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pset4_soln

# pset4_soln - 6.042/18.062J Mathematics for Computer Science...

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6.042/18.062J Mathematics for Computer Science February 22, 2005 Srini Devadas and Eric Lehman Problem Set 4 Solutions Due: Monday, February 28 at 9 PM Problem 1. Prove all of the following statements except for the two that are false; for those, provide counterexamples. Assume n > 1 . When proving each statement, you may assume all its predecessors. (a) a a (mod n ) Solution. Every number divides zero, so n ( a a ) , which means a a (mod n ) . | (b) a b (mod n ) implies b a (mod n ) Solution. The statement a b (mod n ) implies n ( a b ) , which means there is an integer k such that nk = a b . Thus, n ( k ) = b a | , so n ( b a ) as well. This means | b a (mod n ) . (c) a b (mod n ) and b c (mod n ) implies a c (mod n ) Solution. The two assumptions imply n ( a b ) and n ( b c ) . Thus, n divides the | | linear combination ( a b ) + ( b c ) = a c as well. This means n ( a c ) . | (d) a b (mod n ) implies a + c b + c (mod n ) Solution. The first statement implies n | ( a b ) . Rewriting the right side gives n ( a + c ) ( b + c ) , which means a + c b + c (mod n ) . | (e) a b (mod n ) implies ac bc (mod n ) Solution. The first statement implies n ( a b ) . Thus, n also divides c ( a b ) = ac bc . Therefore, ac bc (mod n ) . | (f) ac bc (mod n ) implies a b (mod n ) provided c �≡ 0 (mod n ) .

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