Discrete Mathematics

Transcription

1 Discree Mahemaics Compuer Laboraory Compuer Science Tripos Par 1A Mahemaics Tripos 1A (CS Opion) Peer Robinson Michaelmas 2002 & Len 2003 William Gaes Building JJ Thomson Avenue Cambridge CB3 0FD hp://

2 Peer Robinson, All righs reserved.

3 Inroducion This course will develop he idea of formal proof by way of examples involving simple objecs such as inegers and ses. The maerial enables academic sudy of Compuer Science and will be promoed wih examples from he analysis of algorihms and crypography. Syllabus Inegers Proof. Deducion, conradicion. Inegers, mahemaical inducion. [2 lecures] Facors. Division: highes common facors and leas common muliples. Euclid s algorihm: soluion in inegers of ax + by = c, he complexiy of Euclid s algorihm. Euclid s proof of he infiniy of primes. Exisence and uniqueness of prime facorisaion. Irraionaliy of p ; brief discussion of raional and algebraic numbers. [3 lecures] Modular arihmeic. Solving congruences. Unis modulo m, Euler s oien funcion. Wilson s heorem. Chinese remainder heorem. The Ferma-Euler heorems. Public key crypography. [3 lecures] Ses, relaions and funcions Ses, subses and Boolean operaions. Indicaor (characerisic) funcions and heir algebra. Principle of inclusion-exclusion, wih applicaions o Euler s funcion. Boolean logic. [2 lecures] Binary relaions. Composiion of relaions. Equivalence relaions and quoiens of ses. Closures and Warshall s algorihm. Parial orders and oal orders. Hasse diagrams. Well founded relaions and well ordering. Well founded inducion. [3 lecures] Funcions; Injecive, surjecive and bijecive funcions. Numbers of such funcions beween ses. Soring. The Schröder-Bernsein heorem. Counabiliy. A counable union of counable ses is counable. The uncounabiliy of R. Exisence of ranscendenal numbers. [3 lecures] Objecives On compleing he course, sudens should be able o: Wrie a clear saemen of a problem as a heorem in mahemaical noaion. Prove and disprove asserions using a variey of echniques. Describe, analyse and use Euclid s algorihm. Explain and apply prime facorisaion. Perform calculaions wih modular arihmeic. Use number heory o explain public key crypography. Analyse problems using se heory. Explain and use he principle of inclusion and exclusion. Recognise relaions and discuss heir properies. Describe and analyse Warshall s algorihm. Sae, prove and apply he Schröder-Bernsein heorem. Disinguish counable and uncounable ses. Michaelmas

4 Appropriae books The following books are relevan for he course: NL Biggs: Discree Mahemaics, Oxford Universiy Press, 1989, ISBN , JH Conway & RK Guy: The book of numbers, Springer-Verlag, 1996, ISBN X, A beauiful book deeply suble mahemaics presened in an accessible and exciing way. H Davenpor: The higher arihmeic (6 h ediion), Cambridge Universiy Press, 1992, ISBN , P Giblin: Primes and programming, Cambridge Universiy Press, 1993, ISBN , RL Graham, DE Knuh & O Paashnik: Concree mahemaics (2 nd ediion), Addison Wesley, 1994, ISBN , The ulimae reference book. JF Humphreys & MY Pres: Numbers, groups and codes, Cambridge Universiy Press, 1989, ISBN , Close o he approach in his course. HF Mason: Discree Mahemaics, Wiley, 1993, ISBN , N Nissanke: Inroducory logic and ses for Compuer Scieniss, Addison-Wesley, 1999, ISBN , G Pólya: How o solve i, Penguin, 1990, ISBN , KH Rosen: Discree mahemaics and is applicaions (4 h ediion), McGraw-Hill, 1999, IBN , An excellen book covering a wide range of opics and useful hroughou he course. These noes do no consiue a complee ranscrip of all he lecures and hey are no a subsiue for ex books. They are inended o give a reasonable synopsis of he subjecs discussed, bu hey give neiher complee proofs of all he heorems nor all he background maerial. Michaelmas

5 Proof Wha is a proof? If a heorem is a logical saemen, he proof is mean o convince you ha he saemen is rue. When faced wih a proof you should convince yourself of hree hings: The argumens pu forward are all rue and he sequence follows logically from beginning o end. The argumens are sufficien o prove he heorem. The argumens are all necessary o prove he heorem. A proof has o encompass all he possible cases permied by he saemen of he proof. Usually i will no be possible o work hrough all of hese in urn, so some generaliy will be required. On he oher hand, a single couner-example is sufficien o show ha a heorem is false. Indeed, such a couner-example should be as simple as possible. Good mahemaicians like o avoid effor. This should no be confused wih proof by conradicion. This is an elegan echnique in which we prove a heorem by acceping he possibiliy ha i is no rue. If i is no rue, here mus be a couner-example. Examining his couner-example hen gives rise o a logical inconsisency. If all he inermediae seps are correc, he only explanaion is ha he original assumpion (acceping ha he heorem was no rue) was iself misaken. In oher words, he heorem is rue. Examples Theorem: a n + b n = c n has no soluions. Proof: Lef as an exercise for he reader. Theorem: The whole numbers ha can be expressed as he difference of wo squares are precisely hose ha leave a remainder of 0, 1 or 3 when divided by 4. Proof: Work hrough a sequence of simpler problems. a) Any odd number can be expressed as he difference of wo squares consider (n+1) 2 n 2. b) No even number can be expressed as he difference of wo squares false, consider 4 = c) Any exac muliple of 4 can be expressed as he difference of wo squares consider (n+1) 2 (n 1) 2. d) No odd muliple of wo can be expressed as he difference of wo squares assume rue and find a conradicion by examining cases. Now combine hese resuls. (d) shows ha any difference of wo squares leaves a remainder of 0, 1 or 3 when divided by 4. (a) shows ha a number ha leaves remainder 0 when divided by 4 can be expressed as he difference of wo squares, and (c) shows ha a number ha leaves a remainder of 1 or 3 can. Theorem: 2 is irraional, ha is, i can no be wrien as a fracion y x for whole numbers x and y. x Proof: Assume ha 2 = for whole numbers x and y. Wihou loss of generaliy, we y can assume ha x and y are no boh even and deduce a conradicion. Michaelmas

