Complete Commutative-Ring Qualification, Product Swaps, Powers and Binomial Squares
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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-20261028-r01-finite-commutative-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. Multiplicative Commutativity Is Additional to Complete Ring Qualification
Use the same qualified complete finite unital ring on Fin(8)={0,…,7}, with addition=[[0, 1, 2, 3, 4, 5, 6, 7], [1, 0, 3, 2, 5, 4, 7, 6], [2, 3, 0, 1, 6, 7, 4, 5], [3, 2, 1, 0, 7, 6, 5, 4], [4, 5, 6, 7, 0, 1, 2, 3], [5, 4, 7, 6, 1, 0, 3, 2], [6, 7, 4, 5, 2, 3, 0, 1], [7, 6, 5, 4, 3, 2, 1, 0]] and multiplication=[[0, 0, 0, 0, 0, 0, 0, 0], [0, 1, 0, 1, 0, 1, 0, 1], [0, 2, 0, 2, 0, 2, 0, 2], [0, 3, 0, 3, 0, 3, 0, 3], [0, 0, 2, 2, 4, 4, 6, 6], [0, 1, 2, 3, 4, 5, 6, 7], [0, 2, 2, 0, 4, 6, 6, 4], [0, 3, 2, 1, 4, 7, 6, 5]] in label order. All unital-ring axioms, including distinct two-sided identities, must already hold. The identity labels need not be 0 and 1. Commutativity is a separate multiplication obligation; even a commutative ring may have nonzero zero divisors and need not be a field. Give (n,zero,one,all [a,b,ab,ba] failures with a<b in label order,commutative?,classification). Classification is commutative_ring or unital_ring_not_commutative.
View answer and explanation
(8, 0, 5, [[1, 2, 0, 2], [1, 3, 1, 3], [1, 6, 0, 2], [1, 7, 1, 3], [2, 3, 2, 0], [2, 4, 0, 2], [2, 6, 0, 2], [3, 4, 0, 2], [3, 7, 3, 1], [4, 6, 6, 4], [4, 7, 6, 4], [6, 7, 4, 6]], No, unital_ring_not_commutative)
Compare both multiplication orders for every distinct unordered pair. Off-diagonal failures disprove commutativity; the full ring qualification is a prior premise rather than something repaired by this audit. An exact finite or named-formula slice does not automatically establish an arbitrary infinite theorem. The exact result, retaining the requested order, is (8, 0, 5, [[1, 2, 0, 2], [1, 3, 1, 3], [1, 6, 0, 2], [1, 7, 1, 3], [2, 3, 2, 0], [2, 4, 0, 2], [2, 6, 0, 2], [3, 4, 0, 2], [3, 7, 3, 1], [4, 6, 6, 4], [4, 7, 6, 4], [6, 7, 4, 6]], No, unital_ring_not_commutative).
Reviewed step-by-step derivation
Finite commutative ring · qf-mod28-commring-qualification · candidate
2. Multiplicative Commutativity Is Additional to Complete Ring Qualification
Use the same qualified complete finite unital ring on Fin(8)={0,…,7}, with addition=[[0, 1, 2, 3, 4, 5, 6, 7], [1, 2, 3, 4, 5, 7, 0, 6], [2, 3, 4, 5, 7, 6, 1, 0], [3, 4, 5, 7, 6, 0, 2, 1], [4, 5, 7, 6, 0, 1, 3, 2], [5, 7, 6, 0, 1, 2, 4, 3], [6, 0, 1, 2, 3, 4, 7, 5], [7, 6, 0, 1, 2, 3, 5, 4]] and multiplication=[[0, 0, 0, 0, 0, 0, 0, 0], [0, 1, 2, 3, 4, 5, 6, 7], [0, 2, 4, 7, 0, 2, 7, 4], [0, 3, 7, 1, 4, 6, 5, 2], [0, 4, 0, 4, 0, 4, 4, 0], [0, 5, 2, 6, 4, 1, 3, 7], [0, 6, 7, 5, 4, 3, 1, 2], [0, 7, 4, 2, 0, 7, 2, 4]] in label order. All unital-ring axioms, including distinct two-sided identities, must already hold. The identity labels need not be 0 and 1. Commutativity is a separate multiplication obligation; even a commutative ring may have nonzero zero divisors and need not be a field. Give (n,zero,one,all [a,b,ab,ba] failures with a<b in label order,commutative?,classification). Classification is commutative_ring or unital_ring_not_commutative.
View answer and explanation
(8, 0, 1, [], Yes, commutative_ring)
Compare both multiplication orders for every distinct unordered pair. Off-diagonal failures disprove commutativity; the full ring qualification is a prior premise rather than something repaired by this audit. An exact finite or named-formula slice does not automatically establish an arbitrary infinite theorem. The exact result, retaining the requested order, is (8, 0, 1, [], Yes, commutative_ring).
Reviewed step-by-step derivation
Finite commutative ring · qf-mod28-commring-qualification · candidate
3. Multiplicative Commutativity Is Additional to Complete Ring Qualification
Use the same qualified complete finite unital ring on Fin(8)={0,…,7}, with addition=[[0, 1, 2, 3, 4, 5, 6, 7], [1, 2, 3, 4, 6, 7, 5, 0], [2, 3, 4, 6, 5, 0, 7, 1], [3, 4, 6, 5, 7, 1, 0, 2], [4, 6, 5, 7, 0, 2, 1, 3], [5, 7, 0, 1, 2, 4, 3, 6], [6, 5, 7, 0, 1, 3, 2, 4], [7, 0, 1, 2, 3, 6, 4, 5]] and multiplication=[[0, 0, 0, 0, 0, 0, 0, 0], [0, 1, 2, 3, 4, 5, 6, 7], [0, 2, 4, 5, 0, 4, 2, 5], [0, 3, 5, 1, 4, 2, 7, 6], [0, 4, 0, 4, 0, 0, 4, 4], [0, 5, 4, 2, 0, 4, 5, 2], [0, 6, 2, 7, 4, 5, 1, 3], [0, 7, 5, 6, 4, 2, 3, 1]] in label order. All unital-ring axioms, including distinct two-sided identities, must already hold. The identity labels need not be 0 and 1. Commutativity is a separate multiplication obligation; even a commutative ring may have nonzero zero divisors and need not be a field. Give (n,zero,one,all [a,b,ab,ba] failures with a<b in label order,commutative?,classification). Classification is commutative_ring or unital_ring_not_commutative.
View answer and explanation
(8, 0, 1, [], Yes, commutative_ring)
Compare both multiplication orders for every distinct unordered pair. Off-diagonal failures disprove commutativity; the full ring qualification is a prior premise rather than something repaired by this audit. An exact finite or named-formula slice does not automatically establish an arbitrary infinite theorem. The exact result, retaining the requested order, is (8, 0, 1, [], Yes, commutative_ring).
Reviewed step-by-step derivation
Finite commutative ring · qf-mod28-commring-qualification · candidate
4. Multiplicative Commutativity Is Additional to Complete Ring Qualification
Use the same qualified complete finite unital ring on Fin(8)={0,…,7}, with addition=[[0, 1, 2, 3, 4, 5, 6, 7], [1, 2, 3, 4, 6, 0, 7, 5], [2, 3, 4, 6, 7, 1, 5, 0], [3, 4, 6, 7, 5, 2, 0, 1], [4, 6, 7, 5, 0, 3, 1, 2], [5, 0, 1, 2, 3, 7, 4, 6], [6, 7, 5, 0, 1, 4, 2, 3], [7, 5, 0, 1, 2, 6, 3, 4]] and multiplication=[[0, 0, 0, 0, 0, 0, 0, 0], [0, 1, 2, 3, 4, 5, 6, 7], [0, 2, 4, 7, 0, 7, 2, 4], [0, 3, 7, 1, 4, 6, 5, 2], [0, 4, 0, 4, 0, 4, 4, 0], [0, 5, 7, 6, 4, 1, 3, 2], [0, 6, 2, 5, 4, 3, 1, 7], [0, 7, 4, 2, 0, 2, 7, 4]] in label order. All unital-ring axioms, including distinct two-sided identities, must already hold. The identity labels need not be 0 and 1. Commutativity is a separate multiplication obligation; even a commutative ring may have nonzero zero divisors and need not be a field. Give (n,zero,one,all [a,b,ab,ba] failures with a<b in label order,commutative?,classification). Classification is commutative_ring or unital_ring_not_commutative.
View answer and explanation
(8, 0, 1, [], Yes, commutative_ring)
Compare both multiplication orders for every distinct unordered pair. Off-diagonal failures disprove commutativity; the full ring qualification is a prior premise rather than something repaired by this audit. An exact finite or named-formula slice does not automatically establish an arbitrary infinite theorem. The exact result, retaining the requested order, is (8, 0, 1, [], Yes, commutative_ring).
Reviewed step-by-step derivation
Finite commutative ring · qf-mod28-commring-qualification · candidate
5. Multiplicative Commutativity Is Additional to Complete Ring Qualification
Use the same qualified complete finite unital ring on Fin(8)={0,…,7}, with addition=[[0, 1, 2, 3, 4, 5, 6, 7], [1, 2, 3, 4, 7, 6, 0, 5], [2, 3, 4, 7, 5, 0, 1, 6], [3, 4, 7, 5, 6, 1, 2, 0], [4, 7, 5, 6, 0, 2, 3, 1], [5, 6, 0, 1, 2, 4, 7, 3], [6, 0, 1, 2, 3, 7, 5, 4], [7, 5, 6, 0, 1, 3, 4, 2]] and multiplication=[[0, 0, 0, 0, 0, 0, 0, 0], [0, 1, 2, 3, 4, 5, 6, 7], [0, 2, 4, 5, 0, 4, 5, 2], [0, 3, 5, 1, 4, 2, 7, 6], [0, 4, 0, 4, 0, 0, 4, 4], [0, 5, 4, 2, 0, 4, 2, 5], [0, 6, 5, 7, 4, 2, 1, 3], [0, 7, 2, 6, 4, 5, 3, 1]] in label order. All unital-ring axioms, including distinct two-sided identities, must already hold. The identity labels need not be 0 and 1. Commutativity is a separate multiplication obligation; even a commutative ring may have nonzero zero divisors and need not be a field. Give (n,zero,one,all [a,b,ab,ba] failures with a<b in label order,commutative?,classification). Classification is commutative_ring or unital_ring_not_commutative.
View answer and explanation
(8, 0, 1, [], Yes, commutative_ring)
Compare both multiplication orders for every distinct unordered pair. Off-diagonal failures disprove commutativity; the full ring qualification is a prior premise rather than something repaired by this audit. An exact finite or named-formula slice does not automatically establish an arbitrary infinite theorem. The exact result, retaining the requested order, is (8, 0, 1, [], Yes, commutative_ring).
Reviewed step-by-step derivation
Finite commutative ring · qf-mod28-commring-qualification · candidate
6. Multiplicative Commutativity Is Additional to Complete Ring Qualification
Use the same qualified complete finite unital ring on Fin(8)={0,…,7}, with addition=[[0, 1, 2, 3, 4, 5, 6, 7], [1, 0, 3, 2, 7, 6, 5, 4], [2, 3, 0, 1, 6, 7, 4, 5], [3, 2, 1, 0, 5, 4, 7, 6], [4, 7, 6, 5, 0, 3, 2, 1], [5, 6, 7, 4, 3, 0, 1, 2], [6, 5, 4, 7, 2, 1, 0, 3], [7, 4, 5, 6, 1, 2, 3, 0]] and multiplication=[[0, 0, 0, 0, 0, 0, 0, 0], [0, 1, 0, 1, 0, 1, 0, 1], [0, 2, 0, 2, 0, 2, 0, 2], [0, 3, 0, 3, 0, 3, 0, 3], [0, 0, 2, 2, 4, 6, 6, 4], [0, 3, 2, 1, 4, 7, 6, 5], [0, 2, 2, 0, 4, 4, 6, 6], [0, 1, 2, 3, 4, 5, 6, 7]] in label order. All unital-ring axioms, including distinct two-sided identities, must already hold. The identity labels need not be 0 and 1. Commutativity is a separate multiplication obligation; even a commutative ring may have nonzero zero divisors and need not be a field. Give (n,zero,one,all [a,b,ab,ba] failures with a<b in label order,commutative?,classification). Classification is commutative_ring or unital_ring_not_commutative.
View answer and explanation
(8, 0, 7, [[1, 2, 0, 2], [1, 3, 1, 3], [1, 5, 1, 3], [1, 6, 0, 2], [2, 3, 2, 0], [2, 4, 0, 2], [2, 6, 0, 2], [3, 4, 0, 2], [3, 5, 3, 1], [4, 5, 6, 4], [4, 6, 6, 4], [5, 6, 6, 4]], No, unital_ring_not_commutative)
Compare both multiplication orders for every distinct unordered pair. Off-diagonal failures disprove commutativity; the full ring qualification is a prior premise rather than something repaired by this audit. An exact finite or named-formula slice does not automatically establish an arbitrary infinite theorem. The exact result, retaining the requested order, is (8, 0, 7, [[1, 2, 0, 2], [1, 3, 1, 3], [1, 5, 1, 3], [1, 6, 0, 2], [2, 3, 2, 0], [2, 4, 0, 2], [2, 6, 0, 2], [3, 4, 0, 2], [3, 5, 3, 1], [4, 5, 6, 4], [4, 6, 6, 4], [5, 6, 6, 4]], No, unital_ring_not_commutative).
Reviewed step-by-step derivation
Finite commutative ring · qf-mod28-commring-qualification · candidate
7. Multiplicative Commutativity Is Additional to Complete Ring Qualification
Use the same qualified complete finite unital ring on Fin(8)={0,…,7}, with addition=[[0, 1, 2, 3, 4, 5, 6, 7], [1, 2, 3, 5, 6, 4, 7, 0], [2, 3, 5, 4, 7, 6, 0, 1], [3, 5, 4, 6, 0, 7, 1, 2], [4, 6, 7, 0, 2, 1, 3, 5], [5, 4, 6, 7, 1, 0, 2, 3], [6, 7, 0, 1, 3, 2, 5, 4], [7, 0, 1, 2, 5, 3, 4, 6]] and multiplication=[[0, 0, 0, 0, 0, 0, 0, 0], [0, 1, 2, 3, 4, 5, 6, 7], [0, 2, 5, 6, 2, 0, 5, 6], [0, 3, 6, 1, 7, 5, 2, 4], [0, 4, 2, 7, 1, 5, 6, 3], [0, 5, 0, 5, 5, 0, 0, 5], [0, 6, 5, 2, 6, 0, 5, 2], [0, 7, 6, 4, 3, 5, 2, 1]] in label order. All unital-ring axioms, including distinct two-sided identities, must already hold. The identity labels need not be 0 and 1. Commutativity is a separate multiplication obligation; even a commutative ring may have nonzero zero divisors and need not be a field. Give (n,zero,one,all [a,b,ab,ba] failures with a<b in label order,commutative?,classification). Classification is commutative_ring or unital_ring_not_commutative.
View answer and explanation
(8, 0, 1, [], Yes, commutative_ring)
Compare both multiplication orders for every distinct unordered pair. Off-diagonal failures disprove commutativity; the full ring qualification is a prior premise rather than something repaired by this audit. An exact finite or named-formula slice does not automatically establish an arbitrary infinite theorem. The exact result, retaining the requested order, is (8, 0, 1, [], Yes, commutative_ring).
Reviewed step-by-step derivation
Finite commutative ring · qf-mod28-commring-qualification · candidate
8. Multiplicative Commutativity Is Additional to Complete Ring Qualification
Use the same qualified complete finite unital ring on Fin(8)={0,…,7}, with addition=[[0, 1, 2, 3, 4, 5, 6, 7], [1, 2, 3, 5, 7, 4, 0, 6], [2, 3, 5, 4, 6, 7, 1, 0], [3, 5, 4, 7, 0, 6, 2, 1], [4, 7, 6, 0, 2, 1, 5, 3], [5, 4, 7, 6, 1, 0, 3, 2], [6, 0, 1, 2, 5, 3, 7, 4], [7, 6, 0, 1, 3, 2, 4, 5]] and multiplication=[[0, 0, 0, 0, 0, 0, 0, 0], [0, 1, 2, 3, 4, 5, 6, 7], [0, 2, 5, 7, 2, 0, 7, 5], [0, 3, 7, 1, 6, 5, 4, 2], [0, 4, 2, 6, 1, 5, 3, 7], [0, 5, 0, 5, 5, 0, 5, 0], [0, 6, 7, 4, 3, 5, 1, 2], [0, 7, 5, 2, 7, 0, 2, 5]] in label order. All unital-ring axioms, including distinct two-sided identities, must already hold. The identity labels need not be 0 and 1. Commutativity is a separate multiplication obligation; even a commutative ring may have nonzero zero divisors and need not be a field. Give (n,zero,one,all [a,b,ab,ba] failures with a<b in label order,commutative?,classification). Classification is commutative_ring or unital_ring_not_commutative.
View answer and explanation
(8, 0, 1, [], Yes, commutative_ring)
Compare both multiplication orders for every distinct unordered pair. Off-diagonal failures disprove commutativity; the full ring qualification is a prior premise rather than something repaired by this audit. An exact finite or named-formula slice does not automatically establish an arbitrary infinite theorem. The exact result, retaining the requested order, is (8, 0, 1, [], Yes, commutative_ring).
Reviewed step-by-step derivation
Finite commutative ring · qf-mod28-commring-qualification · candidate
9. Multiplicative Commutativity Is Additional to Complete Ring Qualification
Use the same qualified complete finite unital ring on Fin(8)={0,…,7}, with addition=[[0, 1, 2, 3, 4, 5, 6, 7], [1, 2, 3, 5, 7, 6, 4, 0], [2, 3, 5, 6, 0, 4, 7, 1], [3, 5, 6, 4, 1, 7, 0, 2], [4, 7, 0, 1, 5, 2, 3, 6], [5, 6, 4, 7, 2, 0, 1, 3], [6, 4, 7, 0, 3, 1, 2, 5], [7, 0, 1, 2, 6, 3, 5, 4]] and multiplication=[[0, 0, 0, 0, 0, 0, 0, 0], [0, 1, 2, 3, 4, 5, 6, 7], [0, 2, 5, 4, 5, 0, 2, 4], [0, 3, 4, 1, 2, 5, 7, 6], [0, 4, 5, 2, 5, 0, 4, 2], [0, 5, 0, 5, 0, 0, 5, 5], [0, 6, 2, 7, 4, 5, 1, 3], [0, 7, 4, 6, 2, 5, 3, 1]] in label order. All unital-ring axioms, including distinct two-sided identities, must already hold. The identity labels need not be 0 and 1. Commutativity is a separate multiplication obligation; even a commutative ring may have nonzero zero divisors and need not be a field. Give (n,zero,one,all [a,b,ab,ba] failures with a<b in label order,commutative?,classification). Classification is commutative_ring or unital_ring_not_commutative.
View answer and explanation
(8, 0, 1, [], Yes, commutative_ring)
Compare both multiplication orders for every distinct unordered pair. Off-diagonal failures disprove commutativity; the full ring qualification is a prior premise rather than something repaired by this audit. An exact finite or named-formula slice does not automatically establish an arbitrary infinite theorem. The exact result, retaining the requested order, is (8, 0, 1, [], Yes, commutative_ring).
Reviewed step-by-step derivation
Finite commutative ring · qf-mod28-commring-qualification · candidate
10. Multiplicative Commutativity Is Additional to Complete Ring Qualification
Use the same qualified complete finite unital ring on Fin(8)={0,…,7}, with addition=[[0, 1, 2, 3, 4, 5, 6, 7], [1, 2, 3, 5, 0, 6, 7, 4], [2, 3, 5, 6, 1, 7, 4, 0], [3, 5, 6, 7, 2, 4, 0, 1], [4, 0, 1, 2, 7, 3, 5, 6], [5, 6, 7, 4, 3, 0, 1, 2], [6, 7, 4, 0, 5, 1, 2, 3], [7, 4, 0, 1, 6, 2, 3, 5]] and multiplication=[[0, 0, 0, 0, 0, 0, 0, 0], [0, 1, 2, 3, 4, 5, 6, 7], [0, 2, 5, 7, 7, 0, 2, 5], [0, 3, 7, 1, 6, 5, 4, 2], [0, 4, 7, 6, 1, 5, 3, 2], [0, 5, 0, 5, 5, 0, 5, 0], [0, 6, 2, 4, 3, 5, 1, 7], [0, 7, 5, 2, 2, 0, 7, 5]] in label order. All unital-ring axioms, including distinct two-sided identities, must already hold. The identity labels need not be 0 and 1. Commutativity is a separate multiplication obligation; even a commutative ring may have nonzero zero divisors and need not be a field. Give (n,zero,one,all [a,b,ab,ba] failures with a<b in label order,commutative?,classification). Classification is commutative_ring or unital_ring_not_commutative.
View answer and explanation
(8, 0, 1, [], Yes, commutative_ring)
Compare both multiplication orders for every distinct unordered pair. Off-diagonal failures disprove commutativity; the full ring qualification is a prior premise rather than something repaired by this audit. An exact finite or named-formula slice does not automatically establish an arbitrary infinite theorem. The exact result, retaining the requested order, is (8, 0, 1, [], Yes, commutative_ring).
Reviewed step-by-step derivation
Finite commutative ring · qf-mod28-commring-qualification · candidate
11. Multiplicative Commutativity Is Additional to Complete Ring Qualification
Use the same qualified complete finite unital ring on Fin(8)={0,…,7}, with addition=[[0, 1, 2, 3, 4, 5, 6, 7], [1, 0, 3, 2, 6, 7, 4, 5], [2, 3, 0, 1, 5, 4, 7, 6], [3, 2, 1, 0, 7, 6, 5, 4], [4, 6, 5, 7, 0, 2, 1, 3], [5, 7, 4, 6, 2, 0, 3, 1], [6, 4, 7, 5, 1, 3, 0, 2], [7, 5, 6, 4, 3, 1, 2, 0]] and multiplication=[[0, 0, 0, 0, 0, 0, 0, 0], [0, 1, 0, 1, 0, 0, 1, 1], [0, 2, 0, 2, 0, 0, 2, 2], [0, 3, 0, 3, 0, 0, 3, 3], [0, 2, 2, 0, 4, 5, 5, 4], [0, 0, 2, 2, 4, 5, 4, 5], [0, 3, 2, 1, 4, 5, 7, 6], [0, 1, 2, 3, 4, 5, 6, 7]] in label order. All unital-ring axioms, including distinct two-sided identities, must already hold. The identity labels need not be 0 and 1. Commutativity is a separate multiplication obligation; even a commutative ring may have nonzero zero divisors and need not be a field. Give (n,zero,one,all [a,b,ab,ba] failures with a<b in label order,commutative?,classification). Classification is commutative_ring or unital_ring_not_commutative.
View answer and explanation
(8, 0, 7, [[1, 2, 0, 2], [1, 3, 1, 3], [1, 4, 0, 2], [1, 6, 1, 3], [2, 3, 2, 0], [2, 4, 0, 2], [2, 5, 0, 2], [3, 5, 0, 2], [3, 6, 3, 1], [4, 5, 5, 4], [4, 6, 5, 4], [5, 6, 4, 5]], No, unital_ring_not_commutative)
Compare both multiplication orders for every distinct unordered pair. Off-diagonal failures disprove commutativity; the full ring qualification is a prior premise rather than something repaired by this audit. An exact finite or named-formula slice does not automatically establish an arbitrary infinite theorem. The exact result, retaining the requested order, is (8, 0, 7, [[1, 2, 0, 2], [1, 3, 1, 3], [1, 4, 0, 2], [1, 6, 1, 3], [2, 3, 2, 0], [2, 4, 0, 2], [2, 5, 0, 2], [3, 5, 0, 2], [3, 6, 3, 1], [4, 5, 5, 4], [4, 6, 5, 4], [5, 6, 4, 5]], No, unital_ring_not_commutative).
Reviewed step-by-step derivation
Finite commutative ring · qf-mod28-commring-qualification · candidate
12. Multiplicative Commutativity Is Additional to Complete Ring Qualification
Use the same qualified complete finite unital ring on Fin(8)={0,…,7}, with addition=[[0, 1, 2, 3, 4, 5, 6, 7], [1, 2, 3, 5, 0, 7, 4, 6], [2, 3, 5, 7, 1, 6, 0, 4], [3, 5, 7, 6, 2, 4, 1, 0], [4, 0, 1, 2, 6, 3, 7, 5], [5, 7, 6, 4, 3, 0, 2, 1], [6, 4, 0, 1, 7, 2, 5, 3], [7, 6, 4, 0, 5, 1, 3, 2]] and multiplication=[[0, 0, 0, 0, 0, 0, 0, 0], [0, 1, 2, 3, 4, 5, 6, 7], [0, 2, 5, 6, 6, 0, 5, 2], [0, 3, 6, 1, 7, 5, 2, 4], [0, 4, 6, 7, 1, 5, 2, 3], [0, 5, 0, 5, 5, 0, 0, 5], [0, 6, 5, 2, 2, 0, 5, 6], [0, 7, 2, 4, 3, 5, 6, 1]] in label order. All unital-ring axioms, including distinct two-sided identities, must already hold. The identity labels need not be 0 and 1. Commutativity is a separate multiplication obligation; even a commutative ring may have nonzero zero divisors and need not be a field. Give (n,zero,one,all [a,b,ab,ba] failures with a<b in label order,commutative?,classification). Classification is commutative_ring or unital_ring_not_commutative.
