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)