6 How o solve i Pólya suggess he following four sep plan for problem solving: Undersanding he problem Wha is he unknown? Wha are he daa? Wha is he condiion? Is i possible o saisfy he condiion? Is he condiion sufficien o deermine he unknown? Or is i insufficien? Or redundan? Or conradicory? Draw a figure. Inroduce suiable noaion. Separae he various pars of he condiion. Can you wrie hem down? Devising a plan Find he connecion beween he daa and he unknown. You may be obliged o consider auxiliary problems if an immediae connecion canno be found. You should obain evenually a plan of he soluion. Have you seen i before? Or have you seen he same problem in a slighly differen form? Do you know a relaed problem? Do you know a heorem ha could be useful? Look a he unknown! And ry o hink of a familiar problem having he same or a similar unknown. Here is a problem relaed o yours and solved before. Could you use i? Could you use is resuls? Could you use is mehod? Should you inroduce some auxiliary elemen in order o make is use possible? Could you resae he problem? Could you resae i sill differenly? Go back o definiions. If you canno solve he proposed problem ry o solve firs some relaed problem. Could you imagine a more accessible relaed problem? A more general problem? A more special problem? An analogous problem? Could you solve a par of he problem? Keep only a par of he condiion, drop he oher par; how far is he unknown hen deermined, how can i vary? Could you derive somehing useful from he daa? Could you hink of oher daa appropriae o deermine he unknown? Could you change he unknown or daa, or boh if necessary, so ha he new unknown and he new daa are nearer o each oher? Did you use all he daa? Did you use he whole condiion? Have you aken ino accoun all essenial noions involved in he problem? Carrying ou he plan Carrying ou your plan of he soluion, check each sep. Can you see clearly ha he sep is correc? Can you prove ha i is correc? Looking back Can you check he resul? Can you check he argumen? Can you derive he resul differenly? Can you see i a a glance? Can you use he resul, or he mehod, for some oher problem? Michaelmas

7 Inegers We sar wih he ses of naural numbers, N = {1, 2, 3, }, he naural numbers augmened wih 0, N 0 = {0, 1, 2, }, and inegers, Z = {, -2, -1, 0, 1, 2, }, and will refer o he raional numbers (fracions), Q, and he real numbers R. The curly brackes jus wrap up enumeraions of elemens. The empy se is Ø = {}. We will discuss he noaion for ses more formally in he second half of he course, bu here is enough o ge sared. A paricular value, x, is an elemen of a se X if i is in i. We wrie his wih a sor of Greek epsilon: x X. So -3 Z bu -3 N. One se, X, is a subse of anoher se, Y, if every elemen of X is also an elemen of Y. We wrie his wih a rounded less-han-or-equal sign: X Y. So N N 0 Z Q R. We can also define ses by predicaes or condiions: N = {x Z x > 0}. This noaion is a bi unforunae because we will also use he verical bar o indicae exac divisibiliy: 3 6. So he se of even numbers migh be defined as E = {x Z 2 x}, which is a bi confusing. Sorry. The verical bar is also used in anoher way o coun he number of elemens X in a (finie) se. There are wo paricularly imporan properies of he naural numbers, which urn ou o be equivalen: inducion and well-ordering. Mahemaical inducion Le P(n) be any mahemaical asserion involving he naural number n which may be rue or false. (Think of P as a funcion wih n as an argumen and reurning a Boolean resul.) The principle of mahemaical inducion saes ha, if P(1) is rue, and whenever P(k) is rue hen P(k+1) is rue as well hen P(n) is rue for every naural number n. The wo condiions are known as he base case and he inducive sep, and hey give rise o he conclusion. Examples 1 2 Base case: 1 = ½ L + n = n( n + 1). Inducive sep: Suppose k = ½ k (k+1). Then k + (k+1) = ½ k (k+1) + (k+1) = ½ (k+1) (k+2). Le a n = 2 3n n+1. Then, for all posiive inegers n, a n is exacly divisible by 5. Base case: a 1 = = = 25, which is divisible by 5. Inducive sep: Suppose a k is divisible by 5. Then a k+1 = 2 3(k+1) (k+1)+1 = 8 2 3k k+1 = 5 2 3k a k, which is divisible by 5. Michaelmas

8 If n is a posiive ineger and x and y are any numbers, hen n n n n n ( x+ y) = x x y x y i L L 0 1 n y n n n i i n n where n! is he binomial coefficien, defined o be k k!( n k)! wih 0! = Base case: = = 1, so (x + y) 1 = x + y. 0 1 n n n + 1 Inducive sep: Observe ha + = by direc algebra. k k 1 k Assume he expansion for (x + y) k and muliply i by (x + y) o produce (x + y) k+1 and group erms wih he same powers of x and y in he sum. An alernaive saemen of he principle (known as course of values inducion) saes ha, whenever P(k) can be inferred from he ruh of P(j) for all j < k, hen P(n) is rue for every naural number n. Noe ha P(1) is rue since here are no naural numbers j wih j < 1. Well ordering Any non-empy subse of N conains a smalles elemen. Tha may seem obvious, bu i is no rue for he inegers, raionals or reals. I is also an imporan propery ha will exend o oher ses where each elemen does no have a naural successor and so ordinary inducion can no be used. However, some sor of ordering relaion is sill necessary. Equivalence Well-ordering implies mahemaical inducion. Proof Suppose ha he asserion P(n) saisfies he wo condiions for mahemaical inducion. So P(1) is rue and P(k) implies P(k+1). We need o show ha P(n) is rue for every naural number n. The proof sars raher surprisingly. We give up and suppose ha we can no manage i. In oher words we suppose ha here are some naural numbers n for which P(n) is false. These couner-examples are going o be ineresing, so gaher hem all ogeher in a se. Le S = {x N P(x) is false}. If S is empy hen here are no couner-examples, so P(n) was rue for every naural number n afer all. If S is no empy hen, by well-ordering, i conains a leas elemen. Call i s. Now s 1 since he base case said ha P(1) was rue and so 1 S. Therefore s > 1 and s 1 N. Moreover, P(s 1) mus be rue since s was he smalles couner-example. Bu he inducive sep now says ha P(s) mus be rue and so s S. This is a conradicion, so somehing has gone wrong. The only possibiliy is our original supposiion ha he heorem was no rue. In oher words, S is empy and P(n) is rue for every naural number n. Michaelmas

9 Exercises Prove ha L + n = n( n + 1)(2n + 1) Find he sum of he firs n cubes. Calculae he firs few cases, formulae a general rule and confirm i by inducion. 3. Evaluae he sum 1 2! + 2 3! + 3 n + L +. 4! ( n + 1)! 4. Show ha 7 divides 2 4n n+1 and 13 divides 3 n n-1 for all naural numbers n. 5. The Fibonacci numbers are defined by f 0 = 0, f 1 = 1 and f = f + f for n > 1. n n 1 n 2 Show ha f n = n n for all n 0. Hin: If using inducion, you need o consider wo base cases. 6. Prove ha, for all n N 0 and x R wih x -1, (1 + x) n 1 + n x. 7. A riomino is an L-shaped paern made from hree square iles. A 2 k 2 k chessboard, whose squares are he same size as he iles, has an arbirary square pained purple. Show ha he chessboard can be covered wih riominoes so ha only he purple square is exposed. Hin: Use inducion. The base case has k = 1 and he inducive sep requires you o find four similar bu smaller problems. 8. A prison houses 100 inmaes, one in each of 100 cells, guarded by a oal of 100 warders. One evening, all he cells are locked and he keys lef in he locks. As he firs warder leaves, she urns every key, unlocking all he doors. The second warder urns every second key, re-locking every even numbered cell. The hird warder urns every hird key and so on. Finally he las warder urns he key in jus he las cell. Which doors are lef unlocked and why? Hin: This is a quesion abou division. 9. Le S = {1, 2,, n}. Wrie f ( s) for he sum s S f ( s) and s= 1 s S produc similarly, wih he convenion ha he empy produc f ( s) = 1 o prove ha s S ( 1+ x ) = x. s T S T T Deduce ha 1 ) = ( 1) s S ( x s x. T S T n s φ f ( s) for he. Use inducion 10. [Mahemaical Tripos Par 1A 1988, Paper 6, Quesion 9] Sae he principle of mahemaical inducion. Prove your saemen, assuming ha every non-empy subse of he naural numbers conains a leas elemen. The Maser of Regens College and his wife invie n Fellows and heir spouses o a pary. Afer he pary he Maser asks everyone (including his own wife) how many people hey shook hands wih, and receives 2n + 1 differen answers. Of course, no woman shook hands wih her own husband. Show ha he person who shook he mos hands was no he Maser s wife. How many hands did he Maser shake? Michaelmas

