Complete Finite Ring Tables, Relabeled Identities, Additive Inverses and Powers
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This question bank has corresponding bilingual step-by-step derivations. Source candidate and English ai_candidate states are retained. AI review, finite computation and Lean formal proof are labelled separately.
batch-20261014-advance-20261027-r01-finite-unital-ring-foundations · 2026-10-14 · 100 questions
This English edition is an AI candidate and does not yet have an OCME Atlas (GPT-6) Twin closure receipt.
1. Qualify Both Complete Ring Tables without Assuming Their Identities
Use the same carrier Fin(6)={0,…,5} for complete addition=[[0, 1, 2, 3, 4, 5], [1, 2, 3, 4, 5, 0], [2, 3, 4, 5, 0, 1], [3, 4, 5, 0, 1, 2], [4, 5, 0, 1, 2, 3], [5, 0, 1, 2, 3, 4]] and multiplication=[[0, 0, 0, 0, 0, 0], [0, 1, 2, 3, 4, 5], [0, 2, 4, 1, 2, 4], [0, 3, 0, 3, 0, 3], [0, 4, 2, 0, 4, 2], [0, 5, 4, 3, 2, 1]] tables in label order. Check the additive abelian-group axioms, multiplicative associativity, both distributive laws and a two-sided multiplicative identity distinct from the additive identity. The two identity labels need not be 0 and 1. This protocol excludes the zero ring and does not require multiplicative commutativity or imply a field. Give (n,zero,one,additive_abelian?,multiplication_associative?,left_distributive?,right_distributive?,distinct_identities?,qualified?,first_failure). Missing or nonunique identities and no failure are represented by null. Failure priority is additive_structure, multiplication_associativity, left_distributivity, right_distributivity, then distinct_identities.
View answer and explanation
(6, 0, 1, Yes, No, No, No, Yes, No, multiplication_associativity)
Audit every relevant element pair or triple, retaining the distinct obligation flags even after the first failure. The qualification flag is their conjunction, not a few passing examples. An exact finite or coordinate slice does not automatically establish a general infinite theorem. The exact result, retaining the requested order, is (6, 0, 1, Yes, No, No, No, Yes, No, multiplication_associativity).
Reviewed step-by-step derivation
Finite unital ring · qf-mod27-ring-qualification · candidate
2. Qualify Both Complete Ring Tables without Assuming Their Identities
Use the same carrier Fin(6)={0,…,5} for complete addition=[[0, 1, 2, 3, 4, 5], [1, 2, 3, 5, 0, 4], [2, 3, 5, 4, 1, 0], [3, 5, 4, 0, 2, 1], [4, 0, 1, 2, 5, 3], [5, 4, 0, 1, 3, 2]] and multiplication=[[0, 0, 0, 0, 0, 0], [0, 1, 2, 3, 4, 5], [0, 2, 5, 0, 5, 2], [0, 3, 0, 3, 3, 0], [0, 4, 5, 3, 1, 2], [0, 5, 2, 0, 2, 5]] tables in label order. Check the additive abelian-group axioms, multiplicative associativity, both distributive laws and a two-sided multiplicative identity distinct from the additive identity. The two identity labels need not be 0 and 1. This protocol excludes the zero ring and does not require multiplicative commutativity or imply a field. Give (n,zero,one,additive_abelian?,multiplication_associative?,left_distributive?,right_distributive?,distinct_identities?,qualified?,first_failure). Missing or nonunique identities and no failure are represented by null. Failure priority is additive_structure, multiplication_associativity, left_distributivity, right_distributivity, then distinct_identities.
View answer and explanation
(6, 0, 1, Yes, Yes, Yes, Yes, Yes, Yes, null)
Audit every relevant element pair or triple, retaining the distinct obligation flags even after the first failure. The qualification flag is their conjunction, not a few passing examples. An exact finite or coordinate slice does not automatically establish a general infinite theorem. The exact result, retaining the requested order, is (6, 0, 1, Yes, Yes, Yes, Yes, Yes, Yes, null).
Reviewed step-by-step derivation
Finite unital ring · qf-mod27-ring-qualification · candidate
3. Qualify Both Complete Ring Tables without Assuming Their Identities
Use the same carrier Fin(6)={0,…,5} for complete addition=[[0, 1, 2, 3, 4, 5], [1, 2, 4, 5, 3, 0], [2, 4, 3, 0, 5, 1], [3, 5, 0, 2, 1, 4], [4, 3, 5, 1, 0, 2], [5, 0, 1, 4, 2, 3]] and multiplication=[[0, 0, 0, 0, 0, 0], [0, 1, 2, 3, 4, 5], [0, 2, 3, 2, 0, 3], [0, 3, 2, 3, 0, 2], [0, 4, 0, 0, 4, 4], [0, 5, 3, 2, 4, 1]] tables in label order. Check the additive abelian-group axioms, multiplicative associativity, both distributive laws and a two-sided multiplicative identity distinct from the additive identity. The two identity labels need not be 0 and 1. This protocol excludes the zero ring and does not require multiplicative commutativity or imply a field. Give (n,zero,one,additive_abelian?,multiplication_associative?,left_distributive?,right_distributive?,distinct_identities?,qualified?,first_failure). Missing or nonunique identities and no failure are represented by null. Failure priority is additive_structure, multiplication_associativity, left_distributivity, right_distributivity, then distinct_identities.
View answer and explanation
(6, 0, 1, Yes, Yes, Yes, Yes, Yes, Yes, null)
Audit every relevant element pair or triple, retaining the distinct obligation flags even after the first failure. The qualification flag is their conjunction, not a few passing examples. An exact finite or coordinate slice does not automatically establish a general infinite theorem. The exact result, retaining the requested order, is (6, 0, 1, Yes, Yes, Yes, Yes, Yes, Yes, null).
Reviewed step-by-step derivation
Finite unital ring · qf-mod27-ring-qualification · candidate
4. Qualify Both Complete Ring Tables without Assuming Their Identities
Use the same carrier Fin(6)={0,…,5} for complete addition=[[0, 1, 2, 3, 4, 5], [1, 2, 4, 0, 5, 3], [2, 4, 5, 1, 3, 0], [3, 0, 1, 5, 2, 4], [4, 5, 3, 2, 0, 1], [5, 3, 0, 4, 1, 2]] and multiplication=[[0, 0, 0, 0, 0, 0], [0, 1, 2, 3, 4, 5], [0, 2, 5, 5, 0, 2], [0, 3, 5, 1, 4, 2], [0, 4, 0, 4, 4, 0], [0, 5, 2, 2, 0, 5]] tables in label order. Check the additive abelian-group axioms, multiplicative associativity, both distributive laws and a two-sided multiplicative identity distinct from the additive identity. The two identity labels need not be 0 and 1. This protocol excludes the zero ring and does not require multiplicative commutativity or imply a field. Give (n,zero,one,additive_abelian?,multiplication_associative?,left_distributive?,right_distributive?,distinct_identities?,qualified?,first_failure). Missing or nonunique identities and no failure are represented by null. Failure priority is additive_structure, multiplication_associativity, left_distributivity, right_distributivity, then distinct_identities.
View answer and explanation
(6, 0, 1, Yes, Yes, Yes, Yes, Yes, Yes, null)
Audit every relevant element pair or triple, retaining the distinct obligation flags even after the first failure. The qualification flag is their conjunction, not a few passing examples. An exact finite or coordinate slice does not automatically establish a general infinite theorem. The exact result, retaining the requested order, is (6, 0, 1, Yes, Yes, Yes, Yes, Yes, Yes, null).
Reviewed step-by-step derivation
Finite unital ring · qf-mod27-ring-qualification · candidate
5. Qualify Both Complete Ring Tables without Assuming Their Identities
Use the same carrier Fin(6)={0,…,5} for complete addition=[[0, 1, 2, 3, 4, 5], [1, 2, 5, 4, 0, 3], [2, 5, 3, 0, 1, 4], [3, 4, 0, 2, 5, 1], [4, 0, 1, 5, 3, 2], [5, 3, 4, 1, 2, 0]] and multiplication=[[0, 0, 0, 0, 0, 0], [0, 1, 2, 3, 4, 5], [0, 2, 3, 2, 3, 0], [0, 3, 2, 3, 2, 0], [0, 4, 3, 2, 1, 5], [0, 5, 0, 0, 5, 5]] tables in label order. Check the additive abelian-group axioms, multiplicative associativity, both distributive laws and a two-sided multiplicative identity distinct from the additive identity. The two identity labels need not be 0 and 1. This protocol excludes the zero ring and does not require multiplicative commutativity or imply a field. Give (n,zero,one,additive_abelian?,multiplication_associative?,left_distributive?,right_distributive?,distinct_identities?,qualified?,first_failure). Missing or nonunique identities and no failure are represented by null. Failure priority is additive_structure, multiplication_associativity, left_distributivity, right_distributivity, then distinct_identities.
View answer and explanation
(6, 0, 1, Yes, Yes, Yes, Yes, Yes, Yes, null)
Audit every relevant element pair or triple, retaining the distinct obligation flags even after the first failure. The qualification flag is their conjunction, not a few passing examples. An exact finite or coordinate slice does not automatically establish a general infinite theorem. The exact result, retaining the requested order, is (6, 0, 1, Yes, Yes, Yes, Yes, Yes, Yes, null).
Reviewed step-by-step derivation
Finite unital ring · qf-mod27-ring-qualification · candidate
6. Qualify Both Complete Ring Tables without Assuming Their Identities
Use the same carrier Fin(6)={0,…,5} for complete addition=[[0, 1, 2, 3, 4, 5], [1, 2, 5, 0, 3, 4], [2, 5, 4, 1, 0, 3], [3, 0, 1, 4, 5, 2], [4, 3, 0, 5, 2, 1], [5, 4, 3, 2, 1, 0]] and multiplication=[[0, 0, 0, 0, 0, 0], [0, 1, 2, 3, 4, 5], [0, 2, 4, 5, 2, 0], [0, 3, 4, 1, 2, 5], [0, 4, 2, 2, 4, 0], [0, 5, 0, 5, 0, 5]] tables in label order. Check the additive abelian-group axioms, multiplicative associativity, both distributive laws and a two-sided multiplicative identity distinct from the additive identity. The two identity labels need not be 0 and 1. This protocol excludes the zero ring and does not require multiplicative commutativity or imply a field. Give (n,zero,one,additive_abelian?,multiplication_associative?,left_distributive?,right_distributive?,distinct_identities?,qualified?,first_failure). Missing or nonunique identities and no failure are represented by null. Failure priority is additive_structure, multiplication_associativity, left_distributivity, right_distributivity, then distinct_identities.
View answer and explanation
(6, 0, 1, Yes, No, No, No, Yes, No, multiplication_associativity)
Audit every relevant element pair or triple, retaining the distinct obligation flags even after the first failure. The qualification flag is their conjunction, not a few passing examples. An exact finite or coordinate slice does not automatically establish a general infinite theorem. The exact result, retaining the requested order, is (6, 0, 1, Yes, No, No, No, Yes, No, multiplication_associativity).
Reviewed step-by-step derivation
Finite unital ring · qf-mod27-ring-qualification · candidate
7. Qualify Both Complete Ring Tables without Assuming Their Identities
Use the same carrier Fin(6)={0,…,5} for complete addition=[[0, 1, 2, 3, 4, 5], [1, 3, 4, 2, 5, 0], [2, 4, 0, 5, 1, 3], [3, 2, 5, 4, 0, 1], [4, 5, 1, 0, 3, 2], [5, 0, 3, 1, 2, 4]] and multiplication=[[0, 0, 0, 0, 0, 0], [0, 1, 2, 3, 4, 5], [0, 2, 2, 0, 0, 2], [0, 3, 0, 4, 3, 4], [0, 4, 0, 3, 4, 3], [0, 5, 2, 4, 3, 1]] tables in label order. Check the additive abelian-group axioms, multiplicative associativity, both distributive laws and a two-sided multiplicative identity distinct from the additive identity. The two identity labels need not be 0 and 1. This protocol excludes the zero ring and does not require multiplicative commutativity or imply a field. Give (n,zero,one,additive_abelian?,multiplication_associative?,left_distributive?,right_distributive?,distinct_identities?,qualified?,first_failure). Missing or nonunique identities and no failure are represented by null. Failure priority is additive_structure, multiplication_associativity, left_distributivity, right_distributivity, then distinct_identities.
View answer and explanation
(6, 0, 1, Yes, Yes, Yes, Yes, Yes, Yes, null)
Audit every relevant element pair or triple, retaining the distinct obligation flags even after the first failure. The qualification flag is their conjunction, not a few passing examples. An exact finite or coordinate slice does not automatically establish a general infinite theorem. The exact result, retaining the requested order, is (6, 0, 1, Yes, Yes, Yes, Yes, Yes, Yes, null).
Reviewed step-by-step derivation
Finite unital ring · qf-mod27-ring-qualification · candidate
8. Qualify Both Complete Ring Tables without Assuming Their Identities
Use the same carrier Fin(6)={0,…,5} for complete addition=[[0, 1, 2, 3, 4, 5], [1, 3, 5, 2, 0, 4], [2, 5, 0, 4, 3, 1], [3, 2, 4, 5, 1, 0], [4, 0, 3, 1, 5, 2], [5, 4, 1, 0, 2, 3]] and multiplication=[[0, 0, 0, 0, 0, 0], [0, 1, 2, 3, 4, 5], [0, 2, 2, 0, 2, 0], [0, 3, 0, 5, 5, 3], [0, 4, 2, 5, 1, 3], [0, 5, 0, 3, 3, 5]] tables in label order. Check the additive abelian-group axioms, multiplicative associativity, both distributive laws and a two-sided multiplicative identity distinct from the additive identity. The two identity labels need not be 0 and 1. This protocol excludes the zero ring and does not require multiplicative commutativity or imply a field. Give (n,zero,one,additive_abelian?,multiplication_associative?,left_distributive?,right_distributive?,distinct_identities?,qualified?,first_failure). Missing or nonunique identities and no failure are represented by null. Failure priority is additive_structure, multiplication_associativity, left_distributivity, right_distributivity, then distinct_identities.
View answer and explanation
(6, 0, 1, Yes, Yes, Yes, Yes, Yes, Yes, null)
Audit every relevant element pair or triple, retaining the distinct obligation flags even after the first failure. The qualification flag is their conjunction, not a few passing examples. An exact finite or coordinate slice does not automatically establish a general infinite theorem. The exact result, retaining the requested order, is (6, 0, 1, Yes, Yes, Yes, Yes, Yes, Yes, null).
Reviewed step-by-step derivation
Finite unital ring · qf-mod27-ring-qualification · candidate
9. Qualify Both Complete Ring Tables without Assuming Their Identities
Use the same carrier Fin(6)={0,…,5} for complete addition=[[0, 1, 2, 3, 4, 5], [1, 3, 5, 4, 2, 0], [2, 5, 3, 0, 1, 4], [3, 4, 0, 2, 5, 1], [4, 2, 1, 5, 0, 3], [5, 0, 4, 1, 3, 2]] and multiplication=[[0, 0, 0, 0, 0, 0], [0, 1, 2, 3, 4, 5], [0, 2, 2, 3, 0, 3], [0, 3, 3, 2, 0, 2], [0, 4, 0, 0, 4, 4], [0, 5, 3, 2, 4, 1]] tables in label order. Check the additive abelian-group axioms, multiplicative associativity, both distributive laws and a two-sided multiplicative identity distinct from the additive identity. The two identity labels need not be 0 and 1. This protocol excludes the zero ring and does not require multiplicative commutativity or imply a field. Give (n,zero,one,additive_abelian?,multiplication_associative?,left_distributive?,right_distributive?,distinct_identities?,qualified?,first_failure). Missing or nonunique identities and no failure are represented by null. Failure priority is additive_structure, multiplication_associativity, left_distributivity, right_distributivity, then distinct_identities.
View answer and explanation
(6, 0, 1, Yes, Yes, Yes, Yes, Yes, Yes, null)
Audit every relevant element pair or triple, retaining the distinct obligation flags even after the first failure. The qualification flag is their conjunction, not a few passing examples. An exact finite or coordinate slice does not automatically establish a general infinite theorem. The exact result, retaining the requested order, is (6, 0, 1, Yes, Yes, Yes, Yes, Yes, Yes, null).
Reviewed step-by-step derivation
Finite unital ring · qf-mod27-ring-qualification · candidate
10. Qualify Both Complete Ring Tables without Assuming Their Identities
Use the same carrier Fin(6)={0,…,5} for complete addition=[[0, 1, 2, 3, 4, 5], [1, 3, 0, 4, 5, 2], [2, 0, 5, 1, 3, 4], [3, 4, 1, 5, 2, 0], [4, 5, 3, 2, 0, 1], [5, 2, 4, 0, 1, 3]] and multiplication=[[0, 0, 0, 0, 0, 0], [0, 1, 2, 3, 4, 5], [0, 2, 1, 5, 4, 3], [0, 3, 5, 5, 0, 3], [0, 4, 4, 0, 4, 0], [0, 5, 3, 3, 0, 5]] tables in label order. Check the additive abelian-group axioms, multiplicative associativity, both distributive laws and a two-sided multiplicative identity distinct from the additive identity. The two identity labels need not be 0 and 1. This protocol excludes the zero ring and does not require multiplicative commutativity or imply a field. Give (n,zero,one,additive_abelian?,multiplication_associative?,left_distributive?,right_distributive?,distinct_identities?,qualified?,first_failure). Missing or nonunique identities and no failure are represented by null. Failure priority is additive_structure, multiplication_associativity, left_distributivity, right_distributivity, then distinct_identities.
View answer and explanation
(6, 0, 1, Yes, Yes, Yes, Yes, Yes, Yes, null)
Audit every relevant element pair or triple, retaining the distinct obligation flags even after the first failure. The qualification flag is their conjunction, not a few passing examples. An exact finite or coordinate slice does not automatically establish a general infinite theorem. The exact result, retaining the requested order, is (6, 0, 1, Yes, Yes, Yes, Yes, Yes, Yes, null).
Reviewed step-by-step derivation
Finite unital ring · qf-mod27-ring-qualification · candidate
11. Qualify Both Complete Ring Tables without Assuming Their Identities
Use the same carrier Fin(6)={0,…,5} for complete addition=[[0, 1, 2, 3, 4, 5], [1, 3, 4, 5, 0, 2], [2, 4, 3, 0, 5, 1], [3, 5, 0, 2, 1, 4], [4, 0, 5, 1, 2, 3], [5, 2, 1, 4, 3, 0]] and multiplication=[[0, 0, 0, 0, 0, 0], [0, 1, 2, 3, 4, 5], [0, 2, 2, 4, 3, 0], [0, 3, 3, 2, 2, 0], [0, 4, 3, 2, 1, 5], [0, 5, 0, 0, 5, 5]] tables in label order. Check the additive abelian-group axioms, multiplicative associativity, both distributive laws and a two-sided multiplicative identity distinct from the additive identity. The two identity labels need not be 0 and 1. This protocol excludes the zero ring and does not require multiplicative commutativity or imply a field. Give (n,zero,one,additive_abelian?,multiplication_associative?,left_distributive?,right_distributive?,distinct_identities?,qualified?,first_failure). Missing or nonunique identities and no failure are represented by null. Failure priority is additive_structure, multiplication_associativity, left_distributivity, right_distributivity, then distinct_identities.
