B GIAO THNG VN TI TRNG I HC HNG HI B MN: KHOA HC MY TNH KHOA: CNG NGH THNG TIN BI GING TON RI RC TN HC PHN: TON RI RC M HC PHN: 17203 TRNH O TO: I HC CHNH QUY DNG CHO SV NGNH: CNG NGH THNG TIN
HI PHNG - 2010 MC LC NI DUNGTRANG Chng 1. i cng v logic1 1.1. Php tnh mnh 1 1.1.1. Khi nim v mnh v chn tr1 1.1.2. Cc php ton trn mnh 2 1.2. Biu thc logic5 1.2.1. nh ngha v bng chn tr ca biu thc logic5 1.2.2. S tng ng logic8 1.2.3. Gi tr ca biu thc logic8 1.3. Cc lut logic8 1.3.1. Cc lut logic8 1.3.2. Cc quy tc thay th10 1.3.3. V d p dng11 1.4. Cc dng chun tc12 1.4.1. Chun tc tuyn12 1.4.2. Chun tc hi13 1.5. Quy tc suy din14 1.5.1. i cng v quy tc suy din14 1.5.2. Kim tra mt quy tc suy din 16 1.5.3. Cc quy tc suy din c bn17 1.5.4. Cc v d p dng18 1.6. V t, lng t20 1.6.1. nh ngha v t v v d20 1.6.2. Cc php ton trn v t21 1.6.3. Lng t v mnh c lng t21 1.6.4. Quy tc ph nh mnh c lng t23 1.6.5. Mt s quy tc dng trong suy lun25 Chng 2. Cc phng php chng minh29 2.1. Cc phng php chng minh c bn29 2.1.1. Khi nim v chng minh29 2.1.2. Chng minh trc tip29 2.1.3. Chng minh phn chng31 2.1.4. Chng minh bng cch phn chia trng hp33 NI DUNGTRANG 2.1.5. Phn v d34 2.2. Nguyn l quy np35 2.2.1. i cng v quy np35 2.2.2. Cc nguyn l quy np thng dng36 2.2.3. Cc v d38 Chng 3. Phng php m41 3.1. Tp hp41 3.1.1. Khi nim tp hp41 3.1.2. Quan h bao hm trong v tp hp con 42 3.1.3. Cc php ton trn tp hp43 3.1.4. Tch Decartes ca cc tp hp45 3.2. Cc nguyn l m45 3.2.1. Php m45 3.2.2. Nguyn l cng46 3.2.3. Nguyn l nhn48 3.2.4. Nguyn l b tr52 3.2.5. Nguyn l Dirichlet53 Chng 4. Quan h56 4.1. Quan h hai ngi56 4.1.1. nh ngha quan h v v d56 4.1.2. Cc tnh cht ca quan h57 4.1.3. Biu din quan h58 4.2. Quan h tng ng59 4.2.1. Khi nim quan h tng ng59 4.2.2. Lp tng ng v tp hp tng ng59 4.3. Quan h th t60 4.3.1. Cc nh ngha60 4.3.2. Biu din quan h th t62 4.3.3. Tp hu hn c th t63 4.3.4. Sp xp topo63 4.4. Dn (lattice - tp b chn)65 Chng 5. i s Bool71 5.1. Cc php ton71 NI DUNGTRANG 5.1.1. Cc nh ngha71 5.1.2. Cc tnh cht ca php ton hai ngi73 5.2. i s Bool78 5.2.1. nh nghi v cc tnh cht82 5.2.2. i s Bool v dn80 5.3. Cc cng logic v t hp cc cng logic85 5.3.1. Cc cng logic 85 5.3.2. Mch logic86 5.4. Cc tiu ho cc mch logic91 5.4.1. Bn Karnaugh92 5.4.2. Phng php Quine-McCluskey94 Tn hc phn: Ton ri rc Loi hc phn: 1 B mn ph trch ging dy: Khoa hc my tnh Khoa ph trch: CNTT M hc phn: 17203Tng s TC:2 TS titL thuytThc hnh/XeminaT hcBi tp ln n mn hc 45450000 iu kin tin quyt: Mn hc c th b tr hc t hc k u tin. Mc tiu ca hc phn: Gip sinh vin nm c nhng kin thc c bn v l thuyt t hp, ton logic, h ton mnh . Phng php suy din v chng minh. i s Bool. Ni dung ch yu Gip sinh vin nm c nhng kin thc c bn v l thuyt t hp, h ton mnh , cc phng php m, khi nim quan h, i s Bool v lm c s cho cc mn hc