uses the Full Real Column Span","stem":"Use this same exact integer matrix A=[[1, 10], [2, 20]] for the usual linear map T:R²→R², with rows and columns in the supplied order. Kernel means all real x solving A*x=0; image means the full real column span, not automatically the whole codomain. An empty basis spans the zero space. Rank2 has zero-dimensional kernel and both original columns as an image basis; rank0 has the standard kernel basis and empty image basis. At rank1 the kernel uses a primitive integer normal-equation solution, with its first nonzero coordinate positive, and the image uses the first nonzero original column. All divisions are exact rational operations. Target=[11, 22]. Give (target,rank,image_basis,in_image?,certificate). A rank2 certificate is [unique_solution,exact_coordinates]; rank0 uses zero_image or nonzero_target_zero_image; rank1 uses [column_span_cross,basis_vector,cross_value].","answer":{"kind":"expression","value":"([11, 22], 1, [[1, 2]], Yes, [column_span_cross, [1, 2], 0])","explanation":"At full rank solve A*x=target exactly, allowing noninteger rational coordinates. At rank1 test collinearity with the nonzero column; at rank0 only the zero target is in the image. These are full-real consistency checks, not an integer-only search. The exact result, retaining the requested order, is ([11, 22], 1, [[1, 2]], Yes, [column_span_cross, [1, 2], 0])."},"skills":["Image Membership Uses the Full Real Column Span"],"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-r03-linear-map-image-0011","batch_id":"batch-20261014-advance-20261102-r03-linear-map-image-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"d48f0a5c9bbcd98cbd9b12c4cd0c8724e4877a2ed4071a5b184b9019629964a5","family_id":"qf-mod33-image-membership","seed":202611023011,"semantic_template":"qf-mod33-image-membership/{\"matrix\":[[1,11],[2,22]],\"target\":[12,24]}"},"fields":{"title":"Image Membership Uses the Full Real Column Span","stem":"Use this same exact integer matrix A=[[1, 11], [2, 22]] for the usual linear map T:R²→R², with rows and columns in the supplied order. Kernel means all real x solving A*x=0; image means the full real column span, not automatically the whole codomain. An empty basis spans the zero space. Rank2 has zero-dimensional kernel and both original columns as an image basis; rank0 has the standard kernel basis and empty image basis. At rank1 the kernel uses a primitive integer normal-equation solution, with its first nonzero coordinate positive, and the image uses the first nonzero original column. All divisions are exact rational operations. Target=[12, 24]. Give (target,rank,image_basis,in_image?,certificate). A rank2 certificate is [unique_solution,exact_coordinates]; rank0 uses zero_image or nonzero_target_zero_image; rank1 uses [column_span_cross,basis_vector,cross_value].","answer":{"kind":"expression","value":"([12, 24], 1, [[1, 2]], Yes, [column_span_cross, [1, 2], 0])","explanation":"At full rank solve A*x=target exactly, allowing noninteger rational coordinates. At rank1 test collinearity with the nonzero column; at rank0 only the zero target is in the image. These are full-real consistency checks, not an integer-only search. The exact result, retaining the requested order, is ([12, 24], 1, [[1, 2]], Yes, [column_span_cross, [1, 2], 0])."},"skills":["Image Membership Uses the Full Real Column Span"],"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-r03-linear-map-image-0012","batch_id":"batch-20261014-advance-20261102-r03-linear-map-image-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"d3c10a5011c304a1ad717b9051adbc4f0a94803548812d8831b3819761a652d0","family_id":"qf-mod33-image-membership","seed":202611023012,"semantic_template":"qf-mod33-image-membership/{\"matrix\":[[1,12],[2,24]],\"target\":[13,26]}"},"fields":{"title":"Image Membership Uses the Full Real Column Span","stem":"Use this same exact integer matrix A=[[1, 12], [2, 24]] for the usual linear map T:R²→R², with rows and columns in the supplied order. Kernel means all real x solving A*x=0; image means the full real column span, not automatically the whole codomain. An empty basis spans the zero space. Rank2 has zero-dimensional kernel and both original columns as an image basis; rank0 has the standard kernel basis and empty image basis. At rank1 the kernel uses a primitive integer normal-equation solution, with its first nonzero coordinate positive, and the image uses the first nonzero original column. All divisions are exact rational operations. Target=[13, 26]. Give (target,rank,image_basis,in_image?,certificate). A rank2 certificate is [unique_solution,exact_coordinates]; rank0 uses zero_image or nonzero_target_zero_image; rank1 uses [column_span_cross,basis_vector,cross_value].","answer":{"kind":"expression","value":"([13, 26], 1, [[1, 2]], Yes, [column_span_cross, [1, 2], 0])","explanation":"At full rank solve A*x=target exactly, allowing noninteger rational coordinates. At rank1 test collinearity with the nonzero column; at rank0 only the zero target is in the image. These are full-real consistency checks, not an integer-only search. The exact result, retaining the requested order, is ([13, 26], 1, [[1, 2]], Yes, [column_span_cross, [1, 2], 0])."},"skills":["Image Membership Uses the Full Real Column Span"],"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-r03-linear-map-image-0013","batch_id":"batch-20261014-advance-20261102-r03-linear-map-image-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"afc047ab08c060ae0ed2a67ce6c11d76afd86cc4dc52830706f6ebf962f7ad2c","family_id":"qf-mod33-image-membership","seed":202611023013,"semantic_template":"qf-mod33-image-membership/{\"matrix\":[[1,13],[2,26]],\"target\":[14,28]}"},"fields":{"title":"Image Membership Uses the Full Real Column Span","stem":"Use this same exact integer matrix A=[[1, 13], [2, 26]] for the usual linear map T:R²→R², with rows and columns in the supplied order. Kernel means all real x solving A*x=0; image means the full real column span, not automatically the whole codomain. An empty basis spans the zero space. Rank2 has zero-dimensional kernel and both original columns as an image basis; rank0 has the standard kernel basis and empty image basis. At rank1 the kernel uses a primitive integer normal-equation solution, with its first nonzero coordinate positive, and the image uses the first nonzero original column. All divisions are exact rational operations. Target=[14, 28]. Give (target,rank,image_basis,in_image?,certificate). A rank2 certificate is [unique_solution,exact_coordinates]; rank0 uses zero_image or nonzero_target_zero_image; rank1 uses [column_span_cross,basis_vector,cross_value].","answer":{"kind":"expression","value":"([14, 28], 1, [[1, 2]], Yes, [column_span_cross, [1, 2], 0])","explanation":"At full rank solve A*x=target exactly, allowing noninteger rational coordinates. At rank1 test collinearity with the nonzero column; at rank0 only the zero target is in the image. These are full-real consistency checks, not an integer-only search. The exact result, retaining the requested order, is ([14, 28], 1, [[1, 2]], Yes, [column_span_cross, [1, 2], 0])."},"skills":["Image Membership Uses the Full Real Column Span"],"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-r03-linear-map-image-0014","batch_id":"batch-20261014-advance-20261102-r03-linear-map-image-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"a9040ff2c4ae9e21666643613b2f6809eaf00bf69186233457c59f8478ab258f","family_id":"qf-mod33-image-membership","seed":202611023014,"semantic_template":"qf-mod33-image-membership/{\"matrix\":[[1,14],[2,28]],\"target\":[15,30]}"},"fields":{"title":"Image Membership Uses the Full Real Column Span","stem":"Use this same exact integer matrix A=[[1, 14], [2, 28]] for the usual linear map T:R²→R², with rows and columns in the supplied order. Kernel means all real x solving A*x=0; image means the full real column span, not automatically the whole codomain. An empty basis spans the zero space. Rank2 has zero-dimensional kernel and both original columns as an image basis; rank0 has the standard kernel basis and empty image basis. At rank1 the kernel uses a primitive integer normal-equation solution, with its first nonzero coordinate positive, and the image uses the first nonzero original column. All divisions are exact rational operations. Target=[15, 30]. Give (target,rank,image_basis,in_image?,certificate). A rank2 certificate is [unique_solution,exact_coordinates]; rank0 uses zero_image or nonzero_target_zero_image; rank1 uses [column_span_cross,basis_vector,cross_value].","answer":{"kind":"expression","value":"([15, 30], 1, [[1, 2]], Yes, [column_span_cross, [1, 2], 0])","explanation":"At full rank solve A*x=target exactly, allowing noninteger rational coordinates. At rank1 test collinearity with the nonzero column; at rank0 only the zero target is in the image. These are full-real consistency checks, not an integer-only search. The exact result, retaining the requested order, is ([15, 30], 1, [[1, 2]], Yes, [column_span_cross, [1, 2], 0])."},"skills":["Image Membership Uses the Full Real Column Span"],"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-r03-linear-map-image-0015","batch_id":"batch-20261014-advance-20261102-r03-linear-map-image-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"bf0e39e8c9eb09fc58885ecfa9c0108ab39d7391a7ac7a393d325876576dc3e6","family_id":"qf-mod33-image-membership","seed":202611023015,"semantic_template":"qf-mod33-image-membership/{\"matrix\":[[1,15],[2,30]],\"target\":[16,32]}"},"fields":{"title":"Image Membership Uses the Full Real Column Span","stem":"Use this same exact integer matrix A=[[1, 15], [2, 30]] for the usual linear map T:R²→R², with rows and columns in the supplied order. Kernel means all real x solving A*x=0; image means the full real column span, not automatically the whole codomain. An empty basis spans the zero space. Rank2 has zero-dimensional kernel and both original columns as an image basis; rank0 has the standard kernel basis and empty image basis. At rank1 the kernel uses a primitive integer normal-equation solution, with its first nonzero coordinate positive, and the image uses the first nonzero original column. All divisions are exact rational operations. Target=[16, 32]. Give (target,rank,image_basis,in_image?,certificate). A rank2 certificate is [unique_solution,exact_coordinates]; rank0 uses zero_image or nonzero_target_zero_image; rank1 uses [column_span_cross,basis_vector,cross_value].","answer":{"kind":"expression","value":"([16, 32], 1, [[1, 2]], Yes, [column_span_cross, [1, 2], 0])","explanation":"At full rank solve A*x=target exactly, allowing noninteger rational coordinates. At rank1 test collinearity with the nonzero column; at rank0 only the zero target is in the image. These are full-real consistency checks, not an integer-only search. The exact result, retaining the requested order, is ([16, 32], 1, [[1, 2]], Yes, [column_span_cross, [1, 2], 0])."},"skills":["Image Membership Uses the Full Real Column Span"],"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-r03-linear-map-image-0016","batch_id":"batch-20261014-advance-20261102-r03-linear-map-image-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"218c39bbe664a2b20d0f5ad0964bfbf0851acaf3367e7591733198a0392f2fc0","family_id":"qf-mod33-image-membership","seed":202611023016,"semantic_template":"qf-mod33-image-membership/{\"matrix\":[[1,16],[2,32]],\"target\":[17,34]}"},"fields":{"title":"Image Membership Uses the Full Real Column Span","stem":"Use this same exact integer matrix A=[[1, 16], [2, 32]] for the usual linear map T:R²→R², with rows and columns in the supplied order. Kernel means all real x solving A*x=0; image means the full real column span, not automatically the whole codomain. An empty basis spans the zero space. Rank2 has zero-dimensional kernel and both original columns as an image basis; rank0 has the standard kernel basis and empty image basis. At rank1 the kernel uses a primitive integer normal-equation solution, with its first nonzero coordinate positive, and the image uses the first nonzero original column. All divisions are exact rational operations. Target=[17, 34]. Give (target,rank,image_basis,in_image?,certificate). A rank2 certificate is [unique_solution,exact_coordinates]; rank0 uses zero_image or nonzero_target_zero_image; rank1 uses [column_span_cross,basis_vector,cross_value].","answer":{"kind":"expression","value":"([17, 34], 1, [[1, 2]], Yes, [column_span_cross, [1, 2], 0])","explanation":"At full rank solve A*x=target exactly, allowing noninteger rational coordinates. At rank1 test collinearity with the nonzero column; at rank0 only the zero target is in the image. These are full-real consistency checks, not an integer-only search. The exact result, retaining the requested order, is ([17, 34], 1, [[1, 2]], Yes, [column_span_cross, [1, 2], 0])."},"skills":["Image Membership Uses the Full Real Column Span"],"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-r03-linear-map-image-0017","batch_id":"batch-20261014-advance-20261102-r03-linear-map-image-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"dd8da17a981900080997048596737da0c95220631c51ba6274b1d1c5fd212ca8","family_id":"qf-mod33-image-membership","seed":202611023017,"semantic_template":"qf-mod33-image-membership/{\"matrix\":[[1,17],[2,34]],\"target\":[18,36]}"},"fields":{"title":"Image Membership Uses the Full Real Column Span","stem":"Use this same exact integer matrix A=[[1, 17], [2, 34]] for the usual linear map T:R²→R², with rows and columns in the supplied order. Kernel means all real x solving A*x=0; image means the full real column span, not automatically the whole codomain. An empty basis spans the zero space. Rank2 has zero-dimensional kernel and both original columns as an image basis; rank0 has the standard kernel basis and empty image basis. At rank1 the kernel uses a primitive integer normal-equation solution, with its first nonzero coordinate positive, and the image uses the first nonzero original column. All divisions are exact rational operations. Target=[18, 36]. Give (target,rank,image_basis,in_image?,certificate). A rank2 certificate is [unique_solution,exact_coordinates]; rank0 uses zero_image or nonzero_target_zero_image; rank1 uses [column_span_cross,basis_vector,cross_value].","answer":{"kind":"expression","value":"([18, 36], 1, [[1, 2]], Yes, [column_span_cross, [1, 2], 0])","explanation":"At full rank solve A*x=target exactly, allowing noninteger rational coordinates. At rank1 test collinearity with the nonzero column; at rank0 only the zero target is in the image. These are full-real consistency checks, not an integer-only search. The exact result, retaining the requested order, is ([18, 36], 1, [[1, 2]], Yes, [column_span_cross, [1, 2], 0])."},"skills":["Image Membership Uses the Full Real Column Span"],"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-r03-linear-map-image-0018","batch_id":"batch-20261014-advance-20261102-r03-linear-map-image-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"d1daa9c24fdf0d96975e9c1a323f0ca2169699f65d5dfac6e3a2ad6afe04ad74","family_id":"qf-mod33-image-membership","seed":202611023018,"semantic_template":"qf-mod33-image-membership/{\"matrix\":[[1,18],[2,36]],\"target\":[19,38]}"},"fields":{"title":"Image Membership Uses the Full Real Column Span","stem":"Use this same exact integer matrix A=[[1, 18], [2, 36]] for the usual linear map T:R²→R², with rows and columns in the supplied order. Kernel means all real x solving A*x=0; image means the full real column span, not automatically the whole codomain. An empty basis spans the zero space. Rank2 has zero-dimensional kernel and both original columns as an image basis; rank0 has the standard kernel basis and empty image basis. At rank1 the kernel uses a primitive integer normal-equation solution, with its first nonzero coordinate positive, and the image uses the first nonzero original column. All divisions are exact rational operations. Target=[19, 38]. Give (target,rank,image_basis,in_image?,certificate). A rank2 certificate is [unique_solution,exact_coordinates]; rank0 uses zero_image or nonzero_target_zero_image; rank1 uses [column_span_cross,basis_vector,cross_value].","answer":{"kind":"expression","value":"([19, 38], 1, [[1, 2]], Yes, [column_span_cross, [1, 2], 0])","explanation":"At full rank solve A*x=target exactly, allowing noninteger rational coordinates. At rank1 test collinearity with the nonzero column; at rank0 only the zero target is in the image. These are full-real consistency checks, not an integer-only search. The exact result, retaining the requested order, is ([19, 38], 1, [[1, 2]], Yes, [column_span_cross, [1, 2], 0])."},"skills":["Image Membership Uses the Full Real Column Span"],"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-r03-linear-map-image-0019","batch_id":"batch-20261014-advance-20261102-r03-linear-map-image-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"367d0d3929fa4c29118b62eba1bbc9d8786f56a181ceaaea38fd3224f505a213","family_id":"qf-mod33-image-membership","seed":202611023019,"semantic_template":"qf-mod33-image-membership/{\"matrix\":[[1,19],[2,38]],\"target\":[20,40]}"},"fields":{"title":"Image Membership Uses the Full Real Column Span","stem":"Use this same exact integer matrix A=[[1, 19], [2, 38]] for the usual linear map T:R²→R², with rows and columns in the supplied order. Kernel means all real x solving A*x=0; image means the full real column span, not automatically the whole codomain. An empty basis spans the zero space. Rank2 has zero-dimensional kernel and both original columns as an image basis; rank0 has the standard kernel basis and empty image basis. At rank1 the kernel uses a primitive integer normal-equation solution, with its first nonzero coordinate positive, and the image uses the first nonzero original column. All divisions are exact rational operations. Target=[20, 40]. Give (target,rank,image_basis,in_image?,certificate). A rank2 certificate is [unique_solution,exact_coordinates]; rank0 uses zero_image or nonzero_target_zero_image; rank1 uses [column_span_cross,basis_vector,cross_value].","answer":{"kind":"expression","value":"([20, 40], 1, [[1, 2]], Yes, [column_span_cross, [1, 2], 0])","explanation":"At full rank solve A*x=target exactly, allowing noninteger rational coordinates. At rank1 test collinearity with the nonzero column; at rank0 only the zero target is in the image. These are full-real consistency checks, not an integer-only search. The exact result, retaining the requested order, is ([20, 40], 1, [[1, 2]], Yes, [column_span_cross, [1, 2], 0])."},"skills":["Image Membership Uses the Full Real Column Span"],"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-r03-linear-map-image-0020","batch_id":"batch-20261014-advance-20261102-r03-linear-map-image-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"e4d34a69366627b90559394be54a8f62d335989293fee9591d9eb6bbaf27c19f","family_id":"qf-mod33-image-membership","seed":202611023020,"semantic_template":"qf-mod33-image-membership/{\"matrix\":[[1,20],[2,40]],\"target\":[21,42]}"},"fields":{"title":"Image Membership Uses the Full Real Column Span","stem":"Use this same exact integer matrix A=[[1, 20], [2, 40]] for the usual linear map T:R²→R², with rows and columns in the supplied order. Kernel means all real x solving A*x=0; image means the full real column span, not automatically the whole codomain. An empty basis spans the zero space. Rank2 has zero-dimensional kernel and both original columns as an image basis; rank0 has the standard kernel basis and empty image basis. At rank1 the kernel uses a primitive integer normal-equation solution, with its first nonzero coordinate positive, and the image uses the first nonzero original column. All divisions are exact rational operations. Target=[21, 42]. Give (target,rank,image_basis,in_image?,certificate). A rank2 certificate is [unique_solution,exact_coordinates]; rank0 uses zero_image or nonzero_target_zero_image; rank1 uses [column_span_cross,basis_vector,cross_value].","answer":{"kind":"expression","value":"([21, 42], 1, [[1, 2]], Yes, [column_span_cross, [1, 2], 0])","explanation":"At full rank solve A*x=target exactly, allowing noninteger rational coordinates. At rank1 test collinearity with the nonzero column; at rank0 only the zero target is in the image. These are full-real consistency checks, not an integer-only search. The exact result, retaining the requested order, is ([21, 42], 1, [[1, 2]], Yes, [column_span_cross, [1, 2], 0])."