Database Normal forms

Question

Given a relation R(ABCD) with functional dependencies:

$AB \to CD\\ BC \to D$

Normal form is?

1) 1 NF

2) 2 NF

3) 3 NF

4) BCNF

Answer 1

Answer : 2 (2NF)

Given Relation schema $R(ABCD)$ and $FD$ set $\left \{ AB \rightarrow CD, BC \rightarrow D \right \}$

We can find the Candidate key(s) of the Relation $R$ using the FD Set and the Candidate key of the relation $R$ will be $AB$ .

Hint : If some Attribute(s) are Not on the Right Side of any Non-trivial FD, then those attributes must be the Part of Every Candidate key.

Checking for $2NF$ :

Part of the Candidate key $AB$ i.e. $A$ or $B$ are Not deriving any Non-prime attribute, Hence, No Partial Dependency. Hence, the Relation $R$  is in $2NF$.

Checking for $3NF$ :

Look at the Non-trivial FD $BC \rightarrow D$.. It is a Transitive dependency as the combination of part of the candidate key($B$) along with a non-prime attribute($C$) is deriving a Non-key attribute $D$. Hence, We can say that Non-prime attribute $D$ is Transitively dependent on candidate key $AB$. So, there is Transitive Dependency. Hence, the Relation $R$ is Not in $3NF$.

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Refer here for $2NF$ and $3NF$ : https://gateoverflow.in/219972/normalization?show=220006#a220006