{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261010-advance-20261014-r06-absolute-value-0001","batch_id":"batch-20261010-advance-20261014-r06-absolute-value-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"e1c80f53035a8d2913076adbb09781934c9822ab1b34d95ae834ac4f4699ede1","family_id":"qf-density14-absolute-rational-two-branches","seed":202610146001,"semantic_template":"qf-density14-absolute-rational-two-branches/\"-4/1\""},"fields":{"title":"Normalized Rational Absolute-Value Branch","stem":"Use exact rational absolute values embedded in R; negative representation denominators are normalized before deciding signs. x=-12/3. Give (reduced x, nonnegative or negative branch, abs(x)); zero belongs to the nonnegative branch.","answer":{"kind":"expression","value":"(-4/1, negative, 4/1)","explanation":"Choose the branch by the normalized value’s sign; zero is not negative. A finite rational slice does not prove every real absolute-value property. The exact result, retaining the requested order, is (-4/1, negative, 4/1)."},"skills":["Normalized Rational Absolute-Value Branch"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261010-advance-20261014-r06-absolute-value-0002","batch_id":"batch-20261010-advance-20261014-r06-absolute-value-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"b82f917c018002ef3b8a394fd51e4a473064b7a886d4599b25d1c50544dcb489","family_id":"qf-density14-absolute-rational-two-branches","seed":202610146002,"semantic_template":"qf-density14-absolute-rational-two-branches/\"-11/3\""},"fields":{"title":"Normalized Rational Absolute-Value Branch","stem":"Use exact rational absolute values embedded in R; negative representation denominators are normalized before deciding signs. x=11/-3. Give (reduced x, nonnegative or negative branch, abs(x)); zero belongs to the nonnegative branch.","answer":{"kind":"expression","value":"(-11/3, negative, 11/3)","explanation":"Choose the branch by the normalized value’s sign; zero is not negative. A finite rational slice does not prove every real absolute-value property. The exact result, retaining the requested order, is (-11/3, negative, 11/3)."},"skills":["Normalized Rational Absolute-Value Branch"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261010-advance-20261014-r06-absolute-value-0003","batch_id":"batch-20261010-advance-20261014-r06-absolute-value-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"32e95456df75574a535b06156203a29678da6702fd5d8d5bb6ae992fe98ba1c7","family_id":"qf-density14-absolute-rational-two-branches","seed":202610146003,"semantic_template":"qf-density14-absolute-rational-two-branches/\"-10/3\""},"fields":{"title":"Normalized Rational Absolute-Value Branch","stem":"Use exact rational absolute values embedded in R; negative representation denominators are normalized before deciding signs. x=-10/3. Give (reduced x, nonnegative or negative branch, abs(x)); zero belongs to the nonnegative branch.","answer":{"kind":"expression","value":"(-10/3, negative, 10/3)","explanation":"Choose the branch by the normalized value’s sign; zero is not negative. A finite rational slice does not prove every real absolute-value property. The exact result, retaining the requested order, is (-10/3, negative, 10/3)."},"skills":["Normalized Rational Absolute-Value Branch"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261010-advance-20261014-r06-absolute-value-0004","batch_id":"batch-20261010-advance-20261014-r06-absolute-value-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"17e9f8a7e384b10a41a23e8dd037af8e101e45b384123dee5e53d0c56d165757","family_id":"qf-density14-absolute-rational-two-branches","seed":202610146004,"semantic_template":"qf-density14-absolute-rational-two-branches/\"-3/1\""},"fields":{"title":"Normalized Rational Absolute-Value Branch","stem":"Use exact rational absolute values embedded in R; negative representation denominators are normalized before deciding signs. x=9/-3. Give (reduced x, nonnegative or negative branch, abs(x)); zero belongs to the nonnegative branch.","answer":{"kind":"expression","value":"(-3/1, negative, 3/1)","explanation":"Choose the branch by the normalized value’s sign; zero is not negative. A finite rational slice does not prove every real absolute-value property. The exact result, retaining the requested order, is (-3/1, negative, 3/1)."},"skills":["Normalized Rational Absolute-Value Branch"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261010-advance-20261014-r06-absolute-value-0005","batch_id":"batch-20261010-advance-20261014-r06-absolute-value-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"91ab51ddd15b8060b2f7f2285a334f71cb3baff44bbae3901711676a6c50529f","family_id":"qf-density14-absolute-rational-two-branches","seed":202610146005,"semantic_template":"qf-density14-absolute-rational-two-branches/\"-8/3\""},"fields":{"title":"Normalized Rational Absolute-Value Branch","stem":"Use exact rational absolute values embedded in R; negative representation denominators are normalized before deciding signs. x=-8/3. Give (reduced x, nonnegative or negative branch, abs(x)); zero belongs to the nonnegative branch.","answer":{"kind":"expression","value":"(-8/3, negative, 8/3)","explanation":"Choose the branch by the normalized value’s sign; zero is not negative. A finite rational slice does not prove every real absolute-value property. The exact result, retaining the requested order, is (-8/3, negative, 8/3)."},"skills":["Normalized Rational Absolute-Value Branch"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261010-advance-20261014-r06-absolute-value-0006","batch_id":"batch-20261010-advance-20261014-r06-absolute-value-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"881a9846808a2a6aa46398706809ebcda2db6883be6864c2d876a6036ba36975","family_id":"qf-density14-absolute-rational-two-branches","seed":202610146006,"semantic_template":"qf-density14-absolute-rational-two-branches/\"-7/3\""},"fields":{"title":"Normalized Rational Absolute-Value Branch","stem":"Use exact rational absolute values embedded in R; negative representation denominators are normalized before deciding signs. x=7/-3. Give (reduced x, nonnegative or negative branch, abs(x)); zero belongs to the nonnegative branch.","answer":{"kind":"expression","value":"(-7/3, negative, 7/3)","explanation":"Choose the branch by the normalized value’s sign; zero is not negative. A finite rational slice does not prove every real absolute-value property. The exact result, retaining the requested order, is (-7/3, negative, 7/3)."},"skills":["Normalized Rational Absolute-Value Branch"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261010-advance-20261014-r06-absolute-value-0007","batch_id":"batch-20261010-advance-20261014-r06-absolute-value-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"c05d451afb388b3ac86d6a4c60cc4bd08e772055dd43d410493de022ebfb2e3a","family_id":"qf-density14-absolute-rational-two-branches","seed":202610146007,"semantic_template":"qf-density14-absolute-rational-two-branches/\"-2/1\""},"fields":{"title":"Normalized Rational Absolute-Value Branch","stem":"Use exact rational absolute values embedded in R; negative representation denominators are normalized before deciding signs. x=-6/3. Give (reduced x, nonnegative or negative branch, abs(x)); zero belongs to the nonnegative branch.","answer":{"kind":"expression","value":"(-2/1, negative, 2/1)","explanation":"Choose the branch by the normalized value’s sign; zero is not negative. A finite rational slice does not prove every real absolute-value property. The exact result, retaining the requested order, is (-2/1, negative, 2/1)."},"skills":["Normalized Rational Absolute-Value Branch"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261010-advance-20261014-r06-absolute-value-0008","batch_id":"batch-20261010-advance-20261014-r06-absolute-value-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"5899048d13fa343e35be6488d0f033c7383029efc6d05572c90e0d84cc3957d2","family_id":"qf-density14-absolute-rational-two-branches","seed":202610146008,"semantic_template":"qf-density14-absolute-rational-two-branches/\"-5/3\""},"fields":{"title":"Normalized Rational Absolute-Value Branch","stem":"Use exact rational absolute values embedded in R; negative representation denominators are normalized before deciding signs. x=5/-3. Give (reduced x, nonnegative or negative branch, abs(x)); zero belongs to the nonnegative branch.","answer":{"kind":"expression","value":"(-5/3, negative, 5/3)","explanation":"Choose the branch by the normalized value’s sign; zero is not negative. A finite rational slice does not prove every real absolute-value property. The exact result, retaining the requested order, is (-5/3, negative, 5/3)."},"skills":["Normalized Rational Absolute-Value Branch"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261010-advance-20261014-r06-absolute-value-0009","batch_id":"batch-20261010-advance-20261014-r06-absolute-value-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"eca7c1ad6f2ca05a89f5b3dcb4dece03dce6ff93303269819b586f3a44de33b8","family_id":"qf-density14-absolute-rational-two-branches","seed":202610146009,"semantic_template":"qf-density14-absolute-rational-two-branches/\"-4/3\""},"fields":{"title":"Normalized Rational Absolute-Value Branch","stem":"Use exact rational absolute values embedded in R; negative representation denominators are normalized before deciding signs. x=-4/3. Give (reduced x, nonnegative or negative branch, abs(x)); zero belongs to the nonnegative branch.","answer":{"kind":"expression","value":"(-4/3, negative, 4/3)","explanation":"Choose the branch by the normalized value’s sign; zero is not negative. A finite rational slice does not prove every real absolute-value property. The exact result, retaining the requested order, is (-4/3, negative, 4/3)."},"skills":["Normalized Rational Absolute-Value Branch"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261010-advance-20261014-r06-absolute-value-0010","batch_id":"batch-20261010-advance-20261014-r06-absolute-value-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"14b2b8830b1ab90d1af21dffe53327ba5ba386a8dd76de10002e57e256c0656e","family_id":"qf-density14-absolute-rational-two-branches","seed":202610146010,"semantic_template":"qf-density14-absolute-rational-two-branches/\"-1/1\""},"fields":{"title":"Normalized Rational Absolute-Value Branch","stem":"Use exact rational absolute values embedded in R; negative representation denominators are normalized before deciding signs. x=3/-3. Give (reduced x, nonnegative or negative branch, abs(x)); zero belongs to the nonnegative branch.","answer":{"kind":"expression","value":"(-1/1, negative, 1/1)","explanation":"Choose the branch by the normalized value’s sign; zero is not negative. A finite rational slice does not prove every real absolute-value property. The exact result, retaining the requested order, is (-1/1, negative, 1/1)."},"skills":["Normalized Rational Absolute-Value Branch"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261010-advance-20261014-r06-absolute-value-0011","batch_id":"batch-20261010-advance-20261014-r06-absolute-value-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"9cf0fce505b660caa97c5f4de9bb4f3c1f51bf78ebb721c09ff124d10c93fee2","family_id":"qf-density14-absolute-rational-two-branches","seed":202610146011,"semantic_template":"qf-density14-absolute-rational-two-branches/\"-2/3\""},"fields":{"title":"Normalized Rational Absolute-Value Branch","stem":"Use exact rational absolute values embedded in R; negative representation denominators are normalized before deciding signs. x=-2/3. Give (reduced x, nonnegative or negative branch, abs(x)); zero belongs to the nonnegative branch.","answer":{"kind":"expression","value":"(-2/3, negative, 2/3)","explanation":"Choose the branch by the normalized value’s sign; zero is not negative. A finite rational slice does not prove every real absolute-value property. The exact result, retaining the requested order, is (-2/3, negative, 2/3)."},"skills":["Normalized Rational Absolute-Value Branch"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261010-advance-20261014-r06-absolute-value-0012","batch_id":"batch-20261010-advance-20261014-r06-absolute-value-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"77179120131fa8c6f1ebaad484e3dbdd095cdea963360d355a5aca0cad26b360","family_id":"qf-density14-absolute-rational-two-branches","seed":202610146012,"semantic_template":"qf-density14-absolute-rational-two-branches/\"-1/3\""},"fields":{"title":"Normalized Rational Absolute-Value Branch","stem":"Use exact rational absolute values embedded in R; negative representation denominators are normalized before deciding signs. x=1/-3. Give (reduced x, nonnegative or negative branch, abs(x)); zero belongs to the nonnegative branch.","answer":{"kind":"expression","value":"(-1/3, negative, 1/3)","explanation":"Choose the branch by the normalized value’s sign; zero is not negative. A finite rational slice does not prove every real absolute-value property. The exact result, retaining the requested order, is (-1/3, negative, 1/3)."},"skills":["Normalized Rational Absolute-Value Branch"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261010-advance-20261014-r06-absolute-value-0013","batch_id":"batch-20261010-advance-20261014-r06-absolute-value-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"8ea37bdfa6d2acfd7f2f30c3685627f3177cf1f5586a42ab8e69cc46fbb09048","family_id":"qf-density14-absolute-rational-two-branches","seed":202610146013,"semantic_template":"qf-density14-absolute-rational-two-branches/\"0/1\""},"fields":{"title":"Normalized Rational Absolute-Value Branch","stem":"Use exact rational absolute values embedded in R; negative representation denominators are normalized before deciding signs. x=0/3. Give (reduced x, nonnegative or negative branch, abs(x)); zero belongs to the nonnegative branch.","answer":{"kind":"expression","value":"(0/1, nonnegative, 0/1)","explanation":"Choose the branch by the normalized value’s sign; zero is not negative. A finite rational slice does not prove every real absolute-value property. The exact result, retaining the requested order, is (0/1, nonnegative, 0/1)."},"skills":["Normalized Rational Absolute-Value Branch"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261010-advance-20261014-r06-absolute-value-0014","batch_id":"batch-20261010-advance-20261014-r06-absolute-value-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"5f4b867fc30aae30af4dd8d90e43f8d7c4942413b707a0271202038a1900f26e","family_id":"qf-density14-absolute-rational-two-branches","seed":202610146014,"semantic_template":"qf-density14-absolute-rational-two-branches/\"1/3\""},"fields":{"title":"Normalized Rational Absolute-Value Branch","stem":"Use exact rational absolute values embedded in R; negative representation denominators are normalized before deciding signs. x=-1/-3. Give (reduced x, nonnegative or negative branch, abs(x)); zero belongs to the nonnegative branch.","answer":{"kind":"expression","value":"(1/3, nonnegative, 1/3)","explanation":"Choose the branch by the normalized value’s sign; zero is not negative. A finite rational slice does not prove every real absolute-value property. The exact result, retaining the requested order, is (1/3, nonnegative, 1/3)."