10 Hin: Consider he larges and smalles numbers of people wih whom a gues could shake hands. Wha does his ell you abou he answers ha he Maser received? Wha does his ell you abou he relaionship beween he person who shook mos hands and he person who shook leas? Anoher hin: The las par of he quesion is abou inducion. 11. Complee he proof of equivalence by using inducion o prove ha he naural numbers are well-ordered. Hin: Use conradicion. Suppose ha X is a se of naural numbers which conains no leas elemen. You need o prove ha X is empy. Le L be he se of naural numbers n such ha n is no greaer han or equal o any elemen in X. Show by inducion ha L is he se of naural numbers, so X is, indeed, empy. 12. [No o be aken oo seriously.] Commen on he following alleged proofs by inducion (wih acknowledgemens o Professor JWS Cassels): Le n be a naural number and a j be real numbers for 1 j n. Then a = a for 1 j n, 1 k n. Proof Cerainly rue for n = 1. Assume he resul is rue for n and prove i for n+1. By case n of he resul, we have a 1 = a 2 = L = a n. Applying his o he a j+1 insead of he a j we have a2 = L = a n = an+ 1. Hence a1 = a2 = L = an = an+ 1, which is he resul for n+1. Every naural number n is ineresing. Proof There cerainly are some ineresing naural numbers: 0 is he smalles, 1 is he only naural number whose reciprocal is a naural number, 2 is he smalles prime, 3 is he number of persons in he Triniy, and so on. So, if he saemen were false, here would be a smalles naural number n which is no ineresing. This is a conradicion, since n would be a very ineresing number indeed. Every odd ineger > 1 is prime. Proof The economis s proof runs as follows. 3 is prime, 5 is prime, 7 is prime. Three cases in a row is surely enough. If, however, we imagine an idealised economis who would no be saisfied by his, hen he res of he proof would coninue as follows: Look a he nex odd ineger, 9. Well, i is admiedly no a prime; here mus be some unusual facor of some kind operaing. Le s go on looking a he figures. 11 is prime, 13 is prime. Two more confirmaions, so i mus be rue. Every prime is odd. Proof 3, 5, 7, 11, 13, 17, 19, are all odd. There only remains 2, which mus be he oddes prime of all. n 2 n + 41 is prime for all naural numbers n. Proof The physicis s proof runs as follows. Wrie a compuer program o check successively ha n 2 n + 41 is prime for n = 0, 1, 2, 30. Since quie a number of cases have now been verified using very expensive equipmen, he resul mus be rue. j k Michaelmas

11 Facors The operaions of addiion, muliplicaion and ordering on he inegers have some useful properies. Division Given inegers a and b, we say ha a divides b or a is a facor of b (wrien a b) if b = qa for some ineger q. Moreover, a is a proper divisor of b if a b and a ±1 or ± b. Observaions If a b and b c hen a c. Proof: If a b and b c hen b = q a and c = r b for some q and r. So c = (r q) a and a c. If d a and d b hen d (a x + b y) for any inegers x and y. Proof: If d a and d b hen a = q d and b = r d for some q and r. So a x + b y = q x d + r y d = (q x + r y) d and d (a x + b y). (a x + b y) is called a linear combinaion of a and b (or of x and y). Division algorihm Given a Z and b N, here exis unique inegers q, r Z wih a = bq + r and 0 r < b. q is called he quoien and r is he remainder afer dividing a by b. The laer is wrien as a mod b or, someimes, as a % b. So b a if, and only if, r = 0, ha is, a mod b = 0. Proof: Exisence. Consider R = {a bk k Z and (a bk) 0}. R N 0 and is no empy, so use well ordering o find is smalles elemen, r. r R, so r 0 and we can wrie r = a bq. Now r < b or r b would be a smaller elemen of R. Uniqueness. Suppose a = bq 1 + r 1 and a = bq 2 + r 2 wih 0 r 1 < b and 0 r 2 < b. Then b(q 1 q 2 ) + (r 1 r 2 ) = 0, bu b < (r 1 r 2 ) < b, so r 1 = r 2 and q 1 = q 2. This is no acually an algorihm in he normal sense undersood by compuer scieniss, bu here are algorihms ha implemen division in hardware or sofware. The imporan mahemaical resul is he exisence and uniqueness of quoiens and remainders. Highes common facors Given a, b N, he highes common facor (HCF) or greaes common divisor (GCD) of a and b, wrien as (a, b), is defined o be d N saisfying: d a and d b, and if e a and e b hen e d. The second condiion implies ha e d, bu is a more general expression ha allows he proofs ha follow o be exended easily ino ses oher han he inegers. Observaions The HCF exiss and is unique. Michaelmas

