A relation from a set A to set B is nothing but a subset of the cartesian product of A and B which is denoted by AXB. The types of relations are nothing but their properties. There are different types of relations namely **reflexive, symmetric, transitive and anti symmetric** which are defined and explained as follows through real life examples.

**Reflexive relation:**

A relation R is said to be reflexive over a set A if (a,a) € R for every a € R.

**Example-1:**

If A is the set of all males in a family, then the relation “is brother of” is not reflexive over A. Because any person from the set A cannot be brother of himself.

**Example-2:**

The relation R = {(1,1)(2,2)(3,3)} is reflexive over the set A = {1,2,3}.

**Symmetric relation:**

A relation R is said to be symmetric if (a,b) € R => (b,a) € R

**Example-1:**

If A is the set of all males in a family, then the relation “is brother of” is symmetric over A.

Because if a is brother of b then b is brother of a.

**Example-2:**

If A is the set of mothers and B is the set of children in a family then a relation R on AxB is not symmetric because if a is mother of b then b cannot be mother of a.

**Transitive relation:**

A relation R is said to be symmetric if (a,b) € R, (b,c) € R => (a,c) € R.

**Example-1:**

If A is the set of all males in a family, then the relation “is brother of” is transitive over A.

Because if a is brother of b and b is brother of c then a is brother of c.

**Example-2:**

The relation R = {(1,1)(2,2)(3,3)(1,2)(2,3)} is not transitive over the set A = {1,2,3} because thought (1,2), (2,3) € R , (1,3) is not in R.

**Properties of Union of Sets**

**i) Commutative Law:** The union of two or more sets follows the commutative law i.e., if we have two sets A and B then,

A∪B=B∪A

Example: A = {a, b} and B = {b, c, d}

So, A∪B = {a, b, c, d}

B∪A = {b, c, d, a}

Since, in both the union, the group of elements is same. Therefore, it satisfies commutative law.

A ∪ B = B ∪ A

**ii) Associative Law:** The union operation follows the associative law i.e., if we have three sets A, B and C then

(A ∪ B) ∪ C = A ∪ (B ∪ C)

Example: A = {a, b} and B = {b, c, d} and C = {a, c, e}

(A ∪ B) ∪ C = {a, b, c, d} ∪ {a, c, e} = {a, b, c, d, e}

A ∪ (B ∪ C) = {a, b} ∪ {b, c, d, e} = {a, b, c, d, e}

Hence, associative law proved.

**iii) Identity Law:** The union of an empty set with any set A gives the set itself i.e.,

A ∪ ∅ = A

Suppose, A = {a, b, c} and ∅ = {}

then, A ∪ ∅ = {a, b, c} ∪ {} = {a, b, c}

**iv) Idempotent Law:** The union of any set A with itself gives the set A i.e.,

A ∪ A = A

Suppose, A = {1, 2, 3, 4, 5}

then A ∪ A = {1, 2, 3, 4, 5} ∪ {1, 2, 3, 4, 5} = {1, 2, 3, 4, 5} = A

**v) Domination Law:** The union of a universal set U with its subset A gives the universal set itself.

A ∪ U = U

Suppose, A = {1, 2, 4, 7} and U = {1, 2, 3, 4, 5, 6, 7}

then A ∪ U = {1, 2, 4, 7} ∪ {1, 2, 3, 4, 5, 6, 7} = {1, 2, 3, 4, 5, 6, 7} = U

Hence, proved.

**Intersection of Sets Properties:**

Some Properties of the Operation of Intersection are listed below:

**(i) Commutative law: A ∩ B = B ∩ A**

Consider two sets A = {1, 2, 3, 4, 5, 6} and B = {2, 3, 5, 7}.

Now, A ∩ B = {1, 2, 3, 4, 5, 6} ∩ {2, 3, 5, 7} = {2, 3, 5}

B ∩ A = {2, 3, 5, 7} ∩ {1, 2, 3, 4, 5, 6} = {2, 3, 5}

Therefore, A ∩ B = B ∩ A.

**(ii) Associative law: (A ∩ B) ∩ C = A ∩ (B ∩ C)**

Let A = {1, 2, 3, 4}, B = {3, 4, 5, 6}, and C = {5, 6, 7, 8 }.

Now,

A ∩ B = {1, 2, 3, 4} ∩ {3, 4, 5, 6} = {3, 4}

(A ∩ B) ∩ C = {3, 4} ∩ {5, 6, 7, 8} = { } = φ

And

B ∩ C = {3, 4, 5, 6} ∩ {5, 6, 7, 8} = {5, 6}

A ∩ (B ∩ C) = {1, 2, 3, 4} ∩ {5, 6} = { } = φ

Therefore, (A ∩ B) ∩ C = A ∩ (B ∩ C)

**(iii) Law of φ and U: φ ∩ A = φ, U ∩ A = A**

Consider φ = { } and A = {10, 11, 12}.

φ ∩ A = { } ∩ {10, 11, 12} = { } = φ

Let U = {2, 4, 6, 8, 10, 12, 14, 16, 18, 20} and A = {4, 8, 12, 16, 20}.

U ∩ A = {2, 4, 6, 8, 10, 12, 14, 16, 18, 20} ∩ {4, 8, 12, 16, 20} = {4, 8, 12, 16, 20} = A

**(iv) Idempotent law: A ∩ A = A**

Suppose A = {a, b, c, d, e} such that A ∩ A = {a, b, c, d, e} ∩ {a, b, c, d, e} = {a, b, c, d, e} = A

**(v) Distributive law: A ∩ (B U C) = (A ∩ B) U (A ∩ C)**, i. e., ∩ distributes over U

Let us take three sets A = = {2, 4, 6, 8}, B = {2, 3, 5, 7} and C = {3, 4, 5, 6}.

B U C = {2, 3, 5, 7} U {3, 4, 5, 6} = {2, 3, 4, 5, 6, 7}

A ∩ (B U C) = {2, 4, 6, 8} ∩ {2, 3, 4, 5, 6, 7} = {2, 4, 6}

A ∩ B = {2, 4, 6, 8} ∩ {2, 3, 5, 7} = {2}

A ∩ C = {2, 4, 6, 8} ∩ {3, 4, 5, 6} = {4, 6}

(A ∩ B) U (A ∩ C) = {2} U {4, 6} = {2, 4, 6}

Therefore, A ∩ (B U C) = (A ∩ B) U (A ∩ C)