View answer and explanation
(8, 0, 1, [], Yes, commutative_ring)
Compare both multiplication orders for every distinct unordered pair. Off-diagonal failures disprove commutativity; the full ring qualification is a prior premise rather than something repaired by this audit. An exact finite or named-formula slice does not automatically establish an arbitrary infinite theorem. The exact result, retaining the requested order, is (8, 0, 1, [], Yes, commutative_ring).
Reviewed step-by-step derivation
Finite commutative ring · qf-mod28-commring-qualification · candidate
13. Multiplicative Commutativity Is Additional to Complete Ring Qualification
Use the same qualified complete finite unital ring on Fin(8)={0,…,7}, with addition=[[0, 1, 2, 3, 4, 5, 6, 7], [1, 2, 3, 6, 5, 7, 4, 0], [2, 3, 6, 4, 7, 0, 5, 1], [3, 6, 4, 5, 0, 1, 7, 2], [4, 5, 7, 0, 2, 3, 1, 6], [5, 7, 0, 1, 3, 6, 2, 4], [6, 4, 5, 7, 1, 2, 0, 3], [7, 0, 1, 2, 6, 4, 3, 5]] and multiplication=[[0, 0, 0, 0, 0, 0, 0, 0], [0, 1, 2, 3, 4, 5, 6, 7], [0, 2, 6, 5, 2, 6, 0, 5], [0, 3, 5, 1, 7, 2, 6, 4], [0, 4, 2, 7, 1, 5, 6, 3], [0, 5, 6, 2, 5, 6, 0, 2], [0, 6, 0, 6, 6, 0, 0, 6], [0, 7, 5, 4, 3, 2, 6, 1]] in label order. All unital-ring axioms, including distinct two-sided identities, must already hold. The identity labels need not be 0 and 1. Commutativity is a separate multiplication obligation; even a commutative ring may have nonzero zero divisors and need not be a field. Give (n,zero,one,all [a,b,ab,ba] failures with a<b in label order,commutative?,classification). Classification is commutative_ring or unital_ring_not_commutative.
View answer and explanation
(8, 0, 1, [], Yes, commutative_ring)
Compare both multiplication orders for every distinct unordered pair. Off-diagonal failures disprove commutativity; the full ring qualification is a prior premise rather than something repaired by this audit. An exact finite or named-formula slice does not automatically establish an arbitrary infinite theorem. The exact result, retaining the requested order, is (8, 0, 1, [], Yes, commutative_ring).
Reviewed step-by-step derivation
Finite commutative ring · qf-mod28-commring-qualification · candidate
14. Multiplicative Commutativity Is Additional to Complete Ring Qualification
Use the same qualified complete finite unital ring on Fin(8)={0,…,7}, with addition=[[0, 1, 2, 3, 4, 5, 6, 7], [1, 2, 3, 6, 7, 0, 4, 5], [2, 3, 6, 4, 5, 1, 7, 0], [3, 6, 4, 7, 0, 2, 5, 1], [4, 7, 5, 0, 2, 6, 1, 3], [5, 0, 1, 2, 6, 7, 3, 4], [6, 4, 7, 5, 1, 3, 0, 2], [7, 5, 0, 1, 3, 4, 2, 6]] and multiplication=[[0, 0, 0, 0, 0, 0, 0, 0], [0, 1, 2, 3, 4, 5, 6, 7], [0, 2, 6, 7, 2, 7, 0, 6], [0, 3, 7, 1, 5, 4, 6, 2], [0, 4, 2, 5, 1, 3, 6, 7], [0, 5, 7, 4, 3, 1, 6, 2], [0, 6, 0, 6, 6, 6, 0, 0], [0, 7, 6, 2, 7, 2, 0, 6]] in label order. All unital-ring axioms, including distinct two-sided identities, must already hold. The identity labels need not be 0 and 1. Commutativity is a separate multiplication obligation; even a commutative ring may have nonzero zero divisors and need not be a field. Give (n,zero,one,all [a,b,ab,ba] failures with a<b in label order,commutative?,classification). Classification is commutative_ring or unital_ring_not_commutative.
View answer and explanation
(8, 0, 1, [], Yes, commutative_ring)
Compare both multiplication orders for every distinct unordered pair. Off-diagonal failures disprove commutativity; the full ring qualification is a prior premise rather than something repaired by this audit. An exact finite or named-formula slice does not automatically establish an arbitrary infinite theorem. The exact result, retaining the requested order, is (8, 0, 1, [], Yes, commutative_ring).
Reviewed step-by-step derivation
Finite commutative ring · qf-mod28-commring-qualification · candidate
15. Multiplicative Commutativity Is Additional to Complete Ring Qualification
Use the same qualified complete finite unital ring on Fin(8)={0,…,7}, with addition=[[0, 1, 2, 3, 4, 5, 6, 7], [1, 2, 3, 6, 7, 4, 5, 0], [2, 3, 6, 5, 0, 7, 4, 1], [3, 6, 5, 4, 1, 0, 7, 2], [4, 7, 0, 1, 6, 3, 2, 5], [5, 4, 7, 0, 3, 2, 1, 6], [6, 5, 4, 7, 2, 1, 0, 3], [7, 0, 1, 2, 5, 6, 3, 4]] and multiplication=[[0, 0, 0, 0, 0, 0, 0, 0], [0, 1, 2, 3, 4, 5, 6, 7], [0, 2, 6, 4, 6, 2, 0, 4], [0, 3, 4, 1, 2, 7, 6, 5], [0, 4, 6, 2, 6, 4, 0, 2], [0, 5, 2, 7, 4, 1, 6, 3], [0, 6, 0, 6, 0, 6, 0, 6], [0, 7, 4, 5, 2, 3, 6, 1]] in label order. All unital-ring axioms, including distinct two-sided identities, must already hold. The identity labels need not be 0 and 1. Commutativity is a separate multiplication obligation; even a commutative ring may have nonzero zero divisors and need not be a field. Give (n,zero,one,all [a,b,ab,ba] failures with a<b in label order,commutative?,classification). Classification is commutative_ring or unital_ring_not_commutative.
View answer and explanation
(8, 0, 1, [], Yes, commutative_ring)
Compare both multiplication orders for every distinct unordered pair. Off-diagonal failures disprove commutativity; the full ring qualification is a prior premise rather than something repaired by this audit. An exact finite or named-formula slice does not automatically establish an arbitrary infinite theorem. The exact result, retaining the requested order, is (8, 0, 1, [], Yes, commutative_ring).
Reviewed step-by-step derivation
Finite commutative ring · qf-mod28-commring-qualification · candidate
16. Multiplicative Commutativity Is Additional to Complete Ring Qualification
Use the same qualified complete finite unital ring on Fin(8)={0,…,7}, with addition=[[0, 1, 2, 3, 4, 5, 6, 7], [1, 0, 3, 2, 7, 6, 5, 4], [2, 3, 0, 1, 5, 4, 7, 6], [3, 2, 1, 0, 6, 7, 4, 5], [4, 7, 5, 6, 0, 2, 3, 1], [5, 6, 4, 7, 2, 0, 1, 3], [6, 5, 7, 4, 3, 1, 0, 2], [7, 4, 6, 5, 1, 3, 2, 0]] and multiplication=[[0, 0, 0, 0, 0, 0, 0, 0], [0, 1, 0, 1, 1, 1, 0, 0], [0, 2, 0, 2, 2, 2, 0, 0], [0, 3, 0, 3, 3, 3, 0, 0], [0, 3, 2, 1, 5, 4, 6, 7], [0, 1, 2, 3, 4, 5, 6, 7], [0, 0, 2, 2, 7, 6, 6, 7], [0, 2, 2, 0, 6, 7, 6, 7]] in label order. All unital-ring axioms, including distinct two-sided identities, must already hold. The identity labels need not be 0 and 1. Commutativity is a separate multiplication obligation; even a commutative ring may have nonzero zero divisors and need not be a field. Give (n,zero,one,all [a,b,ab,ba] failures with a<b in label order,commutative?,classification). Classification is commutative_ring or unital_ring_not_commutative.
View answer and explanation
(8, 0, 5, [[1, 2, 0, 2], [1, 3, 1, 3], [1, 4, 1, 3], [1, 7, 0, 2], [2, 3, 2, 0], [2, 6, 0, 2], [2, 7, 0, 2], [3, 4, 3, 1], [3, 6, 0, 2], [4, 6, 6, 7], [4, 7, 7, 6], [6, 7, 7, 6]], No, unital_ring_not_commutative)
Compare both multiplication orders for every distinct unordered pair. Off-diagonal failures disprove commutativity; the full ring qualification is a prior premise rather than something repaired by this audit. An exact finite or named-formula slice does not automatically establish an arbitrary infinite theorem. The exact result, retaining the requested order, is (8, 0, 5, [[1, 2, 0, 2], [1, 3, 1, 3], [1, 4, 1, 3], [1, 7, 0, 2], [2, 3, 2, 0], [2, 6, 0, 2], [2, 7, 0, 2], [3, 4, 3, 1], [3, 6, 0, 2], [4, 6, 6, 7], [4, 7, 7, 6], [6, 7, 7, 6]], No, unital_ring_not_commutative).
Reviewed step-by-step derivation
Finite commutative ring · qf-mod28-commring-qualification · candidate
17. Multiplicative Commutativity Is Additional to Complete Ring Qualification
Use the same qualified complete finite unital ring on Fin(8)={0,…,7}, with addition=[[0, 1, 2, 3, 4, 5, 6, 7], [1, 2, 3, 6, 5, 0, 7, 4], [2, 3, 6, 7, 0, 1, 4, 5], [3, 6, 7, 4, 1, 2, 5, 0], [4, 5, 0, 1, 6, 7, 2, 3], [5, 0, 1, 2, 7, 4, 3, 6], [6, 7, 4, 5, 2, 3, 0, 1], [7, 4, 5, 0, 3, 6, 1, 2]] and multiplication=[[0, 0, 0, 0, 0, 0, 0, 0], [0, 1, 2, 3, 4, 5, 6, 7], [0, 2, 6, 4, 6, 4, 0, 2], [0, 3, 4, 1, 2, 7, 6, 5], [0, 4, 6, 2, 6, 2, 0, 4], [0, 5, 4, 7, 2, 1, 6, 3], [0, 6, 0, 6, 0, 6, 0, 6], [0, 7, 2, 5, 4, 3, 6, 1]] in label order. All unital-ring axioms, including distinct two-sided identities, must already hold. The identity labels need not be 0 and 1. Commutativity is a separate multiplication obligation; even a commutative ring may have nonzero zero divisors and need not be a field. Give (n,zero,one,all [a,b,ab,ba] failures with a<b in label order,commutative?,classification). Classification is commutative_ring or unital_ring_not_commutative.
View answer and explanation
(8, 0, 1, [], Yes, commutative_ring)
Compare both multiplication orders for every distinct unordered pair. Off-diagonal failures disprove commutativity; the full ring qualification is a prior premise rather than something repaired by this audit. An exact finite or named-formula slice does not automatically establish an arbitrary infinite theorem. The exact result, retaining the requested order, is (8, 0, 1, [], Yes, commutative_ring).
Reviewed step-by-step derivation
Finite commutative ring · qf-mod28-commring-qualification · candidate
18. Multiplicative Commutativity Is Additional to Complete Ring Qualification
Use the same qualified complete finite unital ring on Fin(8)={0,…,7}, with addition=[[0, 1, 2, 3, 4, 5, 6, 7], [1, 2, 3, 6, 0, 4, 7, 5], [2, 3, 6, 7, 1, 0, 5, 4], [3, 6, 7, 5, 2, 1, 4, 0], [4, 0, 1, 2, 5, 7, 3, 6], [5, 4, 0, 1, 7, 6, 2, 3], [6, 7, 5, 4, 3, 2, 0, 1], [7, 5, 4, 0, 6, 3, 1, 2]] and multiplication=[[0, 0, 0, 0, 0, 0, 0, 0], [0, 1, 2, 3, 4, 5, 6, 7], [0, 2, 6, 5, 5, 6, 0, 2], [0, 3, 5, 1, 7, 2, 6, 4], [0, 4, 5, 7, 1, 2, 6, 3], [0, 5, 6, 2, 2, 6, 0, 5], [0, 6, 0, 6, 6, 0, 0, 6], [0, 7, 2, 4, 3, 5, 6, 1]] in label order. All unital-ring axioms, including distinct two-sided identities, must already hold. The identity labels need not be 0 and 1. Commutativity is a separate multiplication obligation; even a commutative ring may have nonzero zero divisors and need not be a field. Give (n,zero,one,all [a,b,ab,ba] failures with a<b in label order,commutative?,classification). Classification is commutative_ring or unital_ring_not_commutative.
View answer and explanation
(8, 0, 1, [], Yes, commutative_ring)
Compare both multiplication orders for every distinct unordered pair. Off-diagonal failures disprove commutativity; the full ring qualification is a prior premise rather than something repaired by this audit. An exact finite or named-formula slice does not automatically establish an arbitrary infinite theorem. The exact result, retaining the requested order, is (8, 0, 1, [], Yes, commutative_ring).
Reviewed step-by-step derivation
Finite commutative ring · qf-mod28-commring-qualification · candidate
19. Multiplicative Commutativity Is Additional to Complete Ring Qualification
Use the same qualified complete finite unital ring on Fin(8)={0,…,7}, with addition=[[0, 1, 2, 3, 4, 5, 6, 7], [1, 2, 3, 7, 5, 6, 0, 4], [2, 3, 7, 4, 6, 0, 1, 5], [3, 7, 4, 5, 0, 1, 2, 6], [4, 5, 6, 0, 2, 3, 7, 1], [5, 6, 0, 1, 3, 7, 4, 2], [6, 0, 1, 2, 7, 4, 5, 3], [7, 4, 5, 6, 1, 2, 3, 0]] and multiplication=[[0, 0, 0, 0, 0, 0, 0, 0], [0, 1, 2, 3, 4, 5, 6, 7], [0, 2, 7, 5, 2, 7, 5, 0], [0, 3, 5, 1, 6, 2, 4, 7], [0, 4, 2, 6, 1, 5, 3, 7], [0, 5, 7, 2, 5, 7, 2, 0], [0, 6, 5, 4, 3, 2, 1, 7], [0, 7, 0, 7, 7, 0, 7, 0]] in label order. All unital-ring axioms, including distinct two-sided identities, must already hold. The identity labels need not be 0 and 1. Commutativity is a separate multiplication obligation; even a commutative ring may have nonzero zero divisors and need not be a field. Give (n,zero,one,all [a,b,ab,ba] failures with a<b in label order,commutative?,classification). Classification is commutative_ring or unital_ring_not_commutative.
View answer and explanation
(8, 0, 1, [], Yes, commutative_ring)
Compare both multiplication orders for every distinct unordered pair. Off-diagonal failures disprove commutativity; the full ring qualification is a prior premise rather than something repaired by this audit. An exact finite or named-formula slice does not automatically establish an arbitrary infinite theorem. The exact result, retaining the requested order, is (8, 0, 1, [], Yes, commutative_ring).
Reviewed step-by-step derivation
Finite commutative ring · qf-mod28-commring-qualification · candidate
20. Multiplicative Commutativity Is Additional to Complete Ring Qualification
Use the same qualified complete finite unital ring on Fin(8)={0,…,7}, with addition=[[0, 1, 2, 3, 4, 5, 6, 7], [1, 2, 3, 7, 6, 0, 5, 4], [2, 3, 7, 4, 5, 1, 0, 6], [3, 7, 4, 6, 0, 2, 1, 5], [4, 6, 5, 0, 2, 7, 3, 1], [5, 0, 1, 2, 7, 6, 4, 3], [6, 5, 0, 1, 3, 4, 7, 2], [7, 4, 6, 5, 1, 3, 2, 0]] and multiplication=[[0, 0, 0, 0, 0, 0, 0, 0], [0, 1, 2, 3, 4, 5, 6, 7], [0, 2, 7, 6, 2, 6, 7, 0], [0, 3, 6, 1, 5, 4, 2, 7], [0, 4, 2, 5, 1, 3, 6, 7], [0, 5, 6, 4, 3, 1, 2, 7], [0, 6, 7, 2, 6, 2, 7, 0], [0, 7, 0, 7, 7, 7, 0, 0]] in label order. All unital-ring axioms, including distinct two-sided identities, must already hold. The identity labels need not be 0 and 1. Commutativity is a separate multiplication obligation; even a commutative ring may have nonzero zero divisors and need not be a field. Give (n,zero,one,all [a,b,ab,ba] failures with a<b in label order,commutative?,classification). Classification is commutative_ring or unital_ring_not_commutative.
View answer and explanation
(8, 0, 1, [], Yes, commutative_ring)
Compare both multiplication orders for every distinct unordered pair. Off-diagonal failures disprove commutativity; the full ring qualification is a prior premise rather than something repaired by this audit. An exact finite or named-formula slice does not automatically establish an arbitrary infinite theorem. The exact result, retaining the requested order, is (8, 0, 1, [], Yes, commutative_ring).
Reviewed step-by-step derivation
Finite commutative ring · qf-mod28-commring-qualification · candidate
21. Multiplicative Commutativity Is Additional to Complete Ring Qualification
Use the same qualified complete finite unital ring on Fin(8)={0,…,7}, with addition=[[0, 1, 2, 3, 4, 5, 6, 7], [1, 0, 3, 2, 6, 7, 4, 5], [2, 3, 0, 1, 7, 6, 5, 4], [3, 2, 1, 0, 5, 4, 7, 6], [4, 6, 7, 5, 0, 3, 1, 2], [5, 7, 6, 4, 3, 0, 2, 1], [6, 4, 5, 7, 1, 2, 0, 3], [7, 5, 4, 6, 2, 1, 3, 0]] and multiplication=[[0, 0, 0, 0, 0, 0, 0, 0], [0, 1, 0, 1, 0, 1, 1, 0], [0, 2, 0, 2, 0, 2, 2, 0], [0, 3, 0, 3, 0, 3, 3, 0], [0, 2, 2, 0, 4, 4, 7, 7], [0, 1, 2, 3, 4, 5, 6, 7], [0, 3, 2, 1, 4, 6, 5, 7], [0, 0, 2, 2, 4, 7, 4, 7]] in label order. All unital-ring axioms, including distinct two-sided identities, must already hold. The identity labels need not be 0 and 1. Commutativity is a separate multiplication obligation; even a commutative ring may have nonzero zero divisors and need not be a field. Give (n,zero,one,all [a,b,ab,ba] failures with a<b in label order,commutative?,classification). Classification is commutative_ring or unital_ring_not_commutative.
View answer and explanation
(8, 0, 5, [[1, 2, 0, 2], [1, 3, 1, 3], [1, 4, 0, 2], [1, 6, 1, 3], [2, 3, 2, 0], [2, 4, 0, 2], [2, 7, 0, 2], [3, 6, 3, 1], [3, 7, 0, 2], [4, 6, 7, 4], [4, 7, 7, 4], [6, 7, 7, 4]], No, unital_ring_not_commutative)
Compare both multiplication orders for every distinct unordered pair. Off-diagonal failures disprove commutativity; the full ring qualification is a prior premise rather than something repaired by this audit. An exact finite or named-formula slice does not automatically establish an arbitrary infinite theorem. The exact result, retaining the requested order, is (8, 0, 5, [[1, 2, 0, 2], [1, 3, 1, 3], [1, 4, 0, 2], [1, 6, 1, 3], [2, 3, 2, 0], [2, 4, 0, 2], [2, 7, 0, 2], [3, 6, 3, 1], [3, 7, 0, 2], [4, 6, 7, 4], [4, 7, 7, 4], [6, 7, 7, 4]], No, unital_ring_not_commutative).
Reviewed step-by-step derivation
Finite commutative ring · qf-mod28-commring-qualification · candidate
22. Multiplicative Commutativity Is Additional to Complete Ring Qualification
Use the same qualified complete finite unital ring on Fin(8)={0,…,7}, with addition=[[0, 1, 2, 3, 4, 5, 6, 7], [1, 2, 3, 7, 0, 6, 4, 5], [2, 3, 7, 5, 1, 4, 0, 6], [3, 7, 5, 6, 2, 0, 1, 4], [4, 0, 1, 2, 6, 7, 5, 3], [5, 6, 4, 0, 7, 2, 3, 1], [6, 4, 0, 1, 5, 3, 7, 2], [7, 5, 6, 4, 3, 1, 2, 0]] and multiplication=[[0, 0, 0, 0, 0, 0, 0, 0], [0, 1, 2, 3, 4, 5, 6, 7], [0, 2, 7, 6, 6, 2, 7, 0], [0, 3, 6, 1, 5, 4, 2, 7], [0, 4, 6, 5, 1, 3, 2, 7], [0, 5, 2, 4, 3, 1, 6, 7], [0, 6, 7, 2, 2, 6, 7, 0], [0, 7, 0, 7, 7, 7, 0, 0]] in label order. All unital-ring axioms, including distinct two-sided identities, must already hold. The identity labels need not be 0 and 1. Commutativity is a separate multiplication obligation; even a commutative ring may have nonzero zero divisors and need not be a field. Give (n,zero,one,all [a,b,ab,ba] failures with a<b in label order,commutative?,classification). Classification is commutative_ring or unital_ring_not_commutative.
View answer and explanation
(8, 0, 1, [], Yes, commutative_ring)
Compare both multiplication orders for every distinct unordered pair. Off-diagonal failures disprove commutativity; the full ring qualification is a prior premise rather than something repaired by this audit. An exact finite or named-formula slice does not automatically establish an arbitrary infinite theorem. The exact result, retaining the requested order, is (8, 0, 1, [], Yes, commutative_ring).
Reviewed step-by-step derivation
Finite commutative ring · qf-mod28-commring-qualification · candidate
23. Multiplicative Commutativity Is Additional to Complete Ring Qualification
Use the same qualified complete finite unital ring on Fin(8)={0,…,7}, with addition=[[0, 1, 2, 3, 4, 5, 6, 7], [1, 2, 3, 7, 5, 0, 4, 6], [2, 3, 7, 6, 0, 1, 5, 4], [3, 7, 6, 4, 1, 2, 0, 5], [4, 5, 0, 1, 7, 6, 3, 2], [5, 0, 1, 2, 6, 4, 7, 3], [6, 4, 5, 0, 3, 7, 2, 1], [7, 6, 4, 5, 2, 3, 1, 0]] and multiplication=[[0, 0, 0, 0, 0, 0, 0, 0], [0, 1, 2, 3, 4, 5, 6, 7], [0, 2, 7, 4, 7, 4, 2, 0], [0, 3, 4, 1, 2, 6, 5, 7], [0, 4, 7, 2, 7, 2, 4, 0], [0, 5, 4, 6, 2, 1, 3, 7], [0, 6, 2, 5, 4, 3, 1, 7], [0, 7, 0, 7, 0, 7, 7, 0]] in label order. All unital-ring axioms, including distinct two-sided identities, must already hold. The identity labels need not be 0 and 1. Commutativity is a separate multiplication obligation; even a commutative ring may have nonzero zero divisors and need not be a field. Give (n,zero,one,all [a,b,ab,ba] failures with a<b in label order,commutative?,classification). Classification is commutative_ring or unital_ring_not_commutative.
View answer and explanation
(8, 0, 1, [], Yes, commutative_ring)
Compare both multiplication orders for every distinct unordered pair. Off-diagonal failures disprove commutativity; the full ring qualification is a prior premise rather than something repaired by this audit. An exact finite or named-formula slice does not automatically establish an arbitrary infinite theorem. The exact result, retaining the requested order, is (8, 0, 1, [], Yes, commutative_ring).
Reviewed step-by-step derivation
Finite commutative ring · qf-mod28-commring-qualification · candidate
24. Multiplicative Commutativity Is Additional to Complete Ring Qualification
Use the same qualified complete finite unital ring on Fin(8)={0,…,7}, with addition=[[0, 1, 2, 3, 4, 5, 6, 7], [1, 2, 3, 7, 0, 4, 5, 6], [2, 3, 7, 6, 1, 0, 4, 5], [3, 7, 6, 5, 2, 1, 0, 4], [4, 0, 1, 2, 5, 6, 7, 3], [5, 4, 0, 1, 6, 7, 3, 2], [6, 5, 4, 0, 7, 3, 2, 1], [7, 6, 5, 4, 3, 2, 1, 0]] and multiplication=[[0, 0, 0, 0, 0, 0, 0, 0], [0, 1, 2, 3, 4, 5, 6, 7], [0, 2, 7, 5, 5, 7, 2, 0], [0, 3, 5, 1, 6, 2, 4, 7], [0, 4, 5, 6, 1, 2, 3, 7], [0, 5, 7, 2, 2, 7, 5, 0], [0, 6, 2, 4, 3, 5, 1, 7], [0, 7, 0, 7, 7, 0, 7, 0]] in label order. All unital-ring axioms, including distinct two-sided identities, must already hold. The identity labels need not be 0 and 1. Commutativity is a separate multiplication obligation; even a commutative ring may have nonzero zero divisors and need not be a field. Give (n,zero,one,all [a,b,ab,ba] failures with a<b in label order,commutative?,classification). Classification is commutative_ring or unital_ring_not_commutative.