chuyn ngnh khc. Ni dung chi tit ca hc phn: TN CHNG MC PHN PHI S TIT TSLTTH/XeminaBTKT Chng 1. i cng v logic1616000 1.1. Php tnh mnh 21.1.1. Khi nim v mnh v chn tr1.1.2. Cc php ton trn mnh 1.2. Biu thc logic 21.2.1. nh ngha v bng chn tr ca biu thc logic1.2.2. S tng ng logic1.2.4. Gi tr ca biu thc logic1.3. Cc lut logic 31.3.1. Cc lut logic1.3.2. Cc quy tc thay th1.3.3. V d p dng1.4. Cc dng chun tc 2 TN CHNG MC PHN PHI S TIT TSLTTH/XeminaBTKT 1.4.1. Chun tc tuyn1.4.2. Chun tc hi1.5. Quy tc suy din 31.5.1. i cng v quy tc suy din1.5.2. Kim tra mt quy tc suy din 1.5.3. Cc quy tc suy din c bn1.5.4. Cc v d p dng1.6. V t, lng t 41.6.1. nh ngha v t v v d1.6.2. Cc php ton trn v t1.6.3. Lng t v mnh c lng t1.6.4. Quy tc ph nh mnh c lng t1.6.5. Mt s quy tc dng trong suy lunChng 2. Cc phng php chng minh65001 2.1. Cc phng php chng minh c bn 22.1.1. Khi nim v chng minh2.1.2. Chng minh trc tip2.1.3. Chng minh phn chng2.1.4. Chng minh bng cch phn chia trng hp2.1.5. Phn v d2.2. Nguyn l quy np 32.2.1. i cng v quy np2.2.2. Cc nguyn l quy np thng dng2.2.3. Cc v d1 Chng 3. Phng php m553.1. Tp hp 23.1.1. Khi nim tp hp3.1.2. Quan h bao hm trong v tp hp con 3.1.3. Cc php ton trn tp hp3.1.4. Tch Decartes ca cc tp hp3.2. Cc nguyn l m 33.2.1. Php m TN CHNG MC PHN PHI S TIT TSLTTH/XeminaBTKT 3.2.2. Nguyn l cng3.2.3. Nguyn l nhn3.2.4. Nguyn l b tr3.2.5. Nguyn l Dirichlet1 Chng 4. Quan h774.1. Quan h hai ngi 14.1.1. nh ngha quan h v v d4.1.2. Cc tnh cht ca quan h4.1.3. Biu din quan h4.2. Quan h tng ng 24.2.1. Khi nim quan h tng ng4.2.2. Lp tng ng v tp hp tng ng4.3. Quan h th t 34.3.1. Cc nh ngha4.3.2. Biu din quan h th t4.3.3. Tp hu hn c th t4.3.4. Sp xp topo4.4. Dn (lattice - tp b chn) 1Chng 5. i s Bool1110001 5.1. Cc php ton 25.1.1. Cc nh ngha5.1.2. Cc tnh cht ca php ton hai ngi5.2. i s Bool 25.2.1. nh ngha v cc tnh cht5.2.2. i s Bool v dn5.3. Cc cng logic v t hp cc cng logic 25.4. Cc tiu ho cc mch logic 41 Nhim v ca sinh vin : Thamdccbuithuyttrnhcagiovin,thc,tlmbitpdogiovingiao, tham d cc bi kim tra nh k v cui k. Ti liu hc tp : -KennethRosen,Tonhcrircvngdngtrongtinhc,NXBKHKTHni, 1998. -NguynTThnhvNguyncNgha,Giotrnhtonhcrirc,HBKH ni, 1994. -Phan nh Diu, L thuyt tmt hu hn v thut ton, NXB HTHCN, 1977. -Vng Tt t, Lgic hc i cng, NXB i hc quc gia HN, 2002 Hnh thc v tiu chun nh gi sinh vin:-Hnh thc thi cui k : Thi vit. -Sinh vin phi m bo cc iu kin theo Quy ch ca Nh trng v ca B Thang im: Thang im ch A, B, C, D, F im nh gi hc phn: Z = 0,2X + 0,8Y. 1 CHNG 1 I CNG V LOGIC 1.1. Php tnh mnh 1.1.1. Khi nim v mnh v chn tr Cc i tngc bn m chng ta kho st ylcc pht biu hay cc mnh . Tuy nhin trong chng ny ta ch xt n cc mnh ton hc, v chng ta ni vn tt cc mnh ton hc l cc mnh . l nhng pht biu din t mt tng trn vn v tacthkhngnhmtcchkhchquanlnnghocsai.Tnhchtctyucamt mnh l n ng hoc sai, v khng th vang va sai. Gi trng hoc sai ca mt mnh c gi l chn tr ca mnh . V mt k hiu, ta thng dng cc mu t (nh p, q, r, ...) k hiu cho cc mnh , v chng cng c dng k hiu cho cc bin logic, tc l cc bin ly gi tr ng hoc sai. Chn tr "ng" thng c vit l 1, v chn tr "sai" c vit l 0. V d 1: Cc pht biu sau y l cc mnh (ton hc). 