Database - The Relational Model and Normalization - Discussion
Discussion Forum : The Relational Model and Normalization - General Questions (Q.No. 19)
19.
If attribute A determines both attributes B and C, then it is also true that:
Discussion:
22 comments Page 2 of 3.
Vahini said:
1 decade ago
I think answer is correct. By splitting A -> (B, C) to A ->B, A ->C.
Anubhav Jain said:
1 decade ago
Why you say the answer A -> B. The answer may be right as (B, C) --> A because it shows the functioal independencies.
Tarun V.Nair said:
1 decade ago
Its simple A->BC, so via split rule we can say A->B, A->C so option a is correct.
M,Ajay said:
1 decade ago
From hypothesis we can write A-> (BC) ,
According to the decomposition rule of functional dependency (FD) , we can write the above FD as A->B and A->C. So according to given options A->B is the correct answer.
According to the decomposition rule of functional dependency (FD) , we can write the above FD as A->B and A->C. So according to given options A->B is the correct answer.
Hemant said:
1 decade ago
In simple words it can be said that if from a bigger set of entries i.e. A we can take out B and similarly C, but it is not necessary we obtain A from both B and c. Vice versa not true.
Kamlesh said:
1 decade ago
* Subset Property (Axiom of Reflexivity): If Y is a subset of X, then X → Y
* Augmentation (Axiom of Augmentation): If X → Y, then XZ → YZ
* Transitivity (Axiom of Transitivity): If X → Y and Y → Z, then X → Z
Hence the answer A->B is perfect, any option of A->B or A->c is right, but there is no A->c option is available, so A->B is the right answer.
* Augmentation (Axiom of Augmentation): If X → Y, then XZ → YZ
* Transitivity (Axiom of Transitivity): If X → Y and Y → Z, then X → Z
Hence the answer A->B is perfect, any option of A->B or A->c is right, but there is no A->c option is available, so A->B is the right answer.
Jyothi said:
1 decade ago
Its true priyanka .
I think it might b different ans other than A->B, because its alreday mentioned na that A->B & A->C.Then how the ans can b again A->B
I think it might b different ans other than A->B, because its alreday mentioned na that A->B & A->C.Then how the ans can b again A->B
Deepak Kr. Malakar said:
1 decade ago
I am Not Understand Your Answer.
Why A->B
Why not Other
Why A->B
Why not Other
Priyanka said:
1 decade ago
I think that functional dependency do not say that if A->B then B->A is also true. For instance, if the roll_no of a student determines the name of that student then it is not true that the name of the student will recognize the roll_no of that particular student.
Swati said:
1 decade ago
Still not getting about the answer.
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