View answer and explanation
(8, 0, 1, [], Yes, commutative_ring)
Compare both multiplication orders for every distinct unordered pair. Off-diagonal failures disprove commutativity; the full ring qualification is a prior premise rather than something repaired by this audit. An exact finite or named-formula slice does not automatically establish an arbitrary infinite theorem. The exact result, retaining the requested order, is (8, 0, 1, [], Yes, commutative_ring).
Reviewed step-by-step derivation
Finite commutative ring · qf-mod28-commring-qualification · candidate
25. Multiplicative Commutativity Is Additional to Complete Ring Qualification
Use the same qualified complete finite unital ring on Fin(8)={0,…,7}, with addition=[[0, 1, 2, 3, 4, 5, 6, 7], [1, 2, 4, 5, 3, 6, 7, 0], [2, 4, 3, 6, 5, 7, 0, 1], [3, 5, 6, 0, 7, 1, 2, 4], [4, 3, 5, 7, 6, 0, 1, 2], [5, 6, 7, 1, 0, 2, 4, 3], [6, 7, 0, 2, 1, 4, 3, 5], [7, 0, 1, 4, 2, 3, 5, 6]] and multiplication=[[0, 0, 0, 0, 0, 0, 0, 0], [0, 1, 2, 3, 4, 5, 6, 7], [0, 2, 3, 0, 6, 2, 3, 6], [0, 3, 0, 0, 3, 3, 0, 3], [0, 4, 6, 3, 1, 7, 2, 5], [0, 5, 2, 3, 7, 1, 6, 4], [0, 6, 3, 0, 2, 6, 3, 2], [0, 7, 6, 3, 5, 4, 2, 1]] in label order. All unital-ring axioms, including distinct two-sided identities, must already hold. The identity labels need not be 0 and 1. Commutativity is a separate multiplication obligation; even a commutative ring may have nonzero zero divisors and need not be a field. Give (n,zero,one,all [a,b,ab,ba] failures with a<b in label order,commutative?,classification). Classification is commutative_ring or unital_ring_not_commutative.
View answer and explanation
(8, 0, 1, [], Yes, commutative_ring)
Compare both multiplication orders for every distinct unordered pair. Off-diagonal failures disprove commutativity; the full ring qualification is a prior premise rather than something repaired by this audit. An exact finite or named-formula slice does not automatically establish an arbitrary infinite theorem. The exact result, retaining the requested order, is (8, 0, 1, [], Yes, commutative_ring).
Reviewed step-by-step derivation
Finite commutative ring · qf-mod28-commring-qualification · candidate
26. Swap Two Factors Only after Global Commutative-Ring Qualification
Use the same qualified complete finite unital ring on Fin(8)={0,…,7}, with addition=[[0, 1, 2, 3, 4, 5, 6, 7], [1, 2, 3, 4, 5, 6, 7, 0], [2, 3, 4, 5, 6, 7, 0, 1], [3, 4, 5, 6, 7, 0, 1, 2], [4, 5, 6, 7, 0, 1, 2, 3], [5, 6, 7, 0, 1, 2, 3, 4], [6, 7, 0, 1, 2, 3, 4, 5], [7, 0, 1, 2, 3, 4, 5, 6]] and multiplication=[[0, 0, 0, 0, 0, 0, 0, 0], [0, 1, 2, 3, 4, 5, 6, 7], [0, 2, 4, 6, 0, 2, 4, 6], [0, 3, 6, 1, 4, 7, 2, 5], [0, 4, 0, 4, 0, 4, 0, 4], [0, 5, 2, 7, 4, 1, 6, 3], [0, 6, 4, 2, 0, 6, 4, 2], [0, 7, 6, 5, 4, 3, 2, 1]] in label order. All unital-ring axioms, including distinct two-sided identities, must already hold. The identity labels need not be 0 and 1. Commutativity is a separate multiplication obligation; even a commutative ring may have nonzero zero divisors and need not be a field. Require the complete ring also to be commutative. a=1, b=2. Give (a,b,ab,ba,equal?).
View answer and explanation
(1, 2, 2, 2, Yes)
Look up both ordered products in the same multiplication table. One equal pair alone would not establish whole-ring commutativity. An exact finite or named-formula slice does not automatically establish an arbitrary infinite theorem. The exact result, retaining the requested order, is (1, 2, 2, 2, Yes).
Reviewed step-by-step derivation
Finite commutative ring · qf-mod28-commring-product-swap · candidate
27. Swap Two Factors Only after Global Commutative-Ring Qualification
Use the same qualified complete finite unital ring on Fin(8)={0,…,7}, with addition=[[0, 1, 2, 3, 4, 5, 6, 7], [1, 2, 3, 4, 5, 7, 0, 6], [2, 3, 4, 5, 7, 6, 1, 0], [3, 4, 5, 7, 6, 0, 2, 1], [4, 5, 7, 6, 0, 1, 3, 2], [5, 7, 6, 0, 1, 2, 4, 3], [6, 0, 1, 2, 3, 4, 7, 5], [7, 6, 0, 1, 2, 3, 5, 4]] and multiplication=[[0, 0, 0, 0, 0, 0, 0, 0], [0, 1, 2, 3, 4, 5, 6, 7], [0, 2, 4, 7, 0, 2, 7, 4], [0, 3, 7, 1, 4, 6, 5, 2], [0, 4, 0, 4, 0, 4, 4, 0], [0, 5, 2, 6, 4, 1, 3, 7], [0, 6, 7, 5, 4, 3, 1, 2], [0, 7, 4, 2, 0, 7, 2, 4]] in label order. All unital-ring axioms, including distinct two-sided identities, must already hold. The identity labels need not be 0 and 1. Commutativity is a separate multiplication obligation; even a commutative ring may have nonzero zero divisors and need not be a field. Require the complete ring also to be commutative. a=2, b=5. Give (a,b,ab,ba,equal?).
View answer and explanation
(2, 5, 2, 2, Yes)
Look up both ordered products in the same multiplication table. One equal pair alone would not establish whole-ring commutativity. An exact finite or named-formula slice does not automatically establish an arbitrary infinite theorem. The exact result, retaining the requested order, is (2, 5, 2, 2, Yes).
Reviewed step-by-step derivation
Finite commutative ring · qf-mod28-commring-product-swap · candidate
28. Swap Two Factors Only after Global Commutative-Ring Qualification
Use the same qualified complete finite unital ring on Fin(8)={0,…,7}, with addition=[[0, 1, 2, 3, 4, 5, 6, 7], [1, 2, 3, 4, 6, 7, 5, 0], [2, 3, 4, 6, 5, 0, 7, 1], [3, 4, 6, 5, 7, 1, 0, 2], [4, 6, 5, 7, 0, 2, 1, 3], [5, 7, 0, 1, 2, 4, 3, 6], [6, 5, 7, 0, 1, 3, 2, 4], [7, 0, 1, 2, 3, 6, 4, 5]] and multiplication=[[0, 0, 0, 0, 0, 0, 0, 0], [0, 1, 2, 3, 4, 5, 6, 7], [0, 2, 4, 5, 0, 4, 2, 5], [0, 3, 5, 1, 4, 2, 7, 6], [0, 4, 0, 4, 0, 0, 4, 4], [0, 5, 4, 2, 0, 4, 5, 2], [0, 6, 2, 7, 4, 5, 1, 3], [0, 7, 5, 6, 4, 2, 3, 1]] in label order. All unital-ring axioms, including distinct two-sided identities, must already hold. The identity labels need not be 0 and 1. Commutativity is a separate multiplication obligation; even a commutative ring may have nonzero zero divisors and need not be a field. Require the complete ring also to be commutative. a=3, b=0. Give (a,b,ab,ba,equal?).
View answer and explanation
(3, 0, 0, 0, Yes)
Look up both ordered products in the same multiplication table. One equal pair alone would not establish whole-ring commutativity. An exact finite or named-formula slice does not automatically establish an arbitrary infinite theorem. The exact result, retaining the requested order, is (3, 0, 0, 0, Yes).
Reviewed step-by-step derivation
Finite commutative ring · qf-mod28-commring-product-swap · candidate
29. Swap Two Factors Only after Global Commutative-Ring Qualification
Use the same qualified complete finite unital ring on Fin(8)={0,…,7}, with addition=[[0, 1, 2, 3, 4, 5, 6, 7], [1, 2, 3, 4, 6, 0, 7, 5], [2, 3, 4, 6, 7, 1, 5, 0], [3, 4, 6, 7, 5, 2, 0, 1], [4, 6, 7, 5, 0, 3, 1, 2], [5, 0, 1, 2, 3, 7, 4, 6], [6, 7, 5, 0, 1, 4, 2, 3], [7, 5, 0, 1, 2, 6, 3, 4]] and multiplication=[[0, 0, 0, 0, 0, 0, 0, 0], [0, 1, 2, 3, 4, 5, 6, 7], [0, 2, 4, 7, 0, 7, 2, 4], [0, 3, 7, 1, 4, 6, 5, 2], [0, 4, 0, 4, 0, 4, 4, 0], [0, 5, 7, 6, 4, 1, 3, 2], [0, 6, 2, 5, 4, 3, 1, 7], [0, 7, 4, 2, 0, 2, 7, 4]] in label order. All unital-ring axioms, including distinct two-sided identities, must already hold. The identity labels need not be 0 and 1. Commutativity is a separate multiplication obligation; even a commutative ring may have nonzero zero divisors and need not be a field. Require the complete ring also to be commutative. a=4, b=3. Give (a,b,ab,ba,equal?).
View answer and explanation
(4, 3, 4, 4, Yes)
Look up both ordered products in the same multiplication table. One equal pair alone would not establish whole-ring commutativity. An exact finite or named-formula slice does not automatically establish an arbitrary infinite theorem. The exact result, retaining the requested order, is (4, 3, 4, 4, Yes).
Reviewed step-by-step derivation
Finite commutative ring · qf-mod28-commring-product-swap · candidate
30. Swap Two Factors Only after Global Commutative-Ring Qualification
Use the same qualified complete finite unital ring on Fin(8)={0,…,7}, with addition=[[0, 1, 2, 3, 4, 5, 6, 7], [1, 2, 3, 4, 7, 6, 0, 5], [2, 3, 4, 7, 5, 0, 1, 6], [3, 4, 7, 5, 6, 1, 2, 0], [4, 7, 5, 6, 0, 2, 3, 1], [5, 6, 0, 1, 2, 4, 7, 3], [6, 0, 1, 2, 3, 7, 5, 4], [7, 5, 6, 0, 1, 3, 4, 2]] and multiplication=[[0, 0, 0, 0, 0, 0, 0, 0], [0, 1, 2, 3, 4, 5, 6, 7], [0, 2, 4, 5, 0, 4, 5, 2], [0, 3, 5, 1, 4, 2, 7, 6], [0, 4, 0, 4, 0, 0, 4, 4], [0, 5, 4, 2, 0, 4, 2, 5], [0, 6, 5, 7, 4, 2, 1, 3], [0, 7, 2, 6, 4, 5, 3, 1]] in label order. All unital-ring axioms, including distinct two-sided identities, must already hold. The identity labels need not be 0 and 1. Commutativity is a separate multiplication obligation; even a commutative ring may have nonzero zero divisors and need not be a field. Require the complete ring also to be commutative. a=7, b=5. Give (a,b,ab,ba,equal?).
View answer and explanation
(7, 5, 5, 5, Yes)
Look up both ordered products in the same multiplication table. One equal pair alone would not establish whole-ring commutativity. An exact finite or named-formula slice does not automatically establish an arbitrary infinite theorem. The exact result, retaining the requested order, is (7, 5, 5, 5, Yes).
Reviewed step-by-step derivation
Finite commutative ring · qf-mod28-commring-product-swap · candidate
31. Swap Two Factors Only after Global Commutative-Ring Qualification
Use the same qualified complete finite unital ring on Fin(8)={0,…,7}, with addition=[[0, 1, 2, 3, 4, 5, 6, 7], [1, 2, 3, 4, 7, 0, 5, 6], [2, 3, 4, 7, 6, 1, 0, 5], [3, 4, 7, 6, 5, 2, 1, 0], [4, 7, 6, 5, 0, 3, 2, 1], [5, 0, 1, 2, 3, 6, 7, 4], [6, 5, 0, 1, 2, 7, 4, 3], [7, 6, 5, 0, 1, 4, 3, 2]] and multiplication=[[0, 0, 0, 0, 0, 0, 0, 0], [0, 1, 2, 3, 4, 5, 6, 7], [0, 2, 4, 6, 0, 6, 4, 2], [0, 3, 6, 1, 4, 7, 2, 5], [0, 4, 0, 4, 0, 4, 0, 4], [0, 5, 6, 7, 4, 1, 2, 3], [0, 6, 4, 2, 0, 2, 4, 6], [0, 7, 2, 5, 4, 3, 6, 1]] in label order. All unital-ring axioms, including distinct two-sided identities, must already hold. The identity labels need not be 0 and 1. Commutativity is a separate multiplication obligation; even a commutative ring may have nonzero zero divisors and need not be a field. Require the complete ring also to be commutative. a=6, b=1. Give (a,b,ab,ba,equal?).
View answer and explanation
(6, 1, 6, 6, Yes)
Look up both ordered products in the same multiplication table. One equal pair alone would not establish whole-ring commutativity. An exact finite or named-formula slice does not automatically establish an arbitrary infinite theorem. The exact result, retaining the requested order, is (6, 1, 6, 6, Yes).
Reviewed step-by-step derivation
Finite commutative ring · qf-mod28-commring-product-swap · candidate
32. Swap Two Factors Only after Global Commutative-Ring Qualification
Use the same qualified complete finite unital ring on Fin(8)={0,…,7}, with addition=[[0, 1, 2, 3, 4, 5, 6, 7], [1, 2, 3, 5, 6, 4, 7, 0], [2, 3, 5, 4, 7, 6, 0, 1], [3, 5, 4, 6, 0, 7, 1, 2], [4, 6, 7, 0, 2, 1, 3, 5], [5, 4, 6, 7, 1, 0, 2, 3], [6, 7, 0, 1, 3, 2, 5, 4], [7, 0, 1, 2, 5, 3, 4, 6]] and multiplication=[[0, 0, 0, 0, 0, 0, 0, 0], [0, 1, 2, 3, 4, 5, 6, 7], [0, 2, 5, 6, 2, 0, 5, 6], [0, 3, 6, 1, 7, 5, 2, 4], [0, 4, 2, 7, 1, 5, 6, 3], [0, 5, 0, 5, 5, 0, 0, 5], [0, 6, 5, 2, 6, 0, 5, 2], [0, 7, 6, 4, 3, 5, 2, 1]] in label order. All unital-ring axioms, including distinct two-sided identities, must already hold. The identity labels need not be 0 and 1. Commutativity is a separate multiplication obligation; even a commutative ring may have nonzero zero divisors and need not be a field. Require the complete ring also to be commutative. a=7, b=5. Give (a,b,ab,ba,equal?).
View answer and explanation
(7, 5, 5, 5, Yes)
Look up both ordered products in the same multiplication table. One equal pair alone would not establish whole-ring commutativity. An exact finite or named-formula slice does not automatically establish an arbitrary infinite theorem. The exact result, retaining the requested order, is (7, 5, 5, 5, Yes).
Reviewed step-by-step derivation
Finite commutative ring · qf-mod28-commring-product-swap · candidate
33. Swap Two Factors Only after Global Commutative-Ring Qualification
Use the same qualified complete finite unital ring on Fin(8)={0,…,7}, with addition=[[0, 1, 2, 3, 4, 5, 6, 7], [1, 2, 3, 5, 7, 4, 0, 6], [2, 3, 5, 4, 6, 7, 1, 0], [3, 5, 4, 7, 0, 6, 2, 1], [4, 7, 6, 0, 2, 1, 5, 3], [5, 4, 7, 6, 1, 0, 3, 2], [6, 0, 1, 2, 5, 3, 7, 4], [7, 6, 0, 1, 3, 2, 4, 5]] and multiplication=[[0, 0, 0, 0, 0, 0, 0, 0], [0, 1, 2, 3, 4, 5, 6, 7], [0, 2, 5, 7, 2, 0, 7, 5], [0, 3, 7, 1, 6, 5, 4, 2], [0, 4, 2, 6, 1, 5, 3, 7], [0, 5, 0, 5, 5, 0, 5, 0], [0, 6, 7, 4, 3, 5, 1, 2], [0, 7, 5, 2, 7, 0, 2, 5]] in label order. All unital-ring axioms, including distinct two-sided identities, must already hold. The identity labels need not be 0 and 1. Commutativity is a separate multiplication obligation; even a commutative ring may have nonzero zero divisors and need not be a field. Require the complete ring also to be commutative. a=0, b=6. Give (a,b,ab,ba,equal?).
View answer and explanation
(0, 6, 0, 0, Yes)
Look up both ordered products in the same multiplication table. One equal pair alone would not establish whole-ring commutativity. An exact finite or named-formula slice does not automatically establish an arbitrary infinite theorem. The exact result, retaining the requested order, is (0, 6, 0, 0, Yes).
Reviewed step-by-step derivation
Finite commutative ring · qf-mod28-commring-product-swap · candidate
34. Swap Two Factors Only after Global Commutative-Ring Qualification
Use the same qualified complete finite unital ring on Fin(8)={0,…,7}, with addition=[[0, 1, 2, 3, 4, 5, 6, 7], [1, 2, 3, 5, 7, 6, 4, 0], [2, 3, 5, 6, 0, 4, 7, 1], [3, 5, 6, 4, 1, 7, 0, 2], [4, 7, 0, 1, 5, 2, 3, 6], [5, 6, 4, 7, 2, 0, 1, 3], [6, 4, 7, 0, 3, 1, 2, 5], [7, 0, 1, 2, 6, 3, 5, 4]] and multiplication=[[0, 0, 0, 0, 0, 0, 0, 0], [0, 1, 2, 3, 4, 5, 6, 7], [0, 2, 5, 4, 5, 0, 2, 4], [0, 3, 4, 1, 2, 5, 7, 6], [0, 4, 5, 2, 5, 0, 4, 2], [0, 5, 0, 5, 0, 0, 5, 5], [0, 6, 2, 7, 4, 5, 1, 3], [0, 7, 4, 6, 2, 5, 3, 1]] in label order. All unital-ring axioms, including distinct two-sided identities, must already hold. The identity labels need not be 0 and 1. Commutativity is a separate multiplication obligation; even a commutative ring may have nonzero zero divisors and need not be a field. Require the complete ring also to be commutative. a=1, b=2. Give (a,b,ab,ba,equal?).
View answer and explanation
(1, 2, 2, 2, Yes)
Look up both ordered products in the same multiplication table. One equal pair alone would not establish whole-ring commutativity. An exact finite or named-formula slice does not automatically establish an arbitrary infinite theorem. The exact result, retaining the requested order, is (1, 2, 2, 2, Yes).
Reviewed step-by-step derivation
Finite commutative ring · qf-mod28-commring-product-swap · candidate
35. Swap Two Factors Only after Global Commutative-Ring Qualification
Use the same qualified complete finite unital ring on Fin(8)={0,…,7}, with addition=[[0, 1, 2, 3, 4, 5, 6, 7], [1, 2, 3, 5, 0, 6, 7, 4], [2, 3, 5, 6, 1, 7, 4, 0], [3, 5, 6, 7, 2, 4, 0, 1], [4, 0, 1, 2, 7, 3, 5, 6], [5, 6, 7, 4, 3, 0, 1, 2], [6, 7, 4, 0, 5, 1, 2, 3], [7, 4, 0, 1, 6, 2, 3, 5]] and multiplication=[[0, 0, 0, 0, 0, 0, 0, 0], [0, 1, 2, 3, 4, 5, 6, 7], [0, 2, 5, 7, 7, 0, 2, 5], [0, 3, 7, 1, 6, 5, 4, 2], [0, 4, 7, 6, 1, 5, 3, 2], [0, 5, 0, 5, 5, 0, 5, 0], [0, 6, 2, 4, 3, 5, 1, 7], [0, 7, 5, 2, 2, 0, 7, 5]] in label order. All unital-ring axioms, including distinct two-sided identities, must already hold. The identity labels need not be 0 and 1. Commutativity is a separate multiplication obligation; even a commutative ring may have nonzero zero divisors and need not be a field. Require the complete ring also to be commutative. a=2, b=6. Give (a,b,ab,ba,equal?).
View answer and explanation
(2, 6, 2, 2, Yes)
Look up both ordered products in the same multiplication table. One equal pair alone would not establish whole-ring commutativity. An exact finite or named-formula slice does not automatically establish an arbitrary infinite theorem. The exact result, retaining the requested order, is (2, 6, 2, 2, Yes).
Reviewed step-by-step derivation
Finite commutative ring · qf-mod28-commring-product-swap · candidate
36. Swap Two Factors Only after Global Commutative-Ring Qualification
Use the same qualified complete finite unital ring on Fin(8)={0,…,7}, with addition=[[0, 1, 2, 3, 4, 5, 6, 7], [1, 2, 3, 5, 6, 7, 0, 4], [2, 3, 5, 7, 0, 4, 1, 6], [3, 5, 7, 4, 1, 6, 2, 0], [4, 6, 0, 1, 5, 2, 7, 3], [5, 7, 4, 6, 2, 0, 3, 1], [6, 0, 1, 2, 7, 3, 4, 5], [7, 4, 6, 0, 3, 1, 5, 2]] and multiplication=[[0, 0, 0, 0, 0, 0, 0, 0], [0, 1, 2, 3, 4, 5, 6, 7], [0, 2, 5, 4, 5, 0, 4, 2], [0, 3, 4, 1, 2, 5, 7, 6], [0, 4, 5, 2, 5, 0, 2, 4], [0, 5, 0, 5, 0, 0, 5, 5], [0, 6, 4, 7, 2, 5, 1, 3], [0, 7, 2, 6, 4, 5, 3, 1]] in label order. All unital-ring axioms, including distinct two-sided identities, must already hold. The identity labels need not be 0 and 1. Commutativity is a separate multiplication obligation; even a commutative ring may have nonzero zero divisors and need not be a field. Require the complete ring also to be commutative. a=3, b=0. Give (a,b,ab,ba,equal?).
View answer and explanation
(3, 0, 0, 0, Yes)
Look up both ordered products in the same multiplication table. One equal pair alone would not establish whole-ring commutativity. An exact finite or named-formula slice does not automatically establish an arbitrary infinite theorem. The exact result, retaining the requested order, is (3, 0, 0, 0, Yes).
Reviewed step-by-step derivation
Finite commutative ring · qf-mod28-commring-product-swap · candidate
37. Swap Two Factors Only after Global Commutative-Ring Qualification
Use the same qualified complete finite unital ring on Fin(8)={0,…,7}, with addition=[[0, 1, 2, 3, 4, 5, 6, 7], [1, 2, 3, 5, 0, 7, 4, 6], [2, 3, 5, 7, 1, 6, 0, 4], [3, 5, 7, 6, 2, 4, 1, 0], [4, 0, 1, 2, 6, 3, 7, 5], [5, 7, 6, 4, 3, 0, 2, 1], [6, 4, 0, 1, 7, 2, 5, 3], [7, 6, 4, 0, 5, 1, 3, 2]] and multiplication=[[0, 0, 0, 0, 0, 0, 0, 0], [0, 1, 2, 3, 4, 5, 6, 7], [0, 2, 5, 6, 6, 0, 5, 2], [0, 3, 6, 1, 7, 5, 2, 4], [0, 4, 6, 7, 1, 5, 2, 3], [0, 5, 0, 5, 5, 0, 0, 5], [0, 6, 5, 2, 2, 0, 5, 6], [0, 7, 2, 4, 3, 5, 6, 1]] in label order. All unital-ring axioms, including distinct two-sided identities, must already hold. The identity labels need not be 0 and 1. Commutativity is a separate multiplication obligation; even a commutative ring may have nonzero zero divisors and need not be a field. Require the complete ring also to be commutative. a=5, b=3. Give (a,b,ab,ba,equal?).
View answer and explanation
(5, 3, 5, 5, Yes)
Look up both ordered products in the same multiplication table. One equal pair alone would not establish whole-ring commutativity. An exact finite or named-formula slice does not automatically establish an arbitrary infinite theorem. The exact result, retaining the requested order, is (5, 3, 5, 5, Yes).
