Writig Proofs Misha Lavrov Iductio: Solutios Wester PA ARML Practice March 6, 206. Prove that a 2 2 chessboard with ay oe square removed ca always be covered by shaped tiles. Solutio : We iduct o. For = 0, a 2 2 chessboard with oe square removed is empty, ad therefore already covered. (If the = 0 case is awkwardly trivial, the = case is also easy to verify.) Suppose we ca tile a 2 2 chessboard with ay square removed. Take a 2 2 chessboard with a square missig ad divide it ito four quarters. As show below, place oe tile i the ceter so that each quarter is missig oe square: By the iductio hypothesis, each of the quarters ca ow be tiled, which gives us a way of tilig the 2 2 chessboard. This meas that we kow how to tile the chessboard for all. Solutio 2: The setup ad base case of the proof are similar, but the iductio step is differet. Give a 2 2 chessboard with ay square removed, divide it up ito a 2 2 grid of 2 2 squares. The 2 2 square with a square missig ca be filled by a sigle tile, ad the we are left with a 2 2 grid of 2 2 squares, with oe 2 2 square missig. Usig four tiles, we ca make a copy of the origial tile at twice the scale: If we wat to use the scaled-up tiles to cover the 2 2 grid of 2 2 squares missig, that task is equivalet to coverig a 2 2 chessboard with oe square missig by tiles of the ordiary size. By the iductio hypothesis, this ca be doe; therefore we ca tile the 2 2 chessboard with a square missig as well. By iductio, this is possible for all. 2. Chicke McNuggets come i boxes of 6, 9, ad 20 uggets. Prove that for ay iteger > 43, it is possible to buy exactly uggets with a combiatio of these boxes. We prove six base cases: 44 = 20 + 9 + 9 + 6, so = 44 is possible.
45 = 9 + 9 + 9 + 9 + 9, so = 45 is possible. 46 = 20 + 20 + 6, so = 46 is possible. 47 = 20 + 9 + 9 + 9, so = 47 is possible. 48 = 6 + 6 + 6 + 6 + 6 + 6 + 6 + 6, so = 48 is possible. 49 = 20 + 20 + 9, so = 49 is possible. For all larger, if buyig exactly 6 uggets is possible, the buyig exactly uggets is possible: just buy 6, ad the aother box of 6 uggets. Havig prove the result for six cosecutive values of, the result for all larger follows. If we also allow 4-ugget boxes, the what is the largest iteger such that it s impossible to buy exactly uggets? We ca check that = is impossible. If we buy three boxes, the we get at least 4+4+4 = 2 uggets, so we just eed to check all possibilities for at most two boxes: oe of 4, 6, 9, 4 + 6, 4 + 9, 6 + 9 equal, ad of course buyig a 20-ugget box would put us over the target immediately. We prove that all values of 2 are possible i the same way as before: Buyig exactly 2 is possible sice 2 = 4 + 4 + 4. Buyig exactly 3 is possible sice 3 = 9 + 4. Buyig exactly 4 is possible sice 4 = 6 + 4 + 4. Buyig exactly 5 is possible sice 5 = 9 + 6. From there, we ca iduct o : to buy > 5 uggets, buy 4 uggets ad the a box of 4 more. 3. Fid the mistake i the followig proofs: (a) The first proof is missig a base case. It s true that if =, the = +, but = is ever true for ay. (b) The secod proof does ot prove eough base cases for the iductio step to work. To show that l ad l 2 are the same lie, we say that they both pass through A 2 ad A. However, whe = 3, these are the same poit, which says othig about l ad l 2 ; the iductio step oly holds for 4. 4. Prove that for all, the sum of the first odd umbers is a perfect square. We show that + 3 + + (2 ) = 2 by iductio o. For =, both sides of the equatio say =. Now assume the statemet holds for ; we show it holds for +. If +3+ +(2 ) = 2, the + 3 + + (2 + ) = 2 + (2 + ): we re just addig 2 + to both sides. But 2 + (2 + ) factors as ( + ) 2, which was what we wated. By iductio, the equatio holds for all. 2
5. Prove that for all, 2 + 2 2 + 3 2 + + 2 < 2. We show that 2 + 2 2 + 3 2 + + 2 2 by iductio o. For =, we have 2 2, which is true, sice. Now assume the iequality holds for some ; we ll show it holds for +. Startig from 2 + 2 2 + 3 2 + + 2 2, add to both sides. O the left-had side, we get the sum we wat to boud. O the (+) 2 right-had side, we get 2 + ( + ) 2 = 2 ( + )2 ( + ) 2 = 2 + 2 + + 2 < 2 + +. Therefore the iequality cotiues to hold for + if it holds for. By iductio, the iequality holds for all, ad so i particular, the sum is bouded above by 2. 