},"skills":["Normalized Rational Absolute-Value Branch"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261010-advance-20261014-r06-absolute-value-0015","batch_id":"batch-20261010-advance-20261014-r06-absolute-value-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"afd1c98384601af43cd79dd35adf1c64c61b15f0c9643c80653aec2d8a389166","family_id":"qf-density14-absolute-rational-two-branches","seed":202610146015,"semantic_template":"qf-density14-absolute-rational-two-branches/\"2/3\""},"fields":{"title":"Normalized Rational Absolute-Value Branch","stem":"Use exact rational absolute values embedded in R; negative representation denominators are normalized before deciding signs. x=2/3. Give (reduced x, nonnegative or negative branch, abs(x)); zero belongs to the nonnegative branch.","answer":{"kind":"expression","value":"(2/3, nonnegative, 2/3)","explanation":"Choose the branch by the normalized value’s sign; zero is not negative. A finite rational slice does not prove every real absolute-value property. The exact result, retaining the requested order, is (2/3, nonnegative, 2/3)."},"skills":["Normalized Rational Absolute-Value Branch"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261010-advance-20261014-r06-absolute-value-0016","batch_id":"batch-20261010-advance-20261014-r06-absolute-value-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"d72bf6ce0d68dcd36898a13f726e54baeaef769d31d6d43450549a975aad506b","family_id":"qf-density14-absolute-rational-two-branches","seed":202610146016,"semantic_template":"qf-density14-absolute-rational-two-branches/\"1/1\""},"fields":{"title":"Normalized Rational Absolute-Value Branch","stem":"Use exact rational absolute values embedded in R; negative representation denominators are normalized before deciding signs. x=-3/-3. Give (reduced x, nonnegative or negative branch, abs(x)); zero belongs to the nonnegative branch.","answer":{"kind":"expression","value":"(1/1, nonnegative, 1/1)","explanation":"Choose the branch by the normalized value’s sign; zero is not negative. A finite rational slice does not prove every real absolute-value property. The exact result, retaining the requested order, is (1/1, nonnegative, 1/1)."},"skills":["Normalized Rational Absolute-Value Branch"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261010-advance-20261014-r06-absolute-value-0017","batch_id":"batch-20261010-advance-20261014-r06-absolute-value-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"51b29c94e95946e1d36a5a81d1e2a13f07ab346eb6c8c4088e0a8c64bc054ac8","family_id":"qf-density14-absolute-rational-two-branches","seed":202610146017,"semantic_template":"qf-density14-absolute-rational-two-branches/\"4/3\""},"fields":{"title":"Normalized Rational Absolute-Value Branch","stem":"Use exact rational absolute values embedded in R; negative representation denominators are normalized before deciding signs. x=4/3. Give (reduced x, nonnegative or negative branch, abs(x)); zero belongs to the nonnegative branch.","answer":{"kind":"expression","value":"(4/3, nonnegative, 4/3)","explanation":"Choose the branch by the normalized value’s sign; zero is not negative. A finite rational slice does not prove every real absolute-value property. The exact result, retaining the requested order, is (4/3, nonnegative, 4/3)."},"skills":["Normalized Rational Absolute-Value Branch"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261010-advance-20261014-r06-absolute-value-0018","batch_id":"batch-20261010-advance-20261014-r06-absolute-value-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"71f671d1f83ebfce54d4e8753a43f3c6971c8dcdaca97a4279e51e603c9adb01","family_id":"qf-density14-absolute-rational-two-branches","seed":202610146018,"semantic_template":"qf-density14-absolute-rational-two-branches/\"5/3\""},"fields":{"title":"Normalized Rational Absolute-Value Branch","stem":"Use exact rational absolute values embedded in R; negative representation denominators are normalized before deciding signs. x=-5/-3. Give (reduced x, nonnegative or negative branch, abs(x)); zero belongs to the nonnegative branch.","answer":{"kind":"expression","value":"(5/3, nonnegative, 5/3)","explanation":"Choose the branch by the normalized value’s sign; zero is not negative. A finite rational slice does not prove every real absolute-value property. The exact result, retaining the requested order, is (5/3, nonnegative, 5/3)."},"skills":["Normalized Rational Absolute-Value Branch"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261010-advance-20261014-r06-absolute-value-0019","batch_id":"batch-20261010-advance-20261014-r06-absolute-value-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"cf9705c3e8f7061ca0a0a7f545517b8921210ffa502e81136cc05271ef4acc25","family_id":"qf-density14-absolute-rational-two-branches","seed":202610146019,"semantic_template":"qf-density14-absolute-rational-two-branches/\"2/1\""},"fields":{"title":"Normalized Rational Absolute-Value Branch","stem":"Use exact rational absolute values embedded in R; negative representation denominators are normalized before deciding signs. x=6/3. Give (reduced x, nonnegative or negative branch, abs(x)); zero belongs to the nonnegative branch.","answer":{"kind":"expression","value":"(2/1, nonnegative, 2/1)","explanation":"Choose the branch by the normalized value’s sign; zero is not negative. A finite rational slice does not prove every real absolute-value property. The exact result, retaining the requested order, is (2/1, nonnegative, 2/1)."},"skills":["Normalized Rational Absolute-Value Branch"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261010-advance-20261014-r06-absolute-value-0020","batch_id":"batch-20261010-advance-20261014-r06-absolute-value-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"6cfd38c2913187799de2785d4c3dbdeb925662ab55647ed24cd9e10ef0e4ad39","family_id":"qf-density14-absolute-rational-two-branches","seed":202610146020,"semantic_template":"qf-density14-absolute-rational-two-branches/\"7/3\""},"fields":{"title":"Normalized Rational Absolute-Value Branch","stem":"Use exact rational absolute values embedded in R; negative representation denominators are normalized before deciding signs. x=-7/-3. Give (reduced x, nonnegative or negative branch, abs(x)); zero belongs to the nonnegative branch.","answer":{"kind":"expression","value":"(7/3, nonnegative, 7/3)","explanation":"Choose the branch by the normalized value’s sign; zero is not negative. A finite rational slice does not prove every real absolute-value property. The exact result, retaining the requested order, is (7/3, nonnegative, 7/3)."},"skills":["Normalized Rational Absolute-Value Branch"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261010-advance-20261014-r06-absolute-value-0021","batch_id":"batch-20261010-advance-20261014-r06-absolute-value-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"3f6040ebf87ef5ed4ff529d59c39145d61a143bd3315d787b46f0574f3c0e717","family_id":"qf-density14-absolute-rational-two-branches","seed":202610146021,"semantic_template":"qf-density14-absolute-rational-two-branches/\"8/3\""},"fields":{"title":"Normalized Rational Absolute-Value Branch","stem":"Use exact rational absolute values embedded in R; negative representation denominators are normalized before deciding signs. x=8/3. Give (reduced x, nonnegative or negative branch, abs(x)); zero belongs to the nonnegative branch.","answer":{"kind":"expression","value":"(8/3, nonnegative, 8/3)","explanation":"Choose the branch by the normalized value’s sign; zero is not negative. A finite rational slice does not prove every real absolute-value property. The exact result, retaining the requested order, is (8/3, nonnegative, 8/3)."},"skills":["Normalized Rational Absolute-Value Branch"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261010-advance-20261014-r06-absolute-value-0022","batch_id":"batch-20261010-advance-20261014-r06-absolute-value-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"a49c8a39becca4c8ad0745e68f0fce965393eaabc5854ea45e466e6bc7617f4a","family_id":"qf-density14-absolute-rational-two-branches","seed":202610146022,"semantic_template":"qf-density14-absolute-rational-two-branches/\"3/1\""},"fields":{"title":"Normalized Rational Absolute-Value Branch","stem":"Use exact rational absolute values embedded in R; negative representation denominators are normalized before deciding signs. x=-9/-3. Give (reduced x, nonnegative or negative branch, abs(x)); zero belongs to the nonnegative branch.","answer":{"kind":"expression","value":"(3/1, nonnegative, 3/1)","explanation":"Choose the branch by the normalized value’s sign; zero is not negative. A finite rational slice does not prove every real absolute-value property. The exact result, retaining the requested order, is (3/1, nonnegative, 3/1)."},"skills":["Normalized Rational Absolute-Value Branch"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261010-advance-20261014-r06-absolute-value-0023","batch_id":"batch-20261010-advance-20261014-r06-absolute-value-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"2783a821ce19d3e970ee50ec975a62623c5b654478f9ca668609a2742b51ab0b","family_id":"qf-density14-absolute-rational-two-branches","seed":202610146023,"semantic_template":"qf-density14-absolute-rational-two-branches/\"10/3\""},"fields":{"title":"Normalized Rational Absolute-Value Branch","stem":"Use exact rational absolute values embedded in R; negative representation denominators are normalized before deciding signs. x=10/3. Give (reduced x, nonnegative or negative branch, abs(x)); zero belongs to the nonnegative branch.","answer":{"kind":"expression","value":"(10/3, nonnegative, 10/3)","explanation":"Choose the branch by the normalized value’s sign; zero is not negative. A finite rational slice does not prove every real absolute-value property. The exact result, retaining the requested order, is (10/3, nonnegative, 10/3)."},"skills":["Normalized Rational Absolute-Value Branch"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261010-advance-20261014-r06-absolute-value-0024","batch_id":"batch-20261010-advance-20261014-r06-absolute-value-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"a07554b8bedefaa5c42aab2f4dafcc294c5550fd8d8bc0cddf4d8049a64ecf88","family_id":"qf-density14-absolute-rational-two-branches","seed":202610146024,"semantic_template":"qf-density14-absolute-rational-two-branches/\"11/3\""},"fields":{"title":"Normalized Rational Absolute-Value Branch","stem":"Use exact rational absolute values embedded in R; negative representation denominators are normalized before deciding signs. x=-11/-3. Give (reduced x, nonnegative or negative branch, abs(x)); zero belongs to the nonnegative branch.","answer":{"kind":"expression","value":"(11/3, nonnegative, 11/3)","explanation":"Choose the branch by the normalized value’s sign; zero is not negative. A finite rational slice does not prove every real absolute-value property. The exact result, retaining the requested order, is (11/3, nonnegative, 11/3)."},"skills":["Normalized Rational Absolute-Value Branch"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261010-advance-20261014-r06-absolute-value-0025","batch_id":"batch-20261010-advance-20261014-r06-absolute-value-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"3a4d3f02c7ef094f36b4693be505521302922fbd0b58151b4f6fca0ed2c964bf","family_id":"qf-density14-absolute-rational-two-branches","seed":202610146025,"semantic_template":"qf-density14-absolute-rational-two-branches/\"4/1\""},"fields":{"title":"Normalized Rational Absolute-Value Branch","stem":"Use exact rational absolute values embedded in R; negative representation denominators are normalized before deciding signs. x=12/3. Give (reduced x, nonnegative or negative branch, abs(x)); zero belongs to the nonnegative branch.","answer":{"kind":"expression","value":"(4/1, nonnegative, 4/1)","explanation":"Choose the branch by the normalized value’s sign; zero is not negative. A finite rational slice does not prove every real absolute-value property. The exact result, retaining the requested order, is (4/1, nonnegative, 4/1)."},"skills":["Normalized Rational Absolute-Value Branch"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261010-advance-20261014-r06-absolute-value-0026","batch_id":"batch-20261010-advance-20261014-r06-absolute-value-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"093c66ef6f77a7653665f3e12d97e2767c69b5f7f974760a4b343c195b7685d3","family_id":"qf-density14-absolute-three-point-distances","seed":202610146026,"semantic_template":"qf-density14-absolute-three-point-distances/[\"0/1\",\"0/1\",\"0/1\"]"},"fields":{"title":"Three Rational Distances and Path Defect","stem":"Use exact rational absolute values embedded in R; negative representation denominators are normalized before deciding signs. Ordered points=[0/1, 0/1, 0/1]. Give (d01, d12, d02, d01+d12−d02). Use exact reduced fractions with positive denominators, retaining /1 for integers.","answer":{"kind":"expression","value":"(0/1, 0/1, 0/1, 0/1)","explanation":"Take the absolute difference for each segment separately; a path sum is not an endpoint distance, and finite checks are not a universal triangle-inequality proof. The exact result, retaining the requested order, is (0/1, 0/1, 0/1, 0/1)."},"skills":["Three Rational Distances and Path Defect"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261010-advance-20261014-r06-absolute-value-0027","batch_id":"batch-20261010-advance-20261014-r06-absolute-value-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"38aca324f068272cb3abb36da8670b5fb35d50ee5715edcac5428ad9cc9b7fe3","family_id":"qf-density14-absolute-three-point-distances","seed":202610146027,"semantic_template":"qf-density14-absolute-three-point-distances/[\"1/1\",\"1/1\",\"0/1\"]"},"fields":{"title":"Three Rational Distances and Path Defect","stem":"Use exact rational absolute values embedded in R; negative representation denominators are normalized before deciding signs. Ordered points=[0/1, 1/1, 0/1]. Give (d01, d12, d02, d01+d12−d02). Use exact reduced fractions with positive denominators, retaining /1 for integers.","answer":{"kind":"expression","value":"(1/1, 1/1, 0/1, 2/1)","explanation":"Take the absolute difference for each segment separately; a path sum is not an endpoint distance, and finite checks are not a universal triangle-inequality proof. The exact result, retaining the requested order, is (1/1, 1/1, 0/1, 2/1)."