Bijectivity Foundations
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This question bank has corresponding bilingual step-by-step derivations. Source candidate and English ai_candidate states are retained. AI review, finite computation and Lean formal proof are labelled separately.
batch-20261004-advance-20261005-r03-bijective-foundations · 2026-10-04 · 100 questions
This English edition is an AI candidate and does not yet have an OCME Atlas (GPT-6) Twin closure receipt.
1. Bijectivity of a Finite Total Function
The declared total function f:A→B has A=∅, B=∅, and table ∅. Is f both injective and surjective? Decide bijectivity from the actual table.
View answer and explanation
Yes
Both conditions must hold, equivalently every codomain target must have exactly one permitted preimage. The answer is Yes.
Reviewed step-by-step derivation
Bijectivity and unique preimages · qf-bijective-table-property · candidate
2. Bijectivity of a Finite Total Function
The declared total function f:A→B has A={-1, 0}, B={0, 1}, and table {(-1, 1), (0, 0)}. Is f both injective and surjective? Decide bijectivity from the actual table.
View answer and explanation
Yes
Both conditions must hold, equivalently every codomain target must have exactly one permitted preimage. The answer is Yes.
Reviewed step-by-step derivation
Bijectivity and unique preimages · qf-bijective-table-property · candidate
3. Bijectivity of a Finite Total Function
The declared total function f:A→B has A={-1, 0}, B={1, 3, 5}, and table {(-1, 1), (0, 3)}. Is f both injective and surjective? Decide bijectivity from the actual table.
View answer and explanation
No
Both conditions must hold, equivalently every codomain target must have exactly one permitted preimage. The answer is No.
Reviewed step-by-step derivation
Bijectivity and unique preimages · qf-bijective-table-property · candidate
4. Bijectivity of a Finite Total Function
The declared total function f:A→B has A={-1, 0}, B={0}, and table {(-1, 0), (0, 0)}. Is f both injective and surjective? Decide bijectivity from the actual table.
View answer and explanation
No
Both conditions must hold, equivalently every codomain target must have exactly one permitted preimage. The answer is No.
Reviewed step-by-step derivation
Bijectivity and unique preimages · qf-bijective-table-property · candidate
5. Bijectivity of a Finite Total Function
The declared total function f:A→B has A={-1, 0}, B={-3, -2, -1}, and table {(-1, -1), (0, -1)}. Is f both injective and surjective? Decide bijectivity from the actual table.
View answer and explanation
No
Both conditions must hold, equivalently every codomain target must have exactly one permitted preimage. The answer is No.
Reviewed step-by-step derivation
Bijectivity and unique preimages · qf-bijective-table-property · candidate
6. Bijectivity of a Finite Total Function
The declared total function f:A→B has A={-2, -1, 0}, B={0, 1, 2}, and table {(-2, 1), (-1, 2), (0, 0)}. Is f both injective and surjective? Decide bijectivity from the actual table.
View answer and explanation
Yes
Both conditions must hold, equivalently every codomain target must have exactly one permitted preimage. The answer is Yes.
Reviewed step-by-step derivation
Bijectivity and unique preimages · qf-bijective-table-property · candidate
7. Bijectivity of a Finite Total Function
The declared total function f:A→B has A={-2, -1, 0}, B={1, 3, 5, 7}, and table {(-2, 1), (-1, 3), (0, 5)}. Is f both injective and surjective? Decide bijectivity from the actual table.
View answer and explanation
No
Both conditions must hold, equivalently every codomain target must have exactly one permitted preimage. The answer is No.
Reviewed step-by-step derivation
Bijectivity and unique preimages · qf-bijective-table-property · candidate
8. Bijectivity of a Finite Total Function
The declared total function f:A→B has A={-2, -1, 0}, B={0, 1}, and table {(-2, 0), (-1, 1), (0, 0)}. Is f both injective and surjective? Decide bijectivity from the actual table.
View answer and explanation
No
Both conditions must hold, equivalently every codomain target must have exactly one permitted preimage. The answer is No.
Reviewed step-by-step derivation
Bijectivity and unique preimages · qf-bijective-table-property · candidate
9. Bijectivity of a Finite Total Function
The declared total function f:A→B has A={-2, -1, 0}, B={-4, -3, -2, -1}, and table {(-2, -1), (-1, -1), (0, -2)}. Is f both injective and surjective? Decide bijectivity from the actual table.
View answer and explanation
No
Both conditions must hold, equivalently every codomain target must have exactly one permitted preimage. The answer is No.
Reviewed step-by-step derivation
Bijectivity and unique preimages · qf-bijective-table-property · candidate
10. Bijectivity of a Finite Total Function
The declared total function f:A→B has A={-3, -2, -1, 0}, B={0, 1, 2, 3}, and table {(-3, 1), (-2, 2), (-1, 3), (0, 0)}. Is f both injective and surjective? Decide bijectivity from the actual table.
View answer and explanation
Yes
Both conditions must hold, equivalently every codomain target must have exactly one permitted preimage. The answer is Yes.
Reviewed step-by-step derivation
Bijectivity and unique preimages · qf-bijective-table-property · candidate
11. Bijectivity of a Finite Total Function
The declared total function f:A→B has A={-3, -2, -1, 0}, B={1, 3, 5, 7, 9}, and table {(-3, 1), (-2, 3), (-1, 5), (0, 7)}. Is f both injective and surjective? Decide bijectivity from the actual table.
View answer and explanation
No
Both conditions must hold, equivalently every codomain target must have exactly one permitted preimage. The answer is No.
Reviewed step-by-step derivation
Bijectivity and unique preimages · qf-bijective-table-property · candidate
12. Bijectivity of a Finite Total Function
The declared total function f:A→B has A={-3, -2, -1, 0}, B={0, 1, 2}, and table {(-3, 0), (-2, 1), (-1, 2), (0, 0)}. Is f both injective and surjective? Decide bijectivity from the actual table.
View answer and explanation
No
Both conditions must hold, equivalently every codomain target must have exactly one permitted preimage. The answer is No.
Reviewed step-by-step derivation
Bijectivity and unique preimages · qf-bijective-table-property · candidate
13. Bijectivity of a Finite Total Function
The declared total function f:A→B has A={-3, -2, -1, 0}, B={-5, -4, -3, -2, -1}, and table {(-3, -1), (-2, -1), (-1, -2), (0, -2)}. Is f both injective and surjective? Decide bijectivity from the actual table.
View answer and explanation
No
Both conditions must hold, equivalently every codomain target must have exactly one permitted preimage. The answer is No.
Reviewed step-by-step derivation
Bijectivity and unique preimages · qf-bijective-table-property · candidate
14. Bijectivity of a Finite Total Function
The declared total function f:A→B has A={-4, -3, -2, -1, 0}, B={0, 1, 2, 3, 4}, and table {(-4, 1), (-3, 2), (-2, 3), (-1, 4), (0, 0)}. Is f both injective and surjective? Decide bijectivity from the actual table.
View answer and explanation
Yes
Both conditions must hold, equivalently every codomain target must have exactly one permitted preimage. The answer is Yes.