12 Proof: Exisence. Consider D = {as + b s, Z and (as + b) > 0}. a = a1 + b0 D so D Ø. By well ordering D has a leas elemen, d, and d = as + b for some s and. Use he division algorihm o wrie a = dq + r wih 0 r < d. Now r = a dq = a (as + b)q = a(1 sq) + b( q). If r > 0, hen r D. Bu r < d and d is minimal in D, so r D and r 0. Bu r 0 so r = 0 and d a. d b similarly. Now suppose e a and e b. Say a = fe and b = ge. Then d = as + b = fes + ge = e(fs + g) and e d. Uniqueness. Suppose d 1 and d 2 are boh HCFs saisfying he wo condiions. Then d 1 d 2 and d 2 d 1, so d 1 = d 2. There are inegers x and y wih d = ax + by. Moreover, x and y can be calculaed efficienly. Proof: x = s and y = in he above for exisence. See below for algorihm. If a, b N and a = bq + r for inegers q and r wih 0 r < b, hen (a, b) = (b, r). This will give rise o an efficien algorihm for HCFs. Proof: Suppose d = (a, b) and a = bq + r by he division algorihm. d a and d b so d (a bq) = r. Therefore d (b, r). Bu (b, r) b and (b, r) r, so (b, r) a. Therefore (b, r) (a, b) = d, so (b, r) = d. If a bn and (a, b) = 1, hen a n. Proof: a bn, so wrie bn = aq. (a, b) = 1, so find x and y wih ax + by = 1. Now n = nax + nby = nax + aqy = a(nx + qy), and so a n. If a n, b n and (a, b) = 1, hen ab n. Proof: (a, b) = 1, so wrie n = nax + nby as before. a n, so ab nb and ab nby. b n so ab nax similarly. Hence ab n. a b, = 1. ( a, b) ( a, b) a b Proof: Use conradicion. Suppose, = k > 1. Then k (a, b) a and ( a, b) ( a, b) k (a, b) b, which conradics (a, b) being he highes common facor. We say ha a and b are co-prime if (a, b) = 1. The leas common muliple of a and b is he smalles number m which is exacly divisible by boh a and b. This is someimes wrien as [a, b] and is equal o ab (a, b). Euclid s algorihm Given a, b N, use he division algorihm o wrie: a = q 1 b + r 1 0 r 1 < b b = q 2 r 1 + r 2 0 r 2 < r 1 r 1 = q 3 r 2 + r 3 0 r 3 < r 2 r i-2 = q i r i-1 + r i 0 r i < r i-1 r n-2 = q n r n-1 wih remainder r n = 0 Michaelmas

13 Then (a, b) = (b, r 1 ) = (r 1, r 2 ) = = (r n-2, r n-1 ) = r n-1. Moreover, we can now work backwards hrough he algorihm o calculae he inegers x and y wih (a, b) = ax + by. Alernaively, we can produce he same resul working forwards by observing ha line i is jus he difference of line i 2 and q i imes line i 1. Wrie r -1 = a and r 0 = b, so q i is jus he ineger quoien of r i-2 divided by r i-1. Now express r i = s i a + i b so s -1 = 1, -1 = 0, s 0 = 0 and 0 = 1 and observe ha r i = r i-2 q i r i-1, s i = s i-2 q i s i-1 and i = i-2 q i i-1. Here is a worked example: i q i r i s i i a = 55 = b = 20 = = The las line ells us ha = 0 so 4k.55 11k.20 = 0. This is raher like he finding he complemenary funcion ha solves he homogeneous par of a differenial equaion. The penulimae line ells us ha (55, 20) = 5 = This is raher like finding he paricular soluion for an inhomogeneous differenial equaion. Observaions The signs of s i alernae and hose of i alernae Proof: a, b and all he remainders r i are posiive, so he quoiens q i will be as well. s i-1 i s i i-1 = ( 1) i for i 0, so, in paricular, s i and i are co-prime. Proof: By inducion. Corollary: We have a linear combinaion of s i and i which is equal o 1, so (s i, i ) = 1. s b (, ), a ab ( ab, ). n = n = Proof: Noe ha r n = s n a + n b = 0 and divide hrough by (a, b) o show a b b a b sn = n. Remember, = 1, so sn. ( a, b) ( a, b) ( a, b) ( a, b) ( a, b) Bu (s n, n ) = 1 by he above, so s b n a,. Hence b s b = n a,. b Applicaions ( ) ( ) Given a, b, c Z wih a and b no boh zero, he linear Diophanine equaion ax + by = c has a soluion wih x, y Z if, and only if, (a, b) c. Proof: ( ) (a, b) a and (a, b) b, so (a, b) (ax + by) = c. Michaelmas

14 ( ) Suppose (a, b) c. Wrie c f =. Find s and wih (a, b) = as + b using ( a, b) c afs + bf = ( as + b) f = a, b =, as required. a, b Euclid. Now ( ) ( ) c kb Moreover, any soluion o au + bv = c has u = x and ( a, b) k Z. ka = for some ( a, b) v y + Proof: Suppose ax + by = c and au + bv = c. Then a(x u) + b(y v) = 0, so a a(x u) = b(v y). Divide by (a, b), so ( ) ( ) b x u = ( ) ( v y). a, b a, b a b a Bu, = 1 ( a, b) ( a, b) a, b b( v y) kb Now u = x = x. a a, b, so ( ) ( v y) ( ) ka. Hence v = y +. ( a, b) The general soluion is jus he sum of he paricular soluion u = x and v = y wih he kb ka complemenary funcion u = and v = where k is an arbirary consan. ( a, b) ( a, b) a b can be wrien as he coninued fracion q Proof: Wrie a = q 1 b + r 1 so Bu b = q 2 r 1 + r 2 so Efficiency Euclid s algorihm finds (a, b) in O(log a) seps. b r 1 a b 1 q r q 1 + q 1 = 1 + = 1 +. b b r1 1 rn 2 = q2 +, and so on unil = qn + 0. r r r = q1 + L. 1 q2 + q3 + + q + L Proof: a = q 1 b + r 1 b + r 1 > 2r 1 > 2 2 r 3 > 2 3 r 5 > > 2 k r 2k-1. So r 2k-1 < a / 2 k. In paricular, k > log 2 a implies ha r 2k-1 < 1, so r 2k-1 = 0 and he algorihm has finished. Hence Euclid s algorihm akes a mos 2 log 2 a seps. In fac we can do beer han his. If a > b and b has d digis (o he base 10), hen Euclid s algorihm will ake a mos 5d + 2 seps o find (a, b). I is acually raher hard o say how many seps will be required for any given pair of numbers. So we follow Pólya s advice and ask a differen quesion. Wha is he smalles number ha will require n seps? This will arise when q i = 1 for 1 i < n and q n = 2. Using he earlier noaion, s i = s i-1 + s i-2 and i = i-1 + i-2 so s i = f i and i = f i+1 where f i is he i h Fibonacci number. So, if b < f n, s n < f n and we need fewer han n seps. However, if n = 5d + 2, hen f n > 1.6 n-2 = 1.6 5d > 10 d > b, as required. Of course, his is sill O(log a). n 1 3 Michaelmas

