Math 4707 Spring 2018 (Darij Grinberg): midterm 2 page 1. Math 4707 Spring 2018 (Darij Grinberg): midterm 2 with solutions [preliminary version]

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1 Math 4707 Sprig 08 Darij Griberg: idter page Math 4707 Sprig 08 Darij Griberg: idter with solutios [preliiary versio] Cotets 0.. Coutig first-eve tuples Coutig legal paths geeralizatio of Catala ubers Scary fractios Derageets that are ivolutios Hypergree perutatios Coutig the parts of all copositios Recall the followig: If N, the [] deotes the -eleet set {,,..., }. We use the Iverso bracet otatio. Also, here is a collectio of idetities that we shall use: We have for every N ad N satisfyig <. We have 0 for ay Z ad Z. This is the recurrece relatio of the bioial coefficiets. We have for ay N ad N satisfyig. 3 We have for ay N ad N. This follows by applyig 3 to ad istead of ad. 4

2 Math 4707 Sprig 08 Darij Griberg: idter page We have!!! for ay N ad N satisfyig. This is [Gribe6, Propositio.5]. We have for ay N ad N satisfyig. This follows iediately fro 5, sice! 0. We have for ay Q ad {,, 3,...}. This is the absorptio idetity, ad follows by substitutig for X i [Gribe6, Propositio.3]. Also, it is very easy to chec directly. 7 We have for ay Q ad {,, 3,...}. This is easy to chec. 8 Proof of 8: Let Q ad {,, 3,...}. The, 0 sice {,, 3,...}, so that the fractio is well-defied. We have {,, 3,...}; i other words, is a positive iteger. Hece, ad!!. We have {,, 3,...} N. Thus, the defiitio of yields!! sice ad!!. 9! yields Moreover, N sice {,, 3,...}, so that the defiitio of!! sice.

3 Math 4707 Sprig 08 Darij Griberg: idter page 3 Every N satisfies 0. 0 This is Corollary.6b i the classwor fro Jauary 08, or [Gribe6, Propositio.33 b]. Every N satisfies [ 0]. 0 This is Corollary 3.3 i the classwor fro 4 February 08, or [Gribe6, Propositio.33 c]. If N ad N, ad if S is a -eleet set, the is the uber of all -eleet subsets of S. This is the cobiatorial iterpretatio of the bioial coefficiets. 0.. Coutig first-eve tuples Exercise. Let ad d be two positive itegers. A -tuple x, x,..., x [d] will be called first-eve if its first etry x occurs i it a eve uber of ties i.e., the uber of i [] satisfyig x i x is eve. For exaple, the 3-tuples, 5, ad,, 3 are first-eve, while the 3-tuple 4,, is ot. Prove that the uber of first-eve -tuples i [d] is d d d. Our solutio for this exercise will rely o the followig defiitio: Defiitio 0.. Let N ad d N. Let [d]. A -tuple x, x,..., x [d] will be called -eve if the uber occurs i it a eve uber of ties i.e., the uber of i [] satisfyig x i is eve. For exaple, the 3-tuple, 4, 4 is 4-eve ad 3-eve but ot -eve. Multiplyig this equality by, we obtai.! Coparig this with 9, we obtai. This proves 8.

4 Math 4707 Sprig 08 Darij Griberg: idter page 4 This defiitio geeralizes the cocept of -eve defied i Hoewor set 3. Exercise 5 o Hoewor set 3 claied the followig: Propositio 0.. Let N, ad let d be a positive iteger. The, the uber of -eve -tuples i [d] is d d. The sae arguet proves the followig: Propositio 0.3. Let N, ad let d be a positive iteger. Let [d]. The, the uber of -eve -tuples i [d] is d d. Ideed, Propositio 0. is the particular case of Propositio 0.3 for ; but coversely, Propositio 0.3 ca be derived fro Propositio 0. by reaig as. To ae this rigorous, you ca argue as follows: Proof of Propositio 0.3 setched. There is clearly soe perutatio σ S d such that σ. For exaple, we ca let σ be the traspositio swappig with whe, ad otherwise we ca just set σ id. Fix such a σ. The, there is a bijectio { -eve -tuples i [d] } { -eve -tuples i [d] }, x, x,..., x σ x, σ x,..., σ x. This is well-defied, because the occurreces of the uber i a -tuple x, x,..., x [d] clearly correspod to the occurreces of the uber σ i the -tuple σ x, σ x,..., σ x. This bijectio shows that { -eve -tuples i [d] } { -eve -tuples i [d] } the uber of -eve -tuples i [d] d d by Propositio 0.. I other words, the uber of -eve -tuples i [d] is 0.3. Solutio to Exercise setched. We first ae the followig clai: d d. This proves Propositio Observatio : Let []. The, the uber of -tuples i [d] that are ot -eve is d d. [Proof of Observatio : Propositio 0.3 applied to istead of shows that the uber of -eve -tuples i [d] is d d. Hece, the uber of -tuples i [d] that are ot -eve is d d d sice the total uber of -tuples i [d] is d. I view of d d d

