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Complete Nonzero Inverses, Safe Division and the Finite-Domain Theorem

This batch has passed Schema, deduplication, source-snapshot, and independent-answer recomputation checks; it is still not marked as human-reviewed.

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-20261030-r01-finite-field-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. Audit Every Nonzero Inverse after Complete Commutative-Ring Qualification

Use the same already qualified complete finite commutative unital-ring tables on Fin(5)={0,…,4}, addition=[[0, 1, 2, 3, 4], [1, 2, 3, 4, 0], [2, 3, 4, 0, 1], [3, 4, 0, 1, 2], [4, 0, 1, 2, 3]] and multiplication=[[0, 0, 0, 0, 0], [0, 1, 2, 3, 4], [0, 2, 4, 1, 3], [0, 3, 1, 4, 2], [0, 4, 3, 2, 1]] in carrier-label order. The actual zero and one are distinct and need not be labels 0 and 1. The inverse audit is additional to, not a replacement for, those ring axioms. Division by zero is never defined. A finite field has prime-power order, not necessarily prime order. Give (n,zero,one,missing_inverses,field?). List all nonzero labels without a two-sided inverse increasingly.

View answer and explanation

(5, 0, 1, [], Yes)

Find both actual identities and search all labels for each nonzero element's two-sided inverse. Missing inverses refute field qualification even when the parent is a commutative ring. The exact result, retaining the requested order, is (5, 0, 1, [], Yes).

Reviewed step-by-step derivation

Finite field · qf-mod30-field-qualification · candidate

2. Audit Every Nonzero Inverse after Complete Commutative-Ring Qualification

Use the same already qualified complete finite commutative unital-ring tables on Fin(5)={0,…,4}, addition=[[0, 1, 2, 3, 4], [1, 2, 4, 0, 3], [2, 4, 3, 1, 0], [3, 0, 1, 4, 2], [4, 3, 0, 2, 1]] and multiplication=[[0, 0, 0, 0, 0], [0, 1, 2, 3, 4], [0, 2, 3, 4, 1], [0, 3, 4, 1, 2], [0, 4, 1, 2, 3]] in carrier-label order. The actual zero and one are distinct and need not be labels 0 and 1. The inverse audit is additional to, not a replacement for, those ring axioms. Division by zero is never defined. A finite field has prime-power order, not necessarily prime order. Give (n,zero,one,missing_inverses,field?). List all nonzero labels without a two-sided inverse increasingly.

View answer and explanation

(5, 0, 1, [], Yes)

Find both actual identities and search all labels for each nonzero element's two-sided inverse. Missing inverses refute field qualification even when the parent is a commutative ring. The exact result, retaining the requested order, is (5, 0, 1, [], Yes).

Reviewed step-by-step derivation

Finite field · qf-mod30-field-qualification · candidate

3. Audit Every Nonzero Inverse after Complete Commutative-Ring Qualification

Use the same already qualified complete finite commutative unital-ring tables on Fin(5)={0,…,4}, addition=[[0, 1, 2, 3, 4], [1, 3, 4, 2, 0], [2, 4, 1, 0, 3], [3, 2, 0, 4, 1], [4, 0, 3, 1, 2]] and multiplication=[[0, 0, 0, 0, 0], [0, 1, 2, 3, 4], [0, 2, 4, 1, 3], [0, 3, 1, 4, 2], [0, 4, 3, 2, 1]] in carrier-label order. The actual zero and one are distinct and need not be labels 0 and 1. The inverse audit is additional to, not a replacement for, those ring axioms. Division by zero is never defined. A finite field has prime-power order, not necessarily prime order. Give (n,zero,one,missing_inverses,field?). List all nonzero labels without a two-sided inverse increasingly.

View answer and explanation

(5, 0, 1, [], Yes)

Find both actual identities and search all labels for each nonzero element's two-sided inverse. Missing inverses refute field qualification even when the parent is a commutative ring. The exact result, retaining the requested order, is (5, 0, 1, [], Yes).

Reviewed step-by-step derivation

Finite field · qf-mod30-field-qualification · candidate

4. Audit Every Nonzero Inverse after Complete Commutative-Ring Qualification

Use the same already qualified complete finite commutative unital-ring tables on Fin(5)={0,…,4}, addition=[[0, 1, 2, 3, 4], [1, 3, 0, 4, 2], [2, 0, 4, 1, 3], [3, 4, 1, 2, 0], [4, 2, 3, 0, 1]] and multiplication=[[0, 0, 0, 0, 0], [0, 1, 2, 3, 4], [0, 2, 1, 4, 3], [0, 3, 4, 2, 1], [0, 4, 3, 1, 2]] in carrier-label order. The actual zero and one are distinct and need not be labels 0 and 1. The inverse audit is additional to, not a replacement for, those ring axioms. Division by zero is never defined. A finite field has prime-power order, not necessarily prime order. Give (n,zero,one,missing_inverses,field?). List all nonzero labels without a two-sided inverse increasingly.

View answer and explanation

(5, 0, 1, [], Yes)

Find both actual identities and search all labels for each nonzero element's two-sided inverse. Missing inverses refute field qualification even when the parent is a commutative ring. The exact result, retaining the requested order, is (5, 0, 1, [], Yes).

Reviewed step-by-step derivation

Finite field · qf-mod30-field-qualification · candidate

5. Audit Every Nonzero Inverse after Complete Commutative-Ring Qualification

Use the same already qualified complete finite commutative unital-ring tables on Fin(5)={0,…,4}, addition=[[0, 1, 2, 3, 4], [1, 4, 3, 0, 2], [2, 3, 1, 4, 0], [3, 0, 4, 2, 1], [4, 2, 0, 1, 3]] and multiplication=[[0, 0, 0, 0, 0], [0, 1, 2, 3, 4], [0, 2, 3, 4, 1], [0, 3, 4, 1, 2], [0, 4, 1, 2, 3]] in carrier-label order. The actual zero and one are distinct and need not be labels 0 and 1. The inverse audit is additional to, not a replacement for, those ring axioms. Division by zero is never defined. A finite field has prime-power order, not necessarily prime order. Give (n,zero,one,missing_inverses,field?). List all nonzero labels without a two-sided inverse increasingly.

View answer and explanation

(5, 0, 1, [], Yes)

Find both actual identities and search all labels for each nonzero element's two-sided inverse. Missing inverses refute field qualification even when the parent is a commutative ring. The exact result, retaining the requested order, is (5, 0, 1, [], Yes).

Reviewed step-by-step derivation

Finite field · qf-mod30-field-qualification · candidate

6. Audit Every Nonzero Inverse after Complete Commutative-Ring Qualification

Use the same already qualified complete finite commutative unital-ring tables on Fin(5)={0,…,4}, addition=[[0, 1, 2, 3, 4], [1, 4, 0, 2, 3], [2, 0, 3, 4, 1], [3, 2, 4, 1, 0], [4, 3, 1, 0, 2]] and multiplication=[[0, 0, 0, 0, 0], [0, 1, 2, 3, 4], [0, 2, 1, 4, 3], [0, 3, 4, 2, 1], [0, 4, 3, 1, 2]] in carrier-label order. The actual zero and one are distinct and need not be labels 0 and 1. The inverse audit is additional to, not a replacement for, those ring axioms. Division by zero is never defined. A finite field has prime-power order, not necessarily prime order. Give (n,zero,one,missing_inverses,field?). List all nonzero labels without a two-sided inverse increasingly.

View answer and explanation

(5, 0, 1, [], Yes)

Find both actual identities and search all labels for each nonzero element's two-sided inverse. Missing inverses refute field qualification even when the parent is a commutative ring. The exact result, retaining the requested order, is (5, 0, 1, [], Yes).

Reviewed step-by-step derivation

Finite field · qf-mod30-field-qualification · candidate

7. Audit Every Nonzero Inverse after Complete Commutative-Ring Qualification

Use the same already qualified complete finite commutative unital-ring tables on Fin(5)={0,…,4}, addition=[[0, 1, 2, 3, 4], [1, 4, 3, 0, 2], [2, 3, 1, 4, 0], [3, 0, 4, 2, 1], [4, 2, 0, 1, 3]] and multiplication=[[0, 0, 0, 0, 0], [0, 4, 1, 2, 3], [0, 1, 2, 3, 4], [0, 2, 3, 4, 1], [0, 3, 4, 1, 2]] in carrier-label order. The actual zero and one are distinct and need not be labels 0 and 1. The inverse audit is additional to, not a replacement for, those ring axioms. Division by zero is never defined. A finite field has prime-power order, not necessarily prime order. Give (n,zero,one,missing_inverses,field?). List all nonzero labels without a two-sided inverse increasingly.

View answer and explanation

(5, 0, 2, [], Yes)

Find both actual identities and search all labels for each nonzero element's two-sided inverse. Missing inverses refute field qualification even when the parent is a commutative ring. The exact result, retaining the requested order, is (5, 0, 2, [], Yes).

Reviewed step-by-step derivation

Finite field · qf-mod30-field-qualification · candidate

8. Audit Every Nonzero Inverse after Complete Commutative-Ring Qualification

Use the same already qualified complete finite commutative unital-ring tables on Fin(5)={0,…,4}, addition=[[0, 1, 2, 3, 4], [1, 3, 4, 2, 0], [2, 4, 1, 0, 3], [3, 2, 0, 4, 1], [4, 0, 3, 1, 2]] and multiplication=[[0, 0, 0, 0, 0], [0, 3, 1, 4, 2], [0, 1, 2, 3, 4], [0, 4, 3, 2, 1], [0, 2, 4, 1, 3]] in carrier-label order. The actual zero and one are distinct and need not be labels 0 and 1. The inverse audit is additional to, not a replacement for, those ring axioms. Division by zero is never defined. A finite field has prime-power order, not necessarily prime order. Give (n,zero,one,missing_inverses,field?). List all nonzero labels without a two-sided inverse increasingly.

View answer and explanation

(5, 0, 2, [], Yes)

Find both actual identities and search all labels for each nonzero element's two-sided inverse. Missing inverses refute field qualification even when the parent is a commutative ring. The exact result, retaining the requested order, is (5, 0, 2, [], Yes).

Reviewed step-by-step derivation

Finite field · qf-mod30-field-qualification · candidate

9. Audit Every Nonzero Inverse after Complete Commutative-Ring Qualification

Use the same already qualified complete finite commutative unital-ring tables on Fin(5)={0,…,4}, addition=[[0, 1, 2, 3, 4], [1, 2, 4, 0, 3], [2, 4, 3, 1, 0], [3, 0, 1, 4, 2], [4, 3, 0, 2, 1]] and multiplication=[[0, 0, 0, 0, 0], [0, 4, 1, 2, 3], [0, 1, 2, 3, 4], [0, 2, 3, 4, 1], [0, 3, 4, 1, 2]] in carrier-label order. The actual zero and one are distinct and need not be labels 0 and 1. The inverse audit is additional to, not a replacement for, those ring axioms. Division by zero is never defined. A finite field has prime-power order, not necessarily prime order. Give (n,zero,one,missing_inverses,field?). List all nonzero labels without a two-sided inverse increasingly.

View answer and explanation

(5, 0, 2, [], Yes)

Find both actual identities and search all labels for each nonzero element's two-sided inverse. Missing inverses refute field qualification even when the parent is a commutative ring. The exact result, retaining the requested order, is (5, 0, 2, [], Yes).

Reviewed step-by-step derivation

Finite field · qf-mod30-field-qualification · candidate

10. Audit Every Nonzero Inverse after Complete Commutative-Ring Qualification

Use the same already qualified complete finite commutative unital-ring tables on Fin(5)={0,…,4}, addition=[[0, 1, 2, 3, 4], [1, 4, 0, 2, 3], [2, 0, 3, 4, 1], [3, 2, 4, 1, 0], [4, 3, 1, 0, 2]] and multiplication=[[0, 0, 0, 0, 0], [0, 2, 1, 4, 3], [0, 1, 2, 3, 4], [0, 4, 3, 1, 2], [0, 3, 4, 2, 1]] in carrier-label order. The actual zero and one are distinct and need not be labels 0 and 1. The inverse audit is additional to, not a replacement for, those ring axioms. Division by zero is never defined. A finite field has prime-power order, not necessarily prime order. Give (n,zero,one,missing_inverses,field?). List all nonzero labels without a two-sided inverse increasingly.

View answer and explanation

(5, 0, 2, [], Yes)

Find both actual identities and search all labels for each nonzero element's two-sided inverse. Missing inverses refute field qualification even when the parent is a commutative ring. The exact result, retaining the requested order, is (5, 0, 2, [], Yes).

Reviewed step-by-step derivation

Finite field · qf-mod30-field-qualification · candidate

11. Audit Every Nonzero Inverse after Complete Commutative-Ring Qualification

Use the same already qualified complete finite commutative unital-ring tables on Fin(5)={0,…,4}, addition=[[0, 1, 2, 3, 4], [1, 2, 3, 4, 0], [2, 3, 4, 0, 1], [3, 4, 0, 1, 2], [4, 0, 1, 2, 3]] and multiplication=[[0, 0, 0, 0, 0], [0, 3, 1, 4, 2], [0, 1, 2, 3, 4], [0, 4, 3, 2, 1], [0, 2, 4, 1, 3]] in carrier-label order. The actual zero and one are distinct and need not be labels 0 and 1. The inverse audit is additional to, not a replacement for, those ring axioms. Division by zero is never defined. A finite field has prime-power order, not necessarily prime order. Give (n,zero,one,missing_inverses,field?). List all nonzero labels without a two-sided inverse increasingly.

View answer and explanation

(5, 0, 2, [], Yes)

Find both actual identities and search all labels for each nonzero element's two-sided inverse. Missing inverses refute field qualification even when the parent is a commutative ring. The exact result, retaining the requested order, is (5, 0, 2, [], Yes).

Reviewed step-by-step derivation

Finite field · qf-mod30-field-qualification · candidate

12. Audit Every Nonzero Inverse after Complete Commutative-Ring Qualification

Use the same already qualified complete finite commutative unital-ring tables on Fin(5)={0,…,4}, addition=[[0, 1, 2, 3, 4], [1, 3, 0, 4, 2], [2, 0, 4, 1, 3], [3, 4, 1, 2, 0], [4, 2, 3, 0, 1]] and multiplication=[[0, 0, 0, 0, 0], [0, 2, 1, 4, 3], [0, 1, 2, 3, 4], [0, 4, 3, 1, 2], [0, 3, 4, 2, 1]] in carrier-label order. The actual zero and one are distinct and need not be labels 0 and 1. The inverse audit is additional to, not a replacement for, those ring axioms. Division by zero is never defined. A finite field has prime-power order, not necessarily prime order. Give (n,zero,one,missing_inverses,field?). List all nonzero labels without a two-sided inverse increasingly.

View answer and explanation

(5, 0, 2, [], Yes)

Find both actual identities and search all labels for each nonzero element's two-sided inverse. Missing inverses refute field qualification even when the parent is a commutative ring. The exact result, retaining the requested order, is (5, 0, 2, [], Yes).

Reviewed step-by-step derivation

Finite field · qf-mod30-field-qualification · candidate

13. Audit Every Nonzero Inverse after Complete Commutative-Ring Qualification

Use the same already qualified complete finite commutative unital-ring tables on Fin(5)={0,…,4}, addition=[[0, 1, 2, 3, 4], [1, 4, 0, 2, 3], [2, 0, 3, 4, 1], [3, 2, 4, 1, 0], [4, 3, 1, 0, 2]] and multiplication=[[0, 0, 0, 0, 0], [0, 4, 3, 1, 2], [0, 3, 4, 2, 1], [0, 1, 2, 3, 4], [0, 2, 1, 4, 3]] in carrier-label order. The actual zero and one are distinct and need not be labels 0 and 1. The inverse audit is additional to, not a replacement for, those ring axioms. Division by zero is never defined. A finite field has prime-power order, not necessarily prime order. Give (n,zero,one,missing_inverses,field?). List all nonzero labels without a two-sided inverse increasingly.

View answer and explanation

(5, 0, 3, [], Yes)

Find both actual identities and search all labels for each nonzero element's two-sided inverse. Missing inverses refute field qualification even when the parent is a commutative ring. The exact result, retaining the requested order, is (5, 0, 3, [], Yes).

