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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