made easy test series

Question

If a table $R$ has only one candidate key, then which of the following is always true?

• $R$ is in 2NF, but is not in 3NF
• If $R$ is in 3NF, it is also in BCNF
• $R$ is in 2NF, but may not be in 3NF
• None of these

Comments

but i m talking about the case when only i candidate key is there....the option 2 should be correct

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but ad ->b......means that ad is also a candidate key.......as you have mentioned ab to be a primary key......plzz arnabi boss reply

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Even the CK is one,

There might be partial dependency exist. Identification of NF will be accurate if we are given with FD sets.

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okk....arnabi boss......my first doubt is ....is ab primary key in your relation...as you have underlined it...........

2nd doubt if ac is candidate key.........then how ad->b will be in 3nf.....as b is non key........and ad should be then candidate key.......

Answer 1

option B is correct. As option it says "if" it is in 3NF then surely it will be in BCNF because to be a relation in 3NF not in BCNF atleast two compound keys are needed, so that proper set of one candidate key drives proper subset of other candidate key.

if option B was not there then following points need to be consider:

1.) If candidate key is compound key then there may be chance that proper subset of candidate key drive non prime attribute {known as partial dependency}

2.) If candidate key is simple key , then if can say there is no partial dependency and relation is in 2NF but may or may not be in 3NF.

Comments

but what about option 2....i find it correct......also option 3 is also correct.....then why D option??

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  @himanshu19 thanks

correct option 2....

1. if R is in the 3rd NF then it means for all the FDs ...(case1. either left side is 'key'

case.2.  or right side is prime attributes )

for first case it is already in BCNF....

and if we talk about case2. then there would be another candidate key(more than one)....

therefor option 2. is correct

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but it is said that only one candidate key is there....in you case if C->B...then AC will be new candidate key......but only one candidate key can be there.....

Answer 2

if a relation is in 3NF and there are atmost one key then the relation also is in BCNF

HENCE OPTION B is right

Answer 3

Counter Example  for option a and c:

Consider R(ABCDE) and FD: { AB→ CD, B→ E}

we get CK = { AB} and prime attributes = A,B

2^nd FD gives (PA → Non PA/ Part of the key → Non key), therefore it violates 2NF.  Now check the options 

option a:  R is in 2NF, but is not in 3NF. False

option c: R is in 2NF, but may not be in 3NF False

option b : If R is in 3NF, it is also in BCNF. True . Because to be a relation in 3NF not in BCNF then atleast two compound overlapping keys are needed.