Functional Dependency

Question

Lets say their is R(A,B,C)   and FD :   A⟶B, AB⟶C and A is key. Is it in 3NF? Explain

Answer 1

For a relation to be in 3NF ,  we must have X-->Y , so either X should be a key or Y should be a prime attribute , Now with the given FD , u can re-write them as A-->B and A-->C as (A)⁺=(ABC) , so you can remove B from the functional dependency AB-->C , so that it becomes A-->C , therefore now you are left with A-->BC only and since A is a key so clearly the FD is in 3NF .

Point : whenever You have PQ-->R and (P)⁺=(PQR) , then remove Q from the FD ,since P alone can determine R so Q is redundant here , so your FD reduces to P-->R 

 

Comments

Without changing FD AB->C to A->C then the FD's A-> B and AB->C are not in 2NF  ??

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A⟶B, AB⟶C and A is key : yes it is in 3NF (Actually it is in BCNF) but, AB⟶C, AB⟶DEF, AD ⟶E is not in 3NF. Here AB is key and A,B are prime. AD ⟶ E in this dependency Left hand side is not key & right hand side is not Prime.

Answer 2

3NF defination->

for every non -trivial FD x->y either->1.  X is a key  

                                                        2. Y is a prime attribute

so here R(A,B,C) :A->B and AB->c.. A is a key (given) ..for AB->finding closure ->AB⁺=ABC ..it contains all the attributes so it is also a key. both FD satify 1st condition so this is in 3NF