5 Math 4707 Sprig 08 Darij Griberg: idter page 5 d d, this rewrites as follows: The uber of -tuples i [d] that are ot -eve is d d. This proves Observatio.] We ca costruct each first-eve -tuple x, x,..., x i [d] as follows: First, we choose the value of x. We deote this value by. There are d choices at this step sice this value ust belog to [d]. Next, we choose the -tuple x, x 3,..., x. Note that the etry x ust occur a odd uber of ties i this -tuple x, x 3,..., x because we wat the -tuple x, x,..., x to be first-eve, so that x ust occur a eve uber of ties i this -tuple; but the -tuple x, x 3,..., x is issig its very first occurrece, ad thus ust cotai it a odd uber of ties. I other words, the -tuple x, x 3,..., x ust ot be -eve. Thus, there are d d choices at this step sice Observatio yields that the uber of -tuples i [d] that are ot -eve is d d. Hece, the total uber of first-eve -tuples x, x,..., x i [d] is d d d d d d. This solves Exercise. 0.. Coutig legal paths geeralizatio of Catala ubers Recall the otio of a lattice path, defied i Midter. Lattice paths have up-steps ad right-steps. We say that a poit x, y Z is off-liits if y > x. Thus, the off-liits poits are the oes that lie strictly above the x y diagoal i Cartesia coordiates. A lattice path v 0, v,..., v is said to be legal if oe of the poits v 0, v,..., v is off-liits. For exaple, the lattice path draw fro 0, 0 to 4, 5 draw i the picture

6 Math 4707 Sprig 08 Darij Griberg: idter page 6 is ot legal, sice it cotais the off-liits poit 3, 4. Meawhile, the lattice path fro 0, 0 to 4, 4 draw i the picture is legal. For ay Z ad Z, we let L, be the uber of all legal lattice paths fro 0, 0 to,. For each poit v x, y Z, we let x v deote the x-coordiate x of v, ad we let y v deote the y-coordiate y of v. For exaple, x 5, 9 5 ad y 5, 9 9. The followig facts are easy: Lea 0.4. Let Z ad Z. a If v 0, v,..., v p is a lattice path fro 0, 0 to,, the 0 x v 0 x v x v p ad 0 y v 0 y v y v p. b If v 0, v,..., v p is a lattice path fro 0, 0 to,, the x v i y v i i for each i {0,,..., p}. c If v 0, v,..., v p is a lattice path fro 0, 0 to,, the p. d The lattice path 0, 0 cosistig of the sigle poit 0, 0 is the oly lattice path fro 0, 0 to 0, 0. e We have L 0,0. f We have L, 0 if at least oe of the ubers ad is egative. g We have L, 0 if >. The followig oboxiously log arguet just foralizes the obviousess: Proof of Lea 0.4. a Let v 0, v,..., v p be a lattice path fro 0, 0 to,. Thus, the defiitio of a lattice path shows that v 0 0, 0 ad v p,. Fro v 0 0, 0, we obtai x v 0 0 ad y v 0 0. Fro v p,, we obtai x v p ad y vp. Forally speaig, this lattice path is the list 0, 0,, 0,,,,, 3,, 3,, 3, 3, 3, 4, 4, 4, 5, 4.

