lec_oct_12 (1)

# lec_oct_12 (1) - CS 2800 Discrete Math Oct 12 2011 Lecture...

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CS 2800: Discrete Math Oct. 12, 2011? Lecture Lecturer: John Hopcroft Proof Style Correct ideas do not guarantee solid proofs. Part of math and cs is learning how to communicate and structure arguments. To show why picking up these skills is important, consider the following problem from Assignment 6: If for a given a there exists x such that ax 1 (mod m ) , are a and m necessarily relatively prime? Prove why or why not. Here is a poorly written proof: Yes, the condition implies a and m are relatively prime. Proof. a - 1 a a - 1 bc 1 (mod m ) = a - 1 bc = ebd + 1 = b ( a - 1 c - ed ) = 1 The proof contains most of right ideas for a proof, but is completely unreadable for the following reasons: Variables are introduced with little rhyme or reason. It’s not clear what b , c , d , e are. The presence of a equals sign suggests that they are numbers, but what kind of numbers they are is still left in doubt. Whenever you use variables that are not speciﬁed in the problem itself, you have to clearly deﬁne them. Transition from one equation to another should be obvious. If it is not obvious, there should be explanations. Here equations do not clearly follow from previous equations. In particular, it’s unclear how the second line follows from the ﬁrst. It’s not clear why the last equation proves that a and m are relatively prime. Solution readers are not mind readers or codebreakers. All variables should be clearly deﬁned and all steps should be justiﬁed. The following proof is closer to what we are looking for: Proof. The proof is by contradiction. Assume that a and m are not relatively prime. Let b = gcd( a,m ) > 1. Then there exist c,d N such that a = bc and m = bd . Hence a - 1 bc a - 1 a 1 (mod m ), so we can write a - 1 bc = em + 1 for some e N . Thus a - 1 bc = ebd + 1 = b ( a - 1 c - ed ) = 1 The left hand side of the above equation is divisible by

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lec_oct_12 (1) - CS 2800 Discrete Math Oct 12 2011 Lecture...

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