hmk3_key - COT 3100 Spring 2001 Homework #3 Solution...

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COT 3100 Spring 2001 Homework #3 Solution Assigned: Feb. 20/21 Due: March 7,8 in recitation Problem #1 With A = {1, 2, 3, 4}, let R = {(1, 1), (1, 2), (2, 3), (3, 3), (3, 4), (4, 4)} be a relation on A . Find two relations S and T on A where S T but R S = R T = {(1, 1), (1, 2), (1, 4)}. Ans. One possible solution: S = {(1, 1), (1, 2), (2, 4)}, T = {(2, 1), (2, 2), (1, 4)}. We can get (1, 1) in R S in the following ways: (1, 1)+(1, 1), (1, 2)+(2, 1) We can get (1, 2) in R S in the following ways: (1, 1)+(1, 2), (1, 2)+(2, 2) We can get (1, 4) in R S in the following ways: (1, 1)+(1, 4), (1, 2)+(2, 4) So, any combination will do the job. Problem #2. If A ={ w , x , y , z }, determine the number of relations on A that are (a) reflexive; The total number of pairs we can either include or not is 4 2 -4=16. Any reflexive relation is a subset of this set of 12 elements; we know there exist 2 12 such subsets. (b) symmetric; The number of decisions we can make for any symmetric relation is 4+(16- 4)/2=4+6=10. The number od possible symmetric relations is 2 10. (c) reflexive and symmetric; Now we are free to make (16-4)/2=6 choices about couples of pairs from distinct elements. The number of possible relations is 2 6 . (d) reflexive and contain ( x , y ); We must put 4 self-loops and ( x , y ), i. e. 5 out of total 16 pairs. We can make choices about remaining 11 pairs. This allows to make 2 11 reflexive relations that include ( x , y ). (e) symmetric and contain ( x , y ); For a symmetric relation we have 4 choices for self-loops and 5 for coupled pairs of distinct elements, i. e. 2 9 relations. (f) anti-symmetric; We can have 2 4 3 6 anti-symmetric relations. The reason is that we have 4 self-loops that we can either place in the relation or not. The choices of these are independent from choosing the other items in the relation. For the other items, you pair up each ordered pair of the form (a,b) with (b,a). There are 6 such pairs. For each pair, you may choose to NOT have either in the relation, OR to have only one of them in the relation. This is three distinct choices for each pair. Invoke the multiplication principle to get the final answer.
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(g) anti-symmetric and contain ( x , y ); There are 2 4 3 5 anti-symmetric relations that include ( x , y ). Same reasoning as above except that because you have (x,y), you have 1 choice from (x,y), (y,x), and no element to make. This leaves 5 pairs to choose from. From above, we can see that
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hmk3_key - COT 3100 Spring 2001 Homework #3 Solution...

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