View answer and explanation
(6, 0, 1, Yes, No, No, No, Yes, No, multiplication_associativity)
Audit every relevant element pair or triple, retaining the distinct obligation flags even after the first failure. The qualification flag is their conjunction, not a few passing examples. An exact finite or coordinate slice does not automatically establish a general infinite theorem. The exact result, retaining the requested order, is (6, 0, 1, Yes, No, No, No, Yes, No, multiplication_associativity).
Reviewed step-by-step derivation
Finite unital ring · qf-mod27-ring-qualification · candidate
12. Qualify Both Complete Ring Tables without Assuming Their Identities
Use the same carrier Fin(6)={0,…,5} for complete addition=[[0, 1, 2, 3, 4, 5], [1, 3, 0, 5, 2, 4], [2, 0, 4, 1, 5, 3], [3, 5, 1, 4, 0, 2], [4, 2, 5, 0, 3, 1], [5, 4, 3, 2, 1, 0]] and multiplication=[[0, 0, 0, 0, 0, 0], [0, 1, 2, 3, 4, 5], [0, 2, 1, 4, 3, 5], [0, 3, 4, 4, 3, 0], [0, 4, 3, 3, 4, 0], [0, 5, 5, 0, 0, 5]] tables in label order. Check the additive abelian-group axioms, multiplicative associativity, both distributive laws and a two-sided multiplicative identity distinct from the additive identity. The two identity labels need not be 0 and 1. This protocol excludes the zero ring and does not require multiplicative commutativity or imply a field. Give (n,zero,one,additive_abelian?,multiplication_associative?,left_distributive?,right_distributive?,distinct_identities?,qualified?,first_failure). Missing or nonunique identities and no failure are represented by null. Failure priority is additive_structure, multiplication_associativity, left_distributivity, right_distributivity, then distinct_identities.
View answer and explanation
(6, 0, 1, Yes, Yes, Yes, Yes, Yes, Yes, null)
Audit every relevant element pair or triple, retaining the distinct obligation flags even after the first failure. The qualification flag is their conjunction, not a few passing examples. An exact finite or coordinate slice does not automatically establish a general infinite theorem. The exact result, retaining the requested order, is (6, 0, 1, Yes, Yes, Yes, Yes, Yes, Yes, null).
Reviewed step-by-step derivation
Finite unital ring · qf-mod27-ring-qualification · candidate
13. Qualify Both Complete Ring Tables without Assuming Their Identities
Use the same carrier Fin(6)={0,…,5} for complete addition=[[0, 1, 2, 3, 4, 5], [1, 4, 3, 5, 2, 0], [2, 3, 0, 1, 5, 4], [3, 5, 1, 4, 0, 2], [4, 2, 5, 0, 3, 1], [5, 0, 4, 2, 1, 3]] and multiplication=[[0, 0, 0, 0, 0, 0], [0, 1, 2, 3, 4, 5], [0, 2, 2, 0, 0, 2], [0, 3, 0, 3, 4, 4], [0, 4, 0, 4, 3, 3], [0, 5, 2, 4, 3, 1]] tables in label order. Check the additive abelian-group axioms, multiplicative associativity, both distributive laws and a two-sided multiplicative identity distinct from the additive identity. The two identity labels need not be 0 and 1. This protocol excludes the zero ring and does not require multiplicative commutativity or imply a field. Give (n,zero,one,additive_abelian?,multiplication_associative?,left_distributive?,right_distributive?,distinct_identities?,qualified?,first_failure). Missing or nonunique identities and no failure are represented by null. Failure priority is additive_structure, multiplication_associativity, left_distributivity, right_distributivity, then distinct_identities.
View answer and explanation
(6, 0, 1, Yes, Yes, Yes, Yes, Yes, Yes, null)
Audit every relevant element pair or triple, retaining the distinct obligation flags even after the first failure. The qualification flag is their conjunction, not a few passing examples. An exact finite or coordinate slice does not automatically establish a general infinite theorem. The exact result, retaining the requested order, is (6, 0, 1, Yes, Yes, Yes, Yes, Yes, Yes, null).
Reviewed step-by-step derivation
Finite unital ring · qf-mod27-ring-qualification · candidate
14. Qualify Both Complete Ring Tables without Assuming Their Identities
Use the same carrier Fin(6)={0,…,5} for complete addition=[[0, 1, 2, 3, 4, 5], [1, 4, 5, 0, 2, 3], [2, 5, 0, 4, 3, 1], [3, 0, 4, 5, 1, 2], [4, 2, 3, 1, 5, 0], [5, 3, 1, 2, 0, 4]] and multiplication=[[0, 0, 0, 0, 0, 0], [0, 1, 2, 3, 4, 5], [0, 2, 2, 2, 0, 0], [0, 3, 2, 1, 5, 4], [0, 4, 0, 5, 5, 4], [0, 5, 0, 4, 4, 5]] tables in label order. Check the additive abelian-group axioms, multiplicative associativity, both distributive laws and a two-sided multiplicative identity distinct from the additive identity. The two identity labels need not be 0 and 1. This protocol excludes the zero ring and does not require multiplicative commutativity or imply a field. Give (n,zero,one,additive_abelian?,multiplication_associative?,left_distributive?,right_distributive?,distinct_identities?,qualified?,first_failure). Missing or nonunique identities and no failure are represented by null. Failure priority is additive_structure, multiplication_associativity, left_distributivity, right_distributivity, then distinct_identities.
View answer and explanation
(6, 0, 1, Yes, Yes, Yes, Yes, Yes, Yes, null)
Audit every relevant element pair or triple, retaining the distinct obligation flags even after the first failure. The qualification flag is their conjunction, not a few passing examples. An exact finite or coordinate slice does not automatically establish a general infinite theorem. The exact result, retaining the requested order, is (6, 0, 1, Yes, Yes, Yes, Yes, Yes, Yes, null).
Reviewed step-by-step derivation
Finite unital ring · qf-mod27-ring-qualification · candidate
15. Qualify Both Complete Ring Tables without Assuming Their Identities
Use the same carrier Fin(6)={0,…,5} for complete addition=[[0, 1, 2, 3, 4, 5], [1, 4, 5, 2, 3, 0], [2, 5, 4, 1, 0, 3], [3, 2, 1, 0, 5, 4], [4, 3, 0, 5, 2, 1], [5, 0, 3, 4, 1, 2]] and multiplication=[[0, 0, 0, 0, 0, 0], [0, 1, 2, 3, 4, 5], [0, 2, 2, 0, 4, 4], [0, 3, 0, 3, 0, 3], [0, 4, 4, 0, 2, 2], [0, 5, 4, 3, 2, 1]] tables in label order. Check the additive abelian-group axioms, multiplicative associativity, both distributive laws and a two-sided multiplicative identity distinct from the additive identity. The two identity labels need not be 0 and 1. This protocol excludes the zero ring and does not require multiplicative commutativity or imply a field. Give (n,zero,one,additive_abelian?,multiplication_associative?,left_distributive?,right_distributive?,distinct_identities?,qualified?,first_failure). Missing or nonunique identities and no failure are represented by null. Failure priority is additive_structure, multiplication_associativity, left_distributivity, right_distributivity, then distinct_identities.
View answer and explanation
(6, 0, 1, Yes, Yes, Yes, Yes, Yes, Yes, null)
Audit every relevant element pair or triple, retaining the distinct obligation flags even after the first failure. The qualification flag is their conjunction, not a few passing examples. An exact finite or coordinate slice does not automatically establish a general infinite theorem. The exact result, retaining the requested order, is (6, 0, 1, Yes, Yes, Yes, Yes, Yes, Yes, null).
Reviewed step-by-step derivation
Finite unital ring · qf-mod27-ring-qualification · candidate
16. Qualify Both Complete Ring Tables without Assuming Their Identities
Use the same carrier Fin(6)={0,…,5} for complete addition=[[0, 1, 2, 3, 4, 5], [1, 4, 0, 5, 3, 2], [2, 0, 5, 4, 1, 3], [3, 5, 4, 0, 2, 1], [4, 3, 1, 2, 5, 0], [5, 2, 3, 1, 0, 4]] and multiplication=[[0, 0, 0, 0, 0, 0], [0, 1, 2, 3, 4, 5], [0, 2, 1, 4, 5, 4], [0, 3, 3, 3, 0, 0], [0, 4, 5, 0, 5, 4], [0, 5, 4, 0, 4, 5]] tables in label order. Check the additive abelian-group axioms, multiplicative associativity, both distributive laws and a two-sided multiplicative identity distinct from the additive identity. The two identity labels need not be 0 and 1. This protocol excludes the zero ring and does not require multiplicative commutativity or imply a field. Give (n,zero,one,additive_abelian?,multiplication_associative?,left_distributive?,right_distributive?,distinct_identities?,qualified?,first_failure). Missing or nonunique identities and no failure are represented by null. Failure priority is additive_structure, multiplication_associativity, left_distributivity, right_distributivity, then distinct_identities.
View answer and explanation
(6, 0, 1, Yes, No, No, No, Yes, No, multiplication_associativity)
Audit every relevant element pair or triple, retaining the distinct obligation flags even after the first failure. The qualification flag is their conjunction, not a few passing examples. An exact finite or coordinate slice does not automatically establish a general infinite theorem. The exact result, retaining the requested order, is (6, 0, 1, Yes, No, No, No, Yes, No, multiplication_associativity).
Reviewed step-by-step derivation
Finite unital ring · qf-mod27-ring-qualification · candidate
17. Qualify Both Complete Ring Tables without Assuming Their Identities
Use the same carrier Fin(6)={0,…,5} for complete addition=[[0, 1, 2, 3, 4, 5], [1, 4, 3, 0, 5, 2], [2, 3, 4, 5, 0, 1], [3, 0, 5, 2, 1, 4], [4, 5, 0, 1, 2, 3], [5, 2, 1, 4, 3, 0]] and multiplication=[[0, 0, 0, 0, 0, 0], [0, 1, 2, 3, 4, 5], [0, 2, 2, 4, 4, 0], [0, 3, 4, 1, 2, 5], [0, 4, 4, 2, 2, 0], [0, 5, 0, 5, 0, 5]] tables in label order. Check the additive abelian-group axioms, multiplicative associativity, both distributive laws and a two-sided multiplicative identity distinct from the additive identity. The two identity labels need not be 0 and 1. This protocol excludes the zero ring and does not require multiplicative commutativity or imply a field. Give (n,zero,one,additive_abelian?,multiplication_associative?,left_distributive?,right_distributive?,distinct_identities?,qualified?,first_failure). Missing or nonunique identities and no failure are represented by null. Failure priority is additive_structure, multiplication_associativity, left_distributivity, right_distributivity, then distinct_identities.
View answer and explanation
(6, 0, 1, Yes, Yes, Yes, Yes, Yes, Yes, null)
Audit every relevant element pair or triple, retaining the distinct obligation flags even after the first failure. The qualification flag is their conjunction, not a few passing examples. An exact finite or coordinate slice does not automatically establish a general infinite theorem. The exact result, retaining the requested order, is (6, 0, 1, Yes, Yes, Yes, Yes, Yes, Yes, null).
Reviewed step-by-step derivation
Finite unital ring · qf-mod27-ring-qualification · candidate
18. Qualify Both Complete Ring Tables without Assuming Their Identities
Use the same carrier Fin(6)={0,…,5} for complete addition=[[0, 1, 2, 3, 4, 5], [1, 4, 0, 2, 5, 3], [2, 0, 3, 5, 1, 4], [3, 2, 5, 4, 0, 1], [4, 5, 1, 0, 3, 2], [5, 3, 4, 1, 2, 0]] and multiplication=[[0, 0, 0, 0, 0, 0], [0, 1, 2, 3, 4, 5], [0, 2, 1, 4, 3, 5], [0, 3, 4, 3, 4, 0], [0, 4, 3, 4, 3, 0], [0, 5, 5, 0, 0, 5]] tables in label order. Check the additive abelian-group axioms, multiplicative associativity, both distributive laws and a two-sided multiplicative identity distinct from the additive identity. The two identity labels need not be 0 and 1. This protocol excludes the zero ring and does not require multiplicative commutativity or imply a field. Give (n,zero,one,additive_abelian?,multiplication_associative?,left_distributive?,right_distributive?,distinct_identities?,qualified?,first_failure). Missing or nonunique identities and no failure are represented by null. Failure priority is additive_structure, multiplication_associativity, left_distributivity, right_distributivity, then distinct_identities.
View answer and explanation
(6, 0, 1, Yes, Yes, Yes, Yes, Yes, Yes, null)
Audit every relevant element pair or triple, retaining the distinct obligation flags even after the first failure. The qualification flag is their conjunction, not a few passing examples. An exact finite or coordinate slice does not automatically establish a general infinite theorem. The exact result, retaining the requested order, is (6, 0, 1, Yes, Yes, Yes, Yes, Yes, Yes, null).
Reviewed step-by-step derivation
Finite unital ring · qf-mod27-ring-qualification · candidate
19. Qualify Both Complete Ring Tables without Assuming Their Identities
Use the same carrier Fin(6)={0,…,5} for complete addition=[[0, 1, 2, 3, 4, 5], [1, 5, 3, 4, 0, 2], [2, 3, 0, 1, 5, 4], [3, 4, 1, 5, 2, 0], [4, 0, 5, 2, 3, 1], [5, 2, 4, 0, 1, 3]] and multiplication=[[0, 0, 0, 0, 0, 0], [0, 1, 2, 3, 4, 5], [0, 2, 2, 0, 2, 0], [0, 3, 0, 3, 5, 5], [0, 4, 2, 5, 1, 3], [0, 5, 0, 5, 3, 3]] tables in label order. Check the additive abelian-group axioms, multiplicative associativity, both distributive laws and a two-sided multiplicative identity distinct from the additive identity. The two identity labels need not be 0 and 1. This protocol excludes the zero ring and does not require multiplicative commutativity or imply a field. Give (n,zero,one,additive_abelian?,multiplication_associative?,left_distributive?,right_distributive?,distinct_identities?,qualified?,first_failure). Missing or nonunique identities and no failure are represented by null. Failure priority is additive_structure, multiplication_associativity, left_distributivity, right_distributivity, then distinct_identities.
View answer and explanation
(6, 0, 1, Yes, Yes, Yes, Yes, Yes, Yes, null)
Audit every relevant element pair or triple, retaining the distinct obligation flags even after the first failure. The qualification flag is their conjunction, not a few passing examples. An exact finite or coordinate slice does not automatically establish a general infinite theorem. The exact result, retaining the requested order, is (6, 0, 1, Yes, Yes, Yes, Yes, Yes, Yes, null).
Reviewed step-by-step derivation
Finite unital ring · qf-mod27-ring-qualification · candidate
20. Qualify Both Complete Ring Tables without Assuming Their Identities
Use the same carrier Fin(6)={0,…,5} for complete addition=[[0, 1, 2, 3, 4, 5], [1, 5, 4, 0, 3, 2], [2, 4, 0, 5, 1, 3], [3, 0, 5, 4, 2, 1], [4, 3, 1, 2, 5, 0], [5, 2, 3, 1, 0, 4]] and multiplication=[[0, 0, 0, 0, 0, 0], [0, 1, 2, 3, 4, 5], [0, 2, 2, 2, 0, 0], [0, 3, 2, 1, 5, 4], [0, 4, 0, 5, 4, 5], [0, 5, 0, 4, 5, 4]] tables in label order. Check the additive abelian-group axioms, multiplicative associativity, both distributive laws and a two-sided multiplicative identity distinct from the additive identity. The two identity labels need not be 0 and 1. This protocol excludes the zero ring and does not require multiplicative commutativity or imply a field. Give (n,zero,one,additive_abelian?,multiplication_associative?,left_distributive?,right_distributive?,distinct_identities?,qualified?,first_failure). Missing or nonunique identities and no failure are represented by null. Failure priority is additive_structure, multiplication_associativity, left_distributivity, right_distributivity, then distinct_identities.
View answer and explanation
(6, 0, 1, Yes, Yes, Yes, Yes, Yes, Yes, null)
Audit every relevant element pair or triple, retaining the distinct obligation flags even after the first failure. The qualification flag is their conjunction, not a few passing examples. An exact finite or coordinate slice does not automatically establish a general infinite theorem. The exact result, retaining the requested order, is (6, 0, 1, Yes, Yes, Yes, Yes, Yes, Yes, null).
Reviewed step-by-step derivation
Finite unital ring · qf-mod27-ring-qualification · candidate
21. Qualify Both Complete Ring Tables without Assuming Their Identities
Use the same carrier Fin(6)={0,…,5} for complete addition=[[0, 1, 2, 3, 4, 5], [1, 5, 4, 2, 0, 3], [2, 4, 5, 1, 3, 0], [3, 2, 1, 0, 5, 4], [4, 0, 3, 5, 2, 1], [5, 3, 0, 4, 1, 2]] and multiplication=[[0, 0, 0, 0, 0, 0], [0, 1, 2, 3, 4, 5], [0, 2, 2, 1, 5, 5], [0, 3, 0, 3, 3, 0], [0, 4, 5, 3, 1, 2], [0, 5, 5, 0, 2, 2]] tables in label order. Check the additive abelian-group axioms, multiplicative associativity, both distributive laws and a two-sided multiplicative identity distinct from the additive identity. The two identity labels need not be 0 and 1. This protocol excludes the zero ring and does not require multiplicative commutativity or imply a field. Give (n,zero,one,additive_abelian?,multiplication_associative?,left_distributive?,right_distributive?,distinct_identities?,qualified?,first_failure). Missing or nonunique identities and no failure are represented by null. Failure priority is additive_structure, multiplication_associativity, left_distributivity, right_distributivity, then distinct_identities.
View answer and explanation
(6, 0, 1, Yes, No, No, No, Yes, No, multiplication_associativity)
Audit every relevant element pair or triple, retaining the distinct obligation flags even after the first failure. The qualification flag is their conjunction, not a few passing examples. An exact finite or coordinate slice does not automatically establish a general infinite theorem. The exact result, retaining the requested order, is (6, 0, 1, Yes, No, No, No, Yes, No, multiplication_associativity).
