A table has fields F1, F2, F3, F4, and F5, with the following functional…

2024

A table has fields F1, F2, F3, F4, and F5, with the following functional dependencies:

  • F1 → F3

  • F2 → F4

  • (F1, F2) → F5

In terms of normalization, this table is in:

  1. A.

    1NF

  2. B.

    2NF

  3. C.

    3NF

  4. D.

    None of the mentioned

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Correct answer: A

Concept: Normalization levels are cumulative. First Normal Form (1NF) only requires atomic, single-valued fields with no repeating groups. Second Normal Form (2NF) additionally requires that every non-prime attribute be fully dependent on the WHOLE of every candidate key, with no partial dependency on just part of a composite key. Third Normal Form (3NF) and BCNF add further conditions on top of 2NF. A table's normalization level is the highest one whose requirement is not violated.

Application: Finding the candidate key first, using the closure of attributes:

  1. Start from {F1, F2}.

  2. F1 → F3 adds F3, giving {F1, F2, F3}.

  3. F2 → F4 adds F4, giving {F1, F2, F3, F4}.

  4. (F1, F2) → F5 adds F5, giving {F1, F2, F3, F4, F5} — every field in the table.

  5. Since the closure of (F1, F2) covers all fields (and neither F1 nor F2 alone does), (F1, F2) is the candidate key.

  6. The non-prime attributes are F3, F4, F5. F1 → F3 depends only on F1 — part of the key, not all of it — a partial dependency. F2 → F4 is the same: a partial dependency on just F2.

  7. A single partial dependency on any candidate key is enough to violate the no-partial-dependency requirement, so the table cannot go past the base atomic-field requirement.

Cross-check: (F1, F2) → F5 alone IS a full dependency on the whole key, which confirms (F1, F2) really is the candidate key — but that one clean dependency doesn't offset the other two partial ones; any partial dependency is disqualifying, so the classification still stops at the base level.

Result: The table satisfies only the atomic-values / no-repeating-groups requirement — it is in 1NF.

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