{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261101-r01-vector-subspace-0001","batch_id":"batch-20261014-advance-20261101-r01-vector-subspace-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"e79f36d53f2d21c991a2cbb0d2dd264ee97df5618987d6c2ce3e479a364b7849","family_id":"qf-mod32-subspace-line-membership","seed":202611011001,"semantic_template":"qf-mod32-subspace-line-membership/{\"a\":1,\"b\":-1,\"x\":-12,\"y\":-3}"},"fields":{"title":"Membership in a Homogeneous Real Line Is an Exact Equation Check","stem":"Use R² with its usual real vector operations and the homogeneous equation a*x+b*y=0, with nonzero normal [a,b]=[1, -1]. Selected point=[-12, -3]. Give (point,normal,equation_value,in_line?).","answer":{"kind":"expression","value":"([-12, -3], [1, -1], -9, No)","explanation":"Calculate a*x+b*y exactly. Its being zero decides membership of this point; the universal subspace certificate is a separate algebraic obligation, not a conclusion from one point. The exact result, retaining the requested order, is ([-12, -3], [1, -1], -9, No)."},"skills":["Membership in a Homogeneous Real Line Is an Exact Equation Check"],"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-20261101-r01-vector-subspace-0002","batch_id":"batch-20261014-advance-20261101-r01-vector-subspace-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"729c8326e7e999ba167df0103f8e2121efa84b77ca081366be56448777021eb5","family_id":"qf-mod32-subspace-line-membership","seed":202611011002,"semantic_template":"qf-mod32-subspace-line-membership/{\"a\":2,\"b\":-2,\"x\":-11,\"y\":-2}"},"fields":{"title":"Membership in a Homogeneous Real Line Is an Exact Equation Check","stem":"Use R² with its usual real vector operations and the homogeneous equation a*x+b*y=0, with nonzero normal [a,b]=[2, -2]. Selected point=[-11, -2]. Give (point,normal,equation_value,in_line?).","answer":{"kind":"expression","value":"([-11, -2], [2, -2], -18, No)","explanation":"Calculate a*x+b*y exactly. Its being zero decides membership of this point; the universal subspace certificate is a separate algebraic obligation, not a conclusion from one point. The exact result, retaining the requested order, is ([-11, -2], [2, -2], -18, No)."},"skills":["Membership in a Homogeneous Real Line Is an Exact Equation Check"],"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-20261101-r01-vector-subspace-0003","batch_id":"batch-20261014-advance-20261101-r01-vector-subspace-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"943c8ea36c9c30ed6d3af4e913c1d9b05b5cfe3e016abddb2e9692c7b57c3474","family_id":"qf-mod32-subspace-line-membership","seed":202611011003,"semantic_template":"qf-mod32-subspace-line-membership/{\"a\":3,\"b\":-3,\"x\":-10,\"y\":-1}"},"fields":{"title":"Membership in a Homogeneous Real Line Is an Exact Equation Check","stem":"Use R² with its usual real vector operations and the homogeneous equation a*x+b*y=0, with nonzero normal [a,b]=[3, -3]. Selected point=[-10, -1]. Give (point,normal,equation_value,in_line?).","answer":{"kind":"expression","value":"([-10, -1], [3, -3], -27, No)","explanation":"Calculate a*x+b*y exactly. Its being zero decides membership of this point; the universal subspace certificate is a separate algebraic obligation, not a conclusion from one point. The exact result, retaining the requested order, is ([-10, -1], [3, -3], -27, No)."},"skills":["Membership in a Homogeneous Real Line Is an Exact Equation Check"],"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-20261101-r01-vector-subspace-0004","batch_id":"batch-20261014-advance-20261101-r01-vector-subspace-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"82c7401ce55ea4f1af6ae45447f934d5d80cd93754a9634cb25cf3a6fba6e822","family_id":"qf-mod32-subspace-line-membership","seed":202611011004,"semantic_template":"qf-mod32-subspace-line-membership/{\"a\":4,\"b\":-4,\"x\":-9,\"y\":0}"},"fields":{"title":"Membership in a Homogeneous Real Line Is an Exact Equation Check","stem":"Use R² with its usual real vector operations and the homogeneous equation a*x+b*y=0, with nonzero normal [a,b]=[4, -4]. Selected point=[-9, 0]. Give (point,normal,equation_value,in_line?).","answer":{"kind":"expression","value":"([-9, 0], [4, -4], -36, No)","explanation":"Calculate a*x+b*y exactly. Its being zero decides membership of this point; the universal subspace certificate is a separate algebraic obligation, not a conclusion from one point. The exact result, retaining the requested order, is ([-9, 0], [4, -4], -36, No)."},"skills":["Membership in a Homogeneous Real Line Is an Exact Equation Check"],"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-20261101-r01-vector-subspace-0005","batch_id":"batch-20261014-advance-20261101-r01-vector-subspace-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"79db055c1d0740dedfd3275e5a0e731ec34c483ccea400b118a1f6aeed142de1","family_id":"qf-mod32-subspace-line-membership","seed":202611011005,"semantic_template":"qf-mod32-subspace-line-membership/{\"a\":5,\"b\":-5,\"x\":-8,\"y\":1}"},"fields":{"title":"Membership in a Homogeneous Real Line Is an Exact Equation Check","stem":"Use R² with its usual real vector operations and the homogeneous equation a*x+b*y=0, with nonzero normal [a,b]=[5, -5]. Selected point=[-8, 1]. Give (point,normal,equation_value,in_line?).","answer":{"kind":"expression","value":"([-8, 1], [5, -5], -45, No)","explanation":"Calculate a*x+b*y exactly. Its being zero decides membership of this point; the universal subspace certificate is a separate algebraic obligation, not a conclusion from one point. The exact result, retaining the requested order, is ([-8, 1], [5, -5], -45, No)."},"skills":["Membership in a Homogeneous Real Line Is an Exact Equation Check"],"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-20261101-r01-vector-subspace-0006","batch_id":"batch-20261014-advance-20261101-r01-vector-subspace-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"6a4b427c0eeac57e0475dd25aa850310c13decea2173ae70fab8e21868d3f28f","family_id":"qf-mod32-subspace-line-membership","seed":202611011006,"semantic_template":"qf-mod32-subspace-line-membership/{\"a\":6,\"b\":-1,\"x\":-7,\"y\":2}"},"fields":{"title":"Membership in a Homogeneous Real Line Is an Exact Equation Check","stem":"Use R² with its usual real vector operations and the homogeneous equation a*x+b*y=0, with nonzero normal [a,b]=[6, -1]. Selected point=[-7, 2]. Give (point,normal,equation_value,in_line?).","answer":{"kind":"expression","value":"([-7, 2], [6, -1], -44, No)","explanation":"Calculate a*x+b*y exactly. Its being zero decides membership of this point; the universal subspace certificate is a separate algebraic obligation, not a conclusion from one point. The exact result, retaining the requested order, is ([-7, 2], [6, -1], -44, No)."},"skills":["Membership in a Homogeneous Real Line Is an Exact Equation Check"],"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-20261101-r01-vector-subspace-0007","batch_id":"batch-20261014-advance-20261101-r01-vector-subspace-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"e182bedf42051206265295274dbf384a87ec8c06984312609b989bf160b92cb9","family_id":"qf-mod32-subspace-line-membership","seed":202611011007,"semantic_template":"qf-mod32-subspace-line-membership/{\"a\":7,\"b\":-2,\"x\":-6,\"y\":3}"},"fields":{"title":"Membership in a Homogeneous Real Line Is an Exact Equation Check","stem":"Use R² with its usual real vector operations and the homogeneous equation a*x+b*y=0, with nonzero normal [a,b]=[7, -2]. Selected point=[-6, 3]. Give (point,normal,equation_value,in_line?).","answer":{"kind":"expression","value":"([-6, 3], [7, -2], -48, No)","explanation":"Calculate a*x+b*y exactly. Its being zero decides membership of this point; the universal subspace certificate is a separate algebraic obligation, not a conclusion from one point. The exact result, retaining the requested order, is ([-6, 3], [7, -2], -48, No)."},"skills":["Membership in a Homogeneous Real Line Is an Exact Equation Check"],"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-20261101-r01-vector-subspace-0008","batch_id":"batch-20261014-advance-20261101-r01-vector-subspace-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"9c71ff385e9ea284b5815a0e4538fa7b2add56b0a2c31455b9c683c3ff8c33c1","family_id":"qf-mod32-subspace-line-membership","seed":202611011008,"semantic_template":"qf-mod32-subspace-line-membership/{\"a\":8,\"b\":-3,\"x\":-5,\"y\":-3}"},"fields":{"title":"Membership in a Homogeneous Real Line Is an Exact Equation Check","stem":"Use R² with its usual real vector operations and the homogeneous equation a*x+b*y=0, with nonzero normal [a,b]=[8, -3]. Selected point=[-5, -3]. Give (point,normal,equation_value,in_line?).","answer":{"kind":"expression","value":"([-5, -3], [8, -3], -31, No)","explanation":"Calculate a*x+b*y exactly. Its being zero decides membership of this point; the universal subspace certificate is a separate algebraic obligation, not a conclusion from one point. The exact result, retaining the requested order, is ([-5, -3], [8, -3], -31, No)."},"skills":["Membership in a Homogeneous Real Line Is an Exact Equation Check"],"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-20261101-r01-vector-subspace-0009","batch_id":"batch-20261014-advance-20261101-r01-vector-subspace-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"c10bd2466ce1f33336af3708b4e567db88d034369ae03a47b6bdba38a44058e1","family_id":"qf-mod32-subspace-line-membership","seed":202611011009,"semantic_template":"qf-mod32-subspace-line-membership/{\"a\":9,\"b\":-4,\"x\":-4,\"y\":-2}"},"fields":{"title":"Membership in a Homogeneous Real Line Is an Exact Equation Check","stem":"Use R² with its usual real vector operations and the homogeneous equation a*x+b*y=0, with nonzero normal [a,b]=[9, -4]. Selected point=[-4, -2]. Give (point,normal,equation_value,in_line?).","answer":{"kind":"expression","value":"([-4, -2], [9, -4], -28, No)","explanation":"Calculate a*x+b*y exactly. Its being zero decides membership of this point; the universal subspace certificate is a separate algebraic obligation, not a conclusion from one point. The exact result, retaining the requested order, is ([-4, -2], [9, -4], -28, No)."},"skills":["Membership in a Homogeneous Real Line Is an Exact Equation Check"],"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-20261101-r01-vector-subspace-0010","batch_id":"batch-20261014-advance-20261101-r01-vector-subspace-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"e713c9085f77ec756d89ea8929094b006f8a2077a7565038b9f1c92142726ff7","family_id":"qf-mod32-subspace-line-membership","seed":202611011010,"semantic_template":"qf-mod32-subspace-line-membership/{\"a\":10,\"b\":-5,\"x\":-3,\"y\":-1}"},"fields":{"title":"Membership in a Homogeneous Real Line Is an Exact Equation Check","stem":"Use R² with its usual real vector operations and the homogeneous equation a*x+b*y=0, with nonzero normal [a,b]=[10, -5]. Selected point=[-3, -1]. Give (point,normal,equation_value,in_line?).","answer":{"kind":"expression","value":"([-3, -1], [10, -5], -25, No)","explanation":"Calculate a*x+b*y exactly. Its being zero decides membership of this point; the universal subspace certificate is a separate algebraic obligation, not a conclusion from one point. The exact result, retaining the requested order, is ([-3, -1], [10, -5], -25, No)."},"skills":["Membership in a Homogeneous Real Line Is an Exact Equation Check"],"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-20261101-r01-vector-subspace-0011","batch_id":"batch-20261014-advance-20261101-r01-vector-subspace-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"e1dcc2de7e69a796f30adbc25d0f98b40d8483f0f6a6c975831ff10b297e8a60","family_id":"qf-mod32-subspace-line-membership","seed":202611011011,"semantic_template":"qf-mod32-subspace-line-membership/{\"a\":11,\"b\":-1,\"x\":-2,\"y\":0}"},"fields":{"title":"Membership in a Homogeneous Real Line Is an Exact Equation Check","stem":"Use R² with its usual real vector operations and the homogeneous equation a*x+b*y=0, with nonzero normal [a,b]=[11, -1]. Selected point=[-2, 0]. Give (point,normal,equation_value,in_line?).","answer":{"kind":"expression","value":"([-2, 0], [11, -1], -22, No)","explanation":"Calculate a*x+b*y exactly. Its being zero decides membership of this point; the universal subspace certificate is a separate algebraic obligation, not a conclusion from one point. The exact result, retaining the requested order, is ([-2, 0], [11, -1], -22, No)."},"skills":["Membership in a Homogeneous Real Line Is an Exact Equation Check"],"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-20261101-r01-vector-subspace-0012","batch_id":"batch-20261014-advance-20261101-r01-vector-subspace-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"7d2155b4c39b9abe34f5177f6ef73fd3086416ebbffb369e97c5bdd87d83cf00","family_id":"qf-mod32-subspace-line-membership","seed":202611011012,"semantic_template":"qf-mod32-subspace-line-membership/{\"a\":12,\"b\":-2,\"x\":-1,\"y\":1}"},"fields":{"title":"Membership in a Homogeneous Real Line Is an Exact Equation Check","stem":"Use R² with its usual real vector operations and the homogeneous equation a*x+b*y=0, with nonzero normal [a,b]=[12, -2]. Selected point=[-1, 1]. Give (point,normal,equation_value,in_line?).","answer":{"kind":"expression","value":"([-1, 1], [12, -2], -14, No)","explanation":"Calculate a*x+b*y exactly. Its being zero decides membership of this point; the universal subspace certificate is a separate algebraic obligation, not a conclusion from one point. The exact result, retaining the requested order, is ([-1, 1], [12, -2], -14, No)."},"skills":["Membership in a Homogeneous Real Line Is an Exact Equation Check"],"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-20261101-r01-vector-subspace-0013","batch_id":"batch-20261014-advance-20261101-r01-vector-subspace-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"eeb61cc78b4fa5105d0b668df2ee0b91bf82281c00fcda883ef01f5a797067f9","family_id":"qf-mod32-subspace-line-membership","seed":202611011013,"semantic_template":"qf-mod32-subspace-line-membership/{\"a\":13,\"b\":-3,\"x\":0,\"y\":2}"},"fields":{"title":"Membership in a Homogeneous Real Line Is an Exact Equation Check","stem":"Use R² with its usual real vector operations and the homogeneous equation a*x+b*y=0, with nonzero normal [a,b]=[13, -3]. Selected point=[0, 2]. Give (point,normal,equation_value,in_line?).","answer":{"kind":"expression","value":"([0, 2], [13, -3], -6, No)","explanation":"Calculate a*x+b*y exactly. Its being zero decides membership of this point; the universal subspace certificate is a separate algebraic obligation, not a conclusion from one point. The exact result, retaining the requested order, is ([0, 2], [13, -3], -6, No)."},"skills":["Membership in a Homogeneous Real Line Is an Exact Equation Check"],"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-20261101-r01-vector-subspace-0014","batch_id":"batch-20261014-advance-20261101-r01-vector-subspace-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"5d7b0c5579f4203851f5ba71ee9b6830836ba7505efbdd7a77c8391bf5303948","family_id":"qf-mod32-subspace-line-membership","seed":202611011014,"semantic_template":"qf-mod32-subspace-line-membership/{\"a\":14,\"b\":-4,\"x\":1,\"y\":3}"},"fields":{"title":"Membership in a Homogeneous Real Line Is an Exact Equation Check","stem":"Use R² with its usual real vector operations and the homogeneous equation a*x+b*y=0, with nonzero normal [a,b]=[14, -4]. Selected point=[1, 3]. Give (point,normal,equation_value,in_line?).","answer":{"kind":"expression","value":"([1, 3], [14, -4], 2, No)","explanation":"Calculate a*x+b*y exactly. Its being zero decides membership of this point; the universal subspace certificate is a separate algebraic obligation, not a conclusion from one point. The exact result, retaining the requested order, is ([1, 3], [14, -4], 2, No)."},"skills":["Membership in a Homogeneous Real Line Is an Exact Equation Check"],"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-20261101-r01-vector-subspace-0015","batch_id":"batch-20261014-advance-20261101-r01-vector-subspace-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"b180536f2883b7f43e0bfc3e4334cae26c7e959397d8755019b35ec13f05c1ca","family_id":"qf-mod32-subspace-line-membership","seed":202611011015,"semantic_template":"qf-mod32-subspace-line-membership/{\"a\":15,\"b\":-5,\"x\":2,\"y\":-3}"},"fields":{"title":"Membership in a Homogeneous Real Line Is an Exact Equation Check","stem":"Use R² with its usual real vector operations and the homogeneous equation a*x+b*y=0, with nonzero normal [a,b]=[15, -5]. Selected point=[2, -3]. Give (point,normal,equation_value,in_line?).","answer":{"kind":"expression","value":"([2, -3], [15, -5], 45, No)","explanation":"Calculate a*x+b*y exactly. Its being zero decides membership of this point; the universal subspace certificate is a separate algebraic obligation, not a conclusion from one point. The exact result, retaining the requested order, is ([2, -3], [15, -5], 45, No)."