Reviewed step-by-step derivation
Bijectivity and unique preimages · qf-bijective-table-property · candidate
15. Bijectivity of a Finite Total Function
The declared total function f:A→B has A={-4, -3, -2, -1, 0}, B={1, 3, 5, 7, 9, 11}, and table {(-4, 1), (-3, 3), (-2, 5), (-1, 7), (0, 9)}. Is f both injective and surjective? Decide bijectivity from the actual table.
View answer and explanation
No
Both conditions must hold, equivalently every codomain target must have exactly one permitted preimage. The answer is No.
Reviewed step-by-step derivation
Bijectivity and unique preimages · qf-bijective-table-property · candidate
16. Bijectivity of a Finite Total Function
The declared total function f:A→B has A={-4, -3, -2, -1, 0}, B={0, 1, 2, 3}, and table {(-4, 0), (-3, 1), (-2, 2), (-1, 3), (0, 0)}. Is f both injective and surjective? Decide bijectivity from the actual table.
View answer and explanation
No
Both conditions must hold, equivalently every codomain target must have exactly one permitted preimage. The answer is No.
Reviewed step-by-step derivation
Bijectivity and unique preimages · qf-bijective-table-property · candidate
17. Bijectivity of a Finite Total Function
The declared total function f:A→B has A={-4, -3, -2, -1, 0}, B={-6, -5, -4, -3, -2, -1}, and table {(-4, -1), (-3, -1), (-2, -2), (-1, -2), (0, -3)}. Is f both injective and surjective? Decide bijectivity from the actual table.
View answer and explanation
No
Both conditions must hold, equivalently every codomain target must have exactly one permitted preimage. The answer is No.
Reviewed step-by-step derivation
Bijectivity and unique preimages · qf-bijective-table-property · candidate
18. Bijectivity of a Finite Total Function
The declared total function f:A→B has A={-5, -4, -3, -2, -1, 0}, B={0, 1, 2, 3, 4, 5}, and table {(-5, 1), (-4, 2), (-3, 3), (-2, 4), (-1, 5), (0, 0)}. Is f both injective and surjective? Decide bijectivity from the actual table.
View answer and explanation
Yes
Both conditions must hold, equivalently every codomain target must have exactly one permitted preimage. The answer is Yes.
Reviewed step-by-step derivation
Bijectivity and unique preimages · qf-bijective-table-property · candidate
19. Bijectivity of a Finite Total Function
The declared total function f:A→B has A={-5, -4, -3, -2, -1, 0}, B={1, 3, 5, 7, 9, 11, 13}, and table {(-5, 1), (-4, 3), (-3, 5), (-2, 7), (-1, 9), (0, 11)}. Is f both injective and surjective? Decide bijectivity from the actual table.
View answer and explanation
No
Both conditions must hold, equivalently every codomain target must have exactly one permitted preimage. The answer is No.
Reviewed step-by-step derivation
Bijectivity and unique preimages · qf-bijective-table-property · candidate
20. Bijectivity of a Finite Total Function
The declared total function f:A→B has A={-5, -4, -3, -2, -1, 0}, B={0, 1, 2, 3, 4}, and table {(-5, 0), (-4, 1), (-3, 2), (-2, 3), (-1, 4), (0, 0)}. Is f both injective and surjective? Decide bijectivity from the actual table.
View answer and explanation
No
Both conditions must hold, equivalently every codomain target must have exactly one permitted preimage. The answer is No.
Reviewed step-by-step derivation
Bijectivity and unique preimages · qf-bijective-table-property · candidate
21. Bijectivity of a Finite Total Function
The declared total function f:A→B has A={-5, -4, -3, -2, -1, 0}, B={-7, -6, -5, -4, -3, -2, -1}, and table {(-5, -1), (-4, -1), (-3, -2), (-2, -2), (-1, -3), (0, -3)}. Is f both injective and surjective? Decide bijectivity from the actual table.
View answer and explanation
No
Both conditions must hold, equivalently every codomain target must have exactly one permitted preimage. The answer is No.
Reviewed step-by-step derivation
Bijectivity and unique preimages · qf-bijective-table-property · candidate
22. Bijectivity of a Finite Total Function
The declared total function f:A→B has A={-6, -5, -4, -3, -2, -1, 0}, B={0, 1, 2, 3, 4, 5, 6}, and table {(-6, 1), (-5, 2), (-4, 3), (-3, 4), (-2, 5), (-1, 6), (0, 0)}. Is f both injective and surjective? Decide bijectivity from the actual table.
View answer and explanation
Yes
Both conditions must hold, equivalently every codomain target must have exactly one permitted preimage. The answer is Yes.
Reviewed step-by-step derivation
Bijectivity and unique preimages · qf-bijective-table-property · candidate
23. Bijectivity of a Finite Total Function
The declared total function f:A→B has A={-6, -5, -4, -3, -2, -1, 0}, B={1, 3, 5, 7, 9, 11, 13, 15}, and table {(-6, 1), (-5, 3), (-4, 5), (-3, 7), (-2, 9), (-1, 11), (0, 13)}. Is f both injective and surjective? Decide bijectivity from the actual table.
View answer and explanation
No
Both conditions must hold, equivalently every codomain target must have exactly one permitted preimage. The answer is No.
Reviewed step-by-step derivation
Bijectivity and unique preimages · qf-bijective-table-property · candidate
24. Bijectivity of a Finite Total Function
The declared total function f:A→B has A={-6, -5, -4, -3, -2, -1, 0}, B={0, 1, 2, 3, 4, 5}, and table {(-6, 0), (-5, 1), (-4, 2), (-3, 3), (-2, 4), (-1, 5), (0, 0)}. Is f both injective and surjective? Decide bijectivity from the actual table.
View answer and explanation
No
Both conditions must hold, equivalently every codomain target must have exactly one permitted preimage. The answer is No.
Reviewed step-by-step derivation
Bijectivity and unique preimages · qf-bijective-table-property · candidate
25. Bijectivity of a Finite Total Function
The declared total function f:A→B has A={-6, -5, -4, -3, -2, -1, 0}, B={-8, -7, -6, -5, -4, -3, -2, -1}, and table {(-6, -1), (-5, -1), (-4, -2), (-3, -2), (-2, -3), (-1, -3), (0, -4)}. Is f both injective and surjective? Decide bijectivity from the actual table.
View answer and explanation
No
Both conditions must hold, equivalently every codomain target must have exactly one permitted preimage. The answer is No.
Reviewed step-by-step derivation
Bijectivity and unique preimages · qf-bijective-table-property · candidate
26. Four-Way Function Classification
The declared total function f:A→B has A=∅, B=∅, and table ∅. Classify f as bijective, injective only, surjective only, or neither injective nor surjective.
View answer and explanation
bijective
Check distinct-input collisions and full codomain coverage separately; their two truth values give bijective. Set sizes alone do not classify this supplied function.
Reviewed step-by-step derivation
Bijectivity and unique preimages · qf-bijective-classification · candidate
27. Four-Way Function Classification
The declared total function f:A→B has A={-1, 0}, B={0, 1}, and table {(-1, 1), (0, 0)}. Classify f as bijective, injective only, surjective only, or neither injective nor surjective.
