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+36 votes

Which one of the following statements is $\text{FALSE}$?

  1.  Any relation with two attributes is in $\text{BCNF}$
  2. A relation in which every key has only one attribute is in $2NF$

  3. A prime attribute can be transitively dependent on a key in a $3NF$ relation

  4. A prime attribute can be transitively dependent on a key in a $\text{BCNF}$ relation

asked in Databases by Veteran (59.9k points)
edited by | 4.3k views

4 Answers

+47 votes
Best answer
Any relation with two attributes is in $\text{BCNF}$ => This is true. It is trivial

A relation in which every key has only one attribute is in $2NF$ => This is true. As it is not possible to have Partial Functional Dependency !

A prime attribute can be transitively dependent on a key in a $3NF$ relation => This is true. As For $3NF$ to be voilated we have something like Key => Non Key, Non Key => Non key $3NF$ definition say x->y, either x should be key or y should be prime atrribute. Then we can have something like Key=> Non Key, Non key => Prime Attribute, resulting in Transitive $\text{FD}$ on Prime Attribute, still in $3NF$.

$\text{LHS}$ must be always key, so No Transitive dependency is allowed. So answer => $D$
answered by Boss (43.5k points)
edited by

I have a case for where "A prime attribute can be transitively dependent on a key in a BCNF relation".

R(ABC) - BCNF table

AB -> CD, C->ABD, D->C

then here  Candidate key ={ AB, C, D)

Also,here { D-> C, C-> A

 prime attribute A is transitively dependent on key D in a BCNF table.

so here is the counter case where option D can't be false.

Any comments...!

There must be an involvement of Non Key
It is not necessary to have a Non Key in transitive dependency.

Transitive dependency can be b/w one attribute (key) to another attribute(Prime Attrib) via a key attrib or via a non key. i.e

if Candidate key = { AB, C, D}

then following transitive dependency can be possible:

D -> C -> A

Here C, D both are keys but A is (Prime attrib) part of the key AB.


X -> Z is a transitive dependency if the following three functional dependencies hold true:

  • X->Y
  • Y does not ->X
  • Y->Z

In your Example

AB -> CD, C->ABD, D->C

then here  Candidate key ={ AB, C, D)

D ---> C
& C ---> D (This should not be there )


Ya. I got my mistake.

Thanks  yogi_p Junior

but @yogi_p why there must be invovlement of non-key in transitive dependecy.

why it is not possible

KEY($X)$  $\Rightarrow$ prime atrribute($Y$)

prime attribute($Y$) $\Rightarrow$ prime  attribute($Z$)

here $Z$ is transitivily dependept on  key $X$
@reena_kandari  but is this BCNF relation..?  LHS should be a key always...
does key=>non-key violate 3NF??

@ reena_kandari

Have you got answer to this question

why it is not possible

KEY(X)  ⇒ prime atrribute(YY)

prime attribute(Y) ⇒ prime  attribute(Z)

here Z is transitivily dependept on  key X

+16 votes


Defn from wiki:

The 3NF version of the definition is weaker than Date's BCNF variation, as the former is concerned only with ensuring that non-key attributes are dependent on keys. Prime attributes (which are keys or parts of keys) must not be functionally dependent at all; they each represent a fact about the key in the sense of providing part or all of the key itself. (It should be noted here that this rule applies only to functionally dependent attributes, as applying it to all attributes would implicitly prohibit composite candidate keys, since each part of any such key would violate the "whole key" clause.)

answered by Active (3.3k points)
nonprime-> prime dependency is there in 3NF. so

eg: key->nonprime-> prime dependency can be there in 3NF, so in that case transitive dependency is possible

But the same case is not possible in BCNF
how non prime attribute generate prime attribute ??


candidate key of given relation are : AB, BC..
A, B, C all are prime




C->A //non prime attribute  to prime attribute

I think it is also in 3NF but not in BCNF

candidate key of given relation are : BD.
B, D prime..
A, C  non prime..

AB->C  // is in 2NF (AB is not proper subset of candidate key)
D->C    //  not in 2NF (proper subset of candidate key ---> non key)
C->A   // in 2NF    (non key ---> key key)

given relation is not in 2 NF .. It is in 1 NF..
then how 3rd statement is true?
prime attribute transitivelly depends on key (key is also prime)..
X ----> A
then relation will be in 3 NF iff   A - X is set of prime attribute

so why not for BCNF? I mean prime attribute transitivelly depends on key (key is also prime)..
X ----> A it also satisfies BCNF property. Then why statement 4 is false?

3NF :
let A relation R(X, A, B) X is key and A is prime attribute( but not key i.e. proper subset of candidate key)..
Part of FD (not all) of that relation given below.
X -----> B
B-----> A
{So B must b prime.
A is transitively depends on Key X.}
is in 3NF.

but X-----> B
      B -----> A is not in BCNF because B is not a key. even though it is prime (proper subset of candidate key) but it is not a Super key..

BCNF is more restrictive than 3NF.. Transitive dependencies, not allowed  in BCNF.
Why B must be a prime ??.. A is a prime attribute and X is a key.. Is that not enough to say that relation is in 3 NF.
thnx @Digvijay Panday all doubt clear from your comment :)
Perfect explanation. This answer should be added in GO book.
+4 votes

Answer D)

A) possible. If a relation has 2 attribute, then there are 1:1 dependency,So, it will be in BCNF

B) If every key has one attribute , it must have no partial dependency. So, it will be in 2 NF

C) prime attribute can be transitively dependent on a key - i.e. key dependency will be non prime -> prime . But no non prime to non prime dependency will be there. So, it could be in 3NF

answered by Veteran (108k points)
can u play explain  option c ... what u said I did not get
By defination of 3NF, X->Y, X can Super key or Y is prime attribute, hence prime attribute can be transitevely dependent on a Key.
correct it to prime -> non-prime in your option-C explanation.
0 votes

option d is fase

answered by Boss (25.3k points)

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