},"skills":["Image Membership Uses the Full Real Column Span"],"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-r03-linear-map-image-0021","batch_id":"batch-20261014-advance-20261102-r03-linear-map-image-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"acd7f3c58b5e00e005d285b62ccb2358d165c6c894ada849d6aff8eb1ed0b75b","family_id":"qf-mod33-image-membership","seed":202611023021,"semantic_template":"qf-mod33-image-membership/{\"matrix\":[[1,21],[2,42]],\"target\":[22,44]}"},"fields":{"title":"Image Membership Uses the Full Real Column Span","stem":"Use this same exact integer matrix A=[[1, 21], [2, 42]] for the usual linear map T:R²→R², with rows and columns in the supplied order. Kernel means all real x solving A*x=0; image means the full real column span, not automatically the whole codomain. An empty basis spans the zero space. Rank2 has zero-dimensional kernel and both original columns as an image basis; rank0 has the standard kernel basis and empty image basis. At rank1 the kernel uses a primitive integer normal-equation solution, with its first nonzero coordinate positive, and the image uses the first nonzero original column. All divisions are exact rational operations. Target=[22, 44]. Give (target,rank,image_basis,in_image?,certificate). A rank2 certificate is [unique_solution,exact_coordinates]; rank0 uses zero_image or nonzero_target_zero_image; rank1 uses [column_span_cross,basis_vector,cross_value].","answer":{"kind":"expression","value":"([22, 44], 1, [[1, 2]], Yes, [column_span_cross, [1, 2], 0])","explanation":"At full rank solve A*x=target exactly, allowing noninteger rational coordinates. At rank1 test collinearity with the nonzero column; at rank0 only the zero target is in the image. These are full-real consistency checks, not an integer-only search. The exact result, retaining the requested order, is ([22, 44], 1, [[1, 2]], Yes, [column_span_cross, [1, 2], 0])."},"skills":["Image Membership Uses the Full Real Column Span"],"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-r03-linear-map-image-0022","batch_id":"batch-20261014-advance-20261102-r03-linear-map-image-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"606ff6e3e7791ddc7b071e2b9bfdba03cae71d13eec5730d6a94af2748843667","family_id":"qf-mod33-image-membership","seed":202611023022,"semantic_template":"qf-mod33-image-membership/{\"matrix\":[[1,22],[2,44]],\"target\":[23,46]}"},"fields":{"title":"Image Membership Uses the Full Real Column Span","stem":"Use this same exact integer matrix A=[[1, 22], [2, 44]] for the usual linear map T:R²→R², with rows and columns in the supplied order. Kernel means all real x solving A*x=0; image means the full real column span, not automatically the whole codomain. An empty basis spans the zero space. Rank2 has zero-dimensional kernel and both original columns as an image basis; rank0 has the standard kernel basis and empty image basis. At rank1 the kernel uses a primitive integer normal-equation solution, with its first nonzero coordinate positive, and the image uses the first nonzero original column. All divisions are exact rational operations. Target=[23, 46]. Give (target,rank,image_basis,in_image?,certificate). A rank2 certificate is [unique_solution,exact_coordinates]; rank0 uses zero_image or nonzero_target_zero_image; rank1 uses [column_span_cross,basis_vector,cross_value].","answer":{"kind":"expression","value":"([23, 46], 1, [[1, 2]], Yes, [column_span_cross, [1, 2], 0])","explanation":"At full rank solve A*x=target exactly, allowing noninteger rational coordinates. At rank1 test collinearity with the nonzero column; at rank0 only the zero target is in the image. These are full-real consistency checks, not an integer-only search. The exact result, retaining the requested order, is ([23, 46], 1, [[1, 2]], Yes, [column_span_cross, [1, 2], 0])."},"skills":["Image Membership Uses the Full Real Column Span"],"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-r03-linear-map-image-0023","batch_id":"batch-20261014-advance-20261102-r03-linear-map-image-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"77c6d36bb9e0efc36340990da55dc18aaf5f2c5731dd7da32da186e28967b41d","family_id":"qf-mod33-image-membership","seed":202611023023,"semantic_template":"qf-mod33-image-membership/{\"matrix\":[[1,23],[2,46]],\"target\":[24,48]}"},"fields":{"title":"Image Membership Uses the Full Real Column Span","stem":"Use this same exact integer matrix A=[[1, 23], [2, 46]] for the usual linear map T:R²→R², with rows and columns in the supplied order. Kernel means all real x solving A*x=0; image means the full real column span, not automatically the whole codomain. An empty basis spans the zero space. Rank2 has zero-dimensional kernel and both original columns as an image basis; rank0 has the standard kernel basis and empty image basis. At rank1 the kernel uses a primitive integer normal-equation solution, with its first nonzero coordinate positive, and the image uses the first nonzero original column. All divisions are exact rational operations. Target=[24, 48]. Give (target,rank,image_basis,in_image?,certificate). A rank2 certificate is [unique_solution,exact_coordinates]; rank0 uses zero_image or nonzero_target_zero_image; rank1 uses [column_span_cross,basis_vector,cross_value].","answer":{"kind":"expression","value":"([24, 48], 1, [[1, 2]], Yes, [column_span_cross, [1, 2], 0])","explanation":"At full rank solve A*x=target exactly, allowing noninteger rational coordinates. At rank1 test collinearity with the nonzero column; at rank0 only the zero target is in the image. These are full-real consistency checks, not an integer-only search. The exact result, retaining the requested order, is ([24, 48], 1, [[1, 2]], Yes, [column_span_cross, [1, 2], 0])."},"skills":["Image Membership Uses the Full Real Column Span"],"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-r03-linear-map-image-0024","batch_id":"batch-20261014-advance-20261102-r03-linear-map-image-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"5dcf0985df5c1c206eadf91d19a0694beb20563709ec62e22231210fa7ad37e6","family_id":"qf-mod33-image-membership","seed":202611023024,"semantic_template":"qf-mod33-image-membership/{\"matrix\":[[1,24],[2,48]],\"target\":[25,50]}"},"fields":{"title":"Image Membership Uses the Full Real Column Span","stem":"Use this same exact integer matrix A=[[1, 24], [2, 48]] for the usual linear map T:R²→R², with rows and columns in the supplied order. Kernel means all real x solving A*x=0; image means the full real column span, not automatically the whole codomain. An empty basis spans the zero space. Rank2 has zero-dimensional kernel and both original columns as an image basis; rank0 has the standard kernel basis and empty image basis. At rank1 the kernel uses a primitive integer normal-equation solution, with its first nonzero coordinate positive, and the image uses the first nonzero original column. All divisions are exact rational operations. Target=[25, 50]. Give (target,rank,image_basis,in_image?,certificate). A rank2 certificate is [unique_solution,exact_coordinates]; rank0 uses zero_image or nonzero_target_zero_image; rank1 uses [column_span_cross,basis_vector,cross_value].","answer":{"kind":"expression","value":"([25, 50], 1, [[1, 2]], Yes, [column_span_cross, [1, 2], 0])","explanation":"At full rank solve A*x=target exactly, allowing noninteger rational coordinates. At rank1 test collinearity with the nonzero column; at rank0 only the zero target is in the image. These are full-real consistency checks, not an integer-only search. The exact result, retaining the requested order, is ([25, 50], 1, [[1, 2]], Yes, [column_span_cross, [1, 2], 0])."},"skills":["Image Membership Uses the Full Real Column Span"],"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-r03-linear-map-image-0025","batch_id":"batch-20261014-advance-20261102-r03-linear-map-image-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"21cb677ca1ad6ece8e018fa66d53ec7035ae7916526dbf9f3858010ee412f876","family_id":"qf-mod33-image-membership","seed":202611023025,"semantic_template":"qf-mod33-image-membership/{\"matrix\":[[1,25],[2,50]],\"target\":[26,52]}"},"fields":{"title":"Image Membership Uses the Full Real Column Span","stem":"Use this same exact integer matrix A=[[1, 25], [2, 50]] for the usual linear map T:R²→R², with rows and columns in the supplied order. Kernel means all real x solving A*x=0; image means the full real column span, not automatically the whole codomain. An empty basis spans the zero space. Rank2 has zero-dimensional kernel and both original columns as an image basis; rank0 has the standard kernel basis and empty image basis. At rank1 the kernel uses a primitive integer normal-equation solution, with its first nonzero coordinate positive, and the image uses the first nonzero original column. All divisions are exact rational operations. Target=[26, 52]. Give (target,rank,image_basis,in_image?,certificate). A rank2 certificate is [unique_solution,exact_coordinates]; rank0 uses zero_image or nonzero_target_zero_image; rank1 uses [column_span_cross,basis_vector,cross_value].","answer":{"kind":"expression","value":"([26, 52], 1, [[1, 2]], Yes, [column_span_cross, [1, 2], 0])","explanation":"At full rank solve A*x=target exactly, allowing noninteger rational coordinates. At rank1 test collinearity with the nonzero column; at rank0 only the zero target is in the image. These are full-real consistency checks, not an integer-only search. The exact result, retaining the requested order, is ([26, 52], 1, [[1, 2]], Yes, [column_span_cross, [1, 2], 0])."},"skills":["Image Membership Uses the Full Real Column Span"],"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-r03-linear-map-image-0026","batch_id":"batch-20261014-advance-20261102-r03-linear-map-image-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"bc299fa7d09a615ba230d5312c1f291c5d4e1c54ee5ec00b055a4a7c8faafa86","family_id":"qf-mod33-image-basis","seed":202611023026,"semantic_template":"qf-mod33-image-basis/{\"matrix\":[[1,1],[2,2]]}"},"fields":{"title":"The Image Basis Comes from Original Matrix Columns","stem":"Use this same exact integer matrix A=[[1, 1], [2, 2]] for the usual linear map T:R²→R², with rows and columns in the supplied order. Kernel means all real x solving A*x=0; image means the full real column span, not automatically the whole codomain. An empty basis spans the zero space. Rank2 has zero-dimensional kernel and both original columns as an image basis; rank0 has the standard kernel basis and empty image basis. At rank1 the kernel uses a primitive integer normal-equation solution, with its first nonzero coordinate positive, and the image uses the first nonzero original column. All divisions are exact rational operations. Give (matrix,rank,image_basis,full_real_column_span).","answer":{"kind":"expression","value":"([[1, 1], [2, 2]], 1, [[1, 2]], full_real_column_span)","explanation":"Select the stated original column basis and take its full real span. Do not substitute rows, assume the image equals the codomain, or normalize columns into different output vectors. The exact result, retaining the requested order, is ([[1, 1], [2, 2]], 1, [[1, 2]], full_real_column_span)."},"skills":["The Image Basis Comes from Original Matrix Columns"],"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-r03-linear-map-image-0027","batch_id":"batch-20261014-advance-20261102-r03-linear-map-image-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"21cbd349826c640c2a7d492fd71e6dc7934a3d6746da547a342cf392da1d7c5d","family_id":"qf-mod33-image-basis","seed":202611023027,"semantic_template":"qf-mod33-image-basis/{\"matrix\":[[1,2],[2,4]]}"},"fields":{"title":"The Image Basis Comes from Original Matrix Columns","stem":"Use this same exact integer matrix A=[[1, 2], [2, 4]] for the usual linear map T:R²→R², with rows and columns in the supplied order. Kernel means all real x solving A*x=0; image means the full real column span, not automatically the whole codomain. An empty basis spans the zero space. Rank2 has zero-dimensional kernel and both original columns as an image basis; rank0 has the standard kernel basis and empty image basis. At rank1 the kernel uses a primitive integer normal-equation solution, with its first nonzero coordinate positive, and the image uses the first nonzero original column. All divisions are exact rational operations. Give (matrix,rank,image_basis,full_real_column_span).","answer":{"kind":"expression","value":"([[1, 2], [2, 4]], 1, [[1, 2]], full_real_column_span)","explanation":"Select the stated original column basis and take its full real span. Do not substitute rows, assume the image equals the codomain, or normalize columns into different output vectors. The exact result, retaining the requested order, is ([[1, 2], [2, 4]], 1, [[1, 2]], full_real_column_span)."},"skills":["The Image Basis Comes from Original Matrix Columns"],"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-r03-linear-map-image-0028","batch_id":"batch-20261014-advance-20261102-r03-linear-map-image-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"f879f976b4752bbfdbf98c03240008bc87168050d4ccff66190aeb4402dba139","family_id":"qf-mod33-image-basis","seed":202611023028,"semantic_template":"qf-mod33-image-basis/{\"matrix\":[[1,3],[2,6]]}"},"fields":{"title":"The Image Basis Comes from Original Matrix Columns","stem":"Use this same exact integer matrix A=[[1, 3], [2, 6]] for the usual linear map T:R²→R², with rows and columns in the supplied order. Kernel means all real x solving A*x=0; image means the full real column span, not automatically the whole codomain. An empty basis spans the zero space. Rank2 has zero-dimensional kernel and both original columns as an image basis; rank0 has the standard kernel basis and empty image basis. At rank1 the kernel uses a primitive integer normal-equation solution, with its first nonzero coordinate positive, and the image uses the first nonzero original column. All divisions are exact rational operations. Give (matrix,rank,image_basis,full_real_column_span).","answer":{"kind":"expression","value":"([[1, 3], [2, 6]], 1, [[1, 2]], full_real_column_span)","explanation":"Select the stated original column basis and take its full real span. Do not substitute rows, assume the image equals the codomain, or normalize columns into different output vectors. The exact result, retaining the requested order, is ([[1, 3], [2, 6]], 1, [[1, 2]], full_real_column_span)."},"skills":["The Image Basis Comes from Original Matrix Columns"],"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-r03-linear-map-image-0029","batch_id":"batch-20261014-advance-20261102-r03-linear-map-image-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"070bd8c3803335a05b896c5638b12ebf19990de15b0f3d1e60c0639fcdcd7cda","family_id":"qf-mod33-image-basis","seed":202611023029,"semantic_template":"qf-mod33-image-basis/{\"matrix\":[[1,4],[2,8]]}"},"fields":{"title":"The Image Basis Comes from Original Matrix Columns","stem":"Use this same exact integer matrix A=[[1, 4], [2, 8]] for the usual linear map T:R²→R², with rows and columns in the supplied order. Kernel means all real x solving A*x=0; image means the full real column span, not automatically the whole codomain. An empty basis spans the zero space. Rank2 has zero-dimensional kernel and both original columns as an image basis; rank0 has the standard kernel basis and empty image basis. At rank1 the kernel uses a primitive integer normal-equation solution, with its first nonzero coordinate positive, and the image uses the first nonzero original column. All divisions are exact rational operations. Give (matrix,rank,image_basis,full_real_column_span).","answer":{"kind":"expression","value":"([[1, 4], [2, 8]], 1, [[1, 2]], full_real_column_span)","explanation":"Select the stated original column basis and take its full real span. Do not substitute rows, assume the image equals the codomain, or normalize columns into different output vectors. The exact result, retaining the requested order, is ([[1, 4], [2, 8]], 1, [[1, 2]], full_real_column_span)."},"skills":["The Image Basis Comes from Original Matrix Columns"],"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-r03-linear-map-image-0030","batch_id":"batch-20261014-advance-20261102-r03-linear-map-image-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"084f90fc9ac85c2a917d71e3810c0fc029a0454c23719a8cf2669e68f6fff4a2","family_id":"qf-mod33-image-basis","seed":202611023030,"semantic_template":"qf-mod33-image-basis/{\"matrix\":[[1,5],[2,10]]}"},"fields":{"title":"The Image Basis Comes from Original Matrix Columns","stem":"Use this same exact integer matrix A=[[1, 5], [2, 10]] for the usual linear map T:R²→R², with rows and columns in the supplied order. Kernel means all real x solving A*x=0; image means the full real column span, not automatically the whole codomain. An empty basis spans the zero space. Rank2 has zero-dimensional kernel and both original columns as an image basis; rank0 has the standard kernel basis and empty image basis. At rank1 the kernel uses a primitive integer normal-equation solution, with its first nonzero coordinate positive, and the image uses the first nonzero original column. All divisions are exact rational operations. Give (matrix,rank,image_basis,full_real_column_span).","answer":{"kind":"expression","value":"([[1, 5], [2, 10]], 1, [[1, 2]], full_real_column_span)","explanation":"Select the stated original column basis and take its full real span. Do not substitute rows, assume the image equals the codomain, or normalize columns into different output vectors. The exact result, retaining the requested order, is ([[1, 5], [2, 10]], 1, [[1, 2]], full_real_column_span)."},"skills":["The Image Basis Comes from Original Matrix Columns"],"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-r03-linear-map-image-0031","batch_id":"batch-20261014-advance-20261102-r03-linear-map-image-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"f7b9f1cb9f25d4f49b861381c37b14cdaa287345b003b047825e84dc0f733520","family_id":"qf-mod33-image-basis","seed":202611023031,"semantic_template":"qf-mod33-image-basis/{\"matrix\":[[1,6],[2,12]]}"},"fields":{"title":"The Image Basis Comes from Original Matrix Columns","stem":"Use this same exact integer matrix A=[[1, 6], [2, 12]] for the usual linear map T:R²→R², with rows and columns in the supplied order. Kernel means all real x solving A*x=0; image means the full real column span, not automatically the whole codomain. An empty basis spans the zero space. Rank2 has zero-dimensional kernel and both original columns as an image basis; rank0 has the standard kernel basis and empty image basis. At rank1 the kernel uses a primitive integer normal-equation solution, with its first nonzero coordinate positive, and the image uses the first nonzero original column. All divisions are exact rational operations. Give (matrix,rank,image_basis,full_real_column_span).","answer":{"kind":"expression","value":"([[1, 6], [2, 12]], 1, [[1, 2]], full_real_column_span)","explanation":"Select the stated original column basis and take its full real span. Do not substitute rows, assume the image equals the codomain, or normalize columns into different output vectors. The exact result, retaining the requested order, is ([[1, 6], [2, 12]], 1, [[1, 2]], full_real_column_span)."