Reviewed step-by-step derivation
Finite unital ring · qf-mod27-ring-qualification · candidate
22. Qualify Both Complete Ring Tables without Assuming Their Identities
Use the same carrier Fin(6)={0,…,5} for complete addition=[[0, 1, 2, 3, 4, 5], [1, 5, 0, 4, 2, 3], [2, 0, 4, 5, 3, 1], [3, 4, 5, 0, 1, 2], [4, 2, 3, 1, 5, 0], [5, 3, 1, 2, 0, 4]] and multiplication=[[0, 0, 0, 0, 0, 0], [0, 1, 2, 3, 4, 5], [0, 2, 1, 3, 5, 4], [0, 3, 3, 3, 0, 0], [0, 4, 5, 0, 4, 5], [0, 5, 4, 0, 5, 4]] tables in label order. Check the additive abelian-group axioms, multiplicative associativity, both distributive laws and a two-sided multiplicative identity distinct from the additive identity. The two identity labels need not be 0 and 1. This protocol excludes the zero ring and does not require multiplicative commutativity or imply a field. Give (n,zero,one,additive_abelian?,multiplication_associative?,left_distributive?,right_distributive?,distinct_identities?,qualified?,first_failure). Missing or nonunique identities and no failure are represented by null. Failure priority is additive_structure, multiplication_associativity, left_distributivity, right_distributivity, then distinct_identities.
View answer and explanation
(6, 0, 1, Yes, Yes, Yes, Yes, Yes, Yes, null)
Audit every relevant element pair or triple, retaining the distinct obligation flags even after the first failure. The qualification flag is their conjunction, not a few passing examples. An exact finite or coordinate slice does not automatically establish a general infinite theorem. The exact result, retaining the requested order, is (6, 0, 1, Yes, Yes, Yes, Yes, Yes, Yes, null).
Reviewed step-by-step derivation
Finite unital ring · qf-mod27-ring-qualification · candidate
23. Qualify Both Complete Ring Tables without Assuming Their Identities
Use the same carrier Fin(6)={0,…,5} for complete addition=[[0, 1, 2, 3, 4, 5], [1, 5, 3, 0, 2, 4], [2, 3, 5, 4, 1, 0], [3, 0, 4, 2, 5, 1], [4, 2, 1, 5, 0, 3], [5, 4, 0, 1, 3, 2]] and multiplication=[[0, 0, 0, 0, 0, 0], [0, 1, 2, 3, 4, 5], [0, 2, 2, 5, 0, 5], [0, 3, 5, 1, 4, 2], [0, 4, 0, 4, 4, 0], [0, 5, 5, 2, 0, 2]] tables in label order. Check the additive abelian-group axioms, multiplicative associativity, both distributive laws and a two-sided multiplicative identity distinct from the additive identity. The two identity labels need not be 0 and 1. This protocol excludes the zero ring and does not require multiplicative commutativity or imply a field. Give (n,zero,one,additive_abelian?,multiplication_associative?,left_distributive?,right_distributive?,distinct_identities?,qualified?,first_failure). Missing or nonunique identities and no failure are represented by null. Failure priority is additive_structure, multiplication_associativity, left_distributivity, right_distributivity, then distinct_identities.
View answer and explanation
(6, 0, 1, Yes, Yes, Yes, Yes, Yes, Yes, null)
Audit every relevant element pair or triple, retaining the distinct obligation flags even after the first failure. The qualification flag is their conjunction, not a few passing examples. An exact finite or coordinate slice does not automatically establish a general infinite theorem. The exact result, retaining the requested order, is (6, 0, 1, Yes, Yes, Yes, Yes, Yes, Yes, null).
Reviewed step-by-step derivation
Finite unital ring · qf-mod27-ring-qualification · candidate
24. Qualify Both Complete Ring Tables without Assuming Their Identities
Use the same carrier Fin(6)={0,…,5} for complete addition=[[0, 1, 2, 3, 4, 5], [1, 5, 0, 2, 3, 4], [2, 0, 3, 4, 5, 1], [3, 2, 4, 5, 1, 0], [4, 3, 5, 1, 0, 2], [5, 4, 1, 0, 2, 3]] and multiplication=[[0, 0, 0, 0, 0, 0], [0, 1, 2, 3, 4, 5], [0, 2, 1, 5, 4, 3], [0, 3, 5, 3, 0, 5], [0, 4, 4, 0, 4, 0], [0, 5, 3, 5, 0, 3]] tables in label order. Check the additive abelian-group axioms, multiplicative associativity, both distributive laws and a two-sided multiplicative identity distinct from the additive identity. The two identity labels need not be 0 and 1. This protocol excludes the zero ring and does not require multiplicative commutativity or imply a field. Give (n,zero,one,additive_abelian?,multiplication_associative?,left_distributive?,right_distributive?,distinct_identities?,qualified?,first_failure). Missing or nonunique identities and no failure are represented by null. Failure priority is additive_structure, multiplication_associativity, left_distributivity, right_distributivity, then distinct_identities.
View answer and explanation
(6, 0, 1, Yes, Yes, Yes, Yes, Yes, Yes, null)
Audit every relevant element pair or triple, retaining the distinct obligation flags even after the first failure. The qualification flag is their conjunction, not a few passing examples. An exact finite or coordinate slice does not automatically establish a general infinite theorem. The exact result, retaining the requested order, is (6, 0, 1, Yes, Yes, Yes, Yes, Yes, Yes, null).
Reviewed step-by-step derivation
Finite unital ring · qf-mod27-ring-qualification · candidate
25. Qualify Both Complete Ring Tables without Assuming Their Identities
Use the same carrier Fin(6)={0,…,5} for complete addition=[[0, 1, 2, 3, 4, 5], [1, 4, 3, 5, 0, 2], [2, 3, 1, 4, 5, 0], [3, 5, 4, 0, 2, 1], [4, 0, 5, 2, 1, 3], [5, 2, 0, 1, 3, 4]] and multiplication=[[0, 0, 0, 0, 0, 0], [0, 4, 1, 0, 1, 4], [0, 1, 2, 3, 4, 5], [0, 0, 3, 3, 0, 3], [0, 1, 4, 0, 4, 1], [0, 4, 5, 3, 1, 2]] tables in label order. Check the additive abelian-group axioms, multiplicative associativity, both distributive laws and a two-sided multiplicative identity distinct from the additive identity. The two identity labels need not be 0 and 1. This protocol excludes the zero ring and does not require multiplicative commutativity or imply a field. Give (n,zero,one,additive_abelian?,multiplication_associative?,left_distributive?,right_distributive?,distinct_identities?,qualified?,first_failure). Missing or nonunique identities and no failure are represented by null. Failure priority is additive_structure, multiplication_associativity, left_distributivity, right_distributivity, then distinct_identities.
View answer and explanation
(6, 0, 2, Yes, Yes, Yes, Yes, Yes, Yes, null)
Audit every relevant element pair or triple, retaining the distinct obligation flags even after the first failure. The qualification flag is their conjunction, not a few passing examples. An exact finite or coordinate slice does not automatically establish a general infinite theorem. The exact result, retaining the requested order, is (6, 0, 2, Yes, Yes, Yes, Yes, Yes, Yes, null).
Reviewed step-by-step derivation
Finite unital ring · qf-mod27-ring-qualification · candidate
26. Find an Additive Inverse in an Already Qualified Ring
Use the same carrier Fin(6)={0,…,5} for complete addition=[[0, 1, 2, 3, 4, 5], [1, 2, 3, 4, 5, 0], [2, 3, 4, 5, 0, 1], [3, 4, 5, 0, 1, 2], [4, 5, 0, 1, 2, 3], [5, 0, 1, 2, 3, 4]] and multiplication=[[0, 0, 0, 0, 0, 0], [0, 1, 2, 3, 4, 5], [0, 2, 4, 0, 2, 4], [0, 3, 0, 3, 0, 3], [0, 4, 2, 0, 4, 2], [0, 5, 4, 3, 2, 1]] tables in label order. Check the additive abelian-group axioms, multiplicative associativity, both distributive laws and a two-sided multiplicative identity distinct from the additive identity. The two identity labels need not be 0 and 1. This protocol excludes the zero ring and does not require multiplicative commutativity or imply a field. The tables must qualify first. Selected element=1. Give (element,zero,inverse,left_sum,right_sum).
View answer and explanation
(1, 0, 5, 0, 0)
Use the additive table and its actual zero to find the inverse, then check both orders. A multiplicative inverse is not requested or guaranteed. An exact finite or coordinate slice does not automatically establish a general infinite theorem. The exact result, retaining the requested order, is (1, 0, 5, 0, 0).
Reviewed step-by-step derivation
Finite unital ring · qf-mod27-ring-additive-inverse · candidate
27. Find an Additive Inverse in an Already Qualified Ring
Use the same carrier Fin(6)={0,…,5} for complete addition=[[0, 1, 2, 3, 4, 5], [1, 2, 3, 5, 0, 4], [2, 3, 5, 4, 1, 0], [3, 5, 4, 0, 2, 1], [4, 0, 1, 2, 5, 3], [5, 4, 0, 1, 3, 2]] and multiplication=[[0, 0, 0, 0, 0, 0], [0, 1, 2, 3, 4, 5], [0, 2, 5, 0, 5, 2], [0, 3, 0, 3, 3, 0], [0, 4, 5, 3, 1, 2], [0, 5, 2, 0, 2, 5]] tables in label order. Check the additive abelian-group axioms, multiplicative associativity, both distributive laws and a two-sided multiplicative identity distinct from the additive identity. The two identity labels need not be 0 and 1. This protocol excludes the zero ring and does not require multiplicative commutativity or imply a field. The tables must qualify first. Selected element=0. Give (element,zero,inverse,left_sum,right_sum).
View answer and explanation
(0, 0, 0, 0, 0)
Use the additive table and its actual zero to find the inverse, then check both orders. A multiplicative inverse is not requested or guaranteed. An exact finite or coordinate slice does not automatically establish a general infinite theorem. The exact result, retaining the requested order, is (0, 0, 0, 0, 0).
Reviewed step-by-step derivation
Finite unital ring · qf-mod27-ring-additive-inverse · candidate
28. Find an Additive Inverse in an Already Qualified Ring
Use the same carrier Fin(6)={0,…,5} for complete addition=[[0, 1, 2, 3, 4, 5], [1, 2, 4, 5, 3, 0], [2, 4, 3, 0, 5, 1], [3, 5, 0, 2, 1, 4], [4, 3, 5, 1, 0, 2], [5, 0, 1, 4, 2, 3]] and multiplication=[[0, 0, 0, 0, 0, 0], [0, 1, 2, 3, 4, 5], [0, 2, 3, 2, 0, 3], [0, 3, 2, 3, 0, 2], [0, 4, 0, 0, 4, 4], [0, 5, 3, 2, 4, 1]] tables in label order. Check the additive abelian-group axioms, multiplicative associativity, both distributive laws and a two-sided multiplicative identity distinct from the additive identity. The two identity labels need not be 0 and 1. This protocol excludes the zero ring and does not require multiplicative commutativity or imply a field. The tables must qualify first. Selected element=5. Give (element,zero,inverse,left_sum,right_sum).
View answer and explanation
(5, 0, 1, 0, 0)
Use the additive table and its actual zero to find the inverse, then check both orders. A multiplicative inverse is not requested or guaranteed. An exact finite or coordinate slice does not automatically establish a general infinite theorem. The exact result, retaining the requested order, is (5, 0, 1, 0, 0).
Reviewed step-by-step derivation
Finite unital ring · qf-mod27-ring-additive-inverse · candidate
29. Find an Additive Inverse in an Already Qualified Ring
Use the same carrier Fin(6)={0,…,5} for complete addition=[[0, 1, 2, 3, 4, 5], [1, 2, 4, 0, 5, 3], [2, 4, 5, 1, 3, 0], [3, 0, 1, 5, 2, 4], [4, 5, 3, 2, 0, 1], [5, 3, 0, 4, 1, 2]] and multiplication=[[0, 0, 0, 0, 0, 0], [0, 1, 2, 3, 4, 5], [0, 2, 5, 5, 0, 2], [0, 3, 5, 1, 4, 2], [0, 4, 0, 4, 4, 0], [0, 5, 2, 2, 0, 5]] tables in label order. Check the additive abelian-group axioms, multiplicative associativity, both distributive laws and a two-sided multiplicative identity distinct from the additive identity. The two identity labels need not be 0 and 1. This protocol excludes the zero ring and does not require multiplicative commutativity or imply a field. The tables must qualify first. Selected element=4. Give (element,zero,inverse,left_sum,right_sum).
View answer and explanation
(4, 0, 4, 0, 0)
Use the additive table and its actual zero to find the inverse, then check both orders. A multiplicative inverse is not requested or guaranteed. An exact finite or coordinate slice does not automatically establish a general infinite theorem. The exact result, retaining the requested order, is (4, 0, 4, 0, 0).
Reviewed step-by-step derivation
Finite unital ring · qf-mod27-ring-additive-inverse · candidate
30. Find an Additive Inverse in an Already Qualified Ring
Use the same carrier Fin(6)={0,…,5} for complete addition=[[0, 1, 2, 3, 4, 5], [1, 2, 5, 4, 0, 3], [2, 5, 3, 0, 1, 4], [3, 4, 0, 2, 5, 1], [4, 0, 1, 5, 3, 2], [5, 3, 4, 1, 2, 0]] and multiplication=[[0, 0, 0, 0, 0, 0], [0, 1, 2, 3, 4, 5], [0, 2, 3, 2, 3, 0], [0, 3, 2, 3, 2, 0], [0, 4, 3, 2, 1, 5], [0, 5, 0, 0, 5, 5]] tables in label order. Check the additive abelian-group axioms, multiplicative associativity, both distributive laws and a two-sided multiplicative identity distinct from the additive identity. The two identity labels need not be 0 and 1. This protocol excludes the zero ring and does not require multiplicative commutativity or imply a field. The tables must qualify first. Selected element=3. Give (element,zero,inverse,left_sum,right_sum).
View answer and explanation
(3, 0, 2, 0, 0)
Use the additive table and its actual zero to find the inverse, then check both orders. A multiplicative inverse is not requested or guaranteed. An exact finite or coordinate slice does not automatically establish a general infinite theorem. The exact result, retaining the requested order, is (3, 0, 2, 0, 0).
Reviewed step-by-step derivation
Finite unital ring · qf-mod27-ring-additive-inverse · candidate
31. Find an Additive Inverse in an Already Qualified Ring
Use the same carrier Fin(6)={0,…,5} for complete addition=[[0, 1, 2, 3, 4, 5], [1, 2, 5, 0, 3, 4], [2, 5, 4, 1, 0, 3], [3, 0, 1, 4, 5, 2], [4, 3, 0, 5, 2, 1], [5, 4, 3, 2, 1, 0]] and multiplication=[[0, 0, 0, 0, 0, 0], [0, 1, 2, 3, 4, 5], [0, 2, 4, 4, 2, 0], [0, 3, 4, 1, 2, 5], [0, 4, 2, 2, 4, 0], [0, 5, 0, 5, 0, 5]] tables in label order. Check the additive abelian-group axioms, multiplicative associativity, both distributive laws and a two-sided multiplicative identity distinct from the additive identity. The two identity labels need not be 0 and 1. This protocol excludes the zero ring and does not require multiplicative commutativity or imply a field. The tables must qualify first. Selected element=2. Give (element,zero,inverse,left_sum,right_sum).
View answer and explanation
(2, 0, 4, 0, 0)
Use the additive table and its actual zero to find the inverse, then check both orders. A multiplicative inverse is not requested or guaranteed. An exact finite or coordinate slice does not automatically establish a general infinite theorem. The exact result, retaining the requested order, is (2, 0, 4, 0, 0).
Reviewed step-by-step derivation
Finite unital ring · qf-mod27-ring-additive-inverse · candidate
32. Find an Additive Inverse in an Already Qualified Ring
Use the same carrier Fin(6)={0,…,5} for complete addition=[[0, 1, 2, 3, 4, 5], [1, 3, 4, 2, 5, 0], [2, 4, 0, 5, 1, 3], [3, 2, 5, 4, 0, 1], [4, 5, 1, 0, 3, 2], [5, 0, 3, 1, 2, 4]] and multiplication=[[0, 0, 0, 0, 0, 0], [0, 1, 2, 3, 4, 5], [0, 2, 2, 0, 0, 2], [0, 3, 0, 4, 3, 4], [0, 4, 0, 3, 4, 3], [0, 5, 2, 4, 3, 1]] tables in label order. Check the additive abelian-group axioms, multiplicative associativity, both distributive laws and a two-sided multiplicative identity distinct from the additive identity. The two identity labels need not be 0 and 1. This protocol excludes the zero ring and does not require multiplicative commutativity or imply a field. The tables must qualify first. Selected element=1. Give (element,zero,inverse,left_sum,right_sum).
View answer and explanation
(1, 0, 5, 0, 0)
Use the additive table and its actual zero to find the inverse, then check both orders. A multiplicative inverse is not requested or guaranteed. An exact finite or coordinate slice does not automatically establish a general infinite theorem. The exact result, retaining the requested order, is (1, 0, 5, 0, 0).
Reviewed step-by-step derivation
Finite unital ring · qf-mod27-ring-additive-inverse · candidate
33. Find an Additive Inverse in an Already Qualified Ring
Use the same carrier Fin(6)={0,…,5} for complete addition=[[0, 1, 2, 3, 4, 5], [1, 3, 5, 2, 0, 4], [2, 5, 0, 4, 3, 1], [3, 2, 4, 5, 1, 0], [4, 0, 3, 1, 5, 2], [5, 4, 1, 0, 2, 3]] and multiplication=[[0, 0, 0, 0, 0, 0], [0, 1, 2, 3, 4, 5], [0, 2, 2, 0, 2, 0], [0, 3, 0, 5, 5, 3], [0, 4, 2, 5, 1, 3], [0, 5, 0, 3, 3, 5]] tables in label order. Check the additive abelian-group axioms, multiplicative associativity, both distributive laws and a two-sided multiplicative identity distinct from the additive identity. The two identity labels need not be 0 and 1. This protocol excludes the zero ring and does not require multiplicative commutativity or imply a field. The tables must qualify first. Selected element=0. Give (element,zero,inverse,left_sum,right_sum).
View answer and explanation
(0, 0, 0, 0, 0)
Use the additive table and its actual zero to find the inverse, then check both orders. A multiplicative inverse is not requested or guaranteed. An exact finite or coordinate slice does not automatically establish a general infinite theorem. The exact result, retaining the requested order, is (0, 0, 0, 0, 0).
