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Consider a relation r1(ABC), r2(CDE) and r3(FG) with primary keys A, C and F respectively. Assume that r1 has 150 tupples, r2 has 100 tupples and r3 has 75 tupples. The number of resultant tuple in are ________.

The result of natural join = Cartesian pdt when there are no common attributes isn't it ?
yeah you are right

In $r_1\Join r_2$  $C$ is common ,in which C is foreign key ( assuming that foreign key is present) in r1 so in natural join of r1 and r2 all record of r1 will be present so $r_1\Join r_2$ has maximum no. of tupples 150 .let say $r_1\Join r_2$  as S ,now join of S and r3, in which no element is common( assuming all attributes have diffrent quality as in question nothing is mentioned) so $S\Join r_3$  will become $S \textbf{*}r_3$(cross product) .now every tupple of S will relate to every tupple of $r_3$ ,so maximum no. of tupples =150*75=11250.

Yes!
Then the number of resultant tuples will be less than 11250 right?
I think so.

1 vote