View answer and explanation
bijective
Check distinct-input collisions and full codomain coverage separately; their two truth values give bijective. Set sizes alone do not classify this supplied function.
Reviewed step-by-step derivation
Bijectivity and unique preimages · qf-bijective-classification · candidate
28. Four-Way Function Classification
The declared total function f:A→B has A={-1, 0}, B={1, 3, 5}, and table {(-1, 1), (0, 3)}. Classify f as bijective, injective only, surjective only, or neither injective nor surjective.
View answer and explanation
injective only
Check distinct-input collisions and full codomain coverage separately; their two truth values give injective only. Set sizes alone do not classify this supplied function.
Reviewed step-by-step derivation
Bijectivity and unique preimages · qf-bijective-classification · candidate
29. Four-Way Function Classification
The declared total function f:A→B has A={-1, 0}, B={0}, and table {(-1, 0), (0, 0)}. Classify f as bijective, injective only, surjective only, or neither injective nor surjective.
View answer and explanation
surjective only
Check distinct-input collisions and full codomain coverage separately; their two truth values give surjective only. Set sizes alone do not classify this supplied function.
Reviewed step-by-step derivation
Bijectivity and unique preimages · qf-bijective-classification · candidate
30. Four-Way Function Classification
The declared total function f:A→B has A={-1, 0}, B={-3, -2, -1}, and table {(-1, -1), (0, -1)}. Classify f as bijective, injective only, surjective only, or neither injective nor surjective.
View answer and explanation
neither injective nor surjective
Check distinct-input collisions and full codomain coverage separately; their two truth values give neither injective nor surjective. Set sizes alone do not classify this supplied function.
Reviewed step-by-step derivation
Bijectivity and unique preimages · qf-bijective-classification · candidate
31. Four-Way Function Classification
The declared total function f:A→B has A={-2, -1, 0}, B={0, 1, 2}, and table {(-2, 1), (-1, 2), (0, 0)}. Classify f as bijective, injective only, surjective only, or neither injective nor surjective.
View answer and explanation
bijective
Check distinct-input collisions and full codomain coverage separately; their two truth values give bijective. Set sizes alone do not classify this supplied function.
Reviewed step-by-step derivation
Bijectivity and unique preimages · qf-bijective-classification · candidate
32. Four-Way Function Classification
The declared total function f:A→B has A={-2, -1, 0}, B={1, 3, 5, 7}, and table {(-2, 1), (-1, 3), (0, 5)}. Classify f as bijective, injective only, surjective only, or neither injective nor surjective.
View answer and explanation
injective only
Check distinct-input collisions and full codomain coverage separately; their two truth values give injective only. Set sizes alone do not classify this supplied function.
Reviewed step-by-step derivation
Bijectivity and unique preimages · qf-bijective-classification · candidate
33. Four-Way Function Classification
The declared total function f:A→B has A={-2, -1, 0}, B={0, 1}, and table {(-2, 0), (-1, 1), (0, 0)}. Classify f as bijective, injective only, surjective only, or neither injective nor surjective.
View answer and explanation
surjective only
Check distinct-input collisions and full codomain coverage separately; their two truth values give surjective only. Set sizes alone do not classify this supplied function.
Reviewed step-by-step derivation
Bijectivity and unique preimages · qf-bijective-classification · candidate
34. Four-Way Function Classification
The declared total function f:A→B has A={-2, -1, 0}, B={-4, -3, -2, -1}, and table {(-2, -1), (-1, -1), (0, -2)}. Classify f as bijective, injective only, surjective only, or neither injective nor surjective.
View answer and explanation
neither injective nor surjective
Check distinct-input collisions and full codomain coverage separately; their two truth values give neither injective nor surjective. Set sizes alone do not classify this supplied function.
Reviewed step-by-step derivation
Bijectivity and unique preimages · qf-bijective-classification · candidate
35. Four-Way Function Classification
The declared total function f:A→B has A={-3, -2, -1, 0}, B={0, 1, 2, 3}, and table {(-3, 1), (-2, 2), (-1, 3), (0, 0)}. Classify f as bijective, injective only, surjective only, or neither injective nor surjective.
View answer and explanation
bijective
Check distinct-input collisions and full codomain coverage separately; their two truth values give bijective. Set sizes alone do not classify this supplied function.
Reviewed step-by-step derivation
Bijectivity and unique preimages · qf-bijective-classification · candidate
36. Four-Way Function Classification
The declared total function f:A→B has A={-3, -2, -1, 0}, B={1, 3, 5, 7, 9}, and table {(-3, 1), (-2, 3), (-1, 5), (0, 7)}. Classify f as bijective, injective only, surjective only, or neither injective nor surjective.
View answer and explanation
injective only
Check distinct-input collisions and full codomain coverage separately; their two truth values give injective only. Set sizes alone do not classify this supplied function.
Reviewed step-by-step derivation
Bijectivity and unique preimages · qf-bijective-classification · candidate
37. Four-Way Function Classification
The declared total function f:A→B has A={-3, -2, -1, 0}, B={0, 1, 2}, and table {(-3, 0), (-2, 1), (-1, 2), (0, 0)}. Classify f as bijective, injective only, surjective only, or neither injective nor surjective.
View answer and explanation
surjective only
Check distinct-input collisions and full codomain coverage separately; their two truth values give surjective only. Set sizes alone do not classify this supplied function.
Reviewed step-by-step derivation
Bijectivity and unique preimages · qf-bijective-classification · candidate
38. Four-Way Function Classification
The declared total function f:A→B has A={-3, -2, -1, 0}, B={-5, -4, -3, -2, -1}, and table {(-3, -1), (-2, -1), (-1, -2), (0, -2)}. Classify f as bijective, injective only, surjective only, or neither injective nor surjective.
View answer and explanation
neither injective nor surjective
Check distinct-input collisions and full codomain coverage separately; their two truth values give neither injective nor surjective. Set sizes alone do not classify this supplied function.
Reviewed step-by-step derivation
Bijectivity and unique preimages · qf-bijective-classification · candidate
39. Four-Way Function Classification
The declared total function f:A→B has A={-4, -3, -2, -1, 0}, B={0, 1, 2, 3, 4}, and table {(-4, 1), (-3, 2), (-2, 3), (-1, 4), (0, 0)}. Classify f as bijective, injective only, surjective only, or neither injective nor surjective.
View answer and explanation
bijective
Check distinct-input collisions and full codomain coverage separately; their two truth values give bijective. Set sizes alone do not classify this supplied function.
Reviewed step-by-step derivation
Bijectivity and unique preimages · qf-bijective-classification · candidate
40. Four-Way Function Classification
The declared total function f:A→B has A={-4, -3, -2, -1, 0}, B={1, 3, 5, 7, 9, 11}, and table {(-4, 1), (-3, 3), (-2, 5), (-1, 7), (0, 9)}. Classify f as bijective, injective only, surjective only, or neither injective nor surjective.
View answer and explanation
injective only
Check distinct-input collisions and full codomain coverage separately; their two truth values give injective only. Set sizes alone do not classify this supplied function.
