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course Mth 151
7/19 2
Question: `q4.4.6 star operation [ [1, 3, 5, 7], [3, 1, 7, 5], [5, 7, 1, 3], [7, 5, 3, 1]]
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Your solution:
Using * to represent the operation the table is
* 1 3 5 7
1 1 3 5 7
3 3 1 7 5
5 5 7 1 3
7 7 5 3 1
the operation is closed, since all the results of the operation are from the original set {1,3,5,7}
the operation has an identity, which is 1, because when combined with any number 1 doesn't change that number. We can see this in the table because the row corresponding to 1 just repeats the numbers 1,3,5,7, as does the column beneath 1.
The operation is commutative--order doesn't matter because the table is symmetric about the main diagonal..
the operation has the inverse property because every number can be combined with another number to get the identity 1:
1 * 1 = 1 so 1 is its own inverse;
3 * 3 = 1 so 3 is its own inverse;
5 * 5 = 1 so 5 is its own inverse;
7 * 7 = 1 so 7 is its own inverse.
This property can be seen from the table because the identity 1 appears exactly once in every row.
the operation appears associative, which means that any a, b, c we have (a * b ) * c = a * ( b * c). We would have to check this for every possible combination of a, b, c but, for example, we have (1 *3) *5=3*5=7 and 1*(3*5)=1*7=7, so at least for a = 1, b = 3 and c = 5 the associative property seems to hold. **
3
Ok
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Question: `q4.4.24 a, b, c values that show that a + (b * c) not equal to (a+b) * (a+c).
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Your solution:
a + (b + c) = 2 + (5 + 7) = 2 + 12 = 14 but
(a+b) * (a+c) = (2+5) + (2+7) = 7 + 12 = 19
a = 2, b = 5 and c = 7
3
Ok
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Question: `q4.4.33 venn diagrams to show that union distributes over intersection
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Your solution:
For A U (B ^ C) we would shade all of A in addition to the part of B that overlaps C, while for (A U B) ^ (A U C) we would first shade all of A and B, then all of A and C, and our set would be described by the overlap between these two shadings. We would thus have all of A, plus the overlap between B and C. Thus the result would be the same as for A U (B ^ C).
3
Ok
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