1. 6 l mt s nguyn t. 2. 5 l mt s nguyn t. 3. -3 < 2 4. Tam gic cn c hai gc bng nhau. 5. H2O l mt axt. Cc mnh 2, 3, v 4 trong v d trn l nhng mnh ng. Ni cch khc chn tr ca cc mnh ny l ng. Cc mnh 1, 5 l nhng mnh sai. V d 2 : Cc pht biu sau y khng phi l cc mnh (ton hc) v tnhng sai ca chng khng xc nh. 1. Ai ang c sch? (mt cu hi) 2. Hy ng ca li i! 3. Anh ta rt thng minh. 4. Cho x l mt s nguyn dng. 5. a l mt s chnh phng. 6. x + y = z. Trongvickhostccmnh,ngitacnphnralmhailoi:mnhscp (elementary), mnh phc hp (compound). Mnh s cp l cc "nguyn t" theo ngha l n khng th c phn tch thnh mt hay nhiu (t hai tr ln) mnh thnh phn n gin hn. Cn mnh phc hp l mnh c to thnh t mt hay nhiu mnh khc 2 bng cch s dng cc lin kt logic nh t "khng" dng trong vic ph nh mt mnh , cc t ni: "v", "hay", "hoc", "suy ra", v.v.... V d : Xt cc mnh sau y. p = "15 chia ht cho 3". q = "2 l mt s nguyn t v l mt s l". Ta c p l mt mnh s cp. Nhng q l mt mnh phc hp, v mnh q c to thnh t hai mnh "2 l mt s nguyn t" v"2 l mt s l" nh vo lin kt logic "v". 1.1.2. Cc php ton trn mnh iu m chng ta quan tm y khng phi l xc nh tnh ng hoc sai ca mt mnhscp.Bivnhngmnhnythnglnhngphtbiunilnmttng no trong mt phm vi chuyn mn nht nh. Vn m ta quan tm y l lm th no tnh ton chn tr ca cc mnh phc hp theo cc mnh s cp cu thnh mnh phc hp nh vo cc php ton logic. Cc php ton logic y l cc k hiu c dng thay cho cc t lin kt logic nh "khng","v", "hay", "hoc", "suy ra" hay"nu ... th ...", "nu v ch nu". Cc php ton logic c nh ngha bng bng chn tr (truth table). Bng chn tr ch ra r rng chn tr ca mnh phc hp theo tng trng hp ca cc chn tr ca cc mnh s cp to thnh mnh phc hp. Bng chn tr ca cc php ton logic tt nhin l phn nh ng ngha t nhincacc t lin kt tng ng. V mt t nhinca ngn ng, trong nhiu trng hp c ng mt t nhng c th c ngha khc nhau trong nhng ng cnh khc nhau. Do , bngchntr khngth din t mi ngha c th cca t tng ng vi k hiu php ton. iu ny cho thy rng i s logic l r rng hon chnh theo ngha l n cho tamththnglogicngtincy.islogiccncbitquantrngtrongvicthitk mch cho my tnh. Bng chn tr khng ch dng k ra s lin h chn tr gia mnh phc hp vi chn tr caccmnhscpcuthnhn,mbngchntrcncdngvimcchrng hn: lit k s lin h chn tr gia cc mnh vi cc mnh n gin hn cu thnh chng.1.1.2.1. Php ph nh Cho p l mt mnh , chng ta dng k hiu p ch mnh ph nh ca mnh p. "S ph nh" c nh ngha bi bng chn tr sau y: p p 01 10 3 K hiu c c l "khng" . Trong mt s sch khc, ngi