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Unformatted text preview: MAT 300, Mathematical Structures Dr. L. Mantini Problem Set 5 Solutions Spring 2010 3.4 : 2, 6, 10, 11, 26; 3.5 : 2, 3, 6, 9, 10, 13, 27, 28; 3.6 : 2, 3, 7 3.4.2 Prove that if A B and A C then A B C . Proof : Let x A be arbitrary. We know that A B , so x B . We also know that A C , so therefore x C . But if x B and x C , it follows that x B C . This shows that A B C , so the proof is complete. 3.4.6 (4 points) Prove, for any sets A , B , and C , that A \ ( B C ) = ( A \ B ) ( A \ C ). (Use a string of equivalences.) Proof : For sets A , B , and C , we see that x A \ ( B C ) x A x / ( B C ) x A ( x B x C ) x A ( x / B x / C ) ( x A x / B ) ( x A x / C ) ( x A \ B ) ( x A \ C ) x ( A \ B ) ( A \ C ) . 3.4.10 Prove that for every natural number n , n 3 is even if and only if n is even. Proof : ( ) Assume that n is even, so that n = 2 k for some integer k . Then n 3 = 8 k 3 = 2(4 k 3 ). Since 4 k 3 is clearly an integer, this shows that n 3 is even. ( ) We prove the contrapositive statement. Assume that n is not even. Therefore n is odd. This implies that there is an integer k such that n = 2 k + 1. Therefore, n 3 = (2 k + 1) 3 = 2(4 k 3 + 6 k 2 + 3 k ) + 1. Since 4 k 3 + 6 k 2 + 3 k is clearly an integer, this shows that n 3 is odd and, therefore, not even. 3.4.11 (4 points) (a) The error in the purported proof is that the even integer m and the odd integer n are not assumed to have any other relationship to each other, so you cannot assume that m = 2 k and n = 2 k + 1. We must say that m = 2 k for some integer k and n = 2 j + 1 for some other integer j . (b) The Theorem is false: let m = 2 and n = 7. Then n 2 m 2 = 45 6 = n + m . 3.4.26 (a) Prove that for every integer n , 15  n if and only if 3  n and 5  n . Proof : ( ) Assume that 15  n . Then there is an integer q such that n = 15 q . Then n = 15 q = 5(3 q ), and 3 q is clearly an integer. This shows that 5  n ....
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 Spring '07
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