Database

Question

Consider the following statements.
S1:  If relation R is in 3NF and every key is simple, then R is in BCNF
S2:  If relation R is in 3NF and R has only one key, then R is in BCNF

A).Both S1 and S2 are true.

B).S1 is true S2 is false.

C).S2 is true S1 is false.

D).Both S1 and S2 are false.

Comments

@Doctor

you have solved the main logic of qsn, now further proof is very easy, see,

Now suppose we take the second condition that Y is a prime attribute but X is not a super key. and since every key should be simple Y is a candidate key

 now, see it is written like X->Y(candidate key), now see X is deriving Y and Y is deriving everything, since Y is key, this means X is deriving everything, this in turn means X is superkey.

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Just one last question @Ashwin

"Y is CK then you can able to derive all attributes from X also so X must be a SK".

Just explain this little more. I am not able to digest it. :-(

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Thanks @Joshi .

You are awesome.

Answer 1

if a relation is in 3NF . and it have atmost one campound key then it is always is in BCNF.

and hence both s1 and s2 is true.