973 views

1 Answer

0 0 votes
In Normalisation , whenever we decomposed The relation we need to have a attribute common in both relation and that attribute should serve as CK for any one of the relation .-- to have LOSSLESS DECOMPOSITION

They have decomposed into R1 ( ABC ) and R2( D,E) ---Since no common attribute Therefore it is lossy decomposition .

And it is neither dependency preserving also .

hence option d
Position:
Show:

Related questions

0 0 votes
0 0 answers
91
91 views
krishna_panjiyar 1 asked Jul 27
91 views
Design an ER schema for keeping track of information about votes taken in the U.S. House of Representatives during the current two-year congressional session. The databas...
0 0 votes
0 0 answers
263
263 views
GO Classes asked Feb 17
263 views
Find correct query for counting Teamsize.SQL Query $1$ SQL Query $2$ SQL Query $3$ SQL Query $4$  
1 1 vote
0 0 answers
249
249 views
GO Classes asked Feb 17
249 views
Consider two relations $R$ and $S$ with attributes $p$ and $s$.The following tuple relational calculus (TRC) query is given:\[\{\, t \mid t \in R \ \land\ (\exists z \in...
0 0 votes
0 0 answers
214
214 views
GO Classes asked Feb 17
214 views
Relational Algebra Question.The query was like, $\pi (\sigma\dots\bowtie\dots)\bowtie(\sigma\dots\bowtie\dots)$Asking : no. of tuples in output.zero tuple one tuple two t...