i need the answer quickly Choose True or FalseT/FTRUEFALSEThe foreign key must be inone side table when arelationship is 1:N.At the implemen tation level,the Multi-Valued Attributemust be turned to a newen tity, then make a 1:1relationship between thenew entity and the existingone.The result of minusoperation denoted by S-R willbe the same result for R-S.We can consider the renameoperation as one of settheory operations.Join operation degeneratesinto Cartesian product ifthere is no join condition.Rename is technique used tochange the tables name inthe intermediate and finalresults.We can consider the leftou ter join and the right outerjoin as equijoin plus the non-matching rows.

Defined given statements true or false

Answered: The foreign key must be in one side…

Database System Concepts

7th Edition

ISBN:9780078022159

Author:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan

Publisher:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan

Chapter1: Introduction

Section: Chapter Questions

Problem 1PE

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The multiple choice answers are: - One and only one - one or many - zero or one - zero or many.
a prime attribute be one that appears in at least one candidate key. Let α and β be sets of attributes such that α -> β holds, but β ->α does not hold. Let A be an attribute that is not in α, is not in β, and for which β A holds. We say that A is transitively dependent on α. We can restate our definition of 3NF as follows: A relation schema R is in 3NF with respect to a set F of functional dependencies if there are no nonprime attributes A in R for which A is transitively dependent on a key for R. Show that this new definition is equivalent to the original one.
Every weak entity set can be converted to a strong entity set by simply adding theappropriate attributes. Why, then, do we have weak entity sets?
It is possible to can convert any weak entity set to a strong entity set by simply addingappropriate attributes. Why, then, do we have weak entity sets?
The participation of the weak entity in its relationship with the strong entity that defines it is Select one: O a. Exact Participation O b. Partial Participation O c. Soft Participation O d. Total Participation
Suppose we have three relations r(A, B), s(B, C), and t(B, D), with all attributes declared as not null. Consider the expressions • r natural left outer join (s natural left outer join t), and • ( natural left outer join s) natural left outer join t a. Give instances of relations r, s and t such that in the result of the second expression, attribute C has a null value but attribute D has a non-null value. b. Is the above pattern, with C null and D not null possible in the result of the first expression? Explain why or why not.
Each weak entity collection may be fortified by adding more relevant attributes. The question then becomes, what exactly is the root of our poor entity sets?
Any weak entity set can be turned into a strong entity set by just adding the right properties. So, why are our entity sets so bad?
You are given a relation R with the following attributes R(A,B,C,D,E). Assume that you were given the following functional dependencies: • FD1= AB -> E • FD2: = D -> C (a) What is a valid key for this relation? justify your response. (b) Is this relation in 3NF (Third Normal Form)? Justify your response. If it is not in 3NF, normalize the relation and demonstrate all steps of normalization.
Consider the following functional dependencies that hold on R(ABCDEF): AB-> CD, E -> C, B -> EF Is R in 3NF? Explain your answer in detail and decompose the relation, as necessary, into a collection of relations that are in 3NF. Make sure you first find all the key(s) of R so that you will be able to tell whether an attribute is prime.
2.Consider relation R=(A,B,C,D) with the following FDs: B → C, C > D, and D A a. List all candidate keys of R. b. Is R in 3NF? BCNF? Justify your answer
How do we calculate a functional dependence closure? Which of the following give a good example of the process? add non-candidate keys to the dependencies, e.g. F + non-key attribute get all the inferred dependencies: if a --> b and b--> c, then we can infer a --> c is a functional dependency all the superkey dependencies should be calculated, e.g. F + (b-> Super Key) found unidentified dependencies, if a-->b and a-->c, then b-->c is a functional dependency

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