},"skills":["Membership in a Homogeneous Real Line Is an Exact Equation Check"],"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-20261101-r01-vector-subspace-0016","batch_id":"batch-20261014-advance-20261101-r01-vector-subspace-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"134a6c59ddc306ad4d764287d64e2b293f7f609f3d18e5eb5867f365bf5d13e0","family_id":"qf-mod32-subspace-line-membership","seed":202611011016,"semantic_template":"qf-mod32-subspace-line-membership/{\"a\":16,\"b\":-1,\"x\":3,\"y\":-2}"},"fields":{"title":"Membership in a Homogeneous Real Line Is an Exact Equation Check","stem":"Use R² with its usual real vector operations and the homogeneous equation a*x+b*y=0, with nonzero normal [a,b]=[16, -1]. Selected point=[3, -2]. Give (point,normal,equation_value,in_line?).","answer":{"kind":"expression","value":"([3, -2], [16, -1], 50, No)","explanation":"Calculate a*x+b*y exactly. Its being zero decides membership of this point; the universal subspace certificate is a separate algebraic obligation, not a conclusion from one point. The exact result, retaining the requested order, is ([3, -2], [16, -1], 50, No)."},"skills":["Membership in a Homogeneous Real Line Is an Exact Equation Check"],"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-20261101-r01-vector-subspace-0017","batch_id":"batch-20261014-advance-20261101-r01-vector-subspace-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"eb603225f1e7b755b81c2cb75c0261c7715fa49ac9608dda527013283a6796a7","family_id":"qf-mod32-subspace-line-membership","seed":202611011017,"semantic_template":"qf-mod32-subspace-line-membership/{\"a\":17,\"b\":-2,\"x\":4,\"y\":-1}"},"fields":{"title":"Membership in a Homogeneous Real Line Is an Exact Equation Check","stem":"Use R² with its usual real vector operations and the homogeneous equation a*x+b*y=0, with nonzero normal [a,b]=[17, -2]. Selected point=[4, -1]. Give (point,normal,equation_value,in_line?).","answer":{"kind":"expression","value":"([4, -1], [17, -2], 70, No)","explanation":"Calculate a*x+b*y exactly. Its being zero decides membership of this point; the universal subspace certificate is a separate algebraic obligation, not a conclusion from one point. The exact result, retaining the requested order, is ([4, -1], [17, -2], 70, No)."},"skills":["Membership in a Homogeneous Real Line Is an Exact Equation Check"],"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-20261101-r01-vector-subspace-0018","batch_id":"batch-20261014-advance-20261101-r01-vector-subspace-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"eee4e127683c9fc8803fe83e0ed0a0c6a40a6d3678514815270a889d12483ccd","family_id":"qf-mod32-subspace-line-membership","seed":202611011018,"semantic_template":"qf-mod32-subspace-line-membership/{\"a\":18,\"b\":-3,\"x\":5,\"y\":0}"},"fields":{"title":"Membership in a Homogeneous Real Line Is an Exact Equation Check","stem":"Use R² with its usual real vector operations and the homogeneous equation a*x+b*y=0, with nonzero normal [a,b]=[18, -3]. Selected point=[5, 0]. Give (point,normal,equation_value,in_line?).","answer":{"kind":"expression","value":"([5, 0], [18, -3], 90, No)","explanation":"Calculate a*x+b*y exactly. Its being zero decides membership of this point; the universal subspace certificate is a separate algebraic obligation, not a conclusion from one point. The exact result, retaining the requested order, is ([5, 0], [18, -3], 90, No)."},"skills":["Membership in a Homogeneous Real Line Is an Exact Equation Check"],"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-20261101-r01-vector-subspace-0019","batch_id":"batch-20261014-advance-20261101-r01-vector-subspace-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"d0de1bead0ba5815adc4bf083bb3abe36b04adc1695e1f34906d8a527027604b","family_id":"qf-mod32-subspace-line-membership","seed":202611011019,"semantic_template":"qf-mod32-subspace-line-membership/{\"a\":19,\"b\":-4,\"x\":6,\"y\":1}"},"fields":{"title":"Membership in a Homogeneous Real Line Is an Exact Equation Check","stem":"Use R² with its usual real vector operations and the homogeneous equation a*x+b*y=0, with nonzero normal [a,b]=[19, -4]. Selected point=[6, 1]. Give (point,normal,equation_value,in_line?).","answer":{"kind":"expression","value":"([6, 1], [19, -4], 110, No)","explanation":"Calculate a*x+b*y exactly. Its being zero decides membership of this point; the universal subspace certificate is a separate algebraic obligation, not a conclusion from one point. The exact result, retaining the requested order, is ([6, 1], [19, -4], 110, No)."},"skills":["Membership in a Homogeneous Real Line Is an Exact Equation Check"],"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-20261101-r01-vector-subspace-0020","batch_id":"batch-20261014-advance-20261101-r01-vector-subspace-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"4274bae80ad0380653f8dcf4c863136819e736431f818a33a103f394e581fa6b","family_id":"qf-mod32-subspace-line-membership","seed":202611011020,"semantic_template":"qf-mod32-subspace-line-membership/{\"a\":20,\"b\":-5,\"x\":7,\"y\":2}"},"fields":{"title":"Membership in a Homogeneous Real Line Is an Exact Equation Check","stem":"Use R² with its usual real vector operations and the homogeneous equation a*x+b*y=0, with nonzero normal [a,b]=[20, -5]. Selected point=[7, 2]. Give (point,normal,equation_value,in_line?).","answer":{"kind":"expression","value":"([7, 2], [20, -5], 130, No)","explanation":"Calculate a*x+b*y exactly. Its being zero decides membership of this point; the universal subspace certificate is a separate algebraic obligation, not a conclusion from one point. The exact result, retaining the requested order, is ([7, 2], [20, -5], 130, No)."},"skills":["Membership in a Homogeneous Real Line Is an Exact Equation Check"],"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-20261101-r01-vector-subspace-0021","batch_id":"batch-20261014-advance-20261101-r01-vector-subspace-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"1220e23c8b15374b1587ad58f917efeb8229febbf65e4e2d5577ecfe424b4722","family_id":"qf-mod32-subspace-line-membership","seed":202611011021,"semantic_template":"qf-mod32-subspace-line-membership/{\"a\":21,\"b\":-1,\"x\":8,\"y\":3}"},"fields":{"title":"Membership in a Homogeneous Real Line Is an Exact Equation Check","stem":"Use R² with its usual real vector operations and the homogeneous equation a*x+b*y=0, with nonzero normal [a,b]=[21, -1]. Selected point=[8, 3]. Give (point,normal,equation_value,in_line?).","answer":{"kind":"expression","value":"([8, 3], [21, -1], 165, No)","explanation":"Calculate a*x+b*y exactly. Its being zero decides membership of this point; the universal subspace certificate is a separate algebraic obligation, not a conclusion from one point. The exact result, retaining the requested order, is ([8, 3], [21, -1], 165, No)."},"skills":["Membership in a Homogeneous Real Line Is an Exact Equation Check"],"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-20261101-r01-vector-subspace-0022","batch_id":"batch-20261014-advance-20261101-r01-vector-subspace-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"0f0d6705c1020edfdf3d746be69dc66f2e184c9e47ad006776f3511b5a0d8686","family_id":"qf-mod32-subspace-line-membership","seed":202611011022,"semantic_template":"qf-mod32-subspace-line-membership/{\"a\":22,\"b\":-2,\"x\":9,\"y\":-3}"},"fields":{"title":"Membership in a Homogeneous Real Line Is an Exact Equation Check","stem":"Use R² with its usual real vector operations and the homogeneous equation a*x+b*y=0, with nonzero normal [a,b]=[22, -2]. Selected point=[9, -3]. Give (point,normal,equation_value,in_line?).","answer":{"kind":"expression","value":"([9, -3], [22, -2], 204, No)","explanation":"Calculate a*x+b*y exactly. Its being zero decides membership of this point; the universal subspace certificate is a separate algebraic obligation, not a conclusion from one point. The exact result, retaining the requested order, is ([9, -3], [22, -2], 204, No)."},"skills":["Membership in a Homogeneous Real Line Is an Exact Equation Check"],"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-20261101-r01-vector-subspace-0023","batch_id":"batch-20261014-advance-20261101-r01-vector-subspace-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"461b1f69c8ab0dcbc2d87641a1d465c1398432072ece158154a84f284f75f425","family_id":"qf-mod32-subspace-line-membership","seed":202611011023,"semantic_template":"qf-mod32-subspace-line-membership/{\"a\":23,\"b\":-3,\"x\":10,\"y\":-2}"},"fields":{"title":"Membership in a Homogeneous Real Line Is an Exact Equation Check","stem":"Use R² with its usual real vector operations and the homogeneous equation a*x+b*y=0, with nonzero normal [a,b]=[23, -3]. Selected point=[10, -2]. Give (point,normal,equation_value,in_line?).","answer":{"kind":"expression","value":"([10, -2], [23, -3], 236, No)","explanation":"Calculate a*x+b*y exactly. Its being zero decides membership of this point; the universal subspace certificate is a separate algebraic obligation, not a conclusion from one point. The exact result, retaining the requested order, is ([10, -2], [23, -3], 236, No)."},"skills":["Membership in a Homogeneous Real Line Is an Exact Equation Check"],"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-20261101-r01-vector-subspace-0024","batch_id":"batch-20261014-advance-20261101-r01-vector-subspace-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"995db699748cd3fdc85991bf10719bd6b2a62367f2ceb5e291454ac5f481f4d8","family_id":"qf-mod32-subspace-line-membership","seed":202611011024,"semantic_template":"qf-mod32-subspace-line-membership/{\"a\":24,\"b\":-4,\"x\":11,\"y\":-1}"},"fields":{"title":"Membership in a Homogeneous Real Line Is an Exact Equation Check","stem":"Use R² with its usual real vector operations and the homogeneous equation a*x+b*y=0, with nonzero normal [a,b]=[24, -4]. Selected point=[11, -1]. Give (point,normal,equation_value,in_line?).","answer":{"kind":"expression","value":"([11, -1], [24, -4], 268, No)","explanation":"Calculate a*x+b*y exactly. Its being zero decides membership of this point; the universal subspace certificate is a separate algebraic obligation, not a conclusion from one point. The exact result, retaining the requested order, is ([11, -1], [24, -4], 268, No)."},"skills":["Membership in a Homogeneous Real Line Is an Exact Equation Check"],"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-20261101-r01-vector-subspace-0025","batch_id":"batch-20261014-advance-20261101-r01-vector-subspace-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"754207ad9364801c8fe7f71d95d4a2ae5ea98443e8bea34c9815151d358bfee9","family_id":"qf-mod32-subspace-line-membership","seed":202611011025,"semantic_template":"qf-mod32-subspace-line-membership/{\"a\":25,\"b\":-5,\"x\":12,\"y\":0}"},"fields":{"title":"Membership in a Homogeneous Real Line Is an Exact Equation Check","stem":"Use R² with its usual real vector operations and the homogeneous equation a*x+b*y=0, with nonzero normal [a,b]=[25, -5]. Selected point=[12, 0]. Give (point,normal,equation_value,in_line?).","answer":{"kind":"expression","value":"([12, 0], [25, -5], 300, No)","explanation":"Calculate a*x+b*y exactly. Its being zero decides membership of this point; the universal subspace certificate is a separate algebraic obligation, not a conclusion from one point. The exact result, retaining the requested order, is ([12, 0], [25, -5], 300, No)."},"skills":["Membership in a Homogeneous Real Line Is an Exact Equation Check"],"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-20261101-r01-vector-subspace-0026","batch_id":"batch-20261014-advance-20261101-r01-vector-subspace-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"a24ab03d32ab24f027e82786e907ab1f4b56168cd5745cc0754c5596fd77809b","family_id":"qf-mod32-subspace-universal-line-certificate","seed":202611011026,"semantic_template":"qf-mod32-subspace-universal-line-certificate/{\"a\":1,\"b\":-1}"},"fields":{"title":"A Homogeneous Kernel Has Universal Closure under Real Linear Combinations","stem":"Use the subset U={(x,y) in R²:a*x+b*y=0} with nonzero normal [a,b]=[1, -1] and usual real vector operations. Give (normal,zero_in?,addition_closed?,scalar_closed?,homogeneous_kernel_closed_for_all_real_scalars).","answer":{"kind":"expression","value":"([1, -1], Yes, Yes, Yes, homogeneous_kernel_closed_for_all_real_scalars)","explanation":"The zero vector satisfies the equation. Linearity makes the equation hold for sums of solutions and for every real scalar multiple. This algebraic all-vector argument, not sampled points or a new Lean receipt, supplies the stated certificate. The exact result, retaining the requested order, is ([1, -1], Yes, Yes, Yes, homogeneous_kernel_closed_for_all_real_scalars)."},"skills":["A Homogeneous Kernel Has Universal Closure under Real Linear Combinations"],"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-20261101-r01-vector-subspace-0027","batch_id":"batch-20261014-advance-20261101-r01-vector-subspace-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"d3b9c2cb00d9a2dcdc419061efb804004c7fa9eec16d8aca181158931fe3541f","family_id":"qf-mod32-subspace-universal-line-certificate","seed":202611011027,"semantic_template":"qf-mod32-subspace-universal-line-certificate/{\"a\":2,\"b\":-2}"},"fields":{"title":"A Homogeneous Kernel Has Universal Closure under Real Linear Combinations","stem":"Use the subset U={(x,y) in R²:a*x+b*y=0} with nonzero normal [a,b]=[2, -2] and usual real vector operations. Give (normal,zero_in?,addition_closed?,scalar_closed?,homogeneous_kernel_closed_for_all_real_scalars).","answer":{"kind":"expression","value":"([2, -2], Yes, Yes, Yes, homogeneous_kernel_closed_for_all_real_scalars)","explanation":"The zero vector satisfies the equation. Linearity makes the equation hold for sums of solutions and for every real scalar multiple. This algebraic all-vector argument, not sampled points or a new Lean receipt, supplies the stated certificate. The exact result, retaining the requested order, is ([2, -2], Yes, Yes, Yes, homogeneous_kernel_closed_for_all_real_scalars)."},"skills":["A Homogeneous Kernel Has Universal Closure under Real Linear Combinations"],"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-20261101-r01-vector-subspace-0028","batch_id":"batch-20261014-advance-20261101-r01-vector-subspace-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"4b6b75e0a977393b81ffd87154284a176d73f8f822db0ddd936e575ae422d64f","family_id":"qf-mod32-subspace-universal-line-certificate","seed":202611011028,"semantic_template":"qf-mod32-subspace-universal-line-certificate/{\"a\":3,\"b\":-3}"},"fields":{"title":"A Homogeneous Kernel Has Universal Closure under Real Linear Combinations","stem":"Use the subset U={(x,y) in R²:a*x+b*y=0} with nonzero normal [a,b]=[3, -3] and usual real vector operations. Give (normal,zero_in?,addition_closed?,scalar_closed?,homogeneous_kernel_closed_for_all_real_scalars).","answer":{"kind":"expression","value":"([3, -3], Yes, Yes, Yes, homogeneous_kernel_closed_for_all_real_scalars)","explanation":"The zero vector satisfies the equation. Linearity makes the equation hold for sums of solutions and for every real scalar multiple. This algebraic all-vector argument, not sampled points or a new Lean receipt, supplies the stated certificate. The exact result, retaining the requested order, is ([3, -3], Yes, Yes, Yes, homogeneous_kernel_closed_for_all_real_scalars)."},"skills":["A Homogeneous Kernel Has Universal Closure under Real Linear Combinations"],"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-20261101-r01-vector-subspace-0029","batch_id":"batch-20261014-advance-20261101-r01-vector-subspace-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"02f2dd94254b3700a9369c00885723312be1631cf00555ec4d54f9bf743b1be8","family_id":"qf-mod32-subspace-universal-line-certificate","seed":202611011029,"semantic_template":"qf-mod32-subspace-universal-line-certificate/{\"a\":4,\"b\":-4}"},"fields":{"title":"A Homogeneous Kernel Has Universal Closure under Real Linear Combinations","stem":"Use the subset U={(x,y) in R²:a*x+b*y=0} with nonzero normal [a,b]=[4, -4] and usual real vector operations. Give (normal,zero_in?,addition_closed?,scalar_closed?,homogeneous_kernel_closed_for_all_real_scalars).","answer":{"kind":"expression","value":"([4, -4], Yes, Yes, Yes, homogeneous_kernel_closed_for_all_real_scalars)","explanation":"The zero vector satisfies the equation. Linearity makes the equation hold for sums of solutions and for every real scalar multiple. This algebraic all-vector argument, not sampled points or a new Lean receipt, supplies the stated certificate. The exact result, retaining the requested order, is ([4, -4], Yes, Yes, Yes, homogeneous_kernel_closed_for_all_real_scalars)."