},"skills":["The Image Basis Comes from Original Matrix Columns"],"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-r03-linear-map-image-0032","batch_id":"batch-20261014-advance-20261102-r03-linear-map-image-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"7686feff71c84f50f25a4c771b405913c1a59d0a8abc0ceb4d6deb84dd7136ab","family_id":"qf-mod33-image-basis","seed":202611023032,"semantic_template":"qf-mod33-image-basis/{\"matrix\":[[1,7],[2,14]]}"},"fields":{"title":"The Image Basis Comes from Original Matrix Columns","stem":"Use this same exact integer matrix A=[[1, 7], [2, 14]] for the usual linear map T:R²→R², with rows and columns in the supplied order. Kernel means all real x solving A*x=0; image means the full real column span, not automatically the whole codomain. An empty basis spans the zero space. Rank2 has zero-dimensional kernel and both original columns as an image basis; rank0 has the standard kernel basis and empty image basis. At rank1 the kernel uses a primitive integer normal-equation solution, with its first nonzero coordinate positive, and the image uses the first nonzero original column. All divisions are exact rational operations. Give (matrix,rank,image_basis,full_real_column_span).","answer":{"kind":"expression","value":"([[1, 7], [2, 14]], 1, [[1, 2]], full_real_column_span)","explanation":"Select the stated original column basis and take its full real span. Do not substitute rows, assume the image equals the codomain, or normalize columns into different output vectors. The exact result, retaining the requested order, is ([[1, 7], [2, 14]], 1, [[1, 2]], full_real_column_span)."},"skills":["The Image Basis Comes from Original Matrix Columns"],"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-r03-linear-map-image-0033","batch_id":"batch-20261014-advance-20261102-r03-linear-map-image-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"476405ef03d340dfb298040d189e3f7aab35a6196524dc9d048c5476335b22bb","family_id":"qf-mod33-image-basis","seed":202611023033,"semantic_template":"qf-mod33-image-basis/{\"matrix\":[[1,8],[2,16]]}"},"fields":{"title":"The Image Basis Comes from Original Matrix Columns","stem":"Use this same exact integer matrix A=[[1, 8], [2, 16]] for the usual linear map T:R²→R², with rows and columns in the supplied order. Kernel means all real x solving A*x=0; image means the full real column span, not automatically the whole codomain. An empty basis spans the zero space. Rank2 has zero-dimensional kernel and both original columns as an image basis; rank0 has the standard kernel basis and empty image basis. At rank1 the kernel uses a primitive integer normal-equation solution, with its first nonzero coordinate positive, and the image uses the first nonzero original column. All divisions are exact rational operations. Give (matrix,rank,image_basis,full_real_column_span).","answer":{"kind":"expression","value":"([[1, 8], [2, 16]], 1, [[1, 2]], full_real_column_span)","explanation":"Select the stated original column basis and take its full real span. Do not substitute rows, assume the image equals the codomain, or normalize columns into different output vectors. The exact result, retaining the requested order, is ([[1, 8], [2, 16]], 1, [[1, 2]], full_real_column_span)."},"skills":["The Image Basis Comes from Original Matrix Columns"],"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-r03-linear-map-image-0034","batch_id":"batch-20261014-advance-20261102-r03-linear-map-image-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"3cef8371ee51aee3640a4f3888cf531c202fad372bb6ac24251e72efda0abbd4","family_id":"qf-mod33-image-basis","seed":202611023034,"semantic_template":"qf-mod33-image-basis/{\"matrix\":[[1,9],[2,18]]}"},"fields":{"title":"The Image Basis Comes from Original Matrix Columns","stem":"Use this same exact integer matrix A=[[1, 9], [2, 18]] for the usual linear map T:R²→R², with rows and columns in the supplied order. Kernel means all real x solving A*x=0; image means the full real column span, not automatically the whole codomain. An empty basis spans the zero space. Rank2 has zero-dimensional kernel and both original columns as an image basis; rank0 has the standard kernel basis and empty image basis. At rank1 the kernel uses a primitive integer normal-equation solution, with its first nonzero coordinate positive, and the image uses the first nonzero original column. All divisions are exact rational operations. Give (matrix,rank,image_basis,full_real_column_span).","answer":{"kind":"expression","value":"([[1, 9], [2, 18]], 1, [[1, 2]], full_real_column_span)","explanation":"Select the stated original column basis and take its full real span. Do not substitute rows, assume the image equals the codomain, or normalize columns into different output vectors. The exact result, retaining the requested order, is ([[1, 9], [2, 18]], 1, [[1, 2]], full_real_column_span)."},"skills":["The Image Basis Comes from Original Matrix Columns"],"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-r03-linear-map-image-0035","batch_id":"batch-20261014-advance-20261102-r03-linear-map-image-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"4f1722b97b84ce7e32496e3e378a1e0b8fa1b0b19f10fcf2f19b782fa3a77afc","family_id":"qf-mod33-image-basis","seed":202611023035,"semantic_template":"qf-mod33-image-basis/{\"matrix\":[[1,10],[2,20]]}"},"fields":{"title":"The Image Basis Comes from Original Matrix Columns","stem":"Use this same exact integer matrix A=[[1, 10], [2, 20]] for the usual linear map T:R²→R², with rows and columns in the supplied order. Kernel means all real x solving A*x=0; image means the full real column span, not automatically the whole codomain. An empty basis spans the zero space. Rank2 has zero-dimensional kernel and both original columns as an image basis; rank0 has the standard kernel basis and empty image basis. At rank1 the kernel uses a primitive integer normal-equation solution, with its first nonzero coordinate positive, and the image uses the first nonzero original column. All divisions are exact rational operations. Give (matrix,rank,image_basis,full_real_column_span).","answer":{"kind":"expression","value":"([[1, 10], [2, 20]], 1, [[1, 2]], full_real_column_span)","explanation":"Select the stated original column basis and take its full real span. Do not substitute rows, assume the image equals the codomain, or normalize columns into different output vectors. The exact result, retaining the requested order, is ([[1, 10], [2, 20]], 1, [[1, 2]], full_real_column_span)."},"skills":["The Image Basis Comes from Original Matrix Columns"],"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-r03-linear-map-image-0036","batch_id":"batch-20261014-advance-20261102-r03-linear-map-image-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"3d24abab78a473c844445a089e602ca576f74fa5470c0c4648a012a5fed82d34","family_id":"qf-mod33-image-basis","seed":202611023036,"semantic_template":"qf-mod33-image-basis/{\"matrix\":[[1,11],[2,22]]}"},"fields":{"title":"The Image Basis Comes from Original Matrix Columns","stem":"Use this same exact integer matrix A=[[1, 11], [2, 22]] for the usual linear map T:R²→R², with rows and columns in the supplied order. Kernel means all real x solving A*x=0; image means the full real column span, not automatically the whole codomain. An empty basis spans the zero space. Rank2 has zero-dimensional kernel and both original columns as an image basis; rank0 has the standard kernel basis and empty image basis. At rank1 the kernel uses a primitive integer normal-equation solution, with its first nonzero coordinate positive, and the image uses the first nonzero original column. All divisions are exact rational operations. Give (matrix,rank,image_basis,full_real_column_span).","answer":{"kind":"expression","value":"([[1, 11], [2, 22]], 1, [[1, 2]], full_real_column_span)","explanation":"Select the stated original column basis and take its full real span. Do not substitute rows, assume the image equals the codomain, or normalize columns into different output vectors. The exact result, retaining the requested order, is ([[1, 11], [2, 22]], 1, [[1, 2]], full_real_column_span)."},"skills":["The Image Basis Comes from Original Matrix Columns"],"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-r03-linear-map-image-0037","batch_id":"batch-20261014-advance-20261102-r03-linear-map-image-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"d60b4de1c94295fa03e812b42341728bfdb1dc7eaacdc374d784c23fa3053d64","family_id":"qf-mod33-image-basis","seed":202611023037,"semantic_template":"qf-mod33-image-basis/{\"matrix\":[[1,12],[2,24]]}"},"fields":{"title":"The Image Basis Comes from Original Matrix Columns","stem":"Use this same exact integer matrix A=[[1, 12], [2, 24]] for the usual linear map T:R²→R², with rows and columns in the supplied order. Kernel means all real x solving A*x=0; image means the full real column span, not automatically the whole codomain. An empty basis spans the zero space. Rank2 has zero-dimensional kernel and both original columns as an image basis; rank0 has the standard kernel basis and empty image basis. At rank1 the kernel uses a primitive integer normal-equation solution, with its first nonzero coordinate positive, and the image uses the first nonzero original column. All divisions are exact rational operations. Give (matrix,rank,image_basis,full_real_column_span).","answer":{"kind":"expression","value":"([[1, 12], [2, 24]], 1, [[1, 2]], full_real_column_span)","explanation":"Select the stated original column basis and take its full real span. Do not substitute rows, assume the image equals the codomain, or normalize columns into different output vectors. The exact result, retaining the requested order, is ([[1, 12], [2, 24]], 1, [[1, 2]], full_real_column_span)."},"skills":["The Image Basis Comes from Original Matrix Columns"],"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-r03-linear-map-image-0038","batch_id":"batch-20261014-advance-20261102-r03-linear-map-image-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"88aed1c944a699af26e64c5b5a07f43bb4f62ead860aeab935edaee469cbbb65","family_id":"qf-mod33-image-basis","seed":202611023038,"semantic_template":"qf-mod33-image-basis/{\"matrix\":[[1,13],[2,26]]}"},"fields":{"title":"The Image Basis Comes from Original Matrix Columns","stem":"Use this same exact integer matrix A=[[1, 13], [2, 26]] for the usual linear map T:R²→R², with rows and columns in the supplied order. Kernel means all real x solving A*x=0; image means the full real column span, not automatically the whole codomain. An empty basis spans the zero space. Rank2 has zero-dimensional kernel and both original columns as an image basis; rank0 has the standard kernel basis and empty image basis. At rank1 the kernel uses a primitive integer normal-equation solution, with its first nonzero coordinate positive, and the image uses the first nonzero original column. All divisions are exact rational operations. Give (matrix,rank,image_basis,full_real_column_span).","answer":{"kind":"expression","value":"([[1, 13], [2, 26]], 1, [[1, 2]], full_real_column_span)","explanation":"Select the stated original column basis and take its full real span. Do not substitute rows, assume the image equals the codomain, or normalize columns into different output vectors. The exact result, retaining the requested order, is ([[1, 13], [2, 26]], 1, [[1, 2]], full_real_column_span)."},"skills":["The Image Basis Comes from Original Matrix Columns"],"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-r03-linear-map-image-0039","batch_id":"batch-20261014-advance-20261102-r03-linear-map-image-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"5823099c82771f1d02e5ca4f63bfa101491f66c880a33c06a0d95ba474dd4d6c","family_id":"qf-mod33-image-basis","seed":202611023039,"semantic_template":"qf-mod33-image-basis/{\"matrix\":[[1,14],[2,28]]}"},"fields":{"title":"The Image Basis Comes from Original Matrix Columns","stem":"Use this same exact integer matrix A=[[1, 14], [2, 28]] for the usual linear map T:R²→R², with rows and columns in the supplied order. Kernel means all real x solving A*x=0; image means the full real column span, not automatically the whole codomain. An empty basis spans the zero space. Rank2 has zero-dimensional kernel and both original columns as an image basis; rank0 has the standard kernel basis and empty image basis. At rank1 the kernel uses a primitive integer normal-equation solution, with its first nonzero coordinate positive, and the image uses the first nonzero original column. All divisions are exact rational operations. Give (matrix,rank,image_basis,full_real_column_span).","answer":{"kind":"expression","value":"([[1, 14], [2, 28]], 1, [[1, 2]], full_real_column_span)","explanation":"Select the stated original column basis and take its full real span. Do not substitute rows, assume the image equals the codomain, or normalize columns into different output vectors. The exact result, retaining the requested order, is ([[1, 14], [2, 28]], 1, [[1, 2]], full_real_column_span)."},"skills":["The Image Basis Comes from Original Matrix Columns"],"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-r03-linear-map-image-0040","batch_id":"batch-20261014-advance-20261102-r03-linear-map-image-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"cad945df5e50281cf4013d7af3cc8510f9f2ee29f81668f2bb35987f1d829a5f","family_id":"qf-mod33-image-basis","seed":202611023040,"semantic_template":"qf-mod33-image-basis/{\"matrix\":[[1,15],[2,30]]}"},"fields":{"title":"The Image Basis Comes from Original Matrix Columns","stem":"Use this same exact integer matrix A=[[1, 15], [2, 30]] for the usual linear map T:R²→R², with rows and columns in the supplied order. Kernel means all real x solving A*x=0; image means the full real column span, not automatically the whole codomain. An empty basis spans the zero space. Rank2 has zero-dimensional kernel and both original columns as an image basis; rank0 has the standard kernel basis and empty image basis. At rank1 the kernel uses a primitive integer normal-equation solution, with its first nonzero coordinate positive, and the image uses the first nonzero original column. All divisions are exact rational operations. Give (matrix,rank,image_basis,full_real_column_span).","answer":{"kind":"expression","value":"([[1, 15], [2, 30]], 1, [[1, 2]], full_real_column_span)","explanation":"Select the stated original column basis and take its full real span. Do not substitute rows, assume the image equals the codomain, or normalize columns into different output vectors. The exact result, retaining the requested order, is ([[1, 15], [2, 30]], 1, [[1, 2]], full_real_column_span)."},"skills":["The Image Basis Comes from Original Matrix Columns"],"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-r03-linear-map-image-0041","batch_id":"batch-20261014-advance-20261102-r03-linear-map-image-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"81ceb8bfd19a7de47553280956b122d3d35a9ec6647b8ef737dc28cec421b251","family_id":"qf-mod33-image-basis","seed":202611023041,"semantic_template":"qf-mod33-image-basis/{\"matrix\":[[1,16],[2,32]]}"},"fields":{"title":"The Image Basis Comes from Original Matrix Columns","stem":"Use this same exact integer matrix A=[[1, 16], [2, 32]] for the usual linear map T:R²→R², with rows and columns in the supplied order. Kernel means all real x solving A*x=0; image means the full real column span, not automatically the whole codomain. An empty basis spans the zero space. Rank2 has zero-dimensional kernel and both original columns as an image basis; rank0 has the standard kernel basis and empty image basis. At rank1 the kernel uses a primitive integer normal-equation solution, with its first nonzero coordinate positive, and the image uses the first nonzero original column. All divisions are exact rational operations. Give (matrix,rank,image_basis,full_real_column_span).","answer":{"kind":"expression","value":"([[1, 16], [2, 32]], 1, [[1, 2]], full_real_column_span)","explanation":"Select the stated original column basis and take its full real span. Do not substitute rows, assume the image equals the codomain, or normalize columns into different output vectors. The exact result, retaining the requested order, is ([[1, 16], [2, 32]], 1, [[1, 2]], full_real_column_span)."},"skills":["The Image Basis Comes from Original Matrix Columns"],"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-r03-linear-map-image-0042","batch_id":"batch-20261014-advance-20261102-r03-linear-map-image-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"49a41c6889bf668472da77a2ea300ba70e3abb094928cee48628086fa9bd63a8","family_id":"qf-mod33-image-basis","seed":202611023042,"semantic_template":"qf-mod33-image-basis/{\"matrix\":[[1,17],[2,34]]}"},"fields":{"title":"The Image Basis Comes from Original Matrix Columns","stem":"Use this same exact integer matrix A=[[1, 17], [2, 34]] for the usual linear map T:R²→R², with rows and columns in the supplied order. Kernel means all real x solving A*x=0; image means the full real column span, not automatically the whole codomain. An empty basis spans the zero space. Rank2 has zero-dimensional kernel and both original columns as an image basis; rank0 has the standard kernel basis and empty image basis. At rank1 the kernel uses a primitive integer normal-equation solution, with its first nonzero coordinate positive, and the image uses the first nonzero original column. All divisions are exact rational operations. Give (matrix,rank,image_basis,full_real_column_span).","answer":{"kind":"expression","value":"([[1, 17], [2, 34]], 1, [[1, 2]], full_real_column_span)","explanation":"Select the stated original column basis and take its full real span. Do not substitute rows, assume the image equals the codomain, or normalize columns into different output vectors. The exact result, retaining the requested order, is ([[1, 17], [2, 34]], 1, [[1, 2]], full_real_column_span)."},"skills":["The Image Basis Comes from Original Matrix Columns"],"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-r03-linear-map-image-0043","batch_id":"batch-20261014-advance-20261102-r03-linear-map-image-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"341fdc1f67b1c6919df5f61deb2d8dfc8ad7b04e23ea9a025c50e739205a6ccb","family_id":"qf-mod33-image-basis","seed":202611023043,"semantic_template":"qf-mod33-image-basis/{\"matrix\":[[1,18],[2,36]]}"},"fields":{"title":"The Image Basis Comes from Original Matrix Columns","stem":"Use this same exact integer matrix A=[[1, 18], [2, 36]] for the usual linear map T:R²→R², with rows and columns in the supplied order. Kernel means all real x solving A*x=0; image means the full real column span, not automatically the whole codomain. An empty basis spans the zero space. Rank2 has zero-dimensional kernel and both original columns as an image basis; rank0 has the standard kernel basis and empty image basis. At rank1 the kernel uses a primitive integer normal-equation solution, with its first nonzero coordinate positive, and the image uses the first nonzero original column. All divisions are exact rational operations. Give (matrix,rank,image_basis,full_real_column_span).","answer":{"kind":"expression","value":"([[1, 18], [2, 36]], 1, [[1, 2]], full_real_column_span)","explanation":"Select the stated original column basis and take its full real span. Do not substitute rows, assume the image equals the codomain, or normalize columns into different output vectors. The exact result, retaining the requested order, is ([[1, 18], [2, 36]], 1, [[1, 2]], full_real_column_span)."