},"skills":["Three Rational Distances and Path Defect"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261010-advance-20261014-r06-absolute-value-0028","batch_id":"batch-20261010-advance-20261014-r06-absolute-value-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"7ebf71c4b701c68680c7073cdea0a1ae8dc00e9df80c1d70b86545c24681e28c","family_id":"qf-density14-absolute-three-point-distances","seed":202610146028,"semantic_template":"qf-density14-absolute-three-point-distances/[\"1/1\",\"1/2\",\"3/2\"]"},"fields":{"title":"Three Rational Distances and Path Defect","stem":"Use exact rational absolute values embedded in R; negative representation denominators are normalized before deciding signs. Ordered points=[0/1, 2/2, 3/2]. Give (d01, d12, d02, d01+d12−d02). Use exact reduced fractions with positive denominators, retaining /1 for integers.","answer":{"kind":"expression","value":"(1/1, 1/2, 3/2, 0/1)","explanation":"Take the absolute difference for each segment separately; a path sum is not an endpoint distance, and finite checks are not a universal triangle-inequality proof. The exact result, retaining the requested order, is (1/1, 1/2, 3/2, 0/1)."},"skills":["Three Rational Distances and Path Defect"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261010-advance-20261014-r06-absolute-value-0029","batch_id":"batch-20261010-advance-20261014-r06-absolute-value-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"1a4037bc4c87de98991940b2f06ea9ca38647c7b24b98701973203a3cd3d885b","family_id":"qf-density14-absolute-three-point-distances","seed":202610146029,"semantic_template":"qf-density14-absolute-three-point-distances/[\"3/2\",\"1/2\",\"2/1\"]"},"fields":{"title":"Three Rational Distances and Path Defect","stem":"Use exact rational absolute values embedded in R; negative representation denominators are normalized before deciding signs. Ordered points=[0/1, 3/2, 4/2]. Give (d01, d12, d02, d01+d12−d02). Use exact reduced fractions with positive denominators, retaining /1 for integers.","answer":{"kind":"expression","value":"(3/2, 1/2, 2/1, 0/1)","explanation":"Take the absolute difference for each segment separately; a path sum is not an endpoint distance, and finite checks are not a universal triangle-inequality proof. The exact result, retaining the requested order, is (3/2, 1/2, 2/1, 0/1)."},"skills":["Three Rational Distances and Path Defect"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261010-advance-20261014-r06-absolute-value-0030","batch_id":"batch-20261010-advance-20261014-r06-absolute-value-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"6d676e1ba493e804434a22b19b99ac9f01b86091375281da067ffea4da7dbce6","family_id":"qf-density14-absolute-three-point-distances","seed":202610146030,"semantic_template":"qf-density14-absolute-three-point-distances/[\"2/1\",\"1/2\",\"5/2\"]"},"fields":{"title":"Three Rational Distances and Path Defect","stem":"Use exact rational absolute values embedded in R; negative representation denominators are normalized before deciding signs. Ordered points=[0/1, 4/2, 5/2]. Give (d01, d12, d02, d01+d12−d02). Use exact reduced fractions with positive denominators, retaining /1 for integers.","answer":{"kind":"expression","value":"(2/1, 1/2, 5/2, 0/1)","explanation":"Take the absolute difference for each segment separately; a path sum is not an endpoint distance, and finite checks are not a universal triangle-inequality proof. The exact result, retaining the requested order, is (2/1, 1/2, 5/2, 0/1)."},"skills":["Three Rational Distances and Path Defect"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261010-advance-20261014-r06-absolute-value-0031","batch_id":"batch-20261010-advance-20261014-r06-absolute-value-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"6189987b1c095f920dae71d2c7af9c473e1ef3b50c995bf056d91e596097ac18","family_id":"qf-density14-absolute-three-point-distances","seed":202610146031,"semantic_template":"qf-density14-absolute-three-point-distances/[\"5/2\",\"1/2\",\"3/1\"]"},"fields":{"title":"Three Rational Distances and Path Defect","stem":"Use exact rational absolute values embedded in R; negative representation denominators are normalized before deciding signs. Ordered points=[0/1, 5/2, 6/2]. Give (d01, d12, d02, d01+d12−d02). Use exact reduced fractions with positive denominators, retaining /1 for integers.","answer":{"kind":"expression","value":"(5/2, 1/2, 3/1, 0/1)","explanation":"Take the absolute difference for each segment separately; a path sum is not an endpoint distance, and finite checks are not a universal triangle-inequality proof. The exact result, retaining the requested order, is (5/2, 1/2, 3/1, 0/1)."},"skills":["Three Rational Distances and Path Defect"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261010-advance-20261014-r06-absolute-value-0032","batch_id":"batch-20261010-advance-20261014-r06-absolute-value-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"4ce523ebe8836adc3f06d5db2ea73e819aee49bfbd64d2f33c53ef3c9689d289","family_id":"qf-density14-absolute-three-point-distances","seed":202610146032,"semantic_template":"qf-density14-absolute-three-point-distances/[\"3/1\",\"1/2\",\"7/2\"]"},"fields":{"title":"Three Rational Distances and Path Defect","stem":"Use exact rational absolute values embedded in R; negative representation denominators are normalized before deciding signs. Ordered points=[0/1, 6/2, 7/2]. Give (d01, d12, d02, d01+d12−d02). Use exact reduced fractions with positive denominators, retaining /1 for integers.","answer":{"kind":"expression","value":"(3/1, 1/2, 7/2, 0/1)","explanation":"Take the absolute difference for each segment separately; a path sum is not an endpoint distance, and finite checks are not a universal triangle-inequality proof. The exact result, retaining the requested order, is (3/1, 1/2, 7/2, 0/1)."},"skills":["Three Rational Distances and Path Defect"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261010-advance-20261014-r06-absolute-value-0033","batch_id":"batch-20261010-advance-20261014-r06-absolute-value-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"6b5e5aaf1e61ead4bbd6673bb7b047278c75680edf77f60ccd6777940c9c5d36","family_id":"qf-density14-absolute-three-point-distances","seed":202610146033,"semantic_template":"qf-density14-absolute-three-point-distances/[\"7/2\",\"1/2\",\"4/1\"]"},"fields":{"title":"Three Rational Distances and Path Defect","stem":"Use exact rational absolute values embedded in R; negative representation denominators are normalized before deciding signs. Ordered points=[0/1, 7/2, 8/2]. Give (d01, d12, d02, d01+d12−d02). Use exact reduced fractions with positive denominators, retaining /1 for integers.","answer":{"kind":"expression","value":"(7/2, 1/2, 4/1, 0/1)","explanation":"Take the absolute difference for each segment separately; a path sum is not an endpoint distance, and finite checks are not a universal triangle-inequality proof. The exact result, retaining the requested order, is (7/2, 1/2, 4/1, 0/1)."},"skills":["Three Rational Distances and Path Defect"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261010-advance-20261014-r06-absolute-value-0034","batch_id":"batch-20261010-advance-20261014-r06-absolute-value-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"f246fc40e142b91d6200b9dae3659c7c914ab05756697fa1ddd54c4f4f78a1a4","family_id":"qf-density14-absolute-three-point-distances","seed":202610146034,"semantic_template":"qf-density14-absolute-three-point-distances/[\"4/1\",\"1/2\",\"9/2\"]"},"fields":{"title":"Three Rational Distances and Path Defect","stem":"Use exact rational absolute values embedded in R; negative representation denominators are normalized before deciding signs. Ordered points=[0/1, 8/2, 9/2]. Give (d01, d12, d02, d01+d12−d02). Use exact reduced fractions with positive denominators, retaining /1 for integers.","answer":{"kind":"expression","value":"(4/1, 1/2, 9/2, 0/1)","explanation":"Take the absolute difference for each segment separately; a path sum is not an endpoint distance, and finite checks are not a universal triangle-inequality proof. The exact result, retaining the requested order, is (4/1, 1/2, 9/2, 0/1)."},"skills":["Three Rational Distances and Path Defect"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261010-advance-20261014-r06-absolute-value-0035","batch_id":"batch-20261010-advance-20261014-r06-absolute-value-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"10d4366259760f037b87287385670f43afb770353ba7c9456b94cf71545923b0","family_id":"qf-density14-absolute-three-point-distances","seed":202610146035,"semantic_template":"qf-density14-absolute-three-point-distances/[\"9/2\",\"1/2\",\"5/1\"]"},"fields":{"title":"Three Rational Distances and Path Defect","stem":"Use exact rational absolute values embedded in R; negative representation denominators are normalized before deciding signs. Ordered points=[0/1, 9/2, 10/2]. Give (d01, d12, d02, d01+d12−d02). Use exact reduced fractions with positive denominators, retaining /1 for integers.","answer":{"kind":"expression","value":"(9/2, 1/2, 5/1, 0/1)","explanation":"Take the absolute difference for each segment separately; a path sum is not an endpoint distance, and finite checks are not a universal triangle-inequality proof. The exact result, retaining the requested order, is (9/2, 1/2, 5/1, 0/1)."},"skills":["Three Rational Distances and Path Defect"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261010-advance-20261014-r06-absolute-value-0036","batch_id":"batch-20261010-advance-20261014-r06-absolute-value-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"8e9f415321887c8d12b65d0290197678c4a63fd48051ae250304d85a42453048","family_id":"qf-density14-absolute-three-point-distances","seed":202610146036,"semantic_template":"qf-density14-absolute-three-point-distances/[\"5/1\",\"1/2\",\"11/2\"]"},"fields":{"title":"Three Rational Distances and Path Defect","stem":"Use exact rational absolute values embedded in R; negative representation denominators are normalized before deciding signs. Ordered points=[0/1, 10/2, 11/2]. Give (d01, d12, d02, d01+d12−d02). Use exact reduced fractions with positive denominators, retaining /1 for integers.","answer":{"kind":"expression","value":"(5/1, 1/2, 11/2, 0/1)","explanation":"Take the absolute difference for each segment separately; a path sum is not an endpoint distance, and finite checks are not a universal triangle-inequality proof. The exact result, retaining the requested order, is (5/1, 1/2, 11/2, 0/1)."},"skills":["Three Rational Distances and Path Defect"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261010-advance-20261014-r06-absolute-value-0037","batch_id":"batch-20261010-advance-20261014-r06-absolute-value-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"15a369e7f3310f006977c4787e2871946398c76dbf6bbe9b27b7853cb84e4240","family_id":"qf-density14-absolute-three-point-distances","seed":202610146037,"semantic_template":"qf-density14-absolute-three-point-distances/[\"11/2\",\"1/2\",\"6/1\"]"},"fields":{"title":"Three Rational Distances and Path Defect","stem":"Use exact rational absolute values embedded in R; negative representation denominators are normalized before deciding signs. Ordered points=[0/1, 11/2, 12/2]. Give (d01, d12, d02, d01+d12−d02). Use exact reduced fractions with positive denominators, retaining /1 for integers.","answer":{"kind":"expression","value":"(11/2, 1/2, 6/1, 0/1)","explanation":"Take the absolute difference for each segment separately; a path sum is not an endpoint distance, and finite checks are not a universal triangle-inequality proof. The exact result, retaining the requested order, is (11/2, 1/2, 6/1, 0/1)."},"skills":["Three Rational Distances and Path Defect"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261010-advance-20261014-r06-absolute-value-0038","batch_id":"batch-20261010-advance-20261014-r06-absolute-value-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"9d8dc8e69cb5f472f9b6c174ad938034da57f272dee8ba9a90e5c3497a2e01ce","family_id":"qf-density14-absolute-three-point-distances","seed":202610146038,"semantic_template":"qf-density14-absolute-three-point-distances/[\"6/1\",\"1/2\",\"13/2\"]"},"fields":{"title":"Three Rational Distances and Path Defect","stem":"Use exact rational absolute values embedded in R; negative representation denominators are normalized before deciding signs. Ordered points=[0/1, 12/2, 13/2]. Give (d01, d12, d02, d01+d12−d02). Use exact reduced fractions with positive denominators, retaining /1 for integers.","answer":{"kind":"expression","value":"(6/1, 1/2, 13/2, 0/1)","explanation":"Take the absolute difference for each segment separately; a path sum is not an endpoint distance, and finite checks are not a universal triangle-inequality proof. The exact result, retaining the requested order, is (6/1, 1/2, 13/2, 0/1)."},"skills":["Three Rational Distances and Path Defect"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261010-advance-20261014-r06-absolute-value-0039","batch_id":"batch-20261010-advance-20261014-r06-absolute-value-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"2f376ae98ddd2a1e992b437f2b4acd31de3c9eac5a5093596fe5a49e81554aa5","family_id":"qf-density14-absolute-three-point-distances","seed":202610146039,"semantic_template":"qf-density14-absolute-three-point-distances/[\"13/2\",\"1/2\",\"7/1\"]"},"fields":{"title":"Three Rational Distances and Path Defect","stem":"Use exact rational absolute values embedded in R; negative representation denominators are normalized before deciding signs. Ordered points=[0/1, 13/2, 14/2]. Give (d01, d12, d02, d01+d12−d02). Use exact reduced fractions with positive denominators, retaining /1 for integers.","answer":{"kind":"expression","value":"(13/2, 1/2, 7/1, 0/1)","explanation":"Take the absolute difference for each segment separately; a path sum is not an endpoint distance, and finite checks are not a universal triangle-inequality proof. The exact result, retaining the requested order, is (13/2, 1/2, 7/1, 0/1)."},"skills":["Three Rational Distances and Path Defect"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261010-advance-20261014-r06-absolute-value-0040","batch_id":"batch-20261010-advance-20261014-r06-absolute-value-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"745956a5cd951c1946de616568d7c8366a8c2416ab87362c43403d0970f22fc5","family_id":"qf-density14-absolute-three-point-distances","seed":202610146040,"semantic_template":"qf-density14-absolute-three-point-distances/[\"7/1\",\"1/2\",\"15/2\"]"},"fields":{"title":"Three Rational Distances and Path Defect","stem":"Use exact rational absolute values embedded in R; negative representation denominators are normalized before deciding signs. Ordered points=[0/1, 14/2, 15/2]. Give (d01, d12, d02, d01+d12−d02). Use exact reduced fractions with positive denominators, retaining /1 for integers.","answer":{"kind":"expression","value":"(7/1, 1/2, 15/2, 0/1)","explanation":"Take the absolute difference for each segment separately; a path sum is not an endpoint distance, and finite checks are not a universal triangle-inequality proof. The exact result, retaining the requested order, is (7/1, 1/2, 15/2, 0/1)."