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+1 vote

Is the following FD in 2NF?

AB -> C                            

Where BD is candidate key.

please explain why this FD is in 2NF.

non-prime,prime ----> non-prime is in 2NF. But why?

Bcz.  as per my knowledge

Every non-prime attribute should be fully functionally dependent on the primary key of R

asked in Databases by Boss (7.8k points) | 113 views

what if AB is also candidate key ?
Then it satisfies 2NF
Yes, you are correct, it won't violate 2 NF unless either a or b or d derives any other attribute
Man, I'm still in confusion. What should I conclude then ?
You should check that

1)AB is candidate key or not

if yes then check that from all the possible FD, only single A or B or C is able to derive any attribute

                   if yes then not in 2 NF

                   Else no then relation is in 2NF

Check out this question asked in gate-2008

Here, {Catalog_no, ( Title , Author )} are candidate keys.

1.Catalog_no --> {Title , Author , Publisher , Year } 

2.{Publisher , Title , Year} --> Price

Which implies, Catalog_no --> Price.

Is it the reason for considering the relation to be in 2NF??


yes you are correct it is because of transitive dependency

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