15 Primes A naural number p is prime if p > 1 and p has no proper divisor. Observaions If p is a prime and p ab for a, b N bu p / a, hen p b. Proof: If p / a, hen (p, a) = 1 and so p b. There are infiniely many primes. Proof: Use conradicion. Suppose ha he only primes were p 1, p 2, p 3, p n. Consider N = p 1 p 2 p 3 p n + 1. The smalles number ha divides exacly ino N mus be a prime, bu each of p 1, p 2, p 3, p n leaves remainder 1 when divided ino N. Hence N iself mus be a prime, bu i isn in he lis. If p is a prime hen p is irraional; ha is, i can no be expressed as a raio of wo naural numbers. a Proof: Use conradicion. Suppose p = for a, b N wih (a, b) = 1. Then b p pb 2 = a 2, so p a. Wrie a = pc so pb 2 = p 2 c 2 and p b. Hence p (a, b). Digressions is prime. The Mersenne number 2 n 1 is prime only when n is prime, bu ha is no sufficien. The Ferma number 2 n + 1 is prime only when n is of he form 2 m, bu ha is no sufficien. If p is a Ferma prime, hen i is possible o consruc a regular p gon using only pencil, ruler and compasses. Le Π(x) be he number of primes x. Then Π(x) x / ln x. Prime pair conjecure: There are infiniely many primes p wih p + 2 also prime. Goldbach conjecure: Every even ineger greaer han 2 can be expressed as he sum of wo primes. Fundamenal heorem of arihmeic Every naural number greaer han 1 can be expressed as a produc of primes. Moreover, he expression is unique up o he order of he primes. Proof Exisence. Use conradicion. Le n N be he smalles couner-example. If n is prime, hen we are done. Oherwise n = ab for some a, b N wih a, b < n. Wrie a and b as producs of primes and combine hem o give an expression for n. Uniqueness. Suppose n = p 1 p 2 p 3 p r = q 1 q 2 q 3 q s wih he p i and q j all prime. p 1 n, so p 1 q 1 q 2 q 3 q s. Now, eiher p 1 q 1 or p 1 q 2 q 3 q s. In he laer case, coninue unil p 1 q j for some j. Bu q j is prime, so p 1 = q j. Renumber so j = 1. Now p 1 p 2 p 3 p r = p 1 q 2 q 3 q s so p 2 p 3 p r = q 2 q 3 q s. Coninue in his way unil p r = q s and r = s. Michaelmas

16 Observaion If Exercises r 1 r p r k k m = p K p and s 1 s p min( r1, s1 ) min( r2, s2 ) min( r, k ) (, ) k s m n p1 p2 K pk s k k n = p K p hen = and max( r1, s1 ) max( r2, s2 ) max( r, k ) [, ] k s m n p1 p2 K pk =. 1. Are he following saemens rue or false? (a, b) (c, d) = (ac, bd) (a, b) (a, d) = (a 2, bd) (a, b) = (a, d) = 1 implies ha (a, bd) = 1 2. Prove ha, if x and y are inegers such ha 57x + 44y = 1, hen here is an ineger k such ha x = 17 44k and y = 57k Does he equaion 1992x y = 12 have a soluion in inegers? Find all he ineger soluions o he equaion 1992x y = Find inegers x, y and z such ha 56x + 63y + 72z = A phoocopier charges 7.2p for each copy. However, i only acceps 10p coins and gives no change, alhough unused credi is carried forward. Wha is he smalles number of copies ha mus be made if he user is no o forgo any change? 6. Define he leas common muliple of a and b o be m = [a, b] = ab (a, b). Show ha: a m and b m, and if a n and b n hen m n. 7. Le Z[ 5] = {a + b 5 a, b Z}. Show ha α, β Z[ 5] α+β, α β, α β Z[ 5]. By facoring 4 in Z[ 5] in wo differen ways, show ha here is no analogue of prime facorisaion in Z[ 5]. 8. Show ha here are infiniely many prime numbers of he form 4k + 3. [Hin: Consider N = p n 1.] 9. A Pyhagorean Triad is a riple (a, b, c) wih a, b, c N such ha a 2 + b 2 = c 2. For example, (3, 4, 5) and (5, 12, 13) are Pyhagorean Triads. Check he following: (m(p 2 q 2 ), 2mpq, m(p 2 +q 2 )) is a Pyhagorean Triad for any m, p, q N wih p > q. If (a, b, c) is a Pyhagorean Triad, hen we can wrie a = md, b = me and c = mf where d, e and f are pairwise co-prime (ha is, (d, e) = (e, f) = (f, d) = 1), and exacly one of d and e is even, say e = 2g. Moreover, f + d = 2h and f d = 2i for h, i N. Since g 2 = hi and (h, i) = 1, i follows ha h = p 2 and i = q 2 for p, q N. Hence every Pyhagorean Triad is of he form (m(p 2 q 2 ), 2mpq, m(p 2 +q 2 )). Moreover, differen values of m, p and q give rise o differen values. Michaelmas

17 10. Recall he Fibonacci numbers {f n }. Show, by inducion on k or oherwise, ha f n+k = f k f n+1 + f k-1 f n. Deduce ha f n f ln for all l 1. Show ha (f n, f n-1 ) = 1. Deduce also ha (f m, f n ) = (f m-n, f n ) and hence ha (f m, f n ) = f (m, n). Show ha f m f n f mn if (m, n) = 1. Programming 11. Wrie an ML funcion o facor an ineger ino a lis of prime facors. 12. Wrie an ML funcion o implemen Euclid s algorihm. Given wo inegers a and b, his should reurn a riple (x, y, z) such ha ax + by = z where z is he greaes common divisor of a and b. Michaelmas

18 Modular arihmeic If a, b Z and m N hen we say ha a and b are congruen modulo m if m (a b), and we wrie his as a b (mod m). This equivalen o saying ha here is q Z such ha a = b + qm. Observaions For all a Z and m N we have a a (mod m). Proof: m 0 = (a a). If a b (mod m) hen b a (mod m). Proof: If a b (mod m) hen m (a b), so m (b a), and b a (mod m). If a b (mod m) and b c (mod m), hen a c (mod m). Proof: If a b (mod m) and b c (mod m), hen find r, s Z such ha a b = rm and b c = sm. Then a c = (r + s) m and a c (mod m). If a 1 b 1 (mod m) and a 2 b 2 (mod m) hen a 1 + a 2 b 1 + b 2 (mod m), a 1 a 2 b 1 b 2 (mod m) and a 1 a 2 b 1 b 2 (mod m). Proof: Find q 1, q 2 Z such ha a 1 b 1 = q 1 m and a 2 b 2 = q 2 m. Then (a 1 + a 2 ) (b 1 + b 2 ) = (q 1 + q 2 ) m, so a 1 + a 2 b 1 + b 2 (mod m), (a 1 a 2 ) (b 1 b 2 ) = (q 1 q 2 ) m, so a 1 a 2 b 1 b 2 (mod m), and a 1 a 2 b 1 b 2 = (b 1 + q 1 m)(b 2 + q 2 m) b 1 b 2 = (b 1 q 2 + q 1 b 2 + q 1 q 2 m) m, so a 1 a 2 b 1 b 2 (mod m). However, a b (mod m) does no imply ha x a x b (mod m). For example, consider a = 1, b = 4, m = 3, and x = 2. Examples Here are he addiion and muliplicaion ables modulo 4: and muliplicaion modulo 5: Michaelmas