Reviewed step-by-step derivation
Finite commutative ring · qf-mod28-commring-product-swap · candidate
38. Swap Two Factors Only after Global Commutative-Ring Qualification
Use the same qualified complete finite unital ring on Fin(8)={0,…,7}, with addition=[[0, 1, 2, 3, 4, 5, 6, 7], [1, 2, 3, 6, 5, 7, 4, 0], [2, 3, 6, 4, 7, 0, 5, 1], [3, 6, 4, 5, 0, 1, 7, 2], [4, 5, 7, 0, 2, 3, 1, 6], [5, 7, 0, 1, 3, 6, 2, 4], [6, 4, 5, 7, 1, 2, 0, 3], [7, 0, 1, 2, 6, 4, 3, 5]] and multiplication=[[0, 0, 0, 0, 0, 0, 0, 0], [0, 1, 2, 3, 4, 5, 6, 7], [0, 2, 6, 5, 2, 6, 0, 5], [0, 3, 5, 1, 7, 2, 6, 4], [0, 4, 2, 7, 1, 5, 6, 3], [0, 5, 6, 2, 5, 6, 0, 2], [0, 6, 0, 6, 6, 0, 0, 6], [0, 7, 5, 4, 3, 2, 6, 1]] in label order. All unital-ring axioms, including distinct two-sided identities, must already hold. The identity labels need not be 0 and 1. Commutativity is a separate multiplication obligation; even a commutative ring may have nonzero zero divisors and need not be a field. Require the complete ring also to be commutative. a=4, b=5. Give (a,b,ab,ba,equal?).
View answer and explanation
(4, 5, 5, 5, Yes)
Look up both ordered products in the same multiplication table. One equal pair alone would not establish whole-ring commutativity. An exact finite or named-formula slice does not automatically establish an arbitrary infinite theorem. The exact result, retaining the requested order, is (4, 5, 5, 5, Yes).
Reviewed step-by-step derivation
Finite commutative ring · qf-mod28-commring-product-swap · candidate
39. Swap Two Factors Only after Global Commutative-Ring Qualification
Use the same qualified complete finite unital ring on Fin(8)={0,…,7}, with addition=[[0, 1, 2, 3, 4, 5, 6, 7], [1, 2, 3, 6, 7, 0, 4, 5], [2, 3, 6, 4, 5, 1, 7, 0], [3, 6, 4, 7, 0, 2, 5, 1], [4, 7, 5, 0, 2, 6, 1, 3], [5, 0, 1, 2, 6, 7, 3, 4], [6, 4, 7, 5, 1, 3, 0, 2], [7, 5, 0, 1, 3, 4, 2, 6]] and multiplication=[[0, 0, 0, 0, 0, 0, 0, 0], [0, 1, 2, 3, 4, 5, 6, 7], [0, 2, 6, 7, 2, 7, 0, 6], [0, 3, 7, 1, 5, 4, 6, 2], [0, 4, 2, 5, 1, 3, 6, 7], [0, 5, 7, 4, 3, 1, 6, 2], [0, 6, 0, 6, 6, 6, 0, 0], [0, 7, 6, 2, 7, 2, 0, 6]] in label order. All unital-ring axioms, including distinct two-sided identities, must already hold. The identity labels need not be 0 and 1. Commutativity is a separate multiplication obligation; even a commutative ring may have nonzero zero divisors and need not be a field. Require the complete ring also to be commutative. a=7, b=1. Give (a,b,ab,ba,equal?).
View answer and explanation
(7, 1, 7, 7, Yes)
Look up both ordered products in the same multiplication table. One equal pair alone would not establish whole-ring commutativity. An exact finite or named-formula slice does not automatically establish an arbitrary infinite theorem. The exact result, retaining the requested order, is (7, 1, 7, 7, Yes).
Reviewed step-by-step derivation
Finite commutative ring · qf-mod28-commring-product-swap · candidate
40. Swap Two Factors Only after Global Commutative-Ring Qualification
Use the same qualified complete finite unital ring on Fin(8)={0,…,7}, with addition=[[0, 1, 2, 3, 4, 5, 6, 7], [1, 2, 3, 6, 7, 4, 5, 0], [2, 3, 6, 5, 0, 7, 4, 1], [3, 6, 5, 4, 1, 0, 7, 2], [4, 7, 0, 1, 6, 3, 2, 5], [5, 4, 7, 0, 3, 2, 1, 6], [6, 5, 4, 7, 2, 1, 0, 3], [7, 0, 1, 2, 5, 6, 3, 4]] and multiplication=[[0, 0, 0, 0, 0, 0, 0, 0], [0, 1, 2, 3, 4, 5, 6, 7], [0, 2, 6, 4, 6, 2, 0, 4], [0, 3, 4, 1, 2, 7, 6, 5], [0, 4, 6, 2, 6, 4, 0, 2], [0, 5, 2, 7, 4, 1, 6, 3], [0, 6, 0, 6, 0, 6, 0, 6], [0, 7, 4, 5, 2, 3, 6, 1]] in label order. All unital-ring axioms, including distinct two-sided identities, must already hold. The identity labels need not be 0 and 1. Commutativity is a separate multiplication obligation; even a commutative ring may have nonzero zero divisors and need not be a field. Require the complete ring also to be commutative. a=7, b=6. Give (a,b,ab,ba,equal?).
View answer and explanation
(7, 6, 6, 6, Yes)
Look up both ordered products in the same multiplication table. One equal pair alone would not establish whole-ring commutativity. An exact finite or named-formula slice does not automatically establish an arbitrary infinite theorem. The exact result, retaining the requested order, is (7, 6, 6, 6, Yes).
Reviewed step-by-step derivation
Finite commutative ring · qf-mod28-commring-product-swap · candidate
41. Swap Two Factors Only after Global Commutative-Ring Qualification
Use the same qualified complete finite unital ring on Fin(8)={0,…,7}, with addition=[[0, 1, 2, 3, 4, 5, 6, 7], [1, 2, 3, 6, 0, 7, 5, 4], [2, 3, 6, 5, 1, 4, 7, 0], [3, 6, 5, 7, 2, 0, 4, 1], [4, 0, 1, 2, 7, 6, 3, 5], [5, 7, 4, 0, 6, 2, 1, 3], [6, 5, 7, 4, 3, 1, 0, 2], [7, 4, 0, 1, 5, 3, 2, 6]] and multiplication=[[0, 0, 0, 0, 0, 0, 0, 0], [0, 1, 2, 3, 4, 5, 6, 7], [0, 2, 6, 7, 7, 2, 0, 6], [0, 3, 7, 1, 5, 4, 6, 2], [0, 4, 7, 5, 1, 3, 6, 2], [0, 5, 2, 4, 3, 1, 6, 7], [0, 6, 0, 6, 6, 6, 0, 0], [0, 7, 6, 2, 2, 7, 0, 6]] in label order. All unital-ring axioms, including distinct two-sided identities, must already hold. The identity labels need not be 0 and 1. Commutativity is a separate multiplication obligation; even a commutative ring may have nonzero zero divisors and need not be a field. Require the complete ring also to be commutative. a=0, b=4. Give (a,b,ab,ba,equal?).
View answer and explanation
(0, 4, 0, 0, Yes)
Look up both ordered products in the same multiplication table. One equal pair alone would not establish whole-ring commutativity. An exact finite or named-formula slice does not automatically establish an arbitrary infinite theorem. The exact result, retaining the requested order, is (0, 4, 0, 0, Yes).
Reviewed step-by-step derivation
Finite commutative ring · qf-mod28-commring-product-swap · candidate
42. Swap Two Factors Only after Global Commutative-Ring Qualification
Use the same qualified complete finite unital ring on Fin(8)={0,…,7}, with addition=[[0, 1, 2, 3, 4, 5, 6, 7], [1, 2, 3, 6, 5, 0, 7, 4], [2, 3, 6, 7, 0, 1, 4, 5], [3, 6, 7, 4, 1, 2, 5, 0], [4, 5, 0, 1, 6, 7, 2, 3], [5, 0, 1, 2, 7, 4, 3, 6], [6, 7, 4, 5, 2, 3, 0, 1], [7, 4, 5, 0, 3, 6, 1, 2]] and multiplication=[[0, 0, 0, 0, 0, 0, 0, 0], [0, 1, 2, 3, 4, 5, 6, 7], [0, 2, 6, 4, 6, 4, 0, 2], [0, 3, 4, 1, 2, 7, 6, 5], [0, 4, 6, 2, 6, 2, 0, 4], [0, 5, 4, 7, 2, 1, 6, 3], [0, 6, 0, 6, 0, 6, 0, 6], [0, 7, 2, 5, 4, 3, 6, 1]] in label order. All unital-ring axioms, including distinct two-sided identities, must already hold. The identity labels need not be 0 and 1. Commutativity is a separate multiplication obligation; even a commutative ring may have nonzero zero divisors and need not be a field. Require the complete ring also to be commutative. a=1, b=2. Give (a,b,ab,ba,equal?).
View answer and explanation
(1, 2, 2, 2, Yes)
Look up both ordered products in the same multiplication table. One equal pair alone would not establish whole-ring commutativity. An exact finite or named-formula slice does not automatically establish an arbitrary infinite theorem. The exact result, retaining the requested order, is (1, 2, 2, 2, Yes).
Reviewed step-by-step derivation
Finite commutative ring · qf-mod28-commring-product-swap · candidate
43. Swap Two Factors Only after Global Commutative-Ring Qualification
Use the same qualified complete finite unital ring on Fin(8)={0,…,7}, with addition=[[0, 1, 2, 3, 4, 5, 6, 7], [1, 2, 3, 6, 0, 4, 7, 5], [2, 3, 6, 7, 1, 0, 5, 4], [3, 6, 7, 5, 2, 1, 4, 0], [4, 0, 1, 2, 5, 7, 3, 6], [5, 4, 0, 1, 7, 6, 2, 3], [6, 7, 5, 4, 3, 2, 0, 1], [7, 5, 4, 0, 6, 3, 1, 2]] and multiplication=[[0, 0, 0, 0, 0, 0, 0, 0], [0, 1, 2, 3, 4, 5, 6, 7], [0, 2, 6, 5, 5, 6, 0, 2], [0, 3, 5, 1, 7, 2, 6, 4], [0, 4, 5, 7, 1, 2, 6, 3], [0, 5, 6, 2, 2, 6, 0, 5], [0, 6, 0, 6, 6, 0, 0, 6], [0, 7, 2, 4, 3, 5, 6, 1]] in label order. All unital-ring axioms, including distinct two-sided identities, must already hold. The identity labels need not be 0 and 1. Commutativity is a separate multiplication obligation; even a commutative ring may have nonzero zero divisors and need not be a field. Require the complete ring also to be commutative. a=2, b=7. Give (a,b,ab,ba,equal?).
View answer and explanation
(2, 7, 2, 2, Yes)
Look up both ordered products in the same multiplication table. One equal pair alone would not establish whole-ring commutativity. An exact finite or named-formula slice does not automatically establish an arbitrary infinite theorem. The exact result, retaining the requested order, is (2, 7, 2, 2, Yes).
Reviewed step-by-step derivation
Finite commutative ring · qf-mod28-commring-product-swap · candidate
44. Swap Two Factors Only after Global Commutative-Ring Qualification
Use the same qualified complete finite unital ring on Fin(8)={0,…,7}, with addition=[[0, 1, 2, 3, 4, 5, 6, 7], [1, 2, 3, 7, 5, 6, 0, 4], [2, 3, 7, 4, 6, 0, 1, 5], [3, 7, 4, 5, 0, 1, 2, 6], [4, 5, 6, 0, 2, 3, 7, 1], [5, 6, 0, 1, 3, 7, 4, 2], [6, 0, 1, 2, 7, 4, 5, 3], [7, 4, 5, 6, 1, 2, 3, 0]] and multiplication=[[0, 0, 0, 0, 0, 0, 0, 0], [0, 1, 2, 3, 4, 5, 6, 7], [0, 2, 7, 5, 2, 7, 5, 0], [0, 3, 5, 1, 6, 2, 4, 7], [0, 4, 2, 6, 1, 5, 3, 7], [0, 5, 7, 2, 5, 7, 2, 0], [0, 6, 5, 4, 3, 2, 1, 7], [0, 7, 0, 7, 7, 0, 7, 0]] in label order. All unital-ring axioms, including distinct two-sided identities, must already hold. The identity labels need not be 0 and 1. Commutativity is a separate multiplication obligation; even a commutative ring may have nonzero zero divisors and need not be a field. Require the complete ring also to be commutative. a=3, b=0. Give (a,b,ab,ba,equal?).
View answer and explanation
(3, 0, 0, 0, Yes)
Look up both ordered products in the same multiplication table. One equal pair alone would not establish whole-ring commutativity. An exact finite or named-formula slice does not automatically establish an arbitrary infinite theorem. The exact result, retaining the requested order, is (3, 0, 0, 0, Yes).
Reviewed step-by-step derivation
Finite commutative ring · qf-mod28-commring-product-swap · candidate
45. Swap Two Factors Only after Global Commutative-Ring Qualification
Use the same qualified complete finite unital ring on Fin(8)={0,…,7}, with addition=[[0, 1, 2, 3, 4, 5, 6, 7], [1, 2, 3, 7, 6, 0, 5, 4], [2, 3, 7, 4, 5, 1, 0, 6], [3, 7, 4, 6, 0, 2, 1, 5], [4, 6, 5, 0, 2, 7, 3, 1], [5, 0, 1, 2, 7, 6, 4, 3], [6, 5, 0, 1, 3, 4, 7, 2], [7, 4, 6, 5, 1, 3, 2, 0]] and multiplication=[[0, 0, 0, 0, 0, 0, 0, 0], [0, 1, 2, 3, 4, 5, 6, 7], [0, 2, 7, 6, 2, 6, 7, 0], [0, 3, 6, 1, 5, 4, 2, 7], [0, 4, 2, 5, 1, 3, 6, 7], [0, 5, 6, 4, 3, 1, 2, 7], [0, 6, 7, 2, 6, 2, 7, 0], [0, 7, 0, 7, 7, 7, 0, 0]] in label order. All unital-ring axioms, including distinct two-sided identities, must already hold. The identity labels need not be 0 and 1. Commutativity is a separate multiplication obligation; even a commutative ring may have nonzero zero divisors and need not be a field. Require the complete ring also to be commutative. a=7, b=3. Give (a,b,ab,ba,equal?).
View answer and explanation
(7, 3, 7, 7, Yes)
Look up both ordered products in the same multiplication table. One equal pair alone would not establish whole-ring commutativity. An exact finite or named-formula slice does not automatically establish an arbitrary infinite theorem. The exact result, retaining the requested order, is (7, 3, 7, 7, Yes).
Reviewed step-by-step derivation
Finite commutative ring · qf-mod28-commring-product-swap · candidate
46. Swap Two Factors Only after Global Commutative-Ring Qualification
Use the same qualified complete finite unital ring on Fin(8)={0,…,7}, with addition=[[0, 1, 2, 3, 4, 5, 6, 7], [1, 2, 3, 7, 6, 4, 0, 5], [2, 3, 7, 5, 0, 6, 1, 4], [3, 7, 5, 4, 1, 0, 2, 6], [4, 6, 0, 1, 7, 3, 5, 2], [5, 4, 6, 0, 3, 2, 7, 1], [6, 0, 1, 2, 5, 7, 4, 3], [7, 5, 4, 6, 2, 1, 3, 0]] and multiplication=[[0, 0, 0, 0, 0, 0, 0, 0], [0, 1, 2, 3, 4, 5, 6, 7], [0, 2, 7, 4, 7, 2, 4, 0], [0, 3, 4, 1, 2, 6, 5, 7], [0, 4, 7, 2, 7, 4, 2, 0], [0, 5, 2, 6, 4, 1, 3, 7], [0, 6, 4, 5, 2, 3, 1, 7], [0, 7, 0, 7, 0, 7, 7, 0]] in label order. All unital-ring axioms, including distinct two-sided identities, must already hold. The identity labels need not be 0 and 1. Commutativity is a separate multiplication obligation; even a commutative ring may have nonzero zero divisors and need not be a field. Require the complete ring also to be commutative. a=5, b=4. Give (a,b,ab,ba,equal?).
View answer and explanation
(5, 4, 4, 4, Yes)
Look up both ordered products in the same multiplication table. One equal pair alone would not establish whole-ring commutativity. An exact finite or named-formula slice does not automatically establish an arbitrary infinite theorem. The exact result, retaining the requested order, is (5, 4, 4, 4, Yes).
Reviewed step-by-step derivation
Finite commutative ring · qf-mod28-commring-product-swap · candidate
47. Swap Two Factors Only after Global Commutative-Ring Qualification
Use the same qualified complete finite unital ring on Fin(8)={0,…,7}, with addition=[[0, 1, 2, 3, 4, 5, 6, 7], [1, 2, 3, 7, 0, 6, 4, 5], [2, 3, 7, 5, 1, 4, 0, 6], [3, 7, 5, 6, 2, 0, 1, 4], [4, 0, 1, 2, 6, 7, 5, 3], [5, 6, 4, 0, 7, 2, 3, 1], [6, 4, 0, 1, 5, 3, 7, 2], [7, 5, 6, 4, 3, 1, 2, 0]] and multiplication=[[0, 0, 0, 0, 0, 0, 0, 0], [0, 1, 2, 3, 4, 5, 6, 7], [0, 2, 7, 6, 6, 2, 7, 0], [0, 3, 6, 1, 5, 4, 2, 7], [0, 4, 6, 5, 1, 3, 2, 7], [0, 5, 2, 4, 3, 1, 6, 7], [0, 6, 7, 2, 2, 6, 7, 0], [0, 7, 0, 7, 7, 7, 0, 0]] in label order. All unital-ring axioms, including distinct two-sided identities, must already hold. The identity labels need not be 0 and 1. Commutativity is a separate multiplication obligation; even a commutative ring may have nonzero zero divisors and need not be a field. Require the complete ring also to be commutative. a=6, b=1. Give (a,b,ab,ba,equal?).
View answer and explanation
(6, 1, 6, 6, Yes)
Look up both ordered products in the same multiplication table. One equal pair alone would not establish whole-ring commutativity. An exact finite or named-formula slice does not automatically establish an arbitrary infinite theorem. The exact result, retaining the requested order, is (6, 1, 6, 6, Yes).
Reviewed step-by-step derivation
Finite commutative ring · qf-mod28-commring-product-swap · candidate
48. Swap Two Factors Only after Global Commutative-Ring Qualification
Use the same qualified complete finite unital ring on Fin(8)={0,…,7}, with addition=[[0, 1, 2, 3, 4, 5, 6, 7], [1, 2, 3, 7, 5, 0, 4, 6], [2, 3, 7, 6, 0, 1, 5, 4], [3, 7, 6, 4, 1, 2, 0, 5], [4, 5, 0, 1, 7, 6, 3, 2], [5, 0, 1, 2, 6, 4, 7, 3], [6, 4, 5, 0, 3, 7, 2, 1], [7, 6, 4, 5, 2, 3, 1, 0]] and multiplication=[[0, 0, 0, 0, 0, 0, 0, 0], [0, 1, 2, 3, 4, 5, 6, 7], [0, 2, 7, 4, 7, 4, 2, 0], [0, 3, 4, 1, 2, 6, 5, 7], [0, 4, 7, 2, 7, 2, 4, 0], [0, 5, 4, 6, 2, 1, 3, 7], [0, 6, 2, 5, 4, 3, 1, 7], [0, 7, 0, 7, 0, 7, 7, 0]] in label order. All unital-ring axioms, including distinct two-sided identities, must already hold. The identity labels need not be 0 and 1. Commutativity is a separate multiplication obligation; even a commutative ring may have nonzero zero divisors and need not be a field. Require the complete ring also to be commutative. a=5, b=7. Give (a,b,ab,ba,equal?).
View answer and explanation
(5, 7, 7, 7, Yes)
Look up both ordered products in the same multiplication table. One equal pair alone would not establish whole-ring commutativity. An exact finite or named-formula slice does not automatically establish an arbitrary infinite theorem. The exact result, retaining the requested order, is (5, 7, 7, 7, Yes).
Reviewed step-by-step derivation
Finite commutative ring · qf-mod28-commring-product-swap · candidate
49. Swap Two Factors Only after Global Commutative-Ring Qualification
Use the same qualified complete finite unital ring on Fin(8)={0,…,7}, with addition=[[0, 1, 2, 3, 4, 5, 6, 7], [1, 2, 3, 7, 0, 4, 5, 6], [2, 3, 7, 6, 1, 0, 4, 5], [3, 7, 6, 5, 2, 1, 0, 4], [4, 0, 1, 2, 5, 6, 7, 3], [5, 4, 0, 1, 6, 7, 3, 2], [6, 5, 4, 0, 7, 3, 2, 1], [7, 6, 5, 4, 3, 2, 1, 0]] and multiplication=[[0, 0, 0, 0, 0, 0, 0, 0], [0, 1, 2, 3, 4, 5, 6, 7], [0, 2, 7, 5, 5, 7, 2, 0], [0, 3, 5, 1, 6, 2, 4, 7], [0, 4, 5, 6, 1, 2, 3, 7], [0, 5, 7, 2, 2, 7, 5, 0], [0, 6, 2, 4, 3, 5, 1, 7], [0, 7, 0, 7, 7, 0, 7, 0]] in label order. All unital-ring axioms, including distinct two-sided identities, must already hold. The identity labels need not be 0 and 1. Commutativity is a separate multiplication obligation; even a commutative ring may have nonzero zero divisors and need not be a field. Require the complete ring also to be commutative. a=0, b=4. Give (a,b,ab,ba,equal?).
View answer and explanation
(0, 4, 0, 0, Yes)
Look up both ordered products in the same multiplication table. One equal pair alone would not establish whole-ring commutativity. An exact finite or named-formula slice does not automatically establish an arbitrary infinite theorem. The exact result, retaining the requested order, is (0, 4, 0, 0, Yes).
Reviewed step-by-step derivation
Finite commutative ring · qf-mod28-commring-product-swap · candidate
50. Swap Two Factors Only after Global Commutative-Ring Qualification
Use the same qualified complete finite unital ring on Fin(8)={0,…,7}, with addition=[[0, 1, 2, 3, 4, 5, 6, 7], [1, 2, 4, 5, 3, 6, 7, 0], [2, 4, 3, 6, 5, 7, 0, 1], [3, 5, 6, 0, 7, 1, 2, 4], [4, 3, 5, 7, 6, 0, 1, 2], [5, 6, 7, 1, 0, 2, 4, 3], [6, 7, 0, 2, 1, 4, 3, 5], [7, 0, 1, 4, 2, 3, 5, 6]] and multiplication=[[0, 0, 0, 0, 0, 0, 0, 0], [0, 1, 2, 3, 4, 5, 6, 7], [0, 2, 3, 0, 6, 2, 3, 6], [0, 3, 0, 0, 3, 3, 0, 3], [0, 4, 6, 3, 1, 7, 2, 5], [0, 5, 2, 3, 7, 1, 6, 4], [0, 6, 3, 0, 2, 6, 3, 2], [0, 7, 6, 3, 5, 4, 2, 1]] in label order. All unital-ring axioms, including distinct two-sided identities, must already hold. The identity labels need not be 0 and 1. Commutativity is a separate multiplication obligation; even a commutative ring may have nonzero zero divisors and need not be a field. Require the complete ring also to be commutative. a=1, b=2. Give (a,b,ab,ba,equal?).
View answer and explanation
(1, 2, 2, 2, Yes)
Look up both ordered products in the same multiplication table. One equal pair alone would not establish whole-ring commutativity. An exact finite or named-formula slice does not automatically establish an arbitrary infinite theorem. The exact result, retaining the requested order, is (1, 2, 2, 2, Yes).
Reviewed step-by-step derivation
Finite commutative ring · qf-mod28-commring-product-swap · candidate
51. A Power-of-Product Identity in a Qualified Commutative Ring
Use the same qualified complete finite unital ring on Fin(8)={0,…,7}, with addition=[[0, 1, 2, 3, 4, 5, 6, 7], [1, 2, 3, 4, 5, 6, 7, 0], [2, 3, 4, 5, 6, 7, 0, 1], [3, 4, 5, 6, 7, 0, 1, 2], [4, 5, 6, 7, 0, 1, 2, 3], [5, 6, 7, 0, 1, 2, 3, 4], [6, 7, 0, 1, 2, 3, 4, 5], [7, 0, 1, 2, 3, 4, 5, 6]] and multiplication=[[0, 0, 0, 0, 0, 0, 0, 0], [0, 1, 2, 3, 4, 5, 6, 7], [0, 2, 4, 6, 0, 2, 4, 6], [0, 3, 6, 1, 4, 7, 2, 5], [0, 4, 0, 4, 0, 4, 0, 4], [0, 5, 2, 7, 4, 1, 6, 3], [0, 6, 4, 2, 0, 6, 4, 2], [0, 7, 6, 5, 4, 3, 2, 1]] in label order. All unital-ring axioms, including distinct two-sided identities, must already hold. The identity labels need not be 0 and 1. Commutativity is a separate multiplication obligation; even a commutative ring may have nonzero zero divisors and need not be a field. Require multiplication to commute. a=1, b=2, nonnegative exponent=1. Give (a,b,exponent,ab,lhs=(ab)^exponent,rhs=a^exponent*b^exponent,equal?).
