Intersection of sets A & B has all the elements which are common to set A and set B
It is represented by symbol ∩
Let A = {1, 2, 3, 4 } , B = { 3, 4 , 5, 6}
A ∩ B = {3, 4}
The blue region is A ∩ B
Properties of Intersection
 A ∩ B = B ∩ A (Commutative law).
 (A ∩ B) ∩ C = A ∩ (B ∩ C) (Associative law).
 ∅ ∩ A = ∅, U ∩ A = A (Law of ∅ and U).
 A ∩ A = A (Idempotent law)

A ∩ (B ∪ C) = (A ∩ B) ∪ (A ∩ C) (Distributive law) i. e., ∩ distributes over ∪
or we can also write it as
A ∪ (B ∩ C) = (A ∪ B) ∩ (A ∪ C)
Let us discuss these laws
Let us take sets
Let A = {1, 2, 3, 4} , B = {3, 4, 5, 6}, C = {6, 7, 8}
and Universal set = U = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}
A ∩ B = B ∩ A (Commutative law).
A ∩ B = {1, 2, 3, 4} ∩ {3, 4, 5, 6} = {3, 4}
B ∩ A = {3, 4, 5, 6} ∩ {1, 2, 3, 4} = {3, 4}
∴ A ∩ B = B ∩ A
(A ∩ B) ∩ C = A ∩ (B ∩ C) (Associative law).
A ∩ B = {3, 4}
(A ∩ B ) ∩ C = {3, 4} ∩ {6, 7, 8} = {} = ∅
B ∩ C = {3, 4, 5, 6} ∩ {6, 7, 8} = {6}
A ∩ (B ∩ C) = {1, 2, 3, 4} ∩ {6} = {} = ∅
∴ (A ∩ B) ∩ C = A ∩ (B ∩ C)
∅ ∩ A = ∅, U ∩ A = A (Law of ∅ and U).
In intersection, we have all elements which are common
∅ ∩ A = ∅
Since ∅ has no elements, there will be no common element between ∅ and A
∴ Intersection of ∅ and A will be ∅
∅ ∩ A = {} ∩ {1, 2, 3, 4} = {}
∴ ∅ ∩ A = ∅
U ∩ A = A
Since U has all the elements, the common elements between U and A will be all the elements of set A
∴ Intersection of U and A will be A
U ∩ A = { 1 , 2, 3, 4 , 5, 6, 7, 8, 9, 10} ∩ { 1, 2, 3, 4 }
U ∩ A = {1, 2, 3, 4} = A
∴ U ∩ A = A
A ∩ A = A (Idempotent law)
A ∩ A = {1, 2, 3, 4} ∩ {1, 2, 3, 4}
A ∩ A = {1, 2, 3, 4} = A
∴ A ∩ A = A
A ∩ (B ∪ C) = ( A ∩ B) ∪ (A ∩ C) (Distributive law) i . e., ∩ distributes over ∪
B ∪ C = {3, 4, 5, 6} ∪ {6, 7, 8} = {3, 4, 5, 6, 7, 8}
A ∩ (B ∪ C) = {1, 2, 3, 4 } ∩ { 3, 4, 5, 6, 7, 8} = {3, 4}
A ∩ B = {1, 2, 3, 4} ∩ {3, 4, 5, 6} = {3, 4}
A ∩ C = {1, 2, 3, 4} ∩ {6, 7, 8} = {} = ∅
(A ∩ B) ∪ (A ∩ C) = {3, 4} ∪ ∅ = {3, 4}
∴ A ∩ (B ∪ C) = (A ∩ B) ∪ (A ∩ C)
Let's also prove
A ∪ (B ∩ C) = (A ∪ B) ∩ (A ∪ C) (Distributive Law)
B ∩ C = {3, 4, 5, 6 } ∩ { 6 , 7, 8} = {6}
A ∪ (B ∩ C) = {1, 2, 3, 4} ∪ {6} = {1, 2, 3, 4, 6}
A ∪ B = {1, 2, 3, 4} ∪ {3, 4, 5, 6} = {1, 2, 3, 4, 5, 6}
A ∪ C = {1, 2, 3, 4} ∪ {6, 7, 8} = {1, 2, 3, 4, 6, 7, 8}
(A ∪ B) ∩ (A ∪ C) = {1, 2, 3, 4, 5, 6} ∩ {1, 2, 3, 4, 6, 7, 8} = {1, 2, 3, 4, 6}
∴ A ∪ (B ∩ C) = (A ∪ B) ∩ (A ∪ C)
Let us prove distributive law using Venn Diagram
Get live Maths 1on1 Classs  Class 6 to 12