explain

Question

If R and S are two relation in BCNF the natural join of R and S is also
in BCNF.

above statement true or false.??

Comments

Well, you should answer this :)

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Do same for 3NF too..

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What is the form of 3nf

Answer 1

It should be False, In BCNF there should be no redundancy and in natural join of R and S may be reduncancy although both R and S are in BCNF, if R and S are related to each other.

Comments

eg:

R(table)                                                                                                            

A	B
1	5
2	5
3	6
4	3

  

S (table)

B	D
5	4
6	5
3	4

R⋈S(natural join of R and S)

A	B	D
1	5	4
2	5	4
3	6	5
4	3	4

only ONE  attribute in above table cannot form key ..in  above table key should be compound 

so means it is false...??

Answer 2

False. It is logical, while converting a 3NF relation to BCNF we decompose a 3NF relation into two BCNF relation, and in some cases we get lossless decomposition , so joining 2 BCNF relation is not a BCNF relation