635 views

For a database relation $R(a,b,c,d)$ where the domains of $a,b,c,d$ include only atomic values, only the following functional dependencies and those that can be inferred from them hold:

$a\to c\\b\to d$

the relation is in

1. first normal form but not in second normal form.
2. second normal form but not in third normal form.
3. third normal form.
4. none of these.

### 1 comment

Original Source of this question: GATE 1997

The candidate key for the relation is:  $ab$ ,  because  $ab^{+}=abcd$

Since,  $a \rightarrow c$  and  $b \rightarrow d$ are partial dependencies which are not allowed in $2$NF, the relation is hence in $1$NF.

Option A is correct.
by
A) first normal form but not in second normal form,

( ab)+=abcd..so , ab is a candidate key, so, a→c and b→d both are partial dependencies...and in 2NF it is not Allowed...and in the question it is given that  domains of a,b,c,d include only atomic values...So it is in 1NF..
ANS: A

CANDIATE KEY OF THIS RELATION IS AB

a→c

b→d

BOTH CONTAIN PARTIAL INDEPENDENCIES , AND IN 2ND NORMAL FORM PARTIAL INDEPENDENCIES IS NOT ALLOWED.

SO GIVEN RELATION IS ONLY IN 1ST NORMAL FORM.
by

1
601 views