Reviewed step-by-step derivation
Bijectivity and unique preimages · qf-bijective-classification · candidate
41. Four-Way Function Classification
The declared total function f:A→B has A={-4, -3, -2, -1, 0}, B={0, 1, 2, 3}, and table {(-4, 0), (-3, 1), (-2, 2), (-1, 3), (0, 0)}. Classify f as bijective, injective only, surjective only, or neither injective nor surjective.
View answer and explanation
surjective only
Check distinct-input collisions and full codomain coverage separately; their two truth values give surjective only. Set sizes alone do not classify this supplied function.
Reviewed step-by-step derivation
Bijectivity and unique preimages · qf-bijective-classification · candidate
42. Four-Way Function Classification
The declared total function f:A→B has A={-4, -3, -2, -1, 0}, B={-6, -5, -4, -3, -2, -1}, and table {(-4, -1), (-3, -1), (-2, -2), (-1, -2), (0, -3)}. Classify f as bijective, injective only, surjective only, or neither injective nor surjective.
View answer and explanation
neither injective nor surjective
Check distinct-input collisions and full codomain coverage separately; their two truth values give neither injective nor surjective. Set sizes alone do not classify this supplied function.
Reviewed step-by-step derivation
Bijectivity and unique preimages · qf-bijective-classification · candidate
43. Four-Way Function Classification
The declared total function f:A→B has A={-5, -4, -3, -2, -1, 0}, B={0, 1, 2, 3, 4, 5}, and table {(-5, 1), (-4, 2), (-3, 3), (-2, 4), (-1, 5), (0, 0)}. Classify f as bijective, injective only, surjective only, or neither injective nor surjective.
View answer and explanation
bijective
Check distinct-input collisions and full codomain coverage separately; their two truth values give bijective. Set sizes alone do not classify this supplied function.
Reviewed step-by-step derivation
Bijectivity and unique preimages · qf-bijective-classification · candidate
44. Four-Way Function Classification
The declared total function f:A→B has A={-5, -4, -3, -2, -1, 0}, B={1, 3, 5, 7, 9, 11, 13}, and table {(-5, 1), (-4, 3), (-3, 5), (-2, 7), (-1, 9), (0, 11)}. Classify f as bijective, injective only, surjective only, or neither injective nor surjective.
View answer and explanation
injective only
Check distinct-input collisions and full codomain coverage separately; their two truth values give injective only. Set sizes alone do not classify this supplied function.
Reviewed step-by-step derivation
Bijectivity and unique preimages · qf-bijective-classification · candidate
45. Four-Way Function Classification
The declared total function f:A→B has A={-5, -4, -3, -2, -1, 0}, B={0, 1, 2, 3, 4}, and table {(-5, 0), (-4, 1), (-3, 2), (-2, 3), (-1, 4), (0, 0)}. Classify f as bijective, injective only, surjective only, or neither injective nor surjective.
View answer and explanation
surjective only
Check distinct-input collisions and full codomain coverage separately; their two truth values give surjective only. Set sizes alone do not classify this supplied function.
Reviewed step-by-step derivation
Bijectivity and unique preimages · qf-bijective-classification · candidate
46. Four-Way Function Classification
The declared total function f:A→B has A={-5, -4, -3, -2, -1, 0}, B={-7, -6, -5, -4, -3, -2, -1}, and table {(-5, -1), (-4, -1), (-3, -2), (-2, -2), (-1, -3), (0, -3)}. Classify f as bijective, injective only, surjective only, or neither injective nor surjective.
View answer and explanation
neither injective nor surjective
Check distinct-input collisions and full codomain coverage separately; their two truth values give neither injective nor surjective. Set sizes alone do not classify this supplied function.
Reviewed step-by-step derivation
Bijectivity and unique preimages · qf-bijective-classification · candidate
47. Four-Way Function Classification
The declared total function f:A→B has A={-6, -5, -4, -3, -2, -1, 0}, B={0, 1, 2, 3, 4, 5, 6}, and table {(-6, 1), (-5, 2), (-4, 3), (-3, 4), (-2, 5), (-1, 6), (0, 0)}. Classify f as bijective, injective only, surjective only, or neither injective nor surjective.
View answer and explanation
bijective
Check distinct-input collisions and full codomain coverage separately; their two truth values give bijective. Set sizes alone do not classify this supplied function.
Reviewed step-by-step derivation
Bijectivity and unique preimages · qf-bijective-classification · candidate
48. Four-Way Function Classification
The declared total function f:A→B has A={-6, -5, -4, -3, -2, -1, 0}, B={1, 3, 5, 7, 9, 11, 13, 15}, and table {(-6, 1), (-5, 3), (-4, 5), (-3, 7), (-2, 9), (-1, 11), (0, 13)}. Classify f as bijective, injective only, surjective only, or neither injective nor surjective.
View answer and explanation
injective only
Check distinct-input collisions and full codomain coverage separately; their two truth values give injective only. Set sizes alone do not classify this supplied function.
Reviewed step-by-step derivation
Bijectivity and unique preimages · qf-bijective-classification · candidate
49. Four-Way Function Classification
The declared total function f:A→B has A={-6, -5, -4, -3, -2, -1, 0}, B={0, 1, 2, 3, 4, 5}, and table {(-6, 0), (-5, 1), (-4, 2), (-3, 3), (-2, 4), (-1, 5), (0, 0)}. Classify f as bijective, injective only, surjective only, or neither injective nor surjective.
View answer and explanation
surjective only
Check distinct-input collisions and full codomain coverage separately; their two truth values give surjective only. Set sizes alone do not classify this supplied function.
Reviewed step-by-step derivation
Bijectivity and unique preimages · qf-bijective-classification · candidate
50. Four-Way Function Classification
The declared total function f:A→B has A={-6, -5, -4, -3, -2, -1, 0}, B={-8, -7, -6, -5, -4, -3, -2, -1}, and table {(-6, -1), (-5, -1), (-4, -2), (-3, -2), (-2, -3), (-1, -3), (0, -4)}. Classify f as bijective, injective only, surjective only, or neither injective nor surjective.
View answer and explanation
neither injective nor surjective
Check distinct-input collisions and full codomain coverage separately; their two truth values give neither injective nor surjective. Set sizes alone do not classify this supplied function.
Reviewed step-by-step derivation
Bijectivity and unique preimages · qf-bijective-classification · candidate
51. Complete Codomain Fiber Profile
The declared total function f:A→B has A=∅, B=∅, and table ∅. In increasing target order, list every (b,|f⁻¹({b})|), including zero for targets not attained.
View answer and explanation
∅
Count the permitted preimages of every element of B to obtain ∅. The profile includes zero-size fibers, not only attained targets.
Reviewed step-by-step derivation
Bijectivity and unique preimages · qf-bijective-fiber-profile · candidate
52. Complete Codomain Fiber Profile
The declared total function f:A→B has A={-1, 0}, B={0, 1}, and table {(-1, 1), (0, 0)}. In increasing target order, list every (b,|f⁻¹({b})|), including zero for targets not attained.
View answer and explanation
{(0, 1), (1, 1)}
Count the permitted preimages of every element of B to obtain {(0, 1), (1, 1)}. The profile includes zero-size fibers, not only attained targets.