7 Math 4707 Sprig 08 Darij Griberg: idter page 7 Let i [p]. The, the defiitio of a lattice path shows that the differece vector v i v i is either 0, or, 0 because v 0, v,..., v p is a lattice path. I other words, either vi v i 0, or v i v i, 0. Thus, x v i x v i 3. Now, forget that we fixed i. We thus have prove that x v i x v i for each i [p]. I other words, x v 0 x v x v p. Cobiig this with x v0 0 ad x v p, we obtai Siilarly, 0 x v 0 x v x v p. 0 y v 0 y v y v p. Thus, Lea 0.4 a is prove. b Let v 0, v,..., v p be a lattice path fro 0, 0 to,. Thus, the defiitio of a lattice path shows that v 0 0, 0 ad v p,. Fro v 0 0, 0, we obtai x v 0 0 ad y v 0 0. For each i {0,,..., p}, we defie a iteger z i by z i x v i y v i i. Let i [p]. The, the defiitio of a lattice path shows that the differece vector v i v i is either 0, or, 0 because v 0, v,..., v p is a lattice path. I other words, either vi v i 0, or v i v i, 0. Thus, z i z i 4. I other words, z i z i. 3 Proof. We ow that either v i v i 0, or v i v i, 0. Hece, we are i oe of the followig two cases: Case : We have v i v i 0,. Case : We have v i v i, 0. Let us first cosider Case. I this case, we have v i v i 0,. Thus, x v i v i 0. But subtractio of vectors i Z is doe coordiatewise. Thus, x v i v i x v i x v i. Hece, x v i x v i x v i v i 0, so that x v i x v i. Therefore, x v i x v i x v i. Thus, x v i x v i is prove i Case. Let us ext cosider Case. I this case, we have v i v i, 0. Thus, x v i v i. But subtractio of vectors i Z is doe coordiatewise. Thus, x v i v i x v i x v i. Hece, x v i x v i x v i v i, so that x v i x v i x v i. Therefore, x v i x v i. Thus, x v i x v i is prove i Case. We have ow prove x v i x v i i each of the two Cases ad. Hece, x v i x v i always holds. Qed. 4 Proof. The defiitio of z i yields z i x v i y v i i. But the defiitio of z i yields z i x v i y v i i. We ow that either v i v i 0, or v i v i, 0. Hece, we are i oe of the followig two cases: Case : We have v i v i 0,. Case : We have v i v i, 0. Let us first cosider Case. I this case, we have v i v i 0,. Thus, x v i v i 0 ad y v i v i. But subtractio of vectors i Z is doe coordiatewise. Thus, x v i v i x v i x v i. Hece, x v i x v i x v i v i 0. Hece, x v i x v i. Also, subtractio of vectors i Z is doe coordiatewise. Thus, y v i v i y v i y v i. Hece, y v i y v i y v i v i. Hece, y v i y v i. Thus, z i x v i y v i i x v i y v i i xv i yv i x v i y v i i x v i y v i i z i sice z i x v i y v i i. Thus, z i z i is prove i Case. Siilarly, we ca prove z i z i i Case. We have ow prove z i z i i each of the two Cases ad. Hece, z i z i always holds. Qed.

