Readings for Session 5 – (Continued)
Properties of Union and Intersection of Sets
The following set properties are
given here in preparation for the properties for addition and
multiplication in arithmetic. Note the close similarity between
these properties and their corresponding properties for addition
and multiplication.
General Properties:
A
∪
B =
B
∪
A
and
A
∩
B =
B
∩ A.
Example:
Let
A = {x
: x is a whole number
between 4 and 8} and B
= {x :
x is an even natural
number less than 10}.
Then A
∪
B = {5, 6, 7}
∪
{2, 4, 6, 8} = {2, 4, 5, 6, 7, 8} = {2, 4, 6, 8}
∪
{5, 6, 7} = B
∪
A
and
A
∩ B = {5, 6, 7}
∩ {2, 4, 6, 8} = {6} = {2, 4, 6, 8}
∩ {5, 6, 7} = B
∩ A.
Associative Properties:
The Associative Property
for Union and the
Associative Property for Intersection says that how the sets
are grouped does not change the result.
General Property: (A
∪
B)
∪
C =
A
∪
(B
∪
C) and (A
∩ B)
∩ C =
A
∩ (B
∩ C)
Example:
Let A = {a, n,
t}, B = {t, a, p},
and C = {s, a, p}.
Then
and
(A
∩ B)
∩ C = {a, t}
∩ {s, a,
p} = {a} = {a, n, t}
∩ {a, p}
= A
∩ (B
∩ C)
Identity
Property for Union:
The
Identity Property for Union says that the union of a set and
the empty set is the set, i.e., union of a set with the empty
set includes all the members of the set.
General Property:
A
∪
∅
=
∅
∪
A =
A
Example:
Let A = {3, 7, 11}
and B = {x
: x is a natural
number less than 0}.
Then A
∪
B
= {3, 7, 11}
∪
{ } = {3, 7, 11}.
General Property:
A
∩
∅
=
∅
∩ A =
∅.
Example:
Let A = {3, 7, 11}
and B = {x
: x is a natural
number less than 0}.
Then A
∩ B
= {3, 7, 11}
∩ { } = { }.
What number has a similar property
when multiplying whole numbers? What is the corresponding
property for multiplication of whole numbers?
Distributive
Properties:
The
Distributive Property of
Union over Intersection and the
Distributive Property of
Intersection over Union show two ways of finding results for
certain problems mixing the set operations of union and
intersection.
General Property:
A
∪
(B
∩ C) = (A
∪
B)
∩ (A
∪
C) and
A
∩ (B
∪
C) = (A
∩ B)
∪
(A
∩ C)
Example: Let A
= {a, n, t}, B = {t,
a, p}, and C = {s, a,
p}. Then
A
∪
(B
∩ C) = {a, n, t}
∪
{a, p}
= {p, a, n, t} = {p, a, n, t}
∩ {p, a,
n, t, s} = (A
∪
B)
∩ (A
∪
C)
A
∩ (B
∪
C)
= {a, n, t}
∩ {t, a, p, s} = {a, t} = {a, t}
∪
{a} = (A
∩ B)
∪
(A
∩ C)
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