Timothy (normal form)

Question

Let R be a relation. Which options are correct

a)R will necessarily have a composite key  if R is in BCNF  but not in 4NF

b) If R is in 3NF and if every key of R is simple, then R is in BCNF

c) If R is in BCNF and if R has at least one simple key , then R is in 4NF

d) If R is in 3NF and if its every key  is simple, then R is in 5NF

Comments

b) If R is in 3NF and if every key of R is simple, then R is in BCNF

is correct option.

Answer 1

Option B is right option

BCNF is stricter than 3NF. A table is BCNF if it is in 3NF and for every FD  X->Y, X should be the super key of the table.

Consider a relation R X->Y, Since R is in 3NF, either

1. X is a superkey or
2. Y is a member of a key

Since every key in R is simple, Y has only one attribute and Y itself is a key, which implies that X is always a superkey. Therefore, X->Y does not violate BCNF in either case, which implies that R is in BCNF.

Comments

@Rishi yadav, @Manu Thakur

Can you give explanation why others are incorrect?