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Honors Discrete Mathematics

by: Dino Corwin

Honors Discrete Mathematics MAD 2104

Dino Corwin
GPA 3.8

Jorge Viola-Prioli

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Jorge Viola-Prioli
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This 14 page Class Notes was uploaded by Dino Corwin on Monday October 12, 2015. The Class Notes belongs to MAD 2104 at Florida Atlantic University taught by Jorge Viola-Prioli in Fall. Since its upload, it has received 13 views. For similar materials see /class/221637/mad-2104-florida-atlantic-university in Mathematics Discrete at Florida Atlantic University.

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Date Created: 10/12/15
21INDUCTION Our goal as in the previous section is to prove the validity of a list of statements one for each natural number n This time we will not resort to a proof by contradiction since the method we will discuss involves a direct proof of the form quotifA then B PRINCIPLE OF MATHEMATICAL INDUCTION Induction for short Given propositions Pn for each natural number n Step 1the Basis Step Verify that Pn is true for n 0 Step 2lnductive Hypothesis Assume Pk is true for some k Step 3 Inductive Step Prove that Pk1 is true Once the three steps have been completed Pn is true for all n J Viola Prioli Reason if Pn is not true for all n there exists a least counterexample say k By the Basis Step this k is not 0 so k 2 1 But then k 1 gt O and for k l the proposition is true Now use Step 2 to go from k l to k and conclude that Pk is true This is a contradiction that arises from assuming that Pn is not true for all n REMARKS a Observe that Induction relies on quotVerify Assume Provequot in that order b Notice that from Step 2 to Step 3 we face quotif A then Bquot where A means Pk true and B means Pk1 is true thus quotif Pk is true then Pk1 is true Therefore we will have to unravel A unravel B and produce the link J Viola Prioli c In the Inductive Hypothesis we are not assuming the validity of Pn M E but for one k Actually we are not allowed to assume that Pn holds M Q because that is precisely what we want to prove and as we know we must never assume the conclusion d This version of Induction is also called Simple Induction e Notice the analogy between the method of induction and going up or down an in nite ladder rung by rung one at a time f The notation we use in Step 2 is irrelevant we can use k s t Moreover if we decide to use in Step 2 the value k l then we must use k in Step 3 g If instead of n 2 O we are given n 2 5 say then the Basis Step is the veri cation of Pn for n 5 and not n O J Viola Prioli h Soon we will present a Strong version of Induction Now is time to analize several different applications of this important principle A Prove that V n we have 1 2 n nn12 Proof here Pn means 1 2 n nn12 Basis Step verify that P1 is true In fact both sides equal 1 Inductive Hypothesis Assume Pk is true for some k Therefore 1 2 k kk12 This is our A J Viola Prioli Inductive Steg Prove that Pk1 is true that is prove that 1 2 k k1 k1k22 This is our B Now we must go from A to B Add k1 to both sides ofA get 12kk112kk1 kk12 k1 k1 1 k2 k1k22 which shows that B is true We have nished the proof J Viola Prioli B Prove that for all natural n 7 divides 8n 1 Proof Here Pn means quot7 divides 8 1 Basis Steg Verify that Pn is true for n O This means 7 divides 80 1 that is 7 divides O which is true Inductive Hygothesis assume Pn is true for n k that is 7 divides 8k 1 This is our A Inductive Ste Prove that Pk1 is true that is 7 divides 8k1 1 This is our B Now we must go from A to B Unravel A and get 8k 1 7t for some integer t Unravel B and get 8quot 1 7w for some integer w Next multiply both sides ofA by 8 see why and get 8k1 8 56t J Viola Prioli Now with an eye to B we add 7 to both sides of and obtain 8k1 1 56t 7 78t 1 Call w 8t 1 which is an integer and arrive to B We have completed our proof C Prove that 4n 5 n2 7 if n a 6 Proof Here Pn means 4n s n2 7 if n 2 6quot Basis step verify that Pn is true if n 6 This is clear since it reads 24 s 29 Inductive Hypothesis assume for some k 2 6 is true that 4k s k2 7 This is our A Inductive Step prove that Pk1 is true that is 4k1 s k12 7 This is our B Let us go from A to B A is unraveled already whereas B means 4k4 s k2 2k 6 Observe the following inequalities that start from adding 4 to both sides of A 4k1 4k4 s k2 7 4 k2 3 s k2 6 k2 12 6 s k2 2k 6 because k gt 6 Hence 4k4 s k2 2k 6 showing that B holds true Our proof is complete J Viola Prioli D A group of n people