Normalization TRUE/FALSE Questions with explained proofs for understanding / TRUE/FALSE Questions with answer discussion on Normalization / TRUE/FALSE Questions on Normal Forms Discussion
TRUE/FALSE Questions – Normalization and Normal
Forms
1. For a relation schema R(A, B,
C, D), if AB → C and C → D, then AB is a key.
Ans:
TRUE
Proof:
Try to find (AB)+, ie., closure of (AB).
Step 1: result = AB
Step 2: If you know AB, then
using AB → C, result = ABC
Step 3: If you know AB and C, and
using C → D, the result = ABCD
At this stage, result included
all the attributes of R. Hence, (AB) is a key for R.
2. If a relation is in 3NF, then
it is also in BCNF.
Ans:
FALSE
Proof:
3NF permits Non-key → Key dependency. That is, it accepts a dependency
where a non-key attribute can determine the value of a key attribute. But in
BCNF, it expects only candidate keys to be determiner, not non-key attribute. Hence,
a relation that is in 3NF, not in BCNF too.
3. From A →
B, we can derive AC → BC, which can further lead to A →
BC.
Ans:
FALSE
Proof:
From the question, A determines the value of B. According to Armstrong’s
Augmentation rule AC → BC is true for any attribute C. this
is because C → C is a trivial functional dependency. But, this
augmentation cannot lead to the dependency A → BC where A
alone determine both B and C.
4. BCNF decomposition of a
relation R(A, B, C, D) with functional dependencies A →
B and C → D leads to three relations, R1(A, B), R2(C, D),
and R3(C, A).
Ans:
TRUE
Proof:
According to the rule of BCNF, all the functional dependencies should have keys
as determiners (keys as left hand side attributes of any FDs). For the given
relation R and according to the set of FDs given, the key is AC which is
composite. Hence, we break the relation into a relation R1 with attributes A
and B [from FD A → B], a relation R2 with attributes C
and D [from FD C → D], and a relation R3 with attributes
C and A [to link R1 and R2].
5. A relation R(A, B, C) with the
functional dependencies AB → C and C →
B is in BCNF.
Ans:
FALSE
Proof: From
AB →
C, it is clear that AB is the key. But, one of the key attributes is determined
by a non-key attribute C according to C → B. Hence, the
relation R is not in BCNF.
6. A relation R which is in 3NF
is also in 4NF.
Ans:
FALSE
Proof:
3NF uses the functional dependencies of Single valued facts. That is, they are
about one or more attributes uniquely determine the values of one or more other
attributes. But 4NF is about a different kind of functional dependency called Multi-valued
functional dependencies. That is, functional dependencies about multi-valued
facts.
7. For a relation R with
attribute ABCDE and set of functional dependencies F = {A →
B, BC →
D, D →
E}, the closure for attribute combination {AC} is ABCDE.
Ans:
TRUE
Proof:
Find the closure of (AC)
Step 1: result = AC
Step 2: from A →
B, A can determine the value of B. hence, result = AC U B = ACB.
Step 3: through BC →
D, result becomes ABCD.
Step 4: as we know D, then
through D → E, result = ABCDE.
Hence, the closure of (AC) is
ABCDE.
8. For a relation R with
attribute ABCDE and set of functional dependencies F = {A →
B, BC →
E, ED →
A}, the attribute combination {ACE} is one of the keys.
Ans:
FALSE
Proof:
Let us find the closure of (ACE)
Step 1: result = ACE
Step 2: from A →
B, result = ABCE.
Step 3: from BC →
E, result = ABCE, does not change.
The result is not ABCDE (all the
attributes). Hence, (ACE) is not a key for R.
9. In a relation R(A, B, C, D),
if AC →
CB then A → B.
Ans:
FALSE
Proof: As
per the given FD AC → CB, we can understand both A and C
together identifies the value of B. We cannot say, by removing C from both
sides, A alone can determine B. We would say AC → B.
10. In a relation R with
attributes ABC, if a FD A → B holds, then AC →
B also holds.
Ans:
TRUE
Proof: Attribute A alone can determine B. Adding
another attribute C would not change the determination of B by A.