Reviewed step-by-step derivation
Finite unital ring · qf-mod27-ring-additive-inverse · candidate
34. Find an Additive Inverse in an Already Qualified Ring
Use the same carrier Fin(6)={0,…,5} for complete addition=[[0, 1, 2, 3, 4, 5], [1, 3, 5, 4, 2, 0], [2, 5, 3, 0, 1, 4], [3, 4, 0, 2, 5, 1], [4, 2, 1, 5, 0, 3], [5, 0, 4, 1, 3, 2]] and multiplication=[[0, 0, 0, 0, 0, 0], [0, 1, 2, 3, 4, 5], [0, 2, 2, 3, 0, 3], [0, 3, 3, 2, 0, 2], [0, 4, 0, 0, 4, 4], [0, 5, 3, 2, 4, 1]] tables in label order. Check the additive abelian-group axioms, multiplicative associativity, both distributive laws and a two-sided multiplicative identity distinct from the additive identity. The two identity labels need not be 0 and 1. This protocol excludes the zero ring and does not require multiplicative commutativity or imply a field. The tables must qualify first. Selected element=5. Give (element,zero,inverse,left_sum,right_sum).
View answer and explanation
(5, 0, 1, 0, 0)
Use the additive table and its actual zero to find the inverse, then check both orders. A multiplicative inverse is not requested or guaranteed. An exact finite or coordinate slice does not automatically establish a general infinite theorem. The exact result, retaining the requested order, is (5, 0, 1, 0, 0).
Reviewed step-by-step derivation
Finite unital ring · qf-mod27-ring-additive-inverse · candidate
35. Find an Additive Inverse in an Already Qualified Ring
Use the same carrier Fin(6)={0,…,5} for complete addition=[[0, 1, 2, 3, 4, 5], [1, 3, 0, 4, 5, 2], [2, 0, 5, 1, 3, 4], [3, 4, 1, 5, 2, 0], [4, 5, 3, 2, 0, 1], [5, 2, 4, 0, 1, 3]] and multiplication=[[0, 0, 0, 0, 0, 0], [0, 1, 2, 3, 4, 5], [0, 2, 1, 5, 4, 3], [0, 3, 5, 5, 0, 3], [0, 4, 4, 0, 4, 0], [0, 5, 3, 3, 0, 5]] tables in label order. Check the additive abelian-group axioms, multiplicative associativity, both distributive laws and a two-sided multiplicative identity distinct from the additive identity. The two identity labels need not be 0 and 1. This protocol excludes the zero ring and does not require multiplicative commutativity or imply a field. The tables must qualify first. Selected element=4. Give (element,zero,inverse,left_sum,right_sum).
View answer and explanation
(4, 0, 4, 0, 0)
Use the additive table and its actual zero to find the inverse, then check both orders. A multiplicative inverse is not requested or guaranteed. An exact finite or coordinate slice does not automatically establish a general infinite theorem. The exact result, retaining the requested order, is (4, 0, 4, 0, 0).
Reviewed step-by-step derivation
Finite unital ring · qf-mod27-ring-additive-inverse · candidate
36. Find an Additive Inverse in an Already Qualified Ring
Use the same carrier Fin(6)={0,…,5} for complete addition=[[0, 1, 2, 3, 4, 5], [1, 3, 4, 5, 0, 2], [2, 4, 3, 0, 5, 1], [3, 5, 0, 2, 1, 4], [4, 0, 5, 1, 2, 3], [5, 2, 1, 4, 3, 0]] and multiplication=[[0, 0, 0, 0, 0, 0], [0, 1, 2, 3, 4, 5], [0, 2, 2, 3, 3, 0], [0, 3, 3, 2, 2, 0], [0, 4, 3, 2, 1, 5], [0, 5, 0, 0, 5, 5]] tables in label order. Check the additive abelian-group axioms, multiplicative associativity, both distributive laws and a two-sided multiplicative identity distinct from the additive identity. The two identity labels need not be 0 and 1. This protocol excludes the zero ring and does not require multiplicative commutativity or imply a field. The tables must qualify first. Selected element=3. Give (element,zero,inverse,left_sum,right_sum).
View answer and explanation
(3, 0, 2, 0, 0)
Use the additive table and its actual zero to find the inverse, then check both orders. A multiplicative inverse is not requested or guaranteed. An exact finite or coordinate slice does not automatically establish a general infinite theorem. The exact result, retaining the requested order, is (3, 0, 2, 0, 0).
Reviewed step-by-step derivation
Finite unital ring · qf-mod27-ring-additive-inverse · candidate
37. Find an Additive Inverse in an Already Qualified Ring
Use the same carrier Fin(6)={0,…,5} for complete addition=[[0, 1, 2, 3, 4, 5], [1, 3, 0, 5, 2, 4], [2, 0, 4, 1, 5, 3], [3, 5, 1, 4, 0, 2], [4, 2, 5, 0, 3, 1], [5, 4, 3, 2, 1, 0]] and multiplication=[[0, 0, 0, 0, 0, 0], [0, 1, 2, 3, 4, 5], [0, 2, 1, 4, 3, 5], [0, 3, 4, 4, 3, 0], [0, 4, 3, 3, 4, 0], [0, 5, 5, 0, 0, 5]] tables in label order. Check the additive abelian-group axioms, multiplicative associativity, both distributive laws and a two-sided multiplicative identity distinct from the additive identity. The two identity labels need not be 0 and 1. This protocol excludes the zero ring and does not require multiplicative commutativity or imply a field. The tables must qualify first. Selected element=2. Give (element,zero,inverse,left_sum,right_sum).
View answer and explanation
(2, 0, 1, 0, 0)
Use the additive table and its actual zero to find the inverse, then check both orders. A multiplicative inverse is not requested or guaranteed. An exact finite or coordinate slice does not automatically establish a general infinite theorem. The exact result, retaining the requested order, is (2, 0, 1, 0, 0).
Reviewed step-by-step derivation
Finite unital ring · qf-mod27-ring-additive-inverse · candidate
38. Find an Additive Inverse in an Already Qualified Ring
Use the same carrier Fin(6)={0,…,5} for complete addition=[[0, 1, 2, 3, 4, 5], [1, 4, 3, 5, 2, 0], [2, 3, 0, 1, 5, 4], [3, 5, 1, 4, 0, 2], [4, 2, 5, 0, 3, 1], [5, 0, 4, 2, 1, 3]] and multiplication=[[0, 0, 0, 0, 0, 0], [0, 1, 2, 3, 4, 5], [0, 2, 2, 0, 0, 2], [0, 3, 0, 3, 4, 4], [0, 4, 0, 4, 3, 3], [0, 5, 2, 4, 3, 1]] tables in label order. Check the additive abelian-group axioms, multiplicative associativity, both distributive laws and a two-sided multiplicative identity distinct from the additive identity. The two identity labels need not be 0 and 1. This protocol excludes the zero ring and does not require multiplicative commutativity or imply a field. The tables must qualify first. Selected element=1. Give (element,zero,inverse,left_sum,right_sum).
View answer and explanation
(1, 0, 5, 0, 0)
Use the additive table and its actual zero to find the inverse, then check both orders. A multiplicative inverse is not requested or guaranteed. An exact finite or coordinate slice does not automatically establish a general infinite theorem. The exact result, retaining the requested order, is (1, 0, 5, 0, 0).
Reviewed step-by-step derivation
Finite unital ring · qf-mod27-ring-additive-inverse · candidate
39. Find an Additive Inverse in an Already Qualified Ring
Use the same carrier Fin(6)={0,…,5} for complete addition=[[0, 1, 2, 3, 4, 5], [1, 4, 5, 0, 2, 3], [2, 5, 0, 4, 3, 1], [3, 0, 4, 5, 1, 2], [4, 2, 3, 1, 5, 0], [5, 3, 1, 2, 0, 4]] and multiplication=[[0, 0, 0, 0, 0, 0], [0, 1, 2, 3, 4, 5], [0, 2, 2, 2, 0, 0], [0, 3, 2, 1, 5, 4], [0, 4, 0, 5, 5, 4], [0, 5, 0, 4, 4, 5]] tables in label order. Check the additive abelian-group axioms, multiplicative associativity, both distributive laws and a two-sided multiplicative identity distinct from the additive identity. The two identity labels need not be 0 and 1. This protocol excludes the zero ring and does not require multiplicative commutativity or imply a field. The tables must qualify first. Selected element=0. Give (element,zero,inverse,left_sum,right_sum).
View answer and explanation
(0, 0, 0, 0, 0)
Use the additive table and its actual zero to find the inverse, then check both orders. A multiplicative inverse is not requested or guaranteed. An exact finite or coordinate slice does not automatically establish a general infinite theorem. The exact result, retaining the requested order, is (0, 0, 0, 0, 0).
Reviewed step-by-step derivation
Finite unital ring · qf-mod27-ring-additive-inverse · candidate
40. Find an Additive Inverse in an Already Qualified Ring
Use the same carrier Fin(6)={0,…,5} for complete addition=[[0, 1, 2, 3, 4, 5], [1, 4, 5, 2, 3, 0], [2, 5, 4, 1, 0, 3], [3, 2, 1, 0, 5, 4], [4, 3, 0, 5, 2, 1], [5, 0, 3, 4, 1, 2]] and multiplication=[[0, 0, 0, 0, 0, 0], [0, 1, 2, 3, 4, 5], [0, 2, 2, 0, 4, 4], [0, 3, 0, 3, 0, 3], [0, 4, 4, 0, 2, 2], [0, 5, 4, 3, 2, 1]] tables in label order. Check the additive abelian-group axioms, multiplicative associativity, both distributive laws and a two-sided multiplicative identity distinct from the additive identity. The two identity labels need not be 0 and 1. This protocol excludes the zero ring and does not require multiplicative commutativity or imply a field. The tables must qualify first. Selected element=5. Give (element,zero,inverse,left_sum,right_sum).
View answer and explanation
(5, 0, 1, 0, 0)
Use the additive table and its actual zero to find the inverse, then check both orders. A multiplicative inverse is not requested or guaranteed. An exact finite or coordinate slice does not automatically establish a general infinite theorem. The exact result, retaining the requested order, is (5, 0, 1, 0, 0).
Reviewed step-by-step derivation
Finite unital ring · qf-mod27-ring-additive-inverse · candidate
41. Find an Additive Inverse in an Already Qualified Ring
Use the same carrier Fin(6)={0,…,5} for complete addition=[[0, 1, 2, 3, 4, 5], [1, 4, 0, 5, 3, 2], [2, 0, 5, 4, 1, 3], [3, 5, 4, 0, 2, 1], [4, 3, 1, 2, 5, 0], [5, 2, 3, 1, 0, 4]] and multiplication=[[0, 0, 0, 0, 0, 0], [0, 1, 2, 3, 4, 5], [0, 2, 1, 3, 5, 4], [0, 3, 3, 3, 0, 0], [0, 4, 5, 0, 5, 4], [0, 5, 4, 0, 4, 5]] tables in label order. Check the additive abelian-group axioms, multiplicative associativity, both distributive laws and a two-sided multiplicative identity distinct from the additive identity. The two identity labels need not be 0 and 1. This protocol excludes the zero ring and does not require multiplicative commutativity or imply a field. The tables must qualify first. Selected element=4. Give (element,zero,inverse,left_sum,right_sum).
View answer and explanation
(4, 0, 5, 0, 0)
Use the additive table and its actual zero to find the inverse, then check both orders. A multiplicative inverse is not requested or guaranteed. An exact finite or coordinate slice does not automatically establish a general infinite theorem. The exact result, retaining the requested order, is (4, 0, 5, 0, 0).
Reviewed step-by-step derivation
Finite unital ring · qf-mod27-ring-additive-inverse · candidate
42. Find an Additive Inverse in an Already Qualified Ring
Use the same carrier Fin(6)={0,…,5} for complete addition=[[0, 1, 2, 3, 4, 5], [1, 4, 3, 0, 5, 2], [2, 3, 4, 5, 0, 1], [3, 0, 5, 2, 1, 4], [4, 5, 0, 1, 2, 3], [5, 2, 1, 4, 3, 0]] and multiplication=[[0, 0, 0, 0, 0, 0], [0, 1, 2, 3, 4, 5], [0, 2, 2, 4, 4, 0], [0, 3, 4, 1, 2, 5], [0, 4, 4, 2, 2, 0], [0, 5, 0, 5, 0, 5]] tables in label order. Check the additive abelian-group axioms, multiplicative associativity, both distributive laws and a two-sided multiplicative identity distinct from the additive identity. The two identity labels need not be 0 and 1. This protocol excludes the zero ring and does not require multiplicative commutativity or imply a field. The tables must qualify first. Selected element=3. Give (element,zero,inverse,left_sum,right_sum).
View answer and explanation
(3, 0, 1, 0, 0)
Use the additive table and its actual zero to find the inverse, then check both orders. A multiplicative inverse is not requested or guaranteed. An exact finite or coordinate slice does not automatically establish a general infinite theorem. The exact result, retaining the requested order, is (3, 0, 1, 0, 0).
Reviewed step-by-step derivation
Finite unital ring · qf-mod27-ring-additive-inverse · candidate
43. Find an Additive Inverse in an Already Qualified Ring
Use the same carrier Fin(6)={0,…,5} for complete addition=[[0, 1, 2, 3, 4, 5], [1, 4, 0, 2, 5, 3], [2, 0, 3, 5, 1, 4], [3, 2, 5, 4, 0, 1], [4, 5, 1, 0, 3, 2], [5, 3, 4, 1, 2, 0]] and multiplication=[[0, 0, 0, 0, 0, 0], [0, 1, 2, 3, 4, 5], [0, 2, 1, 4, 3, 5], [0, 3, 4, 3, 4, 0], [0, 4, 3, 4, 3, 0], [0, 5, 5, 0, 0, 5]] tables in label order. Check the additive abelian-group axioms, multiplicative associativity, both distributive laws and a two-sided multiplicative identity distinct from the additive identity. The two identity labels need not be 0 and 1. This protocol excludes the zero ring and does not require multiplicative commutativity or imply a field. The tables must qualify first. Selected element=2. Give (element,zero,inverse,left_sum,right_sum).
View answer and explanation
(2, 0, 1, 0, 0)
Use the additive table and its actual zero to find the inverse, then check both orders. A multiplicative inverse is not requested or guaranteed. An exact finite or coordinate slice does not automatically establish a general infinite theorem. The exact result, retaining the requested order, is (2, 0, 1, 0, 0).
Reviewed step-by-step derivation
Finite unital ring · qf-mod27-ring-additive-inverse · candidate
44. Find an Additive Inverse in an Already Qualified Ring
Use the same carrier Fin(6)={0,…,5} for complete addition=[[0, 1, 2, 3, 4, 5], [1, 5, 3, 4, 0, 2], [2, 3, 0, 1, 5, 4], [3, 4, 1, 5, 2, 0], [4, 0, 5, 2, 3, 1], [5, 2, 4, 0, 1, 3]] and multiplication=[[0, 0, 0, 0, 0, 0], [0, 1, 2, 3, 4, 5], [0, 2, 2, 0, 2, 0], [0, 3, 0, 3, 5, 5], [0, 4, 2, 5, 1, 3], [0, 5, 0, 5, 3, 3]] tables in label order. Check the additive abelian-group axioms, multiplicative associativity, both distributive laws and a two-sided multiplicative identity distinct from the additive identity. The two identity labels need not be 0 and 1. This protocol excludes the zero ring and does not require multiplicative commutativity or imply a field. The tables must qualify first. Selected element=1. Give (element,zero,inverse,left_sum,right_sum).
View answer and explanation
(1, 0, 4, 0, 0)
Use the additive table and its actual zero to find the inverse, then check both orders. A multiplicative inverse is not requested or guaranteed. An exact finite or coordinate slice does not automatically establish a general infinite theorem. The exact result, retaining the requested order, is (1, 0, 4, 0, 0).
Reviewed step-by-step derivation
Finite unital ring · qf-mod27-ring-additive-inverse · candidate
45. Find an Additive Inverse in an Already Qualified Ring
Use the same carrier Fin(6)={0,…,5} for complete addition=[[0, 1, 2, 3, 4, 5], [1, 5, 4, 0, 3, 2], [2, 4, 0, 5, 1, 3], [3, 0, 5, 4, 2, 1], [4, 3, 1, 2, 5, 0], [5, 2, 3, 1, 0, 4]] and multiplication=[[0, 0, 0, 0, 0, 0], [0, 1, 2, 3, 4, 5], [0, 2, 2, 2, 0, 0], [0, 3, 2, 1, 5, 4], [0, 4, 0, 5, 4, 5], [0, 5, 0, 4, 5, 4]] tables in label order. Check the additive abelian-group axioms, multiplicative associativity, both distributive laws and a two-sided multiplicative identity distinct from the additive identity. The two identity labels need not be 0 and 1. This protocol excludes the zero ring and does not require multiplicative commutativity or imply a field. The tables must qualify first. Selected element=0. Give (element,zero,inverse,left_sum,right_sum).
View answer and explanation
(0, 0, 0, 0, 0)
Use the additive table and its actual zero to find the inverse, then check both orders. A multiplicative inverse is not requested or guaranteed. An exact finite or coordinate slice does not automatically establish a general infinite theorem. The exact result, retaining the requested order, is (0, 0, 0, 0, 0).
Reviewed step-by-step derivation
Finite unital ring · qf-mod27-ring-additive-inverse · candidate
46. Find an Additive Inverse in an Already Qualified Ring
Use the same carrier Fin(6)={0,…,5} for complete addition=[[0, 1, 2, 3, 4, 5], [1, 5, 4, 2, 0, 3], [2, 4, 5, 1, 3, 0], [3, 2, 1, 0, 5, 4], [4, 0, 3, 5, 2, 1], [5, 3, 0, 4, 1, 2]] and multiplication=[[0, 0, 0, 0, 0, 0], [0, 1, 2, 3, 4, 5], [0, 2, 2, 0, 5, 5], [0, 3, 0, 3, 3, 0], [0, 4, 5, 3, 1, 2], [0, 5, 5, 0, 2, 2]] tables in label order. Check the additive abelian-group axioms, multiplicative associativity, both distributive laws and a two-sided multiplicative identity distinct from the additive identity. The two identity labels need not be 0 and 1. This protocol excludes the zero ring and does not require multiplicative commutativity or imply a field. The tables must qualify first. Selected element=5. Give (element,zero,inverse,left_sum,right_sum).
View answer and explanation
(5, 0, 2, 0, 0)
Use the additive table and its actual zero to find the inverse, then check both orders. A multiplicative inverse is not requested or guaranteed. An exact finite or coordinate slice does not automatically establish a general infinite theorem. The exact result, retaining the requested order, is (5, 0, 2, 0, 0).
Reviewed step-by-step derivation
Finite unital ring · qf-mod27-ring-additive-inverse · candidate
47. Find an Additive Inverse in an Already Qualified Ring
Use the same carrier Fin(6)={0,…,5} for complete addition=[[0, 1, 2, 3, 4, 5], [1, 5, 0, 4, 2, 3], [2, 0, 4, 5, 3, 1], [3, 4, 5, 0, 1, 2], [4, 2, 3, 1, 5, 0], [5, 3, 1, 2, 0, 4]] and multiplication=[[0, 0, 0, 0, 0, 0], [0, 1, 2, 3, 4, 5], [0, 2, 1, 3, 5, 4], [0, 3, 3, 3, 0, 0], [0, 4, 5, 0, 4, 5], [0, 5, 4, 0, 5, 4]] tables in label order. Check the additive abelian-group axioms, multiplicative associativity, both distributive laws and a two-sided multiplicative identity distinct from the additive identity. The two identity labels need not be 0 and 1. This protocol excludes the zero ring and does not require multiplicative commutativity or imply a field. The tables must qualify first. Selected element=4. Give (element,zero,inverse,left_sum,right_sum).