},"skills":["The Image Basis Comes from Original Matrix Columns"],"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-r03-linear-map-image-0044","batch_id":"batch-20261014-advance-20261102-r03-linear-map-image-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"868cc71af0f370f73472d5c2213c198db4882ca42f048f1113c2f69c69af1504","family_id":"qf-mod33-image-basis","seed":202611023044,"semantic_template":"qf-mod33-image-basis/{\"matrix\":[[1,19],[2,38]]}"},"fields":{"title":"The Image Basis Comes from Original Matrix Columns","stem":"Use this same exact integer matrix A=[[1, 19], [2, 38]] for the usual linear map T:R²→R², with rows and columns in the supplied order. Kernel means all real x solving A*x=0; image means the full real column span, not automatically the whole codomain. An empty basis spans the zero space. Rank2 has zero-dimensional kernel and both original columns as an image basis; rank0 has the standard kernel basis and empty image basis. At rank1 the kernel uses a primitive integer normal-equation solution, with its first nonzero coordinate positive, and the image uses the first nonzero original column. All divisions are exact rational operations. Give (matrix,rank,image_basis,full_real_column_span).","answer":{"kind":"expression","value":"([[1, 19], [2, 38]], 1, [[1, 2]], full_real_column_span)","explanation":"Select the stated original column basis and take its full real span. Do not substitute rows, assume the image equals the codomain, or normalize columns into different output vectors. The exact result, retaining the requested order, is ([[1, 19], [2, 38]], 1, [[1, 2]], full_real_column_span)."},"skills":["The Image Basis Comes from Original Matrix Columns"],"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-r03-linear-map-image-0045","batch_id":"batch-20261014-advance-20261102-r03-linear-map-image-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"034461c6b57c9dbd64d98d67aeed5f210fda2a639edae696f177a9669e6cbefb","family_id":"qf-mod33-image-basis","seed":202611023045,"semantic_template":"qf-mod33-image-basis/{\"matrix\":[[1,20],[2,40]]}"},"fields":{"title":"The Image Basis Comes from Original Matrix Columns","stem":"Use this same exact integer matrix A=[[1, 20], [2, 40]] for the usual linear map T:R²→R², with rows and columns in the supplied order. Kernel means all real x solving A*x=0; image means the full real column span, not automatically the whole codomain. An empty basis spans the zero space. Rank2 has zero-dimensional kernel and both original columns as an image basis; rank0 has the standard kernel basis and empty image basis. At rank1 the kernel uses a primitive integer normal-equation solution, with its first nonzero coordinate positive, and the image uses the first nonzero original column. All divisions are exact rational operations. Give (matrix,rank,image_basis,full_real_column_span).","answer":{"kind":"expression","value":"([[1, 20], [2, 40]], 1, [[1, 2]], full_real_column_span)","explanation":"Select the stated original column basis and take its full real span. Do not substitute rows, assume the image equals the codomain, or normalize columns into different output vectors. The exact result, retaining the requested order, is ([[1, 20], [2, 40]], 1, [[1, 2]], full_real_column_span)."},"skills":["The Image Basis Comes from Original Matrix Columns"],"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-r03-linear-map-image-0046","batch_id":"batch-20261014-advance-20261102-r03-linear-map-image-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"29bc614ae58eb9395bb4aa933b22377d3a64177f9f78c9b381043979ed24277d","family_id":"qf-mod33-image-basis","seed":202611023046,"semantic_template":"qf-mod33-image-basis/{\"matrix\":[[1,21],[2,42]]}"},"fields":{"title":"The Image Basis Comes from Original Matrix Columns","stem":"Use this same exact integer matrix A=[[1, 21], [2, 42]] for the usual linear map T:R²→R², with rows and columns in the supplied order. Kernel means all real x solving A*x=0; image means the full real column span, not automatically the whole codomain. An empty basis spans the zero space. Rank2 has zero-dimensional kernel and both original columns as an image basis; rank0 has the standard kernel basis and empty image basis. At rank1 the kernel uses a primitive integer normal-equation solution, with its first nonzero coordinate positive, and the image uses the first nonzero original column. All divisions are exact rational operations. Give (matrix,rank,image_basis,full_real_column_span).","answer":{"kind":"expression","value":"([[1, 21], [2, 42]], 1, [[1, 2]], full_real_column_span)","explanation":"Select the stated original column basis and take its full real span. Do not substitute rows, assume the image equals the codomain, or normalize columns into different output vectors. The exact result, retaining the requested order, is ([[1, 21], [2, 42]], 1, [[1, 2]], full_real_column_span)."},"skills":["The Image Basis Comes from Original Matrix Columns"],"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-r03-linear-map-image-0047","batch_id":"batch-20261014-advance-20261102-r03-linear-map-image-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"dce040d7ed17cb472e93ba938ac98a2b0b919a12d80bd955d4c39102a4a92dce","family_id":"qf-mod33-image-basis","seed":202611023047,"semantic_template":"qf-mod33-image-basis/{\"matrix\":[[1,22],[2,44]]}"},"fields":{"title":"The Image Basis Comes from Original Matrix Columns","stem":"Use this same exact integer matrix A=[[1, 22], [2, 44]] for the usual linear map T:R²→R², with rows and columns in the supplied order. Kernel means all real x solving A*x=0; image means the full real column span, not automatically the whole codomain. An empty basis spans the zero space. Rank2 has zero-dimensional kernel and both original columns as an image basis; rank0 has the standard kernel basis and empty image basis. At rank1 the kernel uses a primitive integer normal-equation solution, with its first nonzero coordinate positive, and the image uses the first nonzero original column. All divisions are exact rational operations. Give (matrix,rank,image_basis,full_real_column_span).","answer":{"kind":"expression","value":"([[1, 22], [2, 44]], 1, [[1, 2]], full_real_column_span)","explanation":"Select the stated original column basis and take its full real span. Do not substitute rows, assume the image equals the codomain, or normalize columns into different output vectors. The exact result, retaining the requested order, is ([[1, 22], [2, 44]], 1, [[1, 2]], full_real_column_span)."},"skills":["The Image Basis Comes from Original Matrix Columns"],"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-r03-linear-map-image-0048","batch_id":"batch-20261014-advance-20261102-r03-linear-map-image-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"e056e3f4bda4507abb4b8f6cc2cf9b3feaf1f53bf103e69f1854560f37f7bef0","family_id":"qf-mod33-image-basis","seed":202611023048,"semantic_template":"qf-mod33-image-basis/{\"matrix\":[[1,23],[2,46]]}"},"fields":{"title":"The Image Basis Comes from Original Matrix Columns","stem":"Use this same exact integer matrix A=[[1, 23], [2, 46]] for the usual linear map T:R²→R², with rows and columns in the supplied order. Kernel means all real x solving A*x=0; image means the full real column span, not automatically the whole codomain. An empty basis spans the zero space. Rank2 has zero-dimensional kernel and both original columns as an image basis; rank0 has the standard kernel basis and empty image basis. At rank1 the kernel uses a primitive integer normal-equation solution, with its first nonzero coordinate positive, and the image uses the first nonzero original column. All divisions are exact rational operations. Give (matrix,rank,image_basis,full_real_column_span).","answer":{"kind":"expression","value":"([[1, 23], [2, 46]], 1, [[1, 2]], full_real_column_span)","explanation":"Select the stated original column basis and take its full real span. Do not substitute rows, assume the image equals the codomain, or normalize columns into different output vectors. The exact result, retaining the requested order, is ([[1, 23], [2, 46]], 1, [[1, 2]], full_real_column_span)."},"skills":["The Image Basis Comes from Original Matrix Columns"],"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-r03-linear-map-image-0049","batch_id":"batch-20261014-advance-20261102-r03-linear-map-image-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"020c7c1d19ab6b0bcdd8b4e7174ef9cb0e2344532e501bfb3ec83b0a24f261ba","family_id":"qf-mod33-image-basis","seed":202611023049,"semantic_template":"qf-mod33-image-basis/{\"matrix\":[[1,24],[2,48]]}"},"fields":{"title":"The Image Basis Comes from Original Matrix Columns","stem":"Use this same exact integer matrix A=[[1, 24], [2, 48]] for the usual linear map T:R²→R², with rows and columns in the supplied order. Kernel means all real x solving A*x=0; image means the full real column span, not automatically the whole codomain. An empty basis spans the zero space. Rank2 has zero-dimensional kernel and both original columns as an image basis; rank0 has the standard kernel basis and empty image basis. At rank1 the kernel uses a primitive integer normal-equation solution, with its first nonzero coordinate positive, and the image uses the first nonzero original column. All divisions are exact rational operations. Give (matrix,rank,image_basis,full_real_column_span).","answer":{"kind":"expression","value":"([[1, 24], [2, 48]], 1, [[1, 2]], full_real_column_span)","explanation":"Select the stated original column basis and take its full real span. Do not substitute rows, assume the image equals the codomain, or normalize columns into different output vectors. The exact result, retaining the requested order, is ([[1, 24], [2, 48]], 1, [[1, 2]], full_real_column_span)."},"skills":["The Image Basis Comes from Original Matrix Columns"],"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-r03-linear-map-image-0050","batch_id":"batch-20261014-advance-20261102-r03-linear-map-image-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"96a57a78cbe4e2adcbd9c76072a6ba218d1b4cc4440a4071523552b0d3671c43","family_id":"qf-mod33-image-basis","seed":202611023050,"semantic_template":"qf-mod33-image-basis/{\"matrix\":[[1,25],[2,50]]}"},"fields":{"title":"The Image Basis Comes from Original Matrix Columns","stem":"Use this same exact integer matrix A=[[1, 25], [2, 50]] for the usual linear map T:R²→R², with rows and columns in the supplied order. Kernel means all real x solving A*x=0; image means the full real column span, not automatically the whole codomain. An empty basis spans the zero space. Rank2 has zero-dimensional kernel and both original columns as an image basis; rank0 has the standard kernel basis and empty image basis. At rank1 the kernel uses a primitive integer normal-equation solution, with its first nonzero coordinate positive, and the image uses the first nonzero original column. All divisions are exact rational operations. Give (matrix,rank,image_basis,full_real_column_span).","answer":{"kind":"expression","value":"([[1, 25], [2, 50]], 1, [[1, 2]], full_real_column_span)","explanation":"Select the stated original column basis and take its full real span. Do not substitute rows, assume the image equals the codomain, or normalize columns into different output vectors. The exact result, retaining the requested order, is ([[1, 25], [2, 50]], 1, [[1, 2]], full_real_column_span)."},"skills":["The Image Basis Comes from Original Matrix Columns"],"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-r03-linear-map-image-0051","batch_id":"batch-20261014-advance-20261102-r03-linear-map-image-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"8a1b42b9c109b0ccc0e78ff35251b7fade124f38d1da248b38f01459a2bb3983","family_id":"qf-mod33-image-rank","seed":202611023051,"semantic_template":"qf-mod33-image-rank/{\"matrix\":[[1,1],[2,2]]}"},"fields":{"title":"Rank Is the Dimension of the Full Real Image","stem":"Use this same exact integer matrix A=[[1, 1], [2, 2]] for the usual linear map T:R²→R², with rows and columns in the supplied order. Kernel means all real x solving A*x=0; image means the full real column span, not automatically the whole codomain. An empty basis spans the zero space. Rank2 has zero-dimensional kernel and both original columns as an image basis; rank0 has the standard kernel basis and empty image basis. At rank1 the kernel uses a primitive integer normal-equation solution, with its first nonzero coordinate positive, and the image uses the first nonzero original column. All divisions are exact rational operations. Give (rank,basis_count,equal?).","answer":{"kind":"expression","value":"(1, 1, Yes)","explanation":"Count the supplied full image basis and compare with exact matrix rank. A rank0 map has an empty image basis even though its image contains the zero vector. The exact result, retaining the requested order, is (1, 1, Yes)."},"skills":["Rank Is the Dimension of the Full Real Image"],"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-r03-linear-map-image-0052","batch_id":"batch-20261014-advance-20261102-r03-linear-map-image-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"21b1436bbd0ecd86ba78fd4e6dcf2d27f95992a98c3bfd69591e7510ae9e0367","family_id":"qf-mod33-image-rank","seed":202611023052,"semantic_template":"qf-mod33-image-rank/{\"matrix\":[[1,2],[2,4]]}"},"fields":{"title":"Rank Is the Dimension of the Full Real Image","stem":"Use this same exact integer matrix A=[[1, 2], [2, 4]] for the usual linear map T:R²→R², with rows and columns in the supplied order. Kernel means all real x solving A*x=0; image means the full real column span, not automatically the whole codomain. An empty basis spans the zero space. Rank2 has zero-dimensional kernel and both original columns as an image basis; rank0 has the standard kernel basis and empty image basis. At rank1 the kernel uses a primitive integer normal-equation solution, with its first nonzero coordinate positive, and the image uses the first nonzero original column. All divisions are exact rational operations. Give (rank,basis_count,equal?).","answer":{"kind":"expression","value":"(1, 1, Yes)","explanation":"Count the supplied full image basis and compare with exact matrix rank. A rank0 map has an empty image basis even though its image contains the zero vector. The exact result, retaining the requested order, is (1, 1, Yes)."},"skills":["Rank Is the Dimension of the Full Real Image"],"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-r03-linear-map-image-0053","batch_id":"batch-20261014-advance-20261102-r03-linear-map-image-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"e83bf462295b107693c08bc507336d6dc271a6e8fb21a6dd6cf50f98acd9a0f5","family_id":"qf-mod33-image-rank","seed":202611023053,"semantic_template":"qf-mod33-image-rank/{\"matrix\":[[1,3],[2,6]]}"},"fields":{"title":"Rank Is the Dimension of the Full Real Image","stem":"Use this same exact integer matrix A=[[1, 3], [2, 6]] for the usual linear map T:R²→R², with rows and columns in the supplied order. Kernel means all real x solving A*x=0; image means the full real column span, not automatically the whole codomain. An empty basis spans the zero space. Rank2 has zero-dimensional kernel and both original columns as an image basis; rank0 has the standard kernel basis and empty image basis. At rank1 the kernel uses a primitive integer normal-equation solution, with its first nonzero coordinate positive, and the image uses the first nonzero original column. All divisions are exact rational operations. Give (rank,basis_count,equal?).","answer":{"kind":"expression","value":"(1, 1, Yes)","explanation":"Count the supplied full image basis and compare with exact matrix rank. A rank0 map has an empty image basis even though its image contains the zero vector. The exact result, retaining the requested order, is (1, 1, Yes)."},"skills":["Rank Is the Dimension of the Full Real Image"],"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-r03-linear-map-image-0054","batch_id":"batch-20261014-advance-20261102-r03-linear-map-image-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"bbe461ed65d2273c22490c573299c36f4bcf4d4dd8d4c1c32ef6747abef8eaa8","family_id":"qf-mod33-image-rank","seed":202611023054,"semantic_template":"qf-mod33-image-rank/{\"matrix\":[[1,4],[2,8]]}"},"fields":{"title":"Rank Is the Dimension of the Full Real Image","stem":"Use this same exact integer matrix A=[[1, 4], [2, 8]] for the usual linear map T:R²→R², with rows and columns in the supplied order. Kernel means all real x solving A*x=0; image means the full real column span, not automatically the whole codomain. An empty basis spans the zero space. Rank2 has zero-dimensional kernel and both original columns as an image basis; rank0 has the standard kernel basis and empty image basis. At rank1 the kernel uses a primitive integer normal-equation solution, with its first nonzero coordinate positive, and the image uses the first nonzero original column. All divisions are exact rational operations. Give (rank,basis_count,equal?).","answer":{"kind":"expression","value":"(1, 1, Yes)","explanation":"Count the supplied full image basis and compare with exact matrix rank. A rank0 map has an empty image basis even though its image contains the zero vector. The exact result, retaining the requested order, is (1, 1, Yes)."},"skills":["Rank Is the Dimension of the Full Real Image"],"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-r03-linear-map-image-0055","batch_id":"batch-20261014-advance-20261102-r03-linear-map-image-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"90d2fa5ece6f0f176beabe47c47fda4189b0c0bb6c242d7a539fc74d2c7b10f5","family_id":"qf-mod33-image-rank","seed":202611023055,"semantic_template":"qf-mod33-image-rank/{\"matrix\":[[1,5],[2,10]]}"},"fields":{"title":"Rank Is the Dimension of the Full Real Image","stem":"Use this same exact integer matrix A=[[1, 5], [2, 10]] for the usual linear map T:R²→R², with rows and columns in the supplied order. Kernel means all real x solving A*x=0; image means the full real column span, not automatically the whole codomain. An empty basis spans the zero space. Rank2 has zero-dimensional kernel and both original columns as an image basis; rank0 has the standard kernel basis and empty image basis. At rank1 the kernel uses a primitive integer normal-equation solution, with its first nonzero coordinate positive, and the image uses the first nonzero original column. All divisions are exact rational operations. Give (rank,basis_count,equal?).","answer":{"kind":"expression","value":"(1, 1, Yes)","explanation":"Count the supplied full image basis and compare with exact matrix rank. A rank0 map has an empty image basis even though its image contains the zero vector. The exact result, retaining the requested order, is (1, 1, Yes)."},"skills":["Rank Is the Dimension of the Full Real Image"],"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-r03-linear-map-image-0056","batch_id":"batch-20261014-advance-20261102-r03-linear-map-image-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"f43c818d46a3bccaa43e488d26be662a2e60d75744c8cdb86c4b60f35556d1f0","family_id":"qf-mod33-image-rank","seed":202611023056,"semantic_template":"qf-mod33-image-rank/{\"matrix\":[[1,6],[2,12]]}"},"fields":{"title":"Rank Is the Dimension of the Full Real Image","stem":"Use this same exact integer matrix A=[[1, 6], [2, 12]] for the usual linear map T:R²→R², with rows and columns in the supplied order. Kernel means all real x solving A*x=0; image means the full real column span, not automatically the whole codomain. An empty basis spans the zero space. Rank2 has zero-dimensional kernel and both original columns as an image basis; rank0 has the standard kernel basis and empty image basis. At rank1 the kernel uses a primitive integer normal-equation solution, with its first nonzero coordinate positive, and the image uses the first nonzero original column. All divisions are exact rational operations. Give (rank,basis_count,equal?).","answer":{"kind":"expression","value":"(1, 1, Yes)","explanation":"Count the supplied full image basis and compare with exact matrix rank. A rank0 map has an empty image basis even though its image contains the zero vector. The exact result, retaining the requested order, is (1, 1, Yes)."