natural number contains 1 and also contains the successor of any number that it contains then the set is all natural numbers.

Transcription

1 2. Proof by Iductio Purpose of Sectio: To itroduce the Priciple of Mathematical Iductio, both the weak ad the strog versio, ad show how certai types of theorems ca easily be prove usig this techique. Itroductio Mathematical iductio is a method of proof ormally used to prove that a propositio is true for all atural umbers. The method is particularly importat i discrete mathematics, ad oe ofte sees theorems prove by iductio i related areas like computer sciece. The techique is so ituitive ad familiar that it sometimes is used without referece to its use. I 1889 Italia mathematicia Guiseppe Peao ( ) published a list of five axioms, based o the otio of successor that serves as a complete descriptio of the atural umber system. The first four axioms state: 1. 1 is a atural umber. 2. every atural umber has uique successor, which is a atural umber 3. o two atural umbers have the same successor 4. 1 is ot a successor for ay atural umber. Peao s th axiom is called the iductio axiom, states that if a subset of atural umber cotais 1 ad also cotais the successor of ay umber that it cotais the the set is all atural umbers. Thus axioms are sufficiet to derive all the properties of atural umbers. Which would costitute a cotet of etitle a semester course o Number Systems. Margi Note: Do t cofuse mathematical iductio with iductive reasoig associated with the atural scieces. I fact mathematical iductio is a form of deductive reasoig.

2 Mathematical iductio provides a coveiet way to establish that a statemet is true for all atural umbers 1,2, 3,. The followig statemets are prime cadidates for proof by mathematical iductio. For all atural umbers, (2 1) 2 If A is a set cotais elemets, the the collectio of all subsets of A cotai 2 elemets. Priciple of Mathematical Iductio (PMI) Defiitio: A subset S of atural umbers N is called a iductive set if it has the property that wheever S, the + 1 S. It is clear that ad N both are iductive subsets (Prove!). We ca state iductio axiom i the followig form: (PMI): If S is iductive subset of atural umbers N ad 1 S, the S N The Priciple of Mathematical Iductio is a method of proof for verifyig a propositio P(), defied for all atural umbers, is true for all atural umbers. 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. 1. Base Case: Prove that P(1) is true. 2. Iductio Step: Assume P() is true for a arbitrary, the prove P( +1) is true.

3 If oe deotes S { x N P() is true} to verify that P() is true for all atural umbers we ca us Priciple of Mathematical Iductio by: 1. Show that 1 S 2. Show that S is iductive set (that is ( )[ ( S ) (+1 S)] 3. Apply PMI S N Note: There are several; modificatios of the basic iductio proof stated here. For example, there is o reaso the base case starts with P(1). 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 (EPMI). Also, if the iductio step is replaced by the implicatio P() P( + 2), oe would coclude P() true for P(1), P(3),..., P(2 +1),... Sometimes the base case cosists i verifyig more tha oe propositio, maybe P(1), P(2) ad P(3) are required for the iitial step. Example 1: If is a positive iteger, the ( + 1) Proof: We deote P(k) as the statemet to be proved: k(k + 1) P(k): k 2 Base Case: Clearly P(1) is true sice 1(1 + 1) P(1) says 1 2 Iductio Step: We assume P() true for a arbitrary : ( + 1) (assume true) 2 Addig +1 to each side of this equatio, we get: ( + 1) ( + 1) + ( + 1) 2

4 ( + 1) + 2( +1 ) 2 ( + 1) ( + 2) 2 which is the statemet P( +1). Hece, we have prove that P( +1) is true. By iductio the result is prove. END Example 2 (Iductio i Calculus). Prove that for every atural umbers, we have write Proof: P(): d (xe x ) (x + )e dx x We show P() is true for all atural umbers by iductio. Base Step: If 1 ad from the product rule for differetiatio, we ca d(xe x ) dx Iductio Step: Assumig d (xe x ) (x + )e dx x true for a arbitrary, we compute P( +1): d ( + 1) (xe x ) dx (+1) de x x dx + ex (x + 1)e x d d (xe x ) dx ( ) dx d (x + )ex ( dx iductive assumptio) (x + )e x + e x (by product rule) (x + + 1)e x which proves P( +1). Hece the theorem is proved. END The type of iductio itroduced thus far ad preformed i Examples 1 ad 2 is called weak iductio.