View answer and explanation
(4, 0, 5, 0, 0)
Use the additive table and its actual zero to find the inverse, then check both orders. A multiplicative inverse is not requested or guaranteed. An exact finite or coordinate slice does not automatically establish a general infinite theorem. The exact result, retaining the requested order, is (4, 0, 5, 0, 0).
Reviewed step-by-step derivation
Finite unital ring · qf-mod27-ring-additive-inverse · candidate
48. Find an Additive Inverse in an Already Qualified Ring
Use the same carrier Fin(6)={0,…,5} for complete addition=[[0, 1, 2, 3, 4, 5], [1, 5, 3, 0, 2, 4], [2, 3, 5, 4, 1, 0], [3, 0, 4, 2, 5, 1], [4, 2, 1, 5, 0, 3], [5, 4, 0, 1, 3, 2]] and multiplication=[[0, 0, 0, 0, 0, 0], [0, 1, 2, 3, 4, 5], [0, 2, 2, 5, 0, 5], [0, 3, 5, 1, 4, 2], [0, 4, 0, 4, 4, 0], [0, 5, 5, 2, 0, 2]] tables in label order. Check the additive abelian-group axioms, multiplicative associativity, both distributive laws and a two-sided multiplicative identity distinct from the additive identity. The two identity labels need not be 0 and 1. This protocol excludes the zero ring and does not require multiplicative commutativity or imply a field. The tables must qualify first. Selected element=3. Give (element,zero,inverse,left_sum,right_sum).
View answer and explanation
(3, 0, 1, 0, 0)
Use the additive table and its actual zero to find the inverse, then check both orders. A multiplicative inverse is not requested or guaranteed. An exact finite or coordinate slice does not automatically establish a general infinite theorem. The exact result, retaining the requested order, is (3, 0, 1, 0, 0).
Reviewed step-by-step derivation
Finite unital ring · qf-mod27-ring-additive-inverse · candidate
49. Find an Additive Inverse in an Already Qualified Ring
Use the same carrier Fin(6)={0,…,5} for complete addition=[[0, 1, 2, 3, 4, 5], [1, 5, 0, 2, 3, 4], [2, 0, 3, 4, 5, 1], [3, 2, 4, 5, 1, 0], [4, 3, 5, 1, 0, 2], [5, 4, 1, 0, 2, 3]] and multiplication=[[0, 0, 0, 0, 0, 0], [0, 1, 2, 3, 4, 5], [0, 2, 1, 5, 4, 3], [0, 3, 5, 3, 0, 5], [0, 4, 4, 0, 4, 0], [0, 5, 3, 5, 0, 3]] tables in label order. Check the additive abelian-group axioms, multiplicative associativity, both distributive laws and a two-sided multiplicative identity distinct from the additive identity. The two identity labels need not be 0 and 1. This protocol excludes the zero ring and does not require multiplicative commutativity or imply a field. The tables must qualify first. Selected element=2. Give (element,zero,inverse,left_sum,right_sum).
View answer and explanation
(2, 0, 1, 0, 0)
Use the additive table and its actual zero to find the inverse, then check both orders. A multiplicative inverse is not requested or guaranteed. An exact finite or coordinate slice does not automatically establish a general infinite theorem. The exact result, retaining the requested order, is (2, 0, 1, 0, 0).
Reviewed step-by-step derivation
Finite unital ring · qf-mod27-ring-additive-inverse · candidate
50. Find an Additive Inverse in an Already Qualified Ring
Use the same carrier Fin(6)={0,…,5} for complete addition=[[0, 1, 2, 3, 4, 5], [1, 4, 3, 5, 0, 2], [2, 3, 1, 4, 5, 0], [3, 5, 4, 0, 2, 1], [4, 0, 5, 2, 1, 3], [5, 2, 0, 1, 3, 4]] and multiplication=[[0, 0, 0, 0, 0, 0], [0, 4, 1, 0, 1, 4], [0, 1, 2, 3, 4, 5], [0, 0, 3, 3, 0, 3], [0, 1, 4, 0, 4, 1], [0, 4, 5, 3, 1, 2]] tables in label order. Check the additive abelian-group axioms, multiplicative associativity, both distributive laws and a two-sided multiplicative identity distinct from the additive identity. The two identity labels need not be 0 and 1. This protocol excludes the zero ring and does not require multiplicative commutativity or imply a field. The tables must qualify first. Selected element=1. Give (element,zero,inverse,left_sum,right_sum).
View answer and explanation
(1, 0, 4, 0, 0)
Use the additive table and its actual zero to find the inverse, then check both orders. A multiplicative inverse is not requested or guaranteed. An exact finite or coordinate slice does not automatically establish a general infinite theorem. The exact result, retaining the requested order, is (1, 0, 4, 0, 0).
Reviewed step-by-step derivation
Finite unital ring · qf-mod27-ring-additive-inverse · candidate
51. A Nonnegative Ring Power Starts at Its Actual One
Use the same carrier Fin(6)={0,…,5} for complete addition=[[0, 1, 2, 3, 4, 5], [1, 2, 3, 4, 5, 0], [2, 3, 4, 5, 0, 1], [3, 4, 5, 0, 1, 2], [4, 5, 0, 1, 2, 3], [5, 0, 1, 2, 3, 4]] and multiplication=[[0, 0, 0, 0, 0, 0], [0, 1, 2, 3, 4, 5], [0, 2, 4, 0, 2, 4], [0, 3, 0, 3, 0, 3], [0, 4, 2, 0, 4, 2], [0, 5, 4, 3, 2, 1]] tables in label order. Check the additive abelian-group axioms, multiplicative associativity, both distributive laws and a two-sided multiplicative identity distinct from the additive identity. The two identity labels need not be 0 and 1. This protocol excludes the zero ring and does not require multiplicative commutativity or imply a field. The tables must qualify first. Selected element=1, exponent=1. Give (element,exponent,one,result).
View answer and explanation
(1, 1, 1, 1)
Start at the actual multiplicative identity and iterate the multiplication table. Exponent zero returns that identity, including when the selected element is the additive zero; this is the declared algebraic power convention. An exact finite or coordinate slice does not automatically establish a general infinite theorem. The exact result, retaining the requested order, is (1, 1, 1, 1).
Reviewed step-by-step derivation
Finite unital ring · qf-mod27-ring-power · candidate
52. A Nonnegative Ring Power Starts at Its Actual One
Use the same carrier Fin(6)={0,…,5} for complete addition=[[0, 1, 2, 3, 4, 5], [1, 2, 3, 5, 0, 4], [2, 3, 5, 4, 1, 0], [3, 5, 4, 0, 2, 1], [4, 0, 1, 2, 5, 3], [5, 4, 0, 1, 3, 2]] and multiplication=[[0, 0, 0, 0, 0, 0], [0, 1, 2, 3, 4, 5], [0, 2, 5, 0, 5, 2], [0, 3, 0, 3, 3, 0], [0, 4, 5, 3, 1, 2], [0, 5, 2, 0, 2, 5]] tables in label order. Check the additive abelian-group axioms, multiplicative associativity, both distributive laws and a two-sided multiplicative identity distinct from the additive identity. The two identity labels need not be 0 and 1. This protocol excludes the zero ring and does not require multiplicative commutativity or imply a field. The tables must qualify first. Selected element=0, exponent=2. Give (element,exponent,one,result).
View answer and explanation
(0, 2, 1, 0)
Start at the actual multiplicative identity and iterate the multiplication table. Exponent zero returns that identity, including when the selected element is the additive zero; this is the declared algebraic power convention. An exact finite or coordinate slice does not automatically establish a general infinite theorem. The exact result, retaining the requested order, is (0, 2, 1, 0).
Reviewed step-by-step derivation
Finite unital ring · qf-mod27-ring-power · candidate
53. A Nonnegative Ring Power Starts at Its Actual One
Use the same carrier Fin(6)={0,…,5} for complete addition=[[0, 1, 2, 3, 4, 5], [1, 2, 4, 5, 3, 0], [2, 4, 3, 0, 5, 1], [3, 5, 0, 2, 1, 4], [4, 3, 5, 1, 0, 2], [5, 0, 1, 4, 2, 3]] and multiplication=[[0, 0, 0, 0, 0, 0], [0, 1, 2, 3, 4, 5], [0, 2, 3, 2, 0, 3], [0, 3, 2, 3, 0, 2], [0, 4, 0, 0, 4, 4], [0, 5, 3, 2, 4, 1]] tables in label order. Check the additive abelian-group axioms, multiplicative associativity, both distributive laws and a two-sided multiplicative identity distinct from the additive identity. The two identity labels need not be 0 and 1. This protocol excludes the zero ring and does not require multiplicative commutativity or imply a field. The tables must qualify first. Selected element=5, exponent=3. Give (element,exponent,one,result).
View answer and explanation
(5, 3, 1, 5)
Start at the actual multiplicative identity and iterate the multiplication table. Exponent zero returns that identity, including when the selected element is the additive zero; this is the declared algebraic power convention. An exact finite or coordinate slice does not automatically establish a general infinite theorem. The exact result, retaining the requested order, is (5, 3, 1, 5).
Reviewed step-by-step derivation
Finite unital ring · qf-mod27-ring-power · candidate
54. A Nonnegative Ring Power Starts at Its Actual One
Use the same carrier Fin(6)={0,…,5} for complete addition=[[0, 1, 2, 3, 4, 5], [1, 2, 4, 0, 5, 3], [2, 4, 5, 1, 3, 0], [3, 0, 1, 5, 2, 4], [4, 5, 3, 2, 0, 1], [5, 3, 0, 4, 1, 2]] and multiplication=[[0, 0, 0, 0, 0, 0], [0, 1, 2, 3, 4, 5], [0, 2, 5, 5, 0, 2], [0, 3, 5, 1, 4, 2], [0, 4, 0, 4, 4, 0], [0, 5, 2, 2, 0, 5]] tables in label order. Check the additive abelian-group axioms, multiplicative associativity, both distributive laws and a two-sided multiplicative identity distinct from the additive identity. The two identity labels need not be 0 and 1. This protocol excludes the zero ring and does not require multiplicative commutativity or imply a field. The tables must qualify first. Selected element=4, exponent=4. Give (element,exponent,one,result).
View answer and explanation
(4, 4, 1, 4)
Start at the actual multiplicative identity and iterate the multiplication table. Exponent zero returns that identity, including when the selected element is the additive zero; this is the declared algebraic power convention. An exact finite or coordinate slice does not automatically establish a general infinite theorem. The exact result, retaining the requested order, is (4, 4, 1, 4).
Reviewed step-by-step derivation
Finite unital ring · qf-mod27-ring-power · candidate
55. A Nonnegative Ring Power Starts at Its Actual One
Use the same carrier Fin(6)={0,…,5} for complete addition=[[0, 1, 2, 3, 4, 5], [1, 2, 5, 4, 0, 3], [2, 5, 3, 0, 1, 4], [3, 4, 0, 2, 5, 1], [4, 0, 1, 5, 3, 2], [5, 3, 4, 1, 2, 0]] and multiplication=[[0, 0, 0, 0, 0, 0], [0, 1, 2, 3, 4, 5], [0, 2, 3, 2, 3, 0], [0, 3, 2, 3, 2, 0], [0, 4, 3, 2, 1, 5], [0, 5, 0, 0, 5, 5]] tables in label order. Check the additive abelian-group axioms, multiplicative associativity, both distributive laws and a two-sided multiplicative identity distinct from the additive identity. The two identity labels need not be 0 and 1. This protocol excludes the zero ring and does not require multiplicative commutativity or imply a field. The tables must qualify first. Selected element=3, exponent=5. Give (element,exponent,one,result).
View answer and explanation
(3, 5, 1, 3)
Start at the actual multiplicative identity and iterate the multiplication table. Exponent zero returns that identity, including when the selected element is the additive zero; this is the declared algebraic power convention. An exact finite or coordinate slice does not automatically establish a general infinite theorem. The exact result, retaining the requested order, is (3, 5, 1, 3).
Reviewed step-by-step derivation
Finite unital ring · qf-mod27-ring-power · candidate
56. A Nonnegative Ring Power Starts at Its Actual One
Use the same carrier Fin(6)={0,…,5} for complete addition=[[0, 1, 2, 3, 4, 5], [1, 2, 5, 0, 3, 4], [2, 5, 4, 1, 0, 3], [3, 0, 1, 4, 5, 2], [4, 3, 0, 5, 2, 1], [5, 4, 3, 2, 1, 0]] and multiplication=[[0, 0, 0, 0, 0, 0], [0, 1, 2, 3, 4, 5], [0, 2, 4, 4, 2, 0], [0, 3, 4, 1, 2, 5], [0, 4, 2, 2, 4, 0], [0, 5, 0, 5, 0, 5]] tables in label order. Check the additive abelian-group axioms, multiplicative associativity, both distributive laws and a two-sided multiplicative identity distinct from the additive identity. The two identity labels need not be 0 and 1. This protocol excludes the zero ring and does not require multiplicative commutativity or imply a field. The tables must qualify first. Selected element=2, exponent=6. Give (element,exponent,one,result).
View answer and explanation
(2, 6, 1, 4)
Start at the actual multiplicative identity and iterate the multiplication table. Exponent zero returns that identity, including when the selected element is the additive zero; this is the declared algebraic power convention. An exact finite or coordinate slice does not automatically establish a general infinite theorem. The exact result, retaining the requested order, is (2, 6, 1, 4).
Reviewed step-by-step derivation
Finite unital ring · qf-mod27-ring-power · candidate
57. A Nonnegative Ring Power Starts at Its Actual One
Use the same carrier Fin(6)={0,…,5} for complete addition=[[0, 1, 2, 3, 4, 5], [1, 3, 4, 2, 5, 0], [2, 4, 0, 5, 1, 3], [3, 2, 5, 4, 0, 1], [4, 5, 1, 0, 3, 2], [5, 0, 3, 1, 2, 4]] and multiplication=[[0, 0, 0, 0, 0, 0], [0, 1, 2, 3, 4, 5], [0, 2, 2, 0, 0, 2], [0, 3, 0, 4, 3, 4], [0, 4, 0, 3, 4, 3], [0, 5, 2, 4, 3, 1]] tables in label order. Check the additive abelian-group axioms, multiplicative associativity, both distributive laws and a two-sided multiplicative identity distinct from the additive identity. The two identity labels need not be 0 and 1. This protocol excludes the zero ring and does not require multiplicative commutativity or imply a field. The tables must qualify first. Selected element=1, exponent=7. Give (element,exponent,one,result).
View answer and explanation
(1, 7, 1, 1)
Start at the actual multiplicative identity and iterate the multiplication table. Exponent zero returns that identity, including when the selected element is the additive zero; this is the declared algebraic power convention. An exact finite or coordinate slice does not automatically establish a general infinite theorem. The exact result, retaining the requested order, is (1, 7, 1, 1).
Reviewed step-by-step derivation
Finite unital ring · qf-mod27-ring-power · candidate
58. A Nonnegative Ring Power Starts at Its Actual One
Use the same carrier Fin(6)={0,…,5} for complete addition=[[0, 1, 2, 3, 4, 5], [1, 3, 5, 2, 0, 4], [2, 5, 0, 4, 3, 1], [3, 2, 4, 5, 1, 0], [4, 0, 3, 1, 5, 2], [5, 4, 1, 0, 2, 3]] and multiplication=[[0, 0, 0, 0, 0, 0], [0, 1, 2, 3, 4, 5], [0, 2, 2, 0, 2, 0], [0, 3, 0, 5, 5, 3], [0, 4, 2, 5, 1, 3], [0, 5, 0, 3, 3, 5]] tables in label order. Check the additive abelian-group axioms, multiplicative associativity, both distributive laws and a two-sided multiplicative identity distinct from the additive identity. The two identity labels need not be 0 and 1. This protocol excludes the zero ring and does not require multiplicative commutativity or imply a field. The tables must qualify first. Selected element=0, exponent=8. Give (element,exponent,one,result).
View answer and explanation
(0, 8, 1, 0)
Start at the actual multiplicative identity and iterate the multiplication table. Exponent zero returns that identity, including when the selected element is the additive zero; this is the declared algebraic power convention. An exact finite or coordinate slice does not automatically establish a general infinite theorem. The exact result, retaining the requested order, is (0, 8, 1, 0).
Reviewed step-by-step derivation
Finite unital ring · qf-mod27-ring-power · candidate
59. A Nonnegative Ring Power Starts at Its Actual One
Use the same carrier Fin(6)={0,…,5} for complete addition=[[0, 1, 2, 3, 4, 5], [1, 3, 5, 4, 2, 0], [2, 5, 3, 0, 1, 4], [3, 4, 0, 2, 5, 1], [4, 2, 1, 5, 0, 3], [5, 0, 4, 1, 3, 2]] and multiplication=[[0, 0, 0, 0, 0, 0], [0, 1, 2, 3, 4, 5], [0, 2, 2, 3, 0, 3], [0, 3, 3, 2, 0, 2], [0, 4, 0, 0, 4, 4], [0, 5, 3, 2, 4, 1]] tables in label order. Check the additive abelian-group axioms, multiplicative associativity, both distributive laws and a two-sided multiplicative identity distinct from the additive identity. The two identity labels need not be 0 and 1. This protocol excludes the zero ring and does not require multiplicative commutativity or imply a field. The tables must qualify first. Selected element=5, exponent=9. Give (element,exponent,one,result).
View answer and explanation
(5, 9, 1, 5)
Start at the actual multiplicative identity and iterate the multiplication table. Exponent zero returns that identity, including when the selected element is the additive zero; this is the declared algebraic power convention. An exact finite or coordinate slice does not automatically establish a general infinite theorem. The exact result, retaining the requested order, is (5, 9, 1, 5).
Reviewed step-by-step derivation
Finite unital ring · qf-mod27-ring-power · candidate
60. A Nonnegative Ring Power Starts at Its Actual One
Use the same carrier Fin(6)={0,…,5} for complete addition=[[0, 1, 2, 3, 4, 5], [1, 3, 0, 4, 5, 2], [2, 0, 5, 1, 3, 4], [3, 4, 1, 5, 2, 0], [4, 5, 3, 2, 0, 1], [5, 2, 4, 0, 1, 3]] and multiplication=[[0, 0, 0, 0, 0, 0], [0, 1, 2, 3, 4, 5], [0, 2, 1, 5, 4, 3], [0, 3, 5, 5, 0, 3], [0, 4, 4, 0, 4, 0], [0, 5, 3, 3, 0, 5]] tables in label order. Check the additive abelian-group axioms, multiplicative associativity, both distributive laws and a two-sided multiplicative identity distinct from the additive identity. The two identity labels need not be 0 and 1. This protocol excludes the zero ring and does not require multiplicative commutativity or imply a field. The tables must qualify first. Selected element=4, exponent=10. Give (element,exponent,one,result).