ta cn dng cc k hiu sau y ch mnh ph nh ca mt mnh p: ~p,pTrong ct th nht ca bng chn tr, ta lit k y cc trng hp chn tr c th c ca mnh p. ct th hai k ra chn tr tng ng ca mnh p theo tng trng hp chn tr ca mnh p. nh ngha ny phhp vi ng ngha t nhin ca s ph nh : Mnh ph nh p c chn tr l ng (1) khi mnh p c chn tr sai (0), ngc li p c chn tr sai (0) khi p c chn tr ng (1). V d 1 :Nu ta k hiu p l mnh "5 < 3" th p l k hiu cho mnh "5 3". Trong trng hp ny p sai, p ng. Ta c th vit p = 0, p = 1. V d 2 : Ch ra rng ( p) v p lun c c ng chn tr. Gii. Lp bng chn tr ca mnh ( p): p p ( p) 010 101 Trn mi dnggi tr trong bng chn tr ta c chn tr ca p v ( p) u bng nhau (so snh ct 1 v ct 3 trong bng). Vy ( p) v p c c ng chn tr. Ta cng ni rng ( p) tng ng logic vi p. Mnh ( p) thng c vit l p, v iu ny khng c g gy ra s nhm ln. 1.1.2.2. Php hi Cho p v q l hai mnh . Ta k hiu mnh "p hi q" l p. q. Php "v", k hiu l . , c nh ngha bi bng chn tr sau y: pqp . q 000 010 100 111 Chn tr ca mnh p . q ph thuc vo cc chn tr ca 2 mnh p, q. Ta c 4 trng hp chn tr ca p . q ng vi 4 trng hp chn tr ca cp mnh (p,q) l (0,0), (0,1), (1,0), (1,1).Trong4trnghpchcmttrnghpmnhp.qng,ltrnghpp ng v q ng. Qua nh ngha trn ta nhn thy rng cc mnh p . q v q . p lun lun c c ng chn tr, hay tng ng logic. Tuy nhin, trong ngn ng thng thng cc mnh "p v q" v "q v p" i khi c ngha khc nhau theo ng cnh. V d: Cho cc mnh 4 p = "5 > -7",q = "2721 l mt s nguyn t",r = "mt tam gic u cng l mt tam gic cn".Khi ta c :p . q = 0 (p L q sai, tc l c chn tr bng 0, v p = 1 v q = 0), p . r = 1 (p L r ng, tc l c chn tr bng 1, v p = 1 v r = 1). Nhn xt: Bng cch lp bng chn tr, ta c: 1. Cc mnh p v p . p lun c c ng chn tr. 2. Mnh p . p lun c chn tr bng 0 (tc l mt mnh lun sai). Mtmnhphchplunluncchntrlsaitrongmitrnghpchntrcacc mnh s cp to thnh n s c gi l mt s mu thun. 1.1.2.3. Php tuyn Cho p v q l hai mnh . Ta k hiu mnh "p hay q" l p v q. Php "hay", k hiu l v , c nh ngha bi bng chn tr sau y: pqp vq 000 011 101 111 Chntrcamnhpvqphthucvoccchntrca2mnhp,q.Trong4trng hp ch c mt trng hp mnh p v q sai, l trng hp p sai v q sai. Qua nh ngha trn ta nhn thy rng cc mnh p v q v q v p lun lun c c ng chn tr, hay tng ng logic.V d : Cho cc mnh p = "5 > 7",q = "2721 l mt s nguyn t",r = "mt tam gic u cng l mt tam gic cn".Khi ta c :p v q = 0,p v r = 1.Nhn xt : 1. Cho p l mt mnh . Lp bng chn tr ca mnh p v pp ppv p 5 011 101 ta c mnh p v p lun lun ng. 2.Ngitacnsdngphp"hoc"trongviclinktccmnh.Chopvqlhai mnh.Takhiumnh"phocq"lpq.Php"hoc",khiul,cnh ngha bi bng chn tr sau y: pq pq 000 011 101 110 Php "hoc" cn c gi l "hay loi tr". Chn tr ca mnh pq ph thuc vo cc chn tr ca 2 mnh p, q : mnh pq ng khi trong 2 mnh p v q c mt mnh ng, mt mnh sai. 1.1.2.4. Php ko theo Php ko theo, k hiu bi , c a ra m hnh cho loi pht biu iu kin c dng : "nu . . . th . . .". Cho p v q l 2 mnh , ta s vit p q din t pht biu "nu p th q". Php ton ko theo c nh ngha bi bng chn tr sau y: pqp q 001 011 100 111 Mnh p q, c c l "nu p th q", cn c pht biu di cc dng khc sau y: "q nu p". "p ch nu q". "p la` iu kin cho q". "q la` iu kin cn cho p". 