MATHEMATICAL INDUCTION

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1 MATHEMATICAL INDUCTION. Itroductio Mthemtics distiguishes itself from the other scieces i tht it is built upo set of xioms d defiitios, o which ll subsequet theorems rely. All theorems c be derived, or proved, usig the xioms d defiitios, or usig previously estblished theorems. By cotrst, the theories i most other scieces, such s the Newtoi lws of motio i physics, re ofte built upo experimetl evidece d c ever be proved to be true. It is therefore isufficiet to rgue tht mthemticl sttemet is true simply by experimets d observtios. For istce, Fermt (60 665) cojectured tht whe is iteger greter th, the equtio x + y = z dmits o solutios i positive itegers. My ttempts by mthemticis i fidig couter-exmple (i.e. set of positive iteger solutio) eded up i filure. Despite tht, we cot coclude tht Fermt s cojecture ws true without rigorous proof. I fct, it too mthemticis more th three ceturies to fid the proof, which ws filly completed by the Eglish mthemtici Adrew Wiles i 994. To coclude or eve to cojecture tht sttemet is true merely by experimetl evidece c be dgerous. For istce, oe might cojecture tht + 4 is prime for ll turl umbers. Oe c esily verify this: whe =, + 4 = 4 is prime; whe =, + 4 = 43 is prime, d so o. Eve if oe cotiues the experimet util = 0, or eve = 0, oe would ot be ble to fid couter-exmple. However, it is esy to see tht the sttemet is wrog, for whe = 4 the expressio is equl to 4 which defiitely is ot prime. While experimetl evidece is isufficiet to gurtee the truthfuless of sttemet, it is ofte ot possible to verify the sttemet for ll possible cses either. For istce, oe might cojecture tht ( ) = for ll turl umbers. Of course oe esily verifies tht the sttemet is true for the first few (eve the first few hudreds or eve thousds of cses if oe bothers to do so) vlues of. Yet we cot coclude tht the sttemet is true. Mybe it will fil t some uttempted vlues, who ows? It is ot possible to verify the sttemet for ll possible vlues of sice there re ifiitely my of them. So how c we verify the sttemet? A powerful tool is mthemticl iductio. Pge of

2 . The Bsic Priciple Mthemticl Dtbse A logy of the priciple of mthemticl iductio is the gme of domioes. Suppose the domioes re lied up properly, so tht whe oe flls, the successive oe will lso fll. Now by pushig the first domio, the secod will fll; whe the secod flls, the third will fll; d so o. We c see tht ll domioes will ultimtely fll. So the bsic priciple of mthemticl iductio is s follows. To prove tht sttemet holds for ll positive itegers, we first verify tht it holds for =, d the we prove tht if it holds for certi turl umber, it lso holds for +. This is give i the followig. Theorem.. (Priciple of Mthemticl Iductio) Let S ( ) deote sttemet ivolvig vrible. Suppose () S () is true; () if S ( ) is true for some positive iteger, the S+ ( ) is lso true. The S ( ) is true for ll positive itegers. Exmple.. Prove tht ( ) = for ll turl umbers. We shll prove the sttemet usig mthemticl iductio. Clerly, the sttemet holds whe = sice =. Suppose the sttemet holds for some positive iteger. Tht is, ( ) =. Cosider the cse = +. By the bove ssumptio (which we shll cll the iductio hypothesis), we hve ( + ) = ( ) + (+ ) [ ] [ ] = + (+ ) = ( + ) Tht is, the sttemet holds for = + provided tht it holds for =. By the priciple of mthemticl iductio, we coclude tht ( ) = for ll turl umbers. Pge of

3 The priciple of mthemticl iductio c be used to prove wide rge of sttemets ivolvig vribles tht te discrete vlues. Some typicl exmples re show below. Exmple.. Prove tht 3 is divisible by for ll positive itegers. Clerly, 3 = is divisible by. Suppose 3 for some positive iteger. The (3 ) + = = + which is lso divisible by. It follows tht 3 is divisible by for ll positive itegers. Exmple.3. Let x > be rel umber. Prove tht ( + x) + x for ll turl umbers. Clerly, the iequlity holds for = sice ( x) ( x) Suppose ( + x) + x for some positive iteger. For the cse = +, we hve + = + + ( + x)( + x) = + ( + ) x+ x + ( + ) x + ( x) ( x) ( x) + = +. Hece, if the iequlity holds for the cse =, it lso holds for the cse = +. It follows tht ( + x) + x for ll turl umbers. (Questio: Where i the proof did we me use of the fct tht x >?) While we hve illustrted how mthemticl iductio c be used to prove certi sttemets, it should be remred tht my of these sttemets c ctully be proved without usig mthemticl iductio. There re lwys igeious wys to prove those sttemets. The use of mthemticl iductio, however, provides esy d mechicl (though sometimes tedious) wy of provig wide rge of sttemets. Pge 3 of