Reviewed step-by-step derivation
Bijectivity and unique preimages · qf-bijective-fiber-profile · candidate
53. Complete Codomain Fiber Profile
The declared total function f:A→B has A={-1, 0}, B={1, 3, 5}, and table {(-1, 1), (0, 3)}. In increasing target order, list every (b,|f⁻¹({b})|), including zero for targets not attained.
View answer and explanation
{(1, 1), (3, 1), (5, 0)}
Count the permitted preimages of every element of B to obtain {(1, 1), (3, 1), (5, 0)}. The profile includes zero-size fibers, not only attained targets.
Reviewed step-by-step derivation
Bijectivity and unique preimages · qf-bijective-fiber-profile · candidate
54. Complete Codomain Fiber Profile
The declared total function f:A→B has A={-1, 0}, B={0}, and table {(-1, 0), (0, 0)}. In increasing target order, list every (b,|f⁻¹({b})|), including zero for targets not attained.
View answer and explanation
{(0, 2)}
Count the permitted preimages of every element of B to obtain {(0, 2)}. The profile includes zero-size fibers, not only attained targets.
Reviewed step-by-step derivation
Bijectivity and unique preimages · qf-bijective-fiber-profile · candidate
55. Complete Codomain Fiber Profile
The declared total function f:A→B has A={-1, 0}, B={-3, -2, -1}, and table {(-1, -1), (0, -1)}. In increasing target order, list every (b,|f⁻¹({b})|), including zero for targets not attained.
View answer and explanation
{(-3, 0), (-2, 0), (-1, 2)}
Count the permitted preimages of every element of B to obtain {(-3, 0), (-2, 0), (-1, 2)}. The profile includes zero-size fibers, not only attained targets.
Reviewed step-by-step derivation
Bijectivity and unique preimages · qf-bijective-fiber-profile · candidate
56. Complete Codomain Fiber Profile
The declared total function f:A→B has A={-2, -1, 0}, B={0, 1, 2}, and table {(-2, 1), (-1, 2), (0, 0)}. In increasing target order, list every (b,|f⁻¹({b})|), including zero for targets not attained.
View answer and explanation
{(0, 1), (1, 1), (2, 1)}
Count the permitted preimages of every element of B to obtain {(0, 1), (1, 1), (2, 1)}. The profile includes zero-size fibers, not only attained targets.
Reviewed step-by-step derivation
Bijectivity and unique preimages · qf-bijective-fiber-profile · candidate
57. Complete Codomain Fiber Profile
The declared total function f:A→B has A={-2, -1, 0}, B={1, 3, 5, 7}, and table {(-2, 1), (-1, 3), (0, 5)}. In increasing target order, list every (b,|f⁻¹({b})|), including zero for targets not attained.
View answer and explanation
{(1, 1), (3, 1), (5, 1), (7, 0)}
Count the permitted preimages of every element of B to obtain {(1, 1), (3, 1), (5, 1), (7, 0)}. The profile includes zero-size fibers, not only attained targets.
Reviewed step-by-step derivation
Bijectivity and unique preimages · qf-bijective-fiber-profile · candidate
58. Complete Codomain Fiber Profile
The declared total function f:A→B has A={-2, -1, 0}, B={0, 1}, and table {(-2, 0), (-1, 1), (0, 0)}. In increasing target order, list every (b,|f⁻¹({b})|), including zero for targets not attained.
View answer and explanation
{(0, 2), (1, 1)}
Count the permitted preimages of every element of B to obtain {(0, 2), (1, 1)}. The profile includes zero-size fibers, not only attained targets.
Reviewed step-by-step derivation
Bijectivity and unique preimages · qf-bijective-fiber-profile · candidate
59. Complete Codomain Fiber Profile
The declared total function f:A→B has A={-2, -1, 0}, B={-4, -3, -2, -1}, and table {(-2, -1), (-1, -1), (0, -2)}. In increasing target order, list every (b,|f⁻¹({b})|), including zero for targets not attained.
View answer and explanation
{(-4, 0), (-3, 0), (-2, 1), (-1, 2)}
Count the permitted preimages of every element of B to obtain {(-4, 0), (-3, 0), (-2, 1), (-1, 2)}. The profile includes zero-size fibers, not only attained targets.
Reviewed step-by-step derivation
Bijectivity and unique preimages · qf-bijective-fiber-profile · candidate
60. Complete Codomain Fiber Profile
The declared total function f:A→B has A={-3, -2, -1, 0}, B={0, 1, 2, 3}, and table {(-3, 1), (-2, 2), (-1, 3), (0, 0)}. In increasing target order, list every (b,|f⁻¹({b})|), including zero for targets not attained.
View answer and explanation
{(0, 1), (1, 1), (2, 1), (3, 1)}
Count the permitted preimages of every element of B to obtain {(0, 1), (1, 1), (2, 1), (3, 1)}. The profile includes zero-size fibers, not only attained targets.
Reviewed step-by-step derivation
Bijectivity and unique preimages · qf-bijective-fiber-profile · candidate
61. Complete Codomain Fiber Profile
The declared total function f:A→B has A={-3, -2, -1, 0}, B={1, 3, 5, 7, 9}, and table {(-3, 1), (-2, 3), (-1, 5), (0, 7)}. In increasing target order, list every (b,|f⁻¹({b})|), including zero for targets not attained.
View answer and explanation
{(1, 1), (3, 1), (5, 1), (7, 1), (9, 0)}
Count the permitted preimages of every element of B to obtain {(1, 1), (3, 1), (5, 1), (7, 1), (9, 0)}. The profile includes zero-size fibers, not only attained targets.
Reviewed step-by-step derivation
Bijectivity and unique preimages · qf-bijective-fiber-profile · candidate
62. Complete Codomain Fiber Profile
The declared total function f:A→B has A={-3, -2, -1, 0}, B={0, 1, 2}, and table {(-3, 0), (-2, 1), (-1, 2), (0, 0)}. In increasing target order, list every (b,|f⁻¹({b})|), including zero for targets not attained.
View answer and explanation
{(0, 2), (1, 1), (2, 1)}
Count the permitted preimages of every element of B to obtain {(0, 2), (1, 1), (2, 1)}. The profile includes zero-size fibers, not only attained targets.
Reviewed step-by-step derivation
Bijectivity and unique preimages · qf-bijective-fiber-profile · candidate
63. Complete Codomain Fiber Profile
The declared total function f:A→B has A={-3, -2, -1, 0}, B={-5, -4, -3, -2, -1}, and table {(-3, -1), (-2, -1), (-1, -2), (0, -2)}. In increasing target order, list every (b,|f⁻¹({b})|), including zero for targets not attained.
View answer and explanation
{(-5, 0), (-4, 0), (-3, 0), (-2, 2), (-1, 2)}
Count the permitted preimages of every element of B to obtain {(-5, 0), (-4, 0), (-3, 0), (-2, 2), (-1, 2)}. The profile includes zero-size fibers, not only attained targets.
Reviewed step-by-step derivation
Bijectivity and unique preimages · qf-bijective-fiber-profile · candidate
64. Complete Codomain Fiber Profile
The declared total function f:A→B has A={-4, -3, -2, -1, 0}, B={0, 1, 2, 3, 4}, and table {(-4, 1), (-3, 2), (-2, 3), (-1, 4), (0, 0)}. In increasing target order, list every (b,|f⁻¹({b})|), including zero for targets not attained.