},"skills":["Three Rational Distances and Path Defect"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261010-advance-20261014-r06-absolute-value-0041","batch_id":"batch-20261010-advance-20261014-r06-absolute-value-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"127fa4650c5570762a9eafb7f1fee76ce54362f26ead0a4b6b08095b1ff38311","family_id":"qf-density14-absolute-three-point-distances","seed":202610146041,"semantic_template":"qf-density14-absolute-three-point-distances/[\"15/2\",\"1/2\",\"8/1\"]"},"fields":{"title":"Three Rational Distances and Path Defect","stem":"Use exact rational absolute values embedded in R; negative representation denominators are normalized before deciding signs. Ordered points=[0/1, 15/2, 16/2]. Give (d01, d12, d02, d01+d12−d02). Use exact reduced fractions with positive denominators, retaining /1 for integers.","answer":{"kind":"expression","value":"(15/2, 1/2, 8/1, 0/1)","explanation":"Take the absolute difference for each segment separately; a path sum is not an endpoint distance, and finite checks are not a universal triangle-inequality proof. The exact result, retaining the requested order, is (15/2, 1/2, 8/1, 0/1)."},"skills":["Three Rational Distances and Path Defect"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261010-advance-20261014-r06-absolute-value-0042","batch_id":"batch-20261010-advance-20261014-r06-absolute-value-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"bb44e94570a168aeb9409f1af00732fbffed2e261571b8797f3ceff76663af08","family_id":"qf-density14-absolute-three-point-distances","seed":202610146042,"semantic_template":"qf-density14-absolute-three-point-distances/[\"8/1\",\"1/2\",\"17/2\"]"},"fields":{"title":"Three Rational Distances and Path Defect","stem":"Use exact rational absolute values embedded in R; negative representation denominators are normalized before deciding signs. Ordered points=[0/1, 16/2, 17/2]. Give (d01, d12, d02, d01+d12−d02). Use exact reduced fractions with positive denominators, retaining /1 for integers.","answer":{"kind":"expression","value":"(8/1, 1/2, 17/2, 0/1)","explanation":"Take the absolute difference for each segment separately; a path sum is not an endpoint distance, and finite checks are not a universal triangle-inequality proof. The exact result, retaining the requested order, is (8/1, 1/2, 17/2, 0/1)."},"skills":["Three Rational Distances and Path Defect"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261010-advance-20261014-r06-absolute-value-0043","batch_id":"batch-20261010-advance-20261014-r06-absolute-value-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"a8a5668c42f54715cc8004e45ae9b05bbb77a9a81d9400b1c0352b01ae1be2ad","family_id":"qf-density14-absolute-three-point-distances","seed":202610146043,"semantic_template":"qf-density14-absolute-three-point-distances/[\"17/2\",\"1/2\",\"9/1\"]"},"fields":{"title":"Three Rational Distances and Path Defect","stem":"Use exact rational absolute values embedded in R; negative representation denominators are normalized before deciding signs. Ordered points=[0/1, 17/2, 18/2]. Give (d01, d12, d02, d01+d12−d02). Use exact reduced fractions with positive denominators, retaining /1 for integers.","answer":{"kind":"expression","value":"(17/2, 1/2, 9/1, 0/1)","explanation":"Take the absolute difference for each segment separately; a path sum is not an endpoint distance, and finite checks are not a universal triangle-inequality proof. The exact result, retaining the requested order, is (17/2, 1/2, 9/1, 0/1)."},"skills":["Three Rational Distances and Path Defect"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261010-advance-20261014-r06-absolute-value-0044","batch_id":"batch-20261010-advance-20261014-r06-absolute-value-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"d35f4a52141add5dbde61c118d7fb2229b753a8a85bf6f0886c61f7cff72d507","family_id":"qf-density14-absolute-three-point-distances","seed":202610146044,"semantic_template":"qf-density14-absolute-three-point-distances/[\"9/1\",\"1/2\",\"19/2\"]"},"fields":{"title":"Three Rational Distances and Path Defect","stem":"Use exact rational absolute values embedded in R; negative representation denominators are normalized before deciding signs. Ordered points=[0/1, 18/2, 19/2]. Give (d01, d12, d02, d01+d12−d02). Use exact reduced fractions with positive denominators, retaining /1 for integers.","answer":{"kind":"expression","value":"(9/1, 1/2, 19/2, 0/1)","explanation":"Take the absolute difference for each segment separately; a path sum is not an endpoint distance, and finite checks are not a universal triangle-inequality proof. The exact result, retaining the requested order, is (9/1, 1/2, 19/2, 0/1)."},"skills":["Three Rational Distances and Path Defect"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261010-advance-20261014-r06-absolute-value-0045","batch_id":"batch-20261010-advance-20261014-r06-absolute-value-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"5d279993f7afbf4f09d0412c164adb4637b5d5689d07d97e4f4ddf2755f282dd","family_id":"qf-density14-absolute-three-point-distances","seed":202610146045,"semantic_template":"qf-density14-absolute-three-point-distances/[\"19/2\",\"1/2\",\"10/1\"]"},"fields":{"title":"Three Rational Distances and Path Defect","stem":"Use exact rational absolute values embedded in R; negative representation denominators are normalized before deciding signs. Ordered points=[0/1, 19/2, 20/2]. Give (d01, d12, d02, d01+d12−d02). Use exact reduced fractions with positive denominators, retaining /1 for integers.","answer":{"kind":"expression","value":"(19/2, 1/2, 10/1, 0/1)","explanation":"Take the absolute difference for each segment separately; a path sum is not an endpoint distance, and finite checks are not a universal triangle-inequality proof. The exact result, retaining the requested order, is (19/2, 1/2, 10/1, 0/1)."},"skills":["Three Rational Distances and Path Defect"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261010-advance-20261014-r06-absolute-value-0046","batch_id":"batch-20261010-advance-20261014-r06-absolute-value-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"759a49603d31f1affd8cda79709c9b8f4e9ef79625bbd9dcef9ac66d22b5cf4f","family_id":"qf-density14-absolute-three-point-distances","seed":202610146046,"semantic_template":"qf-density14-absolute-three-point-distances/[\"10/1\",\"1/2\",\"21/2\"]"},"fields":{"title":"Three Rational Distances and Path Defect","stem":"Use exact rational absolute values embedded in R; negative representation denominators are normalized before deciding signs. Ordered points=[0/1, 20/2, 21/2]. Give (d01, d12, d02, d01+d12−d02). Use exact reduced fractions with positive denominators, retaining /1 for integers.","answer":{"kind":"expression","value":"(10/1, 1/2, 21/2, 0/1)","explanation":"Take the absolute difference for each segment separately; a path sum is not an endpoint distance, and finite checks are not a universal triangle-inequality proof. The exact result, retaining the requested order, is (10/1, 1/2, 21/2, 0/1)."},"skills":["Three Rational Distances and Path Defect"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261010-advance-20261014-r06-absolute-value-0047","batch_id":"batch-20261010-advance-20261014-r06-absolute-value-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"021b4c0f863a73bd1a83dd9461bdf5dbaa4896b2cc4c69ae2d74d584592e99ba","family_id":"qf-density14-absolute-three-point-distances","seed":202610146047,"semantic_template":"qf-density14-absolute-three-point-distances/[\"21/2\",\"1/2\",\"11/1\"]"},"fields":{"title":"Three Rational Distances and Path Defect","stem":"Use exact rational absolute values embedded in R; negative representation denominators are normalized before deciding signs. Ordered points=[0/1, 21/2, 22/2]. Give (d01, d12, d02, d01+d12−d02). Use exact reduced fractions with positive denominators, retaining /1 for integers.","answer":{"kind":"expression","value":"(21/2, 1/2, 11/1, 0/1)","explanation":"Take the absolute difference for each segment separately; a path sum is not an endpoint distance, and finite checks are not a universal triangle-inequality proof. The exact result, retaining the requested order, is (21/2, 1/2, 11/1, 0/1)."},"skills":["Three Rational Distances and Path Defect"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261010-advance-20261014-r06-absolute-value-0048","batch_id":"batch-20261010-advance-20261014-r06-absolute-value-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"e2020acc6674a9e4a1794f736d6bc1c99d5f93e51483899f82503a837fa98f7d","family_id":"qf-density14-absolute-three-point-distances","seed":202610146048,"semantic_template":"qf-density14-absolute-three-point-distances/[\"11/1\",\"1/2\",\"23/2\"]"},"fields":{"title":"Three Rational Distances and Path Defect","stem":"Use exact rational absolute values embedded in R; negative representation denominators are normalized before deciding signs. Ordered points=[0/1, 22/2, 23/2]. Give (d01, d12, d02, d01+d12−d02). Use exact reduced fractions with positive denominators, retaining /1 for integers.","answer":{"kind":"expression","value":"(11/1, 1/2, 23/2, 0/1)","explanation":"Take the absolute difference for each segment separately; a path sum is not an endpoint distance, and finite checks are not a universal triangle-inequality proof. The exact result, retaining the requested order, is (11/1, 1/2, 23/2, 0/1)."},"skills":["Three Rational Distances and Path Defect"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261010-advance-20261014-r06-absolute-value-0049","batch_id":"batch-20261010-advance-20261014-r06-absolute-value-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"36c145ca45acb6ec098358d780206582878e917489423b0626fabce77b22ef39","family_id":"qf-density14-absolute-three-point-distances","seed":202610146049,"semantic_template":"qf-density14-absolute-three-point-distances/[\"23/2\",\"1/2\",\"12/1\"]"},"fields":{"title":"Three Rational Distances and Path Defect","stem":"Use exact rational absolute values embedded in R; negative representation denominators are normalized before deciding signs. Ordered points=[0/1, 23/2, 24/2]. Give (d01, d12, d02, d01+d12−d02). Use exact reduced fractions with positive denominators, retaining /1 for integers.","answer":{"kind":"expression","value":"(23/2, 1/2, 12/1, 0/1)","explanation":"Take the absolute difference for each segment separately; a path sum is not an endpoint distance, and finite checks are not a universal triangle-inequality proof. The exact result, retaining the requested order, is (23/2, 1/2, 12/1, 0/1)."},"skills":["Three Rational Distances and Path Defect"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261010-advance-20261014-r06-absolute-value-0050","batch_id":"batch-20261010-advance-20261014-r06-absolute-value-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"bcfffb9d6c15e2e7e6d48379b1cfb7f4a2853ade4044acd966e3383cc4bed62d","family_id":"qf-density14-absolute-three-point-distances","seed":202610146050,"semantic_template":"qf-density14-absolute-three-point-distances/[\"12/1\",\"1/2\",\"25/2\"]"},"fields":{"title":"Three Rational Distances and Path Defect","stem":"Use exact rational absolute values embedded in R; negative representation denominators are normalized before deciding signs. Ordered points=[0/1, 24/2, 25/2]. Give (d01, d12, d02, d01+d12−d02). Use exact reduced fractions with positive denominators, retaining /1 for integers.","answer":{"kind":"expression","value":"(12/1, 1/2, 25/2, 0/1)","explanation":"Take the absolute difference for each segment separately; a path sum is not an endpoint distance, and finite checks are not a universal triangle-inequality proof. The exact result, retaining the requested order, is (12/1, 1/2, 25/2, 0/1)."},"skills":["Three Rational Distances and Path Defect"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261010-advance-20261014-r06-absolute-value-0051","batch_id":"batch-20261010-advance-20261014-r06-absolute-value-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"301eb0006f4c560d1d0a2a65c0239f2d32d3a6b48af1956a7ecf0c6bd2a2eba5","family_id":"qf-density14-absolute-closed-radius-mask","seed":202610146051,"semantic_template":"qf-density14-absolute-closed-radius-mask/[\"0/1\",[\"0/1\",\"1/1\",\"1/1\",\"0/1\"]]"},"fields":{"title":"Inclusive Absolute-Value Threshold Mask","stem":"Use exact rational absolute values embedded in R; negative representation denominators are normalized before deciding signs. r=0/1≥0, ordered points=[0/1, 1/1, -1/1, 0/-2]. Give (r, four absolute values, four abs(point)≤r eligibility values), including endpoints.","answer":{"kind":"expression","value":"(0/1, [0/1, 1/1, 1/1, 0/1], [Yes, No, No, Yes])","explanation":"Evaluate each absolute value, then compare with the exact closed threshold; r=0 is valid and finite substitutions are not general inequality solving. The exact result, retaining the requested order, is (0/1, [0/1, 1/1, 1/1, 0/1], [Yes, No, No, Yes])."},"skills":["Inclusive Absolute-Value Threshold Mask"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261010-advance-20261014-r06-absolute-value-0052","batch_id":"batch-20261010-advance-20261014-r06-absolute-value-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"728cfd1a7636706a251d5540fbb76f23eb7ac149baab1d015cf0e723c7420d42","family_id":"qf-density14-absolute-closed-radius-mask","seed":202610146052,"semantic_template":"qf-density14-absolute-closed-radius-mask/[\"1/2\",[\"1/2\",\"1/2\",\"3/2\",\"0/1\"]]"},"fields":{"title":"Inclusive Absolute-Value Threshold Mask","stem":"Use exact rational absolute values embedded in R; negative representation denominators are normalized before deciding signs. r=1/2≥0, ordered points=[-1/2, 1/2, 3/2, 0/1]. Give (r, four absolute values, four abs(point)≤r eligibility values), including endpoints.","answer":{"kind":"expression","value":"(1/2, [1/2, 1/2, 3/2, 0/1], [Yes, Yes, No, Yes])","explanation":"Evaluate each absolute value, then compare with the exact closed threshold; r=0 is valid and finite substitutions are not general inequality solving. The exact result, retaining the requested order, is (1/2, [1/2, 1/2, 3/2, 0/1], [Yes, Yes, No, Yes])."