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+19 votes
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Which of the following is TRUE?

  1. Every relation in 3NF is also in BCNF
  2. A relation R is in 3NF if every non-prime attribute of R is fully functionally dependent on every key of R
  3. Every relation in BCNF is also in 3NF
  4. No relation can be in both BCNF and 3NF
asked in Databases by Boss (18.3k points)
edited by | 1.9k views
+1

@ Bikram sir can you plz explain meaning of option B

0

 Arjun sir  Bikram  sir

B) "A relation R is in 3NF if every non-prime attribute of R is fully functionally dependent on every key of R"

means X->non_prime and X is key hence X is superkey also so it is in BCNF so also 3NF 

plz check 

0

Gate Ranker18 but here ans is c.according to u b is also correct

0

Gate Ranker18 in case B) non-prime-->non prime dependency is possible which is not the case in 3NF(transitive dependency)

0

rajoramanoj  acc to me both B and C are correct 

 reena_kandari it is X-> non_prime not  non-prime-->non prime

+3

 every non-prime attribute of R is fully functionally dependent on every key of R

it  gurantee there should not be the case of prime--->non-prime  other than Key--->non-prime

it does not say anything about  non-prime-->non prime, so it can occure here which for sure does not gurantee 3NF.

0
yes correct :)
0

for 3nf 

LHS must be superkey OR 

RHS must be prime attribute .According to second point  key------>non prime attribute

so here second condition(non prime attribute) of 3 nf fails.lets check first condition,key must be superkey hence 3nf,but nothing about non prime which violates 3 nf 

0
Please explain difference between "fully functionally dependent" and "functionally dependent"

2 Answers

+23 votes
Best answer

(C) Every relation in BCNF is also in 3NF. Straight from definition of BCNF.

answered by Veteran (370k points)
edited by
+1
Isn't B correct if we ignore the multivalued attributes?
0
Why is B incorrect , it also seems correct to me.
+6

Nopes, that does not guarantee 3NF. For 3NF there should not be any FD X->Y such that Y-X is a non-key and X is not a super key.  

+6
Yes, got it .  :)

It should not have transitive dependency.
+1
can anyone explain what option b says!
+27

B. A relation R is in 3NF if every non-prime attribute of R is fully functionally dependent on every key of R

Lets take R(ABCD)  with AB is key 

AB->C and C->D  here AB fully determine D i.e. AB-> CD

According to statement it is in 3NF but it is not. Because C->D form non prime -> non prime.

0
plz elaborate option B
0
B option says nothing about transitive dependency which should not be there in 3nf therefore B option not correct
0
Absolutely Correct Answer...
+3 votes

$\text{1. Every relation in 3NF is also in BCNF}$

For a $3NF$ relation to be in $BCNF$, that has to has satisfied this condition-

$\text{for each dependency }$ $X \rightarrow Y$, $\text{X should be the super key.}$

So, this statement is wrong

 $\text{2. A relation R is in 3NF if every non-prime attribute of R is fully functionally dependent}$ $\text{on every key of R}$

This is a straightforward definition of $2NF$. If a relation is in $1NF$ and in that relation each & every non-prime attributes has to be fully functionally dependent on each key of the relation.

& for a $2NF$ relation to be in $3NF$, that has to satisfy the condition which states, no non-prime attributes should be transitively dependent on candidate key or we can say that, no non-prime attribute should be determined by anything but a super key

Let's take an example:

$R(ABCDE)$ and $F = \{BC \rightarrow ADE, A \rightarrow BCDE, D \rightarrow E\}$

The candidate keys will be = $\{ A, BC\}$

Prime Attribute = $A,B,C$

Non-Prime Attribute = $D,E$

Here, all the non-prime attributes$(D,E)$ are fully functionally dependent on each key of $R$

But here, $E$(non-prime attribute) can be determined by $D$ and $D$ is not the candidate key. So, $E$ is transitively dependent on the candidate keys of $R$ and moreover, $D$ is not the super key also.

$\therefore$ This relation is not in $3NF$. For making it to $3NF$ we have to break the relation into $2$ parts: $R_1(ABCD), F=\{ \{A \rightarrow BCD\}, BC \rightarrow AD\}$ & $R_2(DE), F = \{D \rightarrow E\}$

This option is also wrong.

$\text{3. Every relation in BCNF is also in 3NF}$

Obviously, For a relation to be in $BCNF$, it has to be in $3NF$ first.

$\color{Green}{\therefore \text{This statement is Correct}}$

$\text{4. No relation can be in both BCNF and 3NF}$

Yes, there can be some relation which is in $3NF$ and as well as $BCNF$

and if a relation is in $BCNF$ then it is also in $3NF$

So, this statement is also wrong.

Correct option is $C)$

answered by Boss (15.1k points)
edited by
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