19 Applicaions No ineger congruen o 3 modulo 4 can be expressed as he sum of wo squares. Proof: All squares modulo 4 are congruen o eiher 0 or 1, so he sum of wo squares will be congruen o 0, 1 or 2. No ineger congruen o 7 modulo 8 can be expressed as he sum of hree squares. Proof: All squares modulo 8 are congruen o 0, 1 or 4, so he sum of hree squares will be congruen o 0, 1, 2, 3, 4, 5 or 6. I ranspires ha any ineger can be expressed as he sum of four squares, bu his is harder o prove. 3 5 (2 n n ) Proof: Observe ha 2 3n+1 2, 1, 3, 4, 2, 1, 3, 4, (mod 5) and 3 n+1 3, 4, 2, 1, 3, 4, 2, 1 (mod 5) for n = 0, 1, 2, 3,, so heir sum will be congruen o 0 (mod 5). There is no ineger soluion o x 3 x 2 + x + 1 = 0. Proof: Consider he equaion modulo 2. x 0 could no be a soluion, bu x 1 migh be. This ells us ha any soluion would have o be odd. However, considering he equaion modulo 3 shows ha none of x 0, x 1 or x 2 could be a soluion and so here is no soluion (2 + 1). 5 Proof: Consider p = 641, so p = = and (mod p). Observe also ha p 1 = 640 = = so (mod p). Combine hese o see ha 2 32 = ( 5 4 ) (2 7 ) 4 = (5 2 7 ) 4 ( 1) 4 = 1 (mod p). So p ( ). Solving congruences The residues modulo m are Z m = {0, 1, 2, (m 1)}. Addiion, subracion and muliplicaion all work for residues, bu wha abou division? Given a, c Z and m N, he congruence ax c (mod m) has a soluion for x if, and only if, (a, m) c. Proof ax c (mod m) has a soluion we can find x wih m (ax c) we can find x and y wih ax c = my Michaelmas

20 we can find x and y wih ax my = c (a, m) c by he applicaion of Euclid s algorihm o linear Diophanine equaions. Moreover, by considering he complemenary funcion o he equaion, he soluion is unique modulo m (a, m). Unis In paricular, we can calculae he reciprocal of a modulo m if, and only if, (a, m) = 1. Such values a are called unis modulo m and we wrie U m = {a Z m a is a uni}. Observaions If a, b U m hen ab U m. Proof: If a, b U m hen we can find x, y so ha ax by 1 (mod m). So he produc (ab)(xy) = (ax)(by) 1 (mod m) and ab is a uni. The reciprocal of a modulo m is unique modulo m. Proof: Suppose ax 1 (mod m) and ay 1 (mod m). Then m (ax 1) and m (ay 1), so m ((ax 1) (ay 1)) = a(x y). Bu (m, a) = 1 so m (x y) and x y (mod m). We can calculae unis by using he exended Euclid s algorihm o express (a, m) = 1 as a linear combinaion of a and m. Euler s oien funcion Define ϕ(m) o be he number of naural numbers less han m and co-prime o m, so ϕ(m) is he number of unis modulo m. Given a prime p, observe ϕ(p) = (p 1) and ϕ(p n ) = p n p n-1. Chinese Remainder Theorem Given wo naural numbers m and n wih greaes common divisor 1, here is a simulaneous soluion o he congruences x a (mod m) and x b (mod n) and his soluion is unique (mod mn). Proof Exisence. Use Euclid s algorihm o find s and such ha ms + n = 1. Le c = bms + an. Now n 1 (mod m) so c an a (mod m). Similarly c b (mod n). Uniqueness. Suppose here is a furher soluion d. Observe ha c d 0 (mod m) and c d 0 (mod n), so c d 0 (mod mn) as required. Corollaries Euler s oien funcion is muliplicaive: if (m, n) = 1 hen ϕ(mn) = ϕ(m)ϕ(n). Proof: Given c U m and d U n find e Z mn wih c e (mod m) and d e (mod n). Then e U mn and each such pair (c, d) is linked o a unique e. ϕ ( m) = m (1 1 ). p prime p m Proof: Consider he unique expression of m as a produc of primes. Michaelmas

21 Wilson s heorem If p is a prime, hen (p 1)! -1 (mod p). Proof Associae each of he numbers 1, 2,, p 1 wih is reciprocal (mod p). The reciprocal of a may be he same as a, bu only if a 2 1 (mod p) which requires a = 1 or p 1. Apar from hese, he numbers 2, 3,, p 2 can be paired off so ha he produc of each pair is 1 (mod p). I follows ha 2.3..(p 2) 1 (mod p). Muliply by p 1 1 (mod p) o obain he resul. This proof acually fails if p = 2 or 3, bu hese cases are easily verified independenly. Euler s heorem 1 Given m 2 and a wih (a, m) = 1, hen a ϕ(m) 1 (mod m). Proof Le U m = {x 0 < x < m and (x, m) = 1} be he se of unis modulo m. Say U m = {u 1, u 2, u f } where f = ϕ(m). Muliply each of hese u i by a modulo m. The resuling values are coprime o m, since u i and a are. Moreover hey are disinc, since a is a uni and can be divided, so au i au j (mod m) u i u j (mod m). So hey are jus a permuaion of he f values in U m. Hence he produc (au 1 )(au 2 ) (au f ) u 1 u 2 u f (mod m). Bu u 1, u 2, u f are all unis and so can be divided ou, leaving a f 1 (mod m) as required. Corollary (Ferma s lile heorem) 2 Given a prime p and a no divisible by p, hen a p-1 1 (mod p). Moreover, for any a, a p a (mod p). Observaion This gives a es for primaliy. If a number p does no saisfy a p-1 1 (mod p) for any single value of a, hen p can no be prime. However, passing his es is no sufficien o prove primaliy. Composie numbers ha saisfy he es are called pseudo-prime wih respec o he base a. Carmichael numbers, such as 561, are Ferma pseudo-primes for all possible bases a. p 1 2 The Ferma-Euler es a ± 1(mod p) is sharper bu is sill no sufficien. In paricular, i reveals 561 o be composie bu fails o cach Public key crypography Wih he increasing use of compuer neworks and digial, elecronic communicaions, i becomes imporan o ensure ha messages can be sen securely wih he meaning revealed only o he inended recipien and ha hey can be auhenicaed as having been sen by he real originaor. 1 Humphreys & Pres, p Humphreys & Pres, p 54. Michaelmas