4 3. Vritios of the Bsic Priciple Mthemticl Dtbse There re my vritios to the priciple of mthemticl iductio. The ultimte priciple is the sme, s we hve illustrted with the exmple of domioes, but these vritios llow us to prove much wider rge of sttemets. Theorem 3.. (Priciple of Mthemticl Iductio, Vritio ) Let S ( ) deote sttemet ivolvig vrible. Suppose () S ( 0) is true for some positive iteger 0 ; () if S ( ) is true for some positive iteger 0, the S+ ( ) is lso true. The S ( ) is true for ll positive itegers 0. I some cses sttemet ivolvig vrible holds whe is lrge eough, but does ot hold whe, sy, =. I this cse Theorem. does ot pply, but the bove vritio llows us to prove the sttemet. Exmple 3.. Prove tht First, we chec tht > for ll turl umbers 5. 5 = 3> 5= 5, so the iequlity holds for = 5. Suppose > for some iteger 5. The + = > > ( + ) The lst iequlity holds sice ( ) ( ) 0 + = > wheever 5. Hece, if the iequlity holds for =, it lso holds for = +. By Theorem 3., > for ll turl umbers 5. Sometimes sequece my be defied recursively, d term my deped o some previous terms. I prticulr, it my deped o more th oe previous terms. I this cse Theorem. does ot pply becuse ssumig S ( ) holds for sigle is ot sufficiet. We eed the followig. Pge 4 of

5 Theorem 3.. (Priciple of Mthemticl Iductio, Vritio ) Let S ( ) deote sttemet ivolvig vrible. Suppose () S () d S () re true; () if S ( ) d S+ ( ) re true for some positive iteger, the S+ ( ) is lso true. The S ( ) is true for ll positive itegers. Of course there is o eed to restrict ourselves oly to two levels. Moreover, i the spirit of Theorem 3., there is o eed to strt from =. We leve the formultio s exercise. Exmple 3.. Let { } be sequece of turl umbers such tht = 5, 3 = d + = 5+ 6 for ll turl umbers. Prove tht = + 3 for ll turl umbers. We first chec tht Suppose = + 3 d The = = 5= + 3 d + = 3 = = + for some turl umber ( ) ( ) = = = + 3 Hece, if the formul holds for = d = +, it lso holds for = +. By Theorem 3., = + 3 for ll turl umbers. Sometimes to prove sttemet we eed to cosider the odd cses d eve cses seprtely. To combie them icely ito oe sigle cse, we eed the followig. Theorem 3.3. (Priciple of Mthemticl Iductio, Vritio 3) Let S ( ) deote sttemet ivolvig vrible. Suppose () S () d S () re true; () if S ( ) is true for some positive iteger, the S+ ( ) is lso true. The S ( ) is true for ll positive itegers. Pge 5 of

6 Although Theorem 3. d Theorem 3.3 loo similr, their ture is quite differet. Mthemticl Dtbse Agi there is o eed to restrict ourselves to cosiderig oly two iitil cses, but we do ot bother to go ito the detils. Exmple 3.3. Prove tht for ll turl umbers, there exist distict itegers x, y, z for which x + y + z = 4. For = d =, such itegers exist s = 4 d = 4. Suppose for = (where is some positive iteger), such itegers exist, i.e. for some distict itegers x 0, y 0 d z 0. x + y + z = The for = +, such itegers lso exist becuse (4 x ) + (4 y ) + (4 z ) = By Theorem 3.3, the result follows. I Theorem 3., we remred tht sometimes ssumptio of S ( ) for sigle my ot be sufficiet, so we my eed to ssume the sttemet holds for two vlues (d ccordigly we eed to verify two iitil cses). We lso remred tht there is o eed to restrict ourselves to oly two vlues; we could geerlize to y fiite umber of cses. The followig vritio gives further geerliztio of this, ssumig ll cses from to. Theorem 3.4. (Priciple of Mthemticl Iductio, Vritio 4) Let S ( ) deote sttemet ivolvig vrible. Suppose () S () is true; () if for some positive iteger, S (), S (),, S ( ) re ll true, the S+ ( ) is lso true. The S ( ) is true for ll positive itegers. Exmple 3.4. (APMO 999) Let,, be sequece of rel umbers stisfyig i+ j i + j for ll i, j =,, Prove tht Pge 6 of

