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:
- A.
1NF
- B.
2NF
- C.
3NF
- 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:
Start from {F1, F2}.
F1 → F3 adds F3, giving {F1, F2, F3}.
F2 → F4 adds F4, giving {F1, F2, F3, F4}.
(F1, F2) → F5 adds F5, giving {F1, F2, F3, F4, F5} — every field in the table.
Since the closure of (F1, F2) covers all fields (and neither F1 nor F2 alone does), (F1, F2) is the candidate key.
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.
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.