Nielit Scientist-C 2016 march
August 29, 2024Database-Management-System
August 29, 2024Database-Management-System
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Question 802
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Which one of the following statements about normal forms is FALSE?
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BCNF is stricter than 3NF
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Lossless,dependency preserving decomposition into BCNF is always possible
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Lossless,dependency preserving decomposition into 3NF is always possible
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Any relation with two attributes is BCNF
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Question 802 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: B
Question 802 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.