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Assignment 4 Solutions - MATH 135 Fall 2010 Solution of...

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MATH 135, Fall 2010 Solution of Assignment #4 Problem 1 . Let a,b,c Z . Prove that if ac | bc and c 6 = 0, then a | b . Solution. Since ac | bc , then bc = qac for some integer q . Regrouping, bc - qac = 0 or c ( b - qa ) = 0. Since c 6 = 0, then b - qa = 0 or b = qa . Therefore by definition, a | b . Problem 2 . (a) Let m,n P . Prove that if m 2 6 | n 3 , then m 6 | n . (Hint: Use the contrapositive law.) (b) Show that the converse of the above statement is false; i.e., the statement that if m 6 | n , then m 2 6 | n 3 , is false. (Hint: Use a counterexample.) Solution. (a) We prove the contrapositive: “If m | n , then m 2 | n 3 .” Since m | n , then n = qm for some q Z . Thus, n 3 = ( qm ) 3 = q 3 m 3 . Since q 3 m Z , then m 2 | n 3 by definition. Since the contrapositive is true, then the original statement is true. (b) Let m = 8 and n = 4. It is easy to see that m 6 | n . However, m 2 = 64 = (4) 3 = n 3 . Thus, m 2 ± ± n 2 . Problem 3
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Assignment 4 Solutions - MATH 135 Fall 2010 Solution of...

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