Given E(ABCDE) ABC are candidate keys
Normalize into 2NF and 3NF
As far as 2NF is concerned the solution is quite easy:
E1(BD), E2(DE), E3(ABC)
But with regard to 3NF I think I'm wrong if I say that nothing should be done.
3NF schema is:
Is it correct?
Thanks a lot
So according to @OliverAsmus the 3NF result should be:
But if what I wrote is correct, am I right in thinking that the 3NF (in this particular case) doesn't preserve all the attributes? E doesn't depend upon any key so I got rid of it...
Is that correct? Thanks