Normalization Theory: 3NF

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[@toshendrarustagi9861]

In example, Candidate key is ( id ) only . Then the table is already in 2NF because C. Key is irreducible.

[@ShantanuMallick1992]

how L11 is in 3NF? is it in 2NF?

[@niwanshumaheshwari4534]

It's not clear how you convert it in 2NF at 12.01

[@yashprajapati4061]

In L11 id-> (dist,lot,area) , (dist,lot)-> area , thus area is transitively dependent on id then how L11 is in 3NF ?

[@adityapratapsingh658]

no
(dist,lot) is super key in L11 and it fulfills the second definition of 3NF.

[@shabeebrp3225]

A table that is in 1st normal form and contains only a single key as the primary key is automatically in 2nd normal form. so L should be in 2NF

[@abh03533]

actually NO, if you go by the 2nd functional dependencies, (dist, lot) can be a candidate key too, so all non-key attribute has to be FFD on every attribute of CK for this case too, but 3rd FD says that rate could be decided ONLY by dist, which violates the 2NF conditions.

[@newtanagmukhopadhyay4716]

@@abh03533 dist,lot cannot be the candidate key. a candidate key is minimal superkey. (dist,lot) can be a superkey, but not a candidate key since the number of attributes are 2. id is a candidate key since the number of attributes is 1. please let me know if i am wrong anywhere.

[@anivik9814]

(dist,lot) is a candidate key because it can uniquely identify a tuple (as it determines id) and removing any of the element of (dist,lot) will remove key functionality-which means it is minimal. Sir should have mentioned this.

[@newtanagmukhopadhyay4716]

@@anivik9814 but still it has two attributes right. id has only one. so that is superkey and id is candidate key