},"skills":["Rank Is the Dimension of the Full Real Image"],"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-r03-linear-map-image-0057","batch_id":"batch-20261014-advance-20261102-r03-linear-map-image-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"56e8343d7ede2659e6f81e0ccbee5ce56a3fdae8ffd006dce81f5b9af1ffa35e","family_id":"qf-mod33-image-rank","seed":202611023057,"semantic_template":"qf-mod33-image-rank/{\"matrix\":[[1,7],[2,14]]}"},"fields":{"title":"Rank Is the Dimension of the Full Real Image","stem":"Use this same exact integer matrix A=[[1, 7], [2, 14]] for the usual linear map T:R²→R², with rows and columns in the supplied order. Kernel means all real x solving A*x=0; image means the full real column span, not automatically the whole codomain. An empty basis spans the zero space. Rank2 has zero-dimensional kernel and both original columns as an image basis; rank0 has the standard kernel basis and empty image basis. At rank1 the kernel uses a primitive integer normal-equation solution, with its first nonzero coordinate positive, and the image uses the first nonzero original column. All divisions are exact rational operations. Give (rank,basis_count,equal?).","answer":{"kind":"expression","value":"(1, 1, Yes)","explanation":"Count the supplied full image basis and compare with exact matrix rank. A rank0 map has an empty image basis even though its image contains the zero vector. The exact result, retaining the requested order, is (1, 1, Yes)."},"skills":["Rank Is the Dimension of the Full Real Image"],"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-r03-linear-map-image-0058","batch_id":"batch-20261014-advance-20261102-r03-linear-map-image-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"e2e13a52dcba183fa646f3c0a31fc3e0ada21fd063a34649860ad94cf8bbf9cb","family_id":"qf-mod33-image-rank","seed":202611023058,"semantic_template":"qf-mod33-image-rank/{\"matrix\":[[1,8],[2,16]]}"},"fields":{"title":"Rank Is the Dimension of the Full Real Image","stem":"Use this same exact integer matrix A=[[1, 8], [2, 16]] for the usual linear map T:R²→R², with rows and columns in the supplied order. Kernel means all real x solving A*x=0; image means the full real column span, not automatically the whole codomain. An empty basis spans the zero space. Rank2 has zero-dimensional kernel and both original columns as an image basis; rank0 has the standard kernel basis and empty image basis. At rank1 the kernel uses a primitive integer normal-equation solution, with its first nonzero coordinate positive, and the image uses the first nonzero original column. All divisions are exact rational operations. Give (rank,basis_count,equal?).","answer":{"kind":"expression","value":"(1, 1, Yes)","explanation":"Count the supplied full image basis and compare with exact matrix rank. A rank0 map has an empty image basis even though its image contains the zero vector. The exact result, retaining the requested order, is (1, 1, Yes)."},"skills":["Rank Is the Dimension of the Full Real Image"],"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-r03-linear-map-image-0059","batch_id":"batch-20261014-advance-20261102-r03-linear-map-image-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"f23fab342e434209e6a9c0d5542b4d579a4592e6391faa287f336cc4a0683bca","family_id":"qf-mod33-image-rank","seed":202611023059,"semantic_template":"qf-mod33-image-rank/{\"matrix\":[[1,9],[2,18]]}"},"fields":{"title":"Rank Is the Dimension of the Full Real Image","stem":"Use this same exact integer matrix A=[[1, 9], [2, 18]] for the usual linear map T:R²→R², with rows and columns in the supplied order. Kernel means all real x solving A*x=0; image means the full real column span, not automatically the whole codomain. An empty basis spans the zero space. Rank2 has zero-dimensional kernel and both original columns as an image basis; rank0 has the standard kernel basis and empty image basis. At rank1 the kernel uses a primitive integer normal-equation solution, with its first nonzero coordinate positive, and the image uses the first nonzero original column. All divisions are exact rational operations. Give (rank,basis_count,equal?).","answer":{"kind":"expression","value":"(1, 1, Yes)","explanation":"Count the supplied full image basis and compare with exact matrix rank. A rank0 map has an empty image basis even though its image contains the zero vector. The exact result, retaining the requested order, is (1, 1, Yes)."},"skills":["Rank Is the Dimension of the Full Real Image"],"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-r03-linear-map-image-0060","batch_id":"batch-20261014-advance-20261102-r03-linear-map-image-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"39a25187e0e6e53c93958622a652bd33a235243f65fc4470373585d642b5c531","family_id":"qf-mod33-image-rank","seed":202611023060,"semantic_template":"qf-mod33-image-rank/{\"matrix\":[[1,10],[2,20]]}"},"fields":{"title":"Rank Is the Dimension of the Full Real Image","stem":"Use this same exact integer matrix A=[[1, 10], [2, 20]] for the usual linear map T:R²→R², with rows and columns in the supplied order. Kernel means all real x solving A*x=0; image means the full real column span, not automatically the whole codomain. An empty basis spans the zero space. Rank2 has zero-dimensional kernel and both original columns as an image basis; rank0 has the standard kernel basis and empty image basis. At rank1 the kernel uses a primitive integer normal-equation solution, with its first nonzero coordinate positive, and the image uses the first nonzero original column. All divisions are exact rational operations. Give (rank,basis_count,equal?).","answer":{"kind":"expression","value":"(1, 1, Yes)","explanation":"Count the supplied full image basis and compare with exact matrix rank. A rank0 map has an empty image basis even though its image contains the zero vector. The exact result, retaining the requested order, is (1, 1, Yes)."},"skills":["Rank Is the Dimension of the Full Real Image"],"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-r03-linear-map-image-0061","batch_id":"batch-20261014-advance-20261102-r03-linear-map-image-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"0c370dca263b169c09f3517a0bb903be2e3dcff33fb68e412aade553644acd97","family_id":"qf-mod33-image-rank","seed":202611023061,"semantic_template":"qf-mod33-image-rank/{\"matrix\":[[1,11],[2,22]]}"},"fields":{"title":"Rank Is the Dimension of the Full Real Image","stem":"Use this same exact integer matrix A=[[1, 11], [2, 22]] for the usual linear map T:R²→R², with rows and columns in the supplied order. Kernel means all real x solving A*x=0; image means the full real column span, not automatically the whole codomain. An empty basis spans the zero space. Rank2 has zero-dimensional kernel and both original columns as an image basis; rank0 has the standard kernel basis and empty image basis. At rank1 the kernel uses a primitive integer normal-equation solution, with its first nonzero coordinate positive, and the image uses the first nonzero original column. All divisions are exact rational operations. Give (rank,basis_count,equal?).","answer":{"kind":"expression","value":"(1, 1, Yes)","explanation":"Count the supplied full image basis and compare with exact matrix rank. A rank0 map has an empty image basis even though its image contains the zero vector. The exact result, retaining the requested order, is (1, 1, Yes)."},"skills":["Rank Is the Dimension of the Full Real Image"],"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-r03-linear-map-image-0062","batch_id":"batch-20261014-advance-20261102-r03-linear-map-image-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"5e93e7fd0950c130a18c759684c840965f591eb0016fd72c8f581d105f2539c7","family_id":"qf-mod33-image-rank","seed":202611023062,"semantic_template":"qf-mod33-image-rank/{\"matrix\":[[1,12],[2,24]]}"},"fields":{"title":"Rank Is the Dimension of the Full Real Image","stem":"Use this same exact integer matrix A=[[1, 12], [2, 24]] for the usual linear map T:R²→R², with rows and columns in the supplied order. Kernel means all real x solving A*x=0; image means the full real column span, not automatically the whole codomain. An empty basis spans the zero space. Rank2 has zero-dimensional kernel and both original columns as an image basis; rank0 has the standard kernel basis and empty image basis. At rank1 the kernel uses a primitive integer normal-equation solution, with its first nonzero coordinate positive, and the image uses the first nonzero original column. All divisions are exact rational operations. Give (rank,basis_count,equal?).","answer":{"kind":"expression","value":"(1, 1, Yes)","explanation":"Count the supplied full image basis and compare with exact matrix rank. A rank0 map has an empty image basis even though its image contains the zero vector. The exact result, retaining the requested order, is (1, 1, Yes)."},"skills":["Rank Is the Dimension of the Full Real Image"],"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-r03-linear-map-image-0063","batch_id":"batch-20261014-advance-20261102-r03-linear-map-image-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"0fbf7e25a7c5b637dd20970e9b8c39199911e9adb321382c2a749c39b2d3cfb4","family_id":"qf-mod33-image-rank","seed":202611023063,"semantic_template":"qf-mod33-image-rank/{\"matrix\":[[1,13],[2,26]]}"},"fields":{"title":"Rank Is the Dimension of the Full Real Image","stem":"Use this same exact integer matrix A=[[1, 13], [2, 26]] for the usual linear map T:R²→R², with rows and columns in the supplied order. Kernel means all real x solving A*x=0; image means the full real column span, not automatically the whole codomain. An empty basis spans the zero space. Rank2 has zero-dimensional kernel and both original columns as an image basis; rank0 has the standard kernel basis and empty image basis. At rank1 the kernel uses a primitive integer normal-equation solution, with its first nonzero coordinate positive, and the image uses the first nonzero original column. All divisions are exact rational operations. Give (rank,basis_count,equal?).","answer":{"kind":"expression","value":"(1, 1, Yes)","explanation":"Count the supplied full image basis and compare with exact matrix rank. A rank0 map has an empty image basis even though its image contains the zero vector. The exact result, retaining the requested order, is (1, 1, Yes)."},"skills":["Rank Is the Dimension of the Full Real Image"],"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-r03-linear-map-image-0064","batch_id":"batch-20261014-advance-20261102-r03-linear-map-image-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"2a09376fcf70a026113e6d4997d37186a8f665397806c7bb4ea31cb52de780c3","family_id":"qf-mod33-image-rank","seed":202611023064,"semantic_template":"qf-mod33-image-rank/{\"matrix\":[[1,14],[2,28]]}"},"fields":{"title":"Rank Is the Dimension of the Full Real Image","stem":"Use this same exact integer matrix A=[[1, 14], [2, 28]] for the usual linear map T:R²→R², with rows and columns in the supplied order. Kernel means all real x solving A*x=0; image means the full real column span, not automatically the whole codomain. An empty basis spans the zero space. Rank2 has zero-dimensional kernel and both original columns as an image basis; rank0 has the standard kernel basis and empty image basis. At rank1 the kernel uses a primitive integer normal-equation solution, with its first nonzero coordinate positive, and the image uses the first nonzero original column. All divisions are exact rational operations. Give (rank,basis_count,equal?).","answer":{"kind":"expression","value":"(1, 1, Yes)","explanation":"Count the supplied full image basis and compare with exact matrix rank. A rank0 map has an empty image basis even though its image contains the zero vector. The exact result, retaining the requested order, is (1, 1, Yes)."},"skills":["Rank Is the Dimension of the Full Real Image"],"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-r03-linear-map-image-0065","batch_id":"batch-20261014-advance-20261102-r03-linear-map-image-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"af2e38abac8d9b40aca5b7bb19c2ba982cee61bd0400950c8120c8606a9f2961","family_id":"qf-mod33-image-rank","seed":202611023065,"semantic_template":"qf-mod33-image-rank/{\"matrix\":[[1,15],[2,30]]}"},"fields":{"title":"Rank Is the Dimension of the Full Real Image","stem":"Use this same exact integer matrix A=[[1, 15], [2, 30]] for the usual linear map T:R²→R², with rows and columns in the supplied order. Kernel means all real x solving A*x=0; image means the full real column span, not automatically the whole codomain. An empty basis spans the zero space. Rank2 has zero-dimensional kernel and both original columns as an image basis; rank0 has the standard kernel basis and empty image basis. At rank1 the kernel uses a primitive integer normal-equation solution, with its first nonzero coordinate positive, and the image uses the first nonzero original column. All divisions are exact rational operations. Give (rank,basis_count,equal?).","answer":{"kind":"expression","value":"(1, 1, Yes)","explanation":"Count the supplied full image basis and compare with exact matrix rank. A rank0 map has an empty image basis even though its image contains the zero vector. The exact result, retaining the requested order, is (1, 1, Yes)."},"skills":["Rank Is the Dimension of the Full Real Image"],"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-r03-linear-map-image-0066","batch_id":"batch-20261014-advance-20261102-r03-linear-map-image-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"675f9d23b9065ac2f876b4e364fe1475337177c65ae5d10697103dcd4f3159d4","family_id":"qf-mod33-image-rank","seed":202611023066,"semantic_template":"qf-mod33-image-rank/{\"matrix\":[[1,16],[2,32]]}"},"fields":{"title":"Rank Is the Dimension of the Full Real Image","stem":"Use this same exact integer matrix A=[[1, 16], [2, 32]] for the usual linear map T:R²→R², with rows and columns in the supplied order. Kernel means all real x solving A*x=0; image means the full real column span, not automatically the whole codomain. An empty basis spans the zero space. Rank2 has zero-dimensional kernel and both original columns as an image basis; rank0 has the standard kernel basis and empty image basis. At rank1 the kernel uses a primitive integer normal-equation solution, with its first nonzero coordinate positive, and the image uses the first nonzero original column. All divisions are exact rational operations. Give (rank,basis_count,equal?).","answer":{"kind":"expression","value":"(1, 1, Yes)","explanation":"Count the supplied full image basis and compare with exact matrix rank. A rank0 map has an empty image basis even though its image contains the zero vector. The exact result, retaining the requested order, is (1, 1, Yes)."},"skills":["Rank Is the Dimension of the Full Real Image"],"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-r03-linear-map-image-0067","batch_id":"batch-20261014-advance-20261102-r03-linear-map-image-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"2876f2a57bb8f775ffc359ef33c8e51ab21b27680f2b3ffd815f0248c6bb594d","family_id":"qf-mod33-image-rank","seed":202611023067,"semantic_template":"qf-mod33-image-rank/{\"matrix\":[[1,17],[2,34]]}"},"fields":{"title":"Rank Is the Dimension of the Full Real Image","stem":"Use this same exact integer matrix A=[[1, 17], [2, 34]] for the usual linear map T:R²→R², with rows and columns in the supplied order. Kernel means all real x solving A*x=0; image means the full real column span, not automatically the whole codomain. An empty basis spans the zero space. Rank2 has zero-dimensional kernel and both original columns as an image basis; rank0 has the standard kernel basis and empty image basis. At rank1 the kernel uses a primitive integer normal-equation solution, with its first nonzero coordinate positive, and the image uses the first nonzero original column. All divisions are exact rational operations. Give (rank,basis_count,equal?).","answer":{"kind":"expression","value":"(1, 1, Yes)","explanation":"Count the supplied full image basis and compare with exact matrix rank. A rank0 map has an empty image basis even though its image contains the zero vector. The exact result, retaining the requested order, is (1, 1, Yes)."},"skills":["Rank Is the Dimension of the Full Real Image"],"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-r03-linear-map-image-0068","batch_id":"batch-20261014-advance-20261102-r03-linear-map-image-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"c97d6adb6a1fc84f4afeaea9004df5ea5e1f13240efc79fe365b670f4446531d","family_id":"qf-mod33-image-rank","seed":202611023068,"semantic_template":"qf-mod33-image-rank/{\"matrix\":[[1,18],[2,36]]}"},"fields":{"title":"Rank Is the Dimension of the Full Real Image","stem":"Use this same exact integer matrix A=[[1, 18], [2, 36]] for the usual linear map T:R²→R², with rows and columns in the supplied order. Kernel means all real x solving A*x=0; image means the full real column span, not automatically the whole codomain. An empty basis spans the zero space. Rank2 has zero-dimensional kernel and both original columns as an image basis; rank0 has the standard kernel basis and empty image basis. At rank1 the kernel uses a primitive integer normal-equation solution, with its first nonzero coordinate positive, and the image uses the first nonzero original column. All divisions are exact rational operations. Give (rank,basis_count,equal?).","answer":{"kind":"expression","value":"(1, 1, Yes)","explanation":"Count the supplied full image basis and compare with exact matrix rank. A rank0 map has an empty image basis even though its image contains the zero vector. The exact result, retaining the requested order, is (1, 1, Yes)."},"skills":["Rank Is the Dimension of the Full Real Image"],"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-r03-linear-map-image-0069","batch_id":"batch-20261014-advance-20261102-r03-linear-map-image-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"9309e20696e5264feadbb32398dd95b68524c75d24c0c3d58022dc805d28dfce","family_id":"qf-mod33-image-rank","seed":202611023069,"semantic_template":"qf-mod33-image-rank/{\"matrix\":[[1,19],[2,38]]}"},"fields":{"title":"Rank Is the Dimension of the Full Real Image","stem":"Use this same exact integer matrix A=[[1, 19], [2, 38]] for the usual linear map T:R²→R², with rows and columns in the supplied order. Kernel means all real x solving A*x=0; image means the full real column span, not automatically the whole codomain. An empty basis spans the zero space. Rank2 has zero-dimensional kernel and both original columns as an image basis; rank0 has the standard kernel basis and empty image basis. At rank1 the kernel uses a primitive integer normal-equation solution, with its first nonzero coordinate positive, and the image uses the first nonzero original column. All divisions are exact rational operations. Give (rank,basis_count,equal?).","answer":{"kind":"expression","value":"(1, 1, Yes)","explanation":"Count the supplied full image basis and compare with exact matrix rank. A rank0 map has an empty image basis even though its image contains the zero vector. The exact result, retaining the requested order, is (1, 1, Yes)."