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.

Maybe the 3NF schema is:

E1(ABC), E2(BD) 

Is it correct?

Thanks a lot

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.

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) 

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

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?

I do apologize but I'm unable to work out how to normalize a schema into 3NF

Thanks a lot

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.

Maybe the 3NF schema is:

E1(ABC), E2(BD) 

Is it correct?

Thanks a lot