ISRO2020-50

https://gateoverflow.in/15778/every-attribute-functionally-dependent-primary-relation

Check this.

C depends in b as well right in your last example?

If every non-key attribute functionally dependent on the primary key

It doesnt mention that it can depend on another non key right?

@shashin

what is your view in this question?

@Debapaul

From a->b and b->c we get a->c using transitive property of fd.

Hence a->c must be satisfied in R. 

Hence c is functionally dependent on primary key a.

In 2NF , it says that every non prime attribute must depend on primary key either directly or transitively.

This differentiates 3NF and 2NF.

@SuvasishDutta

Consider this case, say we have a relation R={a,b,c,d,e} where $ab$ forms the primary key

So $ab->a,b,c,d,e$

Now say, $b->c$

here also $c$ is functionally dependent on $ab$ as $ab$ derives $b$

Here it is not even $2nf$ right?

Definition of 3NF:

1. The entity is in second normal form.
2. No non-prime (non-key) attribute is transitively dependent on any key i.e. no non-prime attribute depends on other non-prime attributes. All the non-prime attributes must depend only on the candidate keys.

I don't see anything in the question specifying the underlined bits above. It just says that every non-key is functionally dependent on a key - there could be transitive dependencies along the way. So 2NF is my go-to as well.

I can't think of a way to refute Suvashish's example above either. Another example if you're not convinced:

$R(X, Y, Z),  FD = \{ X \rightarrow Y, Y \rightarrow Z, X \rightarrow Z\}$

Both non-primes Y and Z are functionally dependent on the key X (as given in the question). But Z is also transitively dependent on X through Y (and the question says nothing prohibiting this).

@shashin

what about my comment just above yours?

@shashin

Now check @SuvasishDutta's  comment just above ours.

There also 

c is not functionally dependent on the key now is it ? It is functionally dependent on b which is in turn functionally dependent on the key.

@Debapaul

in your example , R is in 1nf because non prime attribute c is functionally dependent on b which is part of key.

moreover consider only non trivial fd for normal forms.

and in transitive fd, both sides of an fd must be non prime attributes for the relation to be in 2NF.

if lhs is part of key i.e. prime attribute and rhs is no prime attribute , R is in 1NF not in 2NF.

 there could be transitive dependencies along the way

If there is a transitive dependency along the way then again same statement (that i commented for @SuvasishDutta's post) applies here also right?

@SuvasishDutta also tried to prove the same thing that transistive dependency does not hold

@Debapaul 

Can you please explain your question once?

https://gateoverflow.in/15778/every-attribute-functionally-dependent-primary-relation

see this, and see the ans other than the selected one..

I think i have actually said the same in the above comments.

Isro framed the question wrong, best ans should be $1NF$

@Debapaul 

in that answer the relation is in 3 NF not in 1NF. 

In that relation keys are AC and AB.

Since AC->B and C->B.

R: {X→Y,  Y→Z,  X→Z}

In above relation,  in FD Y→Z, Z is clearly Non key attribute and It is NOT dependent on Primary key Because Primary key is X.  So, Y→Z is not correct FD for given statement. Answer has to be 3NF.

Why it is not in 2NF:

Let's take simple example:

R: {X→Y,  Y→Z,  X→Z}

In above relation,  in FD Y→Z, Z is clearly Non key attribute and It is NOT dependent on Primary key Because Primary key is X.

Y→Z is not valid functional dependency for the given statement "If EVERY non-key attribute (like Z) functionally dependent on the primary key (Which is X)". 

Answer has to be 3NF