5 Example 3: For ay positive iteger : Proof: x k k x + x x Lettig P() be the statemet we verify P(): 1 + x + x x x +1-1 x - 1. x +1-1 x - 1. 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(1). P(0): 1 P(1): 1 + x x x - 1 x - 1 x x x - 1 x 2-1 x - 1 x + 1 Iductio Step: Assumig P() is true for ay atural umber, we have 1 + x + x x x +1-1 x - 1. Addig x +1 to each side of this equatio, gives 1 + x + x x + x +1 x +1-1 x x +1 x x +1 (x - 1) x - 1 x x +2 - x +1 x - 1 x x - 1 which says the P( +1) is true. By iductio the theorem is prove. END

6 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 except for very simple cases, this is ot feasible. May theorems are easy to follow, but may are extremely difficult. Whe Adrew Wiles proved Fermat s Last Theorem i 1993, he preseted his results to a group of experts. No oe could verify o the spot if the proof was correct due to its legth (100 pages) ad complexity. Oly after the proof was aalyzed by a committee of six specialists over several moths was the theorem validated. Exteded Priciple of Mathematical Iductio The Priciple of Mathematical Iductio is well suited to deal with propositios that cocers to every atural umber. There are may statemets about the atural umbers that hold for almost all atural umbers (except a few cases). I these situatios we apply the Exteded Priciple of Mathematical Iductio (EPMI) which is simple cosequece of PMI. (EPMI): Let S be iductive suset of atural umbers N such that k S the { N k } S Example 4 (Iequality by Iductio) The iequality 2 > 2 holds for 3. Proof: Let P(): 2 > 2. Base Case: If 3 we have P(3) : > Iductio Step: We assume P(): 2 > 2 for a arbitrary 3. Hece > 2(2) (iductive assumptio) > (by assumptio 3 > 1) 2( + 1).

7 Hece we have prove P( +1) to be true ad by iductio we have that P() true for all 3. END Example : Prove by iductio that for ay odd umber a ad 4, divisible by a is 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 do the trick whereas strog iductio does. Ofte the iductio step of assumig oly the immediate precedig case P() is t sufficiet ad does t provide eough puch. Strog iductio is the opposite extreme: it assumes all previous cases, although i most cases oe oly eed to assume a few cases, maybe P() ad P( 1). Methodology of (Strog) Mathematical Iductio (PSMI) Let S be a set of atural umbers with the followig properties: 1) 1 S. 2) For every N, if {1, 2, 3,... } S, the + 1 S. The S N To verify a propositio P() holds for all atural umbers, the Priciple of (Strog) Mathematical Iductio cosists of successfully carryig out the followig steps. 1. Base Case: Prove that P(1) is true. 2. Iductio Step: Show that for all Ν P(1) P(2) P() P( +1).

8 Note: Rarely whe oe uses strog iductio to prove a theorem does oe actually eed to assume all P(1), P(2),..., P(). Keep i mid that if we ca prove the iductio coclusio P( +1) from a subset of previous propositios, say P( 1), P() or maybe simply P( 2), the we have verified the iductio assumptio, sice (do t forget) a weaker hypothesis implies a stroger theorem. A fudametal result i umber theory is the Fudametal Theorem of Arithmetic, which ca be prove by strog iductio. Example 6 (Proof by Strog Iductio) Every positive iteger >1 is either prime, or ca be writte as the product of prime umbers. Proof: Let deote by P() : the positive iteger >1 is either prime or ca be writte as the product of primes ad let S { x N P(x) } {1}. It is true that 1 S (eve 2 S). Suppose {1, 2, 3,... } S ad + 1 S for some atural umber. The + 1 is ot 1 ad + 1 is ot prime umber. Sice + 1 composite umber the +1 ab, where 1< a < +1 ad 1< b < +1. So, a ad b belogs to {1, 2, 3,... } ad therefore they ca be expressed as a product of primes or are prime. Sice + 1 is product of a ad b it is product of primes ad belogs to S. The cotradictio we obtaied shows that ( {1, 2, 3,... } S ) ( + 1 S) ad by the PSMI S N. END Priciple of Mathematical Iductio ad Well Orderig Priciple Aother property that characterizes the set of Natural Numbers is so called Well-Orderig Priciple (WOP) (also ofte call Least-Natural Number Priciple) which states: (WOP) Every o empty subset of atural umbers has a smallest elemet.

9 Theorem: The Well-Orderig Priciple ad the Priciple of Mathematical Iductio are two equivalet statemets. Proof: The proof of this theorem proceeds i two steps: The first PMI WOP: We start with the proof of PMI WOP. Let assume that for set of Natural umbers N the PMI holds. If T N is a subset of atural umbers with o smallest elemet the it is clear that 1 Τ ( otherwise 1 would be be the smallest elemet i T) If T is ot empty the S N \ T also ot empty (defiitely 1 S ). Next it is easy to see that S is iductive set. Let assume for some N, Τ. The, o of the umbers 1, 2,..., -1 belogs to T, otherwise oe of those umbers would be the smallest umber i T. Note that + 1 caot be i T, or else it would be the smallest i T. This imply that + 1 S. By PMI ow we ca coclude that S N, which leads to a cotradictio S T. The secod WOP PMI (left as a exercise). Iductio proofs are ofte used i computer sciece where recursive fuctios play a importat role. Let cosider very famous example of recursive sequece kow as Fiboacci sequece. Defiitio: The Fiboacci sequece is defied as follows: F 1 1, F 2 1, ad F +1 F + F -1 for all N. With the help of mathematical iductio give recurrece relatio defies Fiboacci umber for all atural umbers. Most of properties of recurretly defie Cocepts are prove via use of oe or other form of iductio. Example:7 Prove by iductio that for each N F [( 1+ ) / 2] [( 1 ) / 2]