},"skills":["Inclusive Absolute-Value Threshold Mask"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261010-advance-20261014-r06-absolute-value-0053","batch_id":"batch-20261010-advance-20261014-r06-absolute-value-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"276a124c8b59da0323bd985fa12a1d2b5ccc62278826abdaf148645e032dbb61","family_id":"qf-density14-absolute-closed-radius-mask","seed":202610146053,"semantic_template":"qf-density14-absolute-closed-radius-mask/[\"1/1\",[\"1/1\",\"1/1\",\"2/1\",\"0/1\"]]"},"fields":{"title":"Inclusive Absolute-Value Threshold Mask","stem":"Use exact rational absolute values embedded in R; negative representation denominators are normalized before deciding signs. r=2/2≥0, ordered points=[-2/2, 2/2, 4/2, 0/1]. Give (r, four absolute values, four abs(point)≤r eligibility values), including endpoints.","answer":{"kind":"expression","value":"(1/1, [1/1, 1/1, 2/1, 0/1], [Yes, Yes, No, Yes])","explanation":"Evaluate each absolute value, then compare with the exact closed threshold; r=0 is valid and finite substitutions are not general inequality solving. The exact result, retaining the requested order, is (1/1, [1/1, 1/1, 2/1, 0/1], [Yes, Yes, No, Yes])."},"skills":["Inclusive Absolute-Value Threshold Mask"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261010-advance-20261014-r06-absolute-value-0054","batch_id":"batch-20261010-advance-20261014-r06-absolute-value-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"8abc76a830416626aa0457fb1f159143a599a16d97b9a1a3356702696aa8c018","family_id":"qf-density14-absolute-closed-radius-mask","seed":202610146054,"semantic_template":"qf-density14-absolute-closed-radius-mask/[\"3/2\",[\"3/2\",\"3/2\",\"5/2\",\"0/1\"]]"},"fields":{"title":"Inclusive Absolute-Value Threshold Mask","stem":"Use exact rational absolute values embedded in R; negative representation denominators are normalized before deciding signs. r=3/2≥0, ordered points=[-3/2, 3/2, 5/2, 0/1]. Give (r, four absolute values, four abs(point)≤r eligibility values), including endpoints.","answer":{"kind":"expression","value":"(3/2, [3/2, 3/2, 5/2, 0/1], [Yes, Yes, No, Yes])","explanation":"Evaluate each absolute value, then compare with the exact closed threshold; r=0 is valid and finite substitutions are not general inequality solving. The exact result, retaining the requested order, is (3/2, [3/2, 3/2, 5/2, 0/1], [Yes, Yes, No, Yes])."},"skills":["Inclusive Absolute-Value Threshold Mask"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261010-advance-20261014-r06-absolute-value-0055","batch_id":"batch-20261010-advance-20261014-r06-absolute-value-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"426e00c33d22bf627cb5e5d6ed6ac0a9780f8fd1820ff44a3aaa57559ef6dcc7","family_id":"qf-density14-absolute-closed-radius-mask","seed":202610146055,"semantic_template":"qf-density14-absolute-closed-radius-mask/[\"2/1\",[\"2/1\",\"2/1\",\"3/1\",\"0/1\"]]"},"fields":{"title":"Inclusive Absolute-Value Threshold Mask","stem":"Use exact rational absolute values embedded in R; negative representation denominators are normalized before deciding signs. r=4/2≥0, ordered points=[-4/2, 4/2, 6/2, 0/1]. Give (r, four absolute values, four abs(point)≤r eligibility values), including endpoints.","answer":{"kind":"expression","value":"(2/1, [2/1, 2/1, 3/1, 0/1], [Yes, Yes, No, Yes])","explanation":"Evaluate each absolute value, then compare with the exact closed threshold; r=0 is valid and finite substitutions are not general inequality solving. The exact result, retaining the requested order, is (2/1, [2/1, 2/1, 3/1, 0/1], [Yes, Yes, No, Yes])."},"skills":["Inclusive Absolute-Value Threshold Mask"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261010-advance-20261014-r06-absolute-value-0056","batch_id":"batch-20261010-advance-20261014-r06-absolute-value-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"19b19e50bb8e749b89bb8c089f04ea0b90cf5641c6b32be30379858e8a16574f","family_id":"qf-density14-absolute-closed-radius-mask","seed":202610146056,"semantic_template":"qf-density14-absolute-closed-radius-mask/[\"5/2\",[\"5/2\",\"5/2\",\"7/2\",\"0/1\"]]"},"fields":{"title":"Inclusive Absolute-Value Threshold Mask","stem":"Use exact rational absolute values embedded in R; negative representation denominators are normalized before deciding signs. r=5/2≥0, ordered points=[-5/2, 5/2, 7/2, 0/1]. Give (r, four absolute values, four abs(point)≤r eligibility values), including endpoints.","answer":{"kind":"expression","value":"(5/2, [5/2, 5/2, 7/2, 0/1], [Yes, Yes, No, Yes])","explanation":"Evaluate each absolute value, then compare with the exact closed threshold; r=0 is valid and finite substitutions are not general inequality solving. The exact result, retaining the requested order, is (5/2, [5/2, 5/2, 7/2, 0/1], [Yes, Yes, No, Yes])."},"skills":["Inclusive Absolute-Value Threshold Mask"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261010-advance-20261014-r06-absolute-value-0057","batch_id":"batch-20261010-advance-20261014-r06-absolute-value-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"50cc868ee369e5958ac9559d25fb17cd558416fc055ee4724d9c6d125dfaa7c3","family_id":"qf-density14-absolute-closed-radius-mask","seed":202610146057,"semantic_template":"qf-density14-absolute-closed-radius-mask/[\"3/1\",[\"3/1\",\"3/1\",\"4/1\",\"0/1\"]]"},"fields":{"title":"Inclusive Absolute-Value Threshold Mask","stem":"Use exact rational absolute values embedded in R; negative representation denominators are normalized before deciding signs. r=6/2≥0, ordered points=[-6/2, 6/2, 8/2, 0/1]. Give (r, four absolute values, four abs(point)≤r eligibility values), including endpoints.","answer":{"kind":"expression","value":"(3/1, [3/1, 3/1, 4/1, 0/1], [Yes, Yes, No, Yes])","explanation":"Evaluate each absolute value, then compare with the exact closed threshold; r=0 is valid and finite substitutions are not general inequality solving. The exact result, retaining the requested order, is (3/1, [3/1, 3/1, 4/1, 0/1], [Yes, Yes, No, Yes])."},"skills":["Inclusive Absolute-Value Threshold Mask"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261010-advance-20261014-r06-absolute-value-0058","batch_id":"batch-20261010-advance-20261014-r06-absolute-value-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"0e31781619d578667f62caf04019eea18b43aca0e6b53e1cc764e5a17042ecea","family_id":"qf-density14-absolute-closed-radius-mask","seed":202610146058,"semantic_template":"qf-density14-absolute-closed-radius-mask/[\"7/2\",[\"7/2\",\"7/2\",\"9/2\",\"0/1\"]]"},"fields":{"title":"Inclusive Absolute-Value Threshold Mask","stem":"Use exact rational absolute values embedded in R; negative representation denominators are normalized before deciding signs. r=7/2≥0, ordered points=[-7/2, 7/2, 9/2, 0/1]. Give (r, four absolute values, four abs(point)≤r eligibility values), including endpoints.","answer":{"kind":"expression","value":"(7/2, [7/2, 7/2, 9/2, 0/1], [Yes, Yes, No, Yes])","explanation":"Evaluate each absolute value, then compare with the exact closed threshold; r=0 is valid and finite substitutions are not general inequality solving. The exact result, retaining the requested order, is (7/2, [7/2, 7/2, 9/2, 0/1], [Yes, Yes, No, Yes])."},"skills":["Inclusive Absolute-Value Threshold Mask"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261010-advance-20261014-r06-absolute-value-0059","batch_id":"batch-20261010-advance-20261014-r06-absolute-value-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"d707eae9f6d6421af58e78459c64ebb2167f6287c57f91b6cef787086c40077d","family_id":"qf-density14-absolute-closed-radius-mask","seed":202610146059,"semantic_template":"qf-density14-absolute-closed-radius-mask/[\"4/1\",[\"4/1\",\"4/1\",\"5/1\",\"0/1\"]]"},"fields":{"title":"Inclusive Absolute-Value Threshold Mask","stem":"Use exact rational absolute values embedded in R; negative representation denominators are normalized before deciding signs. r=8/2≥0, ordered points=[-8/2, 8/2, 10/2, 0/1]. Give (r, four absolute values, four abs(point)≤r eligibility values), including endpoints.","answer":{"kind":"expression","value":"(4/1, [4/1, 4/1, 5/1, 0/1], [Yes, Yes, No, Yes])","explanation":"Evaluate each absolute value, then compare with the exact closed threshold; r=0 is valid and finite substitutions are not general inequality solving. The exact result, retaining the requested order, is (4/1, [4/1, 4/1, 5/1, 0/1], [Yes, Yes, No, Yes])."},"skills":["Inclusive Absolute-Value Threshold Mask"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261010-advance-20261014-r06-absolute-value-0060","batch_id":"batch-20261010-advance-20261014-r06-absolute-value-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"033c9cdf1e1d9b569491b77dead64c6560d02c01933b3929901f51be0b5de485","family_id":"qf-density14-absolute-closed-radius-mask","seed":202610146060,"semantic_template":"qf-density14-absolute-closed-radius-mask/[\"9/2\",[\"9/2\",\"9/2\",\"11/2\",\"0/1\"]]"},"fields":{"title":"Inclusive Absolute-Value Threshold Mask","stem":"Use exact rational absolute values embedded in R; negative representation denominators are normalized before deciding signs. r=9/2≥0, ordered points=[-9/2, 9/2, 11/2, 0/1]. Give (r, four absolute values, four abs(point)≤r eligibility values), including endpoints.","answer":{"kind":"expression","value":"(9/2, [9/2, 9/2, 11/2, 0/1], [Yes, Yes, No, Yes])","explanation":"Evaluate each absolute value, then compare with the exact closed threshold; r=0 is valid and finite substitutions are not general inequality solving. The exact result, retaining the requested order, is (9/2, [9/2, 9/2, 11/2, 0/1], [Yes, Yes, No, Yes])."},"skills":["Inclusive Absolute-Value Threshold Mask"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261010-advance-20261014-r06-absolute-value-0061","batch_id":"batch-20261010-advance-20261014-r06-absolute-value-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"98f4361a328788eeddf5c677bd8aff76fb5c348f533ab8369b5bfc7980092da2","family_id":"qf-density14-absolute-closed-radius-mask","seed":202610146061,"semantic_template":"qf-density14-absolute-closed-radius-mask/[\"5/1\",[\"5/1\",\"5/1\",\"6/1\",\"0/1\"]]"},"fields":{"title":"Inclusive Absolute-Value Threshold Mask","stem":"Use exact rational absolute values embedded in R; negative representation denominators are normalized before deciding signs. r=10/2≥0, ordered points=[-10/2, 10/2, 12/2, 0/1]. Give (r, four absolute values, four abs(point)≤r eligibility values), including endpoints.","answer":{"kind":"expression","value":"(5/1, [5/1, 5/1, 6/1, 0/1], [Yes, Yes, No, Yes])","explanation":"Evaluate each absolute value, then compare with the exact closed threshold; r=0 is valid and finite substitutions are not general inequality solving. The exact result, retaining the requested order, is (5/1, [5/1, 5/1, 6/1, 0/1], [Yes, Yes, No, Yes])."},"skills":["Inclusive Absolute-Value Threshold Mask"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261010-advance-20261014-r06-absolute-value-0062","batch_id":"batch-20261010-advance-20261014-r06-absolute-value-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"927ab82ec24a3410fbc4ee06af37fe7b3cfcc197787b147fe1c1ca25db0a943a","family_id":"qf-density14-absolute-closed-radius-mask","seed":202610146062,"semantic_template":"qf-density14-absolute-closed-radius-mask/[\"11/2\",[\"11/2\",\"11/2\",\"13/2\",\"0/1\"]]"},"fields":{"title":"Inclusive Absolute-Value Threshold Mask","stem":"Use exact rational absolute values embedded in R; negative representation denominators are normalized before deciding signs. r=11/2≥0, ordered points=[-11/2, 11/2, 13/2, 0/1]. Give (r, four absolute values, four abs(point)≤r eligibility values), including endpoints.","answer":{"kind":"expression","value":"(11/2, [11/2, 11/2, 13/2, 0/1], [Yes, Yes, No, Yes])","explanation":"Evaluate each absolute value, then compare with the exact closed threshold; r=0 is valid and finite substitutions are not general inequality solving. The exact result, retaining the requested order, is (11/2, [11/2, 11/2, 13/2, 0/1], [Yes, Yes, No, Yes])."},"skills":["Inclusive Absolute-Value Threshold Mask"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261010-advance-20261014-r06-absolute-value-0063","batch_id":"batch-20261010-advance-20261014-r06-absolute-value-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"022f209c7918f358ffe1064810f71baad0f36ba4f4c94014c73a002b2a248911","family_id":"qf-density14-absolute-closed-radius-mask","seed":202610146063,"semantic_template":"qf-density14-absolute-closed-radius-mask/[\"6/1\",[\"6/1\",\"6/1\",\"7/1\",\"0/1\"]]"},"fields":{"title":"Inclusive Absolute-Value Threshold Mask","stem":"Use exact rational absolute values embedded in R; negative representation denominators are normalized before deciding signs. r=12/2≥0, ordered points=[-12/2, 12/2, 14/2, 0/1]. Give (r, four absolute values, four abs(point)≤r eligibility values), including endpoints.","answer":{"kind":"expression","value":"(6/1, [6/1, 6/1, 7/1, 0/1], [Yes, Yes, No, Yes])","explanation":"Evaluate each absolute value, then compare with the exact closed threshold; r=0 is valid and finite substitutions are not general inequality solving. The exact result, retaining the requested order, is (6/1, [6/1, 6/1, 7/1, 0/1], [Yes, Yes, No, Yes])."},"skills":["Inclusive Absolute-Value Threshold Mask"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261010-advance-20261014-r06-absolute-value-0064","batch_id":"batch-20261010-advance-20261014-r06-absolute-value-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"450ba8d3b17fe92fb82f201804b1a59aa0e44fb5766eeefaf591264cfd7a7c31","family_id":"qf-density14-absolute-closed-radius-mask","seed":202610146064,"semantic_template":"qf-density14-absolute-closed-radius-mask/[\"13/2\",[\"13/2\",\"13/2\",\"15/2\",\"0/1\"]]"},"fields":{"title":"Inclusive Absolute-Value Threshold Mask","stem":"Use exact rational absolute values embedded in R; negative representation denominators are normalized before deciding signs. r=13/2≥0, ordered points=[-13/2, 13/2, 15/2, 0/1]. Give (r, four absolute values, four abs(point)≤r eligibility values), including endpoints.","answer":{"kind":"expression","value":"(13/2, [13/2, 13/2, 15/2, 0/1], [Yes, Yes, No, Yes])","explanation":"Evaluate each absolute value, then compare with the exact closed threshold; r=0 is valid and finite substitutions are not general inequality solving. The exact result, retaining the requested order, is (13/2, [13/2, 13/2, 15/2, 0/1], [Yes, Yes, No, Yes])."