n gt 1 know one different secret each They can communicate by email so it is onedirectional communication Prove that 2n 2 messages suf ce for all the people to know the contents of all the secrets Proof Our basis step start with n 2 It is clear that 2 messages suf ce Assume thus that the statement is true for groups of k people Let us prove it is true for a group of size k1 which means that 2k1 2 2k messages will suf ce For simplicity denote the persons by 1 2 k1 k1 emails 1 1 message Next look at 1 2 k They exchange the information they have in 2k 2 messages according to our inductive hypothesis Crucial observation now 1 2 k know aH the secrets but k1 has not received any message yet so his knowledge is the original only Next 1 emails k1 so one more message here passes the secret and all the group knows the whole secret do you see why The total number of messages sent is 1 2k 2 1 2k which is exactly what we needed to arrive to The proof is complete J Viola Prioli E Consider the sequence de ned by a0 10 and an 2 n an1 for n gt 0 Prove that the nth term an 2n n 10 for all n 2 0 Proof The statement Pn is an 2n n 10 for all nquot Basis Step take nO Then by de nition we know that a0 10 On the other hand the RHS of PO is 20 O 10 which equals 10 Inductive Hypothesis assume Pk is true for some value k That is ak 2k k 10 This is our A Inductive Step prove that Pk1 holds That is ak1 2k1k1 10 This is our B We must go from A to B We have ak1 2k1 ak by de nition 2 k1 2k k 10 by A after replacing 2k1 k1 10 after collecting terms We have arrived to B The proof is now complete J Viola Prioli F DISCUSSION PROBLEMS 1 If n a 3 the sum of the angles of an nsided polygon is rtn 2 2 Hanoi Tower 3 An array of ags We need to introduce next a variation of the Induction Principle particularly to be applied when dealing with recursive sequences IN DUCTION Strong Version Given propositions Pn for each natural number n Step 1the Basis Step Verify that Pn is true for n 0 Step 2lnductive Hypothesis Assume Pn is true for n 1 2 3 k Step 3 Inductive Step Prove that Pk1 is true Then Once the three steps have been completed Pn is true for all n J Viola Prioli Reason exactly as in the one given in the Simple Induction Observe that the only difference between the two versions lies in the Inductive Hypothesis In the strong version we actually have more information than in the simple induction case REMARK Actually as the name indicates the Strong Version is more powerful than the Simple Induction version Thus you may prefer to work only with the Strong Induction method Let us exhibit two illustrations of situations in which the Simple Induction is not useful we have to use the Strong Version FUNDAMENTAL THEOREM OF ARITHMETIC Every natural number n gt 1 is a nite product of primes Proof By Strong Induction Basis Step the proposition is true for n 2 clearly since 2 is a prime J Viola Prioli Inductive Hypothesis Assume the result holds for n 2 3 4 k This is our A To go from A to B we consider n k1 If it is a prime we are done If it is not a prime it must be composite that is n axb for some integers a andbsuchthat 1ltaltn1ltbltn Since n k1 it follows that a s k and similarly b s k By Hypothesis Inductive both a and b factor as a product of primes Thus a p1 pS and b q1 qt with all the factors prime numbers Finally k1 nagtltb p1psxq1 qt p1psq1 qtwhich is a nite product of primes as desired The proof is now complete Exercise analyze why the Simple Induction version would not yield the result J Viola Prioli A nal application is the following PROBLEM Consider the sequence de ned recursively by a0 1 a1 3 and am 8 an 15 an1 for all n gt O Prove that an 3n for all n Solution The statement Pn is an 3 for all n We observe that according to our de nition the sequence is a0 1 a1 3 a2 83 151 9 a3 89 153 72 45 27 and so on 50 actually the proposition is true for n O 1 2 and 3 Basis Step verify that PO and P1 are true We have done this Inductive Hypothesis assume Pn is true for n O1 2 3 k Hence an 3 for n 01 2 3 k This is our A J Viola Prioli Inductive Step prove that Pn is true for n k1 that is prove that am 3quot which is our B To go from A to B we have am 8 ak 15 ak1 by de nition of our sequence 8 3k 153k391 by A valid for k m k l 83kl5 3 3k391 83kl5l3k 3 3quot 3k so we arrived to B This completes our proof EXERCISE Explain why we needed the Strong Version and why two veri cations must be carried out in the Basis Step Important see the Additional Illustrations posted on line Syllabus Section 21 J Viola Prioli


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