Question

SQL

2. Decompose the following relation to conform with 2NF and 3NF. Custoner Retal Relation CR76 CR76 CRSA Tine Murphy P016 CRs6 PO16 Prinvey Key fdi ds id4 fas Spelling out the functional dependencies, we have the following: 2 Cuetomer No 3. Property No CNane - RAddress. Rent, Oner ne. OName > QNane PToPerty No Paddress. Rentrintsh 5Customer No, Rentstart Rent, 9wnex Ne. 9name

Answer 1

GOe have the tan chon๗ dependecie. So o f"st ave fnd fu candidate Propet, No) > Custom e-No,Property-No, CNaNe, PAddmess Rent, Oconen-no, Oname , RentStaoit, RentFini sh C to en No,RenVta t, CName, β operty No, Address, Rent Finish, Rent, No) s the cand: date keu on pmime ate.bule Customen-No, o, Rent Stat ustomen as cuell Ccustomun-No, Rentsto,n evenu column ơ31 attr;bute shoald have atom:c valun canned wit the nelalion on table s in JNF.

ton a yielbion。η table to be in 3NF , the follow. na cod:hon have fo be saisfed : prime ailn.bat Siponkej JP one the llecin condikon on both ehem m e tfien superk So by seeing at the E. D Its+, 2,3, ч duo not hold tha ondiho coe coould decompose 2.9,% and then bu tho nest. dfo table So we haue Toble cName Casto me.. No Table 2 o pen n, ihe table coould be sabdiided Jo held tte condi hon hod Here agoin, in each table, there ธ ‘ candidat desh O. So the subdivided table Ocennno ONameTh tn:s table the pnoblem goes an Table NOI RentStant | Customen-No! CName! Rentfinish Table4 Castomen No Table S e tablto monkede,s,®,6, Gonefe decompoied relasion on tablea 3NF cond:kon Sahsfied cs Scanned with CamScanner

   Moreover since the process directly finds the decomposed tables up to 3NF, so the tables are already in 1NF and 2NF according one condition of 3NF that says that to be in 3NF, the relations or tables must be in 1NF and 2NF.