22 The general approach is o choose some large modulus m and encode blocks of a message as numbers in Z m. Caesar s cypher encodes a message a as a + e (mod m) for some encrypion key, e. This is decoded by calculaing (a + e) e a (mod m). Unforunaely, he code is also easily broken by frequency analysis. Using larger blocks and changing e in some agreed sequence unknown o inercepors gives a one-ime pad, which is secure bu difficul o adminiser. A furher problem is he disribuion of he keys. The key can be any secre shared by he wo paricipans. How can one pass i safely o he oher? The rick is o imagine a box wih wo locks and proceed as follows: The sender (convenionally called Alice) places he secre in he box, locks one of he locks wih her key and sends he locked box o he recipien (convenionally called Bob). Bob locks he second lock wih his key and reurns he box o Alice. Alice unlocks he firs lock and reurns he box o Bob. Bob unlocks he second lock, opens he box and exracs he secre. Noe ha Alice and Bob never have o share heir privae keys wih anyone else bu he box is always securely locked when in ransi beween hem. The rick is o find an arihmeic equivalen of a box wih wo locks. Modular addiion is a possibiliy. Alice and Bob agree on a modular base m (which can be made public) and choose privae values a and b. Alice now sends a shared secre s o Bob as follows: A B: m 1 s + a (mod m) B A: m 2 m 1 + b = s + a + b (mod m) A B: m 3 m 2 a = s + b (mod m) Bob can now recover s. Unforunaely, anyone overhearing he conversaion (radiionally called Eve) can recover s m 1 m 2 + m 3 (mod m). Modular muliplicaion is anoher possibiliy. As long as a and b are co-prime o m, Alice and Bob can calculae muliplicaive inverses and replace he subracions by divisions in he above proocol. The same problem arises and Eve can recover s or, sricly speaking, s (mod m/(m,s)) if (m,s) > 1. However, modular exponeniaion really does work. Diffie-Hellman key exchange 3 Choose a large prime modulus, p. Pick e wih (e, p 1) = 1 and find d such ha de 1 (mod p 1) so de = 1 + (p 1) for some. Then (a e ) d = a ed = a 1+(p-1) = a(a p-1 ) a1 (mod p) = a. We now have a proocol: Alice chooses p and he value e and sends p and he message a e o Bob. Bob picks anoher value f wih inverse g and sends (a e ) f back o Alice. Alice works ou ((a e ) f ) d = ((a e ) d ) f = a f and sends i back o Bob. Bob now works ou (a f ) g back o decode and find a. 3 Davenpor, p 191. Michaelmas

23 Breaking his requires discree logarihms, which is as hard as facoring a large ineger. The RSA code 4 The Rives, Shamir and Adleman (RSA) public key sysem 5 uses Euler s Theorem o provide secure communicaions and digial signaures. Le p and q be wo primes wih produc m so ϕ(m) = (p 1)(q 1). Choose e (he encrypion exponen) relaively prime o ϕ(m) and use Euclid s algorihm o find d (he decrypion exponen) and c such ha ed + ϕ(m)c = 1 so ed 1 (mod ϕ(m)). Now, given a < p, q, (a e ) d = a ed = a 1-ϕ(m)c = a(a ϕ(m) ) -c a1 -c = a (mod m) provided (a, m) = 1, which is ensured by a < p, q. We now have a proocol: Alice picks wo large primes and publishes heir produc m and he value e. Bob encodes a message a as a e (mod m) and sends i o Alice. Alice recovers a by raising he encoded message o he power d (mod m). Anyone inerceping he message knows m and e bu no d which can only be calculaed easily if ϕ(m) is known. However, his is believed o be difficul, a leas as difficul as facoring m. Conversely, if d is known, hen m can be facored as follows: de 1 (mod ϕ(m)), so suppose ha de 1 = nϕ(m). Observe ϕ(m) = (p 1)(q 1) = pq p q + 1, which is slighly smaller han pq = m. So n is slighly greaer han (de 1)/m. Calculaing his fracion and rounding up will give n. Once n is known, ϕ(m) = (de 1)/n. Now m + 1 ϕ(m) = p + q and m = pq, so p and q are he roos of he quadraic equaion x 2 (m + 1 ϕ(m))x + m = 0. The encoding and decoding processes are symmeric and can be performed in eiher order. Thus Alice can prove her ideniy by aking a challenge a and reurning a d (mod m) which anyone can hen decode bu only she could have encoded. Coin-ossing by elephone 6 Le p be a prime of he form 4k + 3 and suppose a x 2 (mod p). Now 4k+ 2 p 1 x = x 1(mod p), so ( a k k+ 4 2 ) x x a(mod p) and x = o he original equaion. So we can calculae square roos mod p. a k +1 is a soluion Le p and q be wo such primes wih produc n and suppose a z 2 (mod n). Now a is also a square modulo boh p and q, say a x 2 (mod p) and a y 2 (mod q). Use he Chinese Remainder Theorem o consruc 4 soluions z ±s, ± (mod n). 2 2 Observe ha, if we know boh s and, i is possible o facor n. s a(mod n), so 2 2 pq = n ( s ) = ( s + )( s ). However, s and are disinc so neiher ( s+ ) nor ( s ) is divisible by n. Wihou loss of generaliy, p ( s+ ) and q ( s ), and we can use Euclid o find p and q as he HCFs of n and ( s+ ) and ( s ) respecively. We now have a proocol: 4 Humphreys & Pres, p R Rives, A Shamir & L Adleman: A mehod for obaining digial signaures and public-key cryposysems, Communicaions ACM 21(2), February 1978, pp Giblin, p 145. Michaelmas

24 Alice picks wo large primes and ells Bob heir produc n. Bob picks s co-prime o n and ells Alice a s 2 (mod n). Alice calculaes he 4 roos, picks one a random and ells Bob. If his is ± s, Bob concedes defea. Oherwise i is ± which allows Bob o facor n and, by so doing, win. Pracical remarks These mahemaical resuls are no sufficien by hemselves o build secure encrypion sysems. Care mus be aken over he acual choice of he prime numbers used and, even more imporanly, over he sysems procedures. The securiy course explores hese issues furher. Exercises 1. Show ha a number is divisible by 9 if, and only if, he sum of is digis is divisible by 9. (This is known as casing ou he 9s.) For example, is no divisible by 9 as = 17 which is no divisible by Find a similar es for divisibiliy by Is i possible o form a sum of numbers using each of he digis 0 o 9 exacly once whose oal is 100? (Tricks like exponeniaion are no allowed.) 4. A digi number is exacly divisible by 99. A new number is formed by reversing he order of is digis. Wha is he probabiliy ha he new number is also exacly divisible by 99? 5. The Inernaional Sandard Book Number (ISBN) found in he fron of many books is a 10 digi code such as (where he hyphens can be ignored). In his case, he 0 indicaes ha he book was published in he UK and some oher English speaking counries, 521 is he publisher (he Cambridge Universiy Press), is he book number and 4 a check digi. The check digi is chosen so ha if he ISBN is d 1 d 2 d 10 9 hen d10 = i= 1 i di (mod11). I may be ha he las digi has o be 10, in which case X is wrien, as in X. 10 Prove ha i= 1 i d i 0(mod11) and verify ha he wo given ISBNs saisfy he congruence. Prove ha he check digi will show up common copying errors caused by inerchanging wo adjacen digis (so, for example, 67 becomes 76) or doubling he wrong one of a riple (so, for example, 667 becomes 677). Why do you hink he modulus 11 was chosen insead of he more naural 10? 6. Show ha he equaion x 5 3x 2 + 2x 1 = 0 has no soluions for x Z. 7. Solve he following congruences: 77x 11 (mod 40) 12y 30 (mod 54) z 13 (mod 21) and 3z 2 (mod 17) 8. A band of 15 piraes acquires a hoard of gold pieces. When hey come o divide up he coins, hey find ha hree are lef over. Their discussion of wha o do wih hese exra coins becomes animaed and, by he ime some semblance of order reurns, here remain only seven piraes capable of making an effecive claim on he hoard. However, when he hoard is divided beween hese seven, i is found ha wo pieces are lef over. There ensues an unforunae repeiion of he earlier disagreemen, bu his does a leas have he consequence ha he four piraes who remain are able o divide he hoard evenly Michaelmas

