hello,
let us solve this problem.
R(a, b, c, d, e)
key (a,b)
F = { {a, b} → {c, d, e}, b → {d, e} }
The above set of functional dependencies are in 1NF because each attribute takes a single value.( If any one of the attribute would take more than one value then it should not be in 1NF.) i.,e only one value can be there in a column of the table , for example if a column exists of name phone_number then it should have only one phone number, not more than one phone number. (it is inherited that a column will have only one value unless and until it is not mentioned to have many values).
So it is in 1NF.
Since the key is (a,b) and we have a partial dependency (i.,e we have dependency {b}->{d,e} in which the left hand side attribute is part of the key but not complete key) so the dependencies are not in 2NF.
Since the dependencies are not in 2NF , so it can not be in 3NF . (so no need to check).
i hope i was able to solve your problem to a greater extent, please feel free to comment your queries, Please consider my efforts and upvote my solution.
Thanku:)
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