8 Math 4707 Sprig 08 Darij Griberg: idter page 8 Now, forget that we fixed i. We thus have prove that z i z i for each i [p]. I other words, z 0 z z p. Now, let i {0,,..., p} be arbitrary. The, fro z 0 z z p, we obtai z i z 0 x v 0 y v by the defiitio of z 0 Hece, 0 z i x v i y v i i by the defiitio of z i. I other words, x v i y v i i. This proves Lea 0.4 b. c Let v 0, v,..., v p be a lattice path fro 0, 0 to,. Thus, the defiitio of a lattice path shows that v 0 0, 0 ad v p,. Fro v p,, we obtai x v p ad y vp. But Lea 0.4 b applied to i p yields x v p y vp p. Hece, p x vp y v p. This proves Lea 0.4 c. d Clearly, the lattice path 0, 0 cotaiig just the sigle poit 0, 0 is a lattice path fro 0, 0 to 0, 0. Let v 0, v,..., v p be a lattice path fro 0, 0 to 0, 0. Thus, the defiitio of a lattice path shows that v 0 0, 0 ad v p 0, 0. But Lea 0.4 c applied to 0 ad 0 istead of ad yields p Hece, v 0, v,..., v p v0, v,..., v 0 v 0 0, 0 sice v 0 0, 0. Now, forget that we fixed v 0, v,..., v p. We thus have show that if v0, v,..., v p is a lattice path fro 0, 0 to 0, 0, the v 0, v,..., v p 0, 0. I other words, every lattice path fro 0, 0 to 0, 0 ust be equal to 0, 0. Hece, the path 0, 0 is the oly lattice path fro 0, 0 to 0, 0 because we already ow that 0, 0 is a lattice path fro 0, 0 to 0, 0. This proves Lea 0.4 d. e We ust prove that L 0,0. I other words, we ust prove that there is exactly oe legal lattice path fro 0, 0 to 0, 0 because L 0,0 was defied as the uber of all legal lattice paths fro 0, 0 to 0, 0. So let us prove this. Clearly, the lattice path 0, 0 cotaiig just the sigle poit 0, 0 is a lattice path fro 0, 0 to 0, 0, ad is legal sice the poit 0, 0 is ot off-liits. Hece, there exists at least oe legal lattice path fro 0, 0 to 0, 0 aely, this lattice path 0, 0. But Lea 0.4 d yields that the lattice path 0, 0 is the oly lattice path fro 0, 0 to 0, 0. Hece, the lattice path 0, 0 is the oly legal lattice path fro 0, 0 to 0, 0 as well. Thus, there is exactly oe legal lattice path fro 0, 0 to 0, 0 aely, 0, 0. I other words, L 0,0 sice L 0,0 is the uber of all legal lattice paths fro 0, 0 to 0, 0. This proves Lea 0.4 e. f Assue that at least oe of the ubers ad is egative. We ust prove that L, 0. I other words, we ust prove that there are o legal lattice paths fro 0, 0 to, because L, was defied as the uber of all legal lattice paths fro 0, 0 to,. So let us prove this. Let v 0, v,..., v p be a legal lattice path fro 0, 0 to,. The, Lea 0.4 a yields 0 x v 0 x v x v p ad 0 y v 0 y v y v p. Hece, 0 ad 0. Thus, is oegative sice 0 ad is oegative sice 0. Hece, both ad are oegative. This cotradicts the fact that at least oe of ad is egative. Now, forget that we fixed v 0, v,..., v p. We thus have obtaied a cotradictio for each legal lattice path v 0, v,..., v p fro 0, 0 to,. Hece, there are o legal lattice paths fro 0, 0 to,. I other words, L, 0 sice L, is the uber of all legal lattice paths fro 0, 0 to,. This proves Lea 0.4 f. g Assue that >. We ust prove that L, 0. I other words, we ust prove that there are o legal lattice paths fro 0, 0 to, because L, was defied as the uber of all legal lattice paths fro 0, 0 to,. So let us prove this.

9 Math 4707 Sprig 08 Darij Griberg: idter page 9 Let v 0, v,..., v p be a legal lattice path fro 0, 0 to,. Thus, the defiitio of a lattice path shows that v 0 0, 0 ad v p,. Moreover, the defiitio of legal shows that oe of the poits v 0, v,..., v p is off-liits sice the lattice path v 0, v,..., v p is legal. Hece, i particular, the poit v p is ot off-liits. But >. Thus, the poit, is off-liits. I other words, the poit v p is off-liits sice v p,. This cotradicts the fact that the poit v p is ot off-liits. Now, forget that we fixed v 0, v,..., v p. We thus have obtaied a cotradictio for each legal lattice path v 0, v,..., v p fro 0, 0 to,. Hece, there are o legal lattice paths fro 0, 0 to,. I other words, L, 0 sice L, is the uber of all legal lattice paths fro 0, 0 to,. This proves Lea 0.4 g. Rear 0.5. Clearly, L, 0 if ay of ad is egative because a lattice path fro 0, 0 to, ca oly exist if both ad are oegative 5. Also, L, 0 if > because if >, the the poit, is off-liits, ad thus a legal lattice path caot ed at,. Exercise. a Prove that L, L, L, for ay Z ad Z satisfyig ad, 0, 0. b Prove that L, for all N ad N satisfyig. [The requireet as opposed to is ot a typo; the equality still holds for, albeit for fairly siple reasos.] c Prove that L, for all N ad N satisfyig. d Prove that L, for ay N. Rear 0.6. Exercise c ca be rewritte as follows: L, for all N ad N satisfyig. This is a particular case of the so-called ballot theore see, e.g., [Reaul07], obtaied by settig a, b ad. Ideed, a legal lattice path fro 0, 0 to, correspods to a way to cout votes for cadidate A ad votes for cadidate B durig a aoyous electio i such a way that cadidate A leads i.e., has ore votes tha cadidate B throughout the