1.1.2.5. Php ko theo hai chiu 6 Php ko theo 2 chiu hay php tng ng, k hiu bi , c a ra m hnh cho loi pht biu iu kin hai chiu c dng : ". . . nu v ch nu . . .". Cho p v q l 2 mnh , ta vit p q din t pht biu "p nu v ch nu q". Php ton tng ng c nh ngha bi bng chn tr sau y: pqp q 001 010 100 111 Mnh p q, c c l "p nu v ch nu q", cn c pht biu di cc dng khc sau y: "p khi v ch khi q". "p la` iu kin cn va` cho q". Mnh p q c chn tr ng (=1) trong cc trng hp p v q c c ng chn tr. 1.1.2.6. u tin ca cc ton t logic. Tngtnhiviccphptonshc,trnhphidngnhiudungoctrongcc biu thc logic, ta a ra mt th t u tin trong vic tnh ton. trn ta c 5 ton t logic: (khng) , . (v), v (hay), (ko theo), (tng ng) u tin mc 1 (cao nht) ., vu tin mc 2 (thp hn) , u tin mc 3 (thp nht) trong , cc ton t lit k trn c ng dng c c ng u tin. V d : 1. p v q c ngha l (( p) v q). 2. p vq r . s c ngha l ((( p) v q) (r . s)). 3. p vq . r l nhp nhng. Cn phi dng cc du ngoc ch r ngha. 1.2. Biu thc logic 1.2.1. nh ngha v bng chn tr ca biu thc logic Trong i s ta c cc biu thc i s c xydng t cc hng s, cc bin v cc php ton. Khi thay thcc bin trong mt biu thc i s bi cc hng s th kt qu thc hin 7 cc php ton trong biu thc s l mt hng s. Trong php tnh mnh ta cng c cc biu thc logic c xy dng t : Cc mnh hay cc gi tr hng. Cc bin mnh . Cc php ton logic, v c cc du ngoc "( )" ch r th t thc hin ca cc php ton. Gi s E, F l 2 biu thc logic, khi y E, E . F, E F, E F cng l cc biu thc logic. V d: Biu din E(p, q, r) = ((( p) vq) (r . s)) l mt biu thc logic trong p, q, r l cc bin mnh . Bng chn tr Bng chn tr ca mt biu thc logic l bng lit k chn tr ca biu thc logic theo cc trng hp v chn tr ca tt c cc bin mnh trong biu thc logic hay theo cc b gi tr ca b bin mnh . Vi mt bin mnh , ta c 2 trng hp l 0 (sai) hoc 1 (ng). Vi 2 bin mnh p, q ta 4 trng hp chn tr ca b bin (p,q) l cc b gi tr (0,0), (0,1), (1,0), v (1,1). Trong trng hp tng qut, vi n bin mnh th ta c 2n trng hp chn tr cho b n bin . V d 1: Bng chn tr ca cc biu thc logic p qv p v q theo cc bin mnh p,q nh sau: pqp q p p v q 00111 01111 10000 11101 V d 2: Bng chn tr ca cc biu thc logic p v( q . r) theo cc bin mnh p, q, r nh sau: Th tpqrq . rp v( q . r) 100000 200100 301000 401111 510001 610101 711001 8 Th tpqrq . rp v( q . r) 811111 1.2.2. S tng ng logic Hai biu thc logic E v F theo cc bin mnh no c gi l tng ng logic khi E v F lun lun c c ng chn tr trong mi trng hp chn tr ca b bin mnh .Khi ta vit: E F c l "E tng ng vi F". Nh vy, theo nh ngha ta