Descri Te Report 1
Descri Te Report 1
Descri Te Report 1
a) Roster or tabular form: In this form of representation we list all the elements of
the set within braces { } and separate them by commas.
Example: If A= set of all odd numbers less than 10 then in the roster from it can be
expressed as A={ 1,3,5,7,9}.
b) Set Builder form: In this form of representation we list the properties fulfilled by
all the elements of the set. We note as {x: x satisfies properties P}. and read as 'the set
of those entire x such that each x has properties P.'
Example: If B= {2, 4, 8, 16, 32}, then the set builder representation will be: B={x: x=2 n,
where n ∈ N and 1≤ n ≥5}
Standard notations:
x∈A x belongs to A or x is an element of set A.
Examples:
1. Let P = {k, l, m, n}
The cardinality of the set P is 4.
Example: Set R of all +ve real numbers less than 1 that can be represented by
the decimal form 0. a1,a2,a3..... Where a1 is an integer such that 0 ≤ ai ≤ 9.
3. Subsets: If every element in a set A is also an element of a set B, then A is
called a subset of B. It can be denoted as A ⊆ B. Here B is called Superset of A.
Properties of Subsets:
Every set is a subset of itself.
The Null Set i.e.∅ is a subset of every set.
If A is a subset of B and B is a subset of C, then A will be the subset of C. If
A⊂B and B⊂ C ⟹ A ⊂ C
A finite set having n elements has 2 n subsets.
4. Proper Subset: If A is a subset of B and A ≠ B then A is said to be a proper
subset of B. If A is a proper subset of B then B is not a subset of A, i.e., there is at
least one element in B which is not in A.
Example:
(i) Let A = {2, 3, 4}
B = {2, 3, 4, 5}
A is a proper subset of B.
(ii) The null ∅ is a proper subset of every set.
5. Improper Subset: If A is a subset of B and A = B, then A is said to be an
improper subset of B.
Example
(i) A = {2, 3, 4}, B = {2, 3, 4}
A is an improper subset of B.
(ii) Every set is an improper subset of itself.
6. Universal Set: If all the sets under investigations are subsets of a fixed set U,
then the set U is called Universal Set.
Example: In the human population studies the universal set consists of all the
people in the world.
7. Null Set or Empty Set: A set having no elements is called a Null set or void
set. It is denoted by∅.
9. Equal Sets: Two sets A and B are said to be equal and written as A = B if both
have the same elements. Therefore, every element which belongs to A is also an
element of the set B and every element which belongs to the set B is also an
element of the set A.
A = B ⟺ {x ϵ A ⟺ x ϵ B}.
If there is some element in set A that does not belong to set B or vice versa then
A ≠ B, i.e., A is not equal to B.
10. Equivalent Sets: If the cardinalities of two sets are equal, they are called
equivalent sets.
Example: If A= {1, 2, 6} and B= {16, 17, 22}, they are equivalent as cardinality of
A is equal to the cardinality of B. i.e. |A|=|B|=3
Example:
R = {a, b, c}
S = {k, p, m}
R and S are disjoint sets.
12. Power Sets: The power of any given set A is the set of all subsets of A and is
denoted by P (A). If A has n elements, then P (A) has 2n elements.
Example: A = {1, 2, 3}
P (A) = {∅, {1}, {2}, {3}, {1, 2}, {1, 3}, {2, 3}, {1, 2, 3}}.
Partitions of a Set:
Let S be a nonempty set. A partition of S is a subdivision of S into
nonoverlapping, nonempty subsets. Speceficially, a partition of S is a
collection {Ai} of nonempty subsets of S such that:
Each a in S belongs to one of the Ai.
The sets of {Ai} are mutually disjoint; that is,
Aj≠ Ak Then Aj ∩ Ak= ∅
The subsets in a partition are called cells.
Fig: Venn diagram of a partition of the rectangular set S of points
into five cells,A1,A2,A3,A4,A5
Venn Diagrams:
Venn diagram is a pictorial representation of sets in which an enclosed
area in the plane represents sets.
Examples:
Operations on Sets
The basic set operations are:
1. Union of Sets: Union of Sets A and B is defined to be the set of all those elements
which belong to A or B or both and is denoted by A∪B.
A∪B = {x: x ∈ A or x ∈ B}
Example: Let A = {1, 2, 3}, B= {3, 4, 5, 6}
A∪B = {1, 2, 3, 4, 5, 6}.
Operations on Sets
2. Intersection of Sets: Intersection of two sets A and B is the set of all those elements
which belong to both A and B and is denoted by A ∩ B.
A ∩ B = {x: x ∈ A and x ∈ B}
Solution:
Since, B ⊂ A ∪ B, therefore A ⊂ A ∪ A
Let x ∈ A ∪ A ⇒ x ∈ A or x ∈ A ⇒ x ∈ A
∴ A ∪ A ⊂ A
As A ∪ A ⊂ A and A ⊂ A ∪ A ⇒ A =A ∪ A. Hence Proved.