View answer and explanation
(1, 2, 1, 2, 2, 2, Yes)
Start each power at the actual multiplicative identity. At exponent zero all powers are the identity. This identity is used under the full commutative-ring premise, not to certify commutativity from one instance. An exact finite or named-formula slice does not automatically establish an arbitrary infinite theorem. The exact result, retaining the requested order, is (1, 2, 1, 2, 2, 2, Yes).
Reviewed step-by-step derivation
Finite commutative ring · qf-mod28-commring-power-product · candidate
52. A Power-of-Product Identity in a Qualified Commutative Ring
Use the same qualified complete finite unital ring on Fin(8)={0,…,7}, with addition=[[0, 1, 2, 3, 4, 5, 6, 7], [1, 2, 3, 4, 5, 7, 0, 6], [2, 3, 4, 5, 7, 6, 1, 0], [3, 4, 5, 7, 6, 0, 2, 1], [4, 5, 7, 6, 0, 1, 3, 2], [5, 7, 6, 0, 1, 2, 4, 3], [6, 0, 1, 2, 3, 4, 7, 5], [7, 6, 0, 1, 2, 3, 5, 4]] and multiplication=[[0, 0, 0, 0, 0, 0, 0, 0], [0, 1, 2, 3, 4, 5, 6, 7], [0, 2, 4, 7, 0, 2, 7, 4], [0, 3, 7, 1, 4, 6, 5, 2], [0, 4, 0, 4, 0, 4, 4, 0], [0, 5, 2, 6, 4, 1, 3, 7], [0, 6, 7, 5, 4, 3, 1, 2], [0, 7, 4, 2, 0, 7, 2, 4]] in label order. All unital-ring axioms, including distinct two-sided identities, must already hold. The identity labels need not be 0 and 1. Commutativity is a separate multiplication obligation; even a commutative ring may have nonzero zero divisors and need not be a field. Require multiplication to commute. a=2, b=5, nonnegative exponent=2. Give (a,b,exponent,ab,lhs=(ab)^exponent,rhs=a^exponent*b^exponent,equal?).
View answer and explanation
(2, 5, 2, 2, 4, 4, Yes)
Start each power at the actual multiplicative identity. At exponent zero all powers are the identity. This identity is used under the full commutative-ring premise, not to certify commutativity from one instance. An exact finite or named-formula slice does not automatically establish an arbitrary infinite theorem. The exact result, retaining the requested order, is (2, 5, 2, 2, 4, 4, Yes).
Reviewed step-by-step derivation
Finite commutative ring · qf-mod28-commring-power-product · candidate
53. A Power-of-Product Identity in a Qualified Commutative Ring
Use the same qualified complete finite unital ring on Fin(8)={0,…,7}, with addition=[[0, 1, 2, 3, 4, 5, 6, 7], [1, 2, 3, 4, 6, 7, 5, 0], [2, 3, 4, 6, 5, 0, 7, 1], [3, 4, 6, 5, 7, 1, 0, 2], [4, 6, 5, 7, 0, 2, 1, 3], [5, 7, 0, 1, 2, 4, 3, 6], [6, 5, 7, 0, 1, 3, 2, 4], [7, 0, 1, 2, 3, 6, 4, 5]] and multiplication=[[0, 0, 0, 0, 0, 0, 0, 0], [0, 1, 2, 3, 4, 5, 6, 7], [0, 2, 4, 5, 0, 4, 2, 5], [0, 3, 5, 1, 4, 2, 7, 6], [0, 4, 0, 4, 0, 0, 4, 4], [0, 5, 4, 2, 0, 4, 5, 2], [0, 6, 2, 7, 4, 5, 1, 3], [0, 7, 5, 6, 4, 2, 3, 1]] in label order. All unital-ring axioms, including distinct two-sided identities, must already hold. The identity labels need not be 0 and 1. Commutativity is a separate multiplication obligation; even a commutative ring may have nonzero zero divisors and need not be a field. Require multiplication to commute. a=3, b=0, nonnegative exponent=3. Give (a,b,exponent,ab,lhs=(ab)^exponent,rhs=a^exponent*b^exponent,equal?).
View answer and explanation
(3, 0, 3, 0, 0, 0, Yes)
Start each power at the actual multiplicative identity. At exponent zero all powers are the identity. This identity is used under the full commutative-ring premise, not to certify commutativity from one instance. An exact finite or named-formula slice does not automatically establish an arbitrary infinite theorem. The exact result, retaining the requested order, is (3, 0, 3, 0, 0, 0, Yes).
Reviewed step-by-step derivation
Finite commutative ring · qf-mod28-commring-power-product · candidate
54. A Power-of-Product Identity in a Qualified Commutative Ring
Use the same qualified complete finite unital ring on Fin(8)={0,…,7}, with addition=[[0, 1, 2, 3, 4, 5, 6, 7], [1, 2, 3, 4, 6, 0, 7, 5], [2, 3, 4, 6, 7, 1, 5, 0], [3, 4, 6, 7, 5, 2, 0, 1], [4, 6, 7, 5, 0, 3, 1, 2], [5, 0, 1, 2, 3, 7, 4, 6], [6, 7, 5, 0, 1, 4, 2, 3], [7, 5, 0, 1, 2, 6, 3, 4]] and multiplication=[[0, 0, 0, 0, 0, 0, 0, 0], [0, 1, 2, 3, 4, 5, 6, 7], [0, 2, 4, 7, 0, 7, 2, 4], [0, 3, 7, 1, 4, 6, 5, 2], [0, 4, 0, 4, 0, 4, 4, 0], [0, 5, 7, 6, 4, 1, 3, 2], [0, 6, 2, 5, 4, 3, 1, 7], [0, 7, 4, 2, 0, 2, 7, 4]] in label order. All unital-ring axioms, including distinct two-sided identities, must already hold. The identity labels need not be 0 and 1. Commutativity is a separate multiplication obligation; even a commutative ring may have nonzero zero divisors and need not be a field. Require multiplication to commute. a=4, b=3, nonnegative exponent=4. Give (a,b,exponent,ab,lhs=(ab)^exponent,rhs=a^exponent*b^exponent,equal?).
View answer and explanation
(4, 3, 4, 4, 0, 0, Yes)
Start each power at the actual multiplicative identity. At exponent zero all powers are the identity. This identity is used under the full commutative-ring premise, not to certify commutativity from one instance. An exact finite or named-formula slice does not automatically establish an arbitrary infinite theorem. The exact result, retaining the requested order, is (4, 3, 4, 4, 0, 0, Yes).
Reviewed step-by-step derivation
Finite commutative ring · qf-mod28-commring-power-product · candidate
55. A Power-of-Product Identity in a Qualified Commutative Ring
Use the same qualified complete finite unital ring on Fin(8)={0,…,7}, with addition=[[0, 1, 2, 3, 4, 5, 6, 7], [1, 2, 3, 4, 7, 6, 0, 5], [2, 3, 4, 7, 5, 0, 1, 6], [3, 4, 7, 5, 6, 1, 2, 0], [4, 7, 5, 6, 0, 2, 3, 1], [5, 6, 0, 1, 2, 4, 7, 3], [6, 0, 1, 2, 3, 7, 5, 4], [7, 5, 6, 0, 1, 3, 4, 2]] and multiplication=[[0, 0, 0, 0, 0, 0, 0, 0], [0, 1, 2, 3, 4, 5, 6, 7], [0, 2, 4, 5, 0, 4, 5, 2], [0, 3, 5, 1, 4, 2, 7, 6], [0, 4, 0, 4, 0, 0, 4, 4], [0, 5, 4, 2, 0, 4, 2, 5], [0, 6, 5, 7, 4, 2, 1, 3], [0, 7, 2, 6, 4, 5, 3, 1]] in label order. All unital-ring axioms, including distinct two-sided identities, must already hold. The identity labels need not be 0 and 1. Commutativity is a separate multiplication obligation; even a commutative ring may have nonzero zero divisors and need not be a field. Require multiplication to commute. a=7, b=5, nonnegative exponent=5. Give (a,b,exponent,ab,lhs=(ab)^exponent,rhs=a^exponent*b^exponent,equal?).
View answer and explanation
(7, 5, 5, 5, 0, 0, Yes)
Start each power at the actual multiplicative identity. At exponent zero all powers are the identity. This identity is used under the full commutative-ring premise, not to certify commutativity from one instance. An exact finite or named-formula slice does not automatically establish an arbitrary infinite theorem. The exact result, retaining the requested order, is (7, 5, 5, 5, 0, 0, Yes).
Reviewed step-by-step derivation
Finite commutative ring · qf-mod28-commring-power-product · candidate
56. A Power-of-Product Identity in a Qualified Commutative Ring
Use the same qualified complete finite unital ring on Fin(8)={0,…,7}, with addition=[[0, 1, 2, 3, 4, 5, 6, 7], [1, 2, 3, 4, 7, 0, 5, 6], [2, 3, 4, 7, 6, 1, 0, 5], [3, 4, 7, 6, 5, 2, 1, 0], [4, 7, 6, 5, 0, 3, 2, 1], [5, 0, 1, 2, 3, 6, 7, 4], [6, 5, 0, 1, 2, 7, 4, 3], [7, 6, 5, 0, 1, 4, 3, 2]] and multiplication=[[0, 0, 0, 0, 0, 0, 0, 0], [0, 1, 2, 3, 4, 5, 6, 7], [0, 2, 4, 6, 0, 6, 4, 2], [0, 3, 6, 1, 4, 7, 2, 5], [0, 4, 0, 4, 0, 4, 0, 4], [0, 5, 6, 7, 4, 1, 2, 3], [0, 6, 4, 2, 0, 2, 4, 6], [0, 7, 2, 5, 4, 3, 6, 1]] in label order. All unital-ring axioms, including distinct two-sided identities, must already hold. The identity labels need not be 0 and 1. Commutativity is a separate multiplication obligation; even a commutative ring may have nonzero zero divisors and need not be a field. Require multiplication to commute. a=6, b=1, nonnegative exponent=6. Give (a,b,exponent,ab,lhs=(ab)^exponent,rhs=a^exponent*b^exponent,equal?).
View answer and explanation
(6, 1, 6, 6, 0, 0, Yes)
Start each power at the actual multiplicative identity. At exponent zero all powers are the identity. This identity is used under the full commutative-ring premise, not to certify commutativity from one instance. An exact finite or named-formula slice does not automatically establish an arbitrary infinite theorem. The exact result, retaining the requested order, is (6, 1, 6, 6, 0, 0, Yes).
Reviewed step-by-step derivation
Finite commutative ring · qf-mod28-commring-power-product · candidate
57. A Power-of-Product Identity in a Qualified Commutative Ring
Use the same qualified complete finite unital ring on Fin(8)={0,…,7}, with addition=[[0, 1, 2, 3, 4, 5, 6, 7], [1, 2, 3, 5, 6, 4, 7, 0], [2, 3, 5, 4, 7, 6, 0, 1], [3, 5, 4, 6, 0, 7, 1, 2], [4, 6, 7, 0, 2, 1, 3, 5], [5, 4, 6, 7, 1, 0, 2, 3], [6, 7, 0, 1, 3, 2, 5, 4], [7, 0, 1, 2, 5, 3, 4, 6]] and multiplication=[[0, 0, 0, 0, 0, 0, 0, 0], [0, 1, 2, 3, 4, 5, 6, 7], [0, 2, 5, 6, 2, 0, 5, 6], [0, 3, 6, 1, 7, 5, 2, 4], [0, 4, 2, 7, 1, 5, 6, 3], [0, 5, 0, 5, 5, 0, 0, 5], [0, 6, 5, 2, 6, 0, 5, 2], [0, 7, 6, 4, 3, 5, 2, 1]] in label order. All unital-ring axioms, including distinct two-sided identities, must already hold. The identity labels need not be 0 and 1. Commutativity is a separate multiplication obligation; even a commutative ring may have nonzero zero divisors and need not be a field. Require multiplication to commute. a=7, b=5, nonnegative exponent=7. Give (a,b,exponent,ab,lhs=(ab)^exponent,rhs=a^exponent*b^exponent,equal?).
View answer and explanation
(7, 5, 7, 5, 0, 0, Yes)
Start each power at the actual multiplicative identity. At exponent zero all powers are the identity. This identity is used under the full commutative-ring premise, not to certify commutativity from one instance. An exact finite or named-formula slice does not automatically establish an arbitrary infinite theorem. The exact result, retaining the requested order, is (7, 5, 7, 5, 0, 0, Yes).
Reviewed step-by-step derivation
Finite commutative ring · qf-mod28-commring-power-product · candidate
58. A Power-of-Product Identity in a Qualified Commutative Ring
Use the same qualified complete finite unital ring on Fin(8)={0,…,7}, with addition=[[0, 1, 2, 3, 4, 5, 6, 7], [1, 2, 3, 5, 7, 4, 0, 6], [2, 3, 5, 4, 6, 7, 1, 0], [3, 5, 4, 7, 0, 6, 2, 1], [4, 7, 6, 0, 2, 1, 5, 3], [5, 4, 7, 6, 1, 0, 3, 2], [6, 0, 1, 2, 5, 3, 7, 4], [7, 6, 0, 1, 3, 2, 4, 5]] and multiplication=[[0, 0, 0, 0, 0, 0, 0, 0], [0, 1, 2, 3, 4, 5, 6, 7], [0, 2, 5, 7, 2, 0, 7, 5], [0, 3, 7, 1, 6, 5, 4, 2], [0, 4, 2, 6, 1, 5, 3, 7], [0, 5, 0, 5, 5, 0, 5, 0], [0, 6, 7, 4, 3, 5, 1, 2], [0, 7, 5, 2, 7, 0, 2, 5]] in label order. All unital-ring axioms, including distinct two-sided identities, must already hold. The identity labels need not be 0 and 1. Commutativity is a separate multiplication obligation; even a commutative ring may have nonzero zero divisors and need not be a field. Require multiplication to commute. a=0, b=6, nonnegative exponent=1. Give (a,b,exponent,ab,lhs=(ab)^exponent,rhs=a^exponent*b^exponent,equal?).
View answer and explanation
(0, 6, 1, 0, 0, 0, Yes)
Start each power at the actual multiplicative identity. At exponent zero all powers are the identity. This identity is used under the full commutative-ring premise, not to certify commutativity from one instance. An exact finite or named-formula slice does not automatically establish an arbitrary infinite theorem. The exact result, retaining the requested order, is (0, 6, 1, 0, 0, 0, Yes).
Reviewed step-by-step derivation
Finite commutative ring · qf-mod28-commring-power-product · candidate
59. A Power-of-Product Identity in a Qualified Commutative Ring
Use the same qualified complete finite unital ring on Fin(8)={0,…,7}, with addition=[[0, 1, 2, 3, 4, 5, 6, 7], [1, 2, 3, 5, 7, 6, 4, 0], [2, 3, 5, 6, 0, 4, 7, 1], [3, 5, 6, 4, 1, 7, 0, 2], [4, 7, 0, 1, 5, 2, 3, 6], [5, 6, 4, 7, 2, 0, 1, 3], [6, 4, 7, 0, 3, 1, 2, 5], [7, 0, 1, 2, 6, 3, 5, 4]] and multiplication=[[0, 0, 0, 0, 0, 0, 0, 0], [0, 1, 2, 3, 4, 5, 6, 7], [0, 2, 5, 4, 5, 0, 2, 4], [0, 3, 4, 1, 2, 5, 7, 6], [0, 4, 5, 2, 5, 0, 4, 2], [0, 5, 0, 5, 0, 0, 5, 5], [0, 6, 2, 7, 4, 5, 1, 3], [0, 7, 4, 6, 2, 5, 3, 1]] in label order. All unital-ring axioms, including distinct two-sided identities, must already hold. The identity labels need not be 0 and 1. Commutativity is a separate multiplication obligation; even a commutative ring may have nonzero zero divisors and need not be a field. Require multiplication to commute. a=1, b=2, nonnegative exponent=2. Give (a,b,exponent,ab,lhs=(ab)^exponent,rhs=a^exponent*b^exponent,equal?).
View answer and explanation
(1, 2, 2, 2, 5, 5, Yes)
Start each power at the actual multiplicative identity. At exponent zero all powers are the identity. This identity is used under the full commutative-ring premise, not to certify commutativity from one instance. An exact finite or named-formula slice does not automatically establish an arbitrary infinite theorem. The exact result, retaining the requested order, is (1, 2, 2, 2, 5, 5, Yes).
Reviewed step-by-step derivation
Finite commutative ring · qf-mod28-commring-power-product · candidate
60. A Power-of-Product Identity in a Qualified Commutative Ring
Use the same qualified complete finite unital ring on Fin(8)={0,…,7}, with addition=[[0, 1, 2, 3, 4, 5, 6, 7], [1, 2, 3, 5, 0, 6, 7, 4], [2, 3, 5, 6, 1, 7, 4, 0], [3, 5, 6, 7, 2, 4, 0, 1], [4, 0, 1, 2, 7, 3, 5, 6], [5, 6, 7, 4, 3, 0, 1, 2], [6, 7, 4, 0, 5, 1, 2, 3], [7, 4, 0, 1, 6, 2, 3, 5]] and multiplication=[[0, 0, 0, 0, 0, 0, 0, 0], [0, 1, 2, 3, 4, 5, 6, 7], [0, 2, 5, 7, 7, 0, 2, 5], [0, 3, 7, 1, 6, 5, 4, 2], [0, 4, 7, 6, 1, 5, 3, 2], [0, 5, 0, 5, 5, 0, 5, 0], [0, 6, 2, 4, 3, 5, 1, 7], [0, 7, 5, 2, 2, 0, 7, 5]] in label order. All unital-ring axioms, including distinct two-sided identities, must already hold. The identity labels need not be 0 and 1. Commutativity is a separate multiplication obligation; even a commutative ring may have nonzero zero divisors and need not be a field. Require multiplication to commute. a=2, b=6, nonnegative exponent=3. Give (a,b,exponent,ab,lhs=(ab)^exponent,rhs=a^exponent*b^exponent,equal?).
View answer and explanation
(2, 6, 3, 2, 0, 0, Yes)
Start each power at the actual multiplicative identity. At exponent zero all powers are the identity. This identity is used under the full commutative-ring premise, not to certify commutativity from one instance. An exact finite or named-formula slice does not automatically establish an arbitrary infinite theorem. The exact result, retaining the requested order, is (2, 6, 3, 2, 0, 0, Yes).
Reviewed step-by-step derivation
Finite commutative ring · qf-mod28-commring-power-product · candidate
61. A Power-of-Product Identity in a Qualified Commutative Ring
Use the same qualified complete finite unital ring on Fin(8)={0,…,7}, with addition=[[0, 1, 2, 3, 4, 5, 6, 7], [1, 2, 3, 5, 6, 7, 0, 4], [2, 3, 5, 7, 0, 4, 1, 6], [3, 5, 7, 4, 1, 6, 2, 0], [4, 6, 0, 1, 5, 2, 7, 3], [5, 7, 4, 6, 2, 0, 3, 1], [6, 0, 1, 2, 7, 3, 4, 5], [7, 4, 6, 0, 3, 1, 5, 2]] and multiplication=[[0, 0, 0, 0, 0, 0, 0, 0], [0, 1, 2, 3, 4, 5, 6, 7], [0, 2, 5, 4, 5, 0, 4, 2], [0, 3, 4, 1, 2, 5, 7, 6], [0, 4, 5, 2, 5, 0, 2, 4], [0, 5, 0, 5, 0, 0, 5, 5], [0, 6, 4, 7, 2, 5, 1, 3], [0, 7, 2, 6, 4, 5, 3, 1]] in label order. All unital-ring axioms, including distinct two-sided identities, must already hold. The identity labels need not be 0 and 1. Commutativity is a separate multiplication obligation; even a commutative ring may have nonzero zero divisors and need not be a field. Require multiplication to commute. a=3, b=0, nonnegative exponent=4. Give (a,b,exponent,ab,lhs=(ab)^exponent,rhs=a^exponent*b^exponent,equal?).
View answer and explanation
(3, 0, 4, 0, 0, 0, Yes)
Start each power at the actual multiplicative identity. At exponent zero all powers are the identity. This identity is used under the full commutative-ring premise, not to certify commutativity from one instance. An exact finite or named-formula slice does not automatically establish an arbitrary infinite theorem. The exact result, retaining the requested order, is (3, 0, 4, 0, 0, 0, Yes).
Reviewed step-by-step derivation
Finite commutative ring · qf-mod28-commring-power-product · candidate
62. A Power-of-Product Identity in a Qualified Commutative Ring
Use the same qualified complete finite unital ring on Fin(8)={0,…,7}, with addition=[[0, 1, 2, 3, 4, 5, 6, 7], [1, 2, 3, 5, 0, 7, 4, 6], [2, 3, 5, 7, 1, 6, 0, 4], [3, 5, 7, 6, 2, 4, 1, 0], [4, 0, 1, 2, 6, 3, 7, 5], [5, 7, 6, 4, 3, 0, 2, 1], [6, 4, 0, 1, 7, 2, 5, 3], [7, 6, 4, 0, 5, 1, 3, 2]] and multiplication=[[0, 0, 0, 0, 0, 0, 0, 0], [0, 1, 2, 3, 4, 5, 6, 7], [0, 2, 5, 6, 6, 0, 5, 2], [0, 3, 6, 1, 7, 5, 2, 4], [0, 4, 6, 7, 1, 5, 2, 3], [0, 5, 0, 5, 5, 0, 0, 5], [0, 6, 5, 2, 2, 0, 5, 6], [0, 7, 2, 4, 3, 5, 6, 1]] in label order. All unital-ring axioms, including distinct two-sided identities, must already hold. The identity labels need not be 0 and 1. Commutativity is a separate multiplication obligation; even a commutative ring may have nonzero zero divisors and need not be a field. Require multiplication to commute. a=5, b=3, nonnegative exponent=5. Give (a,b,exponent,ab,lhs=(ab)^exponent,rhs=a^exponent*b^exponent,equal?).
View answer and explanation
(5, 3, 5, 5, 0, 0, Yes)
Start each power at the actual multiplicative identity. At exponent zero all powers are the identity. This identity is used under the full commutative-ring premise, not to certify commutativity from one instance. An exact finite or named-formula slice does not automatically establish an arbitrary infinite theorem. The exact result, retaining the requested order, is (5, 3, 5, 5, 0, 0, Yes).
Reviewed step-by-step derivation
Finite commutative ring · qf-mod28-commring-power-product · candidate
63. A Power-of-Product Identity in a Qualified Commutative Ring
Use the same qualified complete finite unital ring on Fin(8)={0,…,7}, with addition=[[0, 1, 2, 3, 4, 5, 6, 7], [1, 2, 3, 6, 5, 7, 4, 0], [2, 3, 6, 4, 7, 0, 5, 1], [3, 6, 4, 5, 0, 1, 7, 2], [4, 5, 7, 0, 2, 3, 1, 6], [5, 7, 0, 1, 3, 6, 2, 4], [6, 4, 5, 7, 1, 2, 0, 3], [7, 0, 1, 2, 6, 4, 3, 5]] and multiplication=[[0, 0, 0, 0, 0, 0, 0, 0], [0, 1, 2, 3, 4, 5, 6, 7], [0, 2, 6, 5, 2, 6, 0, 5], [0, 3, 5, 1, 7, 2, 6, 4], [0, 4, 2, 7, 1, 5, 6, 3], [0, 5, 6, 2, 5, 6, 0, 2], [0, 6, 0, 6, 6, 0, 0, 6], [0, 7, 5, 4, 3, 2, 6, 1]] in label order. All unital-ring axioms, including distinct two-sided identities, must already hold. The identity labels need not be 0 and 1. Commutativity is a separate multiplication obligation; even a commutative ring may have nonzero zero divisors and need not be a field. Require multiplication to commute. a=4, b=5, nonnegative exponent=6. Give (a,b,exponent,ab,lhs=(ab)^exponent,rhs=a^exponent*b^exponent,equal?).
View answer and explanation
(4, 5, 6, 5, 0, 0, Yes)
Start each power at the actual multiplicative identity. At exponent zero all powers are the identity. This identity is used under the full commutative-ring premise, not to certify commutativity from one instance. An exact finite or named-formula slice does not automatically establish an arbitrary infinite theorem. The exact result, retaining the requested order, is (4, 5, 6, 5, 0, 0, Yes).