7 for ech positive iteger. Clerly, the iequlity holds for =. Suppose the iequlity holds for =,,, for some positive iteger. The, by ddig the iequlities we get + ( ) , i.e. ( + ) ( ) = ( + ) + ( + ) + + ( + ) +. + It follows tht ( + ) ( + ) Hece , i.e. the iequlity lso holds for = +. + By Theorem 3.4, the result follows. Filly, we itroduce specil vritio, commoly ow s bcwrd iductio. Theorem 3.5. (Bcwrd Iductio) Let S ( ) deote sttemet ivolvig vrible. Suppose () S ( ) is true for ifiitely my turl umbers ; () if S ( ) is true for some positive iteger >, the S ( ) is lso true. The S ( ) is true for ll positive itegers. The most typicl exmple where bcwrd iductio is used is perhps i the proof of the AM-GM iequlity, s show i the exmple below. Pge 7 of

8 Exmple 3.5. (AM-GM iequlity) Prove tht for positive itegers,,,, I other words, the rithmetic me (AM) is lwys greter th or equl to the geometric me (GM). From ( ) +, we obti 0, i.e. the iequlity holds for =. Suppose the iequlity holds whe = for some positive iteger. Cosider the cse =. Usig the cse = d the iductio hypothesis, we hve = = 3 4 i.e. the iequlity lso holds for =. By Theorem., the iequlity holds for ll positive powers of. I other words, coditio () i Theorem 3.5 is stisfied. Agi, we suppose the iequlity holds whe Applyig the substitutio = exercise), we get i.e. the iequlity lso holds whe =. By Theorem 3.5, the iequlity is proved. = for some positive iteger, i.e. d simplifyig (the detils of which re left s Pge 8 of

9 4. Miscelleous Exmples Mthemticl Dtbse Most of the exmples we hve see del with lgebric (i)equlities d problems i umber theory. Oe should ot be misled to thi tht these re the oly res i which the method of mthemticl iductio pplies. I fct, the method is so powerful tht it is useful i lmost every brch of mthemtics. I this sectio we shll see some miscelleous exmples. Exmple 4.. Prove tht ( + ) θ θ θ siθ + si θ + + si θ = si si csc for ll positive itegers. Whe =, the right hd side is θ θ siθsi csc = siθ. So the formul holds for =. Suppose the formul holds for =, i.e. ( + ) θ θ θ siθ + si θ + + si θ = si si csc. Cosider the cse = +. By the iductio hypothesis, siθ + si θ + + si θ + si( + ) θ ( + ) θ θ θ = si si csc + si( + ) θ ( + ) θ θ θ ( + ) θ ( + ) θ = si si csc + si cos ( + ) θ θ θ θ ( + ) θ = si csc si si cos + ( + ) θ θ θ θ ( + ) θ θ ( + ) θ = si csc si + si + + si ( + ) θ θ θ ( + ) θ θ = si csc si si si + [( + ) + ] θ ( + ) θ θ = si si csc By the priciple of mthemticl iductio, the formul holds for ll positive itegers. Pge 9 of

