LEARN MATH STEP BY STEP THROUGH VERY EASY PROCESS

SETS - 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 ∩ B ≠ ϕ. 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 –**

**(1) n(A – B) =
n(A) – n(A ∩ B)**

**=> n(A) = n(A – B) + n(A ∩ B)**

**(2) n(B – A) = n(B) – n(A ∩ B)**

**=> n(B) = n(B – A) + n(A ∩ B)**

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

**(4) n(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.**

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