Section 1.6: Proof by Mathematical Induction

Transcription

1 Sectio.6 Proof by Iductio Sectio.6: Proof by Mathematical Iductio Purpose of Sectio: To itroduce the Priciple of Mathematical Iductio, both weak ad the strog versios, ad show how certai types of theorems ca be prove usig this techique. Itroductio The Priciple of Mathematical Iductio is a method of proof ormally used to prove that a propositio is true for all atural umbers,,3,, although there are may variatios of the basic method. The method is particularly importat i discrete mathematics, ad oe ofte sees theorems prove by iductio i areas like computer sciece. The techique is so ituitive ad familiar that it sometimes is used without referece to its use. For eample, suppose someoe tells you they are goig to color the atural umbers,,3, with some color ad that the umber will be colored blue, ad that if a give umber is colored blue, the the et umber will also be colored blue. Is there ay doubt i your mid that all the umbers will be colored blue? Of course ot. This is the iductio aiom. Ad the good thig is you do t have to proof it. It is a aiom. I 889 Italia mathematicia Guiseppe Peao (858-93) published a list of five aioms which defie the atural umbers. Peao s 5 th aiom is called the iductio aiom, which states that ay property which belogs to ad also to the successor of ay umber which has the property belogs to all umbers.

2 Sectio.6 Proof by Iductio Margi Note: Do t cofuse mathematical iductio with iductive reasoig associated with the atural scieces. Iductive reasoig i the scieces is a scietific method whereby oe iduces geeral priciples from observatios. Mathematical iductio is ot the same thig: it is a deductive form of reasoig used to establish the validity of a propositio for all atural umbers. Mathematical iductio provides a coveiet way to establish that a statemet is true for all atural umbers,,. The followig statemets are prime cadidates for proof by mathematical iductio. For all atural umbers, ( ) = If A is a set cotais elemets, the the collectio of all subsets of A cotai elemets. 3 5 for all atural umbers Here, the is how the method of mathematical iductio works. Mathematical Iductio The Priciple of Mathematical Iductio is a method of proof for verifyig that a propositio P( ) is true for all atural umbers. The methodology for provig theorems by iductio is as follows. Methodology of Mathematical Iductio To verify that a propositio P( ) holds for all atural umbers, the Priciple of Mathematical Iductio cosists of successfully carryig out the followig two steps. i Base Case: Prove that P () is true. i Iductio Step: Assume that P( ) is true for a arbitrary, the prove that P( + ) is true. Note: There are several; modificatios of the basic iductio proof stated here. For eample, there is o reaso the base case starts with P (). If the base case is replaced by the verificatio of P( a ), where " a " is ay iteger (positive or egative), oe would coclude P( ) true for all a. Also, if the

3 Sectio.6 3 Proof by Iductio iductio step is replaced by the implicatio P( ) P( + ), oe would coclude P( ) true for P(), P(3),..., P( + ),... Also, sometimes the base case cosists i verifyig more tha oe propositio, maybe P(), P () ad P (3) are required for the iitial step (look at Eample 7 i this sectio). Margi Note: Iductio works like domios. You tip over the first domio; the first domio tips over the secod oe; the secod oe tips over the third oe; ad so o. You get the idea. Evetually, all domioes are tipped over. Theorem (Famous Idetity by Iductio I ductio) ( + ) If is a positive iteger, the =. Proof: We deote P( ) as the statemet to be proved: ( + ) P( ) : = Base Case: Clearly P() is true sice P () says () = Iductio Step: We assume P( ) true for a arbitrary : ( + ) P( ) : = (assume true) Addig + to each side of this equatio, we get: ( + ) ( + ) = + ( + ) ( + ) + ( + ) = ( + )( + ) = which is the statemet P( + ). Hece, we have prove that P( + ) is true. By iductio the result is prove. The reader ca verify that P() ad P(3) are also true, but that is t relevat to proof by iductio.

4 Sectio.6 4 Proof by Iductio Note: Do we have to prove that the priciple of mathematical iductio holds? The aswer is o. We accept mathematical iductio as a logical aiom i much the same way as we accept the classical rules of Aristotelia logic. Famous Story i Mathematics The idetity ( + ) = ca also be prove usig the idea Gauss had whe he was 9 years old ad impressed his teacher by summig = Ideed: ( ) ( ) ( ) = = 50 0= Theorem (Iductio i Calculus) Prove that for every atural umbers, we have ( e ) d P( ) : = ( + ) e d Proof: We show P( ) is true for all atural umbers by iductio. Base Step: If = ad from the product rule for differetiatio, we ca write ( ) d e d Iductio Step: Assumig true for a arbitrary, we compute d e e ( ) e d = + = +. ( e ) d P( ) : = ( + ) e d

5 Sectio.6 5 Proof by Iductio ( ) d ( ) + d e d e P( + ) : = + d d d d = ( ) e d + = + e + e ( ) ( ) = + + e (iductio assumptio) (product rule) which proves P( + ). Hece the theorem is proved. Theorem 3 (Provig Closed Forms of Series by Iductio) iteger : For ay positive k = k = 0 + Proof: Lettig P( ) be the statemet + P( ) : = we verify Base Case: I this problem the iitial step starts at = 0 due to the way P( ) is defied. It is ot ecessary, but we evaluate both P (0) ad P (). 0 + P(0) : = = = P() : + = = + Iductio Step: Assumig P( ) is true for ay atural umber, we have = Addig + to each side of this equatio, gives

6 Sectio.6 6 Proof by Iductio = ( ) = + = + = which says the P( + ) is true. By iductio the theorem is prove. Margi Note: How do we kow if the proof of a theorem is correct? It would be ice if we could feed a theorem ito a computer ad let the machie verify if the proof, but ecept for very simple cases, this is ot feasible. May theorems are easy to follow, but may are etremely difficult. Whe Adrew Wiles proved Fermat s Last Theorem i 993, he preseted his results to a group of eperts. No oe could verify o the spot if the proof was correct due to its legth (00 pages) ad compleity. Oly after the proof was aalyzed by a committee of si specialists over several moths was the theorem validated. Theorem 4 (Iequality by Iductio) The iequality > holds for 5. Proof: Defiig P( ) : > we prove: Base Case: 5 P (5) : = 3 > 5 = 5. Iductio Step: P ( ) P( ) ( ) +, 5 + for 5. We must show > = +. We write + = > (iductio hypothesis) = + + = > + + = ( + ) 5 (we assume 5) Hece P( + ) is true ad so by iductio P( ) true for all 5.

7 Sectio.6 7 Proof by Iductio Theorem 5 (Tower of Beares Theorem) The Tower of Beares 3 puzzle (or tower of Haoi) cosists movig a collectio of disks from oe peg oto aother, where oe is oly allowed to move oe disk at a time ad o larger disk ca ever be above a smaller disk. The umber of steps required to move disks from oe peg to aother peg (either oe) is. Proof: Let ( ) = the umber of moves for disks is - P We prove the iductio steps: P ( ) is true ( ) P( + ) P i P ( ) is true sice for a sigle disk it takes = step to move oe disk from oe peg to aother. i Assume P ( ) is true. That is, it takes steps to trasfer disks. We ow show it requires + steps to move + disks. Let + disks be stacked o pet A. We move the top disks to oe of the other pegs, say peg C (which takes steps by assumptio), ad the move the largest disk, still sittig o peg A, to peg B (which takes step). 3 Accordig to leged, the Temple at Beares i aciet Idia marked the ceter of the world. Withi the temple priests moved golde disks from oe diamod eedle to aother. God placed 64 gold disks o oe eedle at the time of creatio. It was said that whe the temple priests completed their task the uiverse would come to a ed. Sice it takes 64 - moves to complete the task ad assumig the priest move oe disk per secod, it would take roughly 585 billio years to move the disks from oe eedle to aother..

8 Sectio.6 8 Proof by Iductio Fially, we move the disks sittig o peg C to the top of the largest disk restig o peg B (aother steps). Hece, we have the + disks sittig o pole B i the proper arragemet. We are doe. Addig up these steps, we fid ( ) ( ) umber of steps required to move + disks = + + = = + Hece, we have prove P ( + ) is true so by iductio the umber of steps required to move disks from oe pole to aother is for ay atural umber N. Is Mathematics Based o Logic? I the late 800s ad early 900s a few mathematicias ad logicias like Dedekid, Frege, Hilbert, Russell, ad Whitehead tried to costruct arithmetic from formal logical priciples ad aioms. The Italia logicia Giuseppe Peao (858-93) formulated five aioms from which oe could deduce the atural umbers,, 3,.. This brigs up the philosophical questio of what eactly should be the startig poit of mathematics ad arithmetic? To some mathematical ituitioists like Kroecker ad Poicare, felt the atural umbers are as ituitive ad basic as oe ca get ad should act as the startig poit, rather tha beig formulated o less ituitive logical aioms. Logicias would disagree. The type of iductio discussed thus far is called weak iductio. We ow itroduce the cocept of strog iductio, although techically the two methods are equivalet. Strog Iductio The Priciple of Mathematical Iductio stated thus far is sometimes called weak iductio i cotrast to a variatio of it called strog iductio. The two types of iductio are actually equivalet but sometimes weak iductio does t fit ito the proof i a atural way whereas strog iductio does.

9 Sectio.6 9 Proof by Iductio Methodology of o f (Strog) Mathematical Iductio To verify a propositio P( ) holds for all atural umbers, the Priciple of (Strog) Mathematical Iductio cosists of successfully carryig out the followig steps.. Base Case: Prove that P () is true.. Iductio Step: Show that for all N P() P() P( ) P( + ). A fudametal result i umber theory is the Fudametal al Theorem of Arithmetic, which ca be prove by strog iductio. Note: To prove that a propositio ( ) P is true for all atural umbers does ot mea you have to use iductio, but geerally iductio is the most effective route. Also, if you use some other method i lieu of iductio, you are might be usig some fact i your proof that does require iductio. Theorem 6 (Fudametal Theorem of Arithmetic) Every atural umber ca be writte as the product of prime umbers. For eample 350 = 5 7, 9 = Proof: Deotig we prove: P( ) = ca be writte as the product of prime umbers Base Case: P () holds sice is a prime umber. Iductio Step: For a arbitrary =,3,... we prove P() P(3) P( ) P( + ). Assume P(), P(3), P( ) is true. That is, each atural umber,3, ca be writte as the product of primes. We ow cosider two cases.

10 Sectio.6 0 Proof by Iductio Case : Suppose + is a prime umber. I this case P( + ) is true ad thus the iductio step is prove sice ay prime umber + ca be writte as a product of primes, amely + = +. Case : Suppose + is ot prime, which meas we ca factor + = qp, where clearly both of the factors p, q must be less tha + ad greater ad or equal to. (For eample if + = 5, we would could factor 5 = 5 3 where both factors 3,5 are less tha 5 ad greater tha or equal to ). But sice p, q are less tha P k are true for +, the iductio hypothesis (all ( ) k < + ) states they ca both be writte as the product of primes, say p = p p p q = q q q m But if both p ad q ca be writte as the product of primes so ca ( )( ) + = pq = p p p q q q. m Hece P( + ) is true ad so by the priciple of strog iductio P( ) is true for all. History of Mathematical Iductio Although some elemets of mathematical iductio have bee hited at sice the time of Euclid, oe of the oldest argumet usig iductio goes back to the Italia mathematicia Fracesco Maurolico, who used iductio i 575 to prove that the sum of the first odd atural umbers is. The method was later discovered idepedetly by the Swiss mathematicia Joh Beroulli, ad Frech mathematicias Blaise Pascal (63-66) ad Pierre de Fermat (60-665). Fially, i 889 the Italia logicia Guiseppe Peao (858-93) laid out five aioms for deducig the atural umbers, of which his fifth aiom was the Priciple of Mathematical Iductio. Hece, from that poit of view iductio is a formal aiom of arithmetic. Theorem 7 (Solutio of a Recurrece Relatio) Suppose a sequece u, u,..., u,... is defied by the recursio relatio with iitial coditios: 0 u+ = 3u u u0 =, u = 3 Fid the sequece u, =,,... that satisfies these equatios. Doig a few computatios we fid u = 3u u0 = 5, u 3 = 7, u 4 = 9, u 5 = 33 ad thus a reasoable guess would be u = +. To show P( ) : u = + satisfies the

11 Sectio.6 Proof by Iductio recurrece relatio for all 0, we use strog iductio startig with iitial step = 0.. Base Case: P u = + = (check) 0 (0) : 0 3 Iductio Step: Assumig P(0), P(),..., P( ) true, we ca write u = + = +, u ad from the recurrece relatio, we have u = 3u u + ( ) ( ) = = 3 + = + + which proves P( + ). Hece, by strog iductio we have that P( ) : u = + satisfies the recurrece relatio for all 0. This et eample shows a variatio of the base step from previous eamples. Each problem is differet ad you must adjust the iductio proof accordigly. Theorem 8 (Modifyig the Base Step) You are give two rulers: oe is 3 uits log, the other 5. Show that you ca measure ay uit distace greater tha or equal to 8 usig oly these two rulers. Proof: Let P( ) be the propositio ( ) = ay iteger legth 8 ca be measured with rulers of legth 3 ad 5 P It is useful to see the followig patter that develops.

12 Sectio.6 Proof by Iductio P(8) = P(9) = P(0) = ( ) ( ) ( ) P() = P(8) + 3 = P() = P(9) + 3 = P(3) = P(0) + 3 = This patter will serve as a aid i decidig the base ad iductio steps which is ofte the most difficult part i a iductio proof. Base Step: For the base step, we verify the first three propositios: ( ) P(8) = 5 + 3, P(9) = , P 0 = Iductio Step: We ow prove the iductio step ( ) P(8), P(9),..., P( ) P +, 0 To prove this we make the simple observatio that if P ( ) is true (i.e. a legth of ca be measured with rulers of legth 3 ad 5), the P ( + ) is also true sice a legth of + is 3 uits loger tha. [ For eample P ( 9) = ad P( ) = ( ) + 3. Hece, we kow ( ) P ( 8) is true, ad P ( ) is true sice ( 9) P is true sice P is true, ad so o. ] Hece if P(8), P(9),..., P( ) 0 is true so is P ( + ) ad so by iductio P ( ) is true for all atural umbers. There are several variatios of the basic method of mathematical iductio. Oe such variatio is double iductio. Double Iductio Sometimes we would like to prove a propositio P ( m, ) ivolvig two atural umbers m by iteratig the iductive process. This is doe by performig a iductio o oe of the variables, say m, ad the a iductio

13 Sectio.6 3 Proof by Iductio o. This strategy is called double iductio ad is carried out by the followig steps. ( ) ( ) P ( m + ) ( ) ( + ). Prove P, is true. Prove P m,, 3. Prove P m, P m, for all atural umbers m You ca thik of double iductio as provig P ( m, ) at all poits i the first quadrat of the y -plae with iteger coordiates.

14 Sectio.6 4 Proof by Iductio Problems. (Proof by Iductio) propositios. Prove by weak or strog iductio the followig a) b) ( + )( + ) = 6 3 is divisible by 3 for. c) ( a + ib) = ( + )( a + b) i= 0 d) <! ( 4) u = u +, u = 5 has the solutio u = + 3 for. f) g) ( 4k 3) = ( ) k = h) 5 divides 7 for all positive itegers. i) > cosθ + i siθ = cos θ + isi θ (De Moivre s Theorem) j) ( ) k) si si l) a b a b < < ( a, b positive real umbers) d d m) = ( ) ( ) + l!. (Graphical Iductio) Draw some lies i the plae as show i Figure ad color adjoiig regios either red or blue. Show that it is possible to draw a arbitrary umber of lies i such a way that two adjoiig regios are ot

15 Sectio.6 5 Proof by Iductio colored the same color. Hit: Defie P( ) as the propositio it is possible to draw lies i the desired maer for arbitrary N. 3. (Clever Mary) To prove the idetity k = 0 Figure ( + ) k = Mary evaluated the left-had side of the equatio for = 0,, gettig. 0 F() 0 3 She the fit a polyomial to these three poits, gettig ( + ) f ( ) =. Mary tured this ito her professor. Is her proof 4 valid? 4 This problem is based o a problem i the book A = B by Marko Petkovsek, Doro Zeilberger ad Herbert Wilf. (This amazig book, icidetally, ca be dowloaded free o the iteret.)

16 Sectio.6 6 Proof by Iductio 4. (Proof Proofs without Words) They say a good picture is worth a thousad words, but i mathematics it might be closer to a millio. For the figures below, describe why the figure provides a visual proof of the statemet. a) 3 a + b = c b) ( + ) = c) ( ) = d) y = y + y ( )( )

17 Sectio.6 7 Proof by Iductio e) y y y = ( )( + ) f) a + b ab g) a a a + = + p / q q / p h) ( ) 0 t + t dt =