View answer and explanation
(4, 10, 1, 4)
Start at the actual multiplicative identity and iterate the multiplication table. Exponent zero returns that identity, including when the selected element is the additive zero; this is the declared algebraic power convention. An exact finite or coordinate slice does not automatically establish a general infinite theorem. The exact result, retaining the requested order, is (4, 10, 1, 4).
Reviewed step-by-step derivation
Finite unital ring · qf-mod27-ring-power · candidate
61. A Nonnegative Ring Power Starts at Its Actual One
Use the same carrier Fin(6)={0,…,5} for complete addition=[[0, 1, 2, 3, 4, 5], [1, 3, 4, 5, 0, 2], [2, 4, 3, 0, 5, 1], [3, 5, 0, 2, 1, 4], [4, 0, 5, 1, 2, 3], [5, 2, 1, 4, 3, 0]] and multiplication=[[0, 0, 0, 0, 0, 0], [0, 1, 2, 3, 4, 5], [0, 2, 2, 3, 3, 0], [0, 3, 3, 2, 2, 0], [0, 4, 3, 2, 1, 5], [0, 5, 0, 0, 5, 5]] tables in label order. Check the additive abelian-group axioms, multiplicative associativity, both distributive laws and a two-sided multiplicative identity distinct from the additive identity. The two identity labels need not be 0 and 1. This protocol excludes the zero ring and does not require multiplicative commutativity or imply a field. The tables must qualify first. Selected element=3, exponent=11. Give (element,exponent,one,result).
View answer and explanation
(3, 11, 1, 3)
Start at the actual multiplicative identity and iterate the multiplication table. Exponent zero returns that identity, including when the selected element is the additive zero; this is the declared algebraic power convention. An exact finite or coordinate slice does not automatically establish a general infinite theorem. The exact result, retaining the requested order, is (3, 11, 1, 3).
Reviewed step-by-step derivation
Finite unital ring · qf-mod27-ring-power · candidate
62. A Nonnegative Ring Power Starts at Its Actual One
Use the same carrier Fin(6)={0,…,5} for complete addition=[[0, 1, 2, 3, 4, 5], [1, 3, 0, 5, 2, 4], [2, 0, 4, 1, 5, 3], [3, 5, 1, 4, 0, 2], [4, 2, 5, 0, 3, 1], [5, 4, 3, 2, 1, 0]] and multiplication=[[0, 0, 0, 0, 0, 0], [0, 1, 2, 3, 4, 5], [0, 2, 1, 4, 3, 5], [0, 3, 4, 4, 3, 0], [0, 4, 3, 3, 4, 0], [0, 5, 5, 0, 0, 5]] tables in label order. Check the additive abelian-group axioms, multiplicative associativity, both distributive laws and a two-sided multiplicative identity distinct from the additive identity. The two identity labels need not be 0 and 1. This protocol excludes the zero ring and does not require multiplicative commutativity or imply a field. The tables must qualify first. Selected element=2, exponent=12. Give (element,exponent,one,result).
View answer and explanation
(2, 12, 1, 1)
Start at the actual multiplicative identity and iterate the multiplication table. Exponent zero returns that identity, including when the selected element is the additive zero; this is the declared algebraic power convention. An exact finite or coordinate slice does not automatically establish a general infinite theorem. The exact result, retaining the requested order, is (2, 12, 1, 1).
Reviewed step-by-step derivation
Finite unital ring · qf-mod27-ring-power · candidate
63. A Nonnegative Ring Power Starts at Its Actual One
Use the same carrier Fin(6)={0,…,5} for complete addition=[[0, 1, 2, 3, 4, 5], [1, 4, 3, 5, 2, 0], [2, 3, 0, 1, 5, 4], [3, 5, 1, 4, 0, 2], [4, 2, 5, 0, 3, 1], [5, 0, 4, 2, 1, 3]] and multiplication=[[0, 0, 0, 0, 0, 0], [0, 1, 2, 3, 4, 5], [0, 2, 2, 0, 0, 2], [0, 3, 0, 3, 4, 4], [0, 4, 0, 4, 3, 3], [0, 5, 2, 4, 3, 1]] tables in label order. Check the additive abelian-group axioms, multiplicative associativity, both distributive laws and a two-sided multiplicative identity distinct from the additive identity. The two identity labels need not be 0 and 1. This protocol excludes the zero ring and does not require multiplicative commutativity or imply a field. The tables must qualify first. Selected element=1, exponent=13. Give (element,exponent,one,result).
View answer and explanation
(1, 13, 1, 1)
Start at the actual multiplicative identity and iterate the multiplication table. Exponent zero returns that identity, including when the selected element is the additive zero; this is the declared algebraic power convention. An exact finite or coordinate slice does not automatically establish a general infinite theorem. The exact result, retaining the requested order, is (1, 13, 1, 1).
Reviewed step-by-step derivation
Finite unital ring · qf-mod27-ring-power · candidate
64. A Nonnegative Ring Power Starts at Its Actual One
Use the same carrier Fin(6)={0,…,5} for complete addition=[[0, 1, 2, 3, 4, 5], [1, 4, 5, 0, 2, 3], [2, 5, 0, 4, 3, 1], [3, 0, 4, 5, 1, 2], [4, 2, 3, 1, 5, 0], [5, 3, 1, 2, 0, 4]] and multiplication=[[0, 0, 0, 0, 0, 0], [0, 1, 2, 3, 4, 5], [0, 2, 2, 2, 0, 0], [0, 3, 2, 1, 5, 4], [0, 4, 0, 5, 5, 4], [0, 5, 0, 4, 4, 5]] tables in label order. Check the additive abelian-group axioms, multiplicative associativity, both distributive laws and a two-sided multiplicative identity distinct from the additive identity. The two identity labels need not be 0 and 1. This protocol excludes the zero ring and does not require multiplicative commutativity or imply a field. The tables must qualify first. Selected element=0, exponent=14. Give (element,exponent,one,result).
View answer and explanation
(0, 14, 1, 0)
Start at the actual multiplicative identity and iterate the multiplication table. Exponent zero returns that identity, including when the selected element is the additive zero; this is the declared algebraic power convention. An exact finite or coordinate slice does not automatically establish a general infinite theorem. The exact result, retaining the requested order, is (0, 14, 1, 0).
Reviewed step-by-step derivation
Finite unital ring · qf-mod27-ring-power · candidate
65. A Nonnegative Ring Power Starts at Its Actual One
Use the same carrier Fin(6)={0,…,5} for complete addition=[[0, 1, 2, 3, 4, 5], [1, 4, 5, 2, 3, 0], [2, 5, 4, 1, 0, 3], [3, 2, 1, 0, 5, 4], [4, 3, 0, 5, 2, 1], [5, 0, 3, 4, 1, 2]] and multiplication=[[0, 0, 0, 0, 0, 0], [0, 1, 2, 3, 4, 5], [0, 2, 2, 0, 4, 4], [0, 3, 0, 3, 0, 3], [0, 4, 4, 0, 2, 2], [0, 5, 4, 3, 2, 1]] tables in label order. Check the additive abelian-group axioms, multiplicative associativity, both distributive laws and a two-sided multiplicative identity distinct from the additive identity. The two identity labels need not be 0 and 1. This protocol excludes the zero ring and does not require multiplicative commutativity or imply a field. The tables must qualify first. Selected element=5, exponent=15. Give (element,exponent,one,result).
View answer and explanation
(5, 15, 1, 5)
Start at the actual multiplicative identity and iterate the multiplication table. Exponent zero returns that identity, including when the selected element is the additive zero; this is the declared algebraic power convention. An exact finite or coordinate slice does not automatically establish a general infinite theorem. The exact result, retaining the requested order, is (5, 15, 1, 5).
Reviewed step-by-step derivation
Finite unital ring · qf-mod27-ring-power · candidate
66. A Nonnegative Ring Power Starts at Its Actual One
Use the same carrier Fin(6)={0,…,5} for complete addition=[[0, 1, 2, 3, 4, 5], [1, 4, 0, 5, 3, 2], [2, 0, 5, 4, 1, 3], [3, 5, 4, 0, 2, 1], [4, 3, 1, 2, 5, 0], [5, 2, 3, 1, 0, 4]] and multiplication=[[0, 0, 0, 0, 0, 0], [0, 1, 2, 3, 4, 5], [0, 2, 1, 3, 5, 4], [0, 3, 3, 3, 0, 0], [0, 4, 5, 0, 5, 4], [0, 5, 4, 0, 4, 5]] tables in label order. Check the additive abelian-group axioms, multiplicative associativity, both distributive laws and a two-sided multiplicative identity distinct from the additive identity. The two identity labels need not be 0 and 1. This protocol excludes the zero ring and does not require multiplicative commutativity or imply a field. The tables must qualify first. Selected element=4, exponent=16. Give (element,exponent,one,result).
View answer and explanation
(4, 16, 1, 5)
Start at the actual multiplicative identity and iterate the multiplication table. Exponent zero returns that identity, including when the selected element is the additive zero; this is the declared algebraic power convention. An exact finite or coordinate slice does not automatically establish a general infinite theorem. The exact result, retaining the requested order, is (4, 16, 1, 5).
Reviewed step-by-step derivation
Finite unital ring · qf-mod27-ring-power · candidate
67. A Nonnegative Ring Power Starts at Its Actual One
Use the same carrier Fin(6)={0,…,5} for complete addition=[[0, 1, 2, 3, 4, 5], [1, 4, 3, 0, 5, 2], [2, 3, 4, 5, 0, 1], [3, 0, 5, 2, 1, 4], [4, 5, 0, 1, 2, 3], [5, 2, 1, 4, 3, 0]] and multiplication=[[0, 0, 0, 0, 0, 0], [0, 1, 2, 3, 4, 5], [0, 2, 2, 4, 4, 0], [0, 3, 4, 1, 2, 5], [0, 4, 4, 2, 2, 0], [0, 5, 0, 5, 0, 5]] tables in label order. Check the additive abelian-group axioms, multiplicative associativity, both distributive laws and a two-sided multiplicative identity distinct from the additive identity. The two identity labels need not be 0 and 1. This protocol excludes the zero ring and does not require multiplicative commutativity or imply a field. The tables must qualify first. Selected element=3, exponent=17. Give (element,exponent,one,result).
View answer and explanation
(3, 17, 1, 3)
Start at the actual multiplicative identity and iterate the multiplication table. Exponent zero returns that identity, including when the selected element is the additive zero; this is the declared algebraic power convention. An exact finite or coordinate slice does not automatically establish a general infinite theorem. The exact result, retaining the requested order, is (3, 17, 1, 3).
Reviewed step-by-step derivation
Finite unital ring · qf-mod27-ring-power · candidate
68. A Nonnegative Ring Power Starts at Its Actual One
Use the same carrier Fin(6)={0,…,5} for complete addition=[[0, 1, 2, 3, 4, 5], [1, 4, 0, 2, 5, 3], [2, 0, 3, 5, 1, 4], [3, 2, 5, 4, 0, 1], [4, 5, 1, 0, 3, 2], [5, 3, 4, 1, 2, 0]] and multiplication=[[0, 0, 0, 0, 0, 0], [0, 1, 2, 3, 4, 5], [0, 2, 1, 4, 3, 5], [0, 3, 4, 3, 4, 0], [0, 4, 3, 4, 3, 0], [0, 5, 5, 0, 0, 5]] tables in label order. Check the additive abelian-group axioms, multiplicative associativity, both distributive laws and a two-sided multiplicative identity distinct from the additive identity. The two identity labels need not be 0 and 1. This protocol excludes the zero ring and does not require multiplicative commutativity or imply a field. The tables must qualify first. Selected element=2, exponent=18. Give (element,exponent,one,result).
View answer and explanation
(2, 18, 1, 1)
Start at the actual multiplicative identity and iterate the multiplication table. Exponent zero returns that identity, including when the selected element is the additive zero; this is the declared algebraic power convention. An exact finite or coordinate slice does not automatically establish a general infinite theorem. The exact result, retaining the requested order, is (2, 18, 1, 1).
Reviewed step-by-step derivation
Finite unital ring · qf-mod27-ring-power · candidate
69. A Nonnegative Ring Power Starts at Its Actual One
Use the same carrier Fin(6)={0,…,5} for complete addition=[[0, 1, 2, 3, 4, 5], [1, 5, 3, 4, 0, 2], [2, 3, 0, 1, 5, 4], [3, 4, 1, 5, 2, 0], [4, 0, 5, 2, 3, 1], [5, 2, 4, 0, 1, 3]] and multiplication=[[0, 0, 0, 0, 0, 0], [0, 1, 2, 3, 4, 5], [0, 2, 2, 0, 2, 0], [0, 3, 0, 3, 5, 5], [0, 4, 2, 5, 1, 3], [0, 5, 0, 5, 3, 3]] tables in label order. Check the additive abelian-group axioms, multiplicative associativity, both distributive laws and a two-sided multiplicative identity distinct from the additive identity. The two identity labels need not be 0 and 1. This protocol excludes the zero ring and does not require multiplicative commutativity or imply a field. The tables must qualify first. Selected element=1, exponent=19. Give (element,exponent,one,result).
View answer and explanation
(1, 19, 1, 1)
Start at the actual multiplicative identity and iterate the multiplication table. Exponent zero returns that identity, including when the selected element is the additive zero; this is the declared algebraic power convention. An exact finite or coordinate slice does not automatically establish a general infinite theorem. The exact result, retaining the requested order, is (1, 19, 1, 1).
Reviewed step-by-step derivation
Finite unital ring · qf-mod27-ring-power · candidate
70. A Nonnegative Ring Power Starts at Its Actual One
Use the same carrier Fin(6)={0,…,5} for complete addition=[[0, 1, 2, 3, 4, 5], [1, 5, 4, 0, 3, 2], [2, 4, 0, 5, 1, 3], [3, 0, 5, 4, 2, 1], [4, 3, 1, 2, 5, 0], [5, 2, 3, 1, 0, 4]] and multiplication=[[0, 0, 0, 0, 0, 0], [0, 1, 2, 3, 4, 5], [0, 2, 2, 2, 0, 0], [0, 3, 2, 1, 5, 4], [0, 4, 0, 5, 4, 5], [0, 5, 0, 4, 5, 4]] tables in label order. Check the additive abelian-group axioms, multiplicative associativity, both distributive laws and a two-sided multiplicative identity distinct from the additive identity. The two identity labels need not be 0 and 1. This protocol excludes the zero ring and does not require multiplicative commutativity or imply a field. The tables must qualify first. Selected element=0, exponent=20. Give (element,exponent,one,result).
View answer and explanation
(0, 20, 1, 0)
Start at the actual multiplicative identity and iterate the multiplication table. Exponent zero returns that identity, including when the selected element is the additive zero; this is the declared algebraic power convention. An exact finite or coordinate slice does not automatically establish a general infinite theorem. The exact result, retaining the requested order, is (0, 20, 1, 0).
Reviewed step-by-step derivation
Finite unital ring · qf-mod27-ring-power · candidate
71. A Nonnegative Ring Power Starts at Its Actual One
Use the same carrier Fin(6)={0,…,5} for complete addition=[[0, 1, 2, 3, 4, 5], [1, 5, 4, 2, 0, 3], [2, 4, 5, 1, 3, 0], [3, 2, 1, 0, 5, 4], [4, 0, 3, 5, 2, 1], [5, 3, 0, 4, 1, 2]] and multiplication=[[0, 0, 0, 0, 0, 0], [0, 1, 2, 3, 4, 5], [0, 2, 2, 0, 5, 5], [0, 3, 0, 3, 3, 0], [0, 4, 5, 3, 1, 2], [0, 5, 5, 0, 2, 2]] tables in label order. Check the additive abelian-group axioms, multiplicative associativity, both distributive laws and a two-sided multiplicative identity distinct from the additive identity. The two identity labels need not be 0 and 1. This protocol excludes the zero ring and does not require multiplicative commutativity or imply a field. The tables must qualify first. Selected element=5, exponent=21. Give (element,exponent,one,result).
View answer and explanation
(5, 21, 1, 5)
Start at the actual multiplicative identity and iterate the multiplication table. Exponent zero returns that identity, including when the selected element is the additive zero; this is the declared algebraic power convention. An exact finite or coordinate slice does not automatically establish a general infinite theorem. The exact result, retaining the requested order, is (5, 21, 1, 5).
Reviewed step-by-step derivation
Finite unital ring · qf-mod27-ring-power · candidate
72. A Nonnegative Ring Power Starts at Its Actual One
Use the same carrier Fin(6)={0,…,5} for complete addition=[[0, 1, 2, 3, 4, 5], [1, 5, 0, 4, 2, 3], [2, 0, 4, 5, 3, 1], [3, 4, 5, 0, 1, 2], [4, 2, 3, 1, 5, 0], [5, 3, 1, 2, 0, 4]] and multiplication=[[0, 0, 0, 0, 0, 0], [0, 1, 2, 3, 4, 5], [0, 2, 1, 3, 5, 4], [0, 3, 3, 3, 0, 0], [0, 4, 5, 0, 4, 5], [0, 5, 4, 0, 5, 4]] tables in label order. Check the additive abelian-group axioms, multiplicative associativity, both distributive laws and a two-sided multiplicative identity distinct from the additive identity. The two identity labels need not be 0 and 1. This protocol excludes the zero ring and does not require multiplicative commutativity or imply a field. The tables must qualify first. Selected element=4, exponent=22. Give (element,exponent,one,result).
View answer and explanation
(4, 22, 1, 4)
Start at the actual multiplicative identity and iterate the multiplication table. Exponent zero returns that identity, including when the selected element is the additive zero; this is the declared algebraic power convention. An exact finite or coordinate slice does not automatically establish a general infinite theorem. The exact result, retaining the requested order, is (4, 22, 1, 4).
Reviewed step-by-step derivation
Finite unital ring · qf-mod27-ring-power · candidate
73. A Nonnegative Ring Power Starts at Its Actual One
Use the same carrier Fin(6)={0,…,5} for complete addition=[[0, 1, 2, 3, 4, 5], [1, 5, 3, 0, 2, 4], [2, 3, 5, 4, 1, 0], [3, 0, 4, 2, 5, 1], [4, 2, 1, 5, 0, 3], [5, 4, 0, 1, 3, 2]] and multiplication=[[0, 0, 0, 0, 0, 0], [0, 1, 2, 3, 4, 5], [0, 2, 2, 5, 0, 5], [0, 3, 5, 1, 4, 2], [0, 4, 0, 4, 4, 0], [0, 5, 5, 2, 0, 2]] tables in label order. Check the additive abelian-group axioms, multiplicative associativity, both distributive laws and a two-sided multiplicative identity distinct from the additive identity. The two identity labels need not be 0 and 1. This protocol excludes the zero ring and does not require multiplicative commutativity or imply a field. The tables must qualify first. Selected element=3, exponent=23. Give (element,exponent,one,result).
