pset2soln

# pset2soln - EECS 310 Fall 2011 Instructor Nicole Immorlica...

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Unformatted text preview: EECS 310, Fall 2011 Instructor: Nicole Immorlica Problem Set #2 Due: October 4, 2011 1. (25 points) Clarification: The notation p | n means that p divides n (meaning that p is a divisor of n). Our goal here is to show that for any prime p and natural numbers n,m , if p | nm , then p | n or p | m . We shall do this by using proof by least-counter-example. Note, there are three variables of interest here, and so we will need to apply the well-ordering principle twice . Let P ( p,n,m ) denote the predicate that if p | nm then p | n or p | m . Assume that the theorem is false and let F = { ( p,n,m ) | P ( p,n,m )is false } By our assumption that this theorem is false, F is a non-empty set. Using the well- ordering principle let p be the smallest p contained in any ( p,n,m ) ∈ F . Using the well-ordering principle again, let n be the smallest integer such that ( p ,n ,m ) ∈ F . That is n is the smallest integer that is paired with the smallest prime p such that P ( p ,n ,m ) is false (note the choice here of m doesn’t matter). So this means that p | n m is true but p | n is false and p | m is false. We now will prove several facts about n . (a) (1 point) Prove that n 6 = 0 and n 6 = 1 (b) (8 points) Prove that n is prime. (c) (8 points) Prove that n < p . (d) (8 points) Using these three above facts, prove this theorem. [Hint: Our assump- tion was that p was the smallest prime number that violated the predicate. Since we know that n < p and n is prime, then n cannot violate the predicate. Use this to derive a contradiction with our assumptions] Solution: (a) n 6 = . By definition, p | 0 is true, so n 6 = 0. n 6 = 1 . If this were true, then p | 1 · m → p | m which is a contradiction. (b) n is prime . Assume that n is not prime, then n = cd for some integers c and d where 1 < c,d < n . First we observe that p | n m → p | cdm . Next, since c < n , we know that p | cdm → p | c or p | dm (if this implication were not true, it would contradict our choice of n as the smallest integer which violated our predicate for p ). But p | c cannot be true because if p | c then p | cd so p | n which contradicts our choice of n . Thus it must be that p | dm . Since d < n , p | dm → p | d or p | m . But p | d cannot be true because if p | d then p | cd which means p | n which contradicts our choice of n . Finally p | m cannot be true because this contradicts that our original predicate was false. Since we reach a contradiction in all cases, our assumption that n is not prime must be false. So we conclude that n is prime. (c) n < p . Assume that n > p (it easy to see that n 6 = p ). Write d = n- p . Then md = mn- mp . Since we know that p | mn , we can write mn = kp for some integer k . Then md = kp- mp = p ( k- m ). So we get that p | md . Notice that d = n- p implies that d < n , so we know that p | md implies either p | m or p | d ....
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pset2soln - EECS 310 Fall 2011 Instructor Nicole Immorlica...

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