Unions and IntersectionsDate: 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 + * For four sets, you'll follow the same pattern: add the single "layers," subtract the double "layers," add the triple "layers," and subtract the quadruple "layer." 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/ |
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