10 Exmple 4.. Prove tht ( 3 5) ( 3 5) + + is eve iteger for ll turl umbers. Write f( ) = α + β where α = 3+ 5 d β = 3 5. It is strightforwrd to chec tht f () = 6 d f () = 8 re eve itegers. Suppose f( ) d f( + ) re both eve itegers for some positive iteger. Cosider the cse = +. Note tht α d β re roots of the equtio x 6x+ 4= 0. So α = 6α 4 d β = 6β 4, d thus + + f( + ) = α + β = α (6α 4) + β (6β 4) = α + β α + β = 6 f( + ) 4 f( ) + + 6( ) 4( ) It follows tht f( + ) must lso be eve iteger. By mthemticl iductio, we coclude tht f( ) is eve iteger for ll turl umbers. Exmple 4.3. Let > be iteger. I footbll legue there re tems. Every two tems hve plyed gist ech other exctly oce, d i ech mtch o drw is llowed. Prove tht it is possible to umber the tems,,, i wy such tht tem i bets tem i + for i =,,,. Whe there re oly two tems we simply umber the wiig tem s d the other s. Suppose such umberig is possible i the cse of tems. Cosider the cse with + tems. Te y tems, d umber them to ccordig to the requiremet. This is possible by the iductio hypothesis. Now we try to umber the ( + ) st tem. If it hs ot wo y mtch, the simply umber it s +. This is possible sice it ws bete by the tem umbered. Suppose it hs wo t lest oe mtch, d tem j is the tem with the smllest umber mogst ll the tems tht it hs bete. The we ssig the umber j to this tem, d reumber the origil Pge 0 of

11 tems j to by icresig the umber by. It is esy to see tht this umberig stisfies the requiremet. So if the cse with tems wors, the cse with + tems lso wors. By the priciple of mthemticl iductio, the origil sttemet is proved. Exmple 4.4. Prove tht, give two or more squres, oe c lwys cut them (usig oly compsses, stright edge d scissors) d reform them ito lrge squre. I the cse of two squres, we resort to the followig digrm: We leve it to the reder to wor out how the dotted lies re to be drw d to verify tht such costructios re ideed possible usig compsses d stright edge. Suppose the sttemet is true i the cse of squres. The, i the cse of + squres, we c cut of the squres to form lrge squre, ccordig to the iductio hypothesis. This leves oly two squres, but we hve show tht two squres c be cut to form oe lrge squre. By the priciple of mthemticl iductio, the sttemet is proved. Exmple 4.5. I prty there re prticipts, where is turl umber. Some prticipts she hds with other prticipts. It is ow tht there do ot exist three prticipts who hve she hds with ech other. Prove tht the totl umber of hdshes is ot more th. Whe =, the umber of hdshes is t most =. Suppose tht with people, the totl umber of hdshes is t most uder the give coditio. Cosider the cse = +, i.e. + people. Pic two people who hve she hds with ech other (if o such people exist, the the totl umber of hdshe would be zero), sy A d B. Pge of

12 Uder the iductio hypothesis, there re t most hdshes mog the other people. Now by the give coditio, oe of these people hve she hds with both A d B. So these people hve t most hdshes with A d B. Tig the hdshe betwee A d B ito ccout, the totl umber of hdshes is t most + + = ( + ). By the priciple of mthemticl iductio, the result follows. Exmple 4.6. There re ideticl crs o circulr trc. Amog ll of them, they hve just eough gs for oe cr to complete lp. Show tht there is cr which c complete lp by collectig gs from the other crs o its wy roud the trc i the clocwise directio. The cse = is trivil. Suppose the sttemet holds whe Cosider the cse = +. =. First, observe tht there is cr A which c rech the ext cr B i the clocwise directio, for otherwise the gs for ll crs will ot be ble to complete lp. Now we empty the gs of B ito A d remove A. There re crs left. By the iductio hypothesis, there is cr which c complete lp. Puttig bc cr A, this cr will lso be ble to complete the lp becuse whe it gets to cr A, the gs collected will be eough to get it to cr B. By mthemticl iductio, the sttemet is true for ll positive itegers. 5. Higher Dimesiol Iductio So fr we hve bee delig with sttemets ivolvig sigle vrible. I mthemtics, sttemets ofte ivolve more th oe vribles. With some modifictio, the priciple of mthemticl iductio c still be pplied to prove certi such sttemets. I this sectio we will see exmples of two-dimesiol iductio. Higher-dimesiol iductio c be delt with similrly. We motivted the priciple of (oe-dimesiol) mthemticl iductio usig the exmple of domioes. The trditiol domioes re lied up i stright lie, logous to oe-dimesiol iductio. I recet yers, however, more d more vritios hve bee itroduced, some hvig Pge of