Reviewed step-by-step derivation
Finite commutative ring · qf-mod28-commring-power-product · candidate
64. A Power-of-Product Identity in a Qualified Commutative Ring
Use the same qualified complete finite unital ring on Fin(8)={0,…,7}, with addition=[[0, 1, 2, 3, 4, 5, 6, 7], [1, 2, 3, 6, 7, 0, 4, 5], [2, 3, 6, 4, 5, 1, 7, 0], [3, 6, 4, 7, 0, 2, 5, 1], [4, 7, 5, 0, 2, 6, 1, 3], [5, 0, 1, 2, 6, 7, 3, 4], [6, 4, 7, 5, 1, 3, 0, 2], [7, 5, 0, 1, 3, 4, 2, 6]] and multiplication=[[0, 0, 0, 0, 0, 0, 0, 0], [0, 1, 2, 3, 4, 5, 6, 7], [0, 2, 6, 7, 2, 7, 0, 6], [0, 3, 7, 1, 5, 4, 6, 2], [0, 4, 2, 5, 1, 3, 6, 7], [0, 5, 7, 4, 3, 1, 6, 2], [0, 6, 0, 6, 6, 6, 0, 0], [0, 7, 6, 2, 7, 2, 0, 6]] in label order. All unital-ring axioms, including distinct two-sided identities, must already hold. The identity labels need not be 0 and 1. Commutativity is a separate multiplication obligation; even a commutative ring may have nonzero zero divisors and need not be a field. Require multiplication to commute. a=7, b=1, nonnegative exponent=7. Give (a,b,exponent,ab,lhs=(ab)^exponent,rhs=a^exponent*b^exponent,equal?).
View answer and explanation
(7, 1, 7, 7, 0, 0, Yes)
Start each power at the actual multiplicative identity. At exponent zero all powers are the identity. This identity is used under the full commutative-ring premise, not to certify commutativity from one instance. An exact finite or named-formula slice does not automatically establish an arbitrary infinite theorem. The exact result, retaining the requested order, is (7, 1, 7, 7, 0, 0, Yes).
Reviewed step-by-step derivation
Finite commutative ring · qf-mod28-commring-power-product · candidate
65. A Power-of-Product Identity in a Qualified Commutative Ring
Use the same qualified complete finite unital ring on Fin(8)={0,…,7}, with addition=[[0, 1, 2, 3, 4, 5, 6, 7], [1, 2, 3, 6, 7, 4, 5, 0], [2, 3, 6, 5, 0, 7, 4, 1], [3, 6, 5, 4, 1, 0, 7, 2], [4, 7, 0, 1, 6, 3, 2, 5], [5, 4, 7, 0, 3, 2, 1, 6], [6, 5, 4, 7, 2, 1, 0, 3], [7, 0, 1, 2, 5, 6, 3, 4]] and multiplication=[[0, 0, 0, 0, 0, 0, 0, 0], [0, 1, 2, 3, 4, 5, 6, 7], [0, 2, 6, 4, 6, 2, 0, 4], [0, 3, 4, 1, 2, 7, 6, 5], [0, 4, 6, 2, 6, 4, 0, 2], [0, 5, 2, 7, 4, 1, 6, 3], [0, 6, 0, 6, 0, 6, 0, 6], [0, 7, 4, 5, 2, 3, 6, 1]] in label order. All unital-ring axioms, including distinct two-sided identities, must already hold. The identity labels need not be 0 and 1. Commutativity is a separate multiplication obligation; even a commutative ring may have nonzero zero divisors and need not be a field. Require multiplication to commute. a=7, b=6, nonnegative exponent=1. Give (a,b,exponent,ab,lhs=(ab)^exponent,rhs=a^exponent*b^exponent,equal?).
View answer and explanation
(7, 6, 1, 6, 6, 6, Yes)
Start each power at the actual multiplicative identity. At exponent zero all powers are the identity. This identity is used under the full commutative-ring premise, not to certify commutativity from one instance. An exact finite or named-formula slice does not automatically establish an arbitrary infinite theorem. The exact result, retaining the requested order, is (7, 6, 1, 6, 6, 6, Yes).
Reviewed step-by-step derivation
Finite commutative ring · qf-mod28-commring-power-product · candidate
66. A Power-of-Product Identity in a Qualified Commutative Ring
Use the same qualified complete finite unital ring on Fin(8)={0,…,7}, with addition=[[0, 1, 2, 3, 4, 5, 6, 7], [1, 2, 3, 6, 0, 7, 5, 4], [2, 3, 6, 5, 1, 4, 7, 0], [3, 6, 5, 7, 2, 0, 4, 1], [4, 0, 1, 2, 7, 6, 3, 5], [5, 7, 4, 0, 6, 2, 1, 3], [6, 5, 7, 4, 3, 1, 0, 2], [7, 4, 0, 1, 5, 3, 2, 6]] and multiplication=[[0, 0, 0, 0, 0, 0, 0, 0], [0, 1, 2, 3, 4, 5, 6, 7], [0, 2, 6, 7, 7, 2, 0, 6], [0, 3, 7, 1, 5, 4, 6, 2], [0, 4, 7, 5, 1, 3, 6, 2], [0, 5, 2, 4, 3, 1, 6, 7], [0, 6, 0, 6, 6, 6, 0, 0], [0, 7, 6, 2, 2, 7, 0, 6]] in label order. All unital-ring axioms, including distinct two-sided identities, must already hold. The identity labels need not be 0 and 1. Commutativity is a separate multiplication obligation; even a commutative ring may have nonzero zero divisors and need not be a field. Require multiplication to commute. a=0, b=4, nonnegative exponent=2. Give (a,b,exponent,ab,lhs=(ab)^exponent,rhs=a^exponent*b^exponent,equal?).
View answer and explanation
(0, 4, 2, 0, 0, 0, Yes)
Start each power at the actual multiplicative identity. At exponent zero all powers are the identity. This identity is used under the full commutative-ring premise, not to certify commutativity from one instance. An exact finite or named-formula slice does not automatically establish an arbitrary infinite theorem. The exact result, retaining the requested order, is (0, 4, 2, 0, 0, 0, Yes).
Reviewed step-by-step derivation
Finite commutative ring · qf-mod28-commring-power-product · candidate
67. A Power-of-Product Identity in a Qualified Commutative Ring
Use the same qualified complete finite unital ring on Fin(8)={0,…,7}, with addition=[[0, 1, 2, 3, 4, 5, 6, 7], [1, 2, 3, 6, 5, 0, 7, 4], [2, 3, 6, 7, 0, 1, 4, 5], [3, 6, 7, 4, 1, 2, 5, 0], [4, 5, 0, 1, 6, 7, 2, 3], [5, 0, 1, 2, 7, 4, 3, 6], [6, 7, 4, 5, 2, 3, 0, 1], [7, 4, 5, 0, 3, 6, 1, 2]] and multiplication=[[0, 0, 0, 0, 0, 0, 0, 0], [0, 1, 2, 3, 4, 5, 6, 7], [0, 2, 6, 4, 6, 4, 0, 2], [0, 3, 4, 1, 2, 7, 6, 5], [0, 4, 6, 2, 6, 2, 0, 4], [0, 5, 4, 7, 2, 1, 6, 3], [0, 6, 0, 6, 0, 6, 0, 6], [0, 7, 2, 5, 4, 3, 6, 1]] in label order. All unital-ring axioms, including distinct two-sided identities, must already hold. The identity labels need not be 0 and 1. Commutativity is a separate multiplication obligation; even a commutative ring may have nonzero zero divisors and need not be a field. Require multiplication to commute. a=1, b=2, nonnegative exponent=3. Give (a,b,exponent,ab,lhs=(ab)^exponent,rhs=a^exponent*b^exponent,equal?).
View answer and explanation
(1, 2, 3, 2, 0, 0, Yes)
Start each power at the actual multiplicative identity. At exponent zero all powers are the identity. This identity is used under the full commutative-ring premise, not to certify commutativity from one instance. An exact finite or named-formula slice does not automatically establish an arbitrary infinite theorem. The exact result, retaining the requested order, is (1, 2, 3, 2, 0, 0, Yes).
Reviewed step-by-step derivation
Finite commutative ring · qf-mod28-commring-power-product · candidate
68. A Power-of-Product Identity in a Qualified Commutative Ring
Use the same qualified complete finite unital ring on Fin(8)={0,…,7}, with addition=[[0, 1, 2, 3, 4, 5, 6, 7], [1, 2, 3, 6, 0, 4, 7, 5], [2, 3, 6, 7, 1, 0, 5, 4], [3, 6, 7, 5, 2, 1, 4, 0], [4, 0, 1, 2, 5, 7, 3, 6], [5, 4, 0, 1, 7, 6, 2, 3], [6, 7, 5, 4, 3, 2, 0, 1], [7, 5, 4, 0, 6, 3, 1, 2]] and multiplication=[[0, 0, 0, 0, 0, 0, 0, 0], [0, 1, 2, 3, 4, 5, 6, 7], [0, 2, 6, 5, 5, 6, 0, 2], [0, 3, 5, 1, 7, 2, 6, 4], [0, 4, 5, 7, 1, 2, 6, 3], [0, 5, 6, 2, 2, 6, 0, 5], [0, 6, 0, 6, 6, 0, 0, 6], [0, 7, 2, 4, 3, 5, 6, 1]] in label order. All unital-ring axioms, including distinct two-sided identities, must already hold. The identity labels need not be 0 and 1. Commutativity is a separate multiplication obligation; even a commutative ring may have nonzero zero divisors and need not be a field. Require multiplication to commute. a=2, b=7, nonnegative exponent=4. Give (a,b,exponent,ab,lhs=(ab)^exponent,rhs=a^exponent*b^exponent,equal?).
View answer and explanation
(2, 7, 4, 2, 0, 0, Yes)
Start each power at the actual multiplicative identity. At exponent zero all powers are the identity. This identity is used under the full commutative-ring premise, not to certify commutativity from one instance. An exact finite or named-formula slice does not automatically establish an arbitrary infinite theorem. The exact result, retaining the requested order, is (2, 7, 4, 2, 0, 0, Yes).
Reviewed step-by-step derivation
Finite commutative ring · qf-mod28-commring-power-product · candidate
69. A Power-of-Product Identity in a Qualified Commutative Ring
Use the same qualified complete finite unital ring on Fin(8)={0,…,7}, with addition=[[0, 1, 2, 3, 4, 5, 6, 7], [1, 2, 3, 7, 5, 6, 0, 4], [2, 3, 7, 4, 6, 0, 1, 5], [3, 7, 4, 5, 0, 1, 2, 6], [4, 5, 6, 0, 2, 3, 7, 1], [5, 6, 0, 1, 3, 7, 4, 2], [6, 0, 1, 2, 7, 4, 5, 3], [7, 4, 5, 6, 1, 2, 3, 0]] and multiplication=[[0, 0, 0, 0, 0, 0, 0, 0], [0, 1, 2, 3, 4, 5, 6, 7], [0, 2, 7, 5, 2, 7, 5, 0], [0, 3, 5, 1, 6, 2, 4, 7], [0, 4, 2, 6, 1, 5, 3, 7], [0, 5, 7, 2, 5, 7, 2, 0], [0, 6, 5, 4, 3, 2, 1, 7], [0, 7, 0, 7, 7, 0, 7, 0]] in label order. All unital-ring axioms, including distinct two-sided identities, must already hold. The identity labels need not be 0 and 1. Commutativity is a separate multiplication obligation; even a commutative ring may have nonzero zero divisors and need not be a field. Require multiplication to commute. a=3, b=0, nonnegative exponent=5. Give (a,b,exponent,ab,lhs=(ab)^exponent,rhs=a^exponent*b^exponent,equal?).
View answer and explanation
(3, 0, 5, 0, 0, 0, Yes)
Start each power at the actual multiplicative identity. At exponent zero all powers are the identity. This identity is used under the full commutative-ring premise, not to certify commutativity from one instance. An exact finite or named-formula slice does not automatically establish an arbitrary infinite theorem. The exact result, retaining the requested order, is (3, 0, 5, 0, 0, 0, Yes).
Reviewed step-by-step derivation
Finite commutative ring · qf-mod28-commring-power-product · candidate
70. A Power-of-Product Identity in a Qualified Commutative Ring
Use the same qualified complete finite unital ring on Fin(8)={0,…,7}, with addition=[[0, 1, 2, 3, 4, 5, 6, 7], [1, 2, 3, 7, 6, 0, 5, 4], [2, 3, 7, 4, 5, 1, 0, 6], [3, 7, 4, 6, 0, 2, 1, 5], [4, 6, 5, 0, 2, 7, 3, 1], [5, 0, 1, 2, 7, 6, 4, 3], [6, 5, 0, 1, 3, 4, 7, 2], [7, 4, 6, 5, 1, 3, 2, 0]] and multiplication=[[0, 0, 0, 0, 0, 0, 0, 0], [0, 1, 2, 3, 4, 5, 6, 7], [0, 2, 7, 6, 2, 6, 7, 0], [0, 3, 6, 1, 5, 4, 2, 7], [0, 4, 2, 5, 1, 3, 6, 7], [0, 5, 6, 4, 3, 1, 2, 7], [0, 6, 7, 2, 6, 2, 7, 0], [0, 7, 0, 7, 7, 7, 0, 0]] in label order. All unital-ring axioms, including distinct two-sided identities, must already hold. The identity labels need not be 0 and 1. Commutativity is a separate multiplication obligation; even a commutative ring may have nonzero zero divisors and need not be a field. Require multiplication to commute. a=7, b=3, nonnegative exponent=6. Give (a,b,exponent,ab,lhs=(ab)^exponent,rhs=a^exponent*b^exponent,equal?).
View answer and explanation
(7, 3, 6, 7, 0, 0, Yes)
Start each power at the actual multiplicative identity. At exponent zero all powers are the identity. This identity is used under the full commutative-ring premise, not to certify commutativity from one instance. An exact finite or named-formula slice does not automatically establish an arbitrary infinite theorem. The exact result, retaining the requested order, is (7, 3, 6, 7, 0, 0, Yes).
Reviewed step-by-step derivation
Finite commutative ring · qf-mod28-commring-power-product · candidate
71. A Power-of-Product Identity in a Qualified Commutative Ring
Use the same qualified complete finite unital ring on Fin(8)={0,…,7}, with addition=[[0, 1, 2, 3, 4, 5, 6, 7], [1, 2, 3, 7, 6, 4, 0, 5], [2, 3, 7, 5, 0, 6, 1, 4], [3, 7, 5, 4, 1, 0, 2, 6], [4, 6, 0, 1, 7, 3, 5, 2], [5, 4, 6, 0, 3, 2, 7, 1], [6, 0, 1, 2, 5, 7, 4, 3], [7, 5, 4, 6, 2, 1, 3, 0]] and multiplication=[[0, 0, 0, 0, 0, 0, 0, 0], [0, 1, 2, 3, 4, 5, 6, 7], [0, 2, 7, 4, 7, 2, 4, 0], [0, 3, 4, 1, 2, 6, 5, 7], [0, 4, 7, 2, 7, 4, 2, 0], [0, 5, 2, 6, 4, 1, 3, 7], [0, 6, 4, 5, 2, 3, 1, 7], [0, 7, 0, 7, 0, 7, 7, 0]] in label order. All unital-ring axioms, including distinct two-sided identities, must already hold. The identity labels need not be 0 and 1. Commutativity is a separate multiplication obligation; even a commutative ring may have nonzero zero divisors and need not be a field. Require multiplication to commute. a=5, b=4, nonnegative exponent=7. Give (a,b,exponent,ab,lhs=(ab)^exponent,rhs=a^exponent*b^exponent,equal?).
View answer and explanation
(5, 4, 7, 4, 0, 0, Yes)
Start each power at the actual multiplicative identity. At exponent zero all powers are the identity. This identity is used under the full commutative-ring premise, not to certify commutativity from one instance. An exact finite or named-formula slice does not automatically establish an arbitrary infinite theorem. The exact result, retaining the requested order, is (5, 4, 7, 4, 0, 0, Yes).
Reviewed step-by-step derivation
Finite commutative ring · qf-mod28-commring-power-product · candidate
72. A Power-of-Product Identity in a Qualified Commutative Ring
Use the same qualified complete finite unital ring on Fin(8)={0,…,7}, with addition=[[0, 1, 2, 3, 4, 5, 6, 7], [1, 2, 3, 7, 0, 6, 4, 5], [2, 3, 7, 5, 1, 4, 0, 6], [3, 7, 5, 6, 2, 0, 1, 4], [4, 0, 1, 2, 6, 7, 5, 3], [5, 6, 4, 0, 7, 2, 3, 1], [6, 4, 0, 1, 5, 3, 7, 2], [7, 5, 6, 4, 3, 1, 2, 0]] and multiplication=[[0, 0, 0, 0, 0, 0, 0, 0], [0, 1, 2, 3, 4, 5, 6, 7], [0, 2, 7, 6, 6, 2, 7, 0], [0, 3, 6, 1, 5, 4, 2, 7], [0, 4, 6, 5, 1, 3, 2, 7], [0, 5, 2, 4, 3, 1, 6, 7], [0, 6, 7, 2, 2, 6, 7, 0], [0, 7, 0, 7, 7, 7, 0, 0]] in label order. All unital-ring axioms, including distinct two-sided identities, must already hold. The identity labels need not be 0 and 1. Commutativity is a separate multiplication obligation; even a commutative ring may have nonzero zero divisors and need not be a field. Require multiplication to commute. a=6, b=1, nonnegative exponent=1. Give (a,b,exponent,ab,lhs=(ab)^exponent,rhs=a^exponent*b^exponent,equal?).
View answer and explanation
(6, 1, 1, 6, 6, 6, Yes)
Start each power at the actual multiplicative identity. At exponent zero all powers are the identity. This identity is used under the full commutative-ring premise, not to certify commutativity from one instance. An exact finite or named-formula slice does not automatically establish an arbitrary infinite theorem. The exact result, retaining the requested order, is (6, 1, 1, 6, 6, 6, Yes).
Reviewed step-by-step derivation
Finite commutative ring · qf-mod28-commring-power-product · candidate
73. A Power-of-Product Identity in a Qualified Commutative Ring
Use the same qualified complete finite unital ring on Fin(8)={0,…,7}, with addition=[[0, 1, 2, 3, 4, 5, 6, 7], [1, 2, 3, 7, 5, 0, 4, 6], [2, 3, 7, 6, 0, 1, 5, 4], [3, 7, 6, 4, 1, 2, 0, 5], [4, 5, 0, 1, 7, 6, 3, 2], [5, 0, 1, 2, 6, 4, 7, 3], [6, 4, 5, 0, 3, 7, 2, 1], [7, 6, 4, 5, 2, 3, 1, 0]] and multiplication=[[0, 0, 0, 0, 0, 0, 0, 0], [0, 1, 2, 3, 4, 5, 6, 7], [0, 2, 7, 4, 7, 4, 2, 0], [0, 3, 4, 1, 2, 6, 5, 7], [0, 4, 7, 2, 7, 2, 4, 0], [0, 5, 4, 6, 2, 1, 3, 7], [0, 6, 2, 5, 4, 3, 1, 7], [0, 7, 0, 7, 0, 7, 7, 0]] in label order. All unital-ring axioms, including distinct two-sided identities, must already hold. The identity labels need not be 0 and 1. Commutativity is a separate multiplication obligation; even a commutative ring may have nonzero zero divisors and need not be a field. Require multiplication to commute. a=5, b=7, nonnegative exponent=2. Give (a,b,exponent,ab,lhs=(ab)^exponent,rhs=a^exponent*b^exponent,equal?).
View answer and explanation
(5, 7, 2, 7, 0, 0, Yes)
Start each power at the actual multiplicative identity. At exponent zero all powers are the identity. This identity is used under the full commutative-ring premise, not to certify commutativity from one instance. An exact finite or named-formula slice does not automatically establish an arbitrary infinite theorem. The exact result, retaining the requested order, is (5, 7, 2, 7, 0, 0, Yes).
Reviewed step-by-step derivation
Finite commutative ring · qf-mod28-commring-power-product · candidate
74. A Power-of-Product Identity in a Qualified Commutative Ring
Use the same qualified complete finite unital ring on Fin(8)={0,…,7}, with addition=[[0, 1, 2, 3, 4, 5, 6, 7], [1, 2, 3, 7, 0, 4, 5, 6], [2, 3, 7, 6, 1, 0, 4, 5], [3, 7, 6, 5, 2, 1, 0, 4], [4, 0, 1, 2, 5, 6, 7, 3], [5, 4, 0, 1, 6, 7, 3, 2], [6, 5, 4, 0, 7, 3, 2, 1], [7, 6, 5, 4, 3, 2, 1, 0]] and multiplication=[[0, 0, 0, 0, 0, 0, 0, 0], [0, 1, 2, 3, 4, 5, 6, 7], [0, 2, 7, 5, 5, 7, 2, 0], [0, 3, 5, 1, 6, 2, 4, 7], [0, 4, 5, 6, 1, 2, 3, 7], [0, 5, 7, 2, 2, 7, 5, 0], [0, 6, 2, 4, 3, 5, 1, 7], [0, 7, 0, 7, 7, 0, 7, 0]] in label order. All unital-ring axioms, including distinct two-sided identities, must already hold. The identity labels need not be 0 and 1. Commutativity is a separate multiplication obligation; even a commutative ring may have nonzero zero divisors and need not be a field. Require multiplication to commute. a=0, b=4, nonnegative exponent=3. Give (a,b,exponent,ab,lhs=(ab)^exponent,rhs=a^exponent*b^exponent,equal?).
View answer and explanation
(0, 4, 3, 0, 0, 0, Yes)
Start each power at the actual multiplicative identity. At exponent zero all powers are the identity. This identity is used under the full commutative-ring premise, not to certify commutativity from one instance. An exact finite or named-formula slice does not automatically establish an arbitrary infinite theorem. The exact result, retaining the requested order, is (0, 4, 3, 0, 0, 0, Yes).
Reviewed step-by-step derivation
Finite commutative ring · qf-mod28-commring-power-product · candidate
75. A Power-of-Product Identity in a Qualified Commutative Ring
Use the same qualified complete finite unital ring on Fin(8)={0,…,7}, with addition=[[0, 1, 2, 3, 4, 5, 6, 7], [1, 2, 4, 5, 3, 6, 7, 0], [2, 4, 3, 6, 5, 7, 0, 1], [3, 5, 6, 0, 7, 1, 2, 4], [4, 3, 5, 7, 6, 0, 1, 2], [5, 6, 7, 1, 0, 2, 4, 3], [6, 7, 0, 2, 1, 4, 3, 5], [7, 0, 1, 4, 2, 3, 5, 6]] and multiplication=[[0, 0, 0, 0, 0, 0, 0, 0], [0, 1, 2, 3, 4, 5, 6, 7], [0, 2, 3, 0, 6, 2, 3, 6], [0, 3, 0, 0, 3, 3, 0, 3], [0, 4, 6, 3, 1, 7, 2, 5], [0, 5, 2, 3, 7, 1, 6, 4], [0, 6, 3, 0, 2, 6, 3, 2], [0, 7, 6, 3, 5, 4, 2, 1]] in label order. All unital-ring axioms, including distinct two-sided identities, must already hold. The identity labels need not be 0 and 1. Commutativity is a separate multiplication obligation; even a commutative ring may have nonzero zero divisors and need not be a field. Require multiplication to commute. a=1, b=2, nonnegative exponent=4. Give (a,b,exponent,ab,lhs=(ab)^exponent,rhs=a^exponent*b^exponent,equal?).
View answer and explanation
(1, 2, 4, 2, 0, 0, Yes)
Start each power at the actual multiplicative identity. At exponent zero all powers are the identity. This identity is used under the full commutative-ring premise, not to certify commutativity from one instance. An exact finite or named-formula slice does not automatically establish an arbitrary infinite theorem. The exact result, retaining the requested order, is (1, 2, 4, 2, 0, 0, Yes).
Reviewed step-by-step derivation
Finite commutative ring · qf-mod28-commring-power-product · candidate
76. Compute a Binomial Square Using Ring Operations, Not Label Arithmetic
Use the same qualified complete finite unital ring on Fin(8)={0,…,7}, with addition=[[0, 1, 2, 3, 4, 5, 6, 7], [1, 2, 3, 4, 5, 6, 7, 0], [2, 3, 4, 5, 6, 7, 0, 1], [3, 4, 5, 6, 7, 0, 1, 2], [4, 5, 6, 7, 0, 1, 2, 3], [5, 6, 7, 0, 1, 2, 3, 4], [6, 7, 0, 1, 2, 3, 4, 5], [7, 0, 1, 2, 3, 4, 5, 6]] and multiplication=[[0, 0, 0, 0, 0, 0, 0, 0], [0, 1, 2, 3, 4, 5, 6, 7], [0, 2, 4, 6, 0, 2, 4, 6], [0, 3, 6, 1, 4, 7, 2, 5], [0, 4, 0, 4, 0, 4, 0, 4], [0, 5, 2, 7, 4, 1, 6, 3], [0, 6, 4, 2, 0, 6, 4, 2], [0, 7, 6, 5, 4, 3, 2, 1]] in label order. All unital-ring axioms, including distinct two-sided identities, must already hold. The identity labels need not be 0 and 1. Commutativity is a separate multiplication obligation; even a commutative ring may have nonzero zero divisors and need not be a field. Require commutativity. a=1, b=2. Give (a,b,sum=a+b,lhs=sum*sum,a2,ab,b2,rhs=a2+ab+ab+b2,equal?).
