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Math 2800 . Math Majors Seminar Book 2 that Show t.tk then 2kt Form form list If to for then the some integer r ,B , for R , every Pe R , n Is R B theorem if Every R . , once 1=21 number odd therefore of to . the this answer proving the of odd number has tools way a 2kt an . infinite . n= > , \ ete , Pn we Put ⇒ have . , , a domino effect n=2kfor some : . . of pay The odd at is If . any is every more lez , that and even need We . all n Pa , DDDD= Example be k , that Pi any This shows . Induction ... for namely that : Pz , 2l t I cannot converse P prove obtain we R 2kt . Mathematical of prove e contradiction KETL ) statements can i. even a , prove Zrti of and 21 ( where principle we not is ← ' want We The Zkti 2kt positive 1 nez for is some integer is even integer either if k . even or odd . integer k ; Example We : solution for number of P solution a First stack is that with nti • using without This results for of Towers the P prove . Pz , , 'll B. Pt with puzzle Hanoi all ... , called ( also : easy every n disks n , R , then step # in step strategy touching our in one at once ⇒ Pn+ the more , to the inductive in all largest the move ntl disks n the more ( We . , is there and has . or the disk single R assuming showing are solution a Pn for n disks is from for n disk disks from from the more ( A . 1 and stack stack 2 . to to . ) modification stack stick of hyped Hanoi 2 2 . . puzzle stack disk to n slight 0 1 top there disks ) nil Towers a if that inductive the given , , disk the largest roles of stacks the switch more finally the We , here • ( . a . inductive disks Next • Hanoi puzzle has about the minimum worry of Towers starting initial case bad his case 2 the the prove to however for called is that the disks of ) that want stack solution This . P, is to prove a To Later This . 0 Next moves the prone ) possible number We . by induction every statement the is . show will is d required . ( Again of stacks this uses 0 and 1.) hypothesis K to interchange the roles to disks in state 2 from smallest ( the largest ( top ) bottom ) ' on . on A statement stronger Theory moves following of Hanoi problem : with n disks has a solution using zny . ( Initial proof ( The Towers . the is formula 2 ' 1=1 - . • First move Then use : El is observe ' , from stack that to 0 there is a stack 2) the top 1 move move the the 221 in n to n minimum moves disks as solution using a predicted by move our disks from number of that the n . disk an - , . stack from hypothesis disk puzzle (n+ , ) given moves 221 to stack 1 in slack 0 largest disk from slack 0 to stack then , the more inductive the by Assuming . solved be Finally . Remark can n= . sleep ) ( Inductive puzzle For . single desk the more ) case moves I to stack required . 2 in although 221 moves I 2. . didn't show this . Example odd Is n theorem Proof Now if odd also If is n Remark then even 0=20 If . A = is n this for by n then a is odd two cases as follows k then , so , , must show we nh = that is , nt : =2k+i nti - zfk ) is even . Placing form to ? . . K ) If . dots chords proof . odd is . ktl where zckti ) is integer an a on , we & circle have then . . . . a . . & . and divided the . . & . • . . . . . • . . connecting circle into . • . . • odd is . ' . . DH Ztk n= i . • . . . . for positive integers which is either even works above positive is n n=2k+1 - . • n The . • . ! odd or negative integer then a • : . integer k even . - . D • . Conjecture odd or or some . either is If . 