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

  • $a \rightarrow c$
  • $b \rightarrow d$

This relation is

  1. in first normal form but not in second normal form

  2. in second normal form but not in first normal form

  3. in third normal form

  4. none of the above

asked in Databases by Veteran (52k points) | 3k views

2 Answers

+27 votes
Best answer
Candidate Key is $ab$.
Since all a,b,c,d are atomic so the relation is in 1 NF.

Checking the FDs :

$a\rightarrow c\ $ (Prime derives Non-Prime.)
$b\rightarrow d\ $ (Prime derives Non-Prime.)

Since, there are partial dependencies it is not in 2NF.

a} Answer 1NF but not 2NF
answered by Active (3.6k points)
edited by
+1
by using decomposition in ac bd ab
0

 sonam vyas

by using decomposition in ac bd ab

 what is means?

0
i have a doubt

assume AB is a primary key  and relation (ABCDE)

dependency:

1)A---->BCDE

2)AB---->CDE

3)AB----->BDE

for 1 & 3,is partial dependency exist ?
0
@bikram sir see this pls
0
AB is key

Prime attribute ={A,B}

For partial FD : prime attribute -> non prime

1) is partiall FD

2) and 3) are not partial FDs
0

 Gate Ranker18

see 

for 3,AB---->B(KEY---->PRIME ATTRIBUTE)ALLOWED IN 2NF

AB---->DE NO PARTIAL DEPENDENCY

0
@set2018  2 and 3rd in not Partial Dependency because it is fully dependent Key coz we AB as primary Key.

 

Only 1st in Partial Dependency.
0

@set2018 When A defines everything then A is the PK and AB is a superkey.

3rd FD is needless, there is no parital dependency. You defined the wrong primary key initially.

+5 votes
Answer: A

ab is the candidate key.
answered by Boss (33.8k points)
Answer:

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