Questions On Second Normal Form(2NF)

[@mnaresh3382]

I think the answer is not 6 but instead 5, since we don't really need a separate table for R(WY) bcz it is not violating the rule of 2NF, we can just add these attributes to the table R(PSVW XY) this is going to introduce any issue since PS is the candidate key of this relation and it can uniquely identify all other attributes of that relation and w->y is not partially depended on the candidate key PS, so the correct answer should be "NO & 5"

[@tushaamagrawal3072]

But W itself is determining Y...This is also case of partial dependency since Y is dependent on a non prime attribute which is not even a part of Candidate key. It should have been dependent on PQS. Hence we would need to make a seperate table for it too.

[@mnaresh3382]

I agree the fact that we still have to separate the table, as you said W is not a prime attribute, but it is determining some other non prime attribute, this does not violate 2NF condition but violates 3NF condition, 2NF speaks only for partial dependency but this is a transitive dependency.

[@nguyenquangthai6373]

@@mnaresh3382 yeah I agree with you, watch until the part I have to scroll down to the comment to see if any thought the same

[@EnglsihCommByNawaz]

Thanks Sir! Learn alot from this video!

[@premkumarreddy6893]

thanks

[@varshaeluru9816]

Thank you so much sir

[@Karma-nc3kg]

hello bro in last sloved problem i have a doubt

[@b-43-pawankadwadkar96]

Thanks a lot ❤

[@samandeepkaur6112]

Thank you

[@EnglsihCommByNawaz]

Thanks sir

[@manasvi4586]

thankyou sir