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An instance of a relational scheme $R(A, B, C)$ has distinct values for attribute $A$. Can you conclude that $A$ is a candidate key for $R?$
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77 votes
77 votes

No.

$$\begin{array}{|c|c|c|} \hline \text {A} & \text {B} & \text {C}  \\\hline\text {1} & \text {5} & \text {6} \\\hline\text {2} & \text {4} & \text {7}\\\hline \text {3} & \text {4} & \text {5}\\\hline \end{array}$$

Suppose this is the relational instance at any point of time.

Now we can see that $A\to BC$ holds for this instance, hence $A^+=\{A,B,C\}.$ (For every unique value of $A$, values of $B$ and $C$ are distinct.

But FDs are defined on the schema and not on any instance. So, based on the state of any instance we cannot say what holds for schema (there can be other instances too for $R$). At the best we can say that $A \to BC$ MAY hold for $R.$

PS: If we have a single instance where $A \to BC$ is not holding, it is enough to say $A \to BC$ does not hold for the relation $R.$

edited by
51 votes
51 votes
an instance of a relation is just a snapshot at any instant of time it is not the whole table. it may be possible in that instance of a relation some attributes behave like keys but who knows in other instances of a relation this attribute is not key, some other attributes behave like keys. so without the whole table, we cannot say anything about the key of the rel
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