Reviewed step-by-step derivation

Finite field · qf-mod30-field-qualification · candidate

14. Audit Every Nonzero Inverse after Complete Commutative-Ring Qualification

Use the same already qualified complete finite commutative unital-ring tables on Fin(5)={0,…,4}, addition=[[0, 1, 2, 3, 4], [1, 2, 3, 4, 0], [2, 3, 4, 0, 1], [3, 4, 0, 1, 2], [4, 0, 1, 2, 3]] and multiplication=[[0, 0, 0, 0, 0], [0, 2, 4, 1, 3], [0, 4, 3, 2, 1], [0, 1, 2, 3, 4], [0, 3, 1, 4, 2]] in carrier-label order. The actual zero and one are distinct and need not be labels 0 and 1. The inverse audit is additional to, not a replacement for, those ring axioms. Division by zero is never defined. A finite field has prime-power order, not necessarily prime order. Give (n,zero,one,missing_inverses,field?). List all nonzero labels without a two-sided inverse increasingly.

View answer and explanation

(5, 0, 3, [], Yes)

Find both actual identities and search all labels for each nonzero element's two-sided inverse. Missing inverses refute field qualification even when the parent is a commutative ring. The exact result, retaining the requested order, is (5, 0, 3, [], Yes).

Reviewed step-by-step derivation

Finite field · qf-mod30-field-qualification · candidate

15. Audit Every Nonzero Inverse after Complete Commutative-Ring Qualification

Use the same already qualified complete finite commutative unital-ring tables on Fin(5)={0,…,4}, addition=[[0, 1, 2, 3, 4], [1, 3, 0, 4, 2], [2, 0, 4, 1, 3], [3, 4, 1, 2, 0], [4, 2, 3, 0, 1]] and multiplication=[[0, 0, 0, 0, 0], [0, 4, 3, 1, 2], [0, 3, 4, 2, 1], [0, 1, 2, 3, 4], [0, 2, 1, 4, 3]] in carrier-label order. The actual zero and one are distinct and need not be labels 0 and 1. The inverse audit is additional to, not a replacement for, those ring axioms. Division by zero is never defined. A finite field has prime-power order, not necessarily prime order. Give (n,zero,one,missing_inverses,field?). List all nonzero labels without a two-sided inverse increasingly.

View answer and explanation

(5, 0, 3, [], Yes)

Find both actual identities and search all labels for each nonzero element's two-sided inverse. Missing inverses refute field qualification even when the parent is a commutative ring. The exact result, retaining the requested order, is (5, 0, 3, [], Yes).

Reviewed step-by-step derivation

Finite field · qf-mod30-field-qualification · candidate

16. Audit Every Nonzero Inverse after Complete Commutative-Ring Qualification

Use the same already qualified complete finite commutative unital-ring tables on Fin(5)={0,…,4}, addition=[[0, 1, 2, 3, 4], [1, 4, 3, 0, 2], [2, 3, 1, 4, 0], [3, 0, 4, 2, 1], [4, 2, 0, 1, 3]] and multiplication=[[0, 0, 0, 0, 0], [0, 3, 4, 1, 2], [0, 4, 1, 2, 3], [0, 1, 2, 3, 4], [0, 2, 3, 4, 1]] in carrier-label order. The actual zero and one are distinct and need not be labels 0 and 1. The inverse audit is additional to, not a replacement for, those ring axioms. Division by zero is never defined. A finite field has prime-power order, not necessarily prime order. Give (n,zero,one,missing_inverses,field?). List all nonzero labels without a two-sided inverse increasingly.

View answer and explanation

(5, 0, 3, [], Yes)

Find both actual identities and search all labels for each nonzero element's two-sided inverse. Missing inverses refute field qualification even when the parent is a commutative ring. The exact result, retaining the requested order, is (5, 0, 3, [], Yes).

Reviewed step-by-step derivation

Finite field · qf-mod30-field-qualification · candidate

17. Audit Every Nonzero Inverse after Complete Commutative-Ring Qualification

Use the same already qualified complete finite commutative unital-ring tables on Fin(5)={0,…,4}, addition=[[0, 1, 2, 3, 4], [1, 3, 4, 2, 0], [2, 4, 1, 0, 3], [3, 2, 0, 4, 1], [4, 0, 3, 1, 2]] and multiplication=[[0, 0, 0, 0, 0], [0, 2, 4, 1, 3], [0, 4, 3, 2, 1], [0, 1, 2, 3, 4], [0, 3, 1, 4, 2]] in carrier-label order. The actual zero and one are distinct and need not be labels 0 and 1. The inverse audit is additional to, not a replacement for, those ring axioms. Division by zero is never defined. A finite field has prime-power order, not necessarily prime order. Give (n,zero,one,missing_inverses,field?). List all nonzero labels without a two-sided inverse increasingly.

View answer and explanation

(5, 0, 3, [], Yes)

Find both actual identities and search all labels for each nonzero element's two-sided inverse. Missing inverses refute field qualification even when the parent is a commutative ring. The exact result, retaining the requested order, is (5, 0, 3, [], Yes).

Reviewed step-by-step derivation

Finite field · qf-mod30-field-qualification · candidate

18. Audit Every Nonzero Inverse after Complete Commutative-Ring Qualification

Use the same already qualified complete finite commutative unital-ring tables on Fin(5)={0,…,4}, addition=[[0, 1, 2, 3, 4], [1, 2, 4, 0, 3], [2, 4, 3, 1, 0], [3, 0, 1, 4, 2], [4, 3, 0, 2, 1]] and multiplication=[[0, 0, 0, 0, 0], [0, 3, 4, 1, 2], [0, 4, 1, 2, 3], [0, 1, 2, 3, 4], [0, 2, 3, 4, 1]] in carrier-label order. The actual zero and one are distinct and need not be labels 0 and 1. The inverse audit is additional to, not a replacement for, those ring axioms. Division by zero is never defined. A finite field has prime-power order, not necessarily prime order. Give (n,zero,one,missing_inverses,field?). List all nonzero labels without a two-sided inverse increasingly.

View answer and explanation

(5, 0, 3, [], Yes)

Find both actual identities and search all labels for each nonzero element's two-sided inverse. Missing inverses refute field qualification even when the parent is a commutative ring. The exact result, retaining the requested order, is (5, 0, 3, [], Yes).

Reviewed step-by-step derivation

Finite field · qf-mod30-field-qualification · candidate

19. Audit Every Nonzero Inverse after Complete Commutative-Ring Qualification

Use the same already qualified complete finite commutative unital-ring tables on Fin(5)={0,…,4}, addition=[[0, 1, 2, 3, 4], [1, 3, 0, 4, 2], [2, 0, 4, 1, 3], [3, 4, 1, 2, 0], [4, 2, 3, 0, 1]] and multiplication=[[0, 0, 0, 0, 0], [0, 3, 4, 2, 1], [0, 4, 3, 1, 2], [0, 2, 1, 4, 3], [0, 1, 2, 3, 4]] in carrier-label order. The actual zero and one are distinct and need not be labels 0 and 1. The inverse audit is additional to, not a replacement for, those ring axioms. Division by zero is never defined. A finite field has prime-power order, not necessarily prime order. Give (n,zero,one,missing_inverses,field?). List all nonzero labels without a two-sided inverse increasingly.

View answer and explanation

(5, 0, 4, [], Yes)

Find both actual identities and search all labels for each nonzero element's two-sided inverse. Missing inverses refute field qualification even when the parent is a commutative ring. The exact result, retaining the requested order, is (5, 0, 4, [], Yes).

Reviewed step-by-step derivation

Finite field · qf-mod30-field-qualification · candidate

20. Audit Every Nonzero Inverse after Complete Commutative-Ring Qualification

Use the same already qualified complete finite commutative unital-ring tables on Fin(5)={0,…,4}, addition=[[0, 1, 2, 3, 4], [1, 2, 4, 0, 3], [2, 4, 3, 1, 0], [3, 0, 1, 4, 2], [4, 3, 0, 2, 1]] and multiplication=[[0, 0, 0, 0, 0], [0, 2, 3, 4, 1], [0, 3, 4, 1, 2], [0, 4, 1, 2, 3], [0, 1, 2, 3, 4]] in carrier-label order. The actual zero and one are distinct and need not be labels 0 and 1. The inverse audit is additional to, not a replacement for, those ring axioms. Division by zero is never defined. A finite field has prime-power order, not necessarily prime order. Give (n,zero,one,missing_inverses,field?). List all nonzero labels without a two-sided inverse increasingly.

View answer and explanation

(5, 0, 4, [], Yes)

Find both actual identities and search all labels for each nonzero element's two-sided inverse. Missing inverses refute field qualification even when the parent is a commutative ring. The exact result, retaining the requested order, is (5, 0, 4, [], Yes).

Reviewed step-by-step derivation

Finite field · qf-mod30-field-qualification · candidate

21. Audit Every Nonzero Inverse after Complete Commutative-Ring Qualification

Use the same already qualified complete finite commutative unital-ring tables on Fin(5)={0,…,4}, addition=[[0, 1, 2, 3, 4], [1, 4, 0, 2, 3], [2, 0, 3, 4, 1], [3, 2, 4, 1, 0], [4, 3, 1, 0, 2]] and multiplication=[[0, 0, 0, 0, 0], [0, 3, 4, 2, 1], [0, 4, 3, 1, 2], [0, 2, 1, 4, 3], [0, 1, 2, 3, 4]] in carrier-label order. The actual zero and one are distinct and need not be labels 0 and 1. The inverse audit is additional to, not a replacement for, those ring axioms. Division by zero is never defined. A finite field has prime-power order, not necessarily prime order. Give (n,zero,one,missing_inverses,field?). List all nonzero labels without a two-sided inverse increasingly.

View answer and explanation

(5, 0, 4, [], Yes)

Find both actual identities and search all labels for each nonzero element's two-sided inverse. Missing inverses refute field qualification even when the parent is a commutative ring. The exact result, retaining the requested order, is (5, 0, 4, [], Yes).

Reviewed step-by-step derivation

Finite field · qf-mod30-field-qualification · candidate

22. Audit Every Nonzero Inverse after Complete Commutative-Ring Qualification

Use the same already qualified complete finite commutative unital-ring tables on Fin(5)={0,…,4}, addition=[[0, 1, 2, 3, 4], [1, 3, 4, 2, 0], [2, 4, 1, 0, 3], [3, 2, 0, 4, 1], [4, 0, 3, 1, 2]] and multiplication=[[0, 0, 0, 0, 0], [0, 4, 3, 2, 1], [0, 3, 1, 4, 2], [0, 2, 4, 1, 3], [0, 1, 2, 3, 4]] in carrier-label order. The actual zero and one are distinct and need not be labels 0 and 1. The inverse audit is additional to, not a replacement for, those ring axioms. Division by zero is never defined. A finite field has prime-power order, not necessarily prime order. Give (n,zero,one,missing_inverses,field?). List all nonzero labels without a two-sided inverse increasingly.

View answer and explanation

(5, 0, 4, [], Yes)

Find both actual identities and search all labels for each nonzero element's two-sided inverse. Missing inverses refute field qualification even when the parent is a commutative ring. The exact result, retaining the requested order, is (5, 0, 4, [], Yes).

Reviewed step-by-step derivation

Finite field · qf-mod30-field-qualification · candidate

23. Audit Every Nonzero Inverse after Complete Commutative-Ring Qualification

Use the same already qualified complete finite commutative unital-ring tables on Fin(5)={0,…,4}, addition=[[0, 1, 2, 3, 4], [1, 4, 3, 0, 2], [2, 3, 1, 4, 0], [3, 0, 4, 2, 1], [4, 2, 0, 1, 3]] and multiplication=[[0, 0, 0, 0, 0], [0, 2, 3, 4, 1], [0, 3, 4, 1, 2], [0, 4, 1, 2, 3], [0, 1, 2, 3, 4]] in carrier-label order. The actual zero and one are distinct and need not be labels 0 and 1. The inverse audit is additional to, not a replacement for, those ring axioms. Division by zero is never defined. A finite field has prime-power order, not necessarily prime order. Give (n,zero,one,missing_inverses,field?). List all nonzero labels without a two-sided inverse increasingly.

View answer and explanation

(5, 0, 4, [], Yes)

Find both actual identities and search all labels for each nonzero element's two-sided inverse. Missing inverses refute field qualification even when the parent is a commutative ring. The exact result, retaining the requested order, is (5, 0, 4, [], Yes).

Reviewed step-by-step derivation

Finite field · qf-mod30-field-qualification · candidate

24. Audit Every Nonzero Inverse after Complete Commutative-Ring Qualification

Use the same already qualified complete finite commutative unital-ring tables on Fin(5)={0,…,4}, addition=[[0, 1, 2, 3, 4], [1, 2, 3, 4, 0], [2, 3, 4, 0, 1], [3, 4, 0, 1, 2], [4, 0, 1, 2, 3]] and multiplication=[[0, 0, 0, 0, 0], [0, 4, 3, 2, 1], [0, 3, 1, 4, 2], [0, 2, 4, 1, 3], [0, 1, 2, 3, 4]] in carrier-label order. The actual zero and one are distinct and need not be labels 0 and 1. The inverse audit is additional to, not a replacement for, those ring axioms. Division by zero is never defined. A finite field has prime-power order, not necessarily prime order. Give (n,zero,one,missing_inverses,field?). List all nonzero labels without a two-sided inverse increasingly.

View answer and explanation

(5, 0, 4, [], Yes)

Find both actual identities and search all labels for each nonzero element's two-sided inverse. Missing inverses refute field qualification even when the parent is a commutative ring. The exact result, retaining the requested order, is (5, 0, 4, [], Yes).

Reviewed step-by-step derivation

Finite field · qf-mod30-field-qualification · candidate

25. Audit Every Nonzero Inverse after Complete Commutative-Ring Qualification

Use the same already qualified complete finite commutative unital-ring tables on Fin(5)={0,…,4}, addition=[[2, 0, 3, 4, 1], [0, 1, 2, 3, 4], [3, 2, 4, 1, 0], [4, 3, 1, 0, 2], [1, 4, 0, 2, 3]] and multiplication=[[0, 1, 2, 3, 4], [1, 1, 1, 1, 1], [2, 1, 4, 0, 3], [3, 1, 0, 4, 2], [4, 1, 3, 2, 0]] in carrier-label order. The actual zero and one are distinct and need not be labels 0 and 1. The inverse audit is additional to, not a replacement for, those ring axioms. Division by zero is never defined. A finite field has prime-power order, not necessarily prime order. Give (n,zero,one,missing_inverses,field?). List all nonzero labels without a two-sided inverse increasingly.

View answer and explanation

(5, 1, 0, [], Yes)

Find both actual identities and search all labels for each nonzero element's two-sided inverse. Missing inverses refute field qualification even when the parent is a commutative ring. The exact result, retaining the requested order, is (5, 1, 0, [], Yes).

Reviewed step-by-step derivation

Finite field · qf-mod30-field-qualification · candidate

26. Find a Two-Sided Inverse of the Specified Nonzero Element

Use the same already qualified complete finite commutative unital-ring tables on Fin(5)={0,…,4}, addition=[[0, 1, 2, 3, 4], [1, 2, 3, 4, 0], [2, 3, 4, 0, 1], [3, 4, 0, 1, 2], [4, 0, 1, 2, 3]] and multiplication=[[0, 0, 0, 0, 0], [0, 1, 2, 3, 4], [0, 2, 4, 1, 3], [0, 3, 1, 4, 2], [0, 4, 3, 2, 1]] in carrier-label order. The actual zero and one are distinct and need not be labels 0 and 1. The inverse audit is additional to, not a replacement for, those ring axioms. Division by zero is never defined. A finite field has prime-power order, not necessarily prime order. element=1 is nonzero. Give (element,one,inverse,found?).

View answer and explanation

(1, 1, 1, Yes)

Search carrier labels increasingly and return the first element whose left and right products are the actual multiplicative identity. The field premise provides existence; zero is rejected rather than assigned an inverse. The exact result, retaining the requested order, is (1, 1, 1, Yes).

Reviewed step-by-step derivation

Finite field · qf-mod30-field-inverse · candidate

27. Find a Two-Sided Inverse of the Specified Nonzero Element

Use the same already qualified complete finite commutative unital-ring tables on Fin(5)={0,…,4}, addition=[[0, 1, 2, 3, 4], [1, 2, 4, 0, 3], [2, 4, 3, 1, 0], [3, 0, 1, 4, 2], [4, 3, 0, 2, 1]] and multiplication=[[0, 0, 0, 0, 0], [0, 1, 2, 3, 4], [0, 2, 3, 4, 1], [0, 3, 4, 1, 2], [0, 4, 1, 2, 3]] in carrier-label order. The actual zero and one are distinct and need not be labels 0 and 1. The inverse audit is additional to, not a replacement for, those ring axioms. Division by zero is never defined. A finite field has prime-power order, not necessarily prime order. element=2 is nonzero. Give (element,one,inverse,found?).