View answer and explanation
(3, 23, 1, 3)
Start at the actual multiplicative identity and iterate the multiplication table. Exponent zero returns that identity, including when the selected element is the additive zero; this is the declared algebraic power convention. An exact finite or coordinate slice does not automatically establish a general infinite theorem. The exact result, retaining the requested order, is (3, 23, 1, 3).
Reviewed step-by-step derivation
Finite unital ring · qf-mod27-ring-power · candidate
74. A Nonnegative Ring Power Starts at Its Actual One
Use the same carrier Fin(6)={0,…,5} for complete addition=[[0, 1, 2, 3, 4, 5], [1, 5, 0, 2, 3, 4], [2, 0, 3, 4, 5, 1], [3, 2, 4, 5, 1, 0], [4, 3, 5, 1, 0, 2], [5, 4, 1, 0, 2, 3]] and multiplication=[[0, 0, 0, 0, 0, 0], [0, 1, 2, 3, 4, 5], [0, 2, 1, 5, 4, 3], [0, 3, 5, 3, 0, 5], [0, 4, 4, 0, 4, 0], [0, 5, 3, 5, 0, 3]] tables in label order. Check the additive abelian-group axioms, multiplicative associativity, both distributive laws and a two-sided multiplicative identity distinct from the additive identity. The two identity labels need not be 0 and 1. This protocol excludes the zero ring and does not require multiplicative commutativity or imply a field. The tables must qualify first. Selected element=2, exponent=24. Give (element,exponent,one,result).
View answer and explanation
(2, 24, 1, 1)
Start at the actual multiplicative identity and iterate the multiplication table. Exponent zero returns that identity, including when the selected element is the additive zero; this is the declared algebraic power convention. An exact finite or coordinate slice does not automatically establish a general infinite theorem. The exact result, retaining the requested order, is (2, 24, 1, 1).
Reviewed step-by-step derivation
Finite unital ring · qf-mod27-ring-power · candidate
75. A Nonnegative Ring Power Starts at Its Actual One
Use the same carrier Fin(6)={0,…,5} for complete addition=[[0, 1, 2, 3, 4, 5], [1, 4, 3, 5, 0, 2], [2, 3, 1, 4, 5, 0], [3, 5, 4, 0, 2, 1], [4, 0, 5, 2, 1, 3], [5, 2, 0, 1, 3, 4]] and multiplication=[[0, 0, 0, 0, 0, 0], [0, 4, 1, 0, 1, 4], [0, 1, 2, 3, 4, 5], [0, 0, 3, 3, 0, 3], [0, 1, 4, 0, 4, 1], [0, 4, 5, 3, 1, 2]] tables in label order. Check the additive abelian-group axioms, multiplicative associativity, both distributive laws and a two-sided multiplicative identity distinct from the additive identity. The two identity labels need not be 0 and 1. This protocol excludes the zero ring and does not require multiplicative commutativity or imply a field. The tables must qualify first. Selected element=1, exponent=25. Give (element,exponent,one,result).
View answer and explanation
(1, 25, 2, 1)
Start at the actual multiplicative identity and iterate the multiplication table. Exponent zero returns that identity, including when the selected element is the additive zero; this is the declared algebraic power convention. An exact finite or coordinate slice does not automatically establish a general infinite theorem. The exact result, retaining the requested order, is (1, 25, 2, 1).
Reviewed step-by-step derivation
Finite unital ring · qf-mod27-ring-power · candidate
76. Keep Additive Zero and Multiplicative One Distinct
Use the same carrier Fin(6)={0,…,5} for complete addition=[[0, 1, 2, 3, 4, 5], [1, 2, 3, 4, 5, 0], [2, 3, 4, 5, 0, 1], [3, 4, 5, 0, 1, 2], [4, 5, 0, 1, 2, 3], [5, 0, 1, 2, 3, 4]] and multiplication=[[0, 0, 0, 0, 0, 0], [0, 1, 2, 3, 4, 5], [0, 2, 4, 0, 2, 4], [0, 3, 0, 3, 0, 3], [0, 4, 2, 0, 4, 2], [0, 5, 4, 3, 2, 1]] tables in label order. Check the additive abelian-group axioms, multiplicative associativity, both distributive laws and a two-sided multiplicative identity distinct from the additive identity. The two identity labels need not be 0 and 1. This protocol excludes the zero ring and does not require multiplicative commutativity or imply a field. Require fully qualified tables. Give (zero,one,distinct?).
View answer and explanation
(0, 1, Yes)
Find both two-sided identities from their respective operations. Labels denote carrier elements, not their assumed arithmetic values. An exact finite or coordinate slice does not automatically establish a general infinite theorem. The exact result, retaining the requested order, is (0, 1, Yes).
Reviewed step-by-step derivation
Finite unital ring · qf-mod27-ring-identities · candidate
77. Keep Additive Zero and Multiplicative One Distinct
Use the same carrier Fin(6)={0,…,5} for complete addition=[[0, 1, 2, 3, 4, 5], [1, 2, 3, 5, 0, 4], [2, 3, 5, 4, 1, 0], [3, 5, 4, 0, 2, 1], [4, 0, 1, 2, 5, 3], [5, 4, 0, 1, 3, 2]] and multiplication=[[0, 0, 0, 0, 0, 0], [0, 1, 2, 3, 4, 5], [0, 2, 5, 0, 5, 2], [0, 3, 0, 3, 3, 0], [0, 4, 5, 3, 1, 2], [0, 5, 2, 0, 2, 5]] tables in label order. Check the additive abelian-group axioms, multiplicative associativity, both distributive laws and a two-sided multiplicative identity distinct from the additive identity. The two identity labels need not be 0 and 1. This protocol excludes the zero ring and does not require multiplicative commutativity or imply a field. Require fully qualified tables. Give (zero,one,distinct?).
View answer and explanation
(0, 1, Yes)
Find both two-sided identities from their respective operations. Labels denote carrier elements, not their assumed arithmetic values. An exact finite or coordinate slice does not automatically establish a general infinite theorem. The exact result, retaining the requested order, is (0, 1, Yes).
Reviewed step-by-step derivation
Finite unital ring · qf-mod27-ring-identities · candidate
78. Keep Additive Zero and Multiplicative One Distinct
Use the same carrier Fin(6)={0,…,5} for complete addition=[[0, 1, 2, 3, 4, 5], [1, 2, 4, 5, 3, 0], [2, 4, 3, 0, 5, 1], [3, 5, 0, 2, 1, 4], [4, 3, 5, 1, 0, 2], [5, 0, 1, 4, 2, 3]] and multiplication=[[0, 0, 0, 0, 0, 0], [0, 1, 2, 3, 4, 5], [0, 2, 3, 2, 0, 3], [0, 3, 2, 3, 0, 2], [0, 4, 0, 0, 4, 4], [0, 5, 3, 2, 4, 1]] tables in label order. Check the additive abelian-group axioms, multiplicative associativity, both distributive laws and a two-sided multiplicative identity distinct from the additive identity. The two identity labels need not be 0 and 1. This protocol excludes the zero ring and does not require multiplicative commutativity or imply a field. Require fully qualified tables. Give (zero,one,distinct?).
View answer and explanation
(0, 1, Yes)
Find both two-sided identities from their respective operations. Labels denote carrier elements, not their assumed arithmetic values. An exact finite or coordinate slice does not automatically establish a general infinite theorem. The exact result, retaining the requested order, is (0, 1, Yes).
Reviewed step-by-step derivation
Finite unital ring · qf-mod27-ring-identities · candidate
79. Keep Additive Zero and Multiplicative One Distinct
Use the same carrier Fin(6)={0,…,5} for complete addition=[[0, 1, 2, 3, 4, 5], [1, 2, 4, 0, 5, 3], [2, 4, 5, 1, 3, 0], [3, 0, 1, 5, 2, 4], [4, 5, 3, 2, 0, 1], [5, 3, 0, 4, 1, 2]] and multiplication=[[0, 0, 0, 0, 0, 0], [0, 1, 2, 3, 4, 5], [0, 2, 5, 5, 0, 2], [0, 3, 5, 1, 4, 2], [0, 4, 0, 4, 4, 0], [0, 5, 2, 2, 0, 5]] tables in label order. Check the additive abelian-group axioms, multiplicative associativity, both distributive laws and a two-sided multiplicative identity distinct from the additive identity. The two identity labels need not be 0 and 1. This protocol excludes the zero ring and does not require multiplicative commutativity or imply a field. Require fully qualified tables. Give (zero,one,distinct?).
View answer and explanation
(0, 1, Yes)
Find both two-sided identities from their respective operations. Labels denote carrier elements, not their assumed arithmetic values. An exact finite or coordinate slice does not automatically establish a general infinite theorem. The exact result, retaining the requested order, is (0, 1, Yes).
Reviewed step-by-step derivation
Finite unital ring · qf-mod27-ring-identities · candidate
80. Keep Additive Zero and Multiplicative One Distinct
Use the same carrier Fin(6)={0,…,5} for complete addition=[[0, 1, 2, 3, 4, 5], [1, 2, 5, 4, 0, 3], [2, 5, 3, 0, 1, 4], [3, 4, 0, 2, 5, 1], [4, 0, 1, 5, 3, 2], [5, 3, 4, 1, 2, 0]] and multiplication=[[0, 0, 0, 0, 0, 0], [0, 1, 2, 3, 4, 5], [0, 2, 3, 2, 3, 0], [0, 3, 2, 3, 2, 0], [0, 4, 3, 2, 1, 5], [0, 5, 0, 0, 5, 5]] tables in label order. Check the additive abelian-group axioms, multiplicative associativity, both distributive laws and a two-sided multiplicative identity distinct from the additive identity. The two identity labels need not be 0 and 1. This protocol excludes the zero ring and does not require multiplicative commutativity or imply a field. Require fully qualified tables. Give (zero,one,distinct?).
View answer and explanation
(0, 1, Yes)
Find both two-sided identities from their respective operations. Labels denote carrier elements, not their assumed arithmetic values. An exact finite or coordinate slice does not automatically establish a general infinite theorem. The exact result, retaining the requested order, is (0, 1, Yes).
Reviewed step-by-step derivation
Finite unital ring · qf-mod27-ring-identities · candidate
81. Keep Additive Zero and Multiplicative One Distinct
Use the same carrier Fin(6)={0,…,5} for complete addition=[[0, 1, 2, 3, 4, 5], [1, 2, 5, 0, 3, 4], [2, 5, 4, 1, 0, 3], [3, 0, 1, 4, 5, 2], [4, 3, 0, 5, 2, 1], [5, 4, 3, 2, 1, 0]] and multiplication=[[0, 0, 0, 0, 0, 0], [0, 1, 2, 3, 4, 5], [0, 2, 4, 4, 2, 0], [0, 3, 4, 1, 2, 5], [0, 4, 2, 2, 4, 0], [0, 5, 0, 5, 0, 5]] tables in label order. Check the additive abelian-group axioms, multiplicative associativity, both distributive laws and a two-sided multiplicative identity distinct from the additive identity. The two identity labels need not be 0 and 1. This protocol excludes the zero ring and does not require multiplicative commutativity or imply a field. Require fully qualified tables. Give (zero,one,distinct?).
View answer and explanation
(0, 1, Yes)
Find both two-sided identities from their respective operations. Labels denote carrier elements, not their assumed arithmetic values. An exact finite or coordinate slice does not automatically establish a general infinite theorem. The exact result, retaining the requested order, is (0, 1, Yes).
Reviewed step-by-step derivation
Finite unital ring · qf-mod27-ring-identities · candidate
82. Keep Additive Zero and Multiplicative One Distinct
Use the same carrier Fin(6)={0,…,5} for complete addition=[[0, 1, 2, 3, 4, 5], [1, 3, 4, 2, 5, 0], [2, 4, 0, 5, 1, 3], [3, 2, 5, 4, 0, 1], [4, 5, 1, 0, 3, 2], [5, 0, 3, 1, 2, 4]] and multiplication=[[0, 0, 0, 0, 0, 0], [0, 1, 2, 3, 4, 5], [0, 2, 2, 0, 0, 2], [0, 3, 0, 4, 3, 4], [0, 4, 0, 3, 4, 3], [0, 5, 2, 4, 3, 1]] tables in label order. Check the additive abelian-group axioms, multiplicative associativity, both distributive laws and a two-sided multiplicative identity distinct from the additive identity. The two identity labels need not be 0 and 1. This protocol excludes the zero ring and does not require multiplicative commutativity or imply a field. Require fully qualified tables. Give (zero,one,distinct?).
View answer and explanation
(0, 1, Yes)
Find both two-sided identities from their respective operations. Labels denote carrier elements, not their assumed arithmetic values. An exact finite or coordinate slice does not automatically establish a general infinite theorem. The exact result, retaining the requested order, is (0, 1, Yes).
Reviewed step-by-step derivation
Finite unital ring · qf-mod27-ring-identities · candidate
83. Keep Additive Zero and Multiplicative One Distinct
Use the same carrier Fin(6)={0,…,5} for complete addition=[[0, 1, 2, 3, 4, 5], [1, 3, 5, 2, 0, 4], [2, 5, 0, 4, 3, 1], [3, 2, 4, 5, 1, 0], [4, 0, 3, 1, 5, 2], [5, 4, 1, 0, 2, 3]] and multiplication=[[0, 0, 0, 0, 0, 0], [0, 1, 2, 3, 4, 5], [0, 2, 2, 0, 2, 0], [0, 3, 0, 5, 5, 3], [0, 4, 2, 5, 1, 3], [0, 5, 0, 3, 3, 5]] tables in label order. Check the additive abelian-group axioms, multiplicative associativity, both distributive laws and a two-sided multiplicative identity distinct from the additive identity. The two identity labels need not be 0 and 1. This protocol excludes the zero ring and does not require multiplicative commutativity or imply a field. Require fully qualified tables. Give (zero,one,distinct?).
View answer and explanation
(0, 1, Yes)
Find both two-sided identities from their respective operations. Labels denote carrier elements, not their assumed arithmetic values. An exact finite or coordinate slice does not automatically establish a general infinite theorem. The exact result, retaining the requested order, is (0, 1, Yes).
Reviewed step-by-step derivation
Finite unital ring · qf-mod27-ring-identities · candidate
84. Keep Additive Zero and Multiplicative One Distinct
Use the same carrier Fin(6)={0,…,5} for complete addition=[[0, 1, 2, 3, 4, 5], [1, 3, 5, 4, 2, 0], [2, 5, 3, 0, 1, 4], [3, 4, 0, 2, 5, 1], [4, 2, 1, 5, 0, 3], [5, 0, 4, 1, 3, 2]] and multiplication=[[0, 0, 0, 0, 0, 0], [0, 1, 2, 3, 4, 5], [0, 2, 2, 3, 0, 3], [0, 3, 3, 2, 0, 2], [0, 4, 0, 0, 4, 4], [0, 5, 3, 2, 4, 1]] tables in label order. Check the additive abelian-group axioms, multiplicative associativity, both distributive laws and a two-sided multiplicative identity distinct from the additive identity. The two identity labels need not be 0 and 1. This protocol excludes the zero ring and does not require multiplicative commutativity or imply a field. Require fully qualified tables. Give (zero,one,distinct?).
View answer and explanation
(0, 1, Yes)
Find both two-sided identities from their respective operations. Labels denote carrier elements, not their assumed arithmetic values. An exact finite or coordinate slice does not automatically establish a general infinite theorem. The exact result, retaining the requested order, is (0, 1, Yes).
Reviewed step-by-step derivation
Finite unital ring · qf-mod27-ring-identities · candidate
85. Keep Additive Zero and Multiplicative One Distinct
Use the same carrier Fin(6)={0,…,5} for complete addition=[[0, 1, 2, 3, 4, 5], [1, 3, 0, 4, 5, 2], [2, 0, 5, 1, 3, 4], [3, 4, 1, 5, 2, 0], [4, 5, 3, 2, 0, 1], [5, 2, 4, 0, 1, 3]] and multiplication=[[0, 0, 0, 0, 0, 0], [0, 1, 2, 3, 4, 5], [0, 2, 1, 5, 4, 3], [0, 3, 5, 5, 0, 3], [0, 4, 4, 0, 4, 0], [0, 5, 3, 3, 0, 5]] tables in label order. Check the additive abelian-group axioms, multiplicative associativity, both distributive laws and a two-sided multiplicative identity distinct from the additive identity. The two identity labels need not be 0 and 1. This protocol excludes the zero ring and does not require multiplicative commutativity or imply a field. Require fully qualified tables. Give (zero,one,distinct?).
View answer and explanation
(0, 1, Yes)
Find both two-sided identities from their respective operations. Labels denote carrier elements, not their assumed arithmetic values. An exact finite or coordinate slice does not automatically establish a general infinite theorem. The exact result, retaining the requested order, is (0, 1, Yes).
Reviewed step-by-step derivation
Finite unital ring · qf-mod27-ring-identities · candidate
86. Keep Additive Zero and Multiplicative One Distinct
Use the same carrier Fin(6)={0,…,5} for complete addition=[[0, 1, 2, 3, 4, 5], [1, 3, 4, 5, 0, 2], [2, 4, 3, 0, 5, 1], [3, 5, 0, 2, 1, 4], [4, 0, 5, 1, 2, 3], [5, 2, 1, 4, 3, 0]] and multiplication=[[0, 0, 0, 0, 0, 0], [0, 1, 2, 3, 4, 5], [0, 2, 2, 3, 3, 0], [0, 3, 3, 2, 2, 0], [0, 4, 3, 2, 1, 5], [0, 5, 0, 0, 5, 5]] tables in label order. Check the additive abelian-group axioms, multiplicative associativity, both distributive laws and a two-sided multiplicative identity distinct from the additive identity. The two identity labels need not be 0 and 1. This protocol excludes the zero ring and does not require multiplicative commutativity or imply a field. Require fully qualified tables. Give (zero,one,distinct?).
View answer and explanation
(0, 1, Yes)
Find both two-sided identities from their respective operations. Labels denote carrier elements, not their assumed arithmetic values. An exact finite or coordinate slice does not automatically establish a general infinite theorem. The exact result, retaining the requested order, is (0, 1, Yes).
