COT3100Hmk02Sol

# COT3100Hmk02Sol - COT 3100 Fall 2002 Homework #2 Solutions...

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COT 3100 Fall 2002 Homework #2 Solutions 1) a) {0, 1, 4, 9, 16, 25, 36, 49, 64, 81} b) {4,5} c) {0, 6, 24} 2) a) |U – ((A B) C)| = |U| - |((A B) C)| = 50 - 40 = 10 b) |U – (A (B C))| = |U| - |(A (B C))| = 50 - 7 = 43 c) Can't be determined. Consider the two following situations described in the table below: A B C Example #1 Example #2 0 0 0 10 10 0 0 1 8 8 0 1 0 3 4 0 1 1 5 4 1 0 0 4 3 1 0 1 4 5 1 1 0 9 9 1 1 1 7 7 To read this, each row on the table corresponds to the intersection of three sets. For example, the row corresponding to 0, 1, 1 corresponds to the set ¬ A B C. Each entry in the example columns signifies the number of elements in the set designated by the row. These two examples show that |A C| could be 11 and could be 12. Since both of these are possible, a definite answer can not be determined. d) |A B| = |A| + |B| - |A B| = 24 + 24 - 16 = 32 e) |(C – A) – B| = |((A B) C)| - |A B| = 40 - 32 = 8 3) A ( (A C) B ¬ ( ¬ A

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COT3100Hmk02Sol - COT 3100 Fall 2002 Homework #2 Solutions...

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