View answer and explanation

(2, 1, 4, Yes)

Search carrier labels increasingly and return the first element whose left and right products are the actual multiplicative identity. The field premise provides existence; zero is rejected rather than assigned an inverse. The exact result, retaining the requested order, is (2, 1, 4, Yes).

Reviewed step-by-step derivation

Finite field · qf-mod30-field-inverse · candidate

28. Find a Two-Sided Inverse of the Specified Nonzero Element

Use the same already qualified complete finite commutative unital-ring tables on Fin(5)={0,…,4}, addition=[[0, 1, 2, 3, 4], [1, 3, 4, 2, 0], [2, 4, 1, 0, 3], [3, 2, 0, 4, 1], [4, 0, 3, 1, 2]] and multiplication=[[0, 0, 0, 0, 0], [0, 1, 2, 3, 4], [0, 2, 4, 1, 3], [0, 3, 1, 4, 2], [0, 4, 3, 2, 1]] in carrier-label order. The actual zero and one are distinct and need not be labels 0 and 1. The inverse audit is additional to, not a replacement for, those ring axioms. Division by zero is never defined. A finite field has prime-power order, not necessarily prime order. element=2 is nonzero. Give (element,one,inverse,found?).

View answer and explanation

(2, 1, 3, Yes)

Search carrier labels increasingly and return the first element whose left and right products are the actual multiplicative identity. The field premise provides existence; zero is rejected rather than assigned an inverse. The exact result, retaining the requested order, is (2, 1, 3, Yes).

Reviewed step-by-step derivation

Finite field · qf-mod30-field-inverse · candidate

29. Find a Two-Sided Inverse of the Specified Nonzero Element

Use the same already qualified complete finite commutative unital-ring tables on Fin(5)={0,…,4}, addition=[[0, 1, 2, 3, 4], [1, 3, 0, 4, 2], [2, 0, 4, 1, 3], [3, 4, 1, 2, 0], [4, 2, 3, 0, 1]] and multiplication=[[0, 0, 0, 0, 0], [0, 1, 2, 3, 4], [0, 2, 1, 4, 3], [0, 3, 4, 2, 1], [0, 4, 3, 1, 2]] in carrier-label order. The actual zero and one are distinct and need not be labels 0 and 1. The inverse audit is additional to, not a replacement for, those ring axioms. Division by zero is never defined. A finite field has prime-power order, not necessarily prime order. element=2 is nonzero. Give (element,one,inverse,found?).

View answer and explanation

(2, 1, 2, Yes)

Search carrier labels increasingly and return the first element whose left and right products are the actual multiplicative identity. The field premise provides existence; zero is rejected rather than assigned an inverse. The exact result, retaining the requested order, is (2, 1, 2, Yes).

Reviewed step-by-step derivation

Finite field · qf-mod30-field-inverse · candidate

30. Find a Two-Sided Inverse of the Specified Nonzero Element

Use the same already qualified complete finite commutative unital-ring tables on Fin(5)={0,…,4}, addition=[[0, 1, 2, 3, 4], [1, 4, 3, 0, 2], [2, 3, 1, 4, 0], [3, 0, 4, 2, 1], [4, 2, 0, 1, 3]] and multiplication=[[0, 0, 0, 0, 0], [0, 1, 2, 3, 4], [0, 2, 3, 4, 1], [0, 3, 4, 1, 2], [0, 4, 1, 2, 3]] in carrier-label order. The actual zero and one are distinct and need not be labels 0 and 1. The inverse audit is additional to, not a replacement for, those ring axioms. Division by zero is never defined. A finite field has prime-power order, not necessarily prime order. element=1 is nonzero. Give (element,one,inverse,found?).

View answer and explanation

(1, 1, 1, Yes)

Search carrier labels increasingly and return the first element whose left and right products are the actual multiplicative identity. The field premise provides existence; zero is rejected rather than assigned an inverse. The exact result, retaining the requested order, is (1, 1, 1, Yes).

Reviewed step-by-step derivation

Finite field · qf-mod30-field-inverse · candidate

31. Find a Two-Sided Inverse of the Specified Nonzero Element

Use the same already qualified complete finite commutative unital-ring tables on Fin(5)={0,…,4}, addition=[[0, 1, 2, 3, 4], [1, 4, 0, 2, 3], [2, 0, 3, 4, 1], [3, 2, 4, 1, 0], [4, 3, 1, 0, 2]] and multiplication=[[0, 0, 0, 0, 0], [0, 1, 2, 3, 4], [0, 2, 1, 4, 3], [0, 3, 4, 2, 1], [0, 4, 3, 1, 2]] in carrier-label order. The actual zero and one are distinct and need not be labels 0 and 1. The inverse audit is additional to, not a replacement for, those ring axioms. Division by zero is never defined. A finite field has prime-power order, not necessarily prime order. element=4 is nonzero. Give (element,one,inverse,found?).

View answer and explanation

(4, 1, 3, Yes)

Search carrier labels increasingly and return the first element whose left and right products are the actual multiplicative identity. The field premise provides existence; zero is rejected rather than assigned an inverse. The exact result, retaining the requested order, is (4, 1, 3, Yes).

Reviewed step-by-step derivation

Finite field · qf-mod30-field-inverse · candidate

32. Find a Two-Sided Inverse of the Specified Nonzero Element

Use the same already qualified complete finite commutative unital-ring tables on Fin(5)={0,…,4}, addition=[[0, 1, 2, 3, 4], [1, 4, 3, 0, 2], [2, 3, 1, 4, 0], [3, 0, 4, 2, 1], [4, 2, 0, 1, 3]] and multiplication=[[0, 0, 0, 0, 0], [0, 4, 1, 2, 3], [0, 1, 2, 3, 4], [0, 2, 3, 4, 1], [0, 3, 4, 1, 2]] in carrier-label order. The actual zero and one are distinct and need not be labels 0 and 1. The inverse audit is additional to, not a replacement for, those ring axioms. Division by zero is never defined. A finite field has prime-power order, not necessarily prime order. element=3 is nonzero. Give (element,one,inverse,found?).

View answer and explanation

(3, 2, 1, Yes)

Search carrier labels increasingly and return the first element whose left and right products are the actual multiplicative identity. The field premise provides existence; zero is rejected rather than assigned an inverse. The exact result, retaining the requested order, is (3, 2, 1, Yes).

Reviewed step-by-step derivation

Finite field · qf-mod30-field-inverse · candidate

33. Find a Two-Sided Inverse of the Specified Nonzero Element

Use the same already qualified complete finite commutative unital-ring tables on Fin(5)={0,…,4}, addition=[[0, 1, 2, 3, 4], [1, 3, 4, 2, 0], [2, 4, 1, 0, 3], [3, 2, 0, 4, 1], [4, 0, 3, 1, 2]] and multiplication=[[0, 0, 0, 0, 0], [0, 3, 1, 4, 2], [0, 1, 2, 3, 4], [0, 4, 3, 2, 1], [0, 2, 4, 1, 3]] in carrier-label order. The actual zero and one are distinct and need not be labels 0 and 1. The inverse audit is additional to, not a replacement for, those ring axioms. Division by zero is never defined. A finite field has prime-power order, not necessarily prime order. element=3 is nonzero. Give (element,one,inverse,found?).

View answer and explanation

(3, 2, 3, Yes)

Search carrier labels increasingly and return the first element whose left and right products are the actual multiplicative identity. The field premise provides existence; zero is rejected rather than assigned an inverse. The exact result, retaining the requested order, is (3, 2, 3, Yes).

Reviewed step-by-step derivation

Finite field · qf-mod30-field-inverse · candidate

34. Find a Two-Sided Inverse of the Specified Nonzero Element

Use the same already qualified complete finite commutative unital-ring tables on Fin(5)={0,…,4}, addition=[[0, 1, 2, 3, 4], [1, 2, 4, 0, 3], [2, 4, 3, 1, 0], [3, 0, 1, 4, 2], [4, 3, 0, 2, 1]] and multiplication=[[0, 0, 0, 0, 0], [0, 4, 1, 2, 3], [0, 1, 2, 3, 4], [0, 2, 3, 4, 1], [0, 3, 4, 1, 2]] in carrier-label order. The actual zero and one are distinct and need not be labels 0 and 1. The inverse audit is additional to, not a replacement for, those ring axioms. Division by zero is never defined. A finite field has prime-power order, not necessarily prime order. element=2 is nonzero. Give (element,one,inverse,found?).

View answer and explanation

(2, 2, 2, Yes)

Search carrier labels increasingly and return the first element whose left and right products are the actual multiplicative identity. The field premise provides existence; zero is rejected rather than assigned an inverse. The exact result, retaining the requested order, is (2, 2, 2, Yes).

Reviewed step-by-step derivation

Finite field · qf-mod30-field-inverse · candidate

35. Find a Two-Sided Inverse of the Specified Nonzero Element

Use the same already qualified complete finite commutative unital-ring tables on Fin(5)={0,…,4}, addition=[[0, 1, 2, 3, 4], [1, 4, 0, 2, 3], [2, 0, 3, 4, 1], [3, 2, 4, 1, 0], [4, 3, 1, 0, 2]] and multiplication=[[0, 0, 0, 0, 0], [0, 2, 1, 4, 3], [0, 1, 2, 3, 4], [0, 4, 3, 1, 2], [0, 3, 4, 2, 1]] in carrier-label order. The actual zero and one are distinct and need not be labels 0 and 1. The inverse audit is additional to, not a replacement for, those ring axioms. Division by zero is never defined. A finite field has prime-power order, not necessarily prime order. element=3 is nonzero. Give (element,one,inverse,found?).

View answer and explanation

(3, 2, 4, Yes)

Search carrier labels increasingly and return the first element whose left and right products are the actual multiplicative identity. The field premise provides existence; zero is rejected rather than assigned an inverse. The exact result, retaining the requested order, is (3, 2, 4, Yes).

Reviewed step-by-step derivation

Finite field · qf-mod30-field-inverse · candidate

36. Find a Two-Sided Inverse of the Specified Nonzero Element

Use the same already qualified complete finite commutative unital-ring tables on Fin(5)={0,…,4}, addition=[[0, 1, 2, 3, 4], [1, 2, 3, 4, 0], [2, 3, 4, 0, 1], [3, 4, 0, 1, 2], [4, 0, 1, 2, 3]] and multiplication=[[0, 0, 0, 0, 0], [0, 3, 1, 4, 2], [0, 1, 2, 3, 4], [0, 4, 3, 2, 1], [0, 2, 4, 1, 3]] in carrier-label order. The actual zero and one are distinct and need not be labels 0 and 1. The inverse audit is additional to, not a replacement for, those ring axioms. Division by zero is never defined. A finite field has prime-power order, not necessarily prime order. element=1 is nonzero. Give (element,one,inverse,found?).

View answer and explanation

(1, 2, 4, Yes)

Search carrier labels increasingly and return the first element whose left and right products are the actual multiplicative identity. The field premise provides existence; zero is rejected rather than assigned an inverse. The exact result, retaining the requested order, is (1, 2, 4, Yes).

Reviewed step-by-step derivation

Finite field · qf-mod30-field-inverse · candidate

37. Find a Two-Sided Inverse of the Specified Nonzero Element

Use the same already qualified complete finite commutative unital-ring tables on Fin(5)={0,…,4}, addition=[[0, 1, 2, 3, 4], [1, 3, 0, 4, 2], [2, 0, 4, 1, 3], [3, 4, 1, 2, 0], [4, 2, 3, 0, 1]] and multiplication=[[0, 0, 0, 0, 0], [0, 2, 1, 4, 3], [0, 1, 2, 3, 4], [0, 4, 3, 1, 2], [0, 3, 4, 2, 1]] in carrier-label order. The actual zero and one are distinct and need not be labels 0 and 1. The inverse audit is additional to, not a replacement for, those ring axioms. Division by zero is never defined. A finite field has prime-power order, not necessarily prime order. element=1 is nonzero. Give (element,one,inverse,found?).

View answer and explanation

(1, 2, 1, Yes)

Search carrier labels increasingly and return the first element whose left and right products are the actual multiplicative identity. The field premise provides existence; zero is rejected rather than assigned an inverse. The exact result, retaining the requested order, is (1, 2, 1, Yes).

Reviewed step-by-step derivation

Finite field · qf-mod30-field-inverse · candidate

38. Find a Two-Sided Inverse of the Specified Nonzero Element

Use the same already qualified complete finite commutative unital-ring tables on Fin(5)={0,…,4}, addition=[[0, 1, 2, 3, 4], [1, 4, 0, 2, 3], [2, 0, 3, 4, 1], [3, 2, 4, 1, 0], [4, 3, 1, 0, 2]] and multiplication=[[0, 0, 0, 0, 0], [0, 4, 3, 1, 2], [0, 3, 4, 2, 1], [0, 1, 2, 3, 4], [0, 2, 1, 4, 3]] in carrier-label order. The actual zero and one are distinct and need not be labels 0 and 1. The inverse audit is additional to, not a replacement for, those ring axioms. Division by zero is never defined. A finite field has prime-power order, not necessarily prime order. element=3 is nonzero. Give (element,one,inverse,found?).

View answer and explanation

(3, 3, 3, Yes)

Search carrier labels increasingly and return the first element whose left and right products are the actual multiplicative identity. The field premise provides existence; zero is rejected rather than assigned an inverse. The exact result, retaining the requested order, is (3, 3, 3, Yes).

Reviewed step-by-step derivation

Finite field · qf-mod30-field-inverse · candidate

39. Find a Two-Sided Inverse of the Specified Nonzero Element

Use the same already qualified complete finite commutative unital-ring tables on Fin(5)={0,…,4}, addition=[[0, 1, 2, 3, 4], [1, 2, 3, 4, 0], [2, 3, 4, 0, 1], [3, 4, 0, 1, 2], [4, 0, 1, 2, 3]] and multiplication=[[0, 0, 0, 0, 0], [0, 2, 4, 1, 3], [0, 4, 3, 2, 1], [0, 1, 2, 3, 4], [0, 3, 1, 4, 2]] in carrier-label order. The actual zero and one are distinct and need not be labels 0 and 1. The inverse audit is additional to, not a replacement for, those ring axioms. Division by zero is never defined. A finite field has prime-power order, not necessarily prime order. element=1 is nonzero. Give (element,one,inverse,found?).

View answer and explanation

(1, 3, 4, Yes)

Search carrier labels increasingly and return the first element whose left and right products are the actual multiplicative identity. The field premise provides existence; zero is rejected rather than assigned an inverse. The exact result, retaining the requested order, is (1, 3, 4, Yes).

Reviewed step-by-step derivation

Finite field · qf-mod30-field-inverse · candidate

40. Find a Two-Sided Inverse of the Specified Nonzero Element

Use the same already qualified complete finite commutative unital-ring tables on Fin(5)={0,…,4}, addition=[[0, 1, 2, 3, 4], [1, 3, 0, 4, 2], [2, 0, 4, 1, 3], [3, 4, 1, 2, 0], [4, 2, 3, 0, 1]] and multiplication=[[0, 0, 0, 0, 0], [0, 4, 3, 1, 2], [0, 3, 4, 2, 1], [0, 1, 2, 3, 4], [0, 2, 1, 4, 3]] in carrier-label order. The actual zero and one are distinct and need not be labels 0 and 1. The inverse audit is additional to, not a replacement for, those ring axioms. Division by zero is never defined. A finite field has prime-power order, not necessarily prime order. element=1 is nonzero. Give (element,one,inverse,found?).

View answer and explanation

(1, 3, 2, Yes)

Search carrier labels increasingly and return the first element whose left and right products are the actual multiplicative identity. The field premise provides existence; zero is rejected rather than assigned an inverse. The exact result, retaining the requested order, is (1, 3, 2, Yes).