},"skills":["A Homogeneous Kernel Has Universal Closure under Real Linear Combinations"],"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-20261101-r01-vector-subspace-0030","batch_id":"batch-20261014-advance-20261101-r01-vector-subspace-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"21f71c30af82a00f01e28d2273d73aeef1fff52097d11c522b4d1d3da204d0ed","family_id":"qf-mod32-subspace-universal-line-certificate","seed":202611011030,"semantic_template":"qf-mod32-subspace-universal-line-certificate/{\"a\":5,\"b\":-5}"},"fields":{"title":"A Homogeneous Kernel Has Universal Closure under Real Linear Combinations","stem":"Use the subset U={(x,y) in R²:a*x+b*y=0} with nonzero normal [a,b]=[5, -5] and usual real vector operations. Give (normal,zero_in?,addition_closed?,scalar_closed?,homogeneous_kernel_closed_for_all_real_scalars).","answer":{"kind":"expression","value":"([5, -5], Yes, Yes, Yes, homogeneous_kernel_closed_for_all_real_scalars)","explanation":"The zero vector satisfies the equation. Linearity makes the equation hold for sums of solutions and for every real scalar multiple. This algebraic all-vector argument, not sampled points or a new Lean receipt, supplies the stated certificate. The exact result, retaining the requested order, is ([5, -5], Yes, Yes, Yes, homogeneous_kernel_closed_for_all_real_scalars)."},"skills":["A Homogeneous Kernel Has Universal Closure under Real Linear Combinations"],"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-20261101-r01-vector-subspace-0031","batch_id":"batch-20261014-advance-20261101-r01-vector-subspace-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"7ca20aaa9ec072e5cc87111de7212c406c8f16103fda947dbb03e41490efce8a","family_id":"qf-mod32-subspace-universal-line-certificate","seed":202611011031,"semantic_template":"qf-mod32-subspace-universal-line-certificate/{\"a\":6,\"b\":-6}"},"fields":{"title":"A Homogeneous Kernel Has Universal Closure under Real Linear Combinations","stem":"Use the subset U={(x,y) in R²:a*x+b*y=0} with nonzero normal [a,b]=[6, -6] and usual real vector operations. Give (normal,zero_in?,addition_closed?,scalar_closed?,homogeneous_kernel_closed_for_all_real_scalars).","answer":{"kind":"expression","value":"([6, -6], Yes, Yes, Yes, homogeneous_kernel_closed_for_all_real_scalars)","explanation":"The zero vector satisfies the equation. Linearity makes the equation hold for sums of solutions and for every real scalar multiple. This algebraic all-vector argument, not sampled points or a new Lean receipt, supplies the stated certificate. The exact result, retaining the requested order, is ([6, -6], Yes, Yes, Yes, homogeneous_kernel_closed_for_all_real_scalars)."},"skills":["A Homogeneous Kernel Has Universal Closure under Real Linear Combinations"],"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-20261101-r01-vector-subspace-0032","batch_id":"batch-20261014-advance-20261101-r01-vector-subspace-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"bab77645de89d923f25f47a0a7d19332be0582611b037e80208ae20341f1d19a","family_id":"qf-mod32-subspace-universal-line-certificate","seed":202611011032,"semantic_template":"qf-mod32-subspace-universal-line-certificate/{\"a\":7,\"b\":-7}"},"fields":{"title":"A Homogeneous Kernel Has Universal Closure under Real Linear Combinations","stem":"Use the subset U={(x,y) in R²:a*x+b*y=0} with nonzero normal [a,b]=[7, -7] and usual real vector operations. Give (normal,zero_in?,addition_closed?,scalar_closed?,homogeneous_kernel_closed_for_all_real_scalars).","answer":{"kind":"expression","value":"([7, -7], Yes, Yes, Yes, homogeneous_kernel_closed_for_all_real_scalars)","explanation":"The zero vector satisfies the equation. Linearity makes the equation hold for sums of solutions and for every real scalar multiple. This algebraic all-vector argument, not sampled points or a new Lean receipt, supplies the stated certificate. The exact result, retaining the requested order, is ([7, -7], Yes, Yes, Yes, homogeneous_kernel_closed_for_all_real_scalars)."},"skills":["A Homogeneous Kernel Has Universal Closure under Real Linear Combinations"],"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-20261101-r01-vector-subspace-0033","batch_id":"batch-20261014-advance-20261101-r01-vector-subspace-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"c5ac8f2a0bc5f3ca39d0abd95efd43e21dc922fef587c82171bbb440cf46b779","family_id":"qf-mod32-subspace-universal-line-certificate","seed":202611011033,"semantic_template":"qf-mod32-subspace-universal-line-certificate/{\"a\":8,\"b\":-1}"},"fields":{"title":"A Homogeneous Kernel Has Universal Closure under Real Linear Combinations","stem":"Use the subset U={(x,y) in R²:a*x+b*y=0} with nonzero normal [a,b]=[8, -1] and usual real vector operations. Give (normal,zero_in?,addition_closed?,scalar_closed?,homogeneous_kernel_closed_for_all_real_scalars).","answer":{"kind":"expression","value":"([8, -1], Yes, Yes, Yes, homogeneous_kernel_closed_for_all_real_scalars)","explanation":"The zero vector satisfies the equation. Linearity makes the equation hold for sums of solutions and for every real scalar multiple. This algebraic all-vector argument, not sampled points or a new Lean receipt, supplies the stated certificate. The exact result, retaining the requested order, is ([8, -1], Yes, Yes, Yes, homogeneous_kernel_closed_for_all_real_scalars)."},"skills":["A Homogeneous Kernel Has Universal Closure under Real Linear Combinations"],"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-20261101-r01-vector-subspace-0034","batch_id":"batch-20261014-advance-20261101-r01-vector-subspace-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"8f470a19573f7be353abff513de0878f2faeec535f107f1378806c6c66afd28b","family_id":"qf-mod32-subspace-universal-line-certificate","seed":202611011034,"semantic_template":"qf-mod32-subspace-universal-line-certificate/{\"a\":9,\"b\":-2}"},"fields":{"title":"A Homogeneous Kernel Has Universal Closure under Real Linear Combinations","stem":"Use the subset U={(x,y) in R²:a*x+b*y=0} with nonzero normal [a,b]=[9, -2] and usual real vector operations. Give (normal,zero_in?,addition_closed?,scalar_closed?,homogeneous_kernel_closed_for_all_real_scalars).","answer":{"kind":"expression","value":"([9, -2], Yes, Yes, Yes, homogeneous_kernel_closed_for_all_real_scalars)","explanation":"The zero vector satisfies the equation. Linearity makes the equation hold for sums of solutions and for every real scalar multiple. This algebraic all-vector argument, not sampled points or a new Lean receipt, supplies the stated certificate. The exact result, retaining the requested order, is ([9, -2], Yes, Yes, Yes, homogeneous_kernel_closed_for_all_real_scalars)."},"skills":["A Homogeneous Kernel Has Universal Closure under Real Linear Combinations"],"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-20261101-r01-vector-subspace-0035","batch_id":"batch-20261014-advance-20261101-r01-vector-subspace-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"523f3e00245427a0abd75d3b6f4593b76f8fa3bdc01524545cbc2382f897c128","family_id":"qf-mod32-subspace-universal-line-certificate","seed":202611011035,"semantic_template":"qf-mod32-subspace-universal-line-certificate/{\"a\":10,\"b\":-3}"},"fields":{"title":"A Homogeneous Kernel Has Universal Closure under Real Linear Combinations","stem":"Use the subset U={(x,y) in R²:a*x+b*y=0} with nonzero normal [a,b]=[10, -3] and usual real vector operations. Give (normal,zero_in?,addition_closed?,scalar_closed?,homogeneous_kernel_closed_for_all_real_scalars).","answer":{"kind":"expression","value":"([10, -3], Yes, Yes, Yes, homogeneous_kernel_closed_for_all_real_scalars)","explanation":"The zero vector satisfies the equation. Linearity makes the equation hold for sums of solutions and for every real scalar multiple. This algebraic all-vector argument, not sampled points or a new Lean receipt, supplies the stated certificate. The exact result, retaining the requested order, is ([10, -3], Yes, Yes, Yes, homogeneous_kernel_closed_for_all_real_scalars)."},"skills":["A Homogeneous Kernel Has Universal Closure under Real Linear Combinations"],"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-20261101-r01-vector-subspace-0036","batch_id":"batch-20261014-advance-20261101-r01-vector-subspace-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"240149a2e723e802bf5260a8691b1ac911071031a49baabb00098ef6860ff65f","family_id":"qf-mod32-subspace-universal-line-certificate","seed":202611011036,"semantic_template":"qf-mod32-subspace-universal-line-certificate/{\"a\":11,\"b\":-4}"},"fields":{"title":"A Homogeneous Kernel Has Universal Closure under Real Linear Combinations","stem":"Use the subset U={(x,y) in R²:a*x+b*y=0} with nonzero normal [a,b]=[11, -4] and usual real vector operations. Give (normal,zero_in?,addition_closed?,scalar_closed?,homogeneous_kernel_closed_for_all_real_scalars).","answer":{"kind":"expression","value":"([11, -4], Yes, Yes, Yes, homogeneous_kernel_closed_for_all_real_scalars)","explanation":"The zero vector satisfies the equation. Linearity makes the equation hold for sums of solutions and for every real scalar multiple. This algebraic all-vector argument, not sampled points or a new Lean receipt, supplies the stated certificate. The exact result, retaining the requested order, is ([11, -4], Yes, Yes, Yes, homogeneous_kernel_closed_for_all_real_scalars)."},"skills":["A Homogeneous Kernel Has Universal Closure under Real Linear Combinations"],"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-20261101-r01-vector-subspace-0037","batch_id":"batch-20261014-advance-20261101-r01-vector-subspace-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"b8e571238435823221d90285e914bb7f4234641f2b3e118e154547e5f412903b","family_id":"qf-mod32-subspace-universal-line-certificate","seed":202611011037,"semantic_template":"qf-mod32-subspace-universal-line-certificate/{\"a\":12,\"b\":-5}"},"fields":{"title":"A Homogeneous Kernel Has Universal Closure under Real Linear Combinations","stem":"Use the subset U={(x,y) in R²:a*x+b*y=0} with nonzero normal [a,b]=[12, -5] and usual real vector operations. Give (normal,zero_in?,addition_closed?,scalar_closed?,homogeneous_kernel_closed_for_all_real_scalars).","answer":{"kind":"expression","value":"([12, -5], Yes, Yes, Yes, homogeneous_kernel_closed_for_all_real_scalars)","explanation":"The zero vector satisfies the equation. Linearity makes the equation hold for sums of solutions and for every real scalar multiple. This algebraic all-vector argument, not sampled points or a new Lean receipt, supplies the stated certificate. The exact result, retaining the requested order, is ([12, -5], Yes, Yes, Yes, homogeneous_kernel_closed_for_all_real_scalars)."},"skills":["A Homogeneous Kernel Has Universal Closure under Real Linear Combinations"],"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-20261101-r01-vector-subspace-0038","batch_id":"batch-20261014-advance-20261101-r01-vector-subspace-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"fed409a670237cde4501c37ec835d2feabb001f23a396263c20b3aab108b7cf5","family_id":"qf-mod32-subspace-universal-line-certificate","seed":202611011038,"semantic_template":"qf-mod32-subspace-universal-line-certificate/{\"a\":13,\"b\":-6}"},"fields":{"title":"A Homogeneous Kernel Has Universal Closure under Real Linear Combinations","stem":"Use the subset U={(x,y) in R²:a*x+b*y=0} with nonzero normal [a,b]=[13, -6] and usual real vector operations. Give (normal,zero_in?,addition_closed?,scalar_closed?,homogeneous_kernel_closed_for_all_real_scalars).","answer":{"kind":"expression","value":"([13, -6], Yes, Yes, Yes, homogeneous_kernel_closed_for_all_real_scalars)","explanation":"The zero vector satisfies the equation. Linearity makes the equation hold for sums of solutions and for every real scalar multiple. This algebraic all-vector argument, not sampled points or a new Lean receipt, supplies the stated certificate. The exact result, retaining the requested order, is ([13, -6], Yes, Yes, Yes, homogeneous_kernel_closed_for_all_real_scalars)."},"skills":["A Homogeneous Kernel Has Universal Closure under Real Linear Combinations"],"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-20261101-r01-vector-subspace-0039","batch_id":"batch-20261014-advance-20261101-r01-vector-subspace-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"47236e855100b0bd9cd58b637be81f15687469e51aaac1cc28305917c390ca6c","family_id":"qf-mod32-subspace-universal-line-certificate","seed":202611011039,"semantic_template":"qf-mod32-subspace-universal-line-certificate/{\"a\":14,\"b\":-7}"},"fields":{"title":"A Homogeneous Kernel Has Universal Closure under Real Linear Combinations","stem":"Use the subset U={(x,y) in R²:a*x+b*y=0} with nonzero normal [a,b]=[14, -7] and usual real vector operations. Give (normal,zero_in?,addition_closed?,scalar_closed?,homogeneous_kernel_closed_for_all_real_scalars).","answer":{"kind":"expression","value":"([14, -7], Yes, Yes, Yes, homogeneous_kernel_closed_for_all_real_scalars)","explanation":"The zero vector satisfies the equation. Linearity makes the equation hold for sums of solutions and for every real scalar multiple. This algebraic all-vector argument, not sampled points or a new Lean receipt, supplies the stated certificate. The exact result, retaining the requested order, is ([14, -7], Yes, Yes, Yes, homogeneous_kernel_closed_for_all_real_scalars)."},"skills":["A Homogeneous Kernel Has Universal Closure under Real Linear Combinations"],"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-20261101-r01-vector-subspace-0040","batch_id":"batch-20261014-advance-20261101-r01-vector-subspace-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"e7fb56c8ca98fcc355dbf9c6b61f53d3d926228f8a67c54f71ebe1561ee98c42","family_id":"qf-mod32-subspace-universal-line-certificate","seed":202611011040,"semantic_template":"qf-mod32-subspace-universal-line-certificate/{\"a\":15,\"b\":-1}"},"fields":{"title":"A Homogeneous Kernel Has Universal Closure under Real Linear Combinations","stem":"Use the subset U={(x,y) in R²:a*x+b*y=0} with nonzero normal [a,b]=[15, -1] and usual real vector operations. Give (normal,zero_in?,addition_closed?,scalar_closed?,homogeneous_kernel_closed_for_all_real_scalars).","answer":{"kind":"expression","value":"([15, -1], Yes, Yes, Yes, homogeneous_kernel_closed_for_all_real_scalars)","explanation":"The zero vector satisfies the equation. Linearity makes the equation hold for sums of solutions and for every real scalar multiple. This algebraic all-vector argument, not sampled points or a new Lean receipt, supplies the stated certificate. The exact result, retaining the requested order, is ([15, -1], Yes, Yes, Yes, homogeneous_kernel_closed_for_all_real_scalars)."},"skills":["A Homogeneous Kernel Has Universal Closure under Real Linear Combinations"],"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-20261101-r01-vector-subspace-0041","batch_id":"batch-20261014-advance-20261101-r01-vector-subspace-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"94e682e7f611740ddc969f09e171136b970b5b0bf3359d688b0c45ade680fe67","family_id":"qf-mod32-subspace-universal-line-certificate","seed":202611011041,"semantic_template":"qf-mod32-subspace-universal-line-certificate/{\"a\":16,\"b\":-2}"},"fields":{"title":"A Homogeneous Kernel Has Universal Closure under Real Linear Combinations","stem":"Use the subset U={(x,y) in R²:a*x+b*y=0} with nonzero normal [a,b]=[16, -2] and usual real vector operations. Give (normal,zero_in?,addition_closed?,scalar_closed?,homogeneous_kernel_closed_for_all_real_scalars).","answer":{"kind":"expression","value":"([16, -2], Yes, Yes, Yes, homogeneous_kernel_closed_for_all_real_scalars)","explanation":"The zero vector satisfies the equation. Linearity makes the equation hold for sums of solutions and for every real scalar multiple. This algebraic all-vector argument, not sampled points or a new Lean receipt, supplies the stated certificate. The exact result, retaining the requested order, is ([16, -2], Yes, Yes, Yes, homogeneous_kernel_closed_for_all_real_scalars)."},"skills":["A Homogeneous Kernel Has Universal Closure under Real Linear Combinations"],"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-20261101-r01-vector-subspace-0042","batch_id":"batch-20261014-advance-20261101-r01-vector-subspace-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"7d9f777711779d2159b09f9532145c0ea1c1ce82ed57c2fcde51e43dc7b803fb","family_id":"qf-mod32-subspace-universal-line-certificate","seed":202611011042,"semantic_template":"qf-mod32-subspace-universal-line-certificate/{\"a\":17,\"b\":-3}"},"fields":{"title":"A Homogeneous Kernel Has Universal Closure under Real Linear Combinations","stem":"Use the subset U={(x,y) in R²:a*x+b*y=0} with nonzero normal [a,b]=[17, -3] and usual real vector operations. Give (normal,zero_in?,addition_closed?,scalar_closed?,homogeneous_kernel_closed_for_all_real_scalars).","answer":{"kind":"expression","value":"([17, -3], Yes, Yes, Yes, homogeneous_kernel_closed_for_all_real_scalars)","explanation":"The zero vector satisfies the equation. Linearity makes the equation hold for sums of solutions and for every real scalar multiple. This algebraic all-vector argument, not sampled points or a new Lean receipt, supplies the stated certificate. The exact result, retaining the requested order, is ([17, -3], Yes, Yes, Yes, homogeneous_kernel_closed_for_all_real_scalars)."