25 beween hemselves. Wha is he smalles number of gold pieces ha could have been in he hoard? 7 9. Calculae 20! (mod 23). 10. Calculae (mod 257). 11. Show ha 42 (n 7 n) for all posiive inegers n. 12. An unwise person publishes he RSA enciphering scheme (m, e) = (3901, 1997) via which he wishes o receive messages. You inercep he ransmission Facor m and hence find he deciphering key d such ha de 1 (mod ϕ(m)). Assuming ha each block of four digis encodes wo leers under he map a-z, space,?,!, 0-9 become 00-25, 26, 27, 28, 29-38, decipher he ex. (You may need o wrie and use he programs below.) 13. The previous quesion uses code blocks ha are larger han he wo primes whose produc forms he base. Verify ha a paricular code block which shares a facor wih m sill can be encoded and decoded correcly. Why does his work? is a prime of he form 4k + 3 (wih k = 2) so we can exrac he square roo of a by raising a o he power k + 1 = 3. For example, he square roo of 5 is 5 3 = (mod 11) and we can check ha 4 2 = 16 5 (mod 11). However, he same approach fails o calculae he square roo of 6. Explain. Programming 15. Wrie an ML funcion o calculae he reciprocal of a number o a given modular base. This may well use he funcion for Euclid s algorihm wrien earlier. 16. Wrie an ML funcion o calculae powers of numbers o a given modular base. 7 Humphreys & Pres, p 50. Michaelmas

26 Revision guide The following diagram shows he developmen of he key ideas presened in he firs half of he course: Inegers Well ordering Inducion Division Highes common facors Primes Euclid s algorihm Modular arihmeic Diophanine equaions Chinese Remainder hm Linear congruences Ferma-Euler heorem Coin ossing Diffie-Hellman RSA Michaelmas

27 Ses A se is jus a collecion of objecs, or elemens. We wrie x A when an elemen x is in he se A and x A when i isn. Ses can be finie or infinie. (Indeed, here are many differen infinie sizes.) If hey are finie, you can define hem explicily by lising heir elemens, oherwise a paern or resricion can be used: L = {a, b, c} M = {alpha, bravo, charlie} N = {1, 2, 3, } Z = { -3, -2, -1, 0, 1, 2, 3, } P = {x N x > 1 and 1 < y < x (x, y) = 1} Given a finie se, A, wrie A for he number of elemens in A. Wrie for he empy se, {}. So = 0. One se, A, is a subse of anoher se, B, if every elemen of A is also a member of B. We wrie his wih a rounded less-han-or-equal sign: A B. So N Z. When he conainmen is sric (as in his case), we wrie N Z for a proper subse. Two ses, A and B, are equal if hey conain he same elemens. This will ofen be proved by showing ha each is a subse of he oher: A B and B A. Ses can hemselves be members of oher ses. There is an imporan disincion beween, for example, {a, b, c} and {{a, b, c}}, or beween and { }. Again, paerns can be used: S = {X L a X} = {{a}, {a, b}, {a, c}, {a, b, c}} hen L S bu a S. The power se, P (X), is he se of all subses of X. So, for he se L above: P (L) = {, {a}, {b}, {c}, {a, b}, {a, c}, {b, c}, {a, b, c}} I is imporan o specify he universe of discourse when discussing ses. This is he se of all possible elemens ha migh be considered. I is ofen wrien as Ω. Russell s paradox Consider R = {X X X}. Then L R and S R, bu is R R? Combining ses There are several ways of combining exising ses o make new ones: The complemen of a se, A, is he collecion, A c or A, of elemens (wihin he universe of discourse) ha are no in A. A c = {x Ω x A}. This is a case where i is necessary o be paricularly clear abou he universe. Oher operaions include: Union A B = {x x A OR x B} Inersecion A B = {x x A AND x B} Len

28 Difference A \ B = {x x A AND x B} = A B c Symmeric difference A B = (A \ B) (B \ A) These operaions saisfy various properies: Idempoence A A = A A A = A Complemens A A c = Ω A A c = (A c ) c = A Commuaiviy A B = B A A B = B A Associaiviy A (B C) = (A B) C A (B C) = (A B) C De Morgan s Laws (A B) c = A c B c (A B) c = A c B c Disribuiviy A (B C) = (A B) (A C) A (B C) = (A B) (A C) Empy se A = A A = Universal se A Ω = Ω A Ω = A Absorpion A (A B) = A A (A B) = A Ω Venn diagrams Venn diagrams provide a way of showing combinaions of ses: A A A B C B C B C A A B A B A A A B C B C B C A \ B A B A B C Normal form Symmeric difference can be expressed as a union of differences and each difference can be expressed as an inersecion of ses and complemens of ses. De Morgan s Laws can be used o expand complemened expressions and disribuiviy can be used o expand inersecions ino Len

29 unions. Togeher hese ransformaions allow any expression o be reduced o a union of erms each of which is he inersecion of he underlying ses and heir complemens. This expression is unique (up o he order of he erms and he facors wihin each erm) so wo expressions can be checked for equaliy by reducing hem o his normal form. Pariions A pariion of a se Ω is jus a division of he whole of Ω ino non-overlapping subses. Mahemaically a pariion P of a se Ω is a subse P P (Ω) such ha U S = Ω (P covers Ω) and S P If S, T P hen S T implies ha S = T (he elemens of P are disjoin). Examples: {{a, b, c}} and {{a}, {b, c}} are boh pariions of {a, b, c}. Neiher {a, b, c} nor {{a, b}, {b, c}} are pariions of {a, b, c}. Produc ses The produc of wo ses A and B is he se of pairs of elemens from A and B: A B = {(a, b) a A AND b B} So {a, b} {b, c} = {(a, b), (a, c), (b, b), (b,c)} This can be exended o ordered n-uples: A 1 = A A n = A A n-1 for n > 1 For convenience, we wrie elemens as (a, b, c) raher han (a, (b, c)). This gives he usual noaion for Euclidean space, R 3. Disjoin sums The disjoin sum of wo ses A and B is A + B = ({0} A) ({1} B). So {a, b} + {b, c} = {(0, a), (0, b), (1, b), (1, c)} while {a, b} {b, c} = {a, b, c}. Indicaor funcions Given a se A Ω, define he indicaor or characerisic funcion for A for x Ω by 1 if x A I A ( x) = 0 oherwise. Observaions I A B (x) = MAX (I A (x), I B (x)) I A B (x) = MIN (I A (x), I B (x)) = I A (x) I B (x) I c A (x) = 1 - I A (x) A = {x Ω I A (x) = 1} A = I ( x) x Ω A Len