10 Math 4707 Sprig 08 Darij Griberg: idter page 0 coutig process at least after the first vote has bee couted. If you have such a vote coutig process, you ca costruct the correspodig lattice path as follows: Igore the first vote which is ecessarily a vote for A, sice otherwise A would lose the lead right away. Every tie a vote for A is couted, tae a right-step; every tie a vote for B is couted, tae a up-step. Exercise d is, of course, equivalet to the well-ow fact that the Catala ubers cout Dyc words ad Dyc paths. Vic Reier proved this i two differet ways oce usig geeratig fuctios ad oce cobiatorially i oe of the classes he substituted. We shall solve Exercise b by a ore or less straightforward iductio o. Exercise d will follow fro Exercise b via Exercise c. However, it is rather difficult to prove Exercise d directly by iductio. Thus, if oe wats to prove Exercise d by iductio, oe is ore or less forced to geeralize it to Exercise b. This illustrates a iportat pheoeo i atheatics: A ore geeral stateet is ofte easier to prove tha a less geeral oe particularly whe the proof uses iductio. Thus, geeralizig is a proble-solvig sill. Solutio to Exercise setched. I the followig, the word poit will always ea a pair x, y Z ad will be regarded as a poit i the Euclidea plae R. The word path will always ea a lattice path. Moreover, if ad are two itegers, the path to, shall always ea path fro 0, 0 to,. So, paths start at 0, 0 by default. For ay Z ad Z, we have L, the uber of all legal lattice paths fro 0, 0 to, by the defiitio of L, the uber of all legal paths to, 3 because we abbreviate lattice paths fro 0, 0 to, as paths to,. A step i a path v 0, v,..., v eas a pair of the for v i, v i for soe i []. More precisely, this pair v i, v i will be called the i-th step of the path. We say that a path v 0, v,..., v passes through a poit w if w {v 0, v,..., v }. a Let Z ad Z be such that ad, 0, 0. If the poit, was off-liits, the we would have >, which would cotradict. Thus, the poit, is ot off-liits. The equality 3 applied to istead of yields L, the uber of all legal paths to,. 4 Ay path to, cotais at least oe step sice otherwise, we would have, 0, 0, which would cotradict, 0, 0, ad thus has a last step.

11 Math 4707 Sprig 08 Darij Griberg: idter page This last step ust be either a up-step, or a right-step. Hece, the uber of all legal paths to, the uber of all legal paths to, whose last step is a up-step the uber of all legal paths to, whose last step is a right-step. 5 Let us ow copute the two ubers o the right had side. Ay legal path p to, whose last step is a up-step ust pass through, because this is the poit fro which a up-step leads to,. Thus, this path p cosists of two parts: the first part is a path to, ; the secod part is a sigle up-step fro, to,. Let us deote the first part by L p; this first part L p is still legal because ay off-liits poit o it would also be cotaied i p. Hece, we have defied a ap L : {legal paths to, whose last step is a up-step} {legal paths to, } which siply reoves the last step fro a path. This ap L is a bijectio ideed, the iverse ap siply adds a up-step at the ed of a path 6. Thus, {legal paths to, } {legal paths to, whose last step is a up-step} the uber of all legal paths to, whose last step is a up-step. Coparig this with {legal paths to, } the uber of all legal paths to, L, by 4, we obtai the uber of all legal paths to, whose last step is a up-step L,. 6 Why is this iverse ap well-defied? We ust show that if q is a legal path to,, the addig a up-step at the ed of q results i a legal path to, whose last step is a up-step. It is clear that addig a up-step at the ed of q results i a path to, whose last step is a up-step; let us deote this latter path by q. All we eed to chec is that this ew path q is legal. The path q is legal; i other words, oe of the poits o q is off-liits. Also, the poit, is ot off-liits. Recall that the path q is obtaied by addig a up-step at the ed of q. Thus, the poits o this path q are the poits o q ad the ew poit, which is where the ewly added up-step leads. Sice either the poits o q or the ew poit, are off-liits, we thus coclude that oe of the poits o q is off-liits. I other words, the path q is legal. This is exactly what we wated to show.

