**
Readings for Session 9 – (Continued)
**

**The Language of Sets — Cartesian Product**

(1, 1) |
(1, 2) |
(1, 3) |
(1, 4) |
(1, 5) |

(2, 1) |
(2, 2) |
(2, 3) |
(2, 4) |
(2, 5) |

(3, 1) |
(3, 2) |
(3, 3) |
(3, 4) |
(3, 5) |

We note that the table has 3(5) = 15 small rectangular regions. We
develop this concept in terms of a set operation that will be
used to define multiplication.

**
Ordered Pair:**
An

**
Cartesian Product:**
The

In set-builder notation,
*A*
×
*B* = {(*a,
b*) : *a*
∈
*A* and
*b*
∈ *B*}.

*
Example:*
Let
*A* = {H, T} and
*B* = {1, 2, 3, 4, 5, 6}.

*A*
× *B *
= {(H, 1), (H, 2), (H,
3), (H, 4), (H, 5), (H, 6), (T, 1), (T, 2), (T, 3), (T, 4), (T,
5), (T, 6)}

*B*
× *A *= {(1, H), (2, H), (3, H),
(4, H), (5, H), (6, H), (1, T), (2, T), (3, T), (4, T), (5, T),
(6, T)}

Note that in this case *A*
×
*B *
≠
*B*
×
*A*, i.e., the Cartesian
product is not commutative.

Also, note that
*n*(*A*)
∙ *n*(*B*) = 2(6) = 12 = *n*(*A*
× *B*).

*
Example:*
*
A*
×
∅*
*=
∅
since no ordered pairs can be formed when one of the sets
is empty.

Also, note that *n*(*A*)
∙ *n*(∅)
= 2(0) = 0 = *n*(*A*
×
∅).

*
Cartesian Product Definition for Multiplication of
Whole Numbers.***
**Let *A* and
*B* be two finite sets
with *a* =
*n*(*A*)
and *b* = *n*(*B*).
Then *ab* = *n*(*A*
´
*B*). The
numbers *a* and
*b* are called
*factors* and *ab* is the
*
product.*

Two common methods for
illustrating a Cartesian product are an array and a tree
diagram.

*Example:* A small
village has four streets and five avenues laid out in a
rectangular grid. How many intersections are there?

We have two sets,
streets (*S*) and
avenues (*A*). The elements from the two sets form a list of ordered pairs such
as the intersection of 1^{st} Street and 2^{nd}
Avenue, (1, 2). We have

4(5) = *n*(*S*)
∙
*n*(*A*)
= *n*(*S*
× *A*) = 20.

There are twenty intersections in the small town.

*Example:* In
algebra the rectangular or Cartesian coordinate plane is an
example of the Cartesian product. We consider the set of all the
ordered pairs describing locations in the plane.

*Example:* A
couple is planning their wedding. They have four nieces (Ann,
Betty, Cathy, and Deanne) and three nephews (Ed, Fred, and
Gill). How many different pairings are possible to have one boy
and one girl as a ring bearer and flower girl?

Note that this problem may be considered as either a repeated addition problem or a Cartesian product problem.

Repeated addition: Each niece may be considered to be a set that contains three nephews, so 4(3) = 3 + 3 + 3 + 3 = 12.

Cartesian product: {(A,
E), (A, F), (A, G), (B, E), (B, F), (B, G), (C, E), (C, F), (C,
G), (D, E), (D, F), (D, G)}

4(3) = *n*(nieces)
∙
*n*(nephews) =
*n*(nieces
×
nephews) = 12

**
**

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