2k induction • False even even integer ; . either In some k integer k some . even is 1=20+1 zkti n= Every integer : form prone for n=2k say , follows result The Also odd or is ntl so either We . integer some n=2kfor if even is integer is it has the odd or a is odd , If for positive integer a even nez I positive since supposing even either 2kt n= Every is A of pay The . . • then I . . • . • in regions . • . . . . . ' ; . . . . . . . . . i. i. . . . pairs • theorem If subsets of F# n=2 n= 1 : is S= {0 S={ } : The theorem is ( which is Remark {a : either usually } a , Directproof { = set If empty { } - and valid the in S , S= 1 {a not it's or , , as a } , . This gives Alternatively Proof subset - a Gz . total of , we argue in can by prone of , {1 } { , 0 I , } . . denoted 5=0 This has . the theorem works so induction by or only that in case directly talking abort amttirset } an , then every ) ( value every ; subset is A E 5 . - - . 2×2×2 x. the induction agreement with } 0 20=1 . ... . We 0 . number the then ) , , - ) . or . n } 0 proved either not ( we're . { , { , usually , itself be can { } contains either a. ag choices .no ? ,t; , 1St . { } are set preferable ) } S subsets two the is subset one subsets of ; has o ie . } I , ( elements n " 2 S={ } : n=o S has S set a the fault . . × 2 = n= ISI subsets " induction by on z . formula 20=1 For . the A E S I . . initial case 151=0 , 5- 0 has 1 . Now to have assumed is Bhas We It subsets ' li ) If be A All Reformulated Proof by follows as S S {b } on any b So B E . A . . have n B nti We . 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MEI for " = nay . + ( n+i) Proof n ! I lxzx = ! . ' . Now xn 1 ' ! 2 a 1×2=2 = 3 ! , theorem a . n Itzel = . induction some = >/4 a > 4 n n ! > 2 != ! , , = ( n+D×n 1×01 . I > . 4 ! " > for 24 . some , = (n+Dn ! z5n > @+ I ) ! , We have checked assuming ( ntp ! 1×2×3=6 For ! as by required . induction The . ! " > z×[ 5×2 = It result follows D ' theorem Eg all For nzo 's It , + tat + ... In 2 < . 1<2 . It tz It tz 1++2 < + + z ¥ < 2 tat 's < 2 by induction Proof Assuming It Since . Itty + 1<2 tgt . . + . , In inequality holds the < for 2 for n3o some n=o . , 'ftXy¥fyy¥ that's conclusion " " cz .tn#d@HgetF@giedg tzt tzttgt lttzttattgt is + . . . . z÷± + I < , ¥ 1+1 < , 2 = D required as O . . even : 0=2.0 The empty set I 3=0 The case subsets . . has n=o Not 1 odd subset gives zkti O= : 0 , 2 : ' has no solution k€21 Recall : itself If lskn subset (the initial case of . then . an S has induction ) . 2 " To R⇒R+ prove " Recall : P F # T F F There This . For not is In the Proof Now result The more subjects , learning nzl have we , great example it pore initial • • P, n n > ! 2,3 : , . to conclude ) prove - npzr(P in R ) case ⇒ mathematical induction : ) Pn+ , ; I have we 1 induction initial sequential process a . but for to show the sake of . : the is . illustrate induction positive integer the fact that n follows by induction 1 5 - > - 5 . satisfies n a " , , . mathematics to by case general form of simply true is . ) 0 T assuming follows From This 3>0 statement ( of times T will argument then " number Q all a implication ( conditional any F most than the hypothesis use , 1=2 It P ⇒ F More can If Q T A the gives Pn+ , , is n -5 want we , positive a > integer ntl also is . D . In order to prone R, R . B. Pg " , . all at once , one > . 