Topic = fd's and normalization.

Question

Q) In R(A,B,C,D)

FD's {AB->CD, C->D}

Is it in 3NF?

Ans) Vaise to shortcut (in x->y fd either x should be super key or y should be prime attribute to be in 3nf) ke hisaab se 3 nf main ni lagra in fd c->d, Par definition ke hisaab se ye 3 nf tab ni hota jab D transitively dependent hoti ab pe, par yaha directly bhi dependent hai transitively bhi
To isliye samjh nhi aa rha

Please help

Comments

A relation is said to be in 3rd normal form in DBMS (or 3NF) when it is in the second normal form, but no transitive dependency exists for a non-prime attribute.

 

Concentrate on AB-> C and C -> D 

Please note that transitive depedency exist btw non-prime attributes.

 

With AB-> C and C -> D, you can get AB->CD

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I think this is not in 3nf, because c->d is a fd from non key->non key. In case of 3nf this violates.

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Thanks for answers, I understood clearly now ☺

Answer 1

In 3NF, the  relation R does not allow N.K.A   ---- > N.K.A (Non key Attribute  -----> Non Key Attribute)

but there is C--->D , which is N.K.A ----> N.K.A.

 

Therefore the relation is not in 3NF.

Comments

ha 2nf mein hi fail kr ra hai ye to