12 Math 4707 Sprig 08 Darij Griberg: idter page Siilarly, the uber of all legal paths to, whose last step is a right-step L,. Hece, 5 becoes the uber of all paths to, the uber of all paths to, whose last step is a up-step L, the uber of all paths to, whose last step is a right-step L, L, L, L, L,. Hece, 3 yields L, the uber of all legal paths to, L, L,. This solves Exercise a. b We shall solve Exercise b by strog iductio o : Iductio step: Let N. Assue as the iductio hypothesis that Exercise b holds wheever <. We ust prove that Exercise b holds whe. We have assued that Exercise b holds wheever <. I other words, if N ad N satisfy ad <, the L,. 6 Now, let N ad N be such that ad. We are goig to prove that L,. 7 Ideed, 7 is true whe 7. Hece, for the rest of this proof, we WLOG assue that we do t have. I other words, we have. We have > sice we have but. Sice ad are itegers, this shows that. 7 Proof. Assue that. Thus, >. Hece, Lea 0.4 g shows that L, 0. But 4 yields sice ad. Therefore,. I other words, 0. Coparig this with L, 0, we obtai L,. I other words, 7 holds. Thus, we have prove that 7 is true whe.

13 Math 4707 Sprig 08 Darij Griberg: idter page 3 Furtherore, 7 is true whe, 0, 0 8. Hece, for the rest of this proof, we WLOG assue that we do t have, 0, 0. I other words, we have, 0, 0. Hece, Exercise a yields L, L, L,. 8 Proof. Assue that, 0, 0. Thus, 0 ad 0. Hece, L, L 0,0 by Lea 0.4 e. Coparig this with , sice 0 ad 0 we obtai L,. I other words, 7 holds. Thus, we have prove that 7 is true whe, 0, 0.

14 Math 4707 Sprig 08 Darij Griberg: idter page 4 Moreover, 7 is true whe 0 9. Hece, for the rest of this proof, we WLOG assue that we do t have 0. I other words, we have 0. Sice N, we thus obtai > 0, so that N sice N. Fro > 0, we also obtai N sice N. The ubers N ad N satisfy. Hece, 6 applied to istead of yields ad < < L, sice. O the other had, the ubers N ad N satisfy 9 Proof. Assue that 0. If we had 0, the we would thus have, 0, 0, which would cotradict, 0, 0. Thus, we caot have 0. Hece, 0, so that > 0 sice N ad therefore N. The uber is egative sice < 0. Hece, at least oe of the itegers 0 ad is egative. Thus, Lea 0.4 f applied to istead of yields L, 0. Also, N ad N satisfy < ad. Hece, < 0 6 applied to istead of yields L, 0 0 sice Now,. Coparig this with we obtai L, 7 is true whe 0. 0 L, L, L, , sice 0. I other words, 7 holds. Thus, we have prove that 0

15 Math 4707 Sprig 08 Darij Griberg: idter page 5 sice ad to istead of yields L, sice. But < <. Hece, 6 applied L, L }, L {{}, } {{ }. Coparig this with } {{ } by applied to ad istead of ad }{{ } by applied to ad istead of ad, we obtai L,. Thus, 7 is prove. Now, forget that we fixed ad. We thus have show that if N ad N satisfy ad, the 7 holds. I other words, Exercise b holds whe. This copletes the iductio step. Hece, Exercise b is solved by iductio. c Let N ad N be such that. The, the fractio well-defied sice 0 sice N. Also, 8 applied to ad istead of ad yields sice. Multiplyig this equality by is, we obtai. 8

16 Math 4707 Sprig 08 Darij Griberg: idter page 6 But Exercise b yields L, } {{} by 8 This solves Exercise c. d Let N. The,. Hece, Exercise c applied to yields L,. sice ad. This solves Exercise d Scary fractios Exercise 3. Let, a ad b be three positive itegers such that a b. Prove that b a a b. Exercise 3 ay loo scary, but it is a straightforward exercise o iductio o b. To ae our life a little bit easier, we shall slightly relax the coditio a b to b a so that we ca use the case b a istead of b a as a iductio base: Propositio 0.7. Let be a positive iteger. Let a be a positive iteger such that a. Let b {a, a, a,...}. The, b a a b. I particular, all fractios appearig i this equality are well-defied. Proof of Propositio 0.7. All fractios appearig i Propositio 0.7 are well-defied.