c th kim tra xem 2 biu thc logic c tng ng hay khng bng cch lp bng chn tr ca cc biu thc logic.V d: t bng chn tr ca cc biu thc logic p q v p v q theo cc bin mnh p, q ta c: ( p q ) ( p v q ) 1.2.3. Gi tr ca biu thc logic Mt biu thc logic c to thnh t cc bin logic kt hp vi php ton logic, bi vy nn gi tr biu thc logic cng ch nhn 1 trong 2 gi tr l ng (true hoc 1) hay sai (false hoc 0) t y thuc vo gi tr ca cc bin logic v quy lut ca cc php ton. V d: Xt biu thc logic ( p v q ), nu thay p = 1 v q = 0 ta c: 1 v0= 0 v0 = 0 1.3. Cc lut logic Cc lut logic l c s ta thc hin cc bin i trn mt biu thc logic c c mt biu thc logic mi tng ng logic vi biu thc logic c trc. Mi biu thc logic cho ta mt s khng nh v s tng ng ca 2 biu thc logic. Ta s s dng cc qui tc thay thvcclutlogicbitthchinccphpbinitngngtrnccbiuthc logic. Diy,chngtaslitkramtslutlogicthngcsdngtronglplunv chng minh. Cc lut ny c th c suy ra trc tip t cc bng chn tr ca cc biu thc logic. 1.3.1. Cc lut logic Cc lut v php ph nh p p (lut ph nh ca ph nh) 1 0 0 1 Lut giao hon 9 p . q q . p p v q q v p Lut kt hp p . (q . r) (p . q) . r p v (q v r) (p v q) v r Lut phn b p . (q v r) (p v q) . (p v r) p v (q . r) (p . q) v (p . r) Lut De Morgan (p . q) p v q (p v q) p . q Lut v phn t b p v p 1 p . p 0 Lut ko theo p q p v q Lut tng ng p q (p q) . (q p) Cc lut n gin ca php tuyn p v p p (tnh ly ng ca php tuyn) p v 1 1 (lut ny cn c gi l lut thng tr) p v 0 p (lut ny cn c gi l lut trung ha) p v (p . q) p (lut ny cn c gi l lut hp th) Cc lut n gin ca php hi p . p p (tnh ly ng ca php hi) p . 1 p (lut ny cn c gi l lut trung ha) 10 p . 0 0 (lut ny cn c gi l lut thng tr) p . (p v q) p (lut ny cn c gi l lut hp th) Mt s lut trong cc lut trnh by trn c th c suy ra t cc lut khc. Chng ta cthtmracmttphplutlogictithiumttacthsuyrattccclut logickhc,nhngiunykhngquantrnglmivichngta.Nhngluttrnc chn la lm c s cho chng ta thc hin cc bin i logic, suy lun v chng minh. Tt nhin l cn nhiu lut logic khc m ta khng lit k ra y. Cc lut kt hp trnh by trn cn c gi l tnh cht kt hp ca php ton hi v phptontuyn.Dotnhchtny,tacthvitccbiuthclogichivccbiuthc tuyn di cc dng sau: E1 . E2 . . Em E1 v E2 v v Em v vic tnh ton chn tr c th c thc hinda trn mt s phn b cc cp du ngoc vo biu thc mt cch t y xc nh mt trnh t thc hin cc php ton. V d: Biu thc E1 . E2 . E3 . E4 c th c tnh ton chn tr bi biu thc sau: (E1 . E2) . (E3 . E4) hay c th tnh ton theo biu thc: E1 . ((E2 . E3 ) . E4) 1.3.2. Cc quy tc thay th Di y l cc qui tc cho ta c th suy ra nhng biu thc logicmi hay tm ra cc biu thc logic tng ng vi mt biu thc logic cho trc. Qui tc 1 Trong mt biu thc logic E, nu ta thay th mt biu thc con bi mt biu thc logic tng ng vi biu thc con th ta s c mt biu thc mi E' tng ng vi biu thc E. V d: Cho biu thc logic E = qv p. Thