Section Mathematical Induction and Section Strong Induction and Well-Ordering

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1 Sectio Mathematical Iductio ad Sectio 4. - Strog Iductio ad Well-Orderig A very special rule of iferece! Defiitio: A set S is well ordered if every subset has a least elemet. Note: [0, 1] is ot well ordered sice (0,1] does ot have a least elemet. Examples: N is well ordered (uder the relatio) Ay coutably ifiite set ca be well ordered The least elemet i a subset is determied by a bijectio (list) which exists from N to the coutably ifiite set. Z ca be well ordered but it is ot well ordered uder the relatio (Z has o smallest elemet). The set of fiite strigs over a alphabet usig lexicographic orderig is well ordered. Let P(x) be a predicate over a well ordered set S. The problem is to prove xp(x). Prepared by: David F. McAllister TP , 007 McGraw-Hill

2 The rule of iferece called The (first) priciple of Mathematical Iductio ca sometimes be used to establish the uiversally quatified assertio. I the case that S = N, the atural umbers, the priciple has the followig form. The hypotheses are P(0) P() P( +1) xp(x) ad H1: P(0) H: P() P( +1) for arbitrary. H1 is called The Basis Step. H is called The Iductio (Iductive) Step We first prove that the predicate is true for the smallest elemet of the set S (0 if S = N). We the show if it is true for a elemet x ( if S = N) implies it is true for the ext elemet i the set ( + 1 if S = N). Prepared by: David F. McAllister TP 1999, 007 McGraw-Hill

3 The kowig it is true for the first elemet meas it must be true for the elemet followig the first or the secod elemet kowig it is true for the secod elemet implies it is true for the third ad so forth. Therefore, iductio is equivalet to modus poes applied a coutable umber of times!! It is like a row of domios: If the th domio falls over the (+1)st must fall over so pushig the first oe dow meas all must fall dow. To prove H we ormally use a Direct Proof. Assumig P() to be true for arbitrary is called the Iductio (Iductive) Hypothesis. Example: (a classic) Prepared by: David F. McAllister TP , 007 McGraw-Hill

4 Prove: i = ( +1) I logical otatio we wish to show [ i = Hece, the predicate P() is ( +1) ] i = ( +1). Note: Idetifyig P(x) is ofte the hardest part! We first prove H1: P(0): 0 = 0 i = Now establish H usig a direct proof: State the Iductio Hypotheses : Assume P() is true for arbitrary 0(0 +1) (this looks as if you are assumig the truth of what is to be proved ad hece we have a circular argumet. This is ot the case.) Now use this ad aythig else you kow to establish that P( + 1) must be true. Prepared by: David F. McAllister TP , 007 McGraw-Hill

5 P( + 1) is the assertio +1 i = ( +1)(( +1)+1) (Note: Write dow the assertio P(+1)! Do't make it hard for yourself because you do't kow what it is you are to prove.) But, +1 i = i + ( +1) i=1 usig the property of summatios. Now apply the iductio hypothesis. Note: you must maipulate the assertio P(+1) so that you ca apply the iductio hypothesis P(). If you do ot apply the iductio hypothesis somewhere, it is ot a valid iductio proof. Use the assumptio P() to substitute to get ( +1) for i +1 i = ( +1) +( +1) ad we maipulate the right side to get Prepared by: David F. McAllister TP , 007 McGraw-Hill

6 +1 i = which is exactly P(+1). Hece, we have established H. ( +1)(( +1)+1) We ow say by the Priciple of Mathematical Iductio it follows that P() is true for all or [ i = ( +1) ] Q.E.D. We ca use the Priciple to prove more geeral assertios because N is well ordered. Suppose we wish to prove for some specific iteger k x[ k P(x)] Now we merely chage the basis step to P(k) ad cotiue. Example: Show is O(). Prepared by: David F. McAllister TP , 007 McGraw-Hill

7 Proof: We must fid C ad k such that wheever k (or > k-1) C If we try C = 1, the the assertio is ot true util k = 5. Hece we prove by iductio that for all 5. The assertio becomes [ ] ad the predicate P() is Basis step: P(5): 3x5 + 5 = 0 (5) which establishes the basis step. The iductio hypothesis: assume P(): is true for arbitrary. Use this ad ay other clever thigs you kow to show P(+1). Write dow the assertio P(+1)! P(+1): 3( +1)+ 5 ( +1) Now put it i a form which will allow you to apply the iductio hypothesis. Prepared by: David F. McAllister TP , 007 McGraw-Hill

8 We rewrite the left side as (3 + 5) + 3 ad apply the iductio hypothesis to (3+5) which we assume is less tha. Now we must show that which is true iff which is true iff + 3 ( + 1) = But we have already restricted 5 so 1 must hold. Hece we have established the iductio step ad the assertio must be true for all : [ ] Q.E.D. Note: i doubly quatified assertios of the form m [P(m,)] we ofte assume m (or ) is arbitrary to elimiate a quatifier ad prove the remaiig result usig iductio. Prepared by: David F. McAllister TP , 007 McGraw-Hill

9 Aother Example: All horses are the same color. Proof: We do iductio o the size of sets of horses of the same color. Basis step: The assertio is obviously true for all sets of 0 horses (ad all sets with 1 horse). Iductio step: The iductio hypothesis becomes 'Assume the assertio is true for all sets with horses.' Now show it must be true for all sets of +1 horses. But every set of +1 horses has a overlap of horses which are the same color. + 1 horses X X X X X X X... X X horses horses Hece the set of +1 horses must have the same color. Therefore, all horses have the same color. What's wrog? Prepared by: David F. McAllister TP , 007 McGraw-Hill

10 The Secod Priciple of Mathematical Iductio The rule of iferece becomes: H1: P(0) H: P(0) P(1)... P() P( +1) xp(x) The two rules are equivalet but sometimes the secod is easier to apply. See your text for the classic examples. Prepared by: David F. McAllister TP , 007 McGraw-Hill

A New Constructive Proof of Graham's Theorem and More New Classes of Functionally Complete Functions

A New Constructive Proof of Graham's Theorem and More New Classes of Functionally Complete Functions A New Costructive Proof of Graham's Theorem ad More New Classes of Fuctioally Complete Fuctios Azhou Yag, Ph.D. Zhu-qi Lu, Ph.D. Abstract A -valued two-variable truth fuctio is called fuctioally complete,