**Third Normal Form (3NF)**

A table will be in the third normal form (3NF) when,

1. It is in the Second Normal form.

2. And, it does not have Transitive Dependency.

According to transitive property A determines B and B determine C. So, “A” can determines C through B, but not directly.

**Let explain with Relational table**

In the following table Roll_No is the candidate key which determines ExamType and ExamType determines the MaxMarks. So the following table holds transition property. That’s why it in not in 3NF.

**Removal of Transitive Property**

Transitive property can be removed by dividing the table into its parts.

Let explain with Functional dependency example

**Conditions for Questions of 3NF**

Any functional dependencies (FD) in relation will be 3NF, if it fulfill the following condition

L.H.S of FD should be a candidate or super key

OR

R.H.S of FD is a prime attribute.

**Explanation of 3NF with FD Rules**

** Example 01:**

**Question:** Consider a relation R =(ABCD) and FD= (AB→ C, C→ D). Check Relation is in 3NF or Not?

**Solution:**

- CK ={ AB}
- Prime attribute ={ A, B}
- Non-prime Attribute= {C, D}

**According to first FD (AB→ C) .**The L.H.S of FD contains candidate key so, it is valid for 3rd NF.

**According to Second FD (C→ D).** The L.H.S of FD is not a candidate key or supper key And R.H.S is also a non-prime attribute. As both conditions false, So this FD is not valid for 3rd NF.

**Conclusion:** As above all FD’s of a relation are not valid, So relation is not in 3NF.

** Example 02:**

**Question:** Consider a relation R =(ABCD) and FD= (AB→CD, D→A). Check Relation is in 3NF or Not?

**Solution:**

- First find Candidate key (C.K) = {AB, DB}
- Second Find Prime attributes = {A, B, D}
- Third Find Non-Prime attributes = {C}

**Now check for 3NF**

- For first FD in Relation = AB→CD (As L.H.S contains the Candidate key so this FD is suitable for 3NF).
- For Second FD in Relation = D→A (As L.H.S does not contains the candidate or super key but R.H.S contains the prime attribute so this FD is suitable for 3NF.)

**Result:** As all FD’s of Relation fulfill the conditions of 3NF so this relation is in 3NF