},"skills":["Rank Is the Dimension of the Full Real Image"],"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-r03-linear-map-image-0070","batch_id":"batch-20261014-advance-20261102-r03-linear-map-image-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"b8d8955f90ed37cafdc555af4cead7d559d447f93dc6181eddbf0a6f4ecdefb1","family_id":"qf-mod33-image-rank","seed":202611023070,"semantic_template":"qf-mod33-image-rank/{\"matrix\":[[1,20],[2,40]]}"},"fields":{"title":"Rank Is the Dimension of the Full Real Image","stem":"Use this same exact integer matrix A=[[1, 20], [2, 40]] for the usual linear map T:R²→R², with rows and columns in the supplied order. Kernel means all real x solving A*x=0; image means the full real column span, not automatically the whole codomain. An empty basis spans the zero space. Rank2 has zero-dimensional kernel and both original columns as an image basis; rank0 has the standard kernel basis and empty image basis. At rank1 the kernel uses a primitive integer normal-equation solution, with its first nonzero coordinate positive, and the image uses the first nonzero original column. All divisions are exact rational operations. Give (rank,basis_count,equal?).","answer":{"kind":"expression","value":"(1, 1, Yes)","explanation":"Count the supplied full image basis and compare with exact matrix rank. A rank0 map has an empty image basis even though its image contains the zero vector. The exact result, retaining the requested order, is (1, 1, Yes)."},"skills":["Rank Is the Dimension of the Full Real Image"],"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-r03-linear-map-image-0071","batch_id":"batch-20261014-advance-20261102-r03-linear-map-image-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"595ccddc2b9cc5e4836d86266ee4e8f8dc30b79f059a9f4c1e93616e8ad84945","family_id":"qf-mod33-image-rank","seed":202611023071,"semantic_template":"qf-mod33-image-rank/{\"matrix\":[[1,21],[2,42]]}"},"fields":{"title":"Rank Is the Dimension of the Full Real Image","stem":"Use this same exact integer matrix A=[[1, 21], [2, 42]] for the usual linear map T:R²→R², with rows and columns in the supplied order. Kernel means all real x solving A*x=0; image means the full real column span, not automatically the whole codomain. An empty basis spans the zero space. Rank2 has zero-dimensional kernel and both original columns as an image basis; rank0 has the standard kernel basis and empty image basis. At rank1 the kernel uses a primitive integer normal-equation solution, with its first nonzero coordinate positive, and the image uses the first nonzero original column. All divisions are exact rational operations. Give (rank,basis_count,equal?).","answer":{"kind":"expression","value":"(1, 1, Yes)","explanation":"Count the supplied full image basis and compare with exact matrix rank. A rank0 map has an empty image basis even though its image contains the zero vector. The exact result, retaining the requested order, is (1, 1, Yes)."},"skills":["Rank Is the Dimension of the Full Real Image"],"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-r03-linear-map-image-0072","batch_id":"batch-20261014-advance-20261102-r03-linear-map-image-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"62c8295b0d7e1b7ed38b1258355ac60d0bc01ad7109cd5aa0c7aad68384f4960","family_id":"qf-mod33-image-rank","seed":202611023072,"semantic_template":"qf-mod33-image-rank/{\"matrix\":[[1,22],[2,44]]}"},"fields":{"title":"Rank Is the Dimension of the Full Real Image","stem":"Use this same exact integer matrix A=[[1, 22], [2, 44]] for the usual linear map T:R²→R², with rows and columns in the supplied order. Kernel means all real x solving A*x=0; image means the full real column span, not automatically the whole codomain. An empty basis spans the zero space. Rank2 has zero-dimensional kernel and both original columns as an image basis; rank0 has the standard kernel basis and empty image basis. At rank1 the kernel uses a primitive integer normal-equation solution, with its first nonzero coordinate positive, and the image uses the first nonzero original column. All divisions are exact rational operations. Give (rank,basis_count,equal?).","answer":{"kind":"expression","value":"(1, 1, Yes)","explanation":"Count the supplied full image basis and compare with exact matrix rank. A rank0 map has an empty image basis even though its image contains the zero vector. The exact result, retaining the requested order, is (1, 1, Yes)."},"skills":["Rank Is the Dimension of the Full Real Image"],"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-r03-linear-map-image-0073","batch_id":"batch-20261014-advance-20261102-r03-linear-map-image-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"54e868c9366c6af6df3d63645c27aed63d41985fa717a1b497517b36edb3eecb","family_id":"qf-mod33-image-rank","seed":202611023073,"semantic_template":"qf-mod33-image-rank/{\"matrix\":[[1,23],[2,46]]}"},"fields":{"title":"Rank Is the Dimension of the Full Real Image","stem":"Use this same exact integer matrix A=[[1, 23], [2, 46]] for the usual linear map T:R²→R², with rows and columns in the supplied order. Kernel means all real x solving A*x=0; image means the full real column span, not automatically the whole codomain. An empty basis spans the zero space. Rank2 has zero-dimensional kernel and both original columns as an image basis; rank0 has the standard kernel basis and empty image basis. At rank1 the kernel uses a primitive integer normal-equation solution, with its first nonzero coordinate positive, and the image uses the first nonzero original column. All divisions are exact rational operations. Give (rank,basis_count,equal?).","answer":{"kind":"expression","value":"(1, 1, Yes)","explanation":"Count the supplied full image basis and compare with exact matrix rank. A rank0 map has an empty image basis even though its image contains the zero vector. The exact result, retaining the requested order, is (1, 1, Yes)."},"skills":["Rank Is the Dimension of the Full Real Image"],"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-r03-linear-map-image-0074","batch_id":"batch-20261014-advance-20261102-r03-linear-map-image-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"a6f14ef36e812c6a2e476f6b6786210237cfb98113c29d6748026da927acb205","family_id":"qf-mod33-image-rank","seed":202611023074,"semantic_template":"qf-mod33-image-rank/{\"matrix\":[[1,24],[2,48]]}"},"fields":{"title":"Rank Is the Dimension of the Full Real Image","stem":"Use this same exact integer matrix A=[[1, 24], [2, 48]] for the usual linear map T:R²→R², with rows and columns in the supplied order. Kernel means all real x solving A*x=0; image means the full real column span, not automatically the whole codomain. An empty basis spans the zero space. Rank2 has zero-dimensional kernel and both original columns as an image basis; rank0 has the standard kernel basis and empty image basis. At rank1 the kernel uses a primitive integer normal-equation solution, with its first nonzero coordinate positive, and the image uses the first nonzero original column. All divisions are exact rational operations. Give (rank,basis_count,equal?).","answer":{"kind":"expression","value":"(1, 1, Yes)","explanation":"Count the supplied full image basis and compare with exact matrix rank. A rank0 map has an empty image basis even though its image contains the zero vector. The exact result, retaining the requested order, is (1, 1, Yes)."},"skills":["Rank Is the Dimension of the Full Real Image"],"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-r03-linear-map-image-0075","batch_id":"batch-20261014-advance-20261102-r03-linear-map-image-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"20484f75a7ad56ce2a4489fe0f7199035879105631a3b78e1e5f9f407f218c30","family_id":"qf-mod33-image-rank","seed":202611023075,"semantic_template":"qf-mod33-image-rank/{\"matrix\":[[1,25],[2,50]]}"},"fields":{"title":"Rank Is the Dimension of the Full Real Image","stem":"Use this same exact integer matrix A=[[1, 25], [2, 50]] for the usual linear map T:R²→R², with rows and columns in the supplied order. Kernel means all real x solving A*x=0; image means the full real column span, not automatically the whole codomain. An empty basis spans the zero space. Rank2 has zero-dimensional kernel and both original columns as an image basis; rank0 has the standard kernel basis and empty image basis. At rank1 the kernel uses a primitive integer normal-equation solution, with its first nonzero coordinate positive, and the image uses the first nonzero original column. All divisions are exact rational operations. Give (rank,basis_count,equal?).","answer":{"kind":"expression","value":"(1, 1, Yes)","explanation":"Count the supplied full image basis and compare with exact matrix rank. A rank0 map has an empty image basis even though its image contains the zero vector. The exact result, retaining the requested order, is (1, 1, Yes)."},"skills":["Rank Is the Dimension of the Full Real Image"],"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-r03-linear-map-image-0076","batch_id":"batch-20261014-advance-20261102-r03-linear-map-image-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"f51039e4252696bc41ae345ae7951ec2616c2ca06733c1ec6b32b3146052a4c3","family_id":"qf-mod33-image-bounded-search-boundary","seed":202611023076,"semantic_template":"qf-mod33-image-bounded-search-boundary/{\"bound\":1,\"matrix\":[[2,0],[0,2]],\"target\":[1]}"},"fields":{"title":"Failure to Find an Integer Preimage Does Not Establish Real Nonmembership","stem":"Use this same exact integer matrix A=[[2, 0], [0, 2]] for the usual linear map T:R²→R², with rows and columns in the supplied order. Kernel means all real x solving A*x=0; image means the full real column span, not automatically the whole codomain. An empty basis spans the zero space. Rank2 has zero-dimensional kernel and both original columns as an image basis; rank0 has the standard kernel basis and empty image basis. At rank1 the kernel uses a primitive integer normal-equation solution, with its first nonzero coordinate positive, and the image uses the first nonzero original column. All divisions are exact rational operations. Target=[1, 1], nonnegative bound=1. Search integer pairs x,y in [−bound,bound] in x-then-y increasing order. Give (target,bound,integer_witnesses,global_membership?,boundary).","answer":{"kind":"expression","value":"([1, 1], 1, [], Yes, bounded_failure_not_global_nonmembership)","explanation":"List only preimages in this bounded integer box, then separately test full-real column-span membership. No integer witness together with true real membership yields bounded_failure_not_global_nonmembership; otherwise retain bounded_search_only. A valid noninteger solution can be missed by every integer box. The exact result, retaining the requested order, is ([1, 1], 1, [], Yes, bounded_failure_not_global_nonmembership)."},"skills":["Failure to Find an Integer Preimage Does Not Establish Real Nonmembership"],"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-r03-linear-map-image-0077","batch_id":"batch-20261014-advance-20261102-r03-linear-map-image-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"6df0251e29d797be0b226033a0a6508067c7de27f31eb53149bbedcedfd657af","family_id":"qf-mod33-image-bounded-search-boundary","seed":202611023077,"semantic_template":"qf-mod33-image-bounded-search-boundary/{\"bound\":2,\"matrix\":[[2,0],[0,2]],\"target\":[1,3]}"},"fields":{"title":"Failure to Find an Integer Preimage Does Not Establish Real Nonmembership","stem":"Use this same exact integer matrix A=[[2, 0], [0, 2]] for the usual linear map T:R²→R², with rows and columns in the supplied order. Kernel means all real x solving A*x=0; image means the full real column span, not automatically the whole codomain. An empty basis spans the zero space. Rank2 has zero-dimensional kernel and both original columns as an image basis; rank0 has the standard kernel basis and empty image basis. At rank1 the kernel uses a primitive integer normal-equation solution, with its first nonzero coordinate positive, and the image uses the first nonzero original column. All divisions are exact rational operations. Target=[3, 1], nonnegative bound=2. Search integer pairs x,y in [−bound,bound] in x-then-y increasing order. Give (target,bound,integer_witnesses,global_membership?,boundary).","answer":{"kind":"expression","value":"([3, 1], 2, [], Yes, bounded_failure_not_global_nonmembership)","explanation":"List only preimages in this bounded integer box, then separately test full-real column-span membership. No integer witness together with true real membership yields bounded_failure_not_global_nonmembership; otherwise retain bounded_search_only. A valid noninteger solution can be missed by every integer box. The exact result, retaining the requested order, is ([3, 1], 2, [], Yes, bounded_failure_not_global_nonmembership)."},"skills":["Failure to Find an Integer Preimage Does Not Establish Real Nonmembership"],"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-r03-linear-map-image-0078","batch_id":"batch-20261014-advance-20261102-r03-linear-map-image-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"3497b58c2abb55e0f245c374714477d84b043d90b32a0b0d46ffc5de69a94f67","family_id":"qf-mod33-image-bounded-search-boundary","seed":202611023078,"semantic_template":"qf-mod33-image-bounded-search-boundary/{\"bound\":3,\"matrix\":[[2,0],[0,2]],\"target\":[1,5]}"},"fields":{"title":"Failure to Find an Integer Preimage Does Not Establish Real Nonmembership","stem":"Use this same exact integer matrix A=[[2, 0], [0, 2]] for the usual linear map T:R²→R², with rows and columns in the supplied order. Kernel means all real x solving A*x=0; image means the full real column span, not automatically the whole codomain. An empty basis spans the zero space. Rank2 has zero-dimensional kernel and both original columns as an image basis; rank0 has the standard kernel basis and empty image basis. At rank1 the kernel uses a primitive integer normal-equation solution, with its first nonzero coordinate positive, and the image uses the first nonzero original column. All divisions are exact rational operations. Target=[5, 1], nonnegative bound=3. Search integer pairs x,y in [−bound,bound] in x-then-y increasing order. Give (target,bound,integer_witnesses,global_membership?,boundary).","answer":{"kind":"expression","value":"([5, 1], 3, [], Yes, bounded_failure_not_global_nonmembership)","explanation":"List only preimages in this bounded integer box, then separately test full-real column-span membership. No integer witness together with true real membership yields bounded_failure_not_global_nonmembership; otherwise retain bounded_search_only. A valid noninteger solution can be missed by every integer box. The exact result, retaining the requested order, is ([5, 1], 3, [], Yes, bounded_failure_not_global_nonmembership)."},"skills":["Failure to Find an Integer Preimage Does Not Establish Real Nonmembership"],"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-r03-linear-map-image-0079","batch_id":"batch-20261014-advance-20261102-r03-linear-map-image-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"8a3386e8fbb43614e27e77f0d94589c6956e16920e67292dda24bd81b692e923","family_id":"qf-mod33-image-bounded-search-boundary","seed":202611023079,"semantic_template":"qf-mod33-image-bounded-search-boundary/{\"bound\":4,\"matrix\":[[2,0],[0,2]],\"target\":[1,7]}"},"fields":{"title":"Failure to Find an Integer Preimage Does Not Establish Real Nonmembership","stem":"Use this same exact integer matrix A=[[2, 0], [0, 2]] for the usual linear map T:R²→R², with rows and columns in the supplied order. Kernel means all real x solving A*x=0; image means the full real column span, not automatically the whole codomain. An empty basis spans the zero space. Rank2 has zero-dimensional kernel and both original columns as an image basis; rank0 has the standard kernel basis and empty image basis. At rank1 the kernel uses a primitive integer normal-equation solution, with its first nonzero coordinate positive, and the image uses the first nonzero original column. All divisions are exact rational operations. Target=[7, 1], nonnegative bound=4. Search integer pairs x,y in [−bound,bound] in x-then-y increasing order. Give (target,bound,integer_witnesses,global_membership?,boundary).","answer":{"kind":"expression","value":"([7, 1], 4, [], Yes, bounded_failure_not_global_nonmembership)","explanation":"List only preimages in this bounded integer box, then separately test full-real column-span membership. No integer witness together with true real membership yields bounded_failure_not_global_nonmembership; otherwise retain bounded_search_only. A valid noninteger solution can be missed by every integer box. The exact result, retaining the requested order, is ([7, 1], 4, [], Yes, bounded_failure_not_global_nonmembership)."},"skills":["Failure to Find an Integer Preimage Does Not Establish Real Nonmembership"],"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-r03-linear-map-image-0080","batch_id":"batch-20261014-advance-20261102-r03-linear-map-image-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"36b3933ffe2a9c6d032d43864c922a87f6cab51f68d5487c7465fca1e1476fba","family_id":"qf-mod33-image-bounded-search-boundary","seed":202611023080,"semantic_template":"qf-mod33-image-bounded-search-boundary/{\"bound\":5,\"matrix\":[[2,0],[0,2]],\"target\":[1,9]}"},"fields":{"title":"Failure to Find an Integer Preimage Does Not Establish Real Nonmembership","stem":"Use this same exact integer matrix A=[[2, 0], [0, 2]] for the usual linear map T:R²→R², with rows and columns in the supplied order. Kernel means all real x solving A*x=0; image means the full real column span, not automatically the whole codomain. An empty basis spans the zero space. Rank2 has zero-dimensional kernel and both original columns as an image basis; rank0 has the standard kernel basis and empty image basis. At rank1 the kernel uses a primitive integer normal-equation solution, with its first nonzero coordinate positive, and the image uses the first nonzero original column. All divisions are exact rational operations. Target=[9, 1], nonnegative bound=5. Search integer pairs x,y in [−bound,bound] in x-then-y increasing order. Give (target,bound,integer_witnesses,global_membership?,boundary).","answer":{"kind":"expression","value":"([9, 1], 5, [], Yes, bounded_failure_not_global_nonmembership)","explanation":"List only preimages in this bounded integer box, then separately test full-real column-span membership. No integer witness together with true real membership yields bounded_failure_not_global_nonmembership; otherwise retain bounded_search_only. A valid noninteger solution can be missed by every integer box. The exact result, retaining the requested order, is ([9, 1], 5, [], Yes, bounded_failure_not_global_nonmembership)."