Reviewed step-by-step derivation
Finite unital ring · qf-mod27-ring-identities · candidate
87. Keep Additive Zero and Multiplicative One Distinct
Use the same carrier Fin(6)={0,…,5} for complete addition=[[0, 1, 2, 3, 4, 5], [1, 3, 0, 5, 2, 4], [2, 0, 4, 1, 5, 3], [3, 5, 1, 4, 0, 2], [4, 2, 5, 0, 3, 1], [5, 4, 3, 2, 1, 0]] and multiplication=[[0, 0, 0, 0, 0, 0], [0, 1, 2, 3, 4, 5], [0, 2, 1, 4, 3, 5], [0, 3, 4, 4, 3, 0], [0, 4, 3, 3, 4, 0], [0, 5, 5, 0, 0, 5]] tables in label order. Check the additive abelian-group axioms, multiplicative associativity, both distributive laws and a two-sided multiplicative identity distinct from the additive identity. The two identity labels need not be 0 and 1. This protocol excludes the zero ring and does not require multiplicative commutativity or imply a field. Require fully qualified tables. Give (zero,one,distinct?).
View answer and explanation
(0, 1, Yes)
Find both two-sided identities from their respective operations. Labels denote carrier elements, not their assumed arithmetic values. An exact finite or coordinate slice does not automatically establish a general infinite theorem. The exact result, retaining the requested order, is (0, 1, Yes).
Reviewed step-by-step derivation
Finite unital ring · qf-mod27-ring-identities · candidate
88. Keep Additive Zero and Multiplicative One Distinct
Use the same carrier Fin(6)={0,…,5} for complete addition=[[0, 1, 2, 3, 4, 5], [1, 4, 3, 5, 2, 0], [2, 3, 0, 1, 5, 4], [3, 5, 1, 4, 0, 2], [4, 2, 5, 0, 3, 1], [5, 0, 4, 2, 1, 3]] and multiplication=[[0, 0, 0, 0, 0, 0], [0, 1, 2, 3, 4, 5], [0, 2, 2, 0, 0, 2], [0, 3, 0, 3, 4, 4], [0, 4, 0, 4, 3, 3], [0, 5, 2, 4, 3, 1]] tables in label order. Check the additive abelian-group axioms, multiplicative associativity, both distributive laws and a two-sided multiplicative identity distinct from the additive identity. The two identity labels need not be 0 and 1. This protocol excludes the zero ring and does not require multiplicative commutativity or imply a field. Require fully qualified tables. Give (zero,one,distinct?).
View answer and explanation
(0, 1, Yes)
Find both two-sided identities from their respective operations. Labels denote carrier elements, not their assumed arithmetic values. An exact finite or coordinate slice does not automatically establish a general infinite theorem. The exact result, retaining the requested order, is (0, 1, Yes).
Reviewed step-by-step derivation
Finite unital ring · qf-mod27-ring-identities · candidate
89. Keep Additive Zero and Multiplicative One Distinct
Use the same carrier Fin(6)={0,…,5} for complete addition=[[0, 1, 2, 3, 4, 5], [1, 4, 5, 0, 2, 3], [2, 5, 0, 4, 3, 1], [3, 0, 4, 5, 1, 2], [4, 2, 3, 1, 5, 0], [5, 3, 1, 2, 0, 4]] and multiplication=[[0, 0, 0, 0, 0, 0], [0, 1, 2, 3, 4, 5], [0, 2, 2, 2, 0, 0], [0, 3, 2, 1, 5, 4], [0, 4, 0, 5, 5, 4], [0, 5, 0, 4, 4, 5]] tables in label order. Check the additive abelian-group axioms, multiplicative associativity, both distributive laws and a two-sided multiplicative identity distinct from the additive identity. The two identity labels need not be 0 and 1. This protocol excludes the zero ring and does not require multiplicative commutativity or imply a field. Require fully qualified tables. Give (zero,one,distinct?).
View answer and explanation
(0, 1, Yes)
Find both two-sided identities from their respective operations. Labels denote carrier elements, not their assumed arithmetic values. An exact finite or coordinate slice does not automatically establish a general infinite theorem. The exact result, retaining the requested order, is (0, 1, Yes).
Reviewed step-by-step derivation
Finite unital ring · qf-mod27-ring-identities · candidate
90. Keep Additive Zero and Multiplicative One Distinct
Use the same carrier Fin(6)={0,…,5} for complete addition=[[0, 1, 2, 3, 4, 5], [1, 4, 5, 2, 3, 0], [2, 5, 4, 1, 0, 3], [3, 2, 1, 0, 5, 4], [4, 3, 0, 5, 2, 1], [5, 0, 3, 4, 1, 2]] and multiplication=[[0, 0, 0, 0, 0, 0], [0, 1, 2, 3, 4, 5], [0, 2, 2, 0, 4, 4], [0, 3, 0, 3, 0, 3], [0, 4, 4, 0, 2, 2], [0, 5, 4, 3, 2, 1]] tables in label order. Check the additive abelian-group axioms, multiplicative associativity, both distributive laws and a two-sided multiplicative identity distinct from the additive identity. The two identity labels need not be 0 and 1. This protocol excludes the zero ring and does not require multiplicative commutativity or imply a field. Require fully qualified tables. Give (zero,one,distinct?).
View answer and explanation
(0, 1, Yes)
Find both two-sided identities from their respective operations. Labels denote carrier elements, not their assumed arithmetic values. An exact finite or coordinate slice does not automatically establish a general infinite theorem. The exact result, retaining the requested order, is (0, 1, Yes).
Reviewed step-by-step derivation
Finite unital ring · qf-mod27-ring-identities · candidate
91. Keep Additive Zero and Multiplicative One Distinct
Use the same carrier Fin(6)={0,…,5} for complete addition=[[0, 1, 2, 3, 4, 5], [1, 4, 0, 5, 3, 2], [2, 0, 5, 4, 1, 3], [3, 5, 4, 0, 2, 1], [4, 3, 1, 2, 5, 0], [5, 2, 3, 1, 0, 4]] and multiplication=[[0, 0, 0, 0, 0, 0], [0, 1, 2, 3, 4, 5], [0, 2, 1, 3, 5, 4], [0, 3, 3, 3, 0, 0], [0, 4, 5, 0, 5, 4], [0, 5, 4, 0, 4, 5]] tables in label order. Check the additive abelian-group axioms, multiplicative associativity, both distributive laws and a two-sided multiplicative identity distinct from the additive identity. The two identity labels need not be 0 and 1. This protocol excludes the zero ring and does not require multiplicative commutativity or imply a field. Require fully qualified tables. Give (zero,one,distinct?).
View answer and explanation
(0, 1, Yes)
Find both two-sided identities from their respective operations. Labels denote carrier elements, not their assumed arithmetic values. An exact finite or coordinate slice does not automatically establish a general infinite theorem. The exact result, retaining the requested order, is (0, 1, Yes).
Reviewed step-by-step derivation
Finite unital ring · qf-mod27-ring-identities · candidate
92. Keep Additive Zero and Multiplicative One Distinct
Use the same carrier Fin(6)={0,…,5} for complete addition=[[0, 1, 2, 3, 4, 5], [1, 4, 3, 0, 5, 2], [2, 3, 4, 5, 0, 1], [3, 0, 5, 2, 1, 4], [4, 5, 0, 1, 2, 3], [5, 2, 1, 4, 3, 0]] and multiplication=[[0, 0, 0, 0, 0, 0], [0, 1, 2, 3, 4, 5], [0, 2, 2, 4, 4, 0], [0, 3, 4, 1, 2, 5], [0, 4, 4, 2, 2, 0], [0, 5, 0, 5, 0, 5]] tables in label order. Check the additive abelian-group axioms, multiplicative associativity, both distributive laws and a two-sided multiplicative identity distinct from the additive identity. The two identity labels need not be 0 and 1. This protocol excludes the zero ring and does not require multiplicative commutativity or imply a field. Require fully qualified tables. Give (zero,one,distinct?).
View answer and explanation
(0, 1, Yes)
Find both two-sided identities from their respective operations. Labels denote carrier elements, not their assumed arithmetic values. An exact finite or coordinate slice does not automatically establish a general infinite theorem. The exact result, retaining the requested order, is (0, 1, Yes).
Reviewed step-by-step derivation
Finite unital ring · qf-mod27-ring-identities · candidate
93. Keep Additive Zero and Multiplicative One Distinct
Use the same carrier Fin(6)={0,…,5} for complete addition=[[0, 1, 2, 3, 4, 5], [1, 4, 0, 2, 5, 3], [2, 0, 3, 5, 1, 4], [3, 2, 5, 4, 0, 1], [4, 5, 1, 0, 3, 2], [5, 3, 4, 1, 2, 0]] and multiplication=[[0, 0, 0, 0, 0, 0], [0, 1, 2, 3, 4, 5], [0, 2, 1, 4, 3, 5], [0, 3, 4, 3, 4, 0], [0, 4, 3, 4, 3, 0], [0, 5, 5, 0, 0, 5]] tables in label order. Check the additive abelian-group axioms, multiplicative associativity, both distributive laws and a two-sided multiplicative identity distinct from the additive identity. The two identity labels need not be 0 and 1. This protocol excludes the zero ring and does not require multiplicative commutativity or imply a field. Require fully qualified tables. Give (zero,one,distinct?).
View answer and explanation
(0, 1, Yes)
Find both two-sided identities from their respective operations. Labels denote carrier elements, not their assumed arithmetic values. An exact finite or coordinate slice does not automatically establish a general infinite theorem. The exact result, retaining the requested order, is (0, 1, Yes).
Reviewed step-by-step derivation
Finite unital ring · qf-mod27-ring-identities · candidate
94. Keep Additive Zero and Multiplicative One Distinct
Use the same carrier Fin(6)={0,…,5} for complete addition=[[0, 1, 2, 3, 4, 5], [1, 5, 3, 4, 0, 2], [2, 3, 0, 1, 5, 4], [3, 4, 1, 5, 2, 0], [4, 0, 5, 2, 3, 1], [5, 2, 4, 0, 1, 3]] and multiplication=[[0, 0, 0, 0, 0, 0], [0, 1, 2, 3, 4, 5], [0, 2, 2, 0, 2, 0], [0, 3, 0, 3, 5, 5], [0, 4, 2, 5, 1, 3], [0, 5, 0, 5, 3, 3]] tables in label order. Check the additive abelian-group axioms, multiplicative associativity, both distributive laws and a two-sided multiplicative identity distinct from the additive identity. The two identity labels need not be 0 and 1. This protocol excludes the zero ring and does not require multiplicative commutativity or imply a field. Require fully qualified tables. Give (zero,one,distinct?).
View answer and explanation
(0, 1, Yes)
Find both two-sided identities from their respective operations. Labels denote carrier elements, not their assumed arithmetic values. An exact finite or coordinate slice does not automatically establish a general infinite theorem. The exact result, retaining the requested order, is (0, 1, Yes).
Reviewed step-by-step derivation
Finite unital ring · qf-mod27-ring-identities · candidate
95. Keep Additive Zero and Multiplicative One Distinct
Use the same carrier Fin(6)={0,…,5} for complete addition=[[0, 1, 2, 3, 4, 5], [1, 5, 4, 0, 3, 2], [2, 4, 0, 5, 1, 3], [3, 0, 5, 4, 2, 1], [4, 3, 1, 2, 5, 0], [5, 2, 3, 1, 0, 4]] and multiplication=[[0, 0, 0, 0, 0, 0], [0, 1, 2, 3, 4, 5], [0, 2, 2, 2, 0, 0], [0, 3, 2, 1, 5, 4], [0, 4, 0, 5, 4, 5], [0, 5, 0, 4, 5, 4]] tables in label order. Check the additive abelian-group axioms, multiplicative associativity, both distributive laws and a two-sided multiplicative identity distinct from the additive identity. The two identity labels need not be 0 and 1. This protocol excludes the zero ring and does not require multiplicative commutativity or imply a field. Require fully qualified tables. Give (zero,one,distinct?).
View answer and explanation
(0, 1, Yes)
Find both two-sided identities from their respective operations. Labels denote carrier elements, not their assumed arithmetic values. An exact finite or coordinate slice does not automatically establish a general infinite theorem. The exact result, retaining the requested order, is (0, 1, Yes).
Reviewed step-by-step derivation
Finite unital ring · qf-mod27-ring-identities · candidate
96. Keep Additive Zero and Multiplicative One Distinct
Use the same carrier Fin(6)={0,…,5} for complete addition=[[0, 1, 2, 3, 4, 5], [1, 5, 4, 2, 0, 3], [2, 4, 5, 1, 3, 0], [3, 2, 1, 0, 5, 4], [4, 0, 3, 5, 2, 1], [5, 3, 0, 4, 1, 2]] and multiplication=[[0, 0, 0, 0, 0, 0], [0, 1, 2, 3, 4, 5], [0, 2, 2, 0, 5, 5], [0, 3, 0, 3, 3, 0], [0, 4, 5, 3, 1, 2], [0, 5, 5, 0, 2, 2]] tables in label order. Check the additive abelian-group axioms, multiplicative associativity, both distributive laws and a two-sided multiplicative identity distinct from the additive identity. The two identity labels need not be 0 and 1. This protocol excludes the zero ring and does not require multiplicative commutativity or imply a field. Require fully qualified tables. Give (zero,one,distinct?).
View answer and explanation
(0, 1, Yes)
Find both two-sided identities from their respective operations. Labels denote carrier elements, not their assumed arithmetic values. An exact finite or coordinate slice does not automatically establish a general infinite theorem. The exact result, retaining the requested order, is (0, 1, Yes).
Reviewed step-by-step derivation
Finite unital ring · qf-mod27-ring-identities · candidate
97. Keep Additive Zero and Multiplicative One Distinct
Use the same carrier Fin(6)={0,…,5} for complete addition=[[0, 1, 2, 3, 4, 5], [1, 5, 0, 4, 2, 3], [2, 0, 4, 5, 3, 1], [3, 4, 5, 0, 1, 2], [4, 2, 3, 1, 5, 0], [5, 3, 1, 2, 0, 4]] and multiplication=[[0, 0, 0, 0, 0, 0], [0, 1, 2, 3, 4, 5], [0, 2, 1, 3, 5, 4], [0, 3, 3, 3, 0, 0], [0, 4, 5, 0, 4, 5], [0, 5, 4, 0, 5, 4]] tables in label order. Check the additive abelian-group axioms, multiplicative associativity, both distributive laws and a two-sided multiplicative identity distinct from the additive identity. The two identity labels need not be 0 and 1. This protocol excludes the zero ring and does not require multiplicative commutativity or imply a field. Require fully qualified tables. Give (zero,one,distinct?).
View answer and explanation
(0, 1, Yes)
Find both two-sided identities from their respective operations. Labels denote carrier elements, not their assumed arithmetic values. An exact finite or coordinate slice does not automatically establish a general infinite theorem. The exact result, retaining the requested order, is (0, 1, Yes).
Reviewed step-by-step derivation
Finite unital ring · qf-mod27-ring-identities · candidate
98. Keep Additive Zero and Multiplicative One Distinct
Use the same carrier Fin(6)={0,…,5} for complete addition=[[0, 1, 2, 3, 4, 5], [1, 5, 3, 0, 2, 4], [2, 3, 5, 4, 1, 0], [3, 0, 4, 2, 5, 1], [4, 2, 1, 5, 0, 3], [5, 4, 0, 1, 3, 2]] and multiplication=[[0, 0, 0, 0, 0, 0], [0, 1, 2, 3, 4, 5], [0, 2, 2, 5, 0, 5], [0, 3, 5, 1, 4, 2], [0, 4, 0, 4, 4, 0], [0, 5, 5, 2, 0, 2]] tables in label order. Check the additive abelian-group axioms, multiplicative associativity, both distributive laws and a two-sided multiplicative identity distinct from the additive identity. The two identity labels need not be 0 and 1. This protocol excludes the zero ring and does not require multiplicative commutativity or imply a field. Require fully qualified tables. Give (zero,one,distinct?).
View answer and explanation
(0, 1, Yes)
Find both two-sided identities from their respective operations. Labels denote carrier elements, not their assumed arithmetic values. An exact finite or coordinate slice does not automatically establish a general infinite theorem. The exact result, retaining the requested order, is (0, 1, Yes).
Reviewed step-by-step derivation
Finite unital ring · qf-mod27-ring-identities · candidate
99. Keep Additive Zero and Multiplicative One Distinct
Use the same carrier Fin(6)={0,…,5} for complete addition=[[0, 1, 2, 3, 4, 5], [1, 5, 0, 2, 3, 4], [2, 0, 3, 4, 5, 1], [3, 2, 4, 5, 1, 0], [4, 3, 5, 1, 0, 2], [5, 4, 1, 0, 2, 3]] and multiplication=[[0, 0, 0, 0, 0, 0], [0, 1, 2, 3, 4, 5], [0, 2, 1, 5, 4, 3], [0, 3, 5, 3, 0, 5], [0, 4, 4, 0, 4, 0], [0, 5, 3, 5, 0, 3]] tables in label order. Check the additive abelian-group axioms, multiplicative associativity, both distributive laws and a two-sided multiplicative identity distinct from the additive identity. The two identity labels need not be 0 and 1. This protocol excludes the zero ring and does not require multiplicative commutativity or imply a field. Require fully qualified tables. Give (zero,one,distinct?).
View answer and explanation
(0, 1, Yes)
Find both two-sided identities from their respective operations. Labels denote carrier elements, not their assumed arithmetic values. An exact finite or coordinate slice does not automatically establish a general infinite theorem. The exact result, retaining the requested order, is (0, 1, Yes).
Reviewed step-by-step derivation
Finite unital ring · qf-mod27-ring-identities · candidate
100. Keep Additive Zero and Multiplicative One Distinct
Use the same carrier Fin(6)={0,…,5} for complete addition=[[0, 1, 2, 3, 4, 5], [1, 4, 3, 5, 0, 2], [2, 3, 1, 4, 5, 0], [3, 5, 4, 0, 2, 1], [4, 0, 5, 2, 1, 3], [5, 2, 0, 1, 3, 4]] and multiplication=[[0, 0, 0, 0, 0, 0], [0, 4, 1, 0, 1, 4], [0, 1, 2, 3, 4, 5], [0, 0, 3, 3, 0, 3], [0, 1, 4, 0, 4, 1], [0, 4, 5, 3, 1, 2]] tables in label order. Check the additive abelian-group axioms, multiplicative associativity, both distributive laws and a two-sided multiplicative identity distinct from the additive identity. The two identity labels need not be 0 and 1. This protocol excludes the zero ring and does not require multiplicative commutativity or imply a field. Require fully qualified tables. Give (zero,one,distinct?).
View answer and explanation
(0, 2, Yes)
Find both two-sided identities from their respective operations. Labels denote carrier elements, not their assumed arithmetic values. An exact finite or coordinate slice does not automatically establish a general infinite theorem. The exact result, retaining the requested order, is (0, 2, Yes).
Reviewed step-by-step derivation
Finite unital ring · qf-mod27-ring-identities · candidate
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