5 . Note If : one initial P, ( the R follows P p . . . } P , . - . . , P , and R ) . get you , i and P⇒Pz P, P and . . * case from statements above satisfy the can ; (R^PD⇒ and and B and Pz ; ( P npzn B) ⇒ , Pq ; AND P.P.BR/P5 Example ( print a other than eg To say . . . , pk Every integer is integer an 1 and p ) 4=2×2 7=7 that , n k is 7 1 a . p> 2 can he Factored which is as not divisible product of a by any primes . positive integers . 6=2×3 . n=2 12=2×2×3 product of primes means n =p Rpj pn : , for Some primes p ,p , . , For initial the Now suppose primes case We If i) c Case cii ) If p p. , ntl . . ; pine nti we , , q, ab ,q , integers 2. 3,4, ; can also be factored . done are ie are qe single factor consisting of a ( . prime one a , primes a factor ) ntl . . product = and b :p pi pn Then qqi . as - qe of a of primes . where ab = prime 2 product ,b£{2,3 ; ;n} a where k , l , . , . pzpjipnqqqiiqe factorization prime of nti result follows by induction is = as the factored be Can n . composite ) then . hypothesis ... , is Prime ( not inductive pp nti which is the , , is ntl By and show that must . Case each of the and 72 n which n=2 p = , a . The some ; we'll 60 60 One uses refer textbooks induction order II . can of = = to 6×10 = = other techniques type of result . induction induction (2×2)×(3×5) ] (2×3)×(2×5) stronger prove the prime factors a it call just 4×15 this that Prime This is induction complete or . same except for order the different prime factors of the factorization ( not just induction ) strong as a integers of are . essentially unique Fundamental Theorem of Arithmetic to prove . . ie . up to changing But this of One p the steps then , We a won't is we say following The this today prove k.net/L If Euclid 's Lerma divisible by is are k divides • that • k is equivalent k is eg . | 1 by " = nez ( some 6 : " 7×1 ab such that z by p if dk n= for DEZ some } n n integer d kln Note k 316 is to satisfies ⇒ p. In . and this with is this : a relation between net operation , kfn -5140 . 5/0 . -5/10 . an n write Don't 51.10 sine pine a the symbolic way since . is by . 54T d. divisible is : 6T3 18 integer Every divisible every for divides 3 of factor of a n=dk . divisor be divisible is didn k n a a. . isisadiuismttdskgthk :! • b or p If : since since 010 -10=-2×5 10 045 = f- 2) xf5) 5=()# 0=0 )x5 0=471×0 Given Eg a bek , Find . the , greatest god ( 18,30 ) of Divisors of Divisors 18 30 : -18 -9 -6 , , The gcd -30 Y5Y0Te greatest ( 100 gcd ( 64 ) , of Divisors of Divisors Common 0 ) 63821 : Wolfram However to -200 divisor = . . . integer 's , -2 will give , -1,1 , 63821 has Algorithm -1 ... , . and 0 Arithmetic ) of -1 -2 , , , → it and 18 2,3 -1,1 , 6,9 , 42 , denoted is gcd ( a ,b ) . , 18 3,5 6,10 , 15,30 , 3,6 1,2 , , , of 63821g -63821 of b is 30 6 : gcd ( 18 , 30 ) = 6 . ; 64=21212×2×2×2 100=2×215×5 , divisors and a 63821 . Alpha , -3 , 3. ' , using : Euclid using -6 - Every positive Theorem 4 = , : -2 , -5 , common 0 63821 -3 , : 30 of . Common divisors of 18 and divisor common a . - the integer factorization digits ; 300 digits 1 ... , , z , . , : . . ) unique factorization Finding gcd ( appropriate answer is is m , n ) as is software quickly very . product of primes ( the Fundamental practically computable for large integers a such as Testing only practically feasible infeasible . Maple for for or primality numbers Mathematical also is up to easy or . about Real numbers R : { = It But careful be real define to possible is } numbers real _ numbers is X 100 1.36363636 = X= algorithm an . . 