View answer and explanation
{(0, 1), (1, 1), (2, 1), (3, 1), (4, 1)}
Count the permitted preimages of every element of B to obtain {(0, 1), (1, 1), (2, 1), (3, 1), (4, 1)}. The profile includes zero-size fibers, not only attained targets.
Reviewed step-by-step derivation
Bijectivity and unique preimages · qf-bijective-fiber-profile · candidate
65. Complete Codomain Fiber Profile
The declared total function f:A→B has A={-4, -3, -2, -1, 0}, B={1, 3, 5, 7, 9, 11}, and table {(-4, 1), (-3, 3), (-2, 5), (-1, 7), (0, 9)}. In increasing target order, list every (b,|f⁻¹({b})|), including zero for targets not attained.
View answer and explanation
{(1, 1), (3, 1), (5, 1), (7, 1), (9, 1), (11, 0)}
Count the permitted preimages of every element of B to obtain {(1, 1), (3, 1), (5, 1), (7, 1), (9, 1), (11, 0)}. The profile includes zero-size fibers, not only attained targets.
Reviewed step-by-step derivation
Bijectivity and unique preimages · qf-bijective-fiber-profile · candidate
66. Complete Codomain Fiber Profile
The declared total function f:A→B has A={-4, -3, -2, -1, 0}, B={0, 1, 2, 3}, and table {(-4, 0), (-3, 1), (-2, 2), (-1, 3), (0, 0)}. In increasing target order, list every (b,|f⁻¹({b})|), including zero for targets not attained.
View answer and explanation
{(0, 2), (1, 1), (2, 1), (3, 1)}
Count the permitted preimages of every element of B to obtain {(0, 2), (1, 1), (2, 1), (3, 1)}. The profile includes zero-size fibers, not only attained targets.
Reviewed step-by-step derivation
Bijectivity and unique preimages · qf-bijective-fiber-profile · candidate
67. Complete Codomain Fiber Profile
The declared total function f:A→B has A={-4, -3, -2, -1, 0}, B={-6, -5, -4, -3, -2, -1}, and table {(-4, -1), (-3, -1), (-2, -2), (-1, -2), (0, -3)}. In increasing target order, list every (b,|f⁻¹({b})|), including zero for targets not attained.
View answer and explanation
{(-6, 0), (-5, 0), (-4, 0), (-3, 1), (-2, 2), (-1, 2)}
Count the permitted preimages of every element of B to obtain {(-6, 0), (-5, 0), (-4, 0), (-3, 1), (-2, 2), (-1, 2)}. The profile includes zero-size fibers, not only attained targets.
Reviewed step-by-step derivation
Bijectivity and unique preimages · qf-bijective-fiber-profile · candidate
68. Complete Codomain Fiber Profile
The declared total function f:A→B has A={-5, -4, -3, -2, -1, 0}, B={0, 1, 2, 3, 4, 5}, and table {(-5, 1), (-4, 2), (-3, 3), (-2, 4), (-1, 5), (0, 0)}. In increasing target order, list every (b,|f⁻¹({b})|), including zero for targets not attained.
View answer and explanation
{(0, 1), (1, 1), (2, 1), (3, 1), (4, 1), (5, 1)}
Count the permitted preimages of every element of B to obtain {(0, 1), (1, 1), (2, 1), (3, 1), (4, 1), (5, 1)}. The profile includes zero-size fibers, not only attained targets.
Reviewed step-by-step derivation
Bijectivity and unique preimages · qf-bijective-fiber-profile · candidate
69. Complete Codomain Fiber Profile
The declared total function f:A→B has A={-5, -4, -3, -2, -1, 0}, B={1, 3, 5, 7, 9, 11, 13}, and table {(-5, 1), (-4, 3), (-3, 5), (-2, 7), (-1, 9), (0, 11)}. In increasing target order, list every (b,|f⁻¹({b})|), including zero for targets not attained.
View answer and explanation
{(1, 1), (3, 1), (5, 1), (7, 1), (9, 1), (11, 1), (13, 0)}
Count the permitted preimages of every element of B to obtain {(1, 1), (3, 1), (5, 1), (7, 1), (9, 1), (11, 1), (13, 0)}. The profile includes zero-size fibers, not only attained targets.
Reviewed step-by-step derivation
Bijectivity and unique preimages · qf-bijective-fiber-profile · candidate
70. Complete Codomain Fiber Profile
The declared total function f:A→B has A={-5, -4, -3, -2, -1, 0}, B={0, 1, 2, 3, 4}, and table {(-5, 0), (-4, 1), (-3, 2), (-2, 3), (-1, 4), (0, 0)}. In increasing target order, list every (b,|f⁻¹({b})|), including zero for targets not attained.
View answer and explanation
{(0, 2), (1, 1), (2, 1), (3, 1), (4, 1)}
Count the permitted preimages of every element of B to obtain {(0, 2), (1, 1), (2, 1), (3, 1), (4, 1)}. The profile includes zero-size fibers, not only attained targets.
Reviewed step-by-step derivation
Bijectivity and unique preimages · qf-bijective-fiber-profile · candidate
71. Complete Codomain Fiber Profile
The declared total function f:A→B has A={-5, -4, -3, -2, -1, 0}, B={-7, -6, -5, -4, -3, -2, -1}, and table {(-5, -1), (-4, -1), (-3, -2), (-2, -2), (-1, -3), (0, -3)}. In increasing target order, list every (b,|f⁻¹({b})|), including zero for targets not attained.
View answer and explanation
{(-7, 0), (-6, 0), (-5, 0), (-4, 0), (-3, 2), (-2, 2), (-1, 2)}
Count the permitted preimages of every element of B to obtain {(-7, 0), (-6, 0), (-5, 0), (-4, 0), (-3, 2), (-2, 2), (-1, 2)}. The profile includes zero-size fibers, not only attained targets.
Reviewed step-by-step derivation
Bijectivity and unique preimages · qf-bijective-fiber-profile · candidate
72. Complete Codomain Fiber Profile
The declared total function f:A→B has A={-6, -5, -4, -3, -2, -1, 0}, B={0, 1, 2, 3, 4, 5, 6}, and table {(-6, 1), (-5, 2), (-4, 3), (-3, 4), (-2, 5), (-1, 6), (0, 0)}. In increasing target order, list every (b,|f⁻¹({b})|), including zero for targets not attained.
View answer and explanation
{(0, 1), (1, 1), (2, 1), (3, 1), (4, 1), (5, 1), (6, 1)}
Count the permitted preimages of every element of B to obtain {(0, 1), (1, 1), (2, 1), (3, 1), (4, 1), (5, 1), (6, 1)}. The profile includes zero-size fibers, not only attained targets.
