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A relation R has no composite candidate keys. Which of the following is always true for relation R?
- a)R is in 2NF
- b)R is in 3NF
- c)R is in BCNF
- d)Can’t be predicted
Correct answer is option 'A'. Can you explain this answer?
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A relation R has no composite candidate keys. Which of the following i...
A relation without composite keys will always satisfy the conditions for a relation to be in 2NF.
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A relation R has no composite candidate keys. Which of the following i...
Understanding Composite Candidate Keys
A relation R having no composite candidate keys implies that all candidate keys are simple, meaning each key consists of a single attribute.
Implications for Normal Forms
- First Normal Form (1NF): R is already in 1NF since it contains atomic values, which is a prerequisite for any normal form.
- Second Normal Form (2NF):
- A relation is in 2NF if it is in 1NF and all non-key attributes are fully functionally dependent on the primary key.
- Since R has no composite keys, any non-key attribute must depend on a single attribute (the primary key), fulfilling the full functional dependency requirement.
- Therefore, R is guaranteed to be in 2NF.
Limitations for Higher Normal Forms
- Third Normal Form (3NF):
- While R being in 2NF does not automatically imply it is in 3NF, it may still have transitive dependencies affecting non-key attributes.
- Thus, we cannot conclude that R is in 3NF just based on the absence of composite candidate keys.
- Boyce-Codd Normal Form (BCNF):
- BCNF is stricter than 3NF and requires that every determinant is a candidate key.
- Without additional information on functional dependencies, we cannot determine if R is in BCNF.
Conclusion
Given that R has no composite candidate keys, we can confidently state that:
- R is always in 2NF.
- Other normal forms (3NF and BCNF) cannot be guaranteed without further analysis of functional dependencies.
Thus, the correct answer is option 'A'.
A relation R having no composite candidate keys implies that all candidate keys are simple, meaning each key consists of a single attribute.
Implications for Normal Forms
- First Normal Form (1NF): R is already in 1NF since it contains atomic values, which is a prerequisite for any normal form.
- Second Normal Form (2NF):
- A relation is in 2NF if it is in 1NF and all non-key attributes are fully functionally dependent on the primary key.
- Since R has no composite keys, any non-key attribute must depend on a single attribute (the primary key), fulfilling the full functional dependency requirement.
- Therefore, R is guaranteed to be in 2NF.
Limitations for Higher Normal Forms
- Third Normal Form (3NF):
- While R being in 2NF does not automatically imply it is in 3NF, it may still have transitive dependencies affecting non-key attributes.
- Thus, we cannot conclude that R is in 3NF just based on the absence of composite candidate keys.
- Boyce-Codd Normal Form (BCNF):
- BCNF is stricter than 3NF and requires that every determinant is a candidate key.
- Without additional information on functional dependencies, we cannot determine if R is in BCNF.
Conclusion
Given that R has no composite candidate keys, we can confidently state that:
- R is always in 2NF.
- Other normal forms (3NF and BCNF) cannot be guaranteed without further analysis of functional dependencies.
Thus, the correct answer is option 'A'.
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A relation R has no composite candidate keys. Which of the following is always true for relation R?a)R is in 2NFb)R is in 3NFc)R is in BCNFd)Can’t be predictedCorrect answer is option 'A'. Can you explain this answer?
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