(b) A ∩ A = A
Example 2: Prove Associative Laws:
(A ∩ B) ∩ C = A ∩ (B ∩ C)
Solution:
Let some x ∈ A ∩ (B ∩ C) ⇒ x ∈ A and x ∈ B ∩ C
⇒ x ∈ A and (x ∈ B and x ∈ C) ⇒ x ∈ A and x ∈ B and x ∈ C
⇒ (x ∈ A and x ∈ B) and x ∈ C) ⇒ x ∈ A ∩ B and x ∈ C
⇒ x ∈ (A ∩ B) ∩ C.
Similarly, if some x ∈ A ∩ (B ∩ C), then x ∈ (A ∩ B) ∩ C
Thus, any x ∈ (A ∩ B) ∩ C ⇔ x ∈ A ∩ (B ∩ C).
Hence Proved.
Example3: Prove Commutative Laws
(a) A ∪ B = B ∪ A
Solution:
To Prove
A ∪ B = B ∪ A
A ∪ B = {x: x ∈ A or x ∈ B}
= {x: x ∈ B or x ∈ A}(∵ Order is not preserved in case of sets)
A ∪ B = B ∪ A. Hence Proved.
Example 4: Prove Distributive Laws
(a) A ∪ (B ∩ C) = (A ∪ B) ∩ (A ∪ C)
Solution:
To Prove
Let x ∈ A ∪ (B ∩ C) ⇒ x ∈ A or x ∈ B ∩ C
⇒ (x ∈ A or x ∈ A) or (x ∈ B and x ∈ C)
⇒ (x ∈ A or x ∈ B) and (x ∈ A or x ∈ C)
⇒ x ∈ A ∪ B and x ∈ A ∪ C
⇒ x ∈ (A ∪ B) ∩ (A ∪ C)
Therefore, A ∪ (B ∩ C) ⊂ (A ∪ B) ∩ (A ∪ C)............(i)
Again, Let y ∈ (A ∪ B) ∩ (A ∪ C) ⇒ y ∈ A ∪ B and y ∈ A ∪ C
⇒ (y ∈ A or y ∈ B) and (y ∈ A or y ∈ C)
⇒ (y ∈ A and y ∈ A) or (y ∈ B and y ∈ C)
⇒ y ∈ A or y ∈ B ∩ C
⇒ y ∈ A ∪ (B ∩ C)
Therefore, (A ∪ B) ∩ (A ∪ C) ⊂ A ∪ (B ∩ C)............(ii)
Solution:
To Prove (A ∪B)c=Ac∩ Bc
Let x ∈ (A ∪B)c ⇒ x ∉ A ∪ B (∵ a ∈ A ⇔ a ∉ Ac)
⇒ x ∉ A and x ∉ B
⇒ x ∉ Ac and x ∉ Bc
⇒ x ∉ Ac∩ Bc
Therefore, (A ∪B)c ⊂ Ac∩ Bc............. (i)
Again, let x ∈ Ac∩ Bc ⇒ x ∈ Ac and x ∈ Bc
⇒ x ∉ A and x ∉ B
⇒ x ∉ A ∪ B
⇒ x ∈ (A ∪B)c
Therefore, Ac∩ Bc ⊂ (A ∪B)c............. (ii)
Combining (i) and (ii), we get Ac∩ Bc =(A ∪B)c. Hence Proved.
Solution:
To Prove A ∪ Ac= U
Every set is a subset of U
∴ A ∪ Ac ⊂ U.................. (i)
We have to show that U ⊆ A ∪ Ac
Let x ∈ U ⇒ x ∈ A or x ∉ A
⇒ x ∈ A or x ∈ Ac ⇒ x ∈ A ∪ Ac
∴ U ⊆ A ∪ Ac................... (ii)
From (i) and (ii), we get A ∪ Ac= U. Hence Proved.
(b) A ∩ Ac=∅
Example7: Prove Complement Laws
(a) A ∪ Ac= U
Solution:
To Prove A ∪ Ac= U
Every set is a subset of U
∴ A ∪ Ac ⊂ U.................. (i)
We have to show that U ⊆ A ∪ Ac
Let x ∈ U ⇒ x ∈ A or x ∉ A
⇒ x ∈ A or x ∈ Ac ⇒ x ∈ A ∪ Ac
∴ U ⊆ A ∪ Ac................... (ii)
From (i) and (ii), we get A ∪ Ac= U. Hence Proved.
(b) A ∩ Ac=∅
Example7: Prove Complement Laws
(a) A ∪ Ac= U
Solution:
To Prove A ∪ Ac= U
Every set is a subset of U
• ∴ A ∪ Ac ⊂ U.................. (i)
• We have to show that U ⊆ A ∪ Ac
• Let x ∈ U ⇒ x ∈ A or x ∉ A
• ⇒ x ∈ A or x ∈ Ac ⇒ x ∈ A ∪ Ac
• ∴ U ⊆ A ∪ Ac................... (ii)
• From (i) and (ii), we get A ∪ Ac= U. Hence Proved.