Database-Management-System
October 3, 2023Functional-Dependency
October 3, 2023Database-Management-System
| Question 91 |
Which one of the following statements about normal forms is FALSE?
| BCNF is stricter than 3NF | |
| Lossless, dependency-preserving decomposition into 3NF is always possible
| |
| Lossless, dependency-preserving decomposition into BCNF is always possible | |
| Any relation with two attributes is in BCNF |
Question 91 Explanation:
Option A: BCNF is stricter than 3NF. In this all redundancy based on functional dependency has been removed.
Option B: Lossless, dependency preserving decomposition into 3NF is always possible.
Option C: It is false.
It is not possible to have dependency preserving in BCNF decomposition.
→ Let take an example, 3NF can’t be decomposed into BCNF.
Option D: It is true.
Let consider two attributes (X, Y).
If (X→Y), X is a candidate key. It is in BCNF and vice-versa.
Option B: Lossless, dependency preserving decomposition into 3NF is always possible.
Option C: It is false.
It is not possible to have dependency preserving in BCNF decomposition.
→ Let take an example, 3NF can’t be decomposed into BCNF.
Option D: It is true.
Let consider two attributes (X, Y).
If (X→Y), X is a candidate key. It is in BCNF and vice-versa.
Correct Answer: C
Question 91 Explanation:
Option A: BCNF is stricter than 3NF. In this all redundancy based on functional dependency has been removed.
Option B: Lossless, dependency preserving decomposition into 3NF is always possible.
Option C: It is false.
It is not possible to have dependency preserving in BCNF decomposition.
→ Let take an example, 3NF can’t be decomposed into BCNF.
Option D: It is true.
Let consider two attributes (X, Y).
If (X→Y), X is a candidate key. It is in BCNF and vice-versa.
Option B: Lossless, dependency preserving decomposition into 3NF is always possible.
Option C: It is false.
It is not possible to have dependency preserving in BCNF decomposition.
→ Let take an example, 3NF can’t be decomposed into BCNF.
Option D: It is true.
Let consider two attributes (X, Y).
If (X→Y), X is a candidate key. It is in BCNF and vice-versa.
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