View answer and explanation
(1, 2, 3, 1, 1, 2, 4, 1, Yes)
Use the addition and multiplication tables throughout. The two copies of ab are added in the ring; do not substitute a carrier label called 2 for repeated addition. Zero divisors do not invalidate this identity. An exact finite or named-formula slice does not automatically establish an arbitrary infinite theorem. The exact result, retaining the requested order, is (1, 2, 3, 1, 1, 2, 4, 1, Yes).
Reviewed step-by-step derivation
Finite commutative ring · qf-mod28-commring-binomial-square · candidate
77. Compute a Binomial Square Using Ring Operations, Not Label Arithmetic
Use the same qualified complete finite unital ring on Fin(8)={0,…,7}, with addition=[[0, 1, 2, 3, 4, 5, 6, 7], [1, 2, 3, 4, 5, 7, 0, 6], [2, 3, 4, 5, 7, 6, 1, 0], [3, 4, 5, 7, 6, 0, 2, 1], [4, 5, 7, 6, 0, 1, 3, 2], [5, 7, 6, 0, 1, 2, 4, 3], [6, 0, 1, 2, 3, 4, 7, 5], [7, 6, 0, 1, 2, 3, 5, 4]] and multiplication=[[0, 0, 0, 0, 0, 0, 0, 0], [0, 1, 2, 3, 4, 5, 6, 7], [0, 2, 4, 7, 0, 2, 7, 4], [0, 3, 7, 1, 4, 6, 5, 2], [0, 4, 0, 4, 0, 4, 4, 0], [0, 5, 2, 6, 4, 1, 3, 7], [0, 6, 7, 5, 4, 3, 1, 2], [0, 7, 4, 2, 0, 7, 2, 4]] in label order. All unital-ring axioms, including distinct two-sided identities, must already hold. The identity labels need not be 0 and 1. Commutativity is a separate multiplication obligation; even a commutative ring may have nonzero zero divisors and need not be a field. Require commutativity. a=2, b=5. Give (a,b,sum=a+b,lhs=sum*sum,a2,ab,b2,rhs=a2+ab+ab+b2,equal?).
View answer and explanation
(2, 5, 6, 1, 4, 2, 1, 1, Yes)
Use the addition and multiplication tables throughout. The two copies of ab are added in the ring; do not substitute a carrier label called 2 for repeated addition. Zero divisors do not invalidate this identity. An exact finite or named-formula slice does not automatically establish an arbitrary infinite theorem. The exact result, retaining the requested order, is (2, 5, 6, 1, 4, 2, 1, 1, Yes).
Reviewed step-by-step derivation
Finite commutative ring · qf-mod28-commring-binomial-square · candidate
78. Compute a Binomial Square Using Ring Operations, Not Label Arithmetic
Use the same qualified complete finite unital ring on Fin(8)={0,…,7}, with addition=[[0, 1, 2, 3, 4, 5, 6, 7], [1, 2, 3, 4, 6, 7, 5, 0], [2, 3, 4, 6, 5, 0, 7, 1], [3, 4, 6, 5, 7, 1, 0, 2], [4, 6, 5, 7, 0, 2, 1, 3], [5, 7, 0, 1, 2, 4, 3, 6], [6, 5, 7, 0, 1, 3, 2, 4], [7, 0, 1, 2, 3, 6, 4, 5]] and multiplication=[[0, 0, 0, 0, 0, 0, 0, 0], [0, 1, 2, 3, 4, 5, 6, 7], [0, 2, 4, 5, 0, 4, 2, 5], [0, 3, 5, 1, 4, 2, 7, 6], [0, 4, 0, 4, 0, 0, 4, 4], [0, 5, 4, 2, 0, 4, 5, 2], [0, 6, 2, 7, 4, 5, 1, 3], [0, 7, 5, 6, 4, 2, 3, 1]] in label order. All unital-ring axioms, including distinct two-sided identities, must already hold. The identity labels need not be 0 and 1. Commutativity is a separate multiplication obligation; even a commutative ring may have nonzero zero divisors and need not be a field. Require commutativity. a=3, b=0. Give (a,b,sum=a+b,lhs=sum*sum,a2,ab,b2,rhs=a2+ab+ab+b2,equal?).
View answer and explanation
(3, 0, 3, 1, 1, 0, 0, 1, Yes)
Use the addition and multiplication tables throughout. The two copies of ab are added in the ring; do not substitute a carrier label called 2 for repeated addition. Zero divisors do not invalidate this identity. An exact finite or named-formula slice does not automatically establish an arbitrary infinite theorem. The exact result, retaining the requested order, is (3, 0, 3, 1, 1, 0, 0, 1, Yes).
Reviewed step-by-step derivation
Finite commutative ring · qf-mod28-commring-binomial-square · candidate
79. Compute a Binomial Square Using Ring Operations, Not Label Arithmetic
Use the same qualified complete finite unital ring on Fin(8)={0,…,7}, with addition=[[0, 1, 2, 3, 4, 5, 6, 7], [1, 2, 3, 4, 6, 0, 7, 5], [2, 3, 4, 6, 7, 1, 5, 0], [3, 4, 6, 7, 5, 2, 0, 1], [4, 6, 7, 5, 0, 3, 1, 2], [5, 0, 1, 2, 3, 7, 4, 6], [6, 7, 5, 0, 1, 4, 2, 3], [7, 5, 0, 1, 2, 6, 3, 4]] and multiplication=[[0, 0, 0, 0, 0, 0, 0, 0], [0, 1, 2, 3, 4, 5, 6, 7], [0, 2, 4, 7, 0, 7, 2, 4], [0, 3, 7, 1, 4, 6, 5, 2], [0, 4, 0, 4, 0, 4, 4, 0], [0, 5, 7, 6, 4, 1, 3, 2], [0, 6, 2, 5, 4, 3, 1, 7], [0, 7, 4, 2, 0, 2, 7, 4]] in label order. All unital-ring axioms, including distinct two-sided identities, must already hold. The identity labels need not be 0 and 1. Commutativity is a separate multiplication obligation; even a commutative ring may have nonzero zero divisors and need not be a field. Require commutativity. a=4, b=3. Give (a,b,sum=a+b,lhs=sum*sum,a2,ab,b2,rhs=a2+ab+ab+b2,equal?).
View answer and explanation
(4, 3, 5, 1, 0, 4, 1, 1, Yes)
Use the addition and multiplication tables throughout. The two copies of ab are added in the ring; do not substitute a carrier label called 2 for repeated addition. Zero divisors do not invalidate this identity. An exact finite or named-formula slice does not automatically establish an arbitrary infinite theorem. The exact result, retaining the requested order, is (4, 3, 5, 1, 0, 4, 1, 1, Yes).
Reviewed step-by-step derivation
Finite commutative ring · qf-mod28-commring-binomial-square · candidate
80. Compute a Binomial Square Using Ring Operations, Not Label Arithmetic
Use the same qualified complete finite unital ring on Fin(8)={0,…,7}, with addition=[[0, 1, 2, 3, 4, 5, 6, 7], [1, 2, 3, 4, 7, 6, 0, 5], [2, 3, 4, 7, 5, 0, 1, 6], [3, 4, 7, 5, 6, 1, 2, 0], [4, 7, 5, 6, 0, 2, 3, 1], [5, 6, 0, 1, 2, 4, 7, 3], [6, 0, 1, 2, 3, 7, 5, 4], [7, 5, 6, 0, 1, 3, 4, 2]] and multiplication=[[0, 0, 0, 0, 0, 0, 0, 0], [0, 1, 2, 3, 4, 5, 6, 7], [0, 2, 4, 5, 0, 4, 5, 2], [0, 3, 5, 1, 4, 2, 7, 6], [0, 4, 0, 4, 0, 0, 4, 4], [0, 5, 4, 2, 0, 4, 2, 5], [0, 6, 5, 7, 4, 2, 1, 3], [0, 7, 2, 6, 4, 5, 3, 1]] in label order. All unital-ring axioms, including distinct two-sided identities, must already hold. The identity labels need not be 0 and 1. Commutativity is a separate multiplication obligation; even a commutative ring may have nonzero zero divisors and need not be a field. Require commutativity. a=7, b=5. Give (a,b,sum=a+b,lhs=sum*sum,a2,ab,b2,rhs=a2+ab+ab+b2,equal?).
View answer and explanation
(7, 5, 3, 1, 1, 5, 4, 1, Yes)
Use the addition and multiplication tables throughout. The two copies of ab are added in the ring; do not substitute a carrier label called 2 for repeated addition. Zero divisors do not invalidate this identity. An exact finite or named-formula slice does not automatically establish an arbitrary infinite theorem. The exact result, retaining the requested order, is (7, 5, 3, 1, 1, 5, 4, 1, Yes).
Reviewed step-by-step derivation
Finite commutative ring · qf-mod28-commring-binomial-square · candidate
81. Compute a Binomial Square Using Ring Operations, Not Label Arithmetic
Use the same qualified complete finite unital ring on Fin(8)={0,…,7}, with addition=[[0, 1, 2, 3, 4, 5, 6, 7], [1, 2, 3, 4, 7, 0, 5, 6], [2, 3, 4, 7, 6, 1, 0, 5], [3, 4, 7, 6, 5, 2, 1, 0], [4, 7, 6, 5, 0, 3, 2, 1], [5, 0, 1, 2, 3, 6, 7, 4], [6, 5, 0, 1, 2, 7, 4, 3], [7, 6, 5, 0, 1, 4, 3, 2]] and multiplication=[[0, 0, 0, 0, 0, 0, 0, 0], [0, 1, 2, 3, 4, 5, 6, 7], [0, 2, 4, 6, 0, 6, 4, 2], [0, 3, 6, 1, 4, 7, 2, 5], [0, 4, 0, 4, 0, 4, 0, 4], [0, 5, 6, 7, 4, 1, 2, 3], [0, 6, 4, 2, 0, 2, 4, 6], [0, 7, 2, 5, 4, 3, 6, 1]] in label order. All unital-ring axioms, including distinct two-sided identities, must already hold. The identity labels need not be 0 and 1. Commutativity is a separate multiplication obligation; even a commutative ring may have nonzero zero divisors and need not be a field. Require commutativity. a=6, b=1. Give (a,b,sum=a+b,lhs=sum*sum,a2,ab,b2,rhs=a2+ab+ab+b2,equal?).
View answer and explanation
(6, 1, 5, 1, 4, 6, 1, 1, Yes)
Use the addition and multiplication tables throughout. The two copies of ab are added in the ring; do not substitute a carrier label called 2 for repeated addition. Zero divisors do not invalidate this identity. An exact finite or named-formula slice does not automatically establish an arbitrary infinite theorem. The exact result, retaining the requested order, is (6, 1, 5, 1, 4, 6, 1, 1, Yes).
Reviewed step-by-step derivation
Finite commutative ring · qf-mod28-commring-binomial-square · candidate
82. Compute a Binomial Square Using Ring Operations, Not Label Arithmetic
Use the same qualified complete finite unital ring on Fin(8)={0,…,7}, with addition=[[0, 1, 2, 3, 4, 5, 6, 7], [1, 2, 3, 5, 6, 4, 7, 0], [2, 3, 5, 4, 7, 6, 0, 1], [3, 5, 4, 6, 0, 7, 1, 2], [4, 6, 7, 0, 2, 1, 3, 5], [5, 4, 6, 7, 1, 0, 2, 3], [6, 7, 0, 1, 3, 2, 5, 4], [7, 0, 1, 2, 5, 3, 4, 6]] and multiplication=[[0, 0, 0, 0, 0, 0, 0, 0], [0, 1, 2, 3, 4, 5, 6, 7], [0, 2, 5, 6, 2, 0, 5, 6], [0, 3, 6, 1, 7, 5, 2, 4], [0, 4, 2, 7, 1, 5, 6, 3], [0, 5, 0, 5, 5, 0, 0, 5], [0, 6, 5, 2, 6, 0, 5, 2], [0, 7, 6, 4, 3, 5, 2, 1]] in label order. All unital-ring axioms, including distinct two-sided identities, must already hold. The identity labels need not be 0 and 1. Commutativity is a separate multiplication obligation; even a commutative ring may have nonzero zero divisors and need not be a field. Require commutativity. a=7, b=5. Give (a,b,sum=a+b,lhs=sum*sum,a2,ab,b2,rhs=a2+ab+ab+b2,equal?).
View answer and explanation
(7, 5, 3, 1, 1, 5, 0, 1, Yes)
Use the addition and multiplication tables throughout. The two copies of ab are added in the ring; do not substitute a carrier label called 2 for repeated addition. Zero divisors do not invalidate this identity. An exact finite or named-formula slice does not automatically establish an arbitrary infinite theorem. The exact result, retaining the requested order, is (7, 5, 3, 1, 1, 5, 0, 1, Yes).
Reviewed step-by-step derivation
Finite commutative ring · qf-mod28-commring-binomial-square · candidate
83. Compute a Binomial Square Using Ring Operations, Not Label Arithmetic
Use the same qualified complete finite unital ring on Fin(8)={0,…,7}, with addition=[[0, 1, 2, 3, 4, 5, 6, 7], [1, 2, 3, 5, 7, 4, 0, 6], [2, 3, 5, 4, 6, 7, 1, 0], [3, 5, 4, 7, 0, 6, 2, 1], [4, 7, 6, 0, 2, 1, 5, 3], [5, 4, 7, 6, 1, 0, 3, 2], [6, 0, 1, 2, 5, 3, 7, 4], [7, 6, 0, 1, 3, 2, 4, 5]] and multiplication=[[0, 0, 0, 0, 0, 0, 0, 0], [0, 1, 2, 3, 4, 5, 6, 7], [0, 2, 5, 7, 2, 0, 7, 5], [0, 3, 7, 1, 6, 5, 4, 2], [0, 4, 2, 6, 1, 5, 3, 7], [0, 5, 0, 5, 5, 0, 5, 0], [0, 6, 7, 4, 3, 5, 1, 2], [0, 7, 5, 2, 7, 0, 2, 5]] in label order. All unital-ring axioms, including distinct two-sided identities, must already hold. The identity labels need not be 0 and 1. Commutativity is a separate multiplication obligation; even a commutative ring may have nonzero zero divisors and need not be a field. Require commutativity. a=0, b=6. Give (a,b,sum=a+b,lhs=sum*sum,a2,ab,b2,rhs=a2+ab+ab+b2,equal?).
View answer and explanation
(0, 6, 6, 1, 0, 0, 1, 1, Yes)
Use the addition and multiplication tables throughout. The two copies of ab are added in the ring; do not substitute a carrier label called 2 for repeated addition. Zero divisors do not invalidate this identity. An exact finite or named-formula slice does not automatically establish an arbitrary infinite theorem. The exact result, retaining the requested order, is (0, 6, 6, 1, 0, 0, 1, 1, Yes).
Reviewed step-by-step derivation
Finite commutative ring · qf-mod28-commring-binomial-square · candidate
84. Compute a Binomial Square Using Ring Operations, Not Label Arithmetic
Use the same qualified complete finite unital ring on Fin(8)={0,…,7}, with addition=[[0, 1, 2, 3, 4, 5, 6, 7], [1, 2, 3, 5, 7, 6, 4, 0], [2, 3, 5, 6, 0, 4, 7, 1], [3, 5, 6, 4, 1, 7, 0, 2], [4, 7, 0, 1, 5, 2, 3, 6], [5, 6, 4, 7, 2, 0, 1, 3], [6, 4, 7, 0, 3, 1, 2, 5], [7, 0, 1, 2, 6, 3, 5, 4]] and multiplication=[[0, 0, 0, 0, 0, 0, 0, 0], [0, 1, 2, 3, 4, 5, 6, 7], [0, 2, 5, 4, 5, 0, 2, 4], [0, 3, 4, 1, 2, 5, 7, 6], [0, 4, 5, 2, 5, 0, 4, 2], [0, 5, 0, 5, 0, 0, 5, 5], [0, 6, 2, 7, 4, 5, 1, 3], [0, 7, 4, 6, 2, 5, 3, 1]] in label order. All unital-ring axioms, including distinct two-sided identities, must already hold. The identity labels need not be 0 and 1. Commutativity is a separate multiplication obligation; even a commutative ring may have nonzero zero divisors and need not be a field. Require commutativity. a=1, b=2. Give (a,b,sum=a+b,lhs=sum*sum,a2,ab,b2,rhs=a2+ab+ab+b2,equal?).
View answer and explanation
(1, 2, 3, 1, 1, 2, 5, 1, Yes)
Use the addition and multiplication tables throughout. The two copies of ab are added in the ring; do not substitute a carrier label called 2 for repeated addition. Zero divisors do not invalidate this identity. An exact finite or named-formula slice does not automatically establish an arbitrary infinite theorem. The exact result, retaining the requested order, is (1, 2, 3, 1, 1, 2, 5, 1, Yes).
Reviewed step-by-step derivation
Finite commutative ring · qf-mod28-commring-binomial-square · candidate
85. Compute a Binomial Square Using Ring Operations, Not Label Arithmetic
Use the same qualified complete finite unital ring on Fin(8)={0,…,7}, with addition=[[0, 1, 2, 3, 4, 5, 6, 7], [1, 2, 3, 5, 0, 6, 7, 4], [2, 3, 5, 6, 1, 7, 4, 0], [3, 5, 6, 7, 2, 4, 0, 1], [4, 0, 1, 2, 7, 3, 5, 6], [5, 6, 7, 4, 3, 0, 1, 2], [6, 7, 4, 0, 5, 1, 2, 3], [7, 4, 0, 1, 6, 2, 3, 5]] and multiplication=[[0, 0, 0, 0, 0, 0, 0, 0], [0, 1, 2, 3, 4, 5, 6, 7], [0, 2, 5, 7, 7, 0, 2, 5], [0, 3, 7, 1, 6, 5, 4, 2], [0, 4, 7, 6, 1, 5, 3, 2], [0, 5, 0, 5, 5, 0, 5, 0], [0, 6, 2, 4, 3, 5, 1, 7], [0, 7, 5, 2, 2, 0, 7, 5]] in label order. All unital-ring axioms, including distinct two-sided identities, must already hold. The identity labels need not be 0 and 1. Commutativity is a separate multiplication obligation; even a commutative ring may have nonzero zero divisors and need not be a field. Require commutativity. a=2, b=6. Give (a,b,sum=a+b,lhs=sum*sum,a2,ab,b2,rhs=a2+ab+ab+b2,equal?).
View answer and explanation
(2, 6, 4, 1, 5, 2, 1, 1, Yes)
Use the addition and multiplication tables throughout. The two copies of ab are added in the ring; do not substitute a carrier label called 2 for repeated addition. Zero divisors do not invalidate this identity. An exact finite or named-formula slice does not automatically establish an arbitrary infinite theorem. The exact result, retaining the requested order, is (2, 6, 4, 1, 5, 2, 1, 1, Yes).
Reviewed step-by-step derivation
Finite commutative ring · qf-mod28-commring-binomial-square · candidate
86. Compute a Binomial Square Using Ring Operations, Not Label Arithmetic
Use the same qualified complete finite unital ring on Fin(8)={0,…,7}, with addition=[[0, 1, 2, 3, 4, 5, 6, 7], [1, 2, 3, 5, 6, 7, 0, 4], [2, 3, 5, 7, 0, 4, 1, 6], [3, 5, 7, 4, 1, 6, 2, 0], [4, 6, 0, 1, 5, 2, 7, 3], [5, 7, 4, 6, 2, 0, 3, 1], [6, 0, 1, 2, 7, 3, 4, 5], [7, 4, 6, 0, 3, 1, 5, 2]] and multiplication=[[0, 0, 0, 0, 0, 0, 0, 0], [0, 1, 2, 3, 4, 5, 6, 7], [0, 2, 5, 4, 5, 0, 4, 2], [0, 3, 4, 1, 2, 5, 7, 6], [0, 4, 5, 2, 5, 0, 2, 4], [0, 5, 0, 5, 0, 0, 5, 5], [0, 6, 4, 7, 2, 5, 1, 3], [0, 7, 2, 6, 4, 5, 3, 1]] in label order. All unital-ring axioms, including distinct two-sided identities, must already hold. The identity labels need not be 0 and 1. Commutativity is a separate multiplication obligation; even a commutative ring may have nonzero zero divisors and need not be a field. Require commutativity. a=3, b=0. Give (a,b,sum=a+b,lhs=sum*sum,a2,ab,b2,rhs=a2+ab+ab+b2,equal?).
View answer and explanation
(3, 0, 3, 1, 1, 0, 0, 1, Yes)
Use the addition and multiplication tables throughout. The two copies of ab are added in the ring; do not substitute a carrier label called 2 for repeated addition. Zero divisors do not invalidate this identity. An exact finite or named-formula slice does not automatically establish an arbitrary infinite theorem. The exact result, retaining the requested order, is (3, 0, 3, 1, 1, 0, 0, 1, Yes).
Reviewed step-by-step derivation
Finite commutative ring · qf-mod28-commring-binomial-square · candidate
87. Compute a Binomial Square Using Ring Operations, Not Label Arithmetic
Use the same qualified complete finite unital ring on Fin(8)={0,…,7}, with addition=[[0, 1, 2, 3, 4, 5, 6, 7], [1, 2, 3, 5, 0, 7, 4, 6], [2, 3, 5, 7, 1, 6, 0, 4], [3, 5, 7, 6, 2, 4, 1, 0], [4, 0, 1, 2, 6, 3, 7, 5], [5, 7, 6, 4, 3, 0, 2, 1], [6, 4, 0, 1, 7, 2, 5, 3], [7, 6, 4, 0, 5, 1, 3, 2]] and multiplication=[[0, 0, 0, 0, 0, 0, 0, 0], [0, 1, 2, 3, 4, 5, 6, 7], [0, 2, 5, 6, 6, 0, 5, 2], [0, 3, 6, 1, 7, 5, 2, 4], [0, 4, 6, 7, 1, 5, 2, 3], [0, 5, 0, 5, 5, 0, 0, 5], [0, 6, 5, 2, 2, 0, 5, 6], [0, 7, 2, 4, 3, 5, 6, 1]] in label order. All unital-ring axioms, including distinct two-sided identities, must already hold. The identity labels need not be 0 and 1. Commutativity is a separate multiplication obligation; even a commutative ring may have nonzero zero divisors and need not be a field. Require commutativity. a=5, b=3. Give (a,b,sum=a+b,lhs=sum*sum,a2,ab,b2,rhs=a2+ab+ab+b2,equal?).
View answer and explanation
(5, 3, 4, 1, 0, 5, 1, 1, Yes)
Use the addition and multiplication tables throughout. The two copies of ab are added in the ring; do not substitute a carrier label called 2 for repeated addition. Zero divisors do not invalidate this identity. An exact finite or named-formula slice does not automatically establish an arbitrary infinite theorem. The exact result, retaining the requested order, is (5, 3, 4, 1, 0, 5, 1, 1, Yes).
Reviewed step-by-step derivation
Finite commutative ring · qf-mod28-commring-binomial-square · candidate
88. Compute a Binomial Square Using Ring Operations, Not Label Arithmetic
Use the same qualified complete finite unital ring on Fin(8)={0,…,7}, with addition=[[0, 1, 2, 3, 4, 5, 6, 7], [1, 2, 3, 6, 5, 7, 4, 0], [2, 3, 6, 4, 7, 0, 5, 1], [3, 6, 4, 5, 0, 1, 7, 2], [4, 5, 7, 0, 2, 3, 1, 6], [5, 7, 0, 1, 3, 6, 2, 4], [6, 4, 5, 7, 1, 2, 0, 3], [7, 0, 1, 2, 6, 4, 3, 5]] and multiplication=[[0, 0, 0, 0, 0, 0, 0, 0], [0, 1, 2, 3, 4, 5, 6, 7], [0, 2, 6, 5, 2, 6, 0, 5], [0, 3, 5, 1, 7, 2, 6, 4], [0, 4, 2, 7, 1, 5, 6, 3], [0, 5, 6, 2, 5, 6, 0, 2], [0, 6, 0, 6, 6, 0, 0, 6], [0, 7, 5, 4, 3, 2, 6, 1]] in label order. All unital-ring axioms, including distinct two-sided identities, must already hold. The identity labels need not be 0 and 1. Commutativity is a separate multiplication obligation; even a commutative ring may have nonzero zero divisors and need not be a field. Require commutativity. a=4, b=5. Give (a,b,sum=a+b,lhs=sum*sum,a2,ab,b2,rhs=a2+ab+ab+b2,equal?).
View answer and explanation
(4, 5, 3, 1, 1, 5, 6, 1, Yes)
Use the addition and multiplication tables throughout. The two copies of ab are added in the ring; do not substitute a carrier label called 2 for repeated addition. Zero divisors do not invalidate this identity. An exact finite or named-formula slice does not automatically establish an arbitrary infinite theorem. The exact result, retaining the requested order, is (4, 5, 3, 1, 1, 5, 6, 1, Yes).
Reviewed step-by-step derivation
Finite commutative ring · qf-mod28-commring-binomial-square · candidate
89. Compute a Binomial Square Using Ring Operations, Not Label Arithmetic
Use the same qualified complete finite unital ring on Fin(8)={0,…,7}, with addition=[[0, 1, 2, 3, 4, 5, 6, 7], [1, 2, 3, 6, 7, 0, 4, 5], [2, 3, 6, 4, 5, 1, 7, 0], [3, 6, 4, 7, 0, 2, 5, 1], [4, 7, 5, 0, 2, 6, 1, 3], [5, 0, 1, 2, 6, 7, 3, 4], [6, 4, 7, 5, 1, 3, 0, 2], [7, 5, 0, 1, 3, 4, 2, 6]] and multiplication=[[0, 0, 0, 0, 0, 0, 0, 0], [0, 1, 2, 3, 4, 5, 6, 7], [0, 2, 6, 7, 2, 7, 0, 6], [0, 3, 7, 1, 5, 4, 6, 2], [0, 4, 2, 5, 1, 3, 6, 7], [0, 5, 7, 4, 3, 1, 6, 2], [0, 6, 0, 6, 6, 6, 0, 0], [0, 7, 6, 2, 7, 2, 0, 6]] in label order. All unital-ring axioms, including distinct two-sided identities, must already hold. The identity labels need not be 0 and 1. Commutativity is a separate multiplication obligation; even a commutative ring may have nonzero zero divisors and need not be a field. Require commutativity. a=7, b=1. Give (a,b,sum=a+b,lhs=sum*sum,a2,ab,b2,rhs=a2+ab+ab+b2,equal?).
View answer and explanation
(7, 1, 5, 1, 6, 7, 1, 1, Yes)
Use the addition and multiplication tables throughout. The two copies of ab are added in the ring; do not substitute a carrier label called 2 for repeated addition. Zero divisors do not invalidate this identity. An exact finite or named-formula slice does not automatically establish an arbitrary infinite theorem. The exact result, retaining the requested order, is (7, 1, 5, 1, 6, 7, 1, 1, Yes).
Reviewed step-by-step derivation
Finite commutative ring · qf-mod28-commring-binomial-square · candidate
90. Compute a Binomial Square Using Ring Operations, Not Label Arithmetic
Use the same qualified complete finite unital ring on Fin(8)={0,…,7}, with addition=[[0, 1, 2, 3, 4, 5, 6, 7], [1, 2, 3, 6, 7, 4, 5, 0], [2, 3, 6, 5, 0, 7, 4, 1], [3, 6, 5, 4, 1, 0, 7, 2], [4, 7, 0, 1, 6, 3, 2, 5], [5, 4, 7, 0, 3, 2, 1, 6], [6, 5, 4, 7, 2, 1, 0, 3], [7, 0, 1, 2, 5, 6, 3, 4]] and multiplication=[[0, 0, 0, 0, 0, 0, 0, 0], [0, 1, 2, 3, 4, 5, 6, 7], [0, 2, 6, 4, 6, 2, 0, 4], [0, 3, 4, 1, 2, 7, 6, 5], [0, 4, 6, 2, 6, 4, 0, 2], [0, 5, 2, 7, 4, 1, 6, 3], [0, 6, 0, 6, 0, 6, 0, 6], [0, 7, 4, 5, 2, 3, 6, 1]] in label order. All unital-ring axioms, including distinct two-sided identities, must already hold. The identity labels need not be 0 and 1. Commutativity is a separate multiplication obligation; even a commutative ring may have nonzero zero divisors and need not be a field. Require commutativity. a=7, b=6. Give (a,b,sum=a+b,lhs=sum*sum,a2,ab,b2,rhs=a2+ab+ab+b2,equal?).
View answer and explanation
(7, 6, 3, 1, 1, 6, 0, 1, Yes)
Use the addition and multiplication tables throughout. The two copies of ab are added in the ring; do not substitute a carrier label called 2 for repeated addition. Zero divisors do not invalidate this identity. An exact finite or named-formula slice does not automatically establish an arbitrary infinite theorem. The exact result, retaining the requested order, is (7, 6, 3, 1, 1, 6, 0, 1, Yes).
Reviewed step-by-step derivation
Finite commutative ring · qf-mod28-commring-binomial-square · candidate
91. Compute a Binomial Square Using Ring Operations, Not Label Arithmetic
Use the same qualified complete finite unital ring on Fin(8)={0,…,7}, with addition=[[0, 1, 2, 3, 4, 5, 6, 7], [1, 2, 3, 6, 0, 7, 5, 4], [2, 3, 6, 5, 1, 4, 7, 0], [3, 6, 5, 7, 2, 0, 4, 1], [4, 0, 1, 2, 7, 6, 3, 5], [5, 7, 4, 0, 6, 2, 1, 3], [6, 5, 7, 4, 3, 1, 0, 2], [7, 4, 0, 1, 5, 3, 2, 6]] and multiplication=[[0, 0, 0, 0, 0, 0, 0, 0], [0, 1, 2, 3, 4, 5, 6, 7], [0, 2, 6, 7, 7, 2, 0, 6], [0, 3, 7, 1, 5, 4, 6, 2], [0, 4, 7, 5, 1, 3, 6, 2], [0, 5, 2, 4, 3, 1, 6, 7], [0, 6, 0, 6, 6, 6, 0, 0], [0, 7, 6, 2, 2, 7, 0, 6]] in label order. All unital-ring axioms, including distinct two-sided identities, must already hold. The identity labels need not be 0 and 1. Commutativity is a separate multiplication obligation; even a commutative ring may have nonzero zero divisors and need not be a field. Require commutativity. a=0, b=4. Give (a,b,sum=a+b,lhs=sum*sum,a2,ab,b2,rhs=a2+ab+ab+b2,equal?).
View answer and explanation
(0, 4, 4, 1, 0, 0, 1, 1, Yes)
Use the addition and multiplication tables throughout. The two copies of ab are added in the ring; do not substitute a carrier label called 2 for repeated addition. Zero divisors do not invalidate this identity. An exact finite or named-formula slice does not automatically establish an arbitrary infinite theorem. The exact result, retaining the requested order, is (0, 4, 4, 1, 0, 0, 1, 1, Yes).
Reviewed step-by-step derivation
Finite commutative ring · qf-mod28-commring-binomial-square · candidate
92. Compute a Binomial Square Using Ring Operations, Not Label Arithmetic
Use the same qualified complete finite unital ring on Fin(8)={0,…,7}, with addition=[[0, 1, 2, 3, 4, 5, 6, 7], [1, 2, 3, 6, 5, 0, 7, 4], [2, 3, 6, 7, 0, 1, 4, 5], [3, 6, 7, 4, 1, 2, 5, 0], [4, 5, 0, 1, 6, 7, 2, 3], [5, 0, 1, 2, 7, 4, 3, 6], [6, 7, 4, 5, 2, 3, 0, 1], [7, 4, 5, 0, 3, 6, 1, 2]] and multiplication=[[0, 0, 0, 0, 0, 0, 0, 0], [0, 1, 2, 3, 4, 5, 6, 7], [0, 2, 6, 4, 6, 4, 0, 2], [0, 3, 4, 1, 2, 7, 6, 5], [0, 4, 6, 2, 6, 2, 0, 4], [0, 5, 4, 7, 2, 1, 6, 3], [0, 6, 0, 6, 0, 6, 0, 6], [0, 7, 2, 5, 4, 3, 6, 1]] in label order. All unital-ring axioms, including distinct two-sided identities, must already hold. The identity labels need not be 0 and 1. Commutativity is a separate multiplication obligation; even a commutative ring may have nonzero zero divisors and need not be a field. Require commutativity. a=1, b=2. Give (a,b,sum=a+b,lhs=sum*sum,a2,ab,b2,rhs=a2+ab+ab+b2,equal?).
View answer and explanation
(1, 2, 3, 1, 1, 2, 6, 1, Yes)
Use the addition and multiplication tables throughout. The two copies of ab are added in the ring; do not substitute a carrier label called 2 for repeated addition. Zero divisors do not invalidate this identity. An exact finite or named-formula slice does not automatically establish an arbitrary infinite theorem. The exact result, retaining the requested order, is (1, 2, 3, 1, 1, 2, 6, 1, Yes).
Reviewed step-by-step derivation
Finite commutative ring · qf-mod28-commring-binomial-square · candidate
93. Compute a Binomial Square Using Ring Operations, Not Label Arithmetic
Use the same qualified complete finite unital ring on Fin(8)={0,…,7}, with addition=[[0, 1, 2, 3, 4, 5, 6, 7], [1, 2, 3, 6, 0, 4, 7, 5], [2, 3, 6, 7, 1, 0, 5, 4], [3, 6, 7, 5, 2, 1, 4, 0], [4, 0, 1, 2, 5, 7, 3, 6], [5, 4, 0, 1, 7, 6, 2, 3], [6, 7, 5, 4, 3, 2, 0, 1], [7, 5, 4, 0, 6, 3, 1, 2]] and multiplication=[[0, 0, 0, 0, 0, 0, 0, 0], [0, 1, 2, 3, 4, 5, 6, 7], [0, 2, 6, 5, 5, 6, 0, 2], [0, 3, 5, 1, 7, 2, 6, 4], [0, 4, 5, 7, 1, 2, 6, 3], [0, 5, 6, 2, 2, 6, 0, 5], [0, 6, 0, 6, 6, 0, 0, 6], [0, 7, 2, 4, 3, 5, 6, 1]] in label order. All unital-ring axioms, including distinct two-sided identities, must already hold. The identity labels need not be 0 and 1. Commutativity is a separate multiplication obligation; even a commutative ring may have nonzero zero divisors and need not be a field. Require commutativity. a=2, b=7. Give (a,b,sum=a+b,lhs=sum*sum,a2,ab,b2,rhs=a2+ab+ab+b2,equal?).
View answer and explanation
(2, 7, 4, 1, 6, 2, 1, 1, Yes)
Use the addition and multiplication tables throughout. The two copies of ab are added in the ring; do not substitute a carrier label called 2 for repeated addition. Zero divisors do not invalidate this identity. An exact finite or named-formula slice does not automatically establish an arbitrary infinite theorem. The exact result, retaining the requested order, is (2, 7, 4, 1, 6, 2, 1, 1, Yes).
Reviewed step-by-step derivation
Finite commutative ring · qf-mod28-commring-binomial-square · candidate
94. Compute a Binomial Square Using Ring Operations, Not Label Arithmetic
Use the same qualified complete finite unital ring on Fin(8)={0,…,7}, with addition=[[0, 1, 2, 3, 4, 5, 6, 7], [1, 2, 3, 7, 5, 6, 0, 4], [2, 3, 7, 4, 6, 0, 1, 5], [3, 7, 4, 5, 0, 1, 2, 6], [4, 5, 6, 0, 2, 3, 7, 1], [5, 6, 0, 1, 3, 7, 4, 2], [6, 0, 1, 2, 7, 4, 5, 3], [7, 4, 5, 6, 1, 2, 3, 0]] and multiplication=[[0, 0, 0, 0, 0, 0, 0, 0], [0, 1, 2, 3, 4, 5, 6, 7], [0, 2, 7, 5, 2, 7, 5, 0], [0, 3, 5, 1, 6, 2, 4, 7], [0, 4, 2, 6, 1, 5, 3, 7], [0, 5, 7, 2, 5, 7, 2, 0], [0, 6, 5, 4, 3, 2, 1, 7], [0, 7, 0, 7, 7, 0, 7, 0]] in label order. All unital-ring axioms, including distinct two-sided identities, must already hold. The identity labels need not be 0 and 1. Commutativity is a separate multiplication obligation; even a commutative ring may have nonzero zero divisors and need not be a field. Require commutativity. a=3, b=0. Give (a,b,sum=a+b,lhs=sum*sum,a2,ab,b2,rhs=a2+ab+ab+b2,equal?).
View answer and explanation
(3, 0, 3, 1, 1, 0, 0, 1, Yes)
Use the addition and multiplication tables throughout. The two copies of ab are added in the ring; do not substitute a carrier label called 2 for repeated addition. Zero divisors do not invalidate this identity. An exact finite or named-formula slice does not automatically establish an arbitrary infinite theorem. The exact result, retaining the requested order, is (3, 0, 3, 1, 1, 0, 0, 1, Yes).
Reviewed step-by-step derivation
Finite commutative ring · qf-mod28-commring-binomial-square · candidate
95. Compute a Binomial Square Using Ring Operations, Not Label Arithmetic
Use the same qualified complete finite unital ring on Fin(8)={0,…,7}, with addition=[[0, 1, 2, 3, 4, 5, 6, 7], [1, 2, 3, 7, 6, 0, 5, 4], [2, 3, 7, 4, 5, 1, 0, 6], [3, 7, 4, 6, 0, 2, 1, 5], [4, 6, 5, 0, 2, 7, 3, 1], [5, 0, 1, 2, 7, 6, 4, 3], [6, 5, 0, 1, 3, 4, 7, 2], [7, 4, 6, 5, 1, 3, 2, 0]] and multiplication=[[0, 0, 0, 0, 0, 0, 0, 0], [0, 1, 2, 3, 4, 5, 6, 7], [0, 2, 7, 6, 2, 6, 7, 0], [0, 3, 6, 1, 5, 4, 2, 7], [0, 4, 2, 5, 1, 3, 6, 7], [0, 5, 6, 4, 3, 1, 2, 7], [0, 6, 7, 2, 6, 2, 7, 0], [0, 7, 0, 7, 7, 7, 0, 0]] in label order. All unital-ring axioms, including distinct two-sided identities, must already hold. The identity labels need not be 0 and 1. Commutativity is a separate multiplication obligation; even a commutative ring may have nonzero zero divisors and need not be a field. Require commutativity. a=7, b=3. Give (a,b,sum=a+b,lhs=sum*sum,a2,ab,b2,rhs=a2+ab+ab+b2,equal?).
View answer and explanation
(7, 3, 5, 1, 0, 7, 1, 1, Yes)
Use the addition and multiplication tables throughout. The two copies of ab are added in the ring; do not substitute a carrier label called 2 for repeated addition. Zero divisors do not invalidate this identity. An exact finite or named-formula slice does not automatically establish an arbitrary infinite theorem. The exact result, retaining the requested order, is (7, 3, 5, 1, 0, 7, 1, 1, Yes).
Reviewed step-by-step derivation
Finite commutative ring · qf-mod28-commring-binomial-square · candidate
96. Compute a Binomial Square Using Ring Operations, Not Label Arithmetic
Use the same qualified complete finite unital ring on Fin(8)={0,…,7}, with addition=[[0, 1, 2, 3, 4, 5, 6, 7], [1, 2, 3, 7, 6, 4, 0, 5], [2, 3, 7, 5, 0, 6, 1, 4], [3, 7, 5, 4, 1, 0, 2, 6], [4, 6, 0, 1, 7, 3, 5, 2], [5, 4, 6, 0, 3, 2, 7, 1], [6, 0, 1, 2, 5, 7, 4, 3], [7, 5, 4, 6, 2, 1, 3, 0]] and multiplication=[[0, 0, 0, 0, 0, 0, 0, 0], [0, 1, 2, 3, 4, 5, 6, 7], [0, 2, 7, 4, 7, 2, 4, 0], [0, 3, 4, 1, 2, 6, 5, 7], [0, 4, 7, 2, 7, 4, 2, 0], [0, 5, 2, 6, 4, 1, 3, 7], [0, 6, 4, 5, 2, 3, 1, 7], [0, 7, 0, 7, 0, 7, 7, 0]] in label order. All unital-ring axioms, including distinct two-sided identities, must already hold. The identity labels need not be 0 and 1. Commutativity is a separate multiplication obligation; even a commutative ring may have nonzero zero divisors and need not be a field. Require commutativity. a=5, b=4. Give (a,b,sum=a+b,lhs=sum*sum,a2,ab,b2,rhs=a2+ab+ab+b2,equal?).
View answer and explanation
(5, 4, 3, 1, 1, 4, 7, 1, Yes)
Use the addition and multiplication tables throughout. The two copies of ab are added in the ring; do not substitute a carrier label called 2 for repeated addition. Zero divisors do not invalidate this identity. An exact finite or named-formula slice does not automatically establish an arbitrary infinite theorem. The exact result, retaining the requested order, is (5, 4, 3, 1, 1, 4, 7, 1, Yes).
Reviewed step-by-step derivation
Finite commutative ring · qf-mod28-commring-binomial-square · candidate
97. Compute a Binomial Square Using Ring Operations, Not Label Arithmetic
Use the same qualified complete finite unital ring on Fin(8)={0,…,7}, with addition=[[0, 1, 2, 3, 4, 5, 6, 7], [1, 2, 3, 7, 0, 6, 4, 5], [2, 3, 7, 5, 1, 4, 0, 6], [3, 7, 5, 6, 2, 0, 1, 4], [4, 0, 1, 2, 6, 7, 5, 3], [5, 6, 4, 0, 7, 2, 3, 1], [6, 4, 0, 1, 5, 3, 7, 2], [7, 5, 6, 4, 3, 1, 2, 0]] and multiplication=[[0, 0, 0, 0, 0, 0, 0, 0], [0, 1, 2, 3, 4, 5, 6, 7], [0, 2, 7, 6, 6, 2, 7, 0], [0, 3, 6, 1, 5, 4, 2, 7], [0, 4, 6, 5, 1, 3, 2, 7], [0, 5, 2, 4, 3, 1, 6, 7], [0, 6, 7, 2, 2, 6, 7, 0], [0, 7, 0, 7, 7, 7, 0, 0]] in label order. All unital-ring axioms, including distinct two-sided identities, must already hold. The identity labels need not be 0 and 1. Commutativity is a separate multiplication obligation; even a commutative ring may have nonzero zero divisors and need not be a field. Require commutativity. a=6, b=1. Give (a,b,sum=a+b,lhs=sum*sum,a2,ab,b2,rhs=a2+ab+ab+b2,equal?).
View answer and explanation
(6, 1, 4, 1, 7, 6, 1, 1, Yes)
Use the addition and multiplication tables throughout. The two copies of ab are added in the ring; do not substitute a carrier label called 2 for repeated addition. Zero divisors do not invalidate this identity. An exact finite or named-formula slice does not automatically establish an arbitrary infinite theorem. The exact result, retaining the requested order, is (6, 1, 4, 1, 7, 6, 1, 1, Yes).
Reviewed step-by-step derivation
Finite commutative ring · qf-mod28-commring-binomial-square · candidate
98. Compute a Binomial Square Using Ring Operations, Not Label Arithmetic
Use the same qualified complete finite unital ring on Fin(8)={0,…,7}, with addition=[[0, 1, 2, 3, 4, 5, 6, 7], [1, 2, 3, 7, 5, 0, 4, 6], [2, 3, 7, 6, 0, 1, 5, 4], [3, 7, 6, 4, 1, 2, 0, 5], [4, 5, 0, 1, 7, 6, 3, 2], [5, 0, 1, 2, 6, 4, 7, 3], [6, 4, 5, 0, 3, 7, 2, 1], [7, 6, 4, 5, 2, 3, 1, 0]] and multiplication=[[0, 0, 0, 0, 0, 0, 0, 0], [0, 1, 2, 3, 4, 5, 6, 7], [0, 2, 7, 4, 7, 4, 2, 0], [0, 3, 4, 1, 2, 6, 5, 7], [0, 4, 7, 2, 7, 2, 4, 0], [0, 5, 4, 6, 2, 1, 3, 7], [0, 6, 2, 5, 4, 3, 1, 7], [0, 7, 0, 7, 0, 7, 7, 0]] in label order. All unital-ring axioms, including distinct two-sided identities, must already hold. The identity labels need not be 0 and 1. Commutativity is a separate multiplication obligation; even a commutative ring may have nonzero zero divisors and need not be a field. Require commutativity. a=5, b=7. Give (a,b,sum=a+b,lhs=sum*sum,a2,ab,b2,rhs=a2+ab+ab+b2,equal?).
View answer and explanation
(5, 7, 3, 1, 1, 7, 0, 1, Yes)
Use the addition and multiplication tables throughout. The two copies of ab are added in the ring; do not substitute a carrier label called 2 for repeated addition. Zero divisors do not invalidate this identity. An exact finite or named-formula slice does not automatically establish an arbitrary infinite theorem. The exact result, retaining the requested order, is (5, 7, 3, 1, 1, 7, 0, 1, Yes).
Reviewed step-by-step derivation
Finite commutative ring · qf-mod28-commring-binomial-square · candidate
99. Compute a Binomial Square Using Ring Operations, Not Label Arithmetic
Use the same qualified complete finite unital ring on Fin(8)={0,…,7}, with addition=[[0, 1, 2, 3, 4, 5, 6, 7], [1, 2, 3, 7, 0, 4, 5, 6], [2, 3, 7, 6, 1, 0, 4, 5], [3, 7, 6, 5, 2, 1, 0, 4], [4, 0, 1, 2, 5, 6, 7, 3], [5, 4, 0, 1, 6, 7, 3, 2], [6, 5, 4, 0, 7, 3, 2, 1], [7, 6, 5, 4, 3, 2, 1, 0]] and multiplication=[[0, 0, 0, 0, 0, 0, 0, 0], [0, 1, 2, 3, 4, 5, 6, 7], [0, 2, 7, 5, 5, 7, 2, 0], [0, 3, 5, 1, 6, 2, 4, 7], [0, 4, 5, 6, 1, 2, 3, 7], [0, 5, 7, 2, 2, 7, 5, 0], [0, 6, 2, 4, 3, 5, 1, 7], [0, 7, 0, 7, 7, 0, 7, 0]] in label order. All unital-ring axioms, including distinct two-sided identities, must already hold. The identity labels need not be 0 and 1. Commutativity is a separate multiplication obligation; even a commutative ring may have nonzero zero divisors and need not be a field. Require commutativity. a=0, b=4. Give (a,b,sum=a+b,lhs=sum*sum,a2,ab,b2,rhs=a2+ab+ab+b2,equal?).
View answer and explanation
(0, 4, 4, 1, 0, 0, 1, 1, Yes)
Use the addition and multiplication tables throughout. The two copies of ab are added in the ring; do not substitute a carrier label called 2 for repeated addition. Zero divisors do not invalidate this identity. An exact finite or named-formula slice does not automatically establish an arbitrary infinite theorem. The exact result, retaining the requested order, is (0, 4, 4, 1, 0, 0, 1, 1, Yes).
Reviewed step-by-step derivation
Finite commutative ring · qf-mod28-commring-binomial-square · candidate
100. Compute a Binomial Square Using Ring Operations, Not Label Arithmetic
Use the same qualified complete finite unital ring on Fin(8)={0,…,7}, with addition=[[0, 1, 2, 3, 4, 5, 6, 7], [1, 2, 4, 5, 3, 6, 7, 0], [2, 4, 3, 6, 5, 7, 0, 1], [3, 5, 6, 0, 7, 1, 2, 4], [4, 3, 5, 7, 6, 0, 1, 2], [5, 6, 7, 1, 0, 2, 4, 3], [6, 7, 0, 2, 1, 4, 3, 5], [7, 0, 1, 4, 2, 3, 5, 6]] and multiplication=[[0, 0, 0, 0, 0, 0, 0, 0], [0, 1, 2, 3, 4, 5, 6, 7], [0, 2, 3, 0, 6, 2, 3, 6], [0, 3, 0, 0, 3, 3, 0, 3], [0, 4, 6, 3, 1, 7, 2, 5], [0, 5, 2, 3, 7, 1, 6, 4], [0, 6, 3, 0, 2, 6, 3, 2], [0, 7, 6, 3, 5, 4, 2, 1]] in label order. All unital-ring axioms, including distinct two-sided identities, must already hold. The identity labels need not be 0 and 1. Commutativity is a separate multiplication obligation; even a commutative ring may have nonzero zero divisors and need not be a field. Require commutativity. a=1, b=2. Give (a,b,sum=a+b,lhs=sum*sum,a2,ab,b2,rhs=a2+ab+ab+b2,equal?).
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(1, 2, 4, 1, 1, 2, 3, 1, Yes)
Use the addition and multiplication tables throughout. The two copies of ab are added in the ring; do not substitute a carrier label called 2 for repeated addition. Zero divisors do not invalidate this identity. An exact finite or named-formula slice does not automatically establish an arbitrary infinite theorem. The exact result, retaining the requested order, is (1, 2, 4, 1, 1, 2, 3, 1, Yes).
Reviewed step-by-step derivation
Finite commutative ring · qf-mod28-commring-binomial-square · candidate
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