Reviewed step-by-step derivation

Finite field · qf-mod30-field-inverse · candidate

41. Find a Two-Sided Inverse of the Specified Nonzero Element

Use the same already qualified complete finite commutative unital-ring tables on Fin(5)={0,…,4}, addition=[[0, 1, 2, 3, 4], [1, 4, 3, 0, 2], [2, 3, 1, 4, 0], [3, 0, 4, 2, 1], [4, 2, 0, 1, 3]] and multiplication=[[0, 0, 0, 0, 0], [0, 3, 4, 1, 2], [0, 4, 1, 2, 3], [0, 1, 2, 3, 4], [0, 2, 3, 4, 1]] in carrier-label order. The actual zero and one are distinct and need not be labels 0 and 1. The inverse audit is additional to, not a replacement for, those ring axioms. Division by zero is never defined. A finite field has prime-power order, not necessarily prime order. element=1 is nonzero. Give (element,one,inverse,found?).

View answer and explanation

(1, 3, 1, Yes)

Search carrier labels increasingly and return the first element whose left and right products are the actual multiplicative identity. The field premise provides existence; zero is rejected rather than assigned an inverse. The exact result, retaining the requested order, is (1, 3, 1, Yes).

Reviewed step-by-step derivation

Finite field · qf-mod30-field-inverse · candidate

42. Find a Two-Sided Inverse of the Specified Nonzero Element

Use the same already qualified complete finite commutative unital-ring tables on Fin(5)={0,…,4}, addition=[[0, 1, 2, 3, 4], [1, 3, 4, 2, 0], [2, 4, 1, 0, 3], [3, 2, 0, 4, 1], [4, 0, 3, 1, 2]] and multiplication=[[0, 0, 0, 0, 0], [0, 2, 4, 1, 3], [0, 4, 3, 2, 1], [0, 1, 2, 3, 4], [0, 3, 1, 4, 2]] in carrier-label order. The actual zero and one are distinct and need not be labels 0 and 1. The inverse audit is additional to, not a replacement for, those ring axioms. Division by zero is never defined. A finite field has prime-power order, not necessarily prime order. element=3 is nonzero. Give (element,one,inverse,found?).

View answer and explanation

(3, 3, 3, Yes)

Search carrier labels increasingly and return the first element whose left and right products are the actual multiplicative identity. The field premise provides existence; zero is rejected rather than assigned an inverse. The exact result, retaining the requested order, is (3, 3, 3, Yes).

Reviewed step-by-step derivation

Finite field · qf-mod30-field-inverse · candidate

43. Find a Two-Sided Inverse of the Specified Nonzero Element

Use the same already qualified complete finite commutative unital-ring tables on Fin(5)={0,…,4}, addition=[[0, 1, 2, 3, 4], [1, 2, 4, 0, 3], [2, 4, 3, 1, 0], [3, 0, 1, 4, 2], [4, 3, 0, 2, 1]] and multiplication=[[0, 0, 0, 0, 0], [0, 3, 4, 1, 2], [0, 4, 1, 2, 3], [0, 1, 2, 3, 4], [0, 2, 3, 4, 1]] in carrier-label order. The actual zero and one are distinct and need not be labels 0 and 1. The inverse audit is additional to, not a replacement for, those ring axioms. Division by zero is never defined. A finite field has prime-power order, not necessarily prime order. element=4 is nonzero. Give (element,one,inverse,found?).

View answer and explanation

(4, 3, 2, Yes)

Search carrier labels increasingly and return the first element whose left and right products are the actual multiplicative identity. The field premise provides existence; zero is rejected rather than assigned an inverse. The exact result, retaining the requested order, is (4, 3, 2, Yes).

Reviewed step-by-step derivation

Finite field · qf-mod30-field-inverse · candidate

44. Find a Two-Sided Inverse of the Specified Nonzero Element

Use the same already qualified complete finite commutative unital-ring tables on Fin(5)={0,…,4}, addition=[[0, 1, 2, 3, 4], [1, 3, 0, 4, 2], [2, 0, 4, 1, 3], [3, 4, 1, 2, 0], [4, 2, 3, 0, 1]] and multiplication=[[0, 0, 0, 0, 0], [0, 3, 4, 2, 1], [0, 4, 3, 1, 2], [0, 2, 1, 4, 3], [0, 1, 2, 3, 4]] in carrier-label order. The actual zero and one are distinct and need not be labels 0 and 1. The inverse audit is additional to, not a replacement for, those ring axioms. Division by zero is never defined. A finite field has prime-power order, not necessarily prime order. element=2 is nonzero. Give (element,one,inverse,found?).

View answer and explanation

(2, 4, 1, Yes)

Search carrier labels increasingly and return the first element whose left and right products are the actual multiplicative identity. The field premise provides existence; zero is rejected rather than assigned an inverse. The exact result, retaining the requested order, is (2, 4, 1, Yes).

Reviewed step-by-step derivation

Finite field · qf-mod30-field-inverse · candidate

45. Find a Two-Sided Inverse of the Specified Nonzero Element

Use the same already qualified complete finite commutative unital-ring tables on Fin(5)={0,…,4}, addition=[[0, 1, 2, 3, 4], [1, 2, 4, 0, 3], [2, 4, 3, 1, 0], [3, 0, 1, 4, 2], [4, 3, 0, 2, 1]] and multiplication=[[0, 0, 0, 0, 0], [0, 2, 3, 4, 1], [0, 3, 4, 1, 2], [0, 4, 1, 2, 3], [0, 1, 2, 3, 4]] in carrier-label order. The actual zero and one are distinct and need not be labels 0 and 1. The inverse audit is additional to, not a replacement for, those ring axioms. Division by zero is never defined. A finite field has prime-power order, not necessarily prime order. element=2 is nonzero. Give (element,one,inverse,found?).

View answer and explanation

(2, 4, 2, Yes)

Search carrier labels increasingly and return the first element whose left and right products are the actual multiplicative identity. The field premise provides existence; zero is rejected rather than assigned an inverse. The exact result, retaining the requested order, is (2, 4, 2, Yes).

Reviewed step-by-step derivation

Finite field · qf-mod30-field-inverse · candidate

46. Find a Two-Sided Inverse of the Specified Nonzero Element

Use the same already qualified complete finite commutative unital-ring tables on Fin(5)={0,…,4}, addition=[[0, 1, 2, 3, 4], [1, 4, 0, 2, 3], [2, 0, 3, 4, 1], [3, 2, 4, 1, 0], [4, 3, 1, 0, 2]] and multiplication=[[0, 0, 0, 0, 0], [0, 3, 4, 2, 1], [0, 4, 3, 1, 2], [0, 2, 1, 4, 3], [0, 1, 2, 3, 4]] in carrier-label order. The actual zero and one are distinct and need not be labels 0 and 1. The inverse audit is additional to, not a replacement for, those ring axioms. Division by zero is never defined. A finite field has prime-power order, not necessarily prime order. element=4 is nonzero. Give (element,one,inverse,found?).

View answer and explanation

(4, 4, 4, Yes)

Search carrier labels increasingly and return the first element whose left and right products are the actual multiplicative identity. The field premise provides existence; zero is rejected rather than assigned an inverse. The exact result, retaining the requested order, is (4, 4, 4, Yes).

Reviewed step-by-step derivation

Finite field · qf-mod30-field-inverse · candidate

47. Find a Two-Sided Inverse of the Specified Nonzero Element

Use the same already qualified complete finite commutative unital-ring tables on Fin(5)={0,…,4}, addition=[[0, 1, 2, 3, 4], [1, 3, 4, 2, 0], [2, 4, 1, 0, 3], [3, 2, 0, 4, 1], [4, 0, 3, 1, 2]] and multiplication=[[0, 0, 0, 0, 0], [0, 4, 3, 2, 1], [0, 3, 1, 4, 2], [0, 2, 4, 1, 3], [0, 1, 2, 3, 4]] in carrier-label order. The actual zero and one are distinct and need not be labels 0 and 1. The inverse audit is additional to, not a replacement for, those ring axioms. Division by zero is never defined. A finite field has prime-power order, not necessarily prime order. element=2 is nonzero. Give (element,one,inverse,found?).

View answer and explanation

(2, 4, 3, Yes)

Search carrier labels increasingly and return the first element whose left and right products are the actual multiplicative identity. The field premise provides existence; zero is rejected rather than assigned an inverse. The exact result, retaining the requested order, is (2, 4, 3, Yes).

Reviewed step-by-step derivation

Finite field · qf-mod30-field-inverse · candidate

48. Find a Two-Sided Inverse of the Specified Nonzero Element

Use the same already qualified complete finite commutative unital-ring tables on Fin(5)={0,…,4}, addition=[[0, 1, 2, 3, 4], [1, 4, 3, 0, 2], [2, 3, 1, 4, 0], [3, 0, 4, 2, 1], [4, 2, 0, 1, 3]] and multiplication=[[0, 0, 0, 0, 0], [0, 2, 3, 4, 1], [0, 3, 4, 1, 2], [0, 4, 1, 2, 3], [0, 1, 2, 3, 4]] in carrier-label order. The actual zero and one are distinct and need not be labels 0 and 1. The inverse audit is additional to, not a replacement for, those ring axioms. Division by zero is never defined. A finite field has prime-power order, not necessarily prime order. element=1 is nonzero. Give (element,one,inverse,found?).

View answer and explanation

(1, 4, 3, Yes)

Search carrier labels increasingly and return the first element whose left and right products are the actual multiplicative identity. The field premise provides existence; zero is rejected rather than assigned an inverse. The exact result, retaining the requested order, is (1, 4, 3, Yes).

Reviewed step-by-step derivation

Finite field · qf-mod30-field-inverse · candidate

49. Find a Two-Sided Inverse of the Specified Nonzero Element

Use the same already qualified complete finite commutative unital-ring tables on Fin(5)={0,…,4}, addition=[[0, 1, 2, 3, 4], [1, 2, 3, 4, 0], [2, 3, 4, 0, 1], [3, 4, 0, 1, 2], [4, 0, 1, 2, 3]] and multiplication=[[0, 0, 0, 0, 0], [0, 4, 3, 2, 1], [0, 3, 1, 4, 2], [0, 2, 4, 1, 3], [0, 1, 2, 3, 4]] in carrier-label order. The actual zero and one are distinct and need not be labels 0 and 1. The inverse audit is additional to, not a replacement for, those ring axioms. Division by zero is never defined. A finite field has prime-power order, not necessarily prime order. element=1 is nonzero. Give (element,one,inverse,found?).

View answer and explanation

(1, 4, 1, Yes)

Search carrier labels increasingly and return the first element whose left and right products are the actual multiplicative identity. The field premise provides existence; zero is rejected rather than assigned an inverse. The exact result, retaining the requested order, is (1, 4, 1, Yes).

Reviewed step-by-step derivation

Finite field · qf-mod30-field-inverse · candidate

50. Find a Two-Sided Inverse of the Specified Nonzero Element

Use the same already qualified complete finite commutative unital-ring tables on Fin(5)={0,…,4}, addition=[[2, 0, 3, 4, 1], [0, 1, 2, 3, 4], [3, 2, 4, 1, 0], [4, 3, 1, 0, 2], [1, 4, 0, 2, 3]] and multiplication=[[0, 1, 2, 3, 4], [1, 1, 1, 1, 1], [2, 1, 4, 0, 3], [3, 1, 0, 4, 2], [4, 1, 3, 2, 0]] in carrier-label order. The actual zero and one are distinct and need not be labels 0 and 1. The inverse audit is additional to, not a replacement for, those ring axioms. Division by zero is never defined. A finite field has prime-power order, not necessarily prime order. element=0 is nonzero. Give (element,one,inverse,found?).

View answer and explanation

(0, 0, 0, Yes)

Search carrier labels increasingly and return the first element whose left and right products are the actual multiplicative identity. The field premise provides existence; zero is rejected rather than assigned an inverse. The exact result, retaining the requested order, is (0, 0, 0, Yes).

Reviewed step-by-step derivation

Finite field · qf-mod30-field-inverse · candidate

51. Field Division Uses a Nonzero Denominator and the Same Table

Use the same already qualified complete finite commutative unital-ring tables on Fin(5)={0,…,4}, addition=[[0, 1, 2, 3, 4], [1, 2, 3, 4, 0], [2, 3, 4, 0, 1], [3, 4, 0, 1, 2], [4, 0, 1, 2, 3]] and multiplication=[[0, 0, 0, 0, 0], [0, 1, 2, 3, 4], [0, 2, 4, 1, 3], [0, 3, 1, 4, 2], [0, 4, 3, 2, 1]] in carrier-label order. The actual zero and one are distinct and need not be labels 0 and 1. The inverse audit is additional to, not a replacement for, those ring axioms. Division by zero is never defined. A finite field has prime-power order, not necessarily prime order. Under field qualification, numerator=1, denominator=1 is nonzero. Give (numerator,denominator,inverse,quotient,check,pass?).

View answer and explanation

(1, 1, 1, 1, 1, Yes)

Multiply the numerator by the denominator's inverse, then multiply the quotient by the denominator in the same table to recover the numerator. Ordinary arithmetic on the carrier labels is not this operation. The exact result, retaining the requested order, is (1, 1, 1, 1, 1, Yes).

Reviewed step-by-step derivation

Finite field · qf-mod30-field-division · candidate

52. Field Division Uses a Nonzero Denominator and the Same Table

Use the same already qualified complete finite commutative unital-ring tables on Fin(5)={0,…,4}, addition=[[0, 1, 2, 3, 4], [1, 2, 4, 0, 3], [2, 4, 3, 1, 0], [3, 0, 1, 4, 2], [4, 3, 0, 2, 1]] and multiplication=[[0, 0, 0, 0, 0], [0, 1, 2, 3, 4], [0, 2, 3, 4, 1], [0, 3, 4, 1, 2], [0, 4, 1, 2, 3]] in carrier-label order. The actual zero and one are distinct and need not be labels 0 and 1. The inverse audit is additional to, not a replacement for, those ring axioms. Division by zero is never defined. A finite field has prime-power order, not necessarily prime order. Under field qualification, numerator=2, denominator=2 is nonzero. Give (numerator,denominator,inverse,quotient,check,pass?).

View answer and explanation

(2, 2, 4, 1, 2, Yes)

Multiply the numerator by the denominator's inverse, then multiply the quotient by the denominator in the same table to recover the numerator. Ordinary arithmetic on the carrier labels is not this operation. The exact result, retaining the requested order, is (2, 2, 4, 1, 2, Yes).

Reviewed step-by-step derivation

Finite field · qf-mod30-field-division · candidate

53. Field Division Uses a Nonzero Denominator and the Same Table

Use the same already qualified complete finite commutative unital-ring tables on Fin(5)={0,…,4}, addition=[[0, 1, 2, 3, 4], [1, 3, 4, 2, 0], [2, 4, 1, 0, 3], [3, 2, 0, 4, 1], [4, 0, 3, 1, 2]] and multiplication=[[0, 0, 0, 0, 0], [0, 1, 2, 3, 4], [0, 2, 4, 1, 3], [0, 3, 1, 4, 2], [0, 4, 3, 2, 1]] in carrier-label order. The actual zero and one are distinct and need not be labels 0 and 1. The inverse audit is additional to, not a replacement for, those ring axioms. Division by zero is never defined. A finite field has prime-power order, not necessarily prime order. Under field qualification, numerator=2, denominator=2 is nonzero. Give (numerator,denominator,inverse,quotient,check,pass?).

View answer and explanation

(2, 2, 3, 1, 2, Yes)

Multiply the numerator by the denominator's inverse, then multiply the quotient by the denominator in the same table to recover the numerator. Ordinary arithmetic on the carrier labels is not this operation. The exact result, retaining the requested order, is (2, 2, 3, 1, 2, Yes).

Reviewed step-by-step derivation

Finite field · qf-mod30-field-division · candidate

54. Field Division Uses a Nonzero Denominator and the Same Table

Use the same already qualified complete finite commutative unital-ring tables on Fin(5)={0,…,4}, addition=[[0, 1, 2, 3, 4], [1, 3, 0, 4, 2], [2, 0, 4, 1, 3], [3, 4, 1, 2, 0], [4, 2, 3, 0, 1]] and multiplication=[[0, 0, 0, 0, 0], [0, 1, 2, 3, 4], [0, 2, 1, 4, 3], [0, 3, 4, 2, 1], [0, 4, 3, 1, 2]] in carrier-label order. The actual zero and one are distinct and need not be labels 0 and 1. The inverse audit is additional to, not a replacement for, those ring axioms. Division by zero is never defined. A finite field has prime-power order, not necessarily prime order. Under field qualification, numerator=2, denominator=2 is nonzero. Give (numerator,denominator,inverse,quotient,check,pass?).

