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**COMPLEMENTARY SET**

The complement of a set A with respect to the Universal Set U is difference of U and A. Complement of set A is denoted by ≤ or (A^{C}) ≤ or (A^{¢}). Thus is the set of all the elements of the Universal Set which do not belong to the set A.

* = *U – A* = *{x: x ∈ U and x ∉ A}

we can say that A^{∪} * = *U ≤ (Universal Set) and A ∩ * = *ϕ ≤ Void Set

**Some of the useful properties/operations on sets are as follows:**

- A ∪ U = U
- A ∩ ϕ = ϕ
- ϕ
^{C}= U - U
^{C}= ϕ

**Algebra of Sets:**

**Idempotent**** Law****: **For any set A,

- A ∪ A = A
- A ∩ A = A

**Identity Law:** For any set A,

- A ∪ ϕ = A
- A ∩ U = A

**Commutative Law:** For any two sets A and B

- A ∪ B = B ∪ A
- A ∩ B = B ∩ A

**Associative Law:** For any three sets A, B and C

- ≤ A ∪ B) ∪ C = A ∪ ≤ B ∪ C)
- A ∩ ≤ B ∩ C) = ≤ A ∩ B) ∩ C

**Distributive Law: **For any three sets A, B and C

- A ∪ ≤ B ∩ C) = ≤ A ∪ B) ∩ ≤ A ∪ C)
- A ∩ ≤ B ∪ C) = ≤ A ∩ B) ∪ ≤ A ∩ C)

**De Morgan****’s Law: **For any two sets A and B

- ≤ A ∪ B)¢ = A¢ ∩ B¢
- ≤ A ∩ B)¢ = A¢ ∪ B¢

**POWER SET**

The set of all subsets of a given set A is called the power set A and is denoted by P(A).

P(A)* = *{S: S ⊆ A}

For example, if A *= *{1, 2, 3}, then

P(A) *= *{ ϕ,{1},{2},{3},{1,2},{1,3},{2,3},{1,2,3}}

Clearly, if A has n elements, then its power set P(A) contains exactly 2^{n} elements.

**Some More Results:**

- n (set of elements neither in A nor in B) = n (A¢ ∩ B¢) = n (A ∪ B)¢ = n (U) – n (A ∪ B)
n (A¢ ∪ B¢) = n (A ∩ B)¢ = n ( U) – n (A ∩ B)

n (A Δ B) = n [ (A – B) ∪ (B – A)] = n [(A ∩ B

^{¢}) ∪ (A^{¢}∩ B)] = n (A) + n (B)– 2n (A ∩ B).

The diagrams drawn to represent sets are called Venn

The diagrams drawn to represent sets are called Venn diagrams or Euler -Venn diagrams. Here we represent the universal set U by points within rectangle and the subset A of the set U represented by the interior of a circle. If a set A is a subset of a set B then the circle **1 **representing A is drawn inside the circle representing B. If A and B are no equal but they have some common elements, then to represent A and B by two intersecting circles.

*Illustration -: A class has 175 students. The following table shows the number of students studying one or more of the following subjects in this case*

Subjects | No. of students |

Mathematics | 100 |

Physics | 70 |

Chemistry | 46 |

Mathematics and Physics | 30 |

Mathematics and Chemistry | 28 |

Physics and Chemistry | 23 |

Mathematics, Physics and Chemistry | 18 |

* *

*How many students are enrolled in Mathematics alone, Physics alone and Chemistry alone? Are there students who have not offered any one of these subjects?*

** Solution: **Let P, C, M denotes the sets of students studying Physics, Chemistry and Mathematics respectively.

Let a, b, c, d, e, f, g denote the number of elements ≤ students) contained in the bounded region as shown in the diagram then

a + d + e + g = 70

c + d + f + g = 100

b + e + f + g = 46

d + g = 30

e + g = 23

f + g = 28

g = 18

after solving we get g = 18, f = 10, e = 5, d = 12, a = 35, b = 13 and c = 60

Therefore a + b + c + d + e + f + g = 153

So, the number of students who have not offered any of these three subjects = 175 –153 = 22

Number of students studying Mathematics only, c = 60

Number of students studying Physics only, a = 35

Number of students studying Chemistry only, b = 13.

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