17 Math 4707 Sprig 08 Darij Griberg: idter page 7 [Proof: We have 0 sice is a positive iteger; thus, the fractio is well-defied. Also, a N sice a is a positive iteger ad N sice is a positive iteger ad a sice a. Thus, 6 applied to a ad istead of ad yields a a 0. Hece, the fractio is well-defied. a Also, b {a, a, a,...}, so that b a 0 sice N. Hece, b N sice b 0 ad b {a, a, a,...} Z ad N ad b. Thus, 6 applied to b b ad istead of ad yields 0. Hece, the fractio is well-defied. b Now, let {a, a,..., b}. Thus, a > a 0, so that N sice {a, a,..., b} Z. Also, N. Furtherore, a sice a. Hece, 6 applied to ad istead of ad yields 0. Hece, the fractio is well-defied. Now, forget that we fixed. We thus have show that the fractio {a, a,..., b}. We ow have prove that the fractios, also the fractios is well-defied for each a ad b ad for all {a, a,..., b} are well-defied. I other words, all fractios appearig i Propositio 0.7 are well-defied.] Let us ow prove Propositio 0.7 by iductio o b: Iductio base: Coparig a a epty su0 0 0 with 0, we coclude that a a a a a. I other words, Propositio 0.7 holds for b a. This copletes a the iductio base. Iductio step: Let {a, a, a,...}. Assue that Propositio 0.7 holds for b. We ust prove that Propositio 0.7 holds for b. We have assued that Propositio 0.7 holds for b. I other words, we

18 Math 4707 Sprig 08 Darij Griberg: idter page 8 have. 9 a a I particular, all fractios appearig i this equality are well-defied. We have {,, 3,...} sice is a positive iteger. Hece, 7 applied to ad yields. Multiplyig this equality by, we obtai. 0 Now, a a a a a by 9 by 0 a a. a

19 Math 4707 Sprig 08 Darij Griberg: idter page 9 Also, 8 applied to ad yields. Multi- plyig this equality by, we obtai Coparig this with 0, we obtai Therefore, Hece, becoes a.. a. a ad, i particular, all fractios appearig i this equality are well-defied. I other words, Propositio 0.7 holds for b. This copletes the iductio step. Thus, Propositio 0.7 is prove by iductio. Solutio to Exercise 3. Fro b a a, we obtai b {a, a, a,...} sice b is a iteger. Thus, Propositio 0.7 yields This solves Exercise Derageets that are ivolutios b a a b. Defiitio 0.8. Let σ be a perutatio of a set X. a We say that σ is a derageet if ad oly if each x X satisfies σ x x. b We say that σ is a ivolutio if ad oly if σ σ id that is, each x X satisfies σ σ x x. For exaple, the perutatio α of the set [5] that seds,, 3, 4, 5 to 3, 5,, 4, is a ivolutio it satisfies α α α 3 ad α α α 5 ad 3 siilarly α α x x for all other x [5], but ot a derageet sice α

20 Math 4707 Sprig 08 Darij Griberg: idter page 0 O the other had, the perutatio of the set [6] that seds,, 3, 4, 5, 6 to 3, 4,,, 6, 5 is a derageet it satisfies x x for all x [6], but ot a ivolutio sice. Exercise 4. Let N. Prove the followig: a If is odd, the there exist o derageets of [] that are ivolutios. b If is eve, the the uber of derageets of [] that are ivolutios is! / /!.! [Hit: What does the uber / reid you of?] /! 0.5. Hypergree perutatios Exercise 5. Let N be such that. We shall call a perutatio π S hypergree if it satisfies both π < π ad π < π. a Prove that ay π S satisfyig π ust be hypergree. b Prove that the uber of hypergree perutatios π S that do ot satisfy π is 4!. Here, 4! is uderstood to be 0 whe < 4. [Hit: For b, argue first that if π S is hypergree but does ot satisfy π, the the four ubers,, π, π are distict.] 0.6. Coutig the parts of all copositios Recall that if N, the a copositio of eas a fiite list a, a,..., a of positive itegers such that a a a. For exaple, the copositios of 3 are 3,,,, ad,,. The legth of a copositio a, a,..., a of is defied to be. Exercise 6. Let be a positive iteger. Prove that the su of the legths of all copositios of is. Refereces [Gribe6] Darij Griberg, Notes o the cobiatorial fudaetals of algebra, March [Reaul07] Marc Reault, Four Proofs of the Ballot Theore, Matheatics Magazie, vol. 80, o. 5, 007, pp FourProofsoftheBallotTheore.pdf