},"skills":["A Homogeneous Kernel Has Universal Closure under Real Linear Combinations"],"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-20261101-r01-vector-subspace-0043","batch_id":"batch-20261014-advance-20261101-r01-vector-subspace-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"c50f442d9573d136ff28e73d5d4fb4383574b8345da0658c7570971b9ee156d7","family_id":"qf-mod32-subspace-universal-line-certificate","seed":202611011043,"semantic_template":"qf-mod32-subspace-universal-line-certificate/{\"a\":18,\"b\":-4}"},"fields":{"title":"A Homogeneous Kernel Has Universal Closure under Real Linear Combinations","stem":"Use the subset U={(x,y) in R²:a*x+b*y=0} with nonzero normal [a,b]=[18, -4] and usual real vector operations. Give (normal,zero_in?,addition_closed?,scalar_closed?,homogeneous_kernel_closed_for_all_real_scalars).","answer":{"kind":"expression","value":"([18, -4], Yes, Yes, Yes, homogeneous_kernel_closed_for_all_real_scalars)","explanation":"The zero vector satisfies the equation. Linearity makes the equation hold for sums of solutions and for every real scalar multiple. This algebraic all-vector argument, not sampled points or a new Lean receipt, supplies the stated certificate. The exact result, retaining the requested order, is ([18, -4], Yes, Yes, Yes, homogeneous_kernel_closed_for_all_real_scalars)."},"skills":["A Homogeneous Kernel Has Universal Closure under Real Linear Combinations"],"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-20261101-r01-vector-subspace-0044","batch_id":"batch-20261014-advance-20261101-r01-vector-subspace-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"727ed041e702facef5636843e01adf7ebb5c4ff016345c3bcc5bd17540273309","family_id":"qf-mod32-subspace-universal-line-certificate","seed":202611011044,"semantic_template":"qf-mod32-subspace-universal-line-certificate/{\"a\":19,\"b\":-5}"},"fields":{"title":"A Homogeneous Kernel Has Universal Closure under Real Linear Combinations","stem":"Use the subset U={(x,y) in R²:a*x+b*y=0} with nonzero normal [a,b]=[19, -5] and usual real vector operations. Give (normal,zero_in?,addition_closed?,scalar_closed?,homogeneous_kernel_closed_for_all_real_scalars).","answer":{"kind":"expression","value":"([19, -5], Yes, Yes, Yes, homogeneous_kernel_closed_for_all_real_scalars)","explanation":"The zero vector satisfies the equation. Linearity makes the equation hold for sums of solutions and for every real scalar multiple. This algebraic all-vector argument, not sampled points or a new Lean receipt, supplies the stated certificate. The exact result, retaining the requested order, is ([19, -5], Yes, Yes, Yes, homogeneous_kernel_closed_for_all_real_scalars)."},"skills":["A Homogeneous Kernel Has Universal Closure under Real Linear Combinations"],"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-20261101-r01-vector-subspace-0045","batch_id":"batch-20261014-advance-20261101-r01-vector-subspace-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"6504e8d0a601758a41ae0d124a26a1bd79884efb0c06c6577d4be6b34f0c1a22","family_id":"qf-mod32-subspace-universal-line-certificate","seed":202611011045,"semantic_template":"qf-mod32-subspace-universal-line-certificate/{\"a\":20,\"b\":-6}"},"fields":{"title":"A Homogeneous Kernel Has Universal Closure under Real Linear Combinations","stem":"Use the subset U={(x,y) in R²:a*x+b*y=0} with nonzero normal [a,b]=[20, -6] and usual real vector operations. Give (normal,zero_in?,addition_closed?,scalar_closed?,homogeneous_kernel_closed_for_all_real_scalars).","answer":{"kind":"expression","value":"([20, -6], Yes, Yes, Yes, homogeneous_kernel_closed_for_all_real_scalars)","explanation":"The zero vector satisfies the equation. Linearity makes the equation hold for sums of solutions and for every real scalar multiple. This algebraic all-vector argument, not sampled points or a new Lean receipt, supplies the stated certificate. The exact result, retaining the requested order, is ([20, -6], Yes, Yes, Yes, homogeneous_kernel_closed_for_all_real_scalars)."},"skills":["A Homogeneous Kernel Has Universal Closure under Real Linear Combinations"],"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-20261101-r01-vector-subspace-0046","batch_id":"batch-20261014-advance-20261101-r01-vector-subspace-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"bcc617a0ba79b7eb6972a001a3ff4509f14b55a7a3432eaf1ee1a6a9f256585b","family_id":"qf-mod32-subspace-universal-line-certificate","seed":202611011046,"semantic_template":"qf-mod32-subspace-universal-line-certificate/{\"a\":21,\"b\":-7}"},"fields":{"title":"A Homogeneous Kernel Has Universal Closure under Real Linear Combinations","stem":"Use the subset U={(x,y) in R²:a*x+b*y=0} with nonzero normal [a,b]=[21, -7] and usual real vector operations. Give (normal,zero_in?,addition_closed?,scalar_closed?,homogeneous_kernel_closed_for_all_real_scalars).","answer":{"kind":"expression","value":"([21, -7], Yes, Yes, Yes, homogeneous_kernel_closed_for_all_real_scalars)","explanation":"The zero vector satisfies the equation. Linearity makes the equation hold for sums of solutions and for every real scalar multiple. This algebraic all-vector argument, not sampled points or a new Lean receipt, supplies the stated certificate. The exact result, retaining the requested order, is ([21, -7], Yes, Yes, Yes, homogeneous_kernel_closed_for_all_real_scalars)."},"skills":["A Homogeneous Kernel Has Universal Closure under Real Linear Combinations"],"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-20261101-r01-vector-subspace-0047","batch_id":"batch-20261014-advance-20261101-r01-vector-subspace-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"3cddb3852a8a68ff2ff43dfe76427aab689239c156a9993c75a523f04384327c","family_id":"qf-mod32-subspace-universal-line-certificate","seed":202611011047,"semantic_template":"qf-mod32-subspace-universal-line-certificate/{\"a\":22,\"b\":-1}"},"fields":{"title":"A Homogeneous Kernel Has Universal Closure under Real Linear Combinations","stem":"Use the subset U={(x,y) in R²:a*x+b*y=0} with nonzero normal [a,b]=[22, -1] and usual real vector operations. Give (normal,zero_in?,addition_closed?,scalar_closed?,homogeneous_kernel_closed_for_all_real_scalars).","answer":{"kind":"expression","value":"([22, -1], Yes, Yes, Yes, homogeneous_kernel_closed_for_all_real_scalars)","explanation":"The zero vector satisfies the equation. Linearity makes the equation hold for sums of solutions and for every real scalar multiple. This algebraic all-vector argument, not sampled points or a new Lean receipt, supplies the stated certificate. The exact result, retaining the requested order, is ([22, -1], Yes, Yes, Yes, homogeneous_kernel_closed_for_all_real_scalars)."},"skills":["A Homogeneous Kernel Has Universal Closure under Real Linear Combinations"],"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-20261101-r01-vector-subspace-0048","batch_id":"batch-20261014-advance-20261101-r01-vector-subspace-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"f6ec41a05798e1d7d27b6257894cd81fa6584787061c39ef3086eeee418f78c9","family_id":"qf-mod32-subspace-universal-line-certificate","seed":202611011048,"semantic_template":"qf-mod32-subspace-universal-line-certificate/{\"a\":23,\"b\":-2}"},"fields":{"title":"A Homogeneous Kernel Has Universal Closure under Real Linear Combinations","stem":"Use the subset U={(x,y) in R²:a*x+b*y=0} with nonzero normal [a,b]=[23, -2] and usual real vector operations. Give (normal,zero_in?,addition_closed?,scalar_closed?,homogeneous_kernel_closed_for_all_real_scalars).","answer":{"kind":"expression","value":"([23, -2], Yes, Yes, Yes, homogeneous_kernel_closed_for_all_real_scalars)","explanation":"The zero vector satisfies the equation. Linearity makes the equation hold for sums of solutions and for every real scalar multiple. This algebraic all-vector argument, not sampled points or a new Lean receipt, supplies the stated certificate. The exact result, retaining the requested order, is ([23, -2], Yes, Yes, Yes, homogeneous_kernel_closed_for_all_real_scalars)."},"skills":["A Homogeneous Kernel Has Universal Closure under Real Linear Combinations"],"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-20261101-r01-vector-subspace-0049","batch_id":"batch-20261014-advance-20261101-r01-vector-subspace-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"a8f77ce76d88047b063ff71da5e37b74a1180ecb97ee0b813b236276cb9bb4ba","family_id":"qf-mod32-subspace-universal-line-certificate","seed":202611011049,"semantic_template":"qf-mod32-subspace-universal-line-certificate/{\"a\":24,\"b\":-3}"},"fields":{"title":"A Homogeneous Kernel Has Universal Closure under Real Linear Combinations","stem":"Use the subset U={(x,y) in R²:a*x+b*y=0} with nonzero normal [a,b]=[24, -3] and usual real vector operations. Give (normal,zero_in?,addition_closed?,scalar_closed?,homogeneous_kernel_closed_for_all_real_scalars).","answer":{"kind":"expression","value":"([24, -3], Yes, Yes, Yes, homogeneous_kernel_closed_for_all_real_scalars)","explanation":"The zero vector satisfies the equation. Linearity makes the equation hold for sums of solutions and for every real scalar multiple. This algebraic all-vector argument, not sampled points or a new Lean receipt, supplies the stated certificate. The exact result, retaining the requested order, is ([24, -3], Yes, Yes, Yes, homogeneous_kernel_closed_for_all_real_scalars)."},"skills":["A Homogeneous Kernel Has Universal Closure under Real Linear Combinations"],"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-20261101-r01-vector-subspace-0050","batch_id":"batch-20261014-advance-20261101-r01-vector-subspace-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"f9d4fe2652ae7c58ce5fa302d81f43c7de9f9b26b54c7f426be75ee65c044667","family_id":"qf-mod32-subspace-universal-line-certificate","seed":202611011050,"semantic_template":"qf-mod32-subspace-universal-line-certificate/{\"a\":25,\"b\":-4}"},"fields":{"title":"A Homogeneous Kernel Has Universal Closure under Real Linear Combinations","stem":"Use the subset U={(x,y) in R²:a*x+b*y=0} with nonzero normal [a,b]=[25, -4] and usual real vector operations. Give (normal,zero_in?,addition_closed?,scalar_closed?,homogeneous_kernel_closed_for_all_real_scalars).","answer":{"kind":"expression","value":"([25, -4], Yes, Yes, Yes, homogeneous_kernel_closed_for_all_real_scalars)","explanation":"The zero vector satisfies the equation. Linearity makes the equation hold for sums of solutions and for every real scalar multiple. This algebraic all-vector argument, not sampled points or a new Lean receipt, supplies the stated certificate. The exact result, retaining the requested order, is ([25, -4], Yes, Yes, Yes, homogeneous_kernel_closed_for_all_real_scalars)."},"skills":["A Homogeneous Kernel Has Universal Closure under Real Linear Combinations"],"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-20261101-r01-vector-subspace-0051","batch_id":"batch-20261014-advance-20261101-r01-vector-subspace-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"677f346c501ed7700365dfba7bd5c5b51024d7e5364542811d70bdb7698d5d4c","family_id":"qf-mod32-subspace-finite-set-counterexample","seed":202611011051,"semantic_template":"qf-mod32-subspace-finite-set-counterexample/{\"points\":[[0,0],[1,1]]}"},"fields":{"title":"A Finite Nonzero Real Vector Set Cannot Pass Universal Scalar Closure","stem":"Use the complete nonempty duplicate-free finite point set=[[0, 0], [1, 1]] as a candidate subspace of R² with usual operations. Give (points,witness,subspace?,classification). First prefer [zero_missing,[0,0]]; otherwise use [scalar_failure,scalar,point,image] from the first nonzero point in input order and the first missing integer multiple with scalar 2,3,... . A null witness denotes the singleton zero_subspace.","answer":{"kind":"expression","value":"([[0, 0], [1, 1]], [scalar_failure, 2, [1, 1], [2, 2]], No, not_subspace_exact_counterexample)","explanation":"Missing zero immediately refutes a subspace. If zero is present but a nonzero point occurs, distinct nonzero integer multiples cannot all belong to this finite set, yielding an exact scalar-closure counterexample. Do not infer an arbitrary subspace from finitely many positive samples. The exact result, retaining the requested order, is ([[0, 0], [1, 1]], [scalar_failure, 2, [1, 1], [2, 2]], No, not_subspace_exact_counterexample)."},"skills":["A Finite Nonzero Real Vector Set Cannot Pass Universal Scalar Closure"],"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-20261101-r01-vector-subspace-0052","batch_id":"batch-20261014-advance-20261101-r01-vector-subspace-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"f117b9941f0b346e9defd3cefe9680db41d7001c899b642facc6bc3cece93926","family_id":"qf-mod32-subspace-finite-set-counterexample","seed":202611011052,"semantic_template":"qf-mod32-subspace-finite-set-counterexample/{\"points\":[[0,0],[2,1]]}"},"fields":{"title":"A Finite Nonzero Real Vector Set Cannot Pass Universal Scalar Closure","stem":"Use the complete nonempty duplicate-free finite point set=[[0, 0], [2, 1]] as a candidate subspace of R² with usual operations. Give (points,witness,subspace?,classification). First prefer [zero_missing,[0,0]]; otherwise use [scalar_failure,scalar,point,image] from the first nonzero point in input order and the first missing integer multiple with scalar 2,3,... . A null witness denotes the singleton zero_subspace.","answer":{"kind":"expression","value":"([[0, 0], [2, 1]], [scalar_failure, 2, [2, 1], [4, 2]], No, not_subspace_exact_counterexample)","explanation":"Missing zero immediately refutes a subspace. If zero is present but a nonzero point occurs, distinct nonzero integer multiples cannot all belong to this finite set, yielding an exact scalar-closure counterexample. Do not infer an arbitrary subspace from finitely many positive samples. The exact result, retaining the requested order, is ([[0, 0], [2, 1]], [scalar_failure, 2, [2, 1], [4, 2]], No, not_subspace_exact_counterexample)."},"skills":["A Finite Nonzero Real Vector Set Cannot Pass Universal Scalar Closure"],"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-20261101-r01-vector-subspace-0053","batch_id":"batch-20261014-advance-20261101-r01-vector-subspace-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"ffcfd3ac49e7d743ca1e534157d2bde105c9cfb2ed027260226ba732ef58d0dd","family_id":"qf-mod32-subspace-finite-set-counterexample","seed":202611011053,"semantic_template":"qf-mod32-subspace-finite-set-counterexample/{\"points\":[[0,0],[3,1]]}"},"fields":{"title":"A Finite Nonzero Real Vector Set Cannot Pass Universal Scalar Closure","stem":"Use the complete nonempty duplicate-free finite point set=[[0, 0], [3, 1]] as a candidate subspace of R² with usual operations. Give (points,witness,subspace?,classification). First prefer [zero_missing,[0,0]]; otherwise use [scalar_failure,scalar,point,image] from the first nonzero point in input order and the first missing integer multiple with scalar 2,3,... . A null witness denotes the singleton zero_subspace.","answer":{"kind":"expression","value":"([[0, 0], [3, 1]], [scalar_failure, 2, [3, 1], [6, 2]], No, not_subspace_exact_counterexample)","explanation":"Missing zero immediately refutes a subspace. If zero is present but a nonzero point occurs, distinct nonzero integer multiples cannot all belong to this finite set, yielding an exact scalar-closure counterexample. Do not infer an arbitrary subspace from finitely many positive samples. The exact result, retaining the requested order, is ([[0, 0], [3, 1]], [scalar_failure, 2, [3, 1], [6, 2]], No, not_subspace_exact_counterexample)."},"skills":["A Finite Nonzero Real Vector Set Cannot Pass Universal Scalar Closure"],"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-20261101-r01-vector-subspace-0054","batch_id":"batch-20261014-advance-20261101-r01-vector-subspace-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"ebe0dd79ceb9bb0e2fcea53bdec8d48daedb0fa81baf92caad805ef28d6fa30d","family_id":"qf-mod32-subspace-finite-set-counterexample","seed":202611011054,"semantic_template":"qf-mod32-subspace-finite-set-counterexample/{\"points\":[[0,0],[4,1]]}"},"fields":{"title":"A Finite Nonzero Real Vector Set Cannot Pass Universal Scalar Closure","stem":"Use the complete nonempty duplicate-free finite point set=[[0, 0], [4, 1]] as a candidate subspace of R² with usual operations. Give (points,witness,subspace?,classification). First prefer [zero_missing,[0,0]]; otherwise use [scalar_failure,scalar,point,image] from the first nonzero point in input order and the first missing integer multiple with scalar 2,3,... . A null witness denotes the singleton zero_subspace.","answer":{"kind":"expression","value":"([[0, 0], [4, 1]], [scalar_failure, 2, [4, 1], [8, 2]], No, not_subspace_exact_counterexample)","explanation":"Missing zero immediately refutes a subspace. If zero is present but a nonzero point occurs, distinct nonzero integer multiples cannot all belong to this finite set, yielding an exact scalar-closure counterexample. Do not infer an arbitrary subspace from finitely many positive samples. The exact result, retaining the requested order, is ([[0, 0], [4, 1]], [scalar_failure, 2, [4, 1], [8, 2]], No, not_subspace_exact_counterexample)."