},"skills":["Inclusive Absolute-Value Threshold Mask"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261010-advance-20261014-r06-absolute-value-0065","batch_id":"batch-20261010-advance-20261014-r06-absolute-value-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"e8bdb68bbaed315a9c94c13e3434b3e472b0c4ee479df418aa3536b6b00a56b8","family_id":"qf-density14-absolute-closed-radius-mask","seed":202610146065,"semantic_template":"qf-density14-absolute-closed-radius-mask/[\"7/1\",[\"7/1\",\"7/1\",\"8/1\",\"0/1\"]]"},"fields":{"title":"Inclusive Absolute-Value Threshold Mask","stem":"Use exact rational absolute values embedded in R; negative representation denominators are normalized before deciding signs. r=14/2≥0, ordered points=[-14/2, 14/2, 16/2, 0/1]. Give (r, four absolute values, four abs(point)≤r eligibility values), including endpoints.","answer":{"kind":"expression","value":"(7/1, [7/1, 7/1, 8/1, 0/1], [Yes, Yes, No, Yes])","explanation":"Evaluate each absolute value, then compare with the exact closed threshold; r=0 is valid and finite substitutions are not general inequality solving. The exact result, retaining the requested order, is (7/1, [7/1, 7/1, 8/1, 0/1], [Yes, Yes, No, Yes])."},"skills":["Inclusive Absolute-Value Threshold Mask"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261010-advance-20261014-r06-absolute-value-0066","batch_id":"batch-20261010-advance-20261014-r06-absolute-value-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"8d432186aed655cb650d513a6e2b923722894e5e78a80482d12ea31882fa4c30","family_id":"qf-density14-absolute-closed-radius-mask","seed":202610146066,"semantic_template":"qf-density14-absolute-closed-radius-mask/[\"15/2\",[\"15/2\",\"15/2\",\"17/2\",\"0/1\"]]"},"fields":{"title":"Inclusive Absolute-Value Threshold Mask","stem":"Use exact rational absolute values embedded in R; negative representation denominators are normalized before deciding signs. r=15/2≥0, ordered points=[-15/2, 15/2, 17/2, 0/1]. Give (r, four absolute values, four abs(point)≤r eligibility values), including endpoints.","answer":{"kind":"expression","value":"(15/2, [15/2, 15/2, 17/2, 0/1], [Yes, Yes, No, Yes])","explanation":"Evaluate each absolute value, then compare with the exact closed threshold; r=0 is valid and finite substitutions are not general inequality solving. The exact result, retaining the requested order, is (15/2, [15/2, 15/2, 17/2, 0/1], [Yes, Yes, No, Yes])."},"skills":["Inclusive Absolute-Value Threshold Mask"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261010-advance-20261014-r06-absolute-value-0067","batch_id":"batch-20261010-advance-20261014-r06-absolute-value-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"87c8dc3ff3c7c52c5af54f22d77483f5cc5fb1b0a074c2bc164496e82ba72dd9","family_id":"qf-density14-absolute-closed-radius-mask","seed":202610146067,"semantic_template":"qf-density14-absolute-closed-radius-mask/[\"8/1\",[\"8/1\",\"8/1\",\"9/1\",\"0/1\"]]"},"fields":{"title":"Inclusive Absolute-Value Threshold Mask","stem":"Use exact rational absolute values embedded in R; negative representation denominators are normalized before deciding signs. r=16/2≥0, ordered points=[-16/2, 16/2, 18/2, 0/1]. Give (r, four absolute values, four abs(point)≤r eligibility values), including endpoints.","answer":{"kind":"expression","value":"(8/1, [8/1, 8/1, 9/1, 0/1], [Yes, Yes, No, Yes])","explanation":"Evaluate each absolute value, then compare with the exact closed threshold; r=0 is valid and finite substitutions are not general inequality solving. The exact result, retaining the requested order, is (8/1, [8/1, 8/1, 9/1, 0/1], [Yes, Yes, No, Yes])."},"skills":["Inclusive Absolute-Value Threshold Mask"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261010-advance-20261014-r06-absolute-value-0068","batch_id":"batch-20261010-advance-20261014-r06-absolute-value-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"b1f449520e41492768d0d3df146ae2cb6aa97eca390c5772b29c7650b2f4c194","family_id":"qf-density14-absolute-closed-radius-mask","seed":202610146068,"semantic_template":"qf-density14-absolute-closed-radius-mask/[\"17/2\",[\"17/2\",\"17/2\",\"19/2\",\"0/1\"]]"},"fields":{"title":"Inclusive Absolute-Value Threshold Mask","stem":"Use exact rational absolute values embedded in R; negative representation denominators are normalized before deciding signs. r=17/2≥0, ordered points=[-17/2, 17/2, 19/2, 0/1]. Give (r, four absolute values, four abs(point)≤r eligibility values), including endpoints.","answer":{"kind":"expression","value":"(17/2, [17/2, 17/2, 19/2, 0/1], [Yes, Yes, No, Yes])","explanation":"Evaluate each absolute value, then compare with the exact closed threshold; r=0 is valid and finite substitutions are not general inequality solving. The exact result, retaining the requested order, is (17/2, [17/2, 17/2, 19/2, 0/1], [Yes, Yes, No, Yes])."},"skills":["Inclusive Absolute-Value Threshold Mask"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261010-advance-20261014-r06-absolute-value-0069","batch_id":"batch-20261010-advance-20261014-r06-absolute-value-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"d920b04b0dcc3107f11903f865a64ada1a1c7b2c453c7684b6bd0b544c158c83","family_id":"qf-density14-absolute-closed-radius-mask","seed":202610146069,"semantic_template":"qf-density14-absolute-closed-radius-mask/[\"9/1\",[\"9/1\",\"9/1\",\"10/1\",\"0/1\"]]"},"fields":{"title":"Inclusive Absolute-Value Threshold Mask","stem":"Use exact rational absolute values embedded in R; negative representation denominators are normalized before deciding signs. r=18/2≥0, ordered points=[-18/2, 18/2, 20/2, 0/1]. Give (r, four absolute values, four abs(point)≤r eligibility values), including endpoints.","answer":{"kind":"expression","value":"(9/1, [9/1, 9/1, 10/1, 0/1], [Yes, Yes, No, Yes])","explanation":"Evaluate each absolute value, then compare with the exact closed threshold; r=0 is valid and finite substitutions are not general inequality solving. The exact result, retaining the requested order, is (9/1, [9/1, 9/1, 10/1, 0/1], [Yes, Yes, No, Yes])."},"skills":["Inclusive Absolute-Value Threshold Mask"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261010-advance-20261014-r06-absolute-value-0070","batch_id":"batch-20261010-advance-20261014-r06-absolute-value-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"841a323e4bbb33d3d5e476f8c5fb434e836351b6fd9260d67e8c96b561ab4a61","family_id":"qf-density14-absolute-closed-radius-mask","seed":202610146070,"semantic_template":"qf-density14-absolute-closed-radius-mask/[\"19/2\",[\"19/2\",\"19/2\",\"21/2\",\"0/1\"]]"},"fields":{"title":"Inclusive Absolute-Value Threshold Mask","stem":"Use exact rational absolute values embedded in R; negative representation denominators are normalized before deciding signs. r=19/2≥0, ordered points=[-19/2, 19/2, 21/2, 0/1]. Give (r, four absolute values, four abs(point)≤r eligibility values), including endpoints.","answer":{"kind":"expression","value":"(19/2, [19/2, 19/2, 21/2, 0/1], [Yes, Yes, No, Yes])","explanation":"Evaluate each absolute value, then compare with the exact closed threshold; r=0 is valid and finite substitutions are not general inequality solving. The exact result, retaining the requested order, is (19/2, [19/2, 19/2, 21/2, 0/1], [Yes, Yes, No, Yes])."},"skills":["Inclusive Absolute-Value Threshold Mask"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261010-advance-20261014-r06-absolute-value-0071","batch_id":"batch-20261010-advance-20261014-r06-absolute-value-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"a3718d8efa0a737507f7ed75cd908e41b01992e9dd690aee2262e8727662c9e6","family_id":"qf-density14-absolute-closed-radius-mask","seed":202610146071,"semantic_template":"qf-density14-absolute-closed-radius-mask/[\"10/1\",[\"10/1\",\"10/1\",\"11/1\",\"0/1\"]]"},"fields":{"title":"Inclusive Absolute-Value Threshold Mask","stem":"Use exact rational absolute values embedded in R; negative representation denominators are normalized before deciding signs. r=20/2≥0, ordered points=[-20/2, 20/2, 22/2, 0/1]. Give (r, four absolute values, four abs(point)≤r eligibility values), including endpoints.","answer":{"kind":"expression","value":"(10/1, [10/1, 10/1, 11/1, 0/1], [Yes, Yes, No, Yes])","explanation":"Evaluate each absolute value, then compare with the exact closed threshold; r=0 is valid and finite substitutions are not general inequality solving. The exact result, retaining the requested order, is (10/1, [10/1, 10/1, 11/1, 0/1], [Yes, Yes, No, Yes])."},"skills":["Inclusive Absolute-Value Threshold Mask"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261010-advance-20261014-r06-absolute-value-0072","batch_id":"batch-20261010-advance-20261014-r06-absolute-value-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"7ccf6df6a5c548b156003378f4a0d2f3bb35ac6f6772e7bca2b83b05a5cacc5e","family_id":"qf-density14-absolute-closed-radius-mask","seed":202610146072,"semantic_template":"qf-density14-absolute-closed-radius-mask/[\"21/2\",[\"21/2\",\"21/2\",\"23/2\",\"0/1\"]]"},"fields":{"title":"Inclusive Absolute-Value Threshold Mask","stem":"Use exact rational absolute values embedded in R; negative representation denominators are normalized before deciding signs. r=21/2≥0, ordered points=[-21/2, 21/2, 23/2, 0/1]. Give (r, four absolute values, four abs(point)≤r eligibility values), including endpoints.","answer":{"kind":"expression","value":"(21/2, [21/2, 21/2, 23/2, 0/1], [Yes, Yes, No, Yes])","explanation":"Evaluate each absolute value, then compare with the exact closed threshold; r=0 is valid and finite substitutions are not general inequality solving. The exact result, retaining the requested order, is (21/2, [21/2, 21/2, 23/2, 0/1], [Yes, Yes, No, Yes])."},"skills":["Inclusive Absolute-Value Threshold Mask"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261010-advance-20261014-r06-absolute-value-0073","batch_id":"batch-20261010-advance-20261014-r06-absolute-value-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"b6319fe382756c6c824c5fb42d5a5a945ab3bf01c826b066bf119a3c8d1a94f7","family_id":"qf-density14-absolute-closed-radius-mask","seed":202610146073,"semantic_template":"qf-density14-absolute-closed-radius-mask/[\"11/1\",[\"11/1\",\"11/1\",\"12/1\",\"0/1\"]]"},"fields":{"title":"Inclusive Absolute-Value Threshold Mask","stem":"Use exact rational absolute values embedded in R; negative representation denominators are normalized before deciding signs. r=22/2≥0, ordered points=[-22/2, 22/2, 24/2, 0/1]. Give (r, four absolute values, four abs(point)≤r eligibility values), including endpoints.","answer":{"kind":"expression","value":"(11/1, [11/1, 11/1, 12/1, 0/1], [Yes, Yes, No, Yes])","explanation":"Evaluate each absolute value, then compare with the exact closed threshold; r=0 is valid and finite substitutions are not general inequality solving. The exact result, retaining the requested order, is (11/1, [11/1, 11/1, 12/1, 0/1], [Yes, Yes, No, Yes])."},"skills":["Inclusive Absolute-Value Threshold Mask"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261010-advance-20261014-r06-absolute-value-0074","batch_id":"batch-20261010-advance-20261014-r06-absolute-value-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"75436178bc24a692601b38fba39ce87c165cf239468a711b8482f8ef431ff96a","family_id":"qf-density14-absolute-closed-radius-mask","seed":202610146074,"semantic_template":"qf-density14-absolute-closed-radius-mask/[\"23/2\",[\"23/2\",\"23/2\",\"25/2\",\"0/1\"]]"},"fields":{"title":"Inclusive Absolute-Value Threshold Mask","stem":"Use exact rational absolute values embedded in R; negative representation denominators are normalized before deciding signs. r=23/2≥0, ordered points=[-23/2, 23/2, 25/2, 0/1]. Give (r, four absolute values, four abs(point)≤r eligibility values), including endpoints.","answer":{"kind":"expression","value":"(23/2, [23/2, 23/2, 25/2, 0/1], [Yes, Yes, No, Yes])","explanation":"Evaluate each absolute value, then compare with the exact closed threshold; r=0 is valid and finite substitutions are not general inequality solving. The exact result, retaining the requested order, is (23/2, [23/2, 23/2, 25/2, 0/1], [Yes, Yes, No, Yes])."},"skills":["Inclusive Absolute-Value Threshold Mask"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261010-advance-20261014-r06-absolute-value-0075","batch_id":"batch-20261010-advance-20261014-r06-absolute-value-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"5295d5418d076005a0cdcd6a82355f0587e6166611941b656457cec91f8e492b","family_id":"qf-density14-absolute-closed-radius-mask","seed":202610146075,"semantic_template":"qf-density14-absolute-closed-radius-mask/[\"12/1\",[\"12/1\",\"12/1\",\"13/1\",\"0/1\"]]"},"fields":{"title":"Inclusive Absolute-Value Threshold Mask","stem":"Use exact rational absolute values embedded in R; negative representation denominators are normalized before deciding signs. r=24/2≥0, ordered points=[-24/2, 24/2, 26/2, 0/1]. Give (r, four absolute values, four abs(point)≤r eligibility values), including endpoints.","answer":{"kind":"expression","value":"(12/1, [12/1, 12/1, 13/1, 0/1], [Yes, Yes, No, Yes])","explanation":"Evaluate each absolute value, then compare with the exact closed threshold; r=0 is valid and finite substitutions are not general inequality solving. The exact result, retaining the requested order, is (12/1, [12/1, 12/1, 13/1, 0/1], [Yes, Yes, No, Yes])."},"skills":["Inclusive Absolute-Value Threshold Mask"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261010-advance-20261014-r06-absolute-value-0076","batch_id":"batch-20261010-advance-20261014-r06-absolute-value-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"d6f79f8299c25af316552b731f4eaa4b8b41363cfdee7107c6a6850f3b4626b6","family_id":"qf-density14-absolute-preimage-rational-audit","seed":202610146076,"semantic_template":"qf-density14-absolute-preimage-rational-audit/[\"negative_target\",[\"0/1\",\"1/1\",\"2/1\"]]"},"fields":{"title":"Absolute-Value Target Roots and Candidate Audit","stem":"Use exact rational absolute values embedded in R; negative representation denominators are normalized before deciding signs. Target r=-1/1, ordered queries=[0/1, 1/1, 2/1]. Give (is r nonnegative?, increasing roots of |x|=r, three absolute query values, three equality-to-r values).","answer":{"kind":"expression","value":"(No, [], [0/1, 1/1, 2/1], [No, No, No])","explanation":"A negative target has no roots, zero has only root zero, and a positive target has two signed roots; substitute each candidate exactly. The exact result, retaining the requested order, is (No, [], [0/1, 1/1, 2/1], [No, No, No])."