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?

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.
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venereology answered 4 months ago

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