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What should be correct option for 1. and why...?

in Databases by Boss (10.5k points)
edited by | 147 views

2 Answers

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Answer : B

A is Correct. A Weak entity set in ER model doesn't have any key/Candidate key which can distinguish between any two records uniquely. 

C is correct. We can have multiple Candidate keys in a relation.

D is correct. A Relationship can have its own attributes. 

B is False (a Typo in Option B..It should be Relationship instead of relation)... In ER model, A Relationship between different Entity sets can be 1-1, M-N, 1-N, N-1.

by Boss (26.1k points)
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I think, Option A is also incorrect. Key of a weak entity includes partial key attribute of weak entity + foreign key refering to primary key of a strong entity. Thus, although attributes of a weak entity can't form a primary key themselves, but they are present in the key as a partial key.
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If you read my answer for Option A, I've used the term "In ER model". In RDBMS model, there is No such thing as Weak Entity or Multivalued Attributes... These are present in ER model. So, In ER model, A Weak Entity doesn't have any Key. When we make a Table(Relation) corresponding to this Weak enetity in RDBMS model, then We make Key by merging the partial key and primary key of strong entity.
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You are right. Thanks for correcting me.
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Only is B is the Right Answer 

B:EVERY RELATION CAN NOT BE 1:1 RELATION ,IT CAN ALSO BE 1:M AND M:N OR M:1 RELATION ALSO

by Active (1.8k points)
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