BCNF of a Relation with no FDs

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A simple question that I do not seem to find an answer for online: Is the relation that has no non-trivial functional dependencies considered to be in BCNF form and if not, how do I decompose it? Thank you in advance!

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Alexander Nenartovich On

After additional research, I finally stumbled upon this definition of BCNF:

A relational schema R is considered to be in Boyce–Codd normal form (BCNF) if, for every one of its dependencies X → Y, one of the following conditions holds true:

  • X → Y is a trivial functional dependency (i.e., Y is a subset of X)
  • X is a superkey for schema R

So to answer my own question, every relation that has trivial functional dependencies only is in BCNF by definition.

Perhaps it helps somebody who has a similar question.