View answer and explanation

(2, 2, 2, 1, 2, Yes)

Multiply the numerator by the denominator's inverse, then multiply the quotient by the denominator in the same table to recover the numerator. Ordinary arithmetic on the carrier labels is not this operation. The exact result, retaining the requested order, is (2, 2, 2, 1, 2, Yes).

Reviewed step-by-step derivation

Finite field · qf-mod30-field-division · candidate

55. Field Division Uses a Nonzero Denominator and the Same Table

Use the same already qualified complete finite commutative unital-ring tables on Fin(5)={0,…,4}, addition=[[0, 1, 2, 3, 4], [1, 4, 3, 0, 2], [2, 3, 1, 4, 0], [3, 0, 4, 2, 1], [4, 2, 0, 1, 3]] and multiplication=[[0, 0, 0, 0, 0], [0, 1, 2, 3, 4], [0, 2, 3, 4, 1], [0, 3, 4, 1, 2], [0, 4, 1, 2, 3]] in carrier-label order. The actual zero and one are distinct and need not be labels 0 and 1. The inverse audit is additional to, not a replacement for, those ring axioms. Division by zero is never defined. A finite field has prime-power order, not necessarily prime order. Under field qualification, numerator=0, denominator=1 is nonzero. Give (numerator,denominator,inverse,quotient,check,pass?).

View answer and explanation

(0, 1, 1, 0, 0, Yes)

Multiply the numerator by the denominator's inverse, then multiply the quotient by the denominator in the same table to recover the numerator. Ordinary arithmetic on the carrier labels is not this operation. The exact result, retaining the requested order, is (0, 1, 1, 0, 0, Yes).

Reviewed step-by-step derivation

Finite field · qf-mod30-field-division · candidate

56. Field Division Uses a Nonzero Denominator and the Same Table

Use the same already qualified complete finite commutative unital-ring tables on Fin(5)={0,…,4}, addition=[[0, 1, 2, 3, 4], [1, 4, 0, 2, 3], [2, 0, 3, 4, 1], [3, 2, 4, 1, 0], [4, 3, 1, 0, 2]] and multiplication=[[0, 0, 0, 0, 0], [0, 1, 2, 3, 4], [0, 2, 1, 4, 3], [0, 3, 4, 2, 1], [0, 4, 3, 1, 2]] in carrier-label order. The actual zero and one are distinct and need not be labels 0 and 1. The inverse audit is additional to, not a replacement for, those ring axioms. Division by zero is never defined. A finite field has prime-power order, not necessarily prime order. Under field qualification, numerator=1, denominator=4 is nonzero. Give (numerator,denominator,inverse,quotient,check,pass?).

View answer and explanation

(1, 4, 3, 3, 1, Yes)

Multiply the numerator by the denominator's inverse, then multiply the quotient by the denominator in the same table to recover the numerator. Ordinary arithmetic on the carrier labels is not this operation. The exact result, retaining the requested order, is (1, 4, 3, 3, 1, Yes).

Reviewed step-by-step derivation

Finite field · qf-mod30-field-division · candidate

57. Field Division Uses a Nonzero Denominator and the Same Table

Use the same already qualified complete finite commutative unital-ring tables on Fin(5)={0,…,4}, addition=[[0, 1, 2, 3, 4], [1, 4, 3, 0, 2], [2, 3, 1, 4, 0], [3, 0, 4, 2, 1], [4, 2, 0, 1, 3]] and multiplication=[[0, 0, 0, 0, 0], [0, 4, 1, 2, 3], [0, 1, 2, 3, 4], [0, 2, 3, 4, 1], [0, 3, 4, 1, 2]] in carrier-label order. The actual zero and one are distinct and need not be labels 0 and 1. The inverse audit is additional to, not a replacement for, those ring axioms. Division by zero is never defined. A finite field has prime-power order, not necessarily prime order. Under field qualification, numerator=1, denominator=3 is nonzero. Give (numerator,denominator,inverse,quotient,check,pass?).

View answer and explanation

(1, 3, 1, 4, 1, Yes)

Multiply the numerator by the denominator's inverse, then multiply the quotient by the denominator in the same table to recover the numerator. Ordinary arithmetic on the carrier labels is not this operation. The exact result, retaining the requested order, is (1, 3, 1, 4, 1, Yes).

Reviewed step-by-step derivation

Finite field · qf-mod30-field-division · candidate

58. Field Division Uses a Nonzero Denominator and the Same Table

Use the same already qualified complete finite commutative unital-ring tables on Fin(5)={0,…,4}, addition=[[0, 1, 2, 3, 4], [1, 3, 4, 2, 0], [2, 4, 1, 0, 3], [3, 2, 0, 4, 1], [4, 0, 3, 1, 2]] and multiplication=[[0, 0, 0, 0, 0], [0, 3, 1, 4, 2], [0, 1, 2, 3, 4], [0, 4, 3, 2, 1], [0, 2, 4, 1, 3]] in carrier-label order. The actual zero and one are distinct and need not be labels 0 and 1. The inverse audit is additional to, not a replacement for, those ring axioms. Division by zero is never defined. A finite field has prime-power order, not necessarily prime order. Under field qualification, numerator=4, denominator=3 is nonzero. Give (numerator,denominator,inverse,quotient,check,pass?).

View answer and explanation

(4, 3, 3, 1, 4, Yes)

Multiply the numerator by the denominator's inverse, then multiply the quotient by the denominator in the same table to recover the numerator. Ordinary arithmetic on the carrier labels is not this operation. The exact result, retaining the requested order, is (4, 3, 3, 1, 4, Yes).

Reviewed step-by-step derivation

Finite field · qf-mod30-field-division · candidate

59. Field Division Uses a Nonzero Denominator and the Same Table

Use the same already qualified complete finite commutative unital-ring tables on Fin(5)={0,…,4}, addition=[[0, 1, 2, 3, 4], [1, 2, 4, 0, 3], [2, 4, 3, 1, 0], [3, 0, 1, 4, 2], [4, 3, 0, 2, 1]] and multiplication=[[0, 0, 0, 0, 0], [0, 4, 1, 2, 3], [0, 1, 2, 3, 4], [0, 2, 3, 4, 1], [0, 3, 4, 1, 2]] in carrier-label order. The actual zero and one are distinct and need not be labels 0 and 1. The inverse audit is additional to, not a replacement for, those ring axioms. Division by zero is never defined. A finite field has prime-power order, not necessarily prime order. Under field qualification, numerator=4, denominator=2 is nonzero. Give (numerator,denominator,inverse,quotient,check,pass?).

View answer and explanation

(4, 2, 2, 4, 4, Yes)

Multiply the numerator by the denominator's inverse, then multiply the quotient by the denominator in the same table to recover the numerator. Ordinary arithmetic on the carrier labels is not this operation. The exact result, retaining the requested order, is (4, 2, 2, 4, 4, Yes).

Reviewed step-by-step derivation

Finite field · qf-mod30-field-division · candidate

60. Field Division Uses a Nonzero Denominator and the Same Table

Use the same already qualified complete finite commutative unital-ring tables on Fin(5)={0,…,4}, addition=[[0, 1, 2, 3, 4], [1, 4, 0, 2, 3], [2, 0, 3, 4, 1], [3, 2, 4, 1, 0], [4, 3, 1, 0, 2]] and multiplication=[[0, 0, 0, 0, 0], [0, 2, 1, 4, 3], [0, 1, 2, 3, 4], [0, 4, 3, 1, 2], [0, 3, 4, 2, 1]] in carrier-label order. The actual zero and one are distinct and need not be labels 0 and 1. The inverse audit is additional to, not a replacement for, those ring axioms. Division by zero is never defined. A finite field has prime-power order, not necessarily prime order. Under field qualification, numerator=0, denominator=3 is nonzero. Give (numerator,denominator,inverse,quotient,check,pass?).

View answer and explanation

(0, 3, 4, 0, 0, Yes)

Multiply the numerator by the denominator's inverse, then multiply the quotient by the denominator in the same table to recover the numerator. Ordinary arithmetic on the carrier labels is not this operation. The exact result, retaining the requested order, is (0, 3, 4, 0, 0, Yes).

Reviewed step-by-step derivation

Finite field · qf-mod30-field-division · candidate

61. Field Division Uses a Nonzero Denominator and the Same Table

Use the same already qualified complete finite commutative unital-ring tables on Fin(5)={0,…,4}, addition=[[0, 1, 2, 3, 4], [1, 2, 3, 4, 0], [2, 3, 4, 0, 1], [3, 4, 0, 1, 2], [4, 0, 1, 2, 3]] and multiplication=[[0, 0, 0, 0, 0], [0, 3, 1, 4, 2], [0, 1, 2, 3, 4], [0, 4, 3, 2, 1], [0, 2, 4, 1, 3]] in carrier-label order. The actual zero and one are distinct and need not be labels 0 and 1. The inverse audit is additional to, not a replacement for, those ring axioms. Division by zero is never defined. A finite field has prime-power order, not necessarily prime order. Under field qualification, numerator=2, denominator=1 is nonzero. Give (numerator,denominator,inverse,quotient,check,pass?).

View answer and explanation

(2, 1, 4, 4, 2, Yes)

Multiply the numerator by the denominator's inverse, then multiply the quotient by the denominator in the same table to recover the numerator. Ordinary arithmetic on the carrier labels is not this operation. The exact result, retaining the requested order, is (2, 1, 4, 4, 2, Yes).

Reviewed step-by-step derivation

Finite field · qf-mod30-field-division · candidate

62. Field Division Uses a Nonzero Denominator and the Same Table

Use the same already qualified complete finite commutative unital-ring tables on Fin(5)={0,…,4}, addition=[[0, 1, 2, 3, 4], [1, 3, 0, 4, 2], [2, 0, 4, 1, 3], [3, 4, 1, 2, 0], [4, 2, 3, 0, 1]] and multiplication=[[0, 0, 0, 0, 0], [0, 2, 1, 4, 3], [0, 1, 2, 3, 4], [0, 4, 3, 1, 2], [0, 3, 4, 2, 1]] in carrier-label order. The actual zero and one are distinct and need not be labels 0 and 1. The inverse audit is additional to, not a replacement for, those ring axioms. Division by zero is never defined. A finite field has prime-power order, not necessarily prime order. Under field qualification, numerator=4, denominator=1 is nonzero. Give (numerator,denominator,inverse,quotient,check,pass?).

View answer and explanation

(4, 1, 1, 3, 4, Yes)

Multiply the numerator by the denominator's inverse, then multiply the quotient by the denominator in the same table to recover the numerator. Ordinary arithmetic on the carrier labels is not this operation. The exact result, retaining the requested order, is (4, 1, 1, 3, 4, Yes).

Reviewed step-by-step derivation

Finite field · qf-mod30-field-division · candidate

63. Field Division Uses a Nonzero Denominator and the Same Table

Use the same already qualified complete finite commutative unital-ring tables on Fin(5)={0,…,4}, addition=[[0, 1, 2, 3, 4], [1, 4, 0, 2, 3], [2, 0, 3, 4, 1], [3, 2, 4, 1, 0], [4, 3, 1, 0, 2]] and multiplication=[[0, 0, 0, 0, 0], [0, 4, 3, 1, 2], [0, 3, 4, 2, 1], [0, 1, 2, 3, 4], [0, 2, 1, 4, 3]] in carrier-label order. The actual zero and one are distinct and need not be labels 0 and 1. The inverse audit is additional to, not a replacement for, those ring axioms. Division by zero is never defined. A finite field has prime-power order, not necessarily prime order. Under field qualification, numerator=2, denominator=3 is nonzero. Give (numerator,denominator,inverse,quotient,check,pass?).

View answer and explanation

(2, 3, 3, 2, 2, Yes)

Multiply the numerator by the denominator's inverse, then multiply the quotient by the denominator in the same table to recover the numerator. Ordinary arithmetic on the carrier labels is not this operation. The exact result, retaining the requested order, is (2, 3, 3, 2, 2, Yes).

Reviewed step-by-step derivation

Finite field · qf-mod30-field-division · candidate

64. Field Division Uses a Nonzero Denominator and the Same Table

Use the same already qualified complete finite commutative unital-ring tables on Fin(5)={0,…,4}, addition=[[0, 1, 2, 3, 4], [1, 2, 3, 4, 0], [2, 3, 4, 0, 1], [3, 4, 0, 1, 2], [4, 0, 1, 2, 3]] and multiplication=[[0, 0, 0, 0, 0], [0, 2, 4, 1, 3], [0, 4, 3, 2, 1], [0, 1, 2, 3, 4], [0, 3, 1, 4, 2]] in carrier-label order. The actual zero and one are distinct and need not be labels 0 and 1. The inverse audit is additional to, not a replacement for, those ring axioms. Division by zero is never defined. A finite field has prime-power order, not necessarily prime order. Under field qualification, numerator=2, denominator=1 is nonzero. Give (numerator,denominator,inverse,quotient,check,pass?).

View answer and explanation

(2, 1, 4, 1, 2, Yes)

Multiply the numerator by the denominator's inverse, then multiply the quotient by the denominator in the same table to recover the numerator. Ordinary arithmetic on the carrier labels is not this operation. The exact result, retaining the requested order, is (2, 1, 4, 1, 2, Yes).

Reviewed step-by-step derivation

Finite field · qf-mod30-field-division · candidate

65. Field Division Uses a Nonzero Denominator and the Same Table

Use the same already qualified complete finite commutative unital-ring tables on Fin(5)={0,…,4}, addition=[[0, 1, 2, 3, 4], [1, 3, 0, 4, 2], [2, 0, 4, 1, 3], [3, 4, 1, 2, 0], [4, 2, 3, 0, 1]] and multiplication=[[0, 0, 0, 0, 0], [0, 4, 3, 1, 2], [0, 3, 4, 2, 1], [0, 1, 2, 3, 4], [0, 2, 1, 4, 3]] in carrier-label order. The actual zero and one are distinct and need not be labels 0 and 1. The inverse audit is additional to, not a replacement for, those ring axioms. Division by zero is never defined. A finite field has prime-power order, not necessarily prime order. Under field qualification, numerator=0, denominator=1 is nonzero. Give (numerator,denominator,inverse,quotient,check,pass?).

View answer and explanation

(0, 1, 2, 0, 0, Yes)

Multiply the numerator by the denominator's inverse, then multiply the quotient by the denominator in the same table to recover the numerator. Ordinary arithmetic on the carrier labels is not this operation. The exact result, retaining the requested order, is (0, 1, 2, 0, 0, Yes).

Reviewed step-by-step derivation

Finite field · qf-mod30-field-division · candidate

66. Field Division Uses a Nonzero Denominator and the Same Table

Use the same already qualified complete finite commutative unital-ring tables on Fin(5)={0,…,4}, addition=[[0, 1, 2, 3, 4], [1, 4, 3, 0, 2], [2, 3, 1, 4, 0], [3, 0, 4, 2, 1], [4, 2, 0, 1, 3]] and multiplication=[[0, 0, 0, 0, 0], [0, 3, 4, 1, 2], [0, 4, 1, 2, 3], [0, 1, 2, 3, 4], [0, 2, 3, 4, 1]] in carrier-label order. The actual zero and one are distinct and need not be labels 0 and 1. The inverse audit is additional to, not a replacement for, those ring axioms. Division by zero is never defined. A finite field has prime-power order, not necessarily prime order. Under field qualification, numerator=3, denominator=1 is nonzero. Give (numerator,denominator,inverse,quotient,check,pass?).

View answer and explanation

(3, 1, 1, 1, 3, Yes)

Multiply the numerator by the denominator's inverse, then multiply the quotient by the denominator in the same table to recover the numerator. Ordinary arithmetic on the carrier labels is not this operation. The exact result, retaining the requested order, is (3, 1, 1, 1, 3, Yes).

Reviewed step-by-step derivation

Finite field · qf-mod30-field-division · candidate

67. Field Division Uses a Nonzero Denominator and the Same Table

Use the same already qualified complete finite commutative unital-ring tables on Fin(5)={0,…,4}, addition=[[0, 1, 2, 3, 4], [1, 3, 4, 2, 0], [2, 4, 1, 0, 3], [3, 2, 0, 4, 1], [4, 0, 3, 1, 2]] and multiplication=[[0, 0, 0, 0, 0], [0, 2, 4, 1, 3], [0, 4, 3, 2, 1], [0, 1, 2, 3, 4], [0, 3, 1, 4, 2]] in carrier-label order. The actual zero and one are distinct and need not be labels 0 and 1. The inverse audit is additional to, not a replacement for, those ring axioms. Division by zero is never defined. A finite field has prime-power order, not necessarily prime order. Under field qualification, numerator=4, denominator=3 is nonzero. Give (numerator,denominator,inverse,quotient,check,pass?).