},"skills":["A Finite Nonzero Real Vector Set Cannot Pass Universal Scalar Closure"],"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-20261101-r01-vector-subspace-0055","batch_id":"batch-20261014-advance-20261101-r01-vector-subspace-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"04d40f18b4a82e67d6cc483caf1c14acec00893006f6c31315c69b332b3760b5","family_id":"qf-mod32-subspace-finite-set-counterexample","seed":202611011055,"semantic_template":"qf-mod32-subspace-finite-set-counterexample/{\"points\":[[0,0],[5,1]]}"},"fields":{"title":"A Finite Nonzero Real Vector Set Cannot Pass Universal Scalar Closure","stem":"Use the complete nonempty duplicate-free finite point set=[[0, 0], [5, 1]] as a candidate subspace of R² with usual operations. Give (points,witness,subspace?,classification). First prefer [zero_missing,[0,0]]; otherwise use [scalar_failure,scalar,point,image] from the first nonzero point in input order and the first missing integer multiple with scalar 2,3,... . A null witness denotes the singleton zero_subspace.","answer":{"kind":"expression","value":"([[0, 0], [5, 1]], [scalar_failure, 2, [5, 1], [10, 2]], No, not_subspace_exact_counterexample)","explanation":"Missing zero immediately refutes a subspace. If zero is present but a nonzero point occurs, distinct nonzero integer multiples cannot all belong to this finite set, yielding an exact scalar-closure counterexample. Do not infer an arbitrary subspace from finitely many positive samples. The exact result, retaining the requested order, is ([[0, 0], [5, 1]], [scalar_failure, 2, [5, 1], [10, 2]], No, not_subspace_exact_counterexample)."},"skills":["A Finite Nonzero Real Vector Set Cannot Pass Universal Scalar Closure"],"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-20261101-r01-vector-subspace-0056","batch_id":"batch-20261014-advance-20261101-r01-vector-subspace-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"a3e4132e656be2294ebe8e8be1756ab77aa873ff87f7a8d270822cef2e9da645","family_id":"qf-mod32-subspace-finite-set-counterexample","seed":202611011056,"semantic_template":"qf-mod32-subspace-finite-set-counterexample/{\"points\":[[0,0],[6,1]]}"},"fields":{"title":"A Finite Nonzero Real Vector Set Cannot Pass Universal Scalar Closure","stem":"Use the complete nonempty duplicate-free finite point set=[[0, 0], [6, 1]] as a candidate subspace of R² with usual operations. Give (points,witness,subspace?,classification). First prefer [zero_missing,[0,0]]; otherwise use [scalar_failure,scalar,point,image] from the first nonzero point in input order and the first missing integer multiple with scalar 2,3,... . A null witness denotes the singleton zero_subspace.","answer":{"kind":"expression","value":"([[0, 0], [6, 1]], [scalar_failure, 2, [6, 1], [12, 2]], No, not_subspace_exact_counterexample)","explanation":"Missing zero immediately refutes a subspace. If zero is present but a nonzero point occurs, distinct nonzero integer multiples cannot all belong to this finite set, yielding an exact scalar-closure counterexample. Do not infer an arbitrary subspace from finitely many positive samples. The exact result, retaining the requested order, is ([[0, 0], [6, 1]], [scalar_failure, 2, [6, 1], [12, 2]], No, not_subspace_exact_counterexample)."},"skills":["A Finite Nonzero Real Vector Set Cannot Pass Universal Scalar Closure"],"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-20261101-r01-vector-subspace-0057","batch_id":"batch-20261014-advance-20261101-r01-vector-subspace-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"e08f086f06b45c78a3a761d223a8b2d75085d18072a72addd8f20442faffaf51","family_id":"qf-mod32-subspace-finite-set-counterexample","seed":202611011057,"semantic_template":"qf-mod32-subspace-finite-set-counterexample/{\"points\":[[0,0],[7,1]]}"},"fields":{"title":"A Finite Nonzero Real Vector Set Cannot Pass Universal Scalar Closure","stem":"Use the complete nonempty duplicate-free finite point set=[[0, 0], [7, 1]] as a candidate subspace of R² with usual operations. Give (points,witness,subspace?,classification). First prefer [zero_missing,[0,0]]; otherwise use [scalar_failure,scalar,point,image] from the first nonzero point in input order and the first missing integer multiple with scalar 2,3,... . A null witness denotes the singleton zero_subspace.","answer":{"kind":"expression","value":"([[0, 0], [7, 1]], [scalar_failure, 2, [7, 1], [14, 2]], No, not_subspace_exact_counterexample)","explanation":"Missing zero immediately refutes a subspace. If zero is present but a nonzero point occurs, distinct nonzero integer multiples cannot all belong to this finite set, yielding an exact scalar-closure counterexample. Do not infer an arbitrary subspace from finitely many positive samples. The exact result, retaining the requested order, is ([[0, 0], [7, 1]], [scalar_failure, 2, [7, 1], [14, 2]], No, not_subspace_exact_counterexample)."},"skills":["A Finite Nonzero Real Vector Set Cannot Pass Universal Scalar Closure"],"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-20261101-r01-vector-subspace-0058","batch_id":"batch-20261014-advance-20261101-r01-vector-subspace-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"9a4d9563ea778e6a7d35b7c931879994c8948ccdea91ef90a35315931ef8575b","family_id":"qf-mod32-subspace-finite-set-counterexample","seed":202611011058,"semantic_template":"qf-mod32-subspace-finite-set-counterexample/{\"points\":[[0,0],[8,1]]}"},"fields":{"title":"A Finite Nonzero Real Vector Set Cannot Pass Universal Scalar Closure","stem":"Use the complete nonempty duplicate-free finite point set=[[0, 0], [8, 1]] as a candidate subspace of R² with usual operations. Give (points,witness,subspace?,classification). First prefer [zero_missing,[0,0]]; otherwise use [scalar_failure,scalar,point,image] from the first nonzero point in input order and the first missing integer multiple with scalar 2,3,... . A null witness denotes the singleton zero_subspace.","answer":{"kind":"expression","value":"([[0, 0], [8, 1]], [scalar_failure, 2, [8, 1], [16, 2]], No, not_subspace_exact_counterexample)","explanation":"Missing zero immediately refutes a subspace. If zero is present but a nonzero point occurs, distinct nonzero integer multiples cannot all belong to this finite set, yielding an exact scalar-closure counterexample. Do not infer an arbitrary subspace from finitely many positive samples. The exact result, retaining the requested order, is ([[0, 0], [8, 1]], [scalar_failure, 2, [8, 1], [16, 2]], No, not_subspace_exact_counterexample)."},"skills":["A Finite Nonzero Real Vector Set Cannot Pass Universal Scalar Closure"],"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-20261101-r01-vector-subspace-0059","batch_id":"batch-20261014-advance-20261101-r01-vector-subspace-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"c873d3a616603cb7370153531f05d8d86d65010458d422f3dc1f5fdd9ebd4bb8","family_id":"qf-mod32-subspace-finite-set-counterexample","seed":202611011059,"semantic_template":"qf-mod32-subspace-finite-set-counterexample/{\"points\":[[0,0],[9,1]]}"},"fields":{"title":"A Finite Nonzero Real Vector Set Cannot Pass Universal Scalar Closure","stem":"Use the complete nonempty duplicate-free finite point set=[[0, 0], [9, 1]] as a candidate subspace of R² with usual operations. Give (points,witness,subspace?,classification). First prefer [zero_missing,[0,0]]; otherwise use [scalar_failure,scalar,point,image] from the first nonzero point in input order and the first missing integer multiple with scalar 2,3,... . A null witness denotes the singleton zero_subspace.","answer":{"kind":"expression","value":"([[0, 0], [9, 1]], [scalar_failure, 2, [9, 1], [18, 2]], No, not_subspace_exact_counterexample)","explanation":"Missing zero immediately refutes a subspace. If zero is present but a nonzero point occurs, distinct nonzero integer multiples cannot all belong to this finite set, yielding an exact scalar-closure counterexample. Do not infer an arbitrary subspace from finitely many positive samples. The exact result, retaining the requested order, is ([[0, 0], [9, 1]], [scalar_failure, 2, [9, 1], [18, 2]], No, not_subspace_exact_counterexample)."},"skills":["A Finite Nonzero Real Vector Set Cannot Pass Universal Scalar Closure"],"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-20261101-r01-vector-subspace-0060","batch_id":"batch-20261014-advance-20261101-r01-vector-subspace-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"075320d0ac78eb66242fa714468571a0e30b07f32c3b3ecf5fe600af3669bcd6","family_id":"qf-mod32-subspace-finite-set-counterexample","seed":202611011060,"semantic_template":"qf-mod32-subspace-finite-set-counterexample/{\"points\":[[0,0],[10,1]]}"},"fields":{"title":"A Finite Nonzero Real Vector Set Cannot Pass Universal Scalar Closure","stem":"Use the complete nonempty duplicate-free finite point set=[[0, 0], [10, 1]] as a candidate subspace of R² with usual operations. Give (points,witness,subspace?,classification). First prefer [zero_missing,[0,0]]; otherwise use [scalar_failure,scalar,point,image] from the first nonzero point in input order and the first missing integer multiple with scalar 2,3,... . A null witness denotes the singleton zero_subspace.","answer":{"kind":"expression","value":"([[0, 0], [10, 1]], [scalar_failure, 2, [10, 1], [20, 2]], No, not_subspace_exact_counterexample)","explanation":"Missing zero immediately refutes a subspace. If zero is present but a nonzero point occurs, distinct nonzero integer multiples cannot all belong to this finite set, yielding an exact scalar-closure counterexample. Do not infer an arbitrary subspace from finitely many positive samples. The exact result, retaining the requested order, is ([[0, 0], [10, 1]], [scalar_failure, 2, [10, 1], [20, 2]], No, not_subspace_exact_counterexample)."},"skills":["A Finite Nonzero Real Vector Set Cannot Pass Universal Scalar Closure"],"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-20261101-r01-vector-subspace-0061","batch_id":"batch-20261014-advance-20261101-r01-vector-subspace-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"32abb450e04490068f4a340581abc99e890276e0873fdb6131e2066db4968535","family_id":"qf-mod32-subspace-finite-set-counterexample","seed":202611011061,"semantic_template":"qf-mod32-subspace-finite-set-counterexample/{\"points\":[[0,0],[11,1]]}"},"fields":{"title":"A Finite Nonzero Real Vector Set Cannot Pass Universal Scalar Closure","stem":"Use the complete nonempty duplicate-free finite point set=[[0, 0], [11, 1]] as a candidate subspace of R² with usual operations. Give (points,witness,subspace?,classification). First prefer [zero_missing,[0,0]]; otherwise use [scalar_failure,scalar,point,image] from the first nonzero point in input order and the first missing integer multiple with scalar 2,3,... . A null witness denotes the singleton zero_subspace.","answer":{"kind":"expression","value":"([[0, 0], [11, 1]], [scalar_failure, 2, [11, 1], [22, 2]], No, not_subspace_exact_counterexample)","explanation":"Missing zero immediately refutes a subspace. If zero is present but a nonzero point occurs, distinct nonzero integer multiples cannot all belong to this finite set, yielding an exact scalar-closure counterexample. Do not infer an arbitrary subspace from finitely many positive samples. The exact result, retaining the requested order, is ([[0, 0], [11, 1]], [scalar_failure, 2, [11, 1], [22, 2]], No, not_subspace_exact_counterexample)."},"skills":["A Finite Nonzero Real Vector Set Cannot Pass Universal Scalar Closure"],"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-20261101-r01-vector-subspace-0062","batch_id":"batch-20261014-advance-20261101-r01-vector-subspace-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"19f11d1c89758380bba36657ef4ab8ece74f529d6fd6c44df22f46fd5d0dc394","family_id":"qf-mod32-subspace-finite-set-counterexample","seed":202611011062,"semantic_template":"qf-mod32-subspace-finite-set-counterexample/{\"points\":[[0,0],[12,1]]}"},"fields":{"title":"A Finite Nonzero Real Vector Set Cannot Pass Universal Scalar Closure","stem":"Use the complete nonempty duplicate-free finite point set=[[0, 0], [12, 1]] as a candidate subspace of R² with usual operations. Give (points,witness,subspace?,classification). First prefer [zero_missing,[0,0]]; otherwise use [scalar_failure,scalar,point,image] from the first nonzero point in input order and the first missing integer multiple with scalar 2,3,... . A null witness denotes the singleton zero_subspace.","answer":{"kind":"expression","value":"([[0, 0], [12, 1]], [scalar_failure, 2, [12, 1], [24, 2]], No, not_subspace_exact_counterexample)","explanation":"Missing zero immediately refutes a subspace. If zero is present but a nonzero point occurs, distinct nonzero integer multiples cannot all belong to this finite set, yielding an exact scalar-closure counterexample. Do not infer an arbitrary subspace from finitely many positive samples. The exact result, retaining the requested order, is ([[0, 0], [12, 1]], [scalar_failure, 2, [12, 1], [24, 2]], No, not_subspace_exact_counterexample)."},"skills":["A Finite Nonzero Real Vector Set Cannot Pass Universal Scalar Closure"],"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-20261101-r01-vector-subspace-0063","batch_id":"batch-20261014-advance-20261101-r01-vector-subspace-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"fd28181e480b09087706b232c080bf2d1d75fea6f225e38a50c56eb912c3e4f5","family_id":"qf-mod32-subspace-finite-set-counterexample","seed":202611011063,"semantic_template":"qf-mod32-subspace-finite-set-counterexample/{\"points\":[[0,0],[13,1]]}"},"fields":{"title":"A Finite Nonzero Real Vector Set Cannot Pass Universal Scalar Closure","stem":"Use the complete nonempty duplicate-free finite point set=[[0, 0], [13, 1]] as a candidate subspace of R² with usual operations. Give (points,witness,subspace?,classification). First prefer [zero_missing,[0,0]]; otherwise use [scalar_failure,scalar,point,image] from the first nonzero point in input order and the first missing integer multiple with scalar 2,3,... . A null witness denotes the singleton zero_subspace.","answer":{"kind":"expression","value":"([[0, 0], [13, 1]], [scalar_failure, 2, [13, 1], [26, 2]], No, not_subspace_exact_counterexample)","explanation":"Missing zero immediately refutes a subspace. If zero is present but a nonzero point occurs, distinct nonzero integer multiples cannot all belong to this finite set, yielding an exact scalar-closure counterexample. Do not infer an arbitrary subspace from finitely many positive samples. The exact result, retaining the requested order, is ([[0, 0], [13, 1]], [scalar_failure, 2, [13, 1], [26, 2]], No, not_subspace_exact_counterexample)."},"skills":["A Finite Nonzero Real Vector Set Cannot Pass Universal Scalar Closure"],"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-20261101-r01-vector-subspace-0064","batch_id":"batch-20261014-advance-20261101-r01-vector-subspace-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"a8197600e803e6ed9e73aa5d48de7cacd6afa146adbaa1fa0d270fddd9e97843","family_id":"qf-mod32-subspace-finite-set-counterexample","seed":202611011064,"semantic_template":"qf-mod32-subspace-finite-set-counterexample/{\"points\":[[0,0],[14,1]]}"},"fields":{"title":"A Finite Nonzero Real Vector Set Cannot Pass Universal Scalar Closure","stem":"Use the complete nonempty duplicate-free finite point set=[[0, 0], [14, 1]] as a candidate subspace of R² with usual operations. Give (points,witness,subspace?,classification). First prefer [zero_missing,[0,0]]; otherwise use [scalar_failure,scalar,point,image] from the first nonzero point in input order and the first missing integer multiple with scalar 2,3,... . A null witness denotes the singleton zero_subspace.","answer":{"kind":"expression","value":"([[0, 0], [14, 1]], [scalar_failure, 2, [14, 1], [28, 2]], No, not_subspace_exact_counterexample)","explanation":"Missing zero immediately refutes a subspace. If zero is present but a nonzero point occurs, distinct nonzero integer multiples cannot all belong to this finite set, yielding an exact scalar-closure counterexample. Do not infer an arbitrary subspace from finitely many positive samples. The exact result, retaining the requested order, is ([[0, 0], [14, 1]], [scalar_failure, 2, [14, 1], [28, 2]], No, not_subspace_exact_counterexample)."