Reviewed step-by-step derivation
Bijectivity and unique preimages · qf-bijective-fiber-profile · candidate
73. Complete Codomain Fiber Profile
The declared total function f:A→B has A={-6, -5, -4, -3, -2, -1, 0}, B={1, 3, 5, 7, 9, 11, 13, 15}, and table {(-6, 1), (-5, 3), (-4, 5), (-3, 7), (-2, 9), (-1, 11), (0, 13)}. In increasing target order, list every (b,|f⁻¹({b})|), including zero for targets not attained.
View answer and explanation
{(1, 1), (3, 1), (5, 1), (7, 1), (9, 1), (11, 1), (13, 1), (15, 0)}
Count the permitted preimages of every element of B to obtain {(1, 1), (3, 1), (5, 1), (7, 1), (9, 1), (11, 1), (13, 1), (15, 0)}. The profile includes zero-size fibers, not only attained targets.
Reviewed step-by-step derivation
Bijectivity and unique preimages · qf-bijective-fiber-profile · candidate
74. Complete Codomain Fiber Profile
The declared total function f:A→B has A={-6, -5, -4, -3, -2, -1, 0}, B={0, 1, 2, 3, 4, 5}, and table {(-6, 0), (-5, 1), (-4, 2), (-3, 3), (-2, 4), (-1, 5), (0, 0)}. In increasing target order, list every (b,|f⁻¹({b})|), including zero for targets not attained.
View answer and explanation
{(0, 2), (1, 1), (2, 1), (3, 1), (4, 1), (5, 1)}
Count the permitted preimages of every element of B to obtain {(0, 2), (1, 1), (2, 1), (3, 1), (4, 1), (5, 1)}. The profile includes zero-size fibers, not only attained targets.
Reviewed step-by-step derivation
Bijectivity and unique preimages · qf-bijective-fiber-profile · candidate
75. Complete Codomain Fiber Profile
The declared total function f:A→B has A={-6, -5, -4, -3, -2, -1, 0}, B={-8, -7, -6, -5, -4, -3, -2, -1}, and table {(-6, -1), (-5, -1), (-4, -2), (-3, -2), (-2, -3), (-1, -3), (0, -4)}. In increasing target order, list every (b,|f⁻¹({b})|), including zero for targets not attained.
View answer and explanation
{(-8, 0), (-7, 0), (-6, 0), (-5, 0), (-4, 1), (-3, 2), (-2, 2), (-1, 2)}
Count the permitted preimages of every element of B to obtain {(-8, 0), (-7, 0), (-6, 0), (-5, 0), (-4, 1), (-3, 2), (-2, 2), (-1, 2)}. The profile includes zero-size fibers, not only attained targets.
Reviewed step-by-step derivation
Bijectivity and unique preimages · qf-bijective-fiber-profile · candidate
76. Existence of a Bijection from Finite Cardinalities
The explicit finite cardinalities are |A|=0 and |B|=0, with no particular function table supplied. Does some bijection A→B exist?
View answer and explanation
Yes
Finite sets admit a bijection exactly when their cardinalities agree, giving Yes. This asks about existence rather than testing a given function; two empty sets admit the unique empty bijection.
Reviewed step-by-step derivation
Bijectivity and unique preimages · qf-bijective-cardinality-possibility · candidate
77. Existence of a Bijection from Finite Cardinalities
The explicit finite cardinalities are |A|=0 and |B|=1, with no particular function table supplied. Does some bijection A→B exist?
View answer and explanation
No
Finite sets admit a bijection exactly when their cardinalities agree, giving No. This asks about existence rather than testing a given function; two empty sets admit the unique empty bijection.
Reviewed step-by-step derivation
Bijectivity and unique preimages · qf-bijective-cardinality-possibility · candidate
78. Existence of a Bijection from Finite Cardinalities
The explicit finite cardinalities are |A|=1 and |B|=1, with no particular function table supplied. Does some bijection A→B exist?
View answer and explanation
Yes
Finite sets admit a bijection exactly when their cardinalities agree, giving Yes. This asks about existence rather than testing a given function; two empty sets admit the unique empty bijection.
Reviewed step-by-step derivation
Bijectivity and unique preimages · qf-bijective-cardinality-possibility · candidate
79. Existence of a Bijection from Finite Cardinalities
The explicit finite cardinalities are |A|=1 and |B|=2, with no particular function table supplied. Does some bijection A→B exist?
View answer and explanation
No
Finite sets admit a bijection exactly when their cardinalities agree, giving No. This asks about existence rather than testing a given function; two empty sets admit the unique empty bijection.
Reviewed step-by-step derivation
Bijectivity and unique preimages · qf-bijective-cardinality-possibility · candidate
80. Existence of a Bijection from Finite Cardinalities
The explicit finite cardinalities are |A|=2 and |B|=2, with no particular function table supplied. Does some bijection A→B exist?
View answer and explanation
Yes
Finite sets admit a bijection exactly when their cardinalities agree, giving Yes. This asks about existence rather than testing a given function; two empty sets admit the unique empty bijection.
Reviewed step-by-step derivation
Bijectivity and unique preimages · qf-bijective-cardinality-possibility · candidate
81. Existence of a Bijection from Finite Cardinalities
The explicit finite cardinalities are |A|=2 and |B|=3, with no particular function table supplied. Does some bijection A→B exist?
View answer and explanation
No
Finite sets admit a bijection exactly when their cardinalities agree, giving No. This asks about existence rather than testing a given function; two empty sets admit the unique empty bijection.
Reviewed step-by-step derivation
Bijectivity and unique preimages · qf-bijective-cardinality-possibility · candidate
82. Existence of a Bijection from Finite Cardinalities
The explicit finite cardinalities are |A|=3 and |B|=3, with no particular function table supplied. Does some bijection A→B exist?
View answer and explanation
Yes
Finite sets admit a bijection exactly when their cardinalities agree, giving Yes. This asks about existence rather than testing a given function; two empty sets admit the unique empty bijection.
Reviewed step-by-step derivation
Bijectivity and unique preimages · qf-bijective-cardinality-possibility · candidate
83. Existence of a Bijection from Finite Cardinalities
The explicit finite cardinalities are |A|=3 and |B|=4, with no particular function table supplied. Does some bijection A→B exist?
View answer and explanation
No
Finite sets admit a bijection exactly when their cardinalities agree, giving No. This asks about existence rather than testing a given function; two empty sets admit the unique empty bijection.
Reviewed step-by-step derivation
Bijectivity and unique preimages · qf-bijective-cardinality-possibility · candidate
84. Existence of a Bijection from Finite Cardinalities
The explicit finite cardinalities are |A|=4 and |B|=4, with no particular function table supplied. Does some bijection A→B exist?
View answer and explanation
Yes
Finite sets admit a bijection exactly when their cardinalities agree, giving Yes. This asks about existence rather than testing a given function; two empty sets admit the unique empty bijection.
Reviewed step-by-step derivation
Bijectivity and unique preimages · qf-bijective-cardinality-possibility · candidate
85. Existence of a Bijection from Finite Cardinalities
The explicit finite cardinalities are |A|=4 and |B|=5, with no particular function table supplied. Does some bijection A→B exist?
View answer and explanation
No
Finite sets admit a bijection exactly when their cardinalities agree, giving No. This asks about existence rather than testing a given function; two empty sets admit the unique empty bijection.
