Question

1):- (a)Given F = { a → b, {b, c} → d, {c, d} → e, {a, b} → d }. What is {a}⁺ (i.e. the closure of a)? Show your steps to achieve the answer for full credit.

(b) Given F = { a → b, b → c, {a, b} → d , {b, c} → a }. What is {b}⁺ (i.e. the closure of b)? Show your steps to achieve the answer for full credit.
2)(a) given F = { a → b, c → d, b → c, {a, b} → e }. What is {a}⁺ (i.e. the closure of a)? Show your steps to achieve the answer.

(b) Given R(a, b, c, d, e) with two keys, (a,b) and c, and given the following set of functional dependencies F = { {a, b} → {c, d, e}, c → {a, b, d} }.

Is R in 1NF? Justify your answer.

Is R in 2NF? Justify your answer.

Is R in 3NF? Justify your answer.

Answer 1

1)a)answer) {a}+={a,b,d}

{a}+={a,b (since a->b)

={a,b,d (since a,b->d)

{a}+={a,b,d}

1)b)answer){b}+={b,c,a,d}

{b}+={b,c since b->c

={b,c,a since {b,c}->a

={b,c,a,d since {a,b}->d

={b,c,a,d}

2)a){a}+={a,b,c,d,e}

{a}+={a,b since a->b

={a,b,e since a,b->e

={a,b,e,c since c->b

={a,b,e,c,d since c->d

={a,b,e,c,d}

2)b)answer)

keys are (a,b) and c

1nf:this is always in 1nf because atomic values present in each cell of relation

2NF: nO partial dependencies exist means part of candidate key should not determine anything.

here part of candidate key not determining anything.so it is in 2NF.

IS R in 3NF?

follow atleast one condition in below.

X->Y then

i)X is super key ii)Y is prime attribute

(a,b)->{c,d,e} here {a,b} is super key so it is in 3nf

c->{a,b,d} here c is super key so it is in 3nf

therefore the given relation R is in 3NF