13 the domioes with differet colours rrged to form ice picture. With this more complicted set-up, we eed to determie more crefully how to me sure tht ll domioes fll uder certi coditios. I the sme wy we eed to formulte rule for two-dimesiol iductio. We shll preset two differet versios. Theorem 5.. (Two-dimesiol Iductio, Versio ) Let Sm (, ) deote sttemet ivolvig two vribles, m d. Suppose () S (,) is true; () if S (,) is true for some positive iteger, the S+ (,) is lso true; (3) if Sh (, ) holds for some positive itegers h d, the Sh+ (, ) is lso true. The Sm (, ) is true for ll positive itegers m,. Theorem 5. c be esily uderstood. The first two coditios together imply (by Theorem.) tht Sm (,) is true for ll positive itegers m. Thus, fixig m, this together with coditio (3) imply (by Theorem. gi) tht Sm (, ) holds for ll positive itegers. As result, Sm (, ) holds for ll positive itegers m d, s we desire. Exmple 5.. Let f be fuctio of two vribles, with f (,) = d for ll turl umbers m d. Prove tht for ll positive itegers m d. f( m+, ) = f( m, ) + ( m+ ) f( m, + ) = f( m, ) + ( m+ ) (, ) ( ) ( ) f m = m+ m+ + We first chec tht f (, ) = = (+ ) (+ ) () +. Suppose (, ) = ( + ) ( + ) () + = + for some positive iteger. f The f( +,) = f(,) + ( + ) = ( ) ( ) [ ] [ ] = ( + ) + ( + ) + () + Thus coditios () d () i Theorem 5. re stisfied. Pge 3 of

14 Suppose f( h, ) = ( h+ ) ( h+ ) + for some positive itegers h d. The f( h, + ) = f( h, ) + ( h+ ) = ( h ) ( h ) ( h ) = h+ + h ( ) ( ) ( ) Thus coditio (3) i Theorem 5. is lso stisfied. It follows tht (, ) ( ) ( ) f m = m+ m+ + for ll positive itegers m d. Theorem 5. is essetilly pplyig Theorem. twice. The followig ltertive versio of the priciple of two-dimesiol iductio i some sese reduces two-dimesiol problem ito oe dimesio. Theorem 5.. (Two-dimesiol Iductio, Versio ) Let Sm (, ) deote sttemet ivolvig two vribles, m d. Suppose () S (,) is true; () if for some positive iteger >, Sm (, ) is true wheever m+ =, the Sm (, ) is true wheever m+ = +. The Sm (, ) is true for ll positive itegers m,. To see tht the two coditios gurtee Sm (, ) to be true for ll positive itegers m,, cosider the followig figure: m Pge 4 of

15 Ech dotted lie correspods to fixed vlue of m+. It is esy to see tht viewig m+ s sigle vrible, Theorem 5. is essetilly the sme s the bsic priciple of mthemticl iductio (Theorem.). Exmple 5.. For turl umbers p d q, the Rmsey umber R( p, q ) is defied s the smllest iteger so tht mog y people, there exist p of them who ow ech other, or there exist q of them who do t ow ech other. (We ssume tht if A ows B, the B ows A, d vice vers.) It is ow tht R( p,) = R(, q) = R( p+, q+ ) R( p, q+ ) + R( p+, q) for ll turl umbers p d q. Deduce tht for ll turl umbers p, q, R( p, q). p q C + p First, we chec tht R(, ) = = C +. Assume tht the desired iequlity holds for ll p, q with p+ q =, where is positive iteger. Now cosider R( p, q ) with p+ q = +. If either p = or q =, the desired iequlity follows immeditely. If ot, the otig tht ( p ) + q= p+ ( q ) =, the iductive hypothesis gives R( p, q) R( p, q) + R( p, q ) C + C = C. I other words, the desired iequlity holds wheever p+ q = +. By Theorem 5., the result follows. p+ q 3 p+ q 3 p+ q p p p 6. Joes d Prdoxes I this sectio we preset some iterestig exmples ivolvig mthemticl iductio. My of these exmples hve strog prdox feel. It should be oted tht some of the rgumets re ot correct, but we re ot goig to poit them out explicitly. We hope tht the reder will be ble to judge which rgumets re correct d which re ot. I cse rgumet is wrog, we hope the reder will be ble to spot where the flw lies. Pge 5 of