View answer and explanation

(4, 3, 3, 4, 4, Yes)

Multiply the numerator by the denominator's inverse, then multiply the quotient by the denominator in the same table to recover the numerator. Ordinary arithmetic on the carrier labels is not this operation. The exact result, retaining the requested order, is (4, 3, 3, 4, 4, Yes).

Reviewed step-by-step derivation

Finite field · qf-mod30-field-division · candidate

68. Field Division Uses a Nonzero Denominator and the Same Table

Use the same already qualified complete finite commutative unital-ring tables on Fin(5)={0,…,4}, addition=[[0, 1, 2, 3, 4], [1, 2, 4, 0, 3], [2, 4, 3, 1, 0], [3, 0, 1, 4, 2], [4, 3, 0, 2, 1]] and multiplication=[[0, 0, 0, 0, 0], [0, 3, 4, 1, 2], [0, 4, 1, 2, 3], [0, 1, 2, 3, 4], [0, 2, 3, 4, 1]] in carrier-label order. The actual zero and one are distinct and need not be labels 0 and 1. The inverse audit is additional to, not a replacement for, those ring axioms. Division by zero is never defined. A finite field has prime-power order, not necessarily prime order. Under field qualification, numerator=2, denominator=4 is nonzero. Give (numerator,denominator,inverse,quotient,check,pass?).

View answer and explanation

(2, 4, 2, 1, 2, Yes)

Multiply the numerator by the denominator's inverse, then multiply the quotient by the denominator in the same table to recover the numerator. Ordinary arithmetic on the carrier labels is not this operation. The exact result, retaining the requested order, is (2, 4, 2, 1, 2, Yes).

Reviewed step-by-step derivation

Finite field · qf-mod30-field-division · candidate

69. Field Division Uses a Nonzero Denominator and the Same Table

Use the same already qualified complete finite commutative unital-ring tables on Fin(5)={0,…,4}, addition=[[0, 1, 2, 3, 4], [1, 3, 0, 4, 2], [2, 0, 4, 1, 3], [3, 4, 1, 2, 0], [4, 2, 3, 0, 1]] and multiplication=[[0, 0, 0, 0, 0], [0, 3, 4, 2, 1], [0, 4, 3, 1, 2], [0, 2, 1, 4, 3], [0, 1, 2, 3, 4]] in carrier-label order. The actual zero and one are distinct and need not be labels 0 and 1. The inverse audit is additional to, not a replacement for, those ring axioms. Division by zero is never defined. A finite field has prime-power order, not necessarily prime order. Under field qualification, numerator=3, denominator=2 is nonzero. Give (numerator,denominator,inverse,quotient,check,pass?).

View answer and explanation

(3, 2, 1, 2, 3, Yes)

Multiply the numerator by the denominator's inverse, then multiply the quotient by the denominator in the same table to recover the numerator. Ordinary arithmetic on the carrier labels is not this operation. The exact result, retaining the requested order, is (3, 2, 1, 2, 3, Yes).

Reviewed step-by-step derivation

Finite field · qf-mod30-field-division · candidate

70. Field Division Uses a Nonzero Denominator and the Same Table

Use the same already qualified complete finite commutative unital-ring tables on Fin(5)={0,…,4}, addition=[[0, 1, 2, 3, 4], [1, 2, 4, 0, 3], [2, 4, 3, 1, 0], [3, 0, 1, 4, 2], [4, 3, 0, 2, 1]] and multiplication=[[0, 0, 0, 0, 0], [0, 2, 3, 4, 1], [0, 3, 4, 1, 2], [0, 4, 1, 2, 3], [0, 1, 2, 3, 4]] in carrier-label order. The actual zero and one are distinct and need not be labels 0 and 1. The inverse audit is additional to, not a replacement for, those ring axioms. Division by zero is never defined. A finite field has prime-power order, not necessarily prime order. Under field qualification, numerator=0, denominator=2 is nonzero. Give (numerator,denominator,inverse,quotient,check,pass?).

View answer and explanation

(0, 2, 2, 0, 0, Yes)

Multiply the numerator by the denominator's inverse, then multiply the quotient by the denominator in the same table to recover the numerator. Ordinary arithmetic on the carrier labels is not this operation. The exact result, retaining the requested order, is (0, 2, 2, 0, 0, Yes).

Reviewed step-by-step derivation

Finite field · qf-mod30-field-division · candidate

71. Field Division Uses a Nonzero Denominator and the Same Table

Use the same already qualified complete finite commutative unital-ring tables on Fin(5)={0,…,4}, addition=[[0, 1, 2, 3, 4], [1, 4, 0, 2, 3], [2, 0, 3, 4, 1], [3, 2, 4, 1, 0], [4, 3, 1, 0, 2]] and multiplication=[[0, 0, 0, 0, 0], [0, 3, 4, 2, 1], [0, 4, 3, 1, 2], [0, 2, 1, 4, 3], [0, 1, 2, 3, 4]] in carrier-label order. The actual zero and one are distinct and need not be labels 0 and 1. The inverse audit is additional to, not a replacement for, those ring axioms. Division by zero is never defined. A finite field has prime-power order, not necessarily prime order. Under field qualification, numerator=4, denominator=4 is nonzero. Give (numerator,denominator,inverse,quotient,check,pass?).

View answer and explanation

(4, 4, 4, 4, 4, Yes)

Multiply the numerator by the denominator's inverse, then multiply the quotient by the denominator in the same table to recover the numerator. Ordinary arithmetic on the carrier labels is not this operation. The exact result, retaining the requested order, is (4, 4, 4, 4, 4, Yes).

Reviewed step-by-step derivation

Finite field · qf-mod30-field-division · candidate

72. Field Division Uses a Nonzero Denominator and the Same Table

Use the same already qualified complete finite commutative unital-ring tables on Fin(5)={0,…,4}, addition=[[0, 1, 2, 3, 4], [1, 3, 4, 2, 0], [2, 4, 1, 0, 3], [3, 2, 0, 4, 1], [4, 0, 3, 1, 2]] and multiplication=[[0, 0, 0, 0, 0], [0, 4, 3, 2, 1], [0, 3, 1, 4, 2], [0, 2, 4, 1, 3], [0, 1, 2, 3, 4]] in carrier-label order. The actual zero and one are distinct and need not be labels 0 and 1. The inverse audit is additional to, not a replacement for, those ring axioms. Division by zero is never defined. A finite field has prime-power order, not necessarily prime order. Under field qualification, numerator=2, denominator=2 is nonzero. Give (numerator,denominator,inverse,quotient,check,pass?).

View answer and explanation

(2, 2, 3, 4, 2, Yes)

Multiply the numerator by the denominator's inverse, then multiply the quotient by the denominator in the same table to recover the numerator. Ordinary arithmetic on the carrier labels is not this operation. The exact result, retaining the requested order, is (2, 2, 3, 4, 2, Yes).

Reviewed step-by-step derivation

Finite field · qf-mod30-field-division · candidate

73. Field Division Uses a Nonzero Denominator and the Same Table

Use the same already qualified complete finite commutative unital-ring tables on Fin(5)={0,…,4}, addition=[[0, 1, 2, 3, 4], [1, 4, 3, 0, 2], [2, 3, 1, 4, 0], [3, 0, 4, 2, 1], [4, 2, 0, 1, 3]] and multiplication=[[0, 0, 0, 0, 0], [0, 2, 3, 4, 1], [0, 3, 4, 1, 2], [0, 4, 1, 2, 3], [0, 1, 2, 3, 4]] in carrier-label order. The actual zero and one are distinct and need not be labels 0 and 1. The inverse audit is additional to, not a replacement for, those ring axioms. Division by zero is never defined. A finite field has prime-power order, not necessarily prime order. Under field qualification, numerator=1, denominator=1 is nonzero. Give (numerator,denominator,inverse,quotient,check,pass?).

View answer and explanation

(1, 1, 3, 4, 1, Yes)

Multiply the numerator by the denominator's inverse, then multiply the quotient by the denominator in the same table to recover the numerator. Ordinary arithmetic on the carrier labels is not this operation. The exact result, retaining the requested order, is (1, 1, 3, 4, 1, Yes).

Reviewed step-by-step derivation

Finite field · qf-mod30-field-division · candidate

74. Field Division Uses a Nonzero Denominator and the Same Table

Use the same already qualified complete finite commutative unital-ring tables on Fin(5)={0,…,4}, addition=[[0, 1, 2, 3, 4], [1, 2, 3, 4, 0], [2, 3, 4, 0, 1], [3, 4, 0, 1, 2], [4, 0, 1, 2, 3]] and multiplication=[[0, 0, 0, 0, 0], [0, 4, 3, 2, 1], [0, 3, 1, 4, 2], [0, 2, 4, 1, 3], [0, 1, 2, 3, 4]] in carrier-label order. The actual zero and one are distinct and need not be labels 0 and 1. The inverse audit is additional to, not a replacement for, those ring axioms. Division by zero is never defined. A finite field has prime-power order, not necessarily prime order. Under field qualification, numerator=1, denominator=1 is nonzero. Give (numerator,denominator,inverse,quotient,check,pass?).

View answer and explanation

(1, 1, 1, 4, 1, Yes)

Multiply the numerator by the denominator's inverse, then multiply the quotient by the denominator in the same table to recover the numerator. Ordinary arithmetic on the carrier labels is not this operation. The exact result, retaining the requested order, is (1, 1, 1, 4, 1, Yes).

Reviewed step-by-step derivation

Finite field · qf-mod30-field-division · candidate

75. Field Division Uses a Nonzero Denominator and the Same Table

Use the same already qualified complete finite commutative unital-ring tables on Fin(5)={0,…,4}, addition=[[2, 0, 3, 4, 1], [0, 1, 2, 3, 4], [3, 2, 4, 1, 0], [4, 3, 1, 0, 2], [1, 4, 0, 2, 3]] and multiplication=[[0, 1, 2, 3, 4], [1, 1, 1, 1, 1], [2, 1, 4, 0, 3], [3, 1, 0, 4, 2], [4, 1, 3, 2, 0]] in carrier-label order. The actual zero and one are distinct and need not be labels 0 and 1. The inverse audit is additional to, not a replacement for, those ring axioms. Division by zero is never defined. A finite field has prime-power order, not necessarily prime order. Under field qualification, numerator=1, denominator=0 is nonzero. Give (numerator,denominator,inverse,quotient,check,pass?).

View answer and explanation

(1, 0, 0, 1, 1, Yes)

Multiply the numerator by the denominator's inverse, then multiply the quotient by the denominator in the same table to recover the numerator. Ordinary arithmetic on the carrier labels is not this operation. The exact result, retaining the requested order, is (1, 0, 0, 1, 1, Yes).

Reviewed step-by-step derivation

Finite field · qf-mod30-field-division · candidate

76. Do Not Skip the Finite Injective-to-Surjective Step

Use the same already qualified complete finite commutative unital-ring tables on Fin(5)={0,…,4}, addition=[[0, 1, 2, 3, 4], [1, 2, 3, 4, 0], [2, 3, 4, 0, 1], [3, 4, 0, 1, 2], [4, 0, 1, 2, 3]] and multiplication=[[0, 0, 0, 0, 0], [0, 1, 2, 3, 4], [0, 2, 4, 1, 3], [0, 3, 1, 4, 2], [0, 4, 3, 2, 1]] in carrier-label order. The actual zero and one are distinct and need not be labels 0 and 1. The inverse audit is additional to, not a replacement for, those ring axioms. Division by zero is never defined. A finite field has prime-power order, not necessarily prime order. Give (n,missing,field?,theorem_step,field_order_prime_power_not_necessarily_prime). theorem_step is finite_domain_injective_to_surjective_theorem or missing_nonzero_inverse.

View answer and explanation

(5, [], Yes, finite_domain_injective_to_surjective_theorem, field_order_prime_power_not_necessarily_prime)

Record every missing nonzero inverse. In a finite integral domain, nonzero multiplication is injective and therefore surjective, yielding an inverse. A table with missing inverses does not qualify; this does not invalidate that finite-domain theorem. The exact result, retaining the requested order, is (5, [], Yes, finite_domain_injective_to_surjective_theorem, field_order_prime_power_not_necessarily_prime).

Reviewed step-by-step derivation

Finite field · qf-mod30-field-finite-domain-theorem · candidate

77. Do Not Skip the Finite Injective-to-Surjective Step

Use the same already qualified complete finite commutative unital-ring tables on Fin(5)={0,…,4}, addition=[[0, 1, 2, 3, 4], [1, 2, 4, 0, 3], [2, 4, 3, 1, 0], [3, 0, 1, 4, 2], [4, 3, 0, 2, 1]] and multiplication=[[0, 0, 0, 0, 0], [0, 1, 2, 3, 4], [0, 2, 3, 4, 1], [0, 3, 4, 1, 2], [0, 4, 1, 2, 3]] in carrier-label order. The actual zero and one are distinct and need not be labels 0 and 1. The inverse audit is additional to, not a replacement for, those ring axioms. Division by zero is never defined. A finite field has prime-power order, not necessarily prime order. Give (n,missing,field?,theorem_step,field_order_prime_power_not_necessarily_prime). theorem_step is finite_domain_injective_to_surjective_theorem or missing_nonzero_inverse.

View answer and explanation

(5, [], Yes, finite_domain_injective_to_surjective_theorem, field_order_prime_power_not_necessarily_prime)

Record every missing nonzero inverse. In a finite integral domain, nonzero multiplication is injective and therefore surjective, yielding an inverse. A table with missing inverses does not qualify; this does not invalidate that finite-domain theorem. The exact result, retaining the requested order, is (5, [], Yes, finite_domain_injective_to_surjective_theorem, field_order_prime_power_not_necessarily_prime).

Reviewed step-by-step derivation

Finite field · qf-mod30-field-finite-domain-theorem · candidate

78. Do Not Skip the Finite Injective-to-Surjective Step

Use the same already qualified complete finite commutative unital-ring tables on Fin(5)={0,…,4}, addition=[[0, 1, 2, 3, 4], [1, 3, 4, 2, 0], [2, 4, 1, 0, 3], [3, 2, 0, 4, 1], [4, 0, 3, 1, 2]] and multiplication=[[0, 0, 0, 0, 0], [0, 1, 2, 3, 4], [0, 2, 4, 1, 3], [0, 3, 1, 4, 2], [0, 4, 3, 2, 1]] in carrier-label order. The actual zero and one are distinct and need not be labels 0 and 1. The inverse audit is additional to, not a replacement for, those ring axioms. Division by zero is never defined. A finite field has prime-power order, not necessarily prime order. Give (n,missing,field?,theorem_step,field_order_prime_power_not_necessarily_prime). theorem_step is finite_domain_injective_to_surjective_theorem or missing_nonzero_inverse.

View answer and explanation

(5, [], Yes, finite_domain_injective_to_surjective_theorem, field_order_prime_power_not_necessarily_prime)

Record every missing nonzero inverse. In a finite integral domain, nonzero multiplication is injective and therefore surjective, yielding an inverse. A table with missing inverses does not qualify; this does not invalidate that finite-domain theorem. The exact result, retaining the requested order, is (5, [], Yes, finite_domain_injective_to_surjective_theorem, field_order_prime_power_not_necessarily_prime).

Reviewed step-by-step derivation

Finite field · qf-mod30-field-finite-domain-theorem · candidate

79. Do Not Skip the Finite Injective-to-Surjective Step

Use the same already qualified complete finite commutative unital-ring tables on Fin(5)={0,…,4}, addition=[[0, 1, 2, 3, 4], [1, 3, 0, 4, 2], [2, 0, 4, 1, 3], [3, 4, 1, 2, 0], [4, 2, 3, 0, 1]] and multiplication=[[0, 0, 0, 0, 0], [0, 1, 2, 3, 4], [0, 2, 1, 4, 3], [0, 3, 4, 2, 1], [0, 4, 3, 1, 2]] in carrier-label order. The actual zero and one are distinct and need not be labels 0 and 1. The inverse audit is additional to, not a replacement for, those ring axioms. Division by zero is never defined. A finite field has prime-power order, not necessarily prime order. Give (n,missing,field?,theorem_step,field_order_prime_power_not_necessarily_prime). theorem_step is finite_domain_injective_to_surjective_theorem or missing_nonzero_inverse.

