2 votes 2 votes Is the following FD in 2NF? AB -> C Where BD is candidate key. please explain why this FD is in 2NF. non-prime,prime ----> non-prime is in 2NF. But why? Bcz. as per my knowledge Every non-prime attribute should be fully functionally dependent on the primary key of R Databases database-normalization + – Tuhin Dutta asked Aug 12, 2017 edited Jul 12, 2019 by Cristine Tuhin Dutta 510 views answer comment Share Follow See all 7 Comments See all 7 7 Comments reply Tesla! commented Aug 13, 2017 reply Follow Share Think what if AB is also candidate key ? 0 votes 0 votes Tuhin Dutta commented Aug 13, 2017 reply Follow Share Then it satisfies 2NF 0 votes 0 votes Tesla! commented Aug 13, 2017 reply Follow Share Yes, you are correct, it won't violate 2 NF unless either a or b or d derives any other attribute 0 votes 0 votes Tuhin Dutta commented Aug 13, 2017 reply Follow Share Man, I'm still in confusion. What should I conclude then ? 0 votes 0 votes Tesla! commented Aug 13, 2017 reply Follow Share You should check that 1)AB is candidate key or not if yes then check that from all the possible FD, only single A or B or C is able to derive any attribute if yes then not in 2 NF Else no then relation is in 2NF 0 votes 0 votes Tuhin Dutta commented Aug 13, 2017 reply Follow Share Check out this question asked in gate-2008 https://gateoverflow.in/492/gate2008-69 Here, {Catalog_no, ( Title , Author )} are candidate keys. 1.Catalog_no --> {Title , Author , Publisher , Year } 2.{Publisher , Title , Year} --> Price Which implies, Catalog_no --> Price. Is it the reason for considering the relation to be in 2NF?? 0 votes 0 votes Tesla! commented Aug 13, 2017 reply Follow Share yes you are correct it is because of transitive dependency 1 votes 1 votes Please log in or register to add a comment.