Here in question they have mentioned that the table contains two single valued attributes. But the table to be in 1NF, the attributes should be atomic and single valued. So, can we conclude that S1 is TRUE ???

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+21 votes

Given the following two statements:

**S1:** Every table with two single-valued attributes is in 1NF, 2NF, 3NF and BCNF.

**S2:** $AB \to C$. $D \to E$, $E \to C$ is a minimal cover for the set of functional dependencies $AB \to C$, $D \to E$, $AB \to E$, $E \to C$.

Which one of the following is **CORRECT**?

- S1 is TRUE and S2 is FALSE.
- Both S1 and S2 are TRUE.
- S1 is FALSE and S2 is TRUE.
- Both S1 and S2 are FALSE.

+23 votes

Best answer

+13 votes

Table with 2 single valued attributes will be in 1NF,

2NF because there can not be a partial key dependencies. Key can be 1 attribute or 2 attributes. If one is key then other can be dependent, if 2 attributes make key then it is trivial.

3NF, there is no question of Transitive dependencies.

BCNF, because if at all a dependency exists, the determinant will be a Key.

2NF because there can not be a partial key dependencies. Key can be 1 attribute or 2 attributes. If one is key then other can be dependent, if 2 attributes make key then it is trivial.

3NF, there is no question of Transitive dependencies.

BCNF, because if at all a dependency exists, the determinant will be a Key.

+6 votes

S1 is true bcoz if there are only 2 attributes then relation is always in bcnf.

let R(A B) possible cases are:

(1) {A->B } here A is cand key so BCNF

(2) {B->A} here B is cand key so BCNF

(3) { A->B B->A} here A and B both are cand key so BCNF

(4) no non-trivial FD's here AB is cand key so BCNF

S2 is false bcoz minimal cover should be { D->E , AB->E , E->C }

so ans is **A**

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