16 Oe c dri y mout of wter whe feelig thirsty We will prove the sttemet whe oe feels thirsty oe will be ble to dri drops of wter usig mthemticl iductio. Clerly, the sttemet holds for = becuse oe certily wts to dri some wter whe feelig thirsty. Suppose the sttemet holds for =, i.e. whe oe feels thirsty oe is ble to dri drops of wter. Cosider the cse = +. By ssumptio, whe oe feels thirsty oe is ble to dri drops of wter. Beig thirsty, oe certily is ble to dri oe more drop. So the sttemet holds for = + s well. By the priciple of mthemticl iductio, the sttemet holds for ll turl umbers. I other words, whe oe feels thirsty, oe c dri y mout of wter. I prticulr, oe is ble to swllow up ll the wter i the oces! Everythig i the world is of the sme colour We will prove the sttemet y thigs i the world re of the sme colour usig mthemticl iductio. Clerly, the sttemet holds whe = becuse ythig hs the sme colour s itself. Suppose the sttemet holds whe =, i.e. y thigs i the world re of the sme colour. Now cosider the cse = +. Te + thigs i the world d lie them up i row. By the iductio hypothesis, the first thigs re of the sme colour. By the iductio hypothesis gi, the lst thigs re lso of the sme colour. Therefore, the + thigs re of the sme colour. I other words, the sttemet holds for = + if it holds for =. By the priciple of mthemticl iductio, the sttemet holds for ll positive itegers. Tht is, everythig i the world is of the sme colour! Superpower? There re two people, ech give positive iteger, such tht the two itegers differ by. The rule of the gme requires tht if yoe ows tht if his umber is the smller mog the two, the he should put up his hd. We re goig to prove tht the oe with the smller umber will lwys put up his hd. (Note tht this essetilly mes he hs superpower, for suppose the two umbers re 003 d 004, the oe give 003 oly ows tht the other umber is 00 or 004, but should ot be ble to tell whether it is 00 or 004.) Pge 6 of

17 Suppose the two people re give positive itegers d +. We shll prove tht the oe give the iteger will put up his hd. Note tht this is true whe =, becuse is the smllest positive iteger, so the oe give the umber must ow tht his umber is the smller mog the two. Suppose the sttemet holds whe =. Cosider the cse = +. The the two people re give + d +. The oe with + will thi, The other umber is either or +. Suppose it is, the by the iductio hypothesis, the other perso should put up his hd. But he does t, so the other umber must be +. Cosequetly, the oe with + will put up his hd. By the priciple of mthemticl iductio, the result follows. Wht is the colour of my ht? (Prt ) There re people, ech beig put o ht mog ot fewer th white hts d blc hts. They the queue up i row, so tht everyoe c see oly the hts of those stdig i frot of him. Now strtig from the oe t the bc, we s the questio do you ow the colour of your ht? If the first people ll sy o, the the lst perso (the oe t the frot of the queue) must sy yes. The cse = is trivil, sice there is o blc ht. Suppose the sttemet holds for =. Cosider the cse = +. The people t the bc ll sy o, d we eed to see whether the oe t the frot c tell the colour of his ht. The oe t the frot will thi s follows. Suppose my ht is blc. The discrdig me d my blc ht from cosidertio, there re oly people with t lest white hts d blc hts. By the iductio hypothesis, sice the first people swered o, the oe behid me must sy yes. But ow he sid o. Therefore my ht cot be blc. It must be white. Hece the oe i the frot swers yes, i.e. the sttemet holds for = +. By the priciple of mthemticl iductio, the sttemet is proved. Wht is the colour of my ht? (Prt ) Now suppose the people c see ech other (so ech perso ows the colour of the ht of everybody except himself), d yoe who ows the ht o his hed eed to put up his hd. We will prove tht those with white hts will ll put up their hds. Suppose r people re put o white hts. The for r =, the oe with white ht sees blc hts. Yet there re oly blc hts. So he ows tht his ow ht is white, d he puts up his hd. Pge 7 of