},"skills":["Failure to Find an Integer Preimage Does Not Establish Real Nonmembership"],"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-r03-linear-map-image-0081","batch_id":"batch-20261014-advance-20261102-r03-linear-map-image-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"f1ae3c0ae04dd8aac64c4bb234bd251a325803103e53c008daec642bdd65417e","family_id":"qf-mod33-image-bounded-search-boundary","seed":202611023081,"semantic_template":"qf-mod33-image-bounded-search-boundary/{\"bound\":1,\"matrix\":[[2,0],[0,2]],\"target\":[1,11]}"},"fields":{"title":"Failure to Find an Integer Preimage Does Not Establish Real Nonmembership","stem":"Use this same exact integer matrix A=[[2, 0], [0, 2]] for the usual linear map T:R²→R², with rows and columns in the supplied order. Kernel means all real x solving A*x=0; image means the full real column span, not automatically the whole codomain. An empty basis spans the zero space. Rank2 has zero-dimensional kernel and both original columns as an image basis; rank0 has the standard kernel basis and empty image basis. At rank1 the kernel uses a primitive integer normal-equation solution, with its first nonzero coordinate positive, and the image uses the first nonzero original column. All divisions are exact rational operations. Target=[11, 1], nonnegative bound=1. Search integer pairs x,y in [−bound,bound] in x-then-y increasing order. Give (target,bound,integer_witnesses,global_membership?,boundary).","answer":{"kind":"expression","value":"([11, 1], 1, [], Yes, bounded_failure_not_global_nonmembership)","explanation":"List only preimages in this bounded integer box, then separately test full-real column-span membership. No integer witness together with true real membership yields bounded_failure_not_global_nonmembership; otherwise retain bounded_search_only. A valid noninteger solution can be missed by every integer box. The exact result, retaining the requested order, is ([11, 1], 1, [], Yes, bounded_failure_not_global_nonmembership)."},"skills":["Failure to Find an Integer Preimage Does Not Establish Real Nonmembership"],"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-r03-linear-map-image-0082","batch_id":"batch-20261014-advance-20261102-r03-linear-map-image-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"d88f0c59d19289ca083a0467f4c7be49cdb386ff1e34b67187b36676cbe34237","family_id":"qf-mod33-image-bounded-search-boundary","seed":202611023082,"semantic_template":"qf-mod33-image-bounded-search-boundary/{\"bound\":2,\"matrix\":[[2,0],[0,2]],\"target\":[1,13]}"},"fields":{"title":"Failure to Find an Integer Preimage Does Not Establish Real Nonmembership","stem":"Use this same exact integer matrix A=[[2, 0], [0, 2]] for the usual linear map T:R²→R², with rows and columns in the supplied order. Kernel means all real x solving A*x=0; image means the full real column span, not automatically the whole codomain. An empty basis spans the zero space. Rank2 has zero-dimensional kernel and both original columns as an image basis; rank0 has the standard kernel basis and empty image basis. At rank1 the kernel uses a primitive integer normal-equation solution, with its first nonzero coordinate positive, and the image uses the first nonzero original column. All divisions are exact rational operations. Target=[13, 1], nonnegative bound=2. Search integer pairs x,y in [−bound,bound] in x-then-y increasing order. Give (target,bound,integer_witnesses,global_membership?,boundary).","answer":{"kind":"expression","value":"([13, 1], 2, [], Yes, bounded_failure_not_global_nonmembership)","explanation":"List only preimages in this bounded integer box, then separately test full-real column-span membership. No integer witness together with true real membership yields bounded_failure_not_global_nonmembership; otherwise retain bounded_search_only. A valid noninteger solution can be missed by every integer box. The exact result, retaining the requested order, is ([13, 1], 2, [], Yes, bounded_failure_not_global_nonmembership)."},"skills":["Failure to Find an Integer Preimage Does Not Establish Real Nonmembership"],"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-r03-linear-map-image-0083","batch_id":"batch-20261014-advance-20261102-r03-linear-map-image-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"e00221bec5e95fc6a66cc1bd1decc69b9904b092d6c253b945a399c2d39a75f8","family_id":"qf-mod33-image-bounded-search-boundary","seed":202611023083,"semantic_template":"qf-mod33-image-bounded-search-boundary/{\"bound\":3,\"matrix\":[[2,0],[0,2]],\"target\":[1,15]}"},"fields":{"title":"Failure to Find an Integer Preimage Does Not Establish Real Nonmembership","stem":"Use this same exact integer matrix A=[[2, 0], [0, 2]] for the usual linear map T:R²→R², with rows and columns in the supplied order. Kernel means all real x solving A*x=0; image means the full real column span, not automatically the whole codomain. An empty basis spans the zero space. Rank2 has zero-dimensional kernel and both original columns as an image basis; rank0 has the standard kernel basis and empty image basis. At rank1 the kernel uses a primitive integer normal-equation solution, with its first nonzero coordinate positive, and the image uses the first nonzero original column. All divisions are exact rational operations. Target=[15, 1], nonnegative bound=3. Search integer pairs x,y in [−bound,bound] in x-then-y increasing order. Give (target,bound,integer_witnesses,global_membership?,boundary).","answer":{"kind":"expression","value":"([15, 1], 3, [], Yes, bounded_failure_not_global_nonmembership)","explanation":"List only preimages in this bounded integer box, then separately test full-real column-span membership. No integer witness together with true real membership yields bounded_failure_not_global_nonmembership; otherwise retain bounded_search_only. A valid noninteger solution can be missed by every integer box. The exact result, retaining the requested order, is ([15, 1], 3, [], Yes, bounded_failure_not_global_nonmembership)."},"skills":["Failure to Find an Integer Preimage Does Not Establish Real Nonmembership"],"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-r03-linear-map-image-0084","batch_id":"batch-20261014-advance-20261102-r03-linear-map-image-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"deb485b008f8ccc45b80a6474018cf4fa86192dbf41f19ed20bf7377006ad259","family_id":"qf-mod33-image-bounded-search-boundary","seed":202611023084,"semantic_template":"qf-mod33-image-bounded-search-boundary/{\"bound\":4,\"matrix\":[[2,0],[0,2]],\"target\":[1,17]}"},"fields":{"title":"Failure to Find an Integer Preimage Does Not Establish Real Nonmembership","stem":"Use this same exact integer matrix A=[[2, 0], [0, 2]] for the usual linear map T:R²→R², with rows and columns in the supplied order. Kernel means all real x solving A*x=0; image means the full real column span, not automatically the whole codomain. An empty basis spans the zero space. Rank2 has zero-dimensional kernel and both original columns as an image basis; rank0 has the standard kernel basis and empty image basis. At rank1 the kernel uses a primitive integer normal-equation solution, with its first nonzero coordinate positive, and the image uses the first nonzero original column. All divisions are exact rational operations. Target=[17, 1], nonnegative bound=4. Search integer pairs x,y in [−bound,bound] in x-then-y increasing order. Give (target,bound,integer_witnesses,global_membership?,boundary).","answer":{"kind":"expression","value":"([17, 1], 4, [], Yes, bounded_failure_not_global_nonmembership)","explanation":"List only preimages in this bounded integer box, then separately test full-real column-span membership. No integer witness together with true real membership yields bounded_failure_not_global_nonmembership; otherwise retain bounded_search_only. A valid noninteger solution can be missed by every integer box. The exact result, retaining the requested order, is ([17, 1], 4, [], Yes, bounded_failure_not_global_nonmembership)."},"skills":["Failure to Find an Integer Preimage Does Not Establish Real Nonmembership"],"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-r03-linear-map-image-0085","batch_id":"batch-20261014-advance-20261102-r03-linear-map-image-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"bfa33c7de64ca61a9eafd5f70e0f2387012cfbd90cfe04c2e789129824f6fee6","family_id":"qf-mod33-image-bounded-search-boundary","seed":202611023085,"semantic_template":"qf-mod33-image-bounded-search-boundary/{\"bound\":5,\"matrix\":[[2,0],[0,2]],\"target\":[1,19]}"},"fields":{"title":"Failure to Find an Integer Preimage Does Not Establish Real Nonmembership","stem":"Use this same exact integer matrix A=[[2, 0], [0, 2]] for the usual linear map T:R²→R², with rows and columns in the supplied order. Kernel means all real x solving A*x=0; image means the full real column span, not automatically the whole codomain. An empty basis spans the zero space. Rank2 has zero-dimensional kernel and both original columns as an image basis; rank0 has the standard kernel basis and empty image basis. At rank1 the kernel uses a primitive integer normal-equation solution, with its first nonzero coordinate positive, and the image uses the first nonzero original column. All divisions are exact rational operations. Target=[19, 1], nonnegative bound=5. Search integer pairs x,y in [−bound,bound] in x-then-y increasing order. Give (target,bound,integer_witnesses,global_membership?,boundary).","answer":{"kind":"expression","value":"([19, 1], 5, [], Yes, bounded_failure_not_global_nonmembership)","explanation":"List only preimages in this bounded integer box, then separately test full-real column-span membership. No integer witness together with true real membership yields bounded_failure_not_global_nonmembership; otherwise retain bounded_search_only. A valid noninteger solution can be missed by every integer box. The exact result, retaining the requested order, is ([19, 1], 5, [], Yes, bounded_failure_not_global_nonmembership)."},"skills":["Failure to Find an Integer Preimage Does Not Establish Real Nonmembership"],"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-r03-linear-map-image-0086","batch_id":"batch-20261014-advance-20261102-r03-linear-map-image-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"efcc5f1b3112ce22a3f4b90fac9f80c1ae1137367ca19af85269e224c6e57565","family_id":"qf-mod33-image-bounded-search-boundary","seed":202611023086,"semantic_template":"qf-mod33-image-bounded-search-boundary/{\"bound\":1,\"matrix\":[[2,0],[0,2]],\"target\":[1,21]}"},"fields":{"title":"Failure to Find an Integer Preimage Does Not Establish Real Nonmembership","stem":"Use this same exact integer matrix A=[[2, 0], [0, 2]] for the usual linear map T:R²→R², with rows and columns in the supplied order. Kernel means all real x solving A*x=0; image means the full real column span, not automatically the whole codomain. An empty basis spans the zero space. Rank2 has zero-dimensional kernel and both original columns as an image basis; rank0 has the standard kernel basis and empty image basis. At rank1 the kernel uses a primitive integer normal-equation solution, with its first nonzero coordinate positive, and the image uses the first nonzero original column. All divisions are exact rational operations. Target=[21, 1], nonnegative bound=1. Search integer pairs x,y in [−bound,bound] in x-then-y increasing order. Give (target,bound,integer_witnesses,global_membership?,boundary).","answer":{"kind":"expression","value":"([21, 1], 1, [], Yes, bounded_failure_not_global_nonmembership)","explanation":"List only preimages in this bounded integer box, then separately test full-real column-span membership. No integer witness together with true real membership yields bounded_failure_not_global_nonmembership; otherwise retain bounded_search_only. A valid noninteger solution can be missed by every integer box. The exact result, retaining the requested order, is ([21, 1], 1, [], Yes, bounded_failure_not_global_nonmembership)."},"skills":["Failure to Find an Integer Preimage Does Not Establish Real Nonmembership"],"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-r03-linear-map-image-0087","batch_id":"batch-20261014-advance-20261102-r03-linear-map-image-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"a7e9e5deddae70e0cfbd4eb051e6fc7c4ed932724ec036ddd1b5253c286bd8e5","family_id":"qf-mod33-image-bounded-search-boundary","seed":202611023087,"semantic_template":"qf-mod33-image-bounded-search-boundary/{\"bound\":2,\"matrix\":[[2,0],[0,2]],\"target\":[1,23]}"},"fields":{"title":"Failure to Find an Integer Preimage Does Not Establish Real Nonmembership","stem":"Use this same exact integer matrix A=[[2, 0], [0, 2]] for the usual linear map T:R²→R², with rows and columns in the supplied order. Kernel means all real x solving A*x=0; image means the full real column span, not automatically the whole codomain. An empty basis spans the zero space. Rank2 has zero-dimensional kernel and both original columns as an image basis; rank0 has the standard kernel basis and empty image basis. At rank1 the kernel uses a primitive integer normal-equation solution, with its first nonzero coordinate positive, and the image uses the first nonzero original column. All divisions are exact rational operations. Target=[23, 1], nonnegative bound=2. Search integer pairs x,y in [−bound,bound] in x-then-y increasing order. Give (target,bound,integer_witnesses,global_membership?,boundary).","answer":{"kind":"expression","value":"([23, 1], 2, [], Yes, bounded_failure_not_global_nonmembership)","explanation":"List only preimages in this bounded integer box, then separately test full-real column-span membership. No integer witness together with true real membership yields bounded_failure_not_global_nonmembership; otherwise retain bounded_search_only. A valid noninteger solution can be missed by every integer box. The exact result, retaining the requested order, is ([23, 1], 2, [], Yes, bounded_failure_not_global_nonmembership)."},"skills":["Failure to Find an Integer Preimage Does Not Establish Real Nonmembership"],"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-r03-linear-map-image-0088","batch_id":"batch-20261014-advance-20261102-r03-linear-map-image-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"88fe1d84a8c3a363af9028b86c849d52a8c47b129db233b43218a6a9a66021ba","family_id":"qf-mod33-image-bounded-search-boundary","seed":202611023088,"semantic_template":"qf-mod33-image-bounded-search-boundary/{\"bound\":3,\"matrix\":[[2,0],[0,2]],\"target\":[1,25]}"},"fields":{"title":"Failure to Find an Integer Preimage Does Not Establish Real Nonmembership","stem":"Use this same exact integer matrix A=[[2, 0], [0, 2]] for the usual linear map T:R²→R², with rows and columns in the supplied order. Kernel means all real x solving A*x=0; image means the full real column span, not automatically the whole codomain. An empty basis spans the zero space. Rank2 has zero-dimensional kernel and both original columns as an image basis; rank0 has the standard kernel basis and empty image basis. At rank1 the kernel uses a primitive integer normal-equation solution, with its first nonzero coordinate positive, and the image uses the first nonzero original column. All divisions are exact rational operations. Target=[25, 1], nonnegative bound=3. Search integer pairs x,y in [−bound,bound] in x-then-y increasing order. Give (target,bound,integer_witnesses,global_membership?,boundary).","answer":{"kind":"expression","value":"([25, 1], 3, [], Yes, bounded_failure_not_global_nonmembership)","explanation":"List only preimages in this bounded integer box, then separately test full-real column-span membership. No integer witness together with true real membership yields bounded_failure_not_global_nonmembership; otherwise retain bounded_search_only. A valid noninteger solution can be missed by every integer box. The exact result, retaining the requested order, is ([25, 1], 3, [], Yes, bounded_failure_not_global_nonmembership)."},"skills":["Failure to Find an Integer Preimage Does Not Establish Real Nonmembership"],"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-r03-linear-map-image-0089","batch_id":"batch-20261014-advance-20261102-r03-linear-map-image-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"b094816d8d004d6b03de3d4c81c6296121909c2989432cca1af53c7ab0129550","family_id":"qf-mod33-image-bounded-search-boundary","seed":202611023089,"semantic_template":"qf-mod33-image-bounded-search-boundary/{\"bound\":4,\"matrix\":[[2,0],[0,2]],\"target\":[1,27]}"},"fields":{"title":"Failure to Find an Integer Preimage Does Not Establish Real Nonmembership","stem":"Use this same exact integer matrix A=[[2, 0], [0, 2]] for the usual linear map T:R²→R², with rows and columns in the supplied order. Kernel means all real x solving A*x=0; image means the full real column span, not automatically the whole codomain. An empty basis spans the zero space. Rank2 has zero-dimensional kernel and both original columns as an image basis; rank0 has the standard kernel basis and empty image basis. At rank1 the kernel uses a primitive integer normal-equation solution, with its first nonzero coordinate positive, and the image uses the first nonzero original column. All divisions are exact rational operations. Target=[27, 1], nonnegative bound=4. Search integer pairs x,y in [−bound,bound] in x-then-y increasing order. Give (target,bound,integer_witnesses,global_membership?,boundary).","answer":{"kind":"expression","value":"([27, 1], 4, [], Yes, bounded_failure_not_global_nonmembership)","explanation":"List only preimages in this bounded integer box, then separately test full-real column-span membership. No integer witness together with true real membership yields bounded_failure_not_global_nonmembership; otherwise retain bounded_search_only. A valid noninteger solution can be missed by every integer box. The exact result, retaining the requested order, is ([27, 1], 4, [], Yes, bounded_failure_not_global_nonmembership)."},"skills":["Failure to Find an Integer Preimage Does Not Establish Real Nonmembership"],"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-r03-linear-map-image-0090","batch_id":"batch-20261014-advance-20261102-r03-linear-map-image-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"7b4c5255226717c0e0b76a78cf2f777ace77add0690ac91452fc0e9008a7f42d","family_id":"qf-mod33-image-bounded-search-boundary","seed":202611023090,"semantic_template":"qf-mod33-image-bounded-search-boundary/{\"bound\":5,\"matrix\":[[2,0],[0,2]],\"target\":[1,29]}"},"fields":{"title":"Failure to Find an Integer Preimage Does Not Establish Real Nonmembership","stem":"Use this same exact integer matrix A=[[2, 0], [0, 2]] for the usual linear map T:R²→R², with rows and columns in the supplied order. Kernel means all real x solving A*x=0; image means the full real column span, not automatically the whole codomain. An empty basis spans the zero space. Rank2 has zero-dimensional kernel and both original columns as an image basis; rank0 has the standard kernel basis and empty image basis. At rank1 the kernel uses a primitive integer normal-equation solution, with its first nonzero coordinate positive, and the image uses the first nonzero original column. All divisions are exact rational operations. Target=[29, 1], nonnegative bound=5. Search integer pairs x,y in [−bound,bound] in x-then-y increasing order. Give (target,bound,integer_witnesses,global_membership?,boundary).","answer":{"kind":"expression","value":"([29, 1], 5, [], Yes, bounded_failure_not_global_nonmembership)","explanation":"List only preimages in this bounded integer box, then separately test full-real column-span membership. No integer witness together with true real membership yields bounded_failure_not_global_nonmembership; otherwise retain bounded_search_only. A valid noninteger solution can be missed by every integer box. The exact result, retaining the requested order, is ([29, 1], 5, [], Yes, bounded_failure_not_global_nonmembership)."