@ Bikram sir can you plz explain meaning of option B

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+20 votes

Which of the following is **TRUE**?

- Every relation in 3NF is also in BCNF
- A relation R is in 3NF if every non-prime attribute of R is fully functionally dependent on every key of R
- Every relation in BCNF is also in 3NF
- No relation can be in both BCNF and 3NF

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**

+4

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

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

+24 votes

+8

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.

+31

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.

+6 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 $C)$

0

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

Answer about confusion option B is:- https://gateoverflow.in/33464/3nf-true-false

Answer about confusion option B is:- https://gateoverflow.in/33464/3nf-true-false

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