},"skills":["A Finite Nonzero Real Vector Set Cannot Pass Universal Scalar Closure"],"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-20261101-r01-vector-subspace-0065","batch_id":"batch-20261014-advance-20261101-r01-vector-subspace-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"1e29500e1b21989decb38a23de231d34ff2b9a9d39031deea2a85c66155caabc","family_id":"qf-mod32-subspace-finite-set-counterexample","seed":202611011065,"semantic_template":"qf-mod32-subspace-finite-set-counterexample/{\"points\":[[0,0],[15,1]]}"},"fields":{"title":"A Finite Nonzero Real Vector Set Cannot Pass Universal Scalar Closure","stem":"Use the complete nonempty duplicate-free finite point set=[[0, 0], [15, 1]] as a candidate subspace of R² with usual operations. Give (points,witness,subspace?,classification). First prefer [zero_missing,[0,0]]; otherwise use [scalar_failure,scalar,point,image] from the first nonzero point in input order and the first missing integer multiple with scalar 2,3,... . A null witness denotes the singleton zero_subspace.","answer":{"kind":"expression","value":"([[0, 0], [15, 1]], [scalar_failure, 2, [15, 1], [30, 2]], No, not_subspace_exact_counterexample)","explanation":"Missing zero immediately refutes a subspace. If zero is present but a nonzero point occurs, distinct nonzero integer multiples cannot all belong to this finite set, yielding an exact scalar-closure counterexample. Do not infer an arbitrary subspace from finitely many positive samples. The exact result, retaining the requested order, is ([[0, 0], [15, 1]], [scalar_failure, 2, [15, 1], [30, 2]], No, not_subspace_exact_counterexample)."},"skills":["A Finite Nonzero Real Vector Set Cannot Pass Universal Scalar Closure"],"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-20261101-r01-vector-subspace-0066","batch_id":"batch-20261014-advance-20261101-r01-vector-subspace-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"03bf5eaf7c639b01732139c10c309a54d505a5cc722a7fc07b9847fdbba62d82","family_id":"qf-mod32-subspace-finite-set-counterexample","seed":202611011066,"semantic_template":"qf-mod32-subspace-finite-set-counterexample/{\"points\":[[0,0],[16,1]]}"},"fields":{"title":"A Finite Nonzero Real Vector Set Cannot Pass Universal Scalar Closure","stem":"Use the complete nonempty duplicate-free finite point set=[[0, 0], [16, 1]] as a candidate subspace of R² with usual operations. Give (points,witness,subspace?,classification). First prefer [zero_missing,[0,0]]; otherwise use [scalar_failure,scalar,point,image] from the first nonzero point in input order and the first missing integer multiple with scalar 2,3,... . A null witness denotes the singleton zero_subspace.","answer":{"kind":"expression","value":"([[0, 0], [16, 1]], [scalar_failure, 2, [16, 1], [32, 2]], No, not_subspace_exact_counterexample)","explanation":"Missing zero immediately refutes a subspace. If zero is present but a nonzero point occurs, distinct nonzero integer multiples cannot all belong to this finite set, yielding an exact scalar-closure counterexample. Do not infer an arbitrary subspace from finitely many positive samples. The exact result, retaining the requested order, is ([[0, 0], [16, 1]], [scalar_failure, 2, [16, 1], [32, 2]], No, not_subspace_exact_counterexample)."},"skills":["A Finite Nonzero Real Vector Set Cannot Pass Universal Scalar Closure"],"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-20261101-r01-vector-subspace-0067","batch_id":"batch-20261014-advance-20261101-r01-vector-subspace-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"bf90474a71e3ae62ad7964d08eea284ae30581ce324b539ce5710ca37ec4357b","family_id":"qf-mod32-subspace-finite-set-counterexample","seed":202611011067,"semantic_template":"qf-mod32-subspace-finite-set-counterexample/{\"points\":[[0,0],[17,1]]}"},"fields":{"title":"A Finite Nonzero Real Vector Set Cannot Pass Universal Scalar Closure","stem":"Use the complete nonempty duplicate-free finite point set=[[0, 0], [17, 1]] as a candidate subspace of R² with usual operations. Give (points,witness,subspace?,classification). First prefer [zero_missing,[0,0]]; otherwise use [scalar_failure,scalar,point,image] from the first nonzero point in input order and the first missing integer multiple with scalar 2,3,... . A null witness denotes the singleton zero_subspace.","answer":{"kind":"expression","value":"([[0, 0], [17, 1]], [scalar_failure, 2, [17, 1], [34, 2]], No, not_subspace_exact_counterexample)","explanation":"Missing zero immediately refutes a subspace. If zero is present but a nonzero point occurs, distinct nonzero integer multiples cannot all belong to this finite set, yielding an exact scalar-closure counterexample. Do not infer an arbitrary subspace from finitely many positive samples. The exact result, retaining the requested order, is ([[0, 0], [17, 1]], [scalar_failure, 2, [17, 1], [34, 2]], No, not_subspace_exact_counterexample)."},"skills":["A Finite Nonzero Real Vector Set Cannot Pass Universal Scalar Closure"],"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-20261101-r01-vector-subspace-0068","batch_id":"batch-20261014-advance-20261101-r01-vector-subspace-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"eadeab968971d201a33b19925b7bcc8adf1210b43f88d4423fe8f45d3bb746d4","family_id":"qf-mod32-subspace-finite-set-counterexample","seed":202611011068,"semantic_template":"qf-mod32-subspace-finite-set-counterexample/{\"points\":[[0,0],[18,1]]}"},"fields":{"title":"A Finite Nonzero Real Vector Set Cannot Pass Universal Scalar Closure","stem":"Use the complete nonempty duplicate-free finite point set=[[0, 0], [18, 1]] as a candidate subspace of R² with usual operations. Give (points,witness,subspace?,classification). First prefer [zero_missing,[0,0]]; otherwise use [scalar_failure,scalar,point,image] from the first nonzero point in input order and the first missing integer multiple with scalar 2,3,... . A null witness denotes the singleton zero_subspace.","answer":{"kind":"expression","value":"([[0, 0], [18, 1]], [scalar_failure, 2, [18, 1], [36, 2]], No, not_subspace_exact_counterexample)","explanation":"Missing zero immediately refutes a subspace. If zero is present but a nonzero point occurs, distinct nonzero integer multiples cannot all belong to this finite set, yielding an exact scalar-closure counterexample. Do not infer an arbitrary subspace from finitely many positive samples. The exact result, retaining the requested order, is ([[0, 0], [18, 1]], [scalar_failure, 2, [18, 1], [36, 2]], No, not_subspace_exact_counterexample)."},"skills":["A Finite Nonzero Real Vector Set Cannot Pass Universal Scalar Closure"],"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-20261101-r01-vector-subspace-0069","batch_id":"batch-20261014-advance-20261101-r01-vector-subspace-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"6607ef6a0606f6644da9170e361ae7b29e26bcc2cf97665d9a2e6dca8fb75e9d","family_id":"qf-mod32-subspace-finite-set-counterexample","seed":202611011069,"semantic_template":"qf-mod32-subspace-finite-set-counterexample/{\"points\":[[0,0],[19,1]]}"},"fields":{"title":"A Finite Nonzero Real Vector Set Cannot Pass Universal Scalar Closure","stem":"Use the complete nonempty duplicate-free finite point set=[[0, 0], [19, 1]] as a candidate subspace of R² with usual operations. Give (points,witness,subspace?,classification). First prefer [zero_missing,[0,0]]; otherwise use [scalar_failure,scalar,point,image] from the first nonzero point in input order and the first missing integer multiple with scalar 2,3,... . A null witness denotes the singleton zero_subspace.","answer":{"kind":"expression","value":"([[0, 0], [19, 1]], [scalar_failure, 2, [19, 1], [38, 2]], No, not_subspace_exact_counterexample)","explanation":"Missing zero immediately refutes a subspace. If zero is present but a nonzero point occurs, distinct nonzero integer multiples cannot all belong to this finite set, yielding an exact scalar-closure counterexample. Do not infer an arbitrary subspace from finitely many positive samples. The exact result, retaining the requested order, is ([[0, 0], [19, 1]], [scalar_failure, 2, [19, 1], [38, 2]], No, not_subspace_exact_counterexample)."},"skills":["A Finite Nonzero Real Vector Set Cannot Pass Universal Scalar Closure"],"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-20261101-r01-vector-subspace-0070","batch_id":"batch-20261014-advance-20261101-r01-vector-subspace-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"f141c2f399819b1577073aa4b0c7d26b918727e42ed2d7b69950d18b43a551ba","family_id":"qf-mod32-subspace-finite-set-counterexample","seed":202611011070,"semantic_template":"qf-mod32-subspace-finite-set-counterexample/{\"points\":[[0,0],[20,1]]}"},"fields":{"title":"A Finite Nonzero Real Vector Set Cannot Pass Universal Scalar Closure","stem":"Use the complete nonempty duplicate-free finite point set=[[0, 0], [20, 1]] as a candidate subspace of R² with usual operations. Give (points,witness,subspace?,classification). First prefer [zero_missing,[0,0]]; otherwise use [scalar_failure,scalar,point,image] from the first nonzero point in input order and the first missing integer multiple with scalar 2,3,... . A null witness denotes the singleton zero_subspace.","answer":{"kind":"expression","value":"([[0, 0], [20, 1]], [scalar_failure, 2, [20, 1], [40, 2]], No, not_subspace_exact_counterexample)","explanation":"Missing zero immediately refutes a subspace. If zero is present but a nonzero point occurs, distinct nonzero integer multiples cannot all belong to this finite set, yielding an exact scalar-closure counterexample. Do not infer an arbitrary subspace from finitely many positive samples. The exact result, retaining the requested order, is ([[0, 0], [20, 1]], [scalar_failure, 2, [20, 1], [40, 2]], No, not_subspace_exact_counterexample)."},"skills":["A Finite Nonzero Real Vector Set Cannot Pass Universal Scalar Closure"],"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-20261101-r01-vector-subspace-0071","batch_id":"batch-20261014-advance-20261101-r01-vector-subspace-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"bd932e1bc1c7ea01b87250faa61adb2d32ed5ce880ed107f55866f69490890a1","family_id":"qf-mod32-subspace-finite-set-counterexample","seed":202611011071,"semantic_template":"qf-mod32-subspace-finite-set-counterexample/{\"points\":[[0,0],[21,1]]}"},"fields":{"title":"A Finite Nonzero Real Vector Set Cannot Pass Universal Scalar Closure","stem":"Use the complete nonempty duplicate-free finite point set=[[0, 0], [21, 1]] as a candidate subspace of R² with usual operations. Give (points,witness,subspace?,classification). First prefer [zero_missing,[0,0]]; otherwise use [scalar_failure,scalar,point,image] from the first nonzero point in input order and the first missing integer multiple with scalar 2,3,... . A null witness denotes the singleton zero_subspace.","answer":{"kind":"expression","value":"([[0, 0], [21, 1]], [scalar_failure, 2, [21, 1], [42, 2]], No, not_subspace_exact_counterexample)","explanation":"Missing zero immediately refutes a subspace. If zero is present but a nonzero point occurs, distinct nonzero integer multiples cannot all belong to this finite set, yielding an exact scalar-closure counterexample. Do not infer an arbitrary subspace from finitely many positive samples. The exact result, retaining the requested order, is ([[0, 0], [21, 1]], [scalar_failure, 2, [21, 1], [42, 2]], No, not_subspace_exact_counterexample)."},"skills":["A Finite Nonzero Real Vector Set Cannot Pass Universal Scalar Closure"],"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-20261101-r01-vector-subspace-0072","batch_id":"batch-20261014-advance-20261101-r01-vector-subspace-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"ff43c15821eec1aa32afad1500154811b1e253185cb51471ddcffa14be76a080","family_id":"qf-mod32-subspace-finite-set-counterexample","seed":202611011072,"semantic_template":"qf-mod32-subspace-finite-set-counterexample/{\"points\":[[0,0],[22,1]]}"},"fields":{"title":"A Finite Nonzero Real Vector Set Cannot Pass Universal Scalar Closure","stem":"Use the complete nonempty duplicate-free finite point set=[[0, 0], [22, 1]] as a candidate subspace of R² with usual operations. Give (points,witness,subspace?,classification). First prefer [zero_missing,[0,0]]; otherwise use [scalar_failure,scalar,point,image] from the first nonzero point in input order and the first missing integer multiple with scalar 2,3,... . A null witness denotes the singleton zero_subspace.","answer":{"kind":"expression","value":"([[0, 0], [22, 1]], [scalar_failure, 2, [22, 1], [44, 2]], No, not_subspace_exact_counterexample)","explanation":"Missing zero immediately refutes a subspace. If zero is present but a nonzero point occurs, distinct nonzero integer multiples cannot all belong to this finite set, yielding an exact scalar-closure counterexample. Do not infer an arbitrary subspace from finitely many positive samples. The exact result, retaining the requested order, is ([[0, 0], [22, 1]], [scalar_failure, 2, [22, 1], [44, 2]], No, not_subspace_exact_counterexample)."},"skills":["A Finite Nonzero Real Vector Set Cannot Pass Universal Scalar Closure"],"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-20261101-r01-vector-subspace-0073","batch_id":"batch-20261014-advance-20261101-r01-vector-subspace-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"c05dfdf89e51f4d5fe8d6fcd35adab18b7dcfe2a0cbea999786c0c084c4e875d","family_id":"qf-mod32-subspace-finite-set-counterexample","seed":202611011073,"semantic_template":"qf-mod32-subspace-finite-set-counterexample/{\"points\":[[0,0],[23,1]]}"},"fields":{"title":"A Finite Nonzero Real Vector Set Cannot Pass Universal Scalar Closure","stem":"Use the complete nonempty duplicate-free finite point set=[[0, 0], [23, 1]] as a candidate subspace of R² with usual operations. Give (points,witness,subspace?,classification). First prefer [zero_missing,[0,0]]; otherwise use [scalar_failure,scalar,point,image] from the first nonzero point in input order and the first missing integer multiple with scalar 2,3,... . A null witness denotes the singleton zero_subspace.","answer":{"kind":"expression","value":"([[0, 0], [23, 1]], [scalar_failure, 2, [23, 1], [46, 2]], No, not_subspace_exact_counterexample)","explanation":"Missing zero immediately refutes a subspace. If zero is present but a nonzero point occurs, distinct nonzero integer multiples cannot all belong to this finite set, yielding an exact scalar-closure counterexample. Do not infer an arbitrary subspace from finitely many positive samples. The exact result, retaining the requested order, is ([[0, 0], [23, 1]], [scalar_failure, 2, [23, 1], [46, 2]], No, not_subspace_exact_counterexample)."},"skills":["A Finite Nonzero Real Vector Set Cannot Pass Universal Scalar Closure"],"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-20261101-r01-vector-subspace-0074","batch_id":"batch-20261014-advance-20261101-r01-vector-subspace-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"63ad6892221fafd87b25d3edd59f544eba0acca832bab11af2100e44b3633aae","family_id":"qf-mod32-subspace-finite-set-counterexample","seed":202611011074,"semantic_template":"qf-mod32-subspace-finite-set-counterexample/{\"points\":[[0,0],[24,1]]}"},"fields":{"title":"A Finite Nonzero Real Vector Set Cannot Pass Universal Scalar Closure","stem":"Use the complete nonempty duplicate-free finite point set=[[0, 0], [24, 1]] as a candidate subspace of R² with usual operations. Give (points,witness,subspace?,classification). First prefer [zero_missing,[0,0]]; otherwise use [scalar_failure,scalar,point,image] from the first nonzero point in input order and the first missing integer multiple with scalar 2,3,... . A null witness denotes the singleton zero_subspace.","answer":{"kind":"expression","value":"([[0, 0], [24, 1]], [scalar_failure, 2, [24, 1], [48, 2]], No, not_subspace_exact_counterexample)","explanation":"Missing zero immediately refutes a subspace. If zero is present but a nonzero point occurs, distinct nonzero integer multiples cannot all belong to this finite set, yielding an exact scalar-closure counterexample. Do not infer an arbitrary subspace from finitely many positive samples. The exact result, retaining the requested order, is ([[0, 0], [24, 1]], [scalar_failure, 2, [24, 1], [48, 2]], No, not_subspace_exact_counterexample)."