99x=35⇒ YF 13¥ . . = ex contain not . = I . a . x 1 decimal to a fraction , eg : ... , . . any that Here 0,999999 = ... ×= larger number systems big to 0.999999 = does . 1,5T 9999999 9. = 9¥ R . . algorithm X : ? . 1.36363636 = Px Note numbers . converting every repeating for x= this Applying decimals using real the . 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NHZ Someresistance ) Idea that # 1 0.33333 = . . . . . F. 7 = umber These These numbers . . . one based is i decimal numbers ( a on expansion each . had decimal expansion . . . . a€R can he . 3.141592653589793298 have = unique 1,428571428571428571 = ... . 2.0000000 real have } ' ' 1.9999999 = Every . ' . 1.0000000 0.9999999 = = Note . 0.99999 can 3.1415926535 = 2 argument real a = ¥ fallacious number only the to infinitely many expressed as decimal a . . i. e. finite . digits but . these numbers . . eg they are are . not infinite Second Proof 's that 0.333333 = +3=0.333333 ¥9999999 proof atbz . let : . . =/ . : Reward . . 3 has : " . 1 unique decimal which has two . 9=0.999999 = and ... . . . = 6=1 Find the . a Z acts TE 1,9999991=0.799999 19 zfIIIIa# then aa¥< b means a=b 8 79 ÷ The 9 ± ¥ . . expansion decimal expansions . # 1 a unlike " ' 0.333333 ÷= Third 0.999999 fact that at¥=a . . average of a and b : sets For any # s , does what the cardinality 151 mean ? start Let's . the size same eg IAI : H |{ -45,11 }/= . 1131 = 0 ( of S if there a is Is denoted S ) what it by saying 3,10 }| , of size i.e. IN { 1,2 3,4 :} , : 1 , For finite § Ii{!t sets lie same size the have set F- S S itself 12/1=1 Nl is . } . These two sets : 6 5 ' ' countably infinite the have be listed size same S : with as a iff finite ie . {s - , I 's in .sn and say I 1 { 12,3 in element ) ; , . : IN 0 } ... . S is sequence ; } . only finitely proper the only set . subset may The is converse finite set cannot a also true -1 I Z -2 I I 2 I I I 2 3 4 5 -3 3 I 6 : satisfying 171=151 TES . 21 B bare ... , , 1st infinite an s to its elements if , A ,B . . sets . ' 1*1=11141 . sets . § Iif sometimes or one-to-one correspondence between A " 5 4 Lets be a set We IN 4 3 2 . , , No : A No={0,433,4 . = for too means can IN |S| I 7 4 - 4 I I 8 9 . . - . . . is ||N| |Q/= 0 's , . 's it I F. . ¥ - , Q={o , tg 's ' - , i÷ 3g , ' , - , , :} ? ! ! Theorem - ( , . . I E. ¥ -2g Georg Intact , ¥ It . E , - , ; , ¥ , ' , ¥ , 7 -6 , ÷ I -7 , - , F. as , E ' , 8 , , - , -8 , ¥ , . Es . . I , ... . , ... , . " ... ... , t.t.tt ! ! ! Cantor , ... , 3 is countable . ... R is (o ,D={xtR not : ii. , ... ÷, - , , , ¥ , Is , 6 -5 , he , I , 5 , ÷ , # , % , Iz Is , I , -4 , , iii. ? - , ¥ , I 4 , Is . E. I. ¥ , ÷ , E. E. Is -3g , tg , , ⇒ if , -3 3 E , infinite ) ountably is E. I. ¥ - , he , Q . , -3 , ÷ , ' I ÷ , -2 , ie -2 , E. E. , ÷ 2 , I , 't , ¥ tz its , ¥ in -1 1 , ( : anmtably infinite uncountable ocxci } is . It , ... Suppose ( 0 D= { as each number in this Proof , for G, 0.9 = 0,9 = 92 93 = 94 = 0 . 9,9 , list , 924925 } 0 . 95, 952 - . } Choosing . , decimal a ,= 953954955 95g 9jE{ 0,42 . . . 9⇒ an qsqaq , . expanga , aj " . 93,9329394935936 Gp 09,9%9%4 95 , 94,95 9,69g 9,394 9,5 an ,, 9,93 , ' ' ' is the digit . ' } 9 . jtl decimal ith number ; IN ije , . , of the ... 957 . , ' ! . choose Now chosen that so which in b= an but am bit be case for the ne nth ! 0 an 7 some , Think about this b= . . 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