Summer High School 2009 Aaron Bertram
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1 Summer High School 009 Aaro Bertram 3 Iductio ad Related Stuff Let s thik for a bit about the followig two familiar equatios: Triagle Number Equatio Square Number Equatio: = ( ( = What do they mea? is clearly meat to represet a atural umber, ad the equatios are clearly iteded to be true for all values of I other words, they are each ot oe, but a ifiite umber of equatios We ca easily check as may of them as we like by had, but evidetly we caot check a ifiite umber of equatios by had! How do you prove them? Actually, there are several ways to do this Geometric Proofs: (a Triagle Number Equatio: couts poits arraged i a right triagle with rows, which ca be viewed i either oe of the followig ways: or (depedig upo whether you start coutig at the top or the bottom The two triagles fit together to form a ( + rectagle, which has ( + poits So each triagle has ( + / poits A way of doig this (attributed to Gauss at a very youg age without pictures is to write the sum twice, reversig the order the secod time: ( + ( + + Whe you add them first vertically the horizotally, you get ( + Whe you add them first horizotally, the vertically, you get times the triagle umber
2 (b Square Number Equatio: ( couts the poits of a isosceles triagle: We ca slice the triagle vertically to get two right triagles: ad The two triagles fit together to give a square! Alteratively, the umber of poits of the two right triagles are two cosecutive triagle umbers (for ad So they have: ( + + total poits ( ( + = ( + + = The o-picture versio Break apart each odd umber as the sum of cosecutive atural umbers ad regroup: ( = ( ( + + ( ( + ( = to get the same result ( ( ( Geometric proofs are ice, sice they somehow explai why the equatios are true, which is somethig proofs do t always accomplish! However, they are ot always available Proofs by iductio, o the other had, are useful for a wide variety of problems Defiitio 3 A sequece of ifiitely may equatios (statemets E( are all simultaeously proved by iductio oce: (i The iitial ( = equatio (statmet is proved, ad (ii The followig implicatio is proved for a arbitrary : Assumig E( is true, the E( + is also true
3 Why is this a proof? A proof by iductio proves that the set of atural umbers such that E( is false ca have o miimal elemet because (i says E( is true, ad (ii says that if E( were false, the E( would also be false So E( is ever false by well-orderig Examples: (a Proof by Iductio of the Triagle Number Equatios ( + E( is the equatio = (i E( is the equatio = ( + /, which is true (ii Assumig that E( is true, we have: = + = + ad, addig + to both sides, ( ( + = + + ( +, ad + + ( + = + + ( + ( + ( + = so that E( + is also true This completes the proof by iductio! (b Proof by Iductio of the Square Number Equatios E( is the equatio ( = (i E( is the equatio: =, which is true (ii Assumig that E( is true, we have: ( = ad, addig ( + = +, ( ( + (( + = + ( + = ( + so that E( + is also true This completes the proof by iductio We ext wat to get to the biomial theorem But first, cosider: Defiitio 3 The product of the umbers from to is writte: ad it is called factorial! := 3 Defiitio/Propositio 33 If k < are atural umbers, the: (! := k k!( k! is the umber of ways of choosig k differet objects from objects We call this choose k, ad otice that it is equal to choose k 3
4 4 Proof (of the coutig statemet: The first object you choose ca be ay oe of the, the secod ca be ay of the remaiig, etc up to the kth, so there are: ( ( k + ways of choosig k objects i a particular order O the other had, k! is the umber of ways of orderig k objects, so the total umber of ways of choosig k objects out of (whe order does t matter is: ( ( k + k! sice ( ( k + =!/( k! =! k!( k! A Mior Additio: By covetio, we set 0! =, ad i particular: ( ( = = 0 makes sese There is oe way to choose 0 objects (or from Pascal s Idetity: For all atural umbers k < : ( ( ( + + = k k k Remark: This is a ifiite umber of equatios i two directios sice there is oe equatio for each value of ad each value of k < Oe could write E(k, to stad for each equatio, ad try to prove the whole mess by iductio (it ca be doe! Fortuately, there is a really elegat proof available: Proof: Start with red objects ad oe gree object There are ( + k ways of choosig k objects i all But: The k objects may all be red (i ( k ways, or else Oe of them is gree, ad k are red (i ( k ways QED! Pascal s Volcao: Start with a shell of s i the followig shape:
5 Fill i the volcao by addig pairs of adjacet umbers to produce the umber beeath (ad betwee them: etc Pascal s Theorem: The th row of Pascal s Volcao is: = ( 0 ( ( Proof by Iductio: ( ( k k (i The first row is = ( ( 0 = (ii Assume that the th row of the volcao is: = ( 0 ( ( ( = ( = The the + st row (addig adjacet umbers i the th row is: ( ( ( ( ( ( k k (the s o the left ad right come from the shell By Pascal s Idetity, this is exactly the + st row that we wat QED Fially, we have the Biomial Theorem: ( (x + y = x + x y + ( ( x y + xy + y That is, the coefficiets of (x + y are the th row of Pascal s volcao Proof: Also, by iductio (i The first equatio is: (x + y = x + y Duh!
6 6 (ii Assume the coefficiets of (x + y are the th row of Pascal s volcao (this is the th equatio The: (x + y + = (x + y (x + y = x(x + y + y(x + y = x + + ( x y + ( x y + + xy + x y + ( x y + + ( xy + y + ad combiig coefficiets vertically, we see that as i Pascal s volcao, the adjacet coefficiets of (x + y are added to get the coefficiets of (x + y + Thus the coefficiets match with Pascal s volcao We may ow read off the first few examples from Pascal: (x + y = x + xy + y (x + y 3 = x 3 + 3x y + 3xy + y 3 (x + y 4 = x 4 + 4x 3 y + 6x y + 4xy 3 + y 4 (x + y 5 = x 5 + 5x 4 y + 0x 3 y + 0x y 3 + 5xy 4 + y 5
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