},"skills":["Absolute-Value Target Roots and Candidate Audit"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261010-advance-20261014-r06-absolute-value-0077","batch_id":"batch-20261010-advance-20261014-r06-absolute-value-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"c2abc14717d8aedb7f06fff1873bac6cc843d2bc95f0cbc382ccad6b1f817e25","family_id":"qf-density14-absolute-preimage-rational-audit","seed":202610146077,"semantic_template":"qf-density14-absolute-preimage-rational-audit/[\"0/1\",[\"0/1\",\"1/1\",\"1/1\"]]"},"fields":{"title":"Absolute-Value Target Roots and Candidate Audit","stem":"Use exact rational absolute values embedded in R; negative representation denominators are normalized before deciding signs. Target r=0/1, ordered queries=[0/1, 1/1, -1/1]. Give (is r nonnegative?, increasing roots of |x|=r, three absolute query values, three equality-to-r values).","answer":{"kind":"expression","value":"(Yes, [0/1], [0/1, 1/1, 1/1], [Yes, No, No])","explanation":"A negative target has no roots, zero has only root zero, and a positive target has two signed roots; substitute each candidate exactly. The exact result, retaining the requested order, is (Yes, [0/1], [0/1, 1/1, 1/1], [Yes, No, No])."},"skills":["Absolute-Value Target Roots and Candidate Audit"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261010-advance-20261014-r06-absolute-value-0078","batch_id":"batch-20261010-advance-20261014-r06-absolute-value-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"a74e972e3559812dc4298dccff691494c309f392e1f497e904c1b3288df45a62","family_id":"qf-density14-absolute-preimage-rational-audit","seed":202610146078,"semantic_template":"qf-density14-absolute-preimage-rational-audit/[\"1/2\",[\"1/2\",\"1/2\",\"3/2\"]]"},"fields":{"title":"Absolute-Value Target Roots and Candidate Audit","stem":"Use exact rational absolute values embedded in R; negative representation denominators are normalized before deciding signs. Target r=1/2, ordered queries=[-1/2, 1/2, 3/2]. Give (is r nonnegative?, increasing roots of |x|=r, three absolute query values, three equality-to-r values).","answer":{"kind":"expression","value":"(Yes, [-1/2, 1/2], [1/2, 1/2, 3/2], [Yes, Yes, No])","explanation":"A negative target has no roots, zero has only root zero, and a positive target has two signed roots; substitute each candidate exactly. The exact result, retaining the requested order, is (Yes, [-1/2, 1/2], [1/2, 1/2, 3/2], [Yes, Yes, No])."},"skills":["Absolute-Value Target Roots and Candidate Audit"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261010-advance-20261014-r06-absolute-value-0079","batch_id":"batch-20261010-advance-20261014-r06-absolute-value-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"87786d6bd0dd3b11d01e7fe9e503faaa8781b29502292489466f0e3d298ea8a2","family_id":"qf-density14-absolute-preimage-rational-audit","seed":202610146079,"semantic_template":"qf-density14-absolute-preimage-rational-audit/[\"1/1\",[\"1/1\",\"1/1\",\"2/1\"]]"},"fields":{"title":"Absolute-Value Target Roots and Candidate Audit","stem":"Use exact rational absolute values embedded in R; negative representation denominators are normalized before deciding signs. Target r=2/2, ordered queries=[-2/2, 2/2, 4/2]. Give (is r nonnegative?, increasing roots of |x|=r, three absolute query values, three equality-to-r values).","answer":{"kind":"expression","value":"(Yes, [-1/1, 1/1], [1/1, 1/1, 2/1], [Yes, Yes, No])","explanation":"A negative target has no roots, zero has only root zero, and a positive target has two signed roots; substitute each candidate exactly. The exact result, retaining the requested order, is (Yes, [-1/1, 1/1], [1/1, 1/1, 2/1], [Yes, Yes, No])."},"skills":["Absolute-Value Target Roots and Candidate Audit"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261010-advance-20261014-r06-absolute-value-0080","batch_id":"batch-20261010-advance-20261014-r06-absolute-value-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"1538d8dd0f46b2508bc6538024966dbf1016c3a390aa6f61ae31103eb2b1b1b1","family_id":"qf-density14-absolute-preimage-rational-audit","seed":202610146080,"semantic_template":"qf-density14-absolute-preimage-rational-audit/[\"3/2\",[\"3/2\",\"3/2\",\"5/2\"]]"},"fields":{"title":"Absolute-Value Target Roots and Candidate Audit","stem":"Use exact rational absolute values embedded in R; negative representation denominators are normalized before deciding signs. Target r=3/2, ordered queries=[-3/2, 3/2, 5/2]. Give (is r nonnegative?, increasing roots of |x|=r, three absolute query values, three equality-to-r values).","answer":{"kind":"expression","value":"(Yes, [-3/2, 3/2], [3/2, 3/2, 5/2], [Yes, Yes, No])","explanation":"A negative target has no roots, zero has only root zero, and a positive target has two signed roots; substitute each candidate exactly. The exact result, retaining the requested order, is (Yes, [-3/2, 3/2], [3/2, 3/2, 5/2], [Yes, Yes, No])."},"skills":["Absolute-Value Target Roots and Candidate Audit"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261010-advance-20261014-r06-absolute-value-0081","batch_id":"batch-20261010-advance-20261014-r06-absolute-value-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"56e0df9eb74edd84c343d07d879551eb97aa7f8d9e4c3c8e5bf1f568e1fe58d3","family_id":"qf-density14-absolute-preimage-rational-audit","seed":202610146081,"semantic_template":"qf-density14-absolute-preimage-rational-audit/[\"2/1\",[\"2/1\",\"2/1\",\"3/1\"]]"},"fields":{"title":"Absolute-Value Target Roots and Candidate Audit","stem":"Use exact rational absolute values embedded in R; negative representation denominators are normalized before deciding signs. Target r=4/2, ordered queries=[-4/2, 4/2, 6/2]. Give (is r nonnegative?, increasing roots of |x|=r, three absolute query values, three equality-to-r values).","answer":{"kind":"expression","value":"(Yes, [-2/1, 2/1], [2/1, 2/1, 3/1], [Yes, Yes, No])","explanation":"A negative target has no roots, zero has only root zero, and a positive target has two signed roots; substitute each candidate exactly. The exact result, retaining the requested order, is (Yes, [-2/1, 2/1], [2/1, 2/1, 3/1], [Yes, Yes, No])."},"skills":["Absolute-Value Target Roots and Candidate Audit"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261010-advance-20261014-r06-absolute-value-0082","batch_id":"batch-20261010-advance-20261014-r06-absolute-value-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"a8deacbf0f88d80f8d17cc10eed48288c03b5f1ba7e3b621cb60123e10e21793","family_id":"qf-density14-absolute-preimage-rational-audit","seed":202610146082,"semantic_template":"qf-density14-absolute-preimage-rational-audit/[\"5/2\",[\"5/2\",\"5/2\",\"7/2\"]]"},"fields":{"title":"Absolute-Value Target Roots and Candidate Audit","stem":"Use exact rational absolute values embedded in R; negative representation denominators are normalized before deciding signs. Target r=5/2, ordered queries=[-5/2, 5/2, 7/2]. Give (is r nonnegative?, increasing roots of |x|=r, three absolute query values, three equality-to-r values).","answer":{"kind":"expression","value":"(Yes, [-5/2, 5/2], [5/2, 5/2, 7/2], [Yes, Yes, No])","explanation":"A negative target has no roots, zero has only root zero, and a positive target has two signed roots; substitute each candidate exactly. The exact result, retaining the requested order, is (Yes, [-5/2, 5/2], [5/2, 5/2, 7/2], [Yes, Yes, No])."},"skills":["Absolute-Value Target Roots and Candidate Audit"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261010-advance-20261014-r06-absolute-value-0083","batch_id":"batch-20261010-advance-20261014-r06-absolute-value-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"a04425809895c79cfec97f660cd9971fac62e785a4363967543636a12970ea22","family_id":"qf-density14-absolute-preimage-rational-audit","seed":202610146083,"semantic_template":"qf-density14-absolute-preimage-rational-audit/[\"3/1\",[\"3/1\",\"3/1\",\"4/1\"]]"},"fields":{"title":"Absolute-Value Target Roots and Candidate Audit","stem":"Use exact rational absolute values embedded in R; negative representation denominators are normalized before deciding signs. Target r=6/2, ordered queries=[-6/2, 6/2, 8/2]. Give (is r nonnegative?, increasing roots of |x|=r, three absolute query values, three equality-to-r values).","answer":{"kind":"expression","value":"(Yes, [-3/1, 3/1], [3/1, 3/1, 4/1], [Yes, Yes, No])","explanation":"A negative target has no roots, zero has only root zero, and a positive target has two signed roots; substitute each candidate exactly. The exact result, retaining the requested order, is (Yes, [-3/1, 3/1], [3/1, 3/1, 4/1], [Yes, Yes, No])."},"skills":["Absolute-Value Target Roots and Candidate Audit"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261010-advance-20261014-r06-absolute-value-0084","batch_id":"batch-20261010-advance-20261014-r06-absolute-value-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"632b82ddf620390ee37b85ca176d2ab90c16b5c2b5a8ad8d29303a9bb3013203","family_id":"qf-density14-absolute-preimage-rational-audit","seed":202610146084,"semantic_template":"qf-density14-absolute-preimage-rational-audit/[\"7/2\",[\"7/2\",\"7/2\",\"9/2\"]]"},"fields":{"title":"Absolute-Value Target Roots and Candidate Audit","stem":"Use exact rational absolute values embedded in R; negative representation denominators are normalized before deciding signs. Target r=7/2, ordered queries=[-7/2, 7/2, 9/2]. Give (is r nonnegative?, increasing roots of |x|=r, three absolute query values, three equality-to-r values).","answer":{"kind":"expression","value":"(Yes, [-7/2, 7/2], [7/2, 7/2, 9/2], [Yes, Yes, No])","explanation":"A negative target has no roots, zero has only root zero, and a positive target has two signed roots; substitute each candidate exactly. The exact result, retaining the requested order, is (Yes, [-7/2, 7/2], [7/2, 7/2, 9/2], [Yes, Yes, No])."},"skills":["Absolute-Value Target Roots and Candidate Audit"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261010-advance-20261014-r06-absolute-value-0085","batch_id":"batch-20261010-advance-20261014-r06-absolute-value-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"c44233c79a333c9ffead0711ff85f2ccc569289bc83d206c137678ea8592bf30","family_id":"qf-density14-absolute-preimage-rational-audit","seed":202610146085,"semantic_template":"qf-density14-absolute-preimage-rational-audit/[\"4/1\",[\"4/1\",\"4/1\",\"5/1\"]]"},"fields":{"title":"Absolute-Value Target Roots and Candidate Audit","stem":"Use exact rational absolute values embedded in R; negative representation denominators are normalized before deciding signs. Target r=8/2, ordered queries=[-8/2, 8/2, 10/2]. Give (is r nonnegative?, increasing roots of |x|=r, three absolute query values, three equality-to-r values).","answer":{"kind":"expression","value":"(Yes, [-4/1, 4/1], [4/1, 4/1, 5/1], [Yes, Yes, No])","explanation":"A negative target has no roots, zero has only root zero, and a positive target has two signed roots; substitute each candidate exactly. The exact result, retaining the requested order, is (Yes, [-4/1, 4/1], [4/1, 4/1, 5/1], [Yes, Yes, No])."},"skills":["Absolute-Value Target Roots and Candidate Audit"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261010-advance-20261014-r06-absolute-value-0086","batch_id":"batch-20261010-advance-20261014-r06-absolute-value-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"85d2628403b642583cbb82250929fe40ae4ef9bcb7befeca77678589b0cd18c5","family_id":"qf-density14-absolute-preimage-rational-audit","seed":202610146086,"semantic_template":"qf-density14-absolute-preimage-rational-audit/[\"9/2\",[\"9/2\",\"9/2\",\"11/2\"]]"},"fields":{"title":"Absolute-Value Target Roots and Candidate Audit","stem":"Use exact rational absolute values embedded in R; negative representation denominators are normalized before deciding signs. Target r=9/2, ordered queries=[-9/2, 9/2, 11/2]. Give (is r nonnegative?, increasing roots of |x|=r, three absolute query values, three equality-to-r values).","answer":{"kind":"expression","value":"(Yes, [-9/2, 9/2], [9/2, 9/2, 11/2], [Yes, Yes, No])","explanation":"A negative target has no roots, zero has only root zero, and a positive target has two signed roots; substitute each candidate exactly. The exact result, retaining the requested order, is (Yes, [-9/2, 9/2], [9/2, 9/2, 11/2], [Yes, Yes, No])."},"skills":["Absolute-Value Target Roots and Candidate Audit"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261010-advance-20261014-r06-absolute-value-0087","batch_id":"batch-20261010-advance-20261014-r06-absolute-value-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"06e5f2d830ea7660af323d218f709ee899c4f1b964c11fb44994a76898bb0471","family_id":"qf-density14-absolute-preimage-rational-audit","seed":202610146087,"semantic_template":"qf-density14-absolute-preimage-rational-audit/[\"5/1\",[\"5/1\",\"5/1\",\"6/1\"]]"},"fields":{"title":"Absolute-Value Target Roots and Candidate Audit","stem":"Use exact rational absolute values embedded in R; negative representation denominators are normalized before deciding signs. Target r=10/2, ordered queries=[-10/2, 10/2, 12/2]. Give (is r nonnegative?, increasing roots of |x|=r, three absolute query values, three equality-to-r values).","answer":{"kind":"expression","value":"(Yes, [-5/1, 5/1], [5/1, 5/1, 6/1], [Yes, Yes, No])","explanation":"A negative target has no roots, zero has only root zero, and a positive target has two signed roots; substitute each candidate exactly. The exact result, retaining the requested order, is (Yes, [-5/1, 5/1], [5/1, 5/1, 6/1], [Yes, Yes, No])."},"skills":["Absolute-Value Target Roots and Candidate Audit"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261010-advance-20261014-r06-absolute-value-0088","batch_id":"batch-20261010-advance-20261014-r06-absolute-value-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"89fc306551e1a7910fcd65bb128bd7b561a107d0ee94442def93ba2f783bfaea","family_id":"qf-density14-absolute-preimage-rational-audit","seed":202610146088,"semantic_template":"qf-density14-absolute-preimage-rational-audit/[\"11/2\",[\"11/2\",\"11/2\",\"13/2\"]]"},"fields":{"title":"Absolute-Value Target Roots and Candidate Audit","stem":"Use exact rational absolute values embedded in R; negative representation denominators are normalized before deciding signs. Target r=11/2, ordered queries=[-11/2, 11/2, 13/2]. Give (is r nonnegative?, increasing roots of |x|=r, three absolute query values, three equality-to-r values).","answer":{"kind":"expression","value":"(Yes, [-11/2, 11/2], [11/2, 11/2, 13/2], [Yes, Yes, No])","explanation":"A negative target has no roots, zero has only root zero, and a positive target has two signed roots; substitute each candidate exactly. The exact result, retaining the requested order, is (Yes, [-11/2, 11/2], [11/2, 11/2, 13/2], [Yes, Yes, No])."