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### Unions and Intersections

```
Date: 05/10/99 at 20:32:02
From: justin cole
Subject: Unions and intersections - their relation

In our discussion in class we showed that the number of elements in
sets A, B, and C, their unions, and their intersections, have the
following relation (I have to use u = unions and n = intersections
because those are the only keys that even come close to looking like
the symbols):

N(A u B u C) = N(A)+N(B)+N(C)-N(A n B)-N(B n C)-N(C n A)+N(A n B n C)

Generalize the formula for the four-set situation. Apply the newly
derived formula to find the numbers between (2000,3000) that are
multiples of 3, 5, 7, or 11.

You are my last hope.
Thank you.
```

```
Date: 05/11/99 at 09:17:45
From: Doctor Peterson
Subject: Re: Unions and intersections - their relation

Hi, Justin.

Let's see if we can understand the three-set formula well enough to
allow you to extend the idea. To do that, we'll get a running start by
looking first at two sets.

How are the union and intersection of two sets related? Suppose we
have two sets like this:

*********   +++++++++
***         +++         +++
**          ++   **          ++
*           +       *           +
*           +         *           +
*           +           *           +
*    A      +    AnB    *      B    +
*           +           *           +
*           +         *           +
*           +       *           +
**          ++   **          ++
***         +++         +++
*********   +++++++++

The size of the union is like the area covered by the two circles. If
I paint circle A red and circle B green, I will have covered an area
equal to the size of A and the size of B; but I will have covered the
intersection twice, once with each color. So I will have used a total
amount of paint that is:

size(A) + size(B)

which is the sum of the amount of paint in each layer:

BBBBBBBBBBBBBBBBBBBBBBBBBB
AAAAAAAAAAAAAAAAAAAAAAAA

but can also be thought of as

size(AuB) + size(AnB)

which is the sum of the amount of the bottom layer of paint
everywhere, plus the size of the top layer of paint wherever there are
two layers:

2222222222222
1111111111111111111111111111111111111

So I can calculate the size of the union by subtracting size(AnB) from
the sum of size(A) and size(B):

size(AuB) = size(A) + size(B) - size(AnB)

***+++       ***       +++
*      +     *   *     +   +     +*
*        + = *     * + +     + - +  *
*      +     *   *     +   +     +*
***+++       ***       +++

Now what happens if we add a third circle?

*********   +++++++++
***         +++         +++
**          ++   **          ++
*           +       *           +
*    A      +   AnB   *       B   +
*           +           *           +
*           + xxxxxxxxxx*           +
*          xxx          xxx         +
*       xx  +  AnBnC  *   xx      +
*     x     +       *      x    +
**  x  AnC  ++   **  BnC   x ++
*x*         +++         ++x
x *********   +++++++++  x
x                        x
x                      x
x         C          x
xx                xx
xxx          xxx
xxxxxxxxxx

Side view:

AAAAAAAAAAAAAAAAAAAAAAAA
BBBBBBBBBBBBBBBBBBBBBBBBB
CCCCCCCCCCCCCCCCCCCCCCCC

Now if I paint circle C blue, I have one layer of paint on some parts;
two layers on others; and three layers on AnBnC. In trying to get the
size of the union, if I add the sizes of A, B, and C, I will have
counted some areas twice and others three times. So I have to subtract
the size of the region where there are two colors of paint, and then
subtract the size of the triple overlap:

33333333333
22222222222222222222222
11111111111111111111111111111111111111

The area of the third layer is just AnBnC; but how can I find the area
of the second layer? If I add AnB + BnC + AnC, I will have counted
AnBnC three times. When I subtract that, I'll be taking off AnBnC two
extra times, instead of just one extra time, so I have to add it back
on. So the size of the union is

size(AuBuC) = size(A) + size(B) + size(C)
- (size(AnB) + size(BnC) + size(AnC))
+ size(AnBnC)

***+++       ***       +++
*      +     *   *     +   +
*        +   *     *   +     +     xxxx
*      +  =  *   *  +  +   +  +  x    x
x      x      ***       +++     x      x
x    x                          x    x
xxxx                            xxxx

+*
- +  *     xx       xx
+*    - + x   - x *
++     **

+xx
+  *

"layers," subtract the double "layers," add the triple "layers," and

For the final application, you'll have to consider the sets of numbers
between 2000 and 3000 that are multiples of either 3, 5, 7, or 11 as
A, B, C, and D respectively. To count the size of A, divide 2000 and
3000 by 3; this tells you that there are 1000 numbers divisible by 3
between 1 and 3000 inclusive, and 666 between 1 and 2000, so there are
334 between 2000 and 3000. Do the same for the other sets. Then, for
example, the set of numbers divisible by both 3 and 5 (AnB) is the set
of numbers divisible by 15, so you can use the same method to count
each of the intersections. Then put these results into your formula to
find the size of the union.

- Doctor Peterson, The Math Forum
http://mathforum.org/dr.math/
```
Associated Topics:
High School Sets

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