normalization

Question

proof that a relation with the two attribute is always in BCNF....

Comments

Because it is trivial.

R={A,B} FD={ A->B,B->A}, CD={A,B}

hence in BCNF.

Answer 1

If a relation has only two attributes then it is in BCNF.

We say a relation $R$ is said to be in $BCNF$ if it has functional dependencies of the form $a \rightarrow b$ where either $a \rightarrow b$ is a trivial functional dependency or $'a'$ forms a superkey for the relation $R$.

Proof:

Let $a, b$ be the two attributes in a relation $R$. Then the possible functional dependencies are :

case 1: LHS(Determinant) contains both attributes : i.e. Possible FD is $a b$ $\rightarrow$--- Clearly this is a trivial functional dependency because RHS attributes form subset of LHS attributes .

Case 2 : LHS contains only one attribute :  i.e. The Non-trivial FD's are $a \rightarrow b$ Or $b \rightarrow a$ or Both ---- Clearly in this case, the LHS attribute will be candidate key.

Hence the possible functional dependencies (in case of Only Two attributes) are either trivial or the LHS attributes form a superkey. So, the relation having only two attributes will be in $BCNF$.