},"skills":["A Finite Nonzero Real Vector Set Cannot Pass Universal Scalar Closure"],"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-20261101-r01-vector-subspace-0075","batch_id":"batch-20261014-advance-20261101-r01-vector-subspace-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"12be1e2eefe5ae21d8d18bf196c9fd870d63f5247d3d02c899684aa45f6ae410","family_id":"qf-mod32-subspace-finite-set-counterexample","seed":202611011075,"semantic_template":"qf-mod32-subspace-finite-set-counterexample/{\"points\":[[0,0],[25,1]]}"},"fields":{"title":"A Finite Nonzero Real Vector Set Cannot Pass Universal Scalar Closure","stem":"Use the complete nonempty duplicate-free finite point set=[[0, 0], [25, 1]] as a candidate subspace of R² with usual operations. Give (points,witness,subspace?,classification). First prefer [zero_missing,[0,0]]; otherwise use [scalar_failure,scalar,point,image] from the first nonzero point in input order and the first missing integer multiple with scalar 2,3,... . A null witness denotes the singleton zero_subspace.","answer":{"kind":"expression","value":"([[0, 0], [25, 1]], [scalar_failure, 2, [25, 1], [50, 2]], No, not_subspace_exact_counterexample)","explanation":"Missing zero immediately refutes a subspace. If zero is present but a nonzero point occurs, distinct nonzero integer multiples cannot all belong to this finite set, yielding an exact scalar-closure counterexample. Do not infer an arbitrary subspace from finitely many positive samples. The exact result, retaining the requested order, is ([[0, 0], [25, 1]], [scalar_failure, 2, [25, 1], [50, 2]], No, not_subspace_exact_counterexample)."},"skills":["A Finite Nonzero Real Vector Set Cannot Pass Universal Scalar Closure"],"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-20261101-r01-vector-subspace-0076","batch_id":"batch-20261014-advance-20261101-r01-vector-subspace-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"763e37ae0073d2e077e6616ed10383bc4162d11f6d0af314d46e6e11a6e69d42","family_id":"qf-mod32-subspace-classification","seed":202611011076,"semantic_template":"qf-mod32-subspace-classification/{\"a\":0,\"b\":1,\"c\":-12}"},"fields":{"title":"Separate the Whole Space, Homogeneous Line, Affine Line and Empty Set","stem":"In R² with usual operations consider a*x+b*y=c with [a,b,c]=[0, 1, -12]. Give (equation,classification,subspace?,certificate). Classification is whole_R2, empty_set, homogeneous_line or affine_line_not_through_zero.","answer":{"kind":"expression","value":"([0, 1, -12], affine_line_not_through_zero, No, affine_line_not_through_zero)","explanation":"If both coefficients vanish, c=0 gives the whole space and c≠0 gives the empty set. Otherwise c=0 gives a homogeneous line with a universal linear-equation certificate, and nonzero c gives an affine line missing zero. The exact result, retaining the requested order, is ([0, 1, -12], affine_line_not_through_zero, No, affine_line_not_through_zero)."},"skills":["Separate the Whole Space, Homogeneous Line, Affine Line and Empty Set"],"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-20261101-r01-vector-subspace-0077","batch_id":"batch-20261014-advance-20261101-r01-vector-subspace-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"f301493dd9f059ceb89e1813023df0a843d3e78c63d4871d927cf2b860fe34b0","family_id":"qf-mod32-subspace-classification","seed":202611011077,"semantic_template":"qf-mod32-subspace-classification/{\"a\":1,\"b\":2,\"c\":-11}"},"fields":{"title":"Separate the Whole Space, Homogeneous Line, Affine Line and Empty Set","stem":"In R² with usual operations consider a*x+b*y=c with [a,b,c]=[1, 2, -11]. Give (equation,classification,subspace?,certificate). Classification is whole_R2, empty_set, homogeneous_line or affine_line_not_through_zero.","answer":{"kind":"expression","value":"([1, 2, -11], affine_line_not_through_zero, No, affine_line_not_through_zero)","explanation":"If both coefficients vanish, c=0 gives the whole space and c≠0 gives the empty set. Otherwise c=0 gives a homogeneous line with a universal linear-equation certificate, and nonzero c gives an affine line missing zero. The exact result, retaining the requested order, is ([1, 2, -11], affine_line_not_through_zero, No, affine_line_not_through_zero)."},"skills":["Separate the Whole Space, Homogeneous Line, Affine Line and Empty Set"],"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-20261101-r01-vector-subspace-0078","batch_id":"batch-20261014-advance-20261101-r01-vector-subspace-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"af691dde834e720960d1f102b356490387bb0ee1dc378c12e20f5b621cbf2370","family_id":"qf-mod32-subspace-classification","seed":202611011078,"semantic_template":"qf-mod32-subspace-classification/{\"a\":2,\"b\":3,\"c\":-10}"},"fields":{"title":"Separate the Whole Space, Homogeneous Line, Affine Line and Empty Set","stem":"In R² with usual operations consider a*x+b*y=c with [a,b,c]=[2, 3, -10]. Give (equation,classification,subspace?,certificate). Classification is whole_R2, empty_set, homogeneous_line or affine_line_not_through_zero.","answer":{"kind":"expression","value":"([2, 3, -10], affine_line_not_through_zero, No, affine_line_not_through_zero)","explanation":"If both coefficients vanish, c=0 gives the whole space and c≠0 gives the empty set. Otherwise c=0 gives a homogeneous line with a universal linear-equation certificate, and nonzero c gives an affine line missing zero. The exact result, retaining the requested order, is ([2, 3, -10], affine_line_not_through_zero, No, affine_line_not_through_zero)."},"skills":["Separate the Whole Space, Homogeneous Line, Affine Line and Empty Set"],"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-20261101-r01-vector-subspace-0079","batch_id":"batch-20261014-advance-20261101-r01-vector-subspace-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"c3de809832b4a72ac6a28ffe546c05e89ab30683cda0b59a71fd878ba00ea817","family_id":"qf-mod32-subspace-classification","seed":202611011079,"semantic_template":"qf-mod32-subspace-classification/{\"a\":3,\"b\":4,\"c\":-9}"},"fields":{"title":"Separate the Whole Space, Homogeneous Line, Affine Line and Empty Set","stem":"In R² with usual operations consider a*x+b*y=c with [a,b,c]=[3, 4, -9]. Give (equation,classification,subspace?,certificate). Classification is whole_R2, empty_set, homogeneous_line or affine_line_not_through_zero.","answer":{"kind":"expression","value":"([3, 4, -9], affine_line_not_through_zero, No, affine_line_not_through_zero)","explanation":"If both coefficients vanish, c=0 gives the whole space and c≠0 gives the empty set. Otherwise c=0 gives a homogeneous line with a universal linear-equation certificate, and nonzero c gives an affine line missing zero. The exact result, retaining the requested order, is ([3, 4, -9], affine_line_not_through_zero, No, affine_line_not_through_zero)."},"skills":["Separate the Whole Space, Homogeneous Line, Affine Line and Empty Set"],"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-20261101-r01-vector-subspace-0080","batch_id":"batch-20261014-advance-20261101-r01-vector-subspace-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"0501354d04d10652f4392600cb576803ab487d2ef977ecf030bc77d2678931af","family_id":"qf-mod32-subspace-classification","seed":202611011080,"semantic_template":"qf-mod32-subspace-classification/{\"a\":4,\"b\":5,\"c\":-8}"},"fields":{"title":"Separate the Whole Space, Homogeneous Line, Affine Line and Empty Set","stem":"In R² with usual operations consider a*x+b*y=c with [a,b,c]=[4, 5, -8]. Give (equation,classification,subspace?,certificate). Classification is whole_R2, empty_set, homogeneous_line or affine_line_not_through_zero.","answer":{"kind":"expression","value":"([4, 5, -8], affine_line_not_through_zero, No, affine_line_not_through_zero)","explanation":"If both coefficients vanish, c=0 gives the whole space and c≠0 gives the empty set. Otherwise c=0 gives a homogeneous line with a universal linear-equation certificate, and nonzero c gives an affine line missing zero. The exact result, retaining the requested order, is ([4, 5, -8], affine_line_not_through_zero, No, affine_line_not_through_zero)."},"skills":["Separate the Whole Space, Homogeneous Line, Affine Line and Empty Set"],"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-20261101-r01-vector-subspace-0081","batch_id":"batch-20261014-advance-20261101-r01-vector-subspace-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"253dcd25da77b1b5e814627c6a0cfb5b795499fda835376dabb8cc3d6096fdc6","family_id":"qf-mod32-subspace-classification","seed":202611011081,"semantic_template":"qf-mod32-subspace-classification/{\"a\":0,\"b\":6,\"c\":-7}"},"fields":{"title":"Separate the Whole Space, Homogeneous Line, Affine Line and Empty Set","stem":"In R² with usual operations consider a*x+b*y=c with [a,b,c]=[0, 6, -7]. Give (equation,classification,subspace?,certificate). Classification is whole_R2, empty_set, homogeneous_line or affine_line_not_through_zero.","answer":{"kind":"expression","value":"([0, 6, -7], affine_line_not_through_zero, No, affine_line_not_through_zero)","explanation":"If both coefficients vanish, c=0 gives the whole space and c≠0 gives the empty set. Otherwise c=0 gives a homogeneous line with a universal linear-equation certificate, and nonzero c gives an affine line missing zero. The exact result, retaining the requested order, is ([0, 6, -7], affine_line_not_through_zero, No, affine_line_not_through_zero)."},"skills":["Separate the Whole Space, Homogeneous Line, Affine Line and Empty Set"],"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-20261101-r01-vector-subspace-0082","batch_id":"batch-20261014-advance-20261101-r01-vector-subspace-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"7b9f609815e0b2a3e96e440b2f5abfbe9d781db91dafa99d582bdf7f44802319","family_id":"qf-mod32-subspace-classification","seed":202611011082,"semantic_template":"qf-mod32-subspace-classification/{\"a\":1,\"b\":0,\"c\":-6}"},"fields":{"title":"Separate the Whole Space, Homogeneous Line, Affine Line and Empty Set","stem":"In R² with usual operations consider a*x+b*y=c with [a,b,c]=[1, 0, -6]. Give (equation,classification,subspace?,certificate). Classification is whole_R2, empty_set, homogeneous_line or affine_line_not_through_zero.","answer":{"kind":"expression","value":"([1, 0, -6], affine_line_not_through_zero, No, affine_line_not_through_zero)","explanation":"If both coefficients vanish, c=0 gives the whole space and c≠0 gives the empty set. Otherwise c=0 gives a homogeneous line with a universal linear-equation certificate, and nonzero c gives an affine line missing zero. The exact result, retaining the requested order, is ([1, 0, -6], affine_line_not_through_zero, No, affine_line_not_through_zero)."},"skills":["Separate the Whole Space, Homogeneous Line, Affine Line and Empty Set"],"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-20261101-r01-vector-subspace-0083","batch_id":"batch-20261014-advance-20261101-r01-vector-subspace-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"b6459045d785011ac70982c172f74171f354ca70ff2ecc4b6489673ad00b577e","family_id":"qf-mod32-subspace-classification","seed":202611011083,"semantic_template":"qf-mod32-subspace-classification/{\"a\":2,\"b\":1,\"c\":-5}"},"fields":{"title":"Separate the Whole Space, Homogeneous Line, Affine Line and Empty Set","stem":"In R² with usual operations consider a*x+b*y=c with [a,b,c]=[2, 1, -5]. Give (equation,classification,subspace?,certificate). Classification is whole_R2, empty_set, homogeneous_line or affine_line_not_through_zero.","answer":{"kind":"expression","value":"([2, 1, -5], affine_line_not_through_zero, No, affine_line_not_through_zero)","explanation":"If both coefficients vanish, c=0 gives the whole space and c≠0 gives the empty set. Otherwise c=0 gives a homogeneous line with a universal linear-equation certificate, and nonzero c gives an affine line missing zero. The exact result, retaining the requested order, is ([2, 1, -5], affine_line_not_through_zero, No, affine_line_not_through_zero)."},"skills":["Separate the Whole Space, Homogeneous Line, Affine Line and Empty Set"],"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-20261101-r01-vector-subspace-0084","batch_id":"batch-20261014-advance-20261101-r01-vector-subspace-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"f6e2eb6b12c375ea818215b39370b785eb35ebecda25ff46c1ae24b9ec512369","family_id":"qf-mod32-subspace-classification","seed":202611011084,"semantic_template":"qf-mod32-subspace-classification/{\"a\":3,\"b\":2,\"c\":-4}"},"fields":{"title":"Separate the Whole Space, Homogeneous Line, Affine Line and Empty Set","stem":"In R² with usual operations consider a*x+b*y=c with [a,b,c]=[3, 2, -4]. Give (equation,classification,subspace?,certificate). Classification is whole_R2, empty_set, homogeneous_line or affine_line_not_through_zero.","answer":{"kind":"expression","value":"([3, 2, -4], affine_line_not_through_zero, No, affine_line_not_through_zero)","explanation":"If both coefficients vanish, c=0 gives the whole space and c≠0 gives the empty set. Otherwise c=0 gives a homogeneous line with a universal linear-equation certificate, and nonzero c gives an affine line missing zero. The exact result, retaining the requested order, is ([3, 2, -4], affine_line_not_through_zero, No, affine_line_not_through_zero)."},"skills":["Separate the Whole Space, Homogeneous Line, Affine Line and Empty Set"],"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-20261101-r01-vector-subspace-0085","batch_id":"batch-20261014-advance-20261101-r01-vector-subspace-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"9d6aed8184f7bc9b645a52125909958242312609ae72447924fa198a45458efb","family_id":"qf-mod32-subspace-classification","seed":202611011085,"semantic_template":"qf-mod32-subspace-classification/{\"a\":4,\"b\":3,\"c\":-3}"},"fields":{"title":"Separate the Whole Space, Homogeneous Line, Affine Line and Empty Set","stem":"In R² with usual operations consider a*x+b*y=c with [a,b,c]=[4, 3, -3]. Give (equation,classification,subspace?,certificate). Classification is whole_R2, empty_set, homogeneous_line or affine_line_not_through_zero.","answer":{"kind":"expression","value":"([4, 3, -3], affine_line_not_through_zero, No, affine_line_not_through_zero)","explanation":"If both coefficients vanish, c=0 gives the whole space and c≠0 gives the empty set. Otherwise c=0 gives a homogeneous line with a universal linear-equation certificate, and nonzero c gives an affine line missing zero. The exact result, retaining the requested order, is ([4, 3, -3], affine_line_not_through_zero, No, affine_line_not_through_zero)."},"skills":["Separate the Whole Space, Homogeneous Line, Affine Line and Empty Set"],"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-20261101-r01-vector-subspace-0086","batch_id":"batch-20261014-advance-20261101-r01-vector-subspace-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"4497ff9fc3ad07e3d7841e4700f6bac0ba671bd3ee4fe519fc0e89ae4023382a","family_id":"qf-mod32-subspace-classification","seed":202611011086,"semantic_template":"qf-mod32-subspace-classification/{\"a\":0,\"b\":4,\"c\":-2}"},"fields":{"title":"Separate the Whole Space, Homogeneous Line, Affine Line and Empty Set","stem":"In R² with usual operations consider a*x+b*y=c with [a,b,c]=[0, 4, -2]. Give (equation,classification,subspace?,certificate). Classification is whole_R2, empty_set, homogeneous_line or affine_line_not_through_zero.","answer":{"kind":"expression","value":"([0, 4, -2], affine_line_not_through_zero, No, affine_line_not_through_zero)","explanation":"If both coefficients vanish, c=0 gives the whole space and c≠0 gives the empty set. Otherwise c=0 gives a homogeneous line with a universal linear-equation certificate, and nonzero c gives an affine line missing zero. The exact result, retaining the requested order, is ([0, 4, -2], affine_line_not_through_zero, No, affine_line_not_through_zero)."},"skills":["Separate the Whole Space, Homogeneous Line, Affine Line and Empty Set"],"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-20261101-r01-vector-subspace-0087","batch_id":"batch-20261014-advance-20261101-r01-vector-subspace-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"5dab2357e8e1d3d969c7b21ca53f6f385f302b349f3fbf2a16d263d75ae46927","family_id":"qf-mod32-subspace-classification","seed":202611011087,"semantic_template":"qf-mod32-subspace-classification/{\"a\":1,\"b\":5,\"c\":-1}"},"fields":{"title":"Separate the Whole Space, Homogeneous Line, Affine Line and Empty Set","stem":"In R² with usual operations consider a*x+b*y=c with [a,b,c]=[1, 5, -1]. Give (equation,classification,subspace?,certificate). Classification is whole_R2, empty_set, homogeneous_line or affine_line_not_through_zero.","answer":{"kind":"expression","value":"([1, 5, -1], affine_line_not_through_zero, No, affine_line_not_through_zero)","explanation":"If both coefficients vanish, c=0 gives the whole space and c≠0 gives the empty set. Otherwise c=0 gives a homogeneous line with a universal linear-equation certificate, and nonzero c gives an affine line missing zero. The exact result, retaining the requested order, is ([1, 5, -1], affine_line_not_through_zero, No, affine_line_not_through_zero)."