},"skills":["Failure to Find an Integer Preimage Does Not Establish Real Nonmembership"],"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-r03-linear-map-image-0091","batch_id":"batch-20261014-advance-20261102-r03-linear-map-image-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"1a234cad0a5ce28dcf1547f17d412031b9db481c4cf7ebdc357bc257e987ca69","family_id":"qf-mod33-image-bounded-search-boundary","seed":202611023091,"semantic_template":"qf-mod33-image-bounded-search-boundary/{\"bound\":1,\"matrix\":[[2,0],[0,2]],\"target\":[1,31]}"},"fields":{"title":"Failure to Find an Integer Preimage Does Not Establish Real Nonmembership","stem":"Use this same exact integer matrix A=[[2, 0], [0, 2]] for the usual linear map T:R²→R², with rows and columns in the supplied order. Kernel means all real x solving A*x=0; image means the full real column span, not automatically the whole codomain. An empty basis spans the zero space. Rank2 has zero-dimensional kernel and both original columns as an image basis; rank0 has the standard kernel basis and empty image basis. At rank1 the kernel uses a primitive integer normal-equation solution, with its first nonzero coordinate positive, and the image uses the first nonzero original column. All divisions are exact rational operations. Target=[31, 1], nonnegative bound=1. Search integer pairs x,y in [−bound,bound] in x-then-y increasing order. Give (target,bound,integer_witnesses,global_membership?,boundary).","answer":{"kind":"expression","value":"([31, 1], 1, [], Yes, bounded_failure_not_global_nonmembership)","explanation":"List only preimages in this bounded integer box, then separately test full-real column-span membership. No integer witness together with true real membership yields bounded_failure_not_global_nonmembership; otherwise retain bounded_search_only. A valid noninteger solution can be missed by every integer box. The exact result, retaining the requested order, is ([31, 1], 1, [], Yes, bounded_failure_not_global_nonmembership)."},"skills":["Failure to Find an Integer Preimage Does Not Establish Real Nonmembership"],"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-r03-linear-map-image-0092","batch_id":"batch-20261014-advance-20261102-r03-linear-map-image-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"9f0af206597847b06ad5c93eca2a0a9ab00204d457bdbaafc8eb230182629679","family_id":"qf-mod33-image-bounded-search-boundary","seed":202611023092,"semantic_template":"qf-mod33-image-bounded-search-boundary/{\"bound\":2,\"matrix\":[[2,0],[0,2]],\"target\":[1,33]}"},"fields":{"title":"Failure to Find an Integer Preimage Does Not Establish Real Nonmembership","stem":"Use this same exact integer matrix A=[[2, 0], [0, 2]] for the usual linear map T:R²→R², with rows and columns in the supplied order. Kernel means all real x solving A*x=0; image means the full real column span, not automatically the whole codomain. An empty basis spans the zero space. Rank2 has zero-dimensional kernel and both original columns as an image basis; rank0 has the standard kernel basis and empty image basis. At rank1 the kernel uses a primitive integer normal-equation solution, with its first nonzero coordinate positive, and the image uses the first nonzero original column. All divisions are exact rational operations. Target=[33, 1], nonnegative bound=2. Search integer pairs x,y in [−bound,bound] in x-then-y increasing order. Give (target,bound,integer_witnesses,global_membership?,boundary).","answer":{"kind":"expression","value":"([33, 1], 2, [], Yes, bounded_failure_not_global_nonmembership)","explanation":"List only preimages in this bounded integer box, then separately test full-real column-span membership. No integer witness together with true real membership yields bounded_failure_not_global_nonmembership; otherwise retain bounded_search_only. A valid noninteger solution can be missed by every integer box. The exact result, retaining the requested order, is ([33, 1], 2, [], Yes, bounded_failure_not_global_nonmembership)."},"skills":["Failure to Find an Integer Preimage Does Not Establish Real Nonmembership"],"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-r03-linear-map-image-0093","batch_id":"batch-20261014-advance-20261102-r03-linear-map-image-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"b822e69759ec5d8c651687d02b57c1686b1124247205775cdb8811021df48142","family_id":"qf-mod33-image-bounded-search-boundary","seed":202611023093,"semantic_template":"qf-mod33-image-bounded-search-boundary/{\"bound\":3,\"matrix\":[[2,0],[0,2]],\"target\":[1,35]}"},"fields":{"title":"Failure to Find an Integer Preimage Does Not Establish Real Nonmembership","stem":"Use this same exact integer matrix A=[[2, 0], [0, 2]] for the usual linear map T:R²→R², with rows and columns in the supplied order. Kernel means all real x solving A*x=0; image means the full real column span, not automatically the whole codomain. An empty basis spans the zero space. Rank2 has zero-dimensional kernel and both original columns as an image basis; rank0 has the standard kernel basis and empty image basis. At rank1 the kernel uses a primitive integer normal-equation solution, with its first nonzero coordinate positive, and the image uses the first nonzero original column. All divisions are exact rational operations. Target=[35, 1], nonnegative bound=3. Search integer pairs x,y in [−bound,bound] in x-then-y increasing order. Give (target,bound,integer_witnesses,global_membership?,boundary).","answer":{"kind":"expression","value":"([35, 1], 3, [], Yes, bounded_failure_not_global_nonmembership)","explanation":"List only preimages in this bounded integer box, then separately test full-real column-span membership. No integer witness together with true real membership yields bounded_failure_not_global_nonmembership; otherwise retain bounded_search_only. A valid noninteger solution can be missed by every integer box. The exact result, retaining the requested order, is ([35, 1], 3, [], Yes, bounded_failure_not_global_nonmembership)."},"skills":["Failure to Find an Integer Preimage Does Not Establish Real Nonmembership"],"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-r03-linear-map-image-0094","batch_id":"batch-20261014-advance-20261102-r03-linear-map-image-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"c3fa318ffdd96f86753e38a6363126307a6ee96a50cfd5ef96917c013beb424c","family_id":"qf-mod33-image-bounded-search-boundary","seed":202611023094,"semantic_template":"qf-mod33-image-bounded-search-boundary/{\"bound\":4,\"matrix\":[[2,0],[0,2]],\"target\":[1,37]}"},"fields":{"title":"Failure to Find an Integer Preimage Does Not Establish Real Nonmembership","stem":"Use this same exact integer matrix A=[[2, 0], [0, 2]] for the usual linear map T:R²→R², with rows and columns in the supplied order. Kernel means all real x solving A*x=0; image means the full real column span, not automatically the whole codomain. An empty basis spans the zero space. Rank2 has zero-dimensional kernel and both original columns as an image basis; rank0 has the standard kernel basis and empty image basis. At rank1 the kernel uses a primitive integer normal-equation solution, with its first nonzero coordinate positive, and the image uses the first nonzero original column. All divisions are exact rational operations. Target=[37, 1], nonnegative bound=4. Search integer pairs x,y in [−bound,bound] in x-then-y increasing order. Give (target,bound,integer_witnesses,global_membership?,boundary).","answer":{"kind":"expression","value":"([37, 1], 4, [], Yes, bounded_failure_not_global_nonmembership)","explanation":"List only preimages in this bounded integer box, then separately test full-real column-span membership. No integer witness together with true real membership yields bounded_failure_not_global_nonmembership; otherwise retain bounded_search_only. A valid noninteger solution can be missed by every integer box. The exact result, retaining the requested order, is ([37, 1], 4, [], Yes, bounded_failure_not_global_nonmembership)."},"skills":["Failure to Find an Integer Preimage Does Not Establish Real Nonmembership"],"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-r03-linear-map-image-0095","batch_id":"batch-20261014-advance-20261102-r03-linear-map-image-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"b76e0686466ac3a333a6958aa85938e0bd2a7d7a87621f5b68ad872765f0b512","family_id":"qf-mod33-image-bounded-search-boundary","seed":202611023095,"semantic_template":"qf-mod33-image-bounded-search-boundary/{\"bound\":5,\"matrix\":[[2,0],[0,2]],\"target\":[1,39]}"},"fields":{"title":"Failure to Find an Integer Preimage Does Not Establish Real Nonmembership","stem":"Use this same exact integer matrix A=[[2, 0], [0, 2]] for the usual linear map T:R²→R², with rows and columns in the supplied order. Kernel means all real x solving A*x=0; image means the full real column span, not automatically the whole codomain. An empty basis spans the zero space. Rank2 has zero-dimensional kernel and both original columns as an image basis; rank0 has the standard kernel basis and empty image basis. At rank1 the kernel uses a primitive integer normal-equation solution, with its first nonzero coordinate positive, and the image uses the first nonzero original column. All divisions are exact rational operations. Target=[39, 1], nonnegative bound=5. Search integer pairs x,y in [−bound,bound] in x-then-y increasing order. Give (target,bound,integer_witnesses,global_membership?,boundary).","answer":{"kind":"expression","value":"([39, 1], 5, [], Yes, bounded_failure_not_global_nonmembership)","explanation":"List only preimages in this bounded integer box, then separately test full-real column-span membership. No integer witness together with true real membership yields bounded_failure_not_global_nonmembership; otherwise retain bounded_search_only. A valid noninteger solution can be missed by every integer box. The exact result, retaining the requested order, is ([39, 1], 5, [], Yes, bounded_failure_not_global_nonmembership)."},"skills":["Failure to Find an Integer Preimage Does Not Establish Real Nonmembership"],"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-r03-linear-map-image-0096","batch_id":"batch-20261014-advance-20261102-r03-linear-map-image-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"a2c859ac46ea668df1de5a69d2dc0682ed676af9d2c0a792968d9a3f7f273a34","family_id":"qf-mod33-image-bounded-search-boundary","seed":202611023096,"semantic_template":"qf-mod33-image-bounded-search-boundary/{\"bound\":1,\"matrix\":[[2,0],[0,2]],\"target\":[1,41]}"},"fields":{"title":"Failure to Find an Integer Preimage Does Not Establish Real Nonmembership","stem":"Use this same exact integer matrix A=[[2, 0], [0, 2]] for the usual linear map T:R²→R², with rows and columns in the supplied order. Kernel means all real x solving A*x=0; image means the full real column span, not automatically the whole codomain. An empty basis spans the zero space. Rank2 has zero-dimensional kernel and both original columns as an image basis; rank0 has the standard kernel basis and empty image basis. At rank1 the kernel uses a primitive integer normal-equation solution, with its first nonzero coordinate positive, and the image uses the first nonzero original column. All divisions are exact rational operations. Target=[41, 1], nonnegative bound=1. Search integer pairs x,y in [−bound,bound] in x-then-y increasing order. Give (target,bound,integer_witnesses,global_membership?,boundary).","answer":{"kind":"expression","value":"([41, 1], 1, [], Yes, bounded_failure_not_global_nonmembership)","explanation":"List only preimages in this bounded integer box, then separately test full-real column-span membership. No integer witness together with true real membership yields bounded_failure_not_global_nonmembership; otherwise retain bounded_search_only. A valid noninteger solution can be missed by every integer box. The exact result, retaining the requested order, is ([41, 1], 1, [], Yes, bounded_failure_not_global_nonmembership)."},"skills":["Failure to Find an Integer Preimage Does Not Establish Real Nonmembership"],"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-r03-linear-map-image-0097","batch_id":"batch-20261014-advance-20261102-r03-linear-map-image-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"b5d00df66e7e1ebbed04ba6003df56fbf0f15dc875a22507f72401fc6275e98a","family_id":"qf-mod33-image-bounded-search-boundary","seed":202611023097,"semantic_template":"qf-mod33-image-bounded-search-boundary/{\"bound\":2,\"matrix\":[[2,0],[0,2]],\"target\":[1,43]}"},"fields":{"title":"Failure to Find an Integer Preimage Does Not Establish Real Nonmembership","stem":"Use this same exact integer matrix A=[[2, 0], [0, 2]] for the usual linear map T:R²→R², with rows and columns in the supplied order. Kernel means all real x solving A*x=0; image means the full real column span, not automatically the whole codomain. An empty basis spans the zero space. Rank2 has zero-dimensional kernel and both original columns as an image basis; rank0 has the standard kernel basis and empty image basis. At rank1 the kernel uses a primitive integer normal-equation solution, with its first nonzero coordinate positive, and the image uses the first nonzero original column. All divisions are exact rational operations. Target=[43, 1], nonnegative bound=2. Search integer pairs x,y in [−bound,bound] in x-then-y increasing order. Give (target,bound,integer_witnesses,global_membership?,boundary).","answer":{"kind":"expression","value":"([43, 1], 2, [], Yes, bounded_failure_not_global_nonmembership)","explanation":"List only preimages in this bounded integer box, then separately test full-real column-span membership. No integer witness together with true real membership yields bounded_failure_not_global_nonmembership; otherwise retain bounded_search_only. A valid noninteger solution can be missed by every integer box. The exact result, retaining the requested order, is ([43, 1], 2, [], Yes, bounded_failure_not_global_nonmembership)."},"skills":["Failure to Find an Integer Preimage Does Not Establish Real Nonmembership"],"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-r03-linear-map-image-0098","batch_id":"batch-20261014-advance-20261102-r03-linear-map-image-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"9ddaa19d7f76a9a8efd0500509f5f1d2a7b2945b8eba27f638a5aa9ea0f21ead","family_id":"qf-mod33-image-bounded-search-boundary","seed":202611023098,"semantic_template":"qf-mod33-image-bounded-search-boundary/{\"bound\":3,\"matrix\":[[2,0],[0,2]],\"target\":[1,45]}"},"fields":{"title":"Failure to Find an Integer Preimage Does Not Establish Real Nonmembership","stem":"Use this same exact integer matrix A=[[2, 0], [0, 2]] for the usual linear map T:R²→R², with rows and columns in the supplied order. Kernel means all real x solving A*x=0; image means the full real column span, not automatically the whole codomain. An empty basis spans the zero space. Rank2 has zero-dimensional kernel and both original columns as an image basis; rank0 has the standard kernel basis and empty image basis. At rank1 the kernel uses a primitive integer normal-equation solution, with its first nonzero coordinate positive, and the image uses the first nonzero original column. All divisions are exact rational operations. Target=[45, 1], nonnegative bound=3. Search integer pairs x,y in [−bound,bound] in x-then-y increasing order. Give (target,bound,integer_witnesses,global_membership?,boundary).","answer":{"kind":"expression","value":"([45, 1], 3, [], Yes, bounded_failure_not_global_nonmembership)","explanation":"List only preimages in this bounded integer box, then separately test full-real column-span membership. No integer witness together with true real membership yields bounded_failure_not_global_nonmembership; otherwise retain bounded_search_only. A valid noninteger solution can be missed by every integer box. The exact result, retaining the requested order, is ([45, 1], 3, [], Yes, bounded_failure_not_global_nonmembership)."},"skills":["Failure to Find an Integer Preimage Does Not Establish Real Nonmembership"],"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-r03-linear-map-image-0099","batch_id":"batch-20261014-advance-20261102-r03-linear-map-image-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"3bd17caf9b9e7f382cc0711ec36cb1e2929981daa289034b303eb7caee82e7a9","family_id":"qf-mod33-image-bounded-search-boundary","seed":202611023099,"semantic_template":"qf-mod33-image-bounded-search-boundary/{\"bound\":4,\"matrix\":[[2,0],[0,2]],\"target\":[1,47]}"},"fields":{"title":"Failure to Find an Integer Preimage Does Not Establish Real Nonmembership","stem":"Use this same exact integer matrix A=[[2, 0], [0, 2]] for the usual linear map T:R²→R², with rows and columns in the supplied order. Kernel means all real x solving A*x=0; image means the full real column span, not automatically the whole codomain. An empty basis spans the zero space. Rank2 has zero-dimensional kernel and both original columns as an image basis; rank0 has the standard kernel basis and empty image basis. At rank1 the kernel uses a primitive integer normal-equation solution, with its first nonzero coordinate positive, and the image uses the first nonzero original column. All divisions are exact rational operations. Target=[47, 1], nonnegative bound=4. Search integer pairs x,y in [−bound,bound] in x-then-y increasing order. Give (target,bound,integer_witnesses,global_membership?,boundary).","answer":{"kind":"expression","value":"([47, 1], 4, [], Yes, bounded_failure_not_global_nonmembership)","explanation":"List only preimages in this bounded integer box, then separately test full-real column-span membership. No integer witness together with true real membership yields bounded_failure_not_global_nonmembership; otherwise retain bounded_search_only. A valid noninteger solution can be missed by every integer box. The exact result, retaining the requested order, is ([47, 1], 4, [], Yes, bounded_failure_not_global_nonmembership)."},"skills":["Failure to Find an Integer Preimage Does Not Establish Real Nonmembership"],"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-r03-linear-map-image-0100","batch_id":"batch-20261014-advance-20261102-r03-linear-map-image-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"80828abc74abb97f6b2da56ce4f831599e9b190add67b53e59cfe66ae9caf162","family_id":"qf-mod33-image-bounded-search-boundary","seed":202611023100,"semantic_template":"qf-mod33-image-bounded-search-boundary/{\"bound\":5,\"matrix\":[[2,0],[0,2]],\"target\":[1,49]}"},"fields":{"title":"Failure to Find an Integer Preimage Does Not Establish Real Nonmembership","stem":"Use this same exact integer matrix A=[[2, 0], [0, 2]] for the usual linear map T:R²→R², with rows and columns in the supplied order. Kernel means all real x solving A*x=0; image means the full real column span, not automatically the whole codomain. An empty basis spans the zero space. Rank2 has zero-dimensional kernel and both original columns as an image basis; rank0 has the standard kernel basis and empty image basis. At rank1 the kernel uses a primitive integer normal-equation solution, with its first nonzero coordinate positive, and the image uses the first nonzero original column. All divisions are exact rational operations. Target=[49, 1], nonnegative bound=5. Search integer pairs x,y in [−bound,bound] in x-then-y increasing order. Give (target,bound,integer_witnesses,global_membership?,boundary).","answer":{"kind":"expression","value":"([49, 1], 5, [], Yes, bounded_failure_not_global_nonmembership)","explanation":"List only preimages in this bounded integer box, then separately test full-real column-span membership. No integer witness together with true real membership yields bounded_failure_not_global_nonmembership; otherwise retain bounded_search_only. A valid noninteger solution can be missed by every integer box. The exact result, retaining the requested order, is ([49, 1], 5, [], Yes, bounded_failure_not_global_nonmembership)."},"skills":["Failure to Find an Integer Preimage Does Not Establish Real Nonmembership"],"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."}}