The key is actually AE. The proof is simple, a key for definition is a minimal set of attributes whose closure contains all the attributes of the table. If you calculate the closure of A with respect to the given functional dependencies you will find:
A+ = {ABCD}that does not contain the attribute E. So A is not a key, and E must be present in any key of R. And since:
AE+ = {ABCDE}then AE is a key, and it is minimal (you cannot remove any attribute from it without losing the property of determining all the other attributes of the relation).
manpreet![Tuteehub forum best answer]()
Best Answer
3 years ago
I am not sure whether this is offtopic. However, please read on.
A person whom I know has encountered a problem in the answer key published by CBSE, India. As their key challenging form contains a column for citation/proof, we have to find some authentic book which can be shown as the base of our claim.
I attach the question here, and I request you to verify the answer and suggest us some book/other ways to prove our challenge (She has checked the regular syllabus books, but didn't find something precise related to this problem).
Question:
Identify the minimal key for relational scheme R(A, B, C, D, E) with functional dependencies F = {A->B, B->C, AC->D}
The options are A, AE, BE, and CE
The key shows the answer is A and she says it is AE actually.
Please help us find some proof/way to challenge the key.