View answer and explanation

(5, [], Yes, finite_domain_injective_to_surjective_theorem, field_order_prime_power_not_necessarily_prime)

Record every missing nonzero inverse. In a finite integral domain, nonzero multiplication is injective and therefore surjective, yielding an inverse. A table with missing inverses does not qualify; this does not invalidate that finite-domain theorem. The exact result, retaining the requested order, is (5, [], Yes, finite_domain_injective_to_surjective_theorem, field_order_prime_power_not_necessarily_prime).

Reviewed step-by-step derivation

Finite field · qf-mod30-field-finite-domain-theorem · candidate

80. Do Not Skip the Finite Injective-to-Surjective Step

Use the same already qualified complete finite commutative unital-ring tables on Fin(5)={0,…,4}, addition=[[0, 1, 2, 3, 4], [1, 4, 3, 0, 2], [2, 3, 1, 4, 0], [3, 0, 4, 2, 1], [4, 2, 0, 1, 3]] and multiplication=[[0, 0, 0, 0, 0], [0, 1, 2, 3, 4], [0, 2, 3, 4, 1], [0, 3, 4, 1, 2], [0, 4, 1, 2, 3]] in carrier-label order. The actual zero and one are distinct and need not be labels 0 and 1. The inverse audit is additional to, not a replacement for, those ring axioms. Division by zero is never defined. A finite field has prime-power order, not necessarily prime order. Give (n,missing,field?,theorem_step,field_order_prime_power_not_necessarily_prime). theorem_step is finite_domain_injective_to_surjective_theorem or missing_nonzero_inverse.

View answer and explanation

(5, [], Yes, finite_domain_injective_to_surjective_theorem, field_order_prime_power_not_necessarily_prime)

Record every missing nonzero inverse. In a finite integral domain, nonzero multiplication is injective and therefore surjective, yielding an inverse. A table with missing inverses does not qualify; this does not invalidate that finite-domain theorem. The exact result, retaining the requested order, is (5, [], Yes, finite_domain_injective_to_surjective_theorem, field_order_prime_power_not_necessarily_prime).

Reviewed step-by-step derivation

Finite field · qf-mod30-field-finite-domain-theorem · candidate

81. Do Not Skip the Finite Injective-to-Surjective Step

Use the same already qualified complete finite commutative unital-ring tables on Fin(5)={0,…,4}, addition=[[0, 1, 2, 3, 4], [1, 4, 0, 2, 3], [2, 0, 3, 4, 1], [3, 2, 4, 1, 0], [4, 3, 1, 0, 2]] and multiplication=[[0, 0, 0, 0, 0], [0, 1, 2, 3, 4], [0, 2, 1, 4, 3], [0, 3, 4, 2, 1], [0, 4, 3, 1, 2]] in carrier-label order. The actual zero and one are distinct and need not be labels 0 and 1. The inverse audit is additional to, not a replacement for, those ring axioms. Division by zero is never defined. A finite field has prime-power order, not necessarily prime order. Give (n,missing,field?,theorem_step,field_order_prime_power_not_necessarily_prime). theorem_step is finite_domain_injective_to_surjective_theorem or missing_nonzero_inverse.

View answer and explanation

(5, [], Yes, finite_domain_injective_to_surjective_theorem, field_order_prime_power_not_necessarily_prime)

Record every missing nonzero inverse. In a finite integral domain, nonzero multiplication is injective and therefore surjective, yielding an inverse. A table with missing inverses does not qualify; this does not invalidate that finite-domain theorem. The exact result, retaining the requested order, is (5, [], Yes, finite_domain_injective_to_surjective_theorem, field_order_prime_power_not_necessarily_prime).

Reviewed step-by-step derivation

Finite field · qf-mod30-field-finite-domain-theorem · candidate

82. Do Not Skip the Finite Injective-to-Surjective Step

Use the same already qualified complete finite commutative unital-ring tables on Fin(5)={0,…,4}, addition=[[0, 1, 2, 3, 4], [1, 4, 3, 0, 2], [2, 3, 1, 4, 0], [3, 0, 4, 2, 1], [4, 2, 0, 1, 3]] and multiplication=[[0, 0, 0, 0, 0], [0, 4, 1, 2, 3], [0, 1, 2, 3, 4], [0, 2, 3, 4, 1], [0, 3, 4, 1, 2]] in carrier-label order. The actual zero and one are distinct and need not be labels 0 and 1. The inverse audit is additional to, not a replacement for, those ring axioms. Division by zero is never defined. A finite field has prime-power order, not necessarily prime order. Give (n,missing,field?,theorem_step,field_order_prime_power_not_necessarily_prime). theorem_step is finite_domain_injective_to_surjective_theorem or missing_nonzero_inverse.

View answer and explanation

(5, [], Yes, finite_domain_injective_to_surjective_theorem, field_order_prime_power_not_necessarily_prime)

Record every missing nonzero inverse. In a finite integral domain, nonzero multiplication is injective and therefore surjective, yielding an inverse. A table with missing inverses does not qualify; this does not invalidate that finite-domain theorem. The exact result, retaining the requested order, is (5, [], Yes, finite_domain_injective_to_surjective_theorem, field_order_prime_power_not_necessarily_prime).

Reviewed step-by-step derivation

Finite field · qf-mod30-field-finite-domain-theorem · candidate

83. Do Not Skip the Finite Injective-to-Surjective Step

Use the same already qualified complete finite commutative unital-ring tables on Fin(5)={0,…,4}, addition=[[0, 1, 2, 3, 4], [1, 3, 4, 2, 0], [2, 4, 1, 0, 3], [3, 2, 0, 4, 1], [4, 0, 3, 1, 2]] and multiplication=[[0, 0, 0, 0, 0], [0, 3, 1, 4, 2], [0, 1, 2, 3, 4], [0, 4, 3, 2, 1], [0, 2, 4, 1, 3]] in carrier-label order. The actual zero and one are distinct and need not be labels 0 and 1. The inverse audit is additional to, not a replacement for, those ring axioms. Division by zero is never defined. A finite field has prime-power order, not necessarily prime order. Give (n,missing,field?,theorem_step,field_order_prime_power_not_necessarily_prime). theorem_step is finite_domain_injective_to_surjective_theorem or missing_nonzero_inverse.

View answer and explanation

(5, [], Yes, finite_domain_injective_to_surjective_theorem, field_order_prime_power_not_necessarily_prime)

Record every missing nonzero inverse. In a finite integral domain, nonzero multiplication is injective and therefore surjective, yielding an inverse. A table with missing inverses does not qualify; this does not invalidate that finite-domain theorem. The exact result, retaining the requested order, is (5, [], Yes, finite_domain_injective_to_surjective_theorem, field_order_prime_power_not_necessarily_prime).

Reviewed step-by-step derivation

Finite field · qf-mod30-field-finite-domain-theorem · candidate

84. Do Not Skip the Finite Injective-to-Surjective Step

Use the same already qualified complete finite commutative unital-ring tables on Fin(5)={0,…,4}, addition=[[0, 1, 2, 3, 4], [1, 2, 4, 0, 3], [2, 4, 3, 1, 0], [3, 0, 1, 4, 2], [4, 3, 0, 2, 1]] and multiplication=[[0, 0, 0, 0, 0], [0, 4, 1, 2, 3], [0, 1, 2, 3, 4], [0, 2, 3, 4, 1], [0, 3, 4, 1, 2]] in carrier-label order. The actual zero and one are distinct and need not be labels 0 and 1. The inverse audit is additional to, not a replacement for, those ring axioms. Division by zero is never defined. A finite field has prime-power order, not necessarily prime order. Give (n,missing,field?,theorem_step,field_order_prime_power_not_necessarily_prime). theorem_step is finite_domain_injective_to_surjective_theorem or missing_nonzero_inverse.

View answer and explanation

(5, [], Yes, finite_domain_injective_to_surjective_theorem, field_order_prime_power_not_necessarily_prime)

Record every missing nonzero inverse. In a finite integral domain, nonzero multiplication is injective and therefore surjective, yielding an inverse. A table with missing inverses does not qualify; this does not invalidate that finite-domain theorem. The exact result, retaining the requested order, is (5, [], Yes, finite_domain_injective_to_surjective_theorem, field_order_prime_power_not_necessarily_prime).

Reviewed step-by-step derivation

Finite field · qf-mod30-field-finite-domain-theorem · candidate

85. Do Not Skip the Finite Injective-to-Surjective Step

Use the same already qualified complete finite commutative unital-ring tables on Fin(5)={0,…,4}, addition=[[0, 1, 2, 3, 4], [1, 4, 0, 2, 3], [2, 0, 3, 4, 1], [3, 2, 4, 1, 0], [4, 3, 1, 0, 2]] and multiplication=[[0, 0, 0, 0, 0], [0, 2, 1, 4, 3], [0, 1, 2, 3, 4], [0, 4, 3, 1, 2], [0, 3, 4, 2, 1]] in carrier-label order. The actual zero and one are distinct and need not be labels 0 and 1. The inverse audit is additional to, not a replacement for, those ring axioms. Division by zero is never defined. A finite field has prime-power order, not necessarily prime order. Give (n,missing,field?,theorem_step,field_order_prime_power_not_necessarily_prime). theorem_step is finite_domain_injective_to_surjective_theorem or missing_nonzero_inverse.

View answer and explanation

(5, [], Yes, finite_domain_injective_to_surjective_theorem, field_order_prime_power_not_necessarily_prime)

Record every missing nonzero inverse. In a finite integral domain, nonzero multiplication is injective and therefore surjective, yielding an inverse. A table with missing inverses does not qualify; this does not invalidate that finite-domain theorem. The exact result, retaining the requested order, is (5, [], Yes, finite_domain_injective_to_surjective_theorem, field_order_prime_power_not_necessarily_prime).

Reviewed step-by-step derivation

Finite field · qf-mod30-field-finite-domain-theorem · candidate

86. Do Not Skip the Finite Injective-to-Surjective Step

Use the same already qualified complete finite commutative unital-ring tables on Fin(5)={0,…,4}, addition=[[0, 1, 2, 3, 4], [1, 2, 3, 4, 0], [2, 3, 4, 0, 1], [3, 4, 0, 1, 2], [4, 0, 1, 2, 3]] and multiplication=[[0, 0, 0, 0, 0], [0, 3, 1, 4, 2], [0, 1, 2, 3, 4], [0, 4, 3, 2, 1], [0, 2, 4, 1, 3]] in carrier-label order. The actual zero and one are distinct and need not be labels 0 and 1. The inverse audit is additional to, not a replacement for, those ring axioms. Division by zero is never defined. A finite field has prime-power order, not necessarily prime order. Give (n,missing,field?,theorem_step,field_order_prime_power_not_necessarily_prime). theorem_step is finite_domain_injective_to_surjective_theorem or missing_nonzero_inverse.

View answer and explanation

(5, [], Yes, finite_domain_injective_to_surjective_theorem, field_order_prime_power_not_necessarily_prime)

Record every missing nonzero inverse. In a finite integral domain, nonzero multiplication is injective and therefore surjective, yielding an inverse. A table with missing inverses does not qualify; this does not invalidate that finite-domain theorem. The exact result, retaining the requested order, is (5, [], Yes, finite_domain_injective_to_surjective_theorem, field_order_prime_power_not_necessarily_prime).

Reviewed step-by-step derivation

Finite field · qf-mod30-field-finite-domain-theorem · candidate

87. Do Not Skip the Finite Injective-to-Surjective Step

Use the same already qualified complete finite commutative unital-ring tables on Fin(5)={0,…,4}, addition=[[0, 1, 2, 3, 4], [1, 3, 0, 4, 2], [2, 0, 4, 1, 3], [3, 4, 1, 2, 0], [4, 2, 3, 0, 1]] and multiplication=[[0, 0, 0, 0, 0], [0, 2, 1, 4, 3], [0, 1, 2, 3, 4], [0, 4, 3, 1, 2], [0, 3, 4, 2, 1]] in carrier-label order. The actual zero and one are distinct and need not be labels 0 and 1. The inverse audit is additional to, not a replacement for, those ring axioms. Division by zero is never defined. A finite field has prime-power order, not necessarily prime order. Give (n,missing,field?,theorem_step,field_order_prime_power_not_necessarily_prime). theorem_step is finite_domain_injective_to_surjective_theorem or missing_nonzero_inverse.

View answer and explanation

(5, [], Yes, finite_domain_injective_to_surjective_theorem, field_order_prime_power_not_necessarily_prime)

Record every missing nonzero inverse. In a finite integral domain, nonzero multiplication is injective and therefore surjective, yielding an inverse. A table with missing inverses does not qualify; this does not invalidate that finite-domain theorem. The exact result, retaining the requested order, is (5, [], Yes, finite_domain_injective_to_surjective_theorem, field_order_prime_power_not_necessarily_prime).

Reviewed step-by-step derivation

Finite field · qf-mod30-field-finite-domain-theorem · candidate

88. Do Not Skip the Finite Injective-to-Surjective Step

Use the same already qualified complete finite commutative unital-ring tables on Fin(5)={0,…,4}, addition=[[0, 1, 2, 3, 4], [1, 4, 0, 2, 3], [2, 0, 3, 4, 1], [3, 2, 4, 1, 0], [4, 3, 1, 0, 2]] and multiplication=[[0, 0, 0, 0, 0], [0, 4, 3, 1, 2], [0, 3, 4, 2, 1], [0, 1, 2, 3, 4], [0, 2, 1, 4, 3]] in carrier-label order. The actual zero and one are distinct and need not be labels 0 and 1. The inverse audit is additional to, not a replacement for, those ring axioms. Division by zero is never defined. A finite field has prime-power order, not necessarily prime order. Give (n,missing,field?,theorem_step,field_order_prime_power_not_necessarily_prime). theorem_step is finite_domain_injective_to_surjective_theorem or missing_nonzero_inverse.

View answer and explanation

(5, [], Yes, finite_domain_injective_to_surjective_theorem, field_order_prime_power_not_necessarily_prime)

Record every missing nonzero inverse. In a finite integral domain, nonzero multiplication is injective and therefore surjective, yielding an inverse. A table with missing inverses does not qualify; this does not invalidate that finite-domain theorem. The exact result, retaining the requested order, is (5, [], Yes, finite_domain_injective_to_surjective_theorem, field_order_prime_power_not_necessarily_prime).

Reviewed step-by-step derivation

Finite field · qf-mod30-field-finite-domain-theorem · candidate

89. Do Not Skip the Finite Injective-to-Surjective Step

Use the same already qualified complete finite commutative unital-ring tables on Fin(5)={0,…,4}, addition=[[0, 1, 2, 3, 4], [1, 2, 3, 4, 0], [2, 3, 4, 0, 1], [3, 4, 0, 1, 2], [4, 0, 1, 2, 3]] and multiplication=[[0, 0, 0, 0, 0], [0, 2, 4, 1, 3], [0, 4, 3, 2, 1], [0, 1, 2, 3, 4], [0, 3, 1, 4, 2]] in carrier-label order. The actual zero and one are distinct and need not be labels 0 and 1. The inverse audit is additional to, not a replacement for, those ring axioms. Division by zero is never defined. A finite field has prime-power order, not necessarily prime order. Give (n,missing,field?,theorem_step,field_order_prime_power_not_necessarily_prime). theorem_step is finite_domain_injective_to_surjective_theorem or missing_nonzero_inverse.

View answer and explanation

(5, [], Yes, finite_domain_injective_to_surjective_theorem, field_order_prime_power_not_necessarily_prime)

Record every missing nonzero inverse. In a finite integral domain, nonzero multiplication is injective and therefore surjective, yielding an inverse. A table with missing inverses does not qualify; this does not invalidate that finite-domain theorem. The exact result, retaining the requested order, is (5, [], Yes, finite_domain_injective_to_surjective_theorem, field_order_prime_power_not_necessarily_prime).

Reviewed step-by-step derivation

Finite field · qf-mod30-field-finite-domain-theorem · candidate

90. Do Not Skip the Finite Injective-to-Surjective Step

Use the same already qualified complete finite commutative unital-ring tables on Fin(5)={0,…,4}, addition=[[0, 1, 2, 3, 4], [1, 3, 0, 4, 2], [2, 0, 4, 1, 3], [3, 4, 1, 2, 0], [4, 2, 3, 0, 1]] and multiplication=[[0, 0, 0, 0, 0], [0, 4, 3, 1, 2], [0, 3, 4, 2, 1], [0, 1, 2, 3, 4], [0, 2, 1, 4, 3]] in carrier-label order. The actual zero and one are distinct and need not be labels 0 and 1. The inverse audit is additional to, not a replacement for, those ring axioms. Division by zero is never defined. A finite field has prime-power order, not necessarily prime order. Give (n,missing,field?,theorem_step,field_order_prime_power_not_necessarily_prime). theorem_step is finite_domain_injective_to_surjective_theorem or missing_nonzero_inverse.

