Since it is already in 3NF, why decompose further?

Question

Since the schema is already in 3NF we can keep the relations to 1, cant we? If you think the answer is different please explain how

Comments

Is the given relation even in 2NF?

---

it is in 3NF, it satisfies all the conditions

---

@h4kr

Isn’t $AB \rightarrow D$ partial dependency due to presence of $B \rightarrow D$?

---

A,B & D are PA, so B → D isn’t PD

---

@h4kr Yes right the given relation is already in 3NF

---

the answer given for the above question is 2 btw. They have divided the table into ABC and BD. Why is that so?