# CLASS-11SETS - CARDINAL NUMBERS

CARDINAL NUMBER –

The number of elements in a set is often referred to as the cardinal number of the set. The cardinal number of the set A is abbreviated as n(A). Thus, we have

n(ϕ) = 0,

n{a} = 1,

n{a, b} = 2,

n{a, b, c} = 3,

n{a, b, c, d} = 4,

n{a, b, c, d, e} = 5,

and so on………… A ∩ B                               A ∩ B ∩ C

Suppose we wish to find the number of elements in the union of two sets. Let A & B, be any two sets. There are two possibilities.

Case.1) The sets are disjoint, that is A ∩ B = ϕ

In this case, obviously n(A ∪ B) = n(A) + n(B)

Case.2) The sets are not disjoint, that is, A ∩ Bϕ. In this case, n(A) + n(B) counts the common elements in both sets twice and so exceeds n(A ∪ B) by n(A ∩ B). Therefore –

n(A ∪ B) = n(A) + n(B) – n(A ∩ B)

it is possible to develop a formula similar to the above, for three or more sets, but it is easier to reply on the use of Venn Diagrams to solve such problems.

If, A, B, C are any three sets, then

n(A ∪ B ∪ C) = n(A) + n(B) + n(C) – n(A ∩ B) – n(A ∩ C) – n(B ∩ C) + n(A ∩ B ∩ C)

Also from the two intersecting sets A & B shown in the adjoining Venn Diagram, it is obvious that –

(1n(A – B) = n(A) – n(A ∩ B)

=>  n(A) = n(A – B) + n(A ∩ B)

(2n(B – A) = n(B) – n(A ∩ B)

=>  n(B) = n(B – A) + n(A ∩ B)

(3n(A ∪ B) = n(A – B) + n(B-A) + n(A ∩ B)

(4n(ξ) = n(A) + n(A’) = n(B) + n(B’), where ξ is the universal set.

There are some examples are given below for your better understanding.