},"skills":["Absolute-Value Target Roots and Candidate Audit"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261010-advance-20261014-r06-absolute-value-0089","batch_id":"batch-20261010-advance-20261014-r06-absolute-value-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"9f78bdd8dfb40214021da7966f2b485c9267f5656817abfc628a5df481d47d55","family_id":"qf-density14-absolute-preimage-rational-audit","seed":202610146089,"semantic_template":"qf-density14-absolute-preimage-rational-audit/[\"6/1\",[\"6/1\",\"6/1\",\"7/1\"]]"},"fields":{"title":"Absolute-Value Target Roots and Candidate Audit","stem":"Use exact rational absolute values embedded in R; negative representation denominators are normalized before deciding signs. Target r=12/2, ordered queries=[-12/2, 12/2, 14/2]. Give (is r nonnegative?, increasing roots of |x|=r, three absolute query values, three equality-to-r values).","answer":{"kind":"expression","value":"(Yes, [-6/1, 6/1], [6/1, 6/1, 7/1], [Yes, Yes, No])","explanation":"A negative target has no roots, zero has only root zero, and a positive target has two signed roots; substitute each candidate exactly. The exact result, retaining the requested order, is (Yes, [-6/1, 6/1], [6/1, 6/1, 7/1], [Yes, Yes, No])."},"skills":["Absolute-Value Target Roots and Candidate Audit"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261010-advance-20261014-r06-absolute-value-0090","batch_id":"batch-20261010-advance-20261014-r06-absolute-value-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"5fc4da465b037d2c85ac3855ad27d595e26184857f023ac3a2f5efa97b6a8551","family_id":"qf-density14-absolute-preimage-rational-audit","seed":202610146090,"semantic_template":"qf-density14-absolute-preimage-rational-audit/[\"13/2\",[\"13/2\",\"13/2\",\"15/2\"]]"},"fields":{"title":"Absolute-Value Target Roots and Candidate Audit","stem":"Use exact rational absolute values embedded in R; negative representation denominators are normalized before deciding signs. Target r=13/2, ordered queries=[-13/2, 13/2, 15/2]. Give (is r nonnegative?, increasing roots of |x|=r, three absolute query values, three equality-to-r values).","answer":{"kind":"expression","value":"(Yes, [-13/2, 13/2], [13/2, 13/2, 15/2], [Yes, Yes, No])","explanation":"A negative target has no roots, zero has only root zero, and a positive target has two signed roots; substitute each candidate exactly. The exact result, retaining the requested order, is (Yes, [-13/2, 13/2], [13/2, 13/2, 15/2], [Yes, Yes, No])."},"skills":["Absolute-Value Target Roots and Candidate Audit"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261010-advance-20261014-r06-absolute-value-0091","batch_id":"batch-20261010-advance-20261014-r06-absolute-value-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"268ff7bee94eb8302405bad7fca6a62cd459c7c8e277c65df1716c014ec7e3e3","family_id":"qf-density14-absolute-preimage-rational-audit","seed":202610146091,"semantic_template":"qf-density14-absolute-preimage-rational-audit/[\"7/1\",[\"7/1\",\"7/1\",\"8/1\"]]"},"fields":{"title":"Absolute-Value Target Roots and Candidate Audit","stem":"Use exact rational absolute values embedded in R; negative representation denominators are normalized before deciding signs. Target r=14/2, ordered queries=[-14/2, 14/2, 16/2]. Give (is r nonnegative?, increasing roots of |x|=r, three absolute query values, three equality-to-r values).","answer":{"kind":"expression","value":"(Yes, [-7/1, 7/1], [7/1, 7/1, 8/1], [Yes, Yes, No])","explanation":"A negative target has no roots, zero has only root zero, and a positive target has two signed roots; substitute each candidate exactly. The exact result, retaining the requested order, is (Yes, [-7/1, 7/1], [7/1, 7/1, 8/1], [Yes, Yes, No])."},"skills":["Absolute-Value Target Roots and Candidate Audit"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261010-advance-20261014-r06-absolute-value-0092","batch_id":"batch-20261010-advance-20261014-r06-absolute-value-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"b4bd97de2a1777b1de7d1afe317361ad8d973393ebfc260d228cfddd66c97ed6","family_id":"qf-density14-absolute-preimage-rational-audit","seed":202610146092,"semantic_template":"qf-density14-absolute-preimage-rational-audit/[\"15/2\",[\"15/2\",\"15/2\",\"17/2\"]]"},"fields":{"title":"Absolute-Value Target Roots and Candidate Audit","stem":"Use exact rational absolute values embedded in R; negative representation denominators are normalized before deciding signs. Target r=15/2, ordered queries=[-15/2, 15/2, 17/2]. Give (is r nonnegative?, increasing roots of |x|=r, three absolute query values, three equality-to-r values).","answer":{"kind":"expression","value":"(Yes, [-15/2, 15/2], [15/2, 15/2, 17/2], [Yes, Yes, No])","explanation":"A negative target has no roots, zero has only root zero, and a positive target has two signed roots; substitute each candidate exactly. The exact result, retaining the requested order, is (Yes, [-15/2, 15/2], [15/2, 15/2, 17/2], [Yes, Yes, No])."},"skills":["Absolute-Value Target Roots and Candidate Audit"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261010-advance-20261014-r06-absolute-value-0093","batch_id":"batch-20261010-advance-20261014-r06-absolute-value-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"4ef94c4e47c62e342cb926ae8c0751d94e5a05de26ad46128bc67c55285dbbd4","family_id":"qf-density14-absolute-preimage-rational-audit","seed":202610146093,"semantic_template":"qf-density14-absolute-preimage-rational-audit/[\"8/1\",[\"8/1\",\"8/1\",\"9/1\"]]"},"fields":{"title":"Absolute-Value Target Roots and Candidate Audit","stem":"Use exact rational absolute values embedded in R; negative representation denominators are normalized before deciding signs. Target r=16/2, ordered queries=[-16/2, 16/2, 18/2]. Give (is r nonnegative?, increasing roots of |x|=r, three absolute query values, three equality-to-r values).","answer":{"kind":"expression","value":"(Yes, [-8/1, 8/1], [8/1, 8/1, 9/1], [Yes, Yes, No])","explanation":"A negative target has no roots, zero has only root zero, and a positive target has two signed roots; substitute each candidate exactly. The exact result, retaining the requested order, is (Yes, [-8/1, 8/1], [8/1, 8/1, 9/1], [Yes, Yes, No])."},"skills":["Absolute-Value Target Roots and Candidate Audit"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261010-advance-20261014-r06-absolute-value-0094","batch_id":"batch-20261010-advance-20261014-r06-absolute-value-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"df883631911b081cd5d7929f606bf55d8d6735d3f86358763421fa01b6ceb2e6","family_id":"qf-density14-absolute-preimage-rational-audit","seed":202610146094,"semantic_template":"qf-density14-absolute-preimage-rational-audit/[\"17/2\",[\"17/2\",\"17/2\",\"19/2\"]]"},"fields":{"title":"Absolute-Value Target Roots and Candidate Audit","stem":"Use exact rational absolute values embedded in R; negative representation denominators are normalized before deciding signs. Target r=17/2, ordered queries=[-17/2, 17/2, 19/2]. Give (is r nonnegative?, increasing roots of |x|=r, three absolute query values, three equality-to-r values).","answer":{"kind":"expression","value":"(Yes, [-17/2, 17/2], [17/2, 17/2, 19/2], [Yes, Yes, No])","explanation":"A negative target has no roots, zero has only root zero, and a positive target has two signed roots; substitute each candidate exactly. The exact result, retaining the requested order, is (Yes, [-17/2, 17/2], [17/2, 17/2, 19/2], [Yes, Yes, No])."},"skills":["Absolute-Value Target Roots and Candidate Audit"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261010-advance-20261014-r06-absolute-value-0095","batch_id":"batch-20261010-advance-20261014-r06-absolute-value-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"4a9870af26dc2cfd2ef5798322744211cf9666e11b799be26545a881cd0ad670","family_id":"qf-density14-absolute-preimage-rational-audit","seed":202610146095,"semantic_template":"qf-density14-absolute-preimage-rational-audit/[\"9/1\",[\"9/1\",\"9/1\",\"10/1\"]]"},"fields":{"title":"Absolute-Value Target Roots and Candidate Audit","stem":"Use exact rational absolute values embedded in R; negative representation denominators are normalized before deciding signs. Target r=18/2, ordered queries=[-18/2, 18/2, 20/2]. Give (is r nonnegative?, increasing roots of |x|=r, three absolute query values, three equality-to-r values).","answer":{"kind":"expression","value":"(Yes, [-9/1, 9/1], [9/1, 9/1, 10/1], [Yes, Yes, No])","explanation":"A negative target has no roots, zero has only root zero, and a positive target has two signed roots; substitute each candidate exactly. The exact result, retaining the requested order, is (Yes, [-9/1, 9/1], [9/1, 9/1, 10/1], [Yes, Yes, No])."},"skills":["Absolute-Value Target Roots and Candidate Audit"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261010-advance-20261014-r06-absolute-value-0096","batch_id":"batch-20261010-advance-20261014-r06-absolute-value-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"2c619922ba5223f15188aef571ed4a429ab490360a980fcf8839358dc9943d7a","family_id":"qf-density14-absolute-preimage-rational-audit","seed":202610146096,"semantic_template":"qf-density14-absolute-preimage-rational-audit/[\"19/2\",[\"19/2\",\"19/2\",\"21/2\"]]"},"fields":{"title":"Absolute-Value Target Roots and Candidate Audit","stem":"Use exact rational absolute values embedded in R; negative representation denominators are normalized before deciding signs. Target r=19/2, ordered queries=[-19/2, 19/2, 21/2]. Give (is r nonnegative?, increasing roots of |x|=r, three absolute query values, three equality-to-r values).","answer":{"kind":"expression","value":"(Yes, [-19/2, 19/2], [19/2, 19/2, 21/2], [Yes, Yes, No])","explanation":"A negative target has no roots, zero has only root zero, and a positive target has two signed roots; substitute each candidate exactly. The exact result, retaining the requested order, is (Yes, [-19/2, 19/2], [19/2, 19/2, 21/2], [Yes, Yes, No])."},"skills":["Absolute-Value Target Roots and Candidate Audit"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261010-advance-20261014-r06-absolute-value-0097","batch_id":"batch-20261010-advance-20261014-r06-absolute-value-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"6d2536f91c9ad4e22070d1f387ad5bd38786a43165d6fb0374a9a233ee95aa5c","family_id":"qf-density14-absolute-preimage-rational-audit","seed":202610146097,"semantic_template":"qf-density14-absolute-preimage-rational-audit/[\"10/1\",[\"10/1\",\"10/1\",\"11/1\"]]"},"fields":{"title":"Absolute-Value Target Roots and Candidate Audit","stem":"Use exact rational absolute values embedded in R; negative representation denominators are normalized before deciding signs. Target r=20/2, ordered queries=[-20/2, 20/2, 22/2]. Give (is r nonnegative?, increasing roots of |x|=r, three absolute query values, three equality-to-r values).","answer":{"kind":"expression","value":"(Yes, [-10/1, 10/1], [10/1, 10/1, 11/1], [Yes, Yes, No])","explanation":"A negative target has no roots, zero has only root zero, and a positive target has two signed roots; substitute each candidate exactly. The exact result, retaining the requested order, is (Yes, [-10/1, 10/1], [10/1, 10/1, 11/1], [Yes, Yes, No])."},"skills":["Absolute-Value Target Roots and Candidate Audit"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261010-advance-20261014-r06-absolute-value-0098","batch_id":"batch-20261010-advance-20261014-r06-absolute-value-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"c7bc84f1416673d32acda4dc28e42be11b47b619c656a409c0813a499e90b40c","family_id":"qf-density14-absolute-preimage-rational-audit","seed":202610146098,"semantic_template":"qf-density14-absolute-preimage-rational-audit/[\"21/2\",[\"21/2\",\"21/2\",\"23/2\"]]"},"fields":{"title":"Absolute-Value Target Roots and Candidate Audit","stem":"Use exact rational absolute values embedded in R; negative representation denominators are normalized before deciding signs. Target r=21/2, ordered queries=[-21/2, 21/2, 23/2]. Give (is r nonnegative?, increasing roots of |x|=r, three absolute query values, three equality-to-r values).","answer":{"kind":"expression","value":"(Yes, [-21/2, 21/2], [21/2, 21/2, 23/2], [Yes, Yes, No])","explanation":"A negative target has no roots, zero has only root zero, and a positive target has two signed roots; substitute each candidate exactly. The exact result, retaining the requested order, is (Yes, [-21/2, 21/2], [21/2, 21/2, 23/2], [Yes, Yes, No])."},"skills":["Absolute-Value Target Roots and Candidate Audit"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261010-advance-20261014-r06-absolute-value-0099","batch_id":"batch-20261010-advance-20261014-r06-absolute-value-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"c8f6c9d03c04deb795e7887d30d2a130bcf99d4b5e78770af122cc43f28762e8","family_id":"qf-density14-absolute-preimage-rational-audit","seed":202610146099,"semantic_template":"qf-density14-absolute-preimage-rational-audit/[\"11/1\",[\"11/1\",\"11/1\",\"12/1\"]]"},"fields":{"title":"Absolute-Value Target Roots and Candidate Audit","stem":"Use exact rational absolute values embedded in R; negative representation denominators are normalized before deciding signs. Target r=22/2, ordered queries=[-22/2, 22/2, 24/2]. Give (is r nonnegative?, increasing roots of |x|=r, three absolute query values, three equality-to-r values).","answer":{"kind":"expression","value":"(Yes, [-11/1, 11/1], [11/1, 11/1, 12/1], [Yes, Yes, No])","explanation":"A negative target has no roots, zero has only root zero, and a positive target has two signed roots; substitute each candidate exactly. The exact result, retaining the requested order, is (Yes, [-11/1, 11/1], [11/1, 11/1, 12/1], [Yes, Yes, No])."},"skills":["Absolute-Value Target Roots and Candidate Audit"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261010-advance-20261014-r06-absolute-value-0100","batch_id":"batch-20261010-advance-20261014-r06-absolute-value-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"19fde4cff8bb29d5dde01a93f0394192f6f3971f92307148e3f13e73a518aa79","family_id":"qf-density14-absolute-preimage-rational-audit","seed":202610146100,"semantic_template":"qf-density14-absolute-preimage-rational-audit/[\"23/2\",[\"23/2\",\"23/2\",\"25/2\"]]"},"fields":{"title":"Absolute-Value Target Roots and Candidate Audit","stem":"Use exact rational absolute values embedded in R; negative representation denominators are normalized before deciding signs. Target r=23/2, ordered queries=[-23/2, 23/2, 25/2]. Give (is r nonnegative?, increasing roots of |x|=r, three absolute query values, three equality-to-r values).","answer":{"kind":"expression","value":"(Yes, [-23/2, 23/2], [23/2, 23/2, 25/2], [Yes, Yes, No])","explanation":"A negative target has no roots, zero has only root zero, and a positive target has two signed roots; substitute each candidate exactly. The exact result, retaining the requested order, is (Yes, [-23/2, 23/2], [23/2, 23/2, 25/2], [Yes, Yes, No])."},"skills":["Absolute-Value Target Roots and Candidate Audit"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}