6. I a chess touramet, each pair of players played exactly oe game, ad icredibly, oe of them eded i draws. Prove that there is a participat i the touramet (call him Bobby) such that every other player either lost their game with Bobby, or lost a game with someoe else that lost to Bobby. We iduct o, the umber of participats i the touramet. For =, the statemet is trivial, ad so we let this be our base case. Suppose that the statemet holds for players; let s try to show that it holds for + players. Thik of the resultig touramet as oe played by players, with a extra player the showig up ad playig all of them. By the iductio hypothesis, the first players ca be divided ito 3 groups: {Bobby} S T, where Bobby is (oe possible choice of) the participat we call Bobby, S is the set of players that lose to Bobby, ad T is the set of players that beat Bobby (ad must therefore lose to someoe i S). If the extra player loses to Bobby, the they ca also be added to S, ad the claim cotiues to hold. If the extra player wis agaist Bobby, but loses to someoe i S, the they ca be added to T, ad the claim cotiues to hold. The remaiig case is whe the extra player wis agaist Bobby ad wis agaist every player i S. I that case, partitio the + players ito three ew groups: {ew player}, S := S {Bobby}, ad T. The everyoe i S lost their game with the ew player; everyoe i T lost a game to someoe i S (i particular, to someoe i S.) So the claim cotiues to hold, with the ew player becomig Bobby i place of the old Bobby. By iductio, the claim holds for all. 7. (VTRMC 202) Defie a sequece (a ) for a positive iteger iductively by a = ad a = d,d< a, where the product rages over all proper divisors d of. d 3
Thus a 2 = 2, a 3 = 3, a 4 = 2, etc. Fid a 999000. We will prove that the followig geeral formula holds: if is prime, a = p if = p k for some prime p, if has at least two distict prime factors. Sice 999000 = 2 3 3 3 5 3 37, this meas that a 999000 =. We prove the formula by a uusual iductio o. Our base case will be every prime. The we will show that if the formula holds for all proper divisors of, it must also hold for. Suppose that is prime. The its oly proper divisor is, so a = a case is show. = =. The base Now, suppose that the formula holds for all proper divisors of. There are two cases to cosider. First, if is a prime power if = p k for some prime p, the its proper divisors are, p, p 2,..., p k. So we have a = = a a p a p 2 a p k p p = = p. pk If is ot a prime power, the it has a prime factorizatio p e pe 2 2 pem k. For all its divisors d, a d = uless d is a power of a prime, by the iductive hypothesis, so we ca igore a d i the formula. If d is a power of a prime, the it has the form d = p k i for k e i, ad a d = p i. There are exactly e i proper divisors d such that a d = p i : they are p i, p 2 i,..., pe i i. So the product of a d over all divisors d picks up a factor of exactly p e i i for every prime p i i the factorizatio of. This meas that it exactly equals, ad the formula tells us that a = =. By iductio, the formula holds for all, ad i particular for = 999000. 8. A biary sequece such as 0000 is writte o a blackboard. I a step, you are allowed to chage the first umber (from 0 to or vice versa) or the umber after the first occurrece of. (Startig with 0000, you could get to 000 or to 00000.) Prove that you ca chage ay sequece to ay other sequece of the same legth. We prove the statemet Ay sequece of legth ca be chaged to ay other sequece of legth by iductio o. The base case, =, is easy to prove: we ca chage 0 to or to 0 by a applicatio of the first rule. Now assume that the statemet holds for some. A corollary is that i a sequece of legth +, we ca chage the first bits to aythig we like: the same rules that go betwee two -bit sequeces let us go betwee the first bits of two loger sequeces. Take two arbitrary sequeces x 0 x x ad y 0 y y. We ll show how to get from oe to the other. If x = y, the we ca use the -bit method (which we kow exists by the iductio hypothesis) to chage x 0 x x to y 0 y y. 4
If x y, the we ca use the -bit method to chage the first bits of x 0 x x to be 000 0, so that we get the sequece 000 0x. The secod rule (which lets us chage the bit after the first ), ca be used to chage this to 00 0y. Now we ca use the -bit method agai to chage the first bits to y 0 y y, ad we are left with the target sequece. By iductio, we ca chage ay sequece to ay other sequece of the same legth, regardless of the legth. 5