10 Proof: Base case: F 1 [( 1+ ) / 2] [( 1 ) / 2] 1 1 [( 1+ ) ( 1 ) Iductio Step: Let assume that F [( 1 + ) / 2 ] [( 1 ) / 2 ] is true for all itegers {1, 2, 3,..., }. Next we ote that the followig two relatios are true [( 1+ ) / 2] 2 1+ ( 1+ ) / 2 ad [( 1 ) / 2] 2 1+ ( 1 ) / 2. We pressed with simplifyig [( 1+ ) / 2] [( 1 ) / 1] {[( 1+ )/ 2] ( 1+ )/ 2) [( 1 )/ 2] ( 1 )/ 2) 1 1 [( 1+ ) / 2] [ 1+ ( 1+ ) / 2) [( 1 ) / 2] [ 1+ ( 1 ) / 2)] 1 1 ( 1+ )/ 2) + ( 1+ )/ 2) {[( 1 )/ 2] + [( 1 )/ 2)] } 1 1 ( 1+ )/ 2) [( 1 )/ 2] } + {[( 1+ )/ 2] [( 1 )/ 2] } 1 1 [( 1+ ) / 2] [( 1 ) / 1] [( 1+ ) / 2] [( 1 ) / 1] + F + F F END

11 Example 8 (Solutio of a Recurrece Relatio) Suppose a sequece u 0, u 1,..., u,... is defied by the recursio relatio with iitial coditios: u + 1 3u 2u - 1 ; u 0 2, ad u 1 3. Fid the sequece u, 1, 2,... that satisfies these equatios. Solutio: Doig a few computatios we fid u 2 3u 1 2u 0, u 3 7, u 4 9, u 33 ad thus a reasoable guess would be u To show P(): u satisfies the recurrece relatio for all 0, we use strog iductio startig with iitial step 0. Base Case: P(0) : u ad P(1) : u (check). Iductio Step: Assumig P(0), P(1),..., P() true, we ca write u 2 + 1, u ad from the recurrece relatio, we have u + 1 3u 2u - 1 3(2 + 1 ) - 2( ) which proves P( +1). Hece, by strog iductio we have that P(): u satisfies the recurrece relatio for all 0. This ext example shows a variatio of the base step from previous examples. Each problem is differet ad you must adjust the iductio proof accordigly. Example 9 (Modifyig the Base Step) You are give two rulers: oe is 3 uits log, the other. Show that you ca measure ay uit distace greater tha or equal to 8 usig oly these two rulers.

12 Proof: Let P() be the propositio that ay uit legth greater tha or equal to 8 ca be measured usig the two rulers. Base Step: For the base step, we verify the first three propositios: P(8) + 3, P(9) , P(10) + Iductio Step: For arbitrary >10 we assume P(1), P(2),..., P() (or a subset thereof) ad prove P( +1). However, It is ot ecessary to assume all previous propositios, oly P( 2) sice the relatio P( +1) P( 2) + 3 is sufficiet to show that P( + 3) holds.

13 Problems 1. (Proof by Iductio) Prove by weak or strog iductio the followig propositios. [( + 1)] 2 a) b) 3 is divisible by 3 for 1. c) 2 <! (>) d) 2 > 2 ( > ) e) divides 7 2 for all positive itegers. f) ( cosθ + isiθ ) k coskθ + isikθ ( de Moivre s Formula). g) sikx k six for all atural umbers k. h) is divisible by 64 for ay atural Problem 2. a) Complete the proof of the Theorem (WOP PMI). b) Prove that PMI ad EPMI are equivalet. c) Prove that PMI ad PSMI are equivalet. Problem 3. Let F 1, F 2,..., F,... be the Fiboacci sequece. Proof by iductio that for each atural umber : a) F 1 + F 3 +F...+ F 2 1 F 2 b) F 2 + F 4 +F F 2 F 2+1-1

14 c) F 2 1+ F 2 2 +F F 2 F F +1 d) F F 2 F 2+1 e) F F 2-1 F 2 f) F +2 F F 2 +1 (-1) +1 g) F 1 F 2 + F 2 F F 2-1 F 2 F 2 2