1873_solutions

# 2 of r is an associative commutative binary operation

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Unformatted text preview: he event that m  1 S we know that S 1, 2, , m and so card S m  m  1. Suppose now that m  1 S. Using the fact that that S is a proper subset of the set 1, 2, , m  1 we choose an integer k 1, 2, , m S. Since S ß SÞ k m1 and since card S Þ k m1  m1 we see that card S  m  1. 2. Prove that if A is a finite set and B is a proper subset of A then we do not have A ß B. Using the fact that A is a nonempty finite set we choose a positive integer n such that A ß 1, 2, , n . Choose a one-one function f from A onto 1, 2, , n . Since B is a proper subset of A the set f B must be a proper subset of 1, 2, , n . If we had A ß B then we would have f B ß B ß A ß 1, 2, , n which is impossible by Exercise 1. Therefore we can’t have A ß B. 3. Prove that if A and B are finite sets and A B then card A Þ B  card A  card B . Solution: We use mathematical induction. Suppose that A is a finite set and define m  card A . For each nonnegative integer n we take p n to be the assertion that if B is any finite set disjoint from A and satisfying card B  n then the set A Þ B is finite and card A Þ B  m  n. 54 The assertion p 0 is obvious. We shall now prove that p 1 is true. Suppose that B is any set disjoint from A and satisfying card B  1. We can express B in the form u . Choose a one-one function f from A onto 1, 2, , m and extend the function f to A Þ B by defining f u  m  1. Since this extension of f is a one-one function from A Þ B onto 1, 2, , m, m  1 , we have card A Þ B  m  1. Now suppose that n is any positive integer for which the assertion p n is true and that B is a set disjoint from A and satisfying card B  n  1. Choose a member u of the set B. Since the assertion p 1 is true and u is disjoint from the finite set B u we have card B  card B u 1 which tells us that card B u  n. Therefore, since the assertion p n is true we have card A Þ B u  mn and since p 1 is true and since u is disjoint from A Þ B u we have card A Þ B  card AÞ B u Þu  mn1 4. Given that A and B are finite sets and that card A  m and card B  n, prove that A Þ B is finite and that card A Þ B m  n. The point is that A Þ B  A Þ B A and that since B A is a subset of B we have card B A card B . 5. Given that A and B are finite sets and that card A  m and card B  n, prove that A card A B  mn. B is finite and that Solution: We use mathematical induction. Suppose that A is a finite set and that card A  m. For each nonnegative integer n we take p n to be the assertion that whenever B is a finite set and card B  n then the set A B is finite and card A B  mn. The assertion p 0 is obvious because, if B is empty then A B is empty. The assertion p 1 is obvious because, if B  u then A B ß A. Now suppose that n is any positive integer for which the assertion p n is true and suppose that card B  n  1. Choose a member u of the set B. Since card B u  n we have card A B u  mn and since the assertion p 1 gives us card A u  m and since th...
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## This note was uploaded on 11/26/2012 for the course MATH 2313 taught by Professor Staff during the Fall '08 term at Texas El Paso.

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