THIRD NORMAL FORM (3NF) IN NORMALIZATION || NORMAL FORMS || 1NF, 2NF, 3NF, BCNF, 4NF & 5NF || DBMS

INTRODUCTION TO NORMALIZATION
1. 1NF - FIRST NORMAL FORM
2. 2NF - SECOND NORMAL FORM
3. 3NF - THIRD NORMAL FORM
4. BCNF - BOYCE CODD NORMAL FORM
5. 4NF - FOURTH NORMAL FORM
6. 5NF - FIFTH NORMAL FORM
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[@adventure_6043]

Me too believe you for my semester. You are an excellent teacher

[@jeevankumar659]

We have to really appreciate the hard work and dedication you have put on these videos for students. Keep going sir, hopefully you will reach your goal.

[@amannmahapatra6118]

i always trust you for my semester subjects

[@hiteshperni4157]

Sir,it was amazing.i can understand ur lectures.......

[@amishadas3958]

Thank u very much 💖🥰

[@doggcript3949]

Thank you sir, you help me a lot!

[@amannmahapatra6118]

Hello sir thank you for your contribution

[@anishkumarmishra9620]

All about your teaching good sir . But in last u didn't explained how transitive dependency is removed after decomposing. Same as for 2nf and for this too !!

[@pavankumarpadavala6369]

Thanks🙏🙏🙏

[@fog2640]

finished watching

[@bokinalavijay6876]

Sir we always supports u sir

[@smithagv6296]

Please share BCNF 4NF and 5NF

[@rayagirihemanthkumar4797]

Sir the table which you have considered, I guess it's not in 2NF because I found Partial Functional Dependency in it (St_id , Course) ---> (Fee) {Its FD } Now breaking it:- (St_id) --> Fee is "FD" (Course) ---> Fee is also "FD" So the table is having partial Dependency then how can we solve it to get 3NF ???????

[@anvinalias5259]

May be too late😅 You are right about the partial dependency. I guess, even though proceeded with 3nf, with out the condition for 2nf satisfied, the decomposition of the table satisfied both condition for 2nf as-well-as 3nf.

[@raghulgunasekar2422]

Yes you are right!

[@Jai0804]

Yes iam having the same doubt

[@feliciazuko]

that is not partial dependency because St_id itself is a candidate key (it can uniquely identify the tuples) so (St_id,course) can't be a candidate key as CK is supposed to be minimal super key. And for partial dependency you need prime attribute -> non prime attribute. In this case, Std-id-> fee is not PD as Std_id is itself CK and not part of CK. And in course-> fee , course is not a prime attribute, it is a non prime attribute. So, this is transitive dependency and not partial dependency. This prolly is a bad explanation but I had the same doubt and after researching a bit I figured it out. Hope it helps

[@killer4791]

Nope. You are not. There's no partial dependency. What is a partial dependency? As simple as its name is, in a full dependency, if part of the attributes is having a functial dependency, then its partial dependency. Now, ----->stu_id and Fee has a functional dependency ----->ASWELL, the course and Fee has a functional dependency. So, it's not a partial dependency. The word "Partial" should be given importance. Partial means partial.

[@yon1623]

is FEE > Course also, the example is not a good one. also 2nf is not complete

[@sivakrishna8550]

dom properties meedha cheyyandi sir

[@malugurashmitha5241]

Sir explain 4nf and 5bf sir