18 Now suppose our ssertio is true whe people re put o white hts. Cosider the cse r = +. A perso with white ht sees other people i white ht d he will thi, If my ht is blc, the tig myself d my ht wy from cosidertio, there re people, t lest white hts d blc hts. Amog them, of them re put o white hts. By the iductio hypothesis, those with white hts should hve put up their hds. But they do t. So my ht is ot blc; it is white. Cosequetly, he puts up his hd. All others with white hts will do the sme. By the priciple of mthemticl iductio, the sttemet is proved. Two situtios revisited Go bc to the exmple where two people re give two cosecutive positive itegers. Now isted of puttig up hds t y time, suppose oe c put up hds oly t certi fixed time, sy whe light flshes, d tht the light flshes every miute. The, suppose the two people re give d +, the oe give the umber will put up his hd t the -th miute. The proof is essetilly the sme s tht give i Superpower bove, d we leve it s exercise. Similrly, i the situtio of Wht is the colour of my ht? (Prt ), we c prove tht if oe is llowed to put up his hd oly whe the light flshes, d tht if r people re put o white hts, the the people with white hts will put up their hds t the r-th miute. All professors resig The followig iterestig exmple is modified from exercise problem i Clculus by M. Spiv. There re 7 professors i the Deprtmet of Mthemtics of certi uiversity. They hve meetig every wee. There is rule syig tht if y professor fids mistes i his ow ppers, the he must resig. For log o professor hs ever resiged. This does ot me tht o professor hs ever mde y miste i their ppers. Rther, everyoe hs mde mistes, d every other professor discovers tht. I other words, everyoe ows tht every other professor hs mde mistes, but does ot ow his ow miste. O the lst meetig of yer, reserch ssistt sid, I must tell you oe thig. Amog the professors, t lest oe hs mde mistes i his ppers, d ws discovered by someoe else. Accordig to the ht colour exmple, ll professors resig t the 7th meetig of the subsequet yer. We leve the detils s exercise. Pge 8 of

19 Oe iterestig questio is tht, wht the reserch ssistt poits out is somethig tht every professor ows. Why would obody resig without his remr, d why would every professor resig fter he mde this well-ow remr? 7. Exercises. Prove by mthemticl iductio tht the followig sttemets hold for ll positive itegers. () (b) = ( + )(+ ) 6 ( + )( + )(3+ ) ( + ) = (c) 4007 is divisible by 003. (d) is divisible by (e) (f) > (g)! +! + +! = ( + )! ( + ) θ θ θ (h) cosθ + cos θ + + cos θ = si cos csc. To pply the priciple of mthemticl iductio we eed to verify two coditios, mely, the sttemet holds for =, d tht if the sttemet holds for = it lso holds for = +. C you thi of (wrog) sttemet i which the secod coditio is stisfied but the first oe is ot? Tht is, c you costruct sttemet S ( ) such tht if S ( ) true, the S+ ( ) must be true, yet S () is ot true? 3. Prove tht for ll turl umbers, Wht is the sigificce of the bove result o the covergece of the series? 4. The Lucs sequece, 3, 4, 7,, 8, 9, is defied by Pge 9 of

20 = = 3 = + for 3 Prove, by mthemticl iductio, tht < (.75) for ll positive itegers. Mthemticl Dtbse 5. From pc of 5 plyig crds oe extrcts the 6 red crds d pirs them up rdomly. The bc sides of ech pir of crds re the glued together, resultig i 3 crds with both sides beig the frot. Prove tht it is lwys possible to flip the crds so tht the 3 sides fcig upwrd re A,, 3,, 0, J, Q, K. 6. Aswer the questio t the ed of Exmple Formulte ew vritio of the priciple of mthemticl iductio by combiig Theorem 3. d Theorem Complete Exmple 3.5 by worig out the omitted detils i the lst step. 9. Complete Exmple 4.4 by worig out the omitted detils i the cse of two squres. 0. Complete the proofs of the two problems i Two situtios revisited i Sectio 6.. I the subsectio etitled All professors resig i Sectio 6, expli (usig the result i the situtio of ht colours ) why ll professors resig t the 7th meetig i the subsequet yer. Also swer the questio t the ed of the subsectio.. The Fibocci sequece is defied s x 0 = 0, x = d x+ = x+ + x for ll o-egtive itegers. Prove tht () xm = xr xm r + xrxm r for ll itegers m d 0 r m ; (b) + x d divides x d for ll positive itegers d d. Pge 0 of

21 3. Amog the my exmples i this set of otes prove usig mthemticl iductio, try to give other proof which does ot me use of mthemticl iductio. (Some my ot be possible or my be very difficult to prove without usig iductio; just try s my s you c.) Pge of