},"skills":["Separate the Whole Space, Homogeneous Line, Affine Line and Empty Set"],"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-20261101-r01-vector-subspace-0088","batch_id":"batch-20261014-advance-20261101-r01-vector-subspace-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"eb346d252f6f832905133845244e558c5277336ad30431446186ce5a26cc88fe","family_id":"qf-mod32-subspace-classification","seed":202611011088,"semantic_template":"qf-mod32-subspace-classification/{\"a\":2,\"b\":6,\"c\":0}"},"fields":{"title":"Separate the Whole Space, Homogeneous Line, Affine Line and Empty Set","stem":"In R² with usual operations consider a*x+b*y=c with [a,b,c]=[2, 6, 0]. Give (equation,classification,subspace?,certificate). Classification is whole_R2, empty_set, homogeneous_line or affine_line_not_through_zero.","answer":{"kind":"expression","value":"([2, 6, 0], homogeneous_line, Yes, universal_homogeneous_equation_certificate)","explanation":"If both coefficients vanish, c=0 gives the whole space and c≠0 gives the empty set. Otherwise c=0 gives a homogeneous line with a universal linear-equation certificate, and nonzero c gives an affine line missing zero. The exact result, retaining the requested order, is ([2, 6, 0], homogeneous_line, Yes, universal_homogeneous_equation_certificate)."},"skills":["Separate the Whole Space, Homogeneous Line, Affine Line and Empty Set"],"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-20261101-r01-vector-subspace-0089","batch_id":"batch-20261014-advance-20261101-r01-vector-subspace-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"9a1a0ffb6757298c4cb9bd7e238951d87638ae024c5593c30389d7fe6c5a2b08","family_id":"qf-mod32-subspace-classification","seed":202611011089,"semantic_template":"qf-mod32-subspace-classification/{\"a\":3,\"b\":0,\"c\":1}"},"fields":{"title":"Separate the Whole Space, Homogeneous Line, Affine Line and Empty Set","stem":"In R² with usual operations consider a*x+b*y=c with [a,b,c]=[3, 0, 1]. Give (equation,classification,subspace?,certificate). Classification is whole_R2, empty_set, homogeneous_line or affine_line_not_through_zero.","answer":{"kind":"expression","value":"([3, 0, 1], affine_line_not_through_zero, No, affine_line_not_through_zero)","explanation":"If both coefficients vanish, c=0 gives the whole space and c≠0 gives the empty set. Otherwise c=0 gives a homogeneous line with a universal linear-equation certificate, and nonzero c gives an affine line missing zero. The exact result, retaining the requested order, is ([3, 0, 1], affine_line_not_through_zero, No, affine_line_not_through_zero)."},"skills":["Separate the Whole Space, Homogeneous Line, Affine Line and Empty Set"],"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-20261101-r01-vector-subspace-0090","batch_id":"batch-20261014-advance-20261101-r01-vector-subspace-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"43338b1166d61d61008d2991c6dcc5665e45d9c1f6975471d2dce72b1b5f258d","family_id":"qf-mod32-subspace-classification","seed":202611011090,"semantic_template":"qf-mod32-subspace-classification/{\"a\":4,\"b\":1,\"c\":2}"},"fields":{"title":"Separate the Whole Space, Homogeneous Line, Affine Line and Empty Set","stem":"In R² with usual operations consider a*x+b*y=c with [a,b,c]=[4, 1, 2]. Give (equation,classification,subspace?,certificate). Classification is whole_R2, empty_set, homogeneous_line or affine_line_not_through_zero.","answer":{"kind":"expression","value":"([4, 1, 2], affine_line_not_through_zero, No, affine_line_not_through_zero)","explanation":"If both coefficients vanish, c=0 gives the whole space and c≠0 gives the empty set. Otherwise c=0 gives a homogeneous line with a universal linear-equation certificate, and nonzero c gives an affine line missing zero. The exact result, retaining the requested order, is ([4, 1, 2], affine_line_not_through_zero, No, affine_line_not_through_zero)."},"skills":["Separate the Whole Space, Homogeneous Line, Affine Line and Empty Set"],"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-20261101-r01-vector-subspace-0091","batch_id":"batch-20261014-advance-20261101-r01-vector-subspace-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"fcde30a46a4566f43d21229e97c02e3dde6d17761ce0f3e90f7999367395b59f","family_id":"qf-mod32-subspace-classification","seed":202611011091,"semantic_template":"qf-mod32-subspace-classification/{\"a\":0,\"b\":2,\"c\":3}"},"fields":{"title":"Separate the Whole Space, Homogeneous Line, Affine Line and Empty Set","stem":"In R² with usual operations consider a*x+b*y=c with [a,b,c]=[0, 2, 3]. Give (equation,classification,subspace?,certificate). Classification is whole_R2, empty_set, homogeneous_line or affine_line_not_through_zero.","answer":{"kind":"expression","value":"([0, 2, 3], affine_line_not_through_zero, No, affine_line_not_through_zero)","explanation":"If both coefficients vanish, c=0 gives the whole space and c≠0 gives the empty set. Otherwise c=0 gives a homogeneous line with a universal linear-equation certificate, and nonzero c gives an affine line missing zero. The exact result, retaining the requested order, is ([0, 2, 3], affine_line_not_through_zero, No, affine_line_not_through_zero)."},"skills":["Separate the Whole Space, Homogeneous Line, Affine Line and Empty Set"],"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-20261101-r01-vector-subspace-0092","batch_id":"batch-20261014-advance-20261101-r01-vector-subspace-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"964e8a523dd3aedab6fc5d1ff472381023b177a6d4139e41c91c3d1ca31e1efa","family_id":"qf-mod32-subspace-classification","seed":202611011092,"semantic_template":"qf-mod32-subspace-classification/{\"a\":1,\"b\":3,\"c\":4}"},"fields":{"title":"Separate the Whole Space, Homogeneous Line, Affine Line and Empty Set","stem":"In R² with usual operations consider a*x+b*y=c with [a,b,c]=[1, 3, 4]. Give (equation,classification,subspace?,certificate). Classification is whole_R2, empty_set, homogeneous_line or affine_line_not_through_zero.","answer":{"kind":"expression","value":"([1, 3, 4], affine_line_not_through_zero, No, affine_line_not_through_zero)","explanation":"If both coefficients vanish, c=0 gives the whole space and c≠0 gives the empty set. Otherwise c=0 gives a homogeneous line with a universal linear-equation certificate, and nonzero c gives an affine line missing zero. The exact result, retaining the requested order, is ([1, 3, 4], affine_line_not_through_zero, No, affine_line_not_through_zero)."},"skills":["Separate the Whole Space, Homogeneous Line, Affine Line and Empty Set"],"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-20261101-r01-vector-subspace-0093","batch_id":"batch-20261014-advance-20261101-r01-vector-subspace-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"10b6726ac5faec436b2ea71fed28b9f52d2fbbee317e3de67388531540205087","family_id":"qf-mod32-subspace-classification","seed":202611011093,"semantic_template":"qf-mod32-subspace-classification/{\"a\":2,\"b\":4,\"c\":5}"},"fields":{"title":"Separate the Whole Space, Homogeneous Line, Affine Line and Empty Set","stem":"In R² with usual operations consider a*x+b*y=c with [a,b,c]=[2, 4, 5]. Give (equation,classification,subspace?,certificate). Classification is whole_R2, empty_set, homogeneous_line or affine_line_not_through_zero.","answer":{"kind":"expression","value":"([2, 4, 5], affine_line_not_through_zero, No, affine_line_not_through_zero)","explanation":"If both coefficients vanish, c=0 gives the whole space and c≠0 gives the empty set. Otherwise c=0 gives a homogeneous line with a universal linear-equation certificate, and nonzero c gives an affine line missing zero. The exact result, retaining the requested order, is ([2, 4, 5], affine_line_not_through_zero, No, affine_line_not_through_zero)."},"skills":["Separate the Whole Space, Homogeneous Line, Affine Line and Empty Set"],"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-20261101-r01-vector-subspace-0094","batch_id":"batch-20261014-advance-20261101-r01-vector-subspace-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"9a5e1b4f3d0d4de21e00903c38979f557471808ba59e9ebcc37416409e27132e","family_id":"qf-mod32-subspace-classification","seed":202611011094,"semantic_template":"qf-mod32-subspace-classification/{\"a\":3,\"b\":5,\"c\":6}"},"fields":{"title":"Separate the Whole Space, Homogeneous Line, Affine Line and Empty Set","stem":"In R² with usual operations consider a*x+b*y=c with [a,b,c]=[3, 5, 6]. Give (equation,classification,subspace?,certificate). Classification is whole_R2, empty_set, homogeneous_line or affine_line_not_through_zero.","answer":{"kind":"expression","value":"([3, 5, 6], affine_line_not_through_zero, No, affine_line_not_through_zero)","explanation":"If both coefficients vanish, c=0 gives the whole space and c≠0 gives the empty set. Otherwise c=0 gives a homogeneous line with a universal linear-equation certificate, and nonzero c gives an affine line missing zero. The exact result, retaining the requested order, is ([3, 5, 6], affine_line_not_through_zero, No, affine_line_not_through_zero)."},"skills":["Separate the Whole Space, Homogeneous Line, Affine Line and Empty Set"],"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-20261101-r01-vector-subspace-0095","batch_id":"batch-20261014-advance-20261101-r01-vector-subspace-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"41626fb77ab5b123e60e5477e4ce71cac12f200c9dfe78f78d0dff3fafe76e2e","family_id":"qf-mod32-subspace-classification","seed":202611011095,"semantic_template":"qf-mod32-subspace-classification/{\"a\":4,\"b\":6,\"c\":7}"},"fields":{"title":"Separate the Whole Space, Homogeneous Line, Affine Line and Empty Set","stem":"In R² with usual operations consider a*x+b*y=c with [a,b,c]=[4, 6, 7]. Give (equation,classification,subspace?,certificate). Classification is whole_R2, empty_set, homogeneous_line or affine_line_not_through_zero.","answer":{"kind":"expression","value":"([4, 6, 7], affine_line_not_through_zero, No, affine_line_not_through_zero)","explanation":"If both coefficients vanish, c=0 gives the whole space and c≠0 gives the empty set. Otherwise c=0 gives a homogeneous line with a universal linear-equation certificate, and nonzero c gives an affine line missing zero. The exact result, retaining the requested order, is ([4, 6, 7], affine_line_not_through_zero, No, affine_line_not_through_zero)."},"skills":["Separate the Whole Space, Homogeneous Line, Affine Line and Empty Set"],"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-20261101-r01-vector-subspace-0096","batch_id":"batch-20261014-advance-20261101-r01-vector-subspace-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"7e8c9efb6d33875f476f93c352dbf4fa9aa312141f4da95dade33070e7a4417d","family_id":"qf-mod32-subspace-classification","seed":202611011096,"semantic_template":"qf-mod32-subspace-classification/{\"a\":0,\"b\":0,\"c\":8}"},"fields":{"title":"Separate the Whole Space, Homogeneous Line, Affine Line and Empty Set","stem":"In R² with usual operations consider a*x+b*y=c with [a,b,c]=[0, 0, 8]. Give (equation,classification,subspace?,certificate). Classification is whole_R2, empty_set, homogeneous_line or affine_line_not_through_zero.","answer":{"kind":"expression","value":"([0, 0, 8], empty_set, No, empty_set)","explanation":"If both coefficients vanish, c=0 gives the whole space and c≠0 gives the empty set. Otherwise c=0 gives a homogeneous line with a universal linear-equation certificate, and nonzero c gives an affine line missing zero. The exact result, retaining the requested order, is ([0, 0, 8], empty_set, No, empty_set)."},"skills":["Separate the Whole Space, Homogeneous Line, Affine Line and Empty Set"],"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-20261101-r01-vector-subspace-0097","batch_id":"batch-20261014-advance-20261101-r01-vector-subspace-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"34756f5e6279f866fc070fdcbe8bfadeb02245f6c50cb6b30489a23ce6b2f2af","family_id":"qf-mod32-subspace-classification","seed":202611011097,"semantic_template":"qf-mod32-subspace-classification/{\"a\":1,\"b\":1,\"c\":9}"},"fields":{"title":"Separate the Whole Space, Homogeneous Line, Affine Line and Empty Set","stem":"In R² with usual operations consider a*x+b*y=c with [a,b,c]=[1, 1, 9]. Give (equation,classification,subspace?,certificate). Classification is whole_R2, empty_set, homogeneous_line or affine_line_not_through_zero.","answer":{"kind":"expression","value":"([1, 1, 9], affine_line_not_through_zero, No, affine_line_not_through_zero)","explanation":"If both coefficients vanish, c=0 gives the whole space and c≠0 gives the empty set. Otherwise c=0 gives a homogeneous line with a universal linear-equation certificate, and nonzero c gives an affine line missing zero. The exact result, retaining the requested order, is ([1, 1, 9], affine_line_not_through_zero, No, affine_line_not_through_zero)."},"skills":["Separate the Whole Space, Homogeneous Line, Affine Line and Empty Set"],"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-20261101-r01-vector-subspace-0098","batch_id":"batch-20261014-advance-20261101-r01-vector-subspace-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"c3bb52a7ddafbbe686c471896fc4f969adac904050212f9399e76075c8eb35b0","family_id":"qf-mod32-subspace-classification","seed":202611011098,"semantic_template":"qf-mod32-subspace-classification/{\"a\":2,\"b\":2,\"c\":10}"},"fields":{"title":"Separate the Whole Space, Homogeneous Line, Affine Line and Empty Set","stem":"In R² with usual operations consider a*x+b*y=c with [a,b,c]=[2, 2, 10]. Give (equation,classification,subspace?,certificate). Classification is whole_R2, empty_set, homogeneous_line or affine_line_not_through_zero.","answer":{"kind":"expression","value":"([2, 2, 10], affine_line_not_through_zero, No, affine_line_not_through_zero)","explanation":"If both coefficients vanish, c=0 gives the whole space and c≠0 gives the empty set. Otherwise c=0 gives a homogeneous line with a universal linear-equation certificate, and nonzero c gives an affine line missing zero. The exact result, retaining the requested order, is ([2, 2, 10], affine_line_not_through_zero, No, affine_line_not_through_zero)."},"skills":["Separate the Whole Space, Homogeneous Line, Affine Line and Empty Set"],"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-20261101-r01-vector-subspace-0099","batch_id":"batch-20261014-advance-20261101-r01-vector-subspace-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"69e55359fcee3fd06326c8a95b4bd870282fea4e15a083a5eaa2e5abc5681023","family_id":"qf-mod32-subspace-classification","seed":202611011099,"semantic_template":"qf-mod32-subspace-classification/{\"a\":3,\"b\":3,\"c\":11}"},"fields":{"title":"Separate the Whole Space, Homogeneous Line, Affine Line and Empty Set","stem":"In R² with usual operations consider a*x+b*y=c with [a,b,c]=[3, 3, 11]. Give (equation,classification,subspace?,certificate). Classification is whole_R2, empty_set, homogeneous_line or affine_line_not_through_zero.","answer":{"kind":"expression","value":"([3, 3, 11], affine_line_not_through_zero, No, affine_line_not_through_zero)","explanation":"If both coefficients vanish, c=0 gives the whole space and c≠0 gives the empty set. Otherwise c=0 gives a homogeneous line with a universal linear-equation certificate, and nonzero c gives an affine line missing zero. The exact result, retaining the requested order, is ([3, 3, 11], affine_line_not_through_zero, No, affine_line_not_through_zero)."},"skills":["Separate the Whole Space, Homogeneous Line, Affine Line and Empty Set"],"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-20261101-r01-vector-subspace-0100","batch_id":"batch-20261014-advance-20261101-r01-vector-subspace-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"cb8da1350cbf1f4c2e1089f43d2f3381a6511c06ba7c4ae4abaaf6b4858a8b64","family_id":"qf-mod32-subspace-classification","seed":202611011100,"semantic_template":"qf-mod32-subspace-classification/{\"a\":4,\"b\":4,\"c\":12}"},"fields":{"title":"Separate the Whole Space, Homogeneous Line, Affine Line and Empty Set","stem":"In R² with usual operations consider a*x+b*y=c with [a,b,c]=[4, 4, 12]. Give (equation,classification,subspace?,certificate). Classification is whole_R2, empty_set, homogeneous_line or affine_line_not_through_zero.","answer":{"kind":"expression","value":"([4, 4, 12], affine_line_not_through_zero, No, affine_line_not_through_zero)","explanation":"If both coefficients vanish, c=0 gives the whole space and c≠0 gives the empty set. Otherwise c=0 gives a homogeneous line with a universal linear-equation certificate, and nonzero c gives an affine line missing zero. The exact result, retaining the requested order, is ([4, 4, 12], affine_line_not_through_zero, No, affine_line_not_through_zero)."},"skills":["Separate the Whole Space, Homogeneous Line, Affine Line and Empty Set"],"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."}}