Given E(ABCDE) ABC are candidate keys

Normalize into 2NF and 3NF

As far as 2NF is concerned the solution is quite easy:



Maybe the 3NF schema is:

E1(ABC), E2(BD) 

Is it correct?

Thanks a lot

So according to @OliverAsmus the 3NF result should be:

E1(ABC), E2(BD) 


Is that correct? Thanks






Do you mean that there are three candidate keys: A, B and C? If so, your FDs are A-> BCDE, B-> ACDE and C-> ABDE. If there are no other FDs connecting these attributes, you can stop there: you are already in BCNF (and also 3NF).

Your Answer