Reviewed step-by-step derivation
Bijectivity and unique preimages · qf-bijective-cardinality-possibility · candidate
86. Existence of a Bijection from Finite Cardinalities
The explicit finite cardinalities are |A|=5 and |B|=5, with no particular function table supplied. Does some bijection A→B exist?
View answer and explanation
Yes
Finite sets admit a bijection exactly when their cardinalities agree, giving Yes. This asks about existence rather than testing a given function; two empty sets admit the unique empty bijection.
Reviewed step-by-step derivation
Bijectivity and unique preimages · qf-bijective-cardinality-possibility · candidate
87. Existence of a Bijection from Finite Cardinalities
The explicit finite cardinalities are |A|=5 and |B|=6, with no particular function table supplied. Does some bijection A→B exist?
View answer and explanation
No
Finite sets admit a bijection exactly when their cardinalities agree, giving No. This asks about existence rather than testing a given function; two empty sets admit the unique empty bijection.
Reviewed step-by-step derivation
Bijectivity and unique preimages · qf-bijective-cardinality-possibility · candidate
88. Existence of a Bijection from Finite Cardinalities
The explicit finite cardinalities are |A|=6 and |B|=6, with no particular function table supplied. Does some bijection A→B exist?
View answer and explanation
Yes
Finite sets admit a bijection exactly when their cardinalities agree, giving Yes. This asks about existence rather than testing a given function; two empty sets admit the unique empty bijection.
Reviewed step-by-step derivation
Bijectivity and unique preimages · qf-bijective-cardinality-possibility · candidate
89. Existence of a Bijection from Finite Cardinalities
The explicit finite cardinalities are |A|=6 and |B|=7, with no particular function table supplied. Does some bijection A→B exist?
View answer and explanation
No
Finite sets admit a bijection exactly when their cardinalities agree, giving No. This asks about existence rather than testing a given function; two empty sets admit the unique empty bijection.
Reviewed step-by-step derivation
Bijectivity and unique preimages · qf-bijective-cardinality-possibility · candidate
90. Existence of a Bijection from Finite Cardinalities
The explicit finite cardinalities are |A|=7 and |B|=7, with no particular function table supplied. Does some bijection A→B exist?
View answer and explanation
Yes
Finite sets admit a bijection exactly when their cardinalities agree, giving Yes. This asks about existence rather than testing a given function; two empty sets admit the unique empty bijection.
Reviewed step-by-step derivation
Bijectivity and unique preimages · qf-bijective-cardinality-possibility · candidate
91. Existence of a Bijection from Finite Cardinalities
The explicit finite cardinalities are |A|=7 and |B|=8, with no particular function table supplied. Does some bijection A→B exist?
View answer and explanation
No
Finite sets admit a bijection exactly when their cardinalities agree, giving No. This asks about existence rather than testing a given function; two empty sets admit the unique empty bijection.
Reviewed step-by-step derivation
Bijectivity and unique preimages · qf-bijective-cardinality-possibility · candidate
92. Existence of a Bijection from Finite Cardinalities
The explicit finite cardinalities are |A|=8 and |B|=8, with no particular function table supplied. Does some bijection A→B exist?
View answer and explanation
Yes
Finite sets admit a bijection exactly when their cardinalities agree, giving Yes. This asks about existence rather than testing a given function; two empty sets admit the unique empty bijection.
Reviewed step-by-step derivation
Bijectivity and unique preimages · qf-bijective-cardinality-possibility · candidate
93. Existence of a Bijection from Finite Cardinalities
The explicit finite cardinalities are |A|=8 and |B|=9, with no particular function table supplied. Does some bijection A→B exist?
View answer and explanation
No
Finite sets admit a bijection exactly when their cardinalities agree, giving No. This asks about existence rather than testing a given function; two empty sets admit the unique empty bijection.
Reviewed step-by-step derivation
Bijectivity and unique preimages · qf-bijective-cardinality-possibility · candidate
94. Existence of a Bijection from Finite Cardinalities
The explicit finite cardinalities are |A|=9 and |B|=9, with no particular function table supplied. Does some bijection A→B exist?
View answer and explanation
Yes
Finite sets admit a bijection exactly when their cardinalities agree, giving Yes. This asks about existence rather than testing a given function; two empty sets admit the unique empty bijection.
Reviewed step-by-step derivation
Bijectivity and unique preimages · qf-bijective-cardinality-possibility · candidate
95. Existence of a Bijection from Finite Cardinalities
The explicit finite cardinalities are |A|=9 and |B|=10, with no particular function table supplied. Does some bijection A→B exist?
View answer and explanation
No
Finite sets admit a bijection exactly when their cardinalities agree, giving No. This asks about existence rather than testing a given function; two empty sets admit the unique empty bijection.
Reviewed step-by-step derivation
Bijectivity and unique preimages · qf-bijective-cardinality-possibility · candidate
96. Existence of a Bijection from Finite Cardinalities
The explicit finite cardinalities are |A|=10 and |B|=10, with no particular function table supplied. Does some bijection A→B exist?
View answer and explanation
Yes
Finite sets admit a bijection exactly when their cardinalities agree, giving Yes. This asks about existence rather than testing a given function; two empty sets admit the unique empty bijection.
Reviewed step-by-step derivation
Bijectivity and unique preimages · qf-bijective-cardinality-possibility · candidate
97. Existence of a Bijection from Finite Cardinalities
The explicit finite cardinalities are |A|=10 and |B|=11, with no particular function table supplied. Does some bijection A→B exist?
View answer and explanation
No
Finite sets admit a bijection exactly when their cardinalities agree, giving No. This asks about existence rather than testing a given function; two empty sets admit the unique empty bijection.
Reviewed step-by-step derivation
Bijectivity and unique preimages · qf-bijective-cardinality-possibility · candidate
98. Existence of a Bijection from Finite Cardinalities
The explicit finite cardinalities are |A|=11 and |B|=11, with no particular function table supplied. Does some bijection A→B exist?
View answer and explanation
Yes
Finite sets admit a bijection exactly when their cardinalities agree, giving Yes. This asks about existence rather than testing a given function; two empty sets admit the unique empty bijection.
Reviewed step-by-step derivation
Bijectivity and unique preimages · qf-bijective-cardinality-possibility · candidate
99. Existence of a Bijection from Finite Cardinalities
The explicit finite cardinalities are |A|=11 and |B|=12, with no particular function table supplied. Does some bijection A→B exist?
View answer and explanation
No
Finite sets admit a bijection exactly when their cardinalities agree, giving No. This asks about existence rather than testing a given function; two empty sets admit the unique empty bijection.
Reviewed step-by-step derivation
Bijectivity and unique preimages · qf-bijective-cardinality-possibility · candidate
100. Existence of a Bijection from Finite Cardinalities
The explicit finite cardinalities are |A|=12 and |B|=12, with no particular function table supplied. Does some bijection A→B exist?
View answer and explanation
Yes
Finite sets admit a bijection exactly when their cardinalities agree, giving Yes. This asks about existence rather than testing a given function; two empty sets admit the unique empty bijection.
Reviewed step-by-step derivation
Bijectivity and unique preimages · qf-bijective-cardinality-possibility · candidate
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