Let the schema R = (A,B,C) and the set F = {A → B,C → B}...

Let the schema R = (A,B,C) and the set F = {A → B,C → B} of FDs be given. Is R in 3NF? Why or why not?

Solutions

Expert Solution

Solution :

From the given functional dependencies, we can say that candidate key of R is AC. As the closure of AC is {A,B,C}, i.e all attributes of R.
Given relation R is not in 3NF as it is not in 2NF because of two partial dependencies A->B and C->B.

Checking for 2NF :

A->B and C->B both are partial dependencies. Hence the relation R is not in 2NF.
Converting to 2NF : To convert R in 2NF decompose relation R into two relations as given below :
R1(A,B) with functional dependency A->B and candidate key A.
R2(C,B) with functional dependency C->B and candidate key C.
Both relations R1 and R2 are in 2NF.

Checking for 3NF :
Decomposed relations dont have transitive dependency. Hence both are in 3NF.

Hence R is not in 3NF but by decomposing it as given above R can be converted to 3NF.
if you have any doubts then you can ask in comment section if you find the solution helpful then upvote the answer. Thank you.

venereology answered 1 day ago

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