View answer and explanation

(5, [], Yes, finite_domain_injective_to_surjective_theorem, field_order_prime_power_not_necessarily_prime)

Record every missing nonzero inverse. In a finite integral domain, nonzero multiplication is injective and therefore surjective, yielding an inverse. A table with missing inverses does not qualify; this does not invalidate that finite-domain theorem. The exact result, retaining the requested order, is (5, [], Yes, finite_domain_injective_to_surjective_theorem, field_order_prime_power_not_necessarily_prime).

Reviewed step-by-step derivation

Finite field · qf-mod30-field-finite-domain-theorem · candidate

91. Do Not Skip the Finite Injective-to-Surjective Step

Use the same already qualified complete finite commutative unital-ring tables on Fin(5)={0,…,4}, addition=[[0, 1, 2, 3, 4], [1, 4, 3, 0, 2], [2, 3, 1, 4, 0], [3, 0, 4, 2, 1], [4, 2, 0, 1, 3]] and multiplication=[[0, 0, 0, 0, 0], [0, 3, 4, 1, 2], [0, 4, 1, 2, 3], [0, 1, 2, 3, 4], [0, 2, 3, 4, 1]] in carrier-label order. The actual zero and one are distinct and need not be labels 0 and 1. The inverse audit is additional to, not a replacement for, those ring axioms. Division by zero is never defined. A finite field has prime-power order, not necessarily prime order. Give (n,missing,field?,theorem_step,field_order_prime_power_not_necessarily_prime). theorem_step is finite_domain_injective_to_surjective_theorem or missing_nonzero_inverse.

View answer and explanation

(5, [], Yes, finite_domain_injective_to_surjective_theorem, field_order_prime_power_not_necessarily_prime)

Record every missing nonzero inverse. In a finite integral domain, nonzero multiplication is injective and therefore surjective, yielding an inverse. A table with missing inverses does not qualify; this does not invalidate that finite-domain theorem. The exact result, retaining the requested order, is (5, [], Yes, finite_domain_injective_to_surjective_theorem, field_order_prime_power_not_necessarily_prime).

Reviewed step-by-step derivation

Finite field · qf-mod30-field-finite-domain-theorem · candidate

92. Do Not Skip the Finite Injective-to-Surjective Step

Use the same already qualified complete finite commutative unital-ring tables on Fin(5)={0,…,4}, addition=[[0, 1, 2, 3, 4], [1, 3, 4, 2, 0], [2, 4, 1, 0, 3], [3, 2, 0, 4, 1], [4, 0, 3, 1, 2]] and multiplication=[[0, 0, 0, 0, 0], [0, 2, 4, 1, 3], [0, 4, 3, 2, 1], [0, 1, 2, 3, 4], [0, 3, 1, 4, 2]] in carrier-label order. The actual zero and one are distinct and need not be labels 0 and 1. The inverse audit is additional to, not a replacement for, those ring axioms. Division by zero is never defined. A finite field has prime-power order, not necessarily prime order. Give (n,missing,field?,theorem_step,field_order_prime_power_not_necessarily_prime). theorem_step is finite_domain_injective_to_surjective_theorem or missing_nonzero_inverse.

View answer and explanation

(5, [], Yes, finite_domain_injective_to_surjective_theorem, field_order_prime_power_not_necessarily_prime)

Record every missing nonzero inverse. In a finite integral domain, nonzero multiplication is injective and therefore surjective, yielding an inverse. A table with missing inverses does not qualify; this does not invalidate that finite-domain theorem. The exact result, retaining the requested order, is (5, [], Yes, finite_domain_injective_to_surjective_theorem, field_order_prime_power_not_necessarily_prime).

Reviewed step-by-step derivation

Finite field · qf-mod30-field-finite-domain-theorem · candidate

93. Do Not Skip the Finite Injective-to-Surjective Step

Use the same already qualified complete finite commutative unital-ring tables on Fin(5)={0,…,4}, addition=[[0, 1, 2, 3, 4], [1, 2, 4, 0, 3], [2, 4, 3, 1, 0], [3, 0, 1, 4, 2], [4, 3, 0, 2, 1]] and multiplication=[[0, 0, 0, 0, 0], [0, 3, 4, 1, 2], [0, 4, 1, 2, 3], [0, 1, 2, 3, 4], [0, 2, 3, 4, 1]] in carrier-label order. The actual zero and one are distinct and need not be labels 0 and 1. The inverse audit is additional to, not a replacement for, those ring axioms. Division by zero is never defined. A finite field has prime-power order, not necessarily prime order. Give (n,missing,field?,theorem_step,field_order_prime_power_not_necessarily_prime). theorem_step is finite_domain_injective_to_surjective_theorem or missing_nonzero_inverse.

View answer and explanation

(5, [], Yes, finite_domain_injective_to_surjective_theorem, field_order_prime_power_not_necessarily_prime)

Record every missing nonzero inverse. In a finite integral domain, nonzero multiplication is injective and therefore surjective, yielding an inverse. A table with missing inverses does not qualify; this does not invalidate that finite-domain theorem. The exact result, retaining the requested order, is (5, [], Yes, finite_domain_injective_to_surjective_theorem, field_order_prime_power_not_necessarily_prime).

Reviewed step-by-step derivation

Finite field · qf-mod30-field-finite-domain-theorem · candidate

94. Do Not Skip the Finite Injective-to-Surjective Step

Use the same already qualified complete finite commutative unital-ring tables on Fin(5)={0,…,4}, addition=[[0, 1, 2, 3, 4], [1, 3, 0, 4, 2], [2, 0, 4, 1, 3], [3, 4, 1, 2, 0], [4, 2, 3, 0, 1]] and multiplication=[[0, 0, 0, 0, 0], [0, 3, 4, 2, 1], [0, 4, 3, 1, 2], [0, 2, 1, 4, 3], [0, 1, 2, 3, 4]] in carrier-label order. The actual zero and one are distinct and need not be labels 0 and 1. The inverse audit is additional to, not a replacement for, those ring axioms. Division by zero is never defined. A finite field has prime-power order, not necessarily prime order. Give (n,missing,field?,theorem_step,field_order_prime_power_not_necessarily_prime). theorem_step is finite_domain_injective_to_surjective_theorem or missing_nonzero_inverse.

View answer and explanation

(5, [], Yes, finite_domain_injective_to_surjective_theorem, field_order_prime_power_not_necessarily_prime)

Record every missing nonzero inverse. In a finite integral domain, nonzero multiplication is injective and therefore surjective, yielding an inverse. A table with missing inverses does not qualify; this does not invalidate that finite-domain theorem. The exact result, retaining the requested order, is (5, [], Yes, finite_domain_injective_to_surjective_theorem, field_order_prime_power_not_necessarily_prime).

Reviewed step-by-step derivation

Finite field · qf-mod30-field-finite-domain-theorem · candidate

95. Do Not Skip the Finite Injective-to-Surjective Step

Use the same already qualified complete finite commutative unital-ring tables on Fin(5)={0,…,4}, addition=[[0, 1, 2, 3, 4], [1, 2, 4, 0, 3], [2, 4, 3, 1, 0], [3, 0, 1, 4, 2], [4, 3, 0, 2, 1]] and multiplication=[[0, 0, 0, 0, 0], [0, 2, 3, 4, 1], [0, 3, 4, 1, 2], [0, 4, 1, 2, 3], [0, 1, 2, 3, 4]] in carrier-label order. The actual zero and one are distinct and need not be labels 0 and 1. The inverse audit is additional to, not a replacement for, those ring axioms. Division by zero is never defined. A finite field has prime-power order, not necessarily prime order. Give (n,missing,field?,theorem_step,field_order_prime_power_not_necessarily_prime). theorem_step is finite_domain_injective_to_surjective_theorem or missing_nonzero_inverse.

View answer and explanation

(5, [], Yes, finite_domain_injective_to_surjective_theorem, field_order_prime_power_not_necessarily_prime)

Record every missing nonzero inverse. In a finite integral domain, nonzero multiplication is injective and therefore surjective, yielding an inverse. A table with missing inverses does not qualify; this does not invalidate that finite-domain theorem. The exact result, retaining the requested order, is (5, [], Yes, finite_domain_injective_to_surjective_theorem, field_order_prime_power_not_necessarily_prime).

Reviewed step-by-step derivation

Finite field · qf-mod30-field-finite-domain-theorem · candidate

96. Do Not Skip the Finite Injective-to-Surjective Step

Use the same already qualified complete finite commutative unital-ring tables on Fin(5)={0,…,4}, addition=[[0, 1, 2, 3, 4], [1, 4, 0, 2, 3], [2, 0, 3, 4, 1], [3, 2, 4, 1, 0], [4, 3, 1, 0, 2]] and multiplication=[[0, 0, 0, 0, 0], [0, 3, 4, 2, 1], [0, 4, 3, 1, 2], [0, 2, 1, 4, 3], [0, 1, 2, 3, 4]] in carrier-label order. The actual zero and one are distinct and need not be labels 0 and 1. The inverse audit is additional to, not a replacement for, those ring axioms. Division by zero is never defined. A finite field has prime-power order, not necessarily prime order. Give (n,missing,field?,theorem_step,field_order_prime_power_not_necessarily_prime). theorem_step is finite_domain_injective_to_surjective_theorem or missing_nonzero_inverse.

View answer and explanation

(5, [], Yes, finite_domain_injective_to_surjective_theorem, field_order_prime_power_not_necessarily_prime)

Record every missing nonzero inverse. In a finite integral domain, nonzero multiplication is injective and therefore surjective, yielding an inverse. A table with missing inverses does not qualify; this does not invalidate that finite-domain theorem. The exact result, retaining the requested order, is (5, [], Yes, finite_domain_injective_to_surjective_theorem, field_order_prime_power_not_necessarily_prime).

Reviewed step-by-step derivation

Finite field · qf-mod30-field-finite-domain-theorem · candidate

97. Do Not Skip the Finite Injective-to-Surjective Step

Use the same already qualified complete finite commutative unital-ring tables on Fin(5)={0,…,4}, addition=[[0, 1, 2, 3, 4], [1, 3, 4, 2, 0], [2, 4, 1, 0, 3], [3, 2, 0, 4, 1], [4, 0, 3, 1, 2]] and multiplication=[[0, 0, 0, 0, 0], [0, 4, 3, 2, 1], [0, 3, 1, 4, 2], [0, 2, 4, 1, 3], [0, 1, 2, 3, 4]] in carrier-label order. The actual zero and one are distinct and need not be labels 0 and 1. The inverse audit is additional to, not a replacement for, those ring axioms. Division by zero is never defined. A finite field has prime-power order, not necessarily prime order. Give (n,missing,field?,theorem_step,field_order_prime_power_not_necessarily_prime). theorem_step is finite_domain_injective_to_surjective_theorem or missing_nonzero_inverse.

View answer and explanation

(5, [], Yes, finite_domain_injective_to_surjective_theorem, field_order_prime_power_not_necessarily_prime)

Record every missing nonzero inverse. In a finite integral domain, nonzero multiplication is injective and therefore surjective, yielding an inverse. A table with missing inverses does not qualify; this does not invalidate that finite-domain theorem. The exact result, retaining the requested order, is (5, [], Yes, finite_domain_injective_to_surjective_theorem, field_order_prime_power_not_necessarily_prime).

Reviewed step-by-step derivation

Finite field · qf-mod30-field-finite-domain-theorem · candidate

98. Do Not Skip the Finite Injective-to-Surjective Step

Use the same already qualified complete finite commutative unital-ring tables on Fin(5)={0,…,4}, addition=[[0, 1, 2, 3, 4], [1, 4, 3, 0, 2], [2, 3, 1, 4, 0], [3, 0, 4, 2, 1], [4, 2, 0, 1, 3]] and multiplication=[[0, 0, 0, 0, 0], [0, 2, 3, 4, 1], [0, 3, 4, 1, 2], [0, 4, 1, 2, 3], [0, 1, 2, 3, 4]] in carrier-label order. The actual zero and one are distinct and need not be labels 0 and 1. The inverse audit is additional to, not a replacement for, those ring axioms. Division by zero is never defined. A finite field has prime-power order, not necessarily prime order. Give (n,missing,field?,theorem_step,field_order_prime_power_not_necessarily_prime). theorem_step is finite_domain_injective_to_surjective_theorem or missing_nonzero_inverse.

View answer and explanation

(5, [], Yes, finite_domain_injective_to_surjective_theorem, field_order_prime_power_not_necessarily_prime)

Record every missing nonzero inverse. In a finite integral domain, nonzero multiplication is injective and therefore surjective, yielding an inverse. A table with missing inverses does not qualify; this does not invalidate that finite-domain theorem. The exact result, retaining the requested order, is (5, [], Yes, finite_domain_injective_to_surjective_theorem, field_order_prime_power_not_necessarily_prime).

Reviewed step-by-step derivation

Finite field · qf-mod30-field-finite-domain-theorem · candidate

99. Do Not Skip the Finite Injective-to-Surjective Step

Use the same already qualified complete finite commutative unital-ring tables on Fin(5)={0,…,4}, addition=[[0, 1, 2, 3, 4], [1, 2, 3, 4, 0], [2, 3, 4, 0, 1], [3, 4, 0, 1, 2], [4, 0, 1, 2, 3]] and multiplication=[[0, 0, 0, 0, 0], [0, 4, 3, 2, 1], [0, 3, 1, 4, 2], [0, 2, 4, 1, 3], [0, 1, 2, 3, 4]] in carrier-label order. The actual zero and one are distinct and need not be labels 0 and 1. The inverse audit is additional to, not a replacement for, those ring axioms. Division by zero is never defined. A finite field has prime-power order, not necessarily prime order. Give (n,missing,field?,theorem_step,field_order_prime_power_not_necessarily_prime). theorem_step is finite_domain_injective_to_surjective_theorem or missing_nonzero_inverse.

View answer and explanation

(5, [], Yes, finite_domain_injective_to_surjective_theorem, field_order_prime_power_not_necessarily_prime)

Record every missing nonzero inverse. In a finite integral domain, nonzero multiplication is injective and therefore surjective, yielding an inverse. A table with missing inverses does not qualify; this does not invalidate that finite-domain theorem. The exact result, retaining the requested order, is (5, [], Yes, finite_domain_injective_to_surjective_theorem, field_order_prime_power_not_necessarily_prime).

Reviewed step-by-step derivation

Finite field · qf-mod30-field-finite-domain-theorem · candidate

100. Do Not Skip the Finite Injective-to-Surjective Step

Use the same already qualified complete finite commutative unital-ring tables on Fin(5)={0,…,4}, addition=[[2, 0, 3, 4, 1], [0, 1, 2, 3, 4], [3, 2, 4, 1, 0], [4, 3, 1, 0, 2], [1, 4, 0, 2, 3]] and multiplication=[[0, 1, 2, 3, 4], [1, 1, 1, 1, 1], [2, 1, 4, 0, 3], [3, 1, 0, 4, 2], [4, 1, 3, 2, 0]] in carrier-label order. The actual zero and one are distinct and need not be labels 0 and 1. The inverse audit is additional to, not a replacement for, those ring axioms. Division by zero is never defined. A finite field has prime-power order, not necessarily prime order. Give (n,missing,field?,theorem_step,field_order_prime_power_not_necessarily_prime). theorem_step is finite_domain_injective_to_surjective_theorem or missing_nonzero_inverse.

View answer and explanation

(5, [], Yes, finite_domain_injective_to_surjective_theorem, field_order_prime_power_not_necessarily_prime)

Record every missing nonzero inverse. In a finite integral domain, nonzero multiplication is injective and therefore surjective, yielding an inverse. A table with missing inverses does not qualify; this does not invalidate that finite-domain theorem. The exact result, retaining the requested order, is (5, [], Yes, finite_domain_injective_to_surjective_theorem, field_order_prime_power_not_necessarily_prime).

Reviewed step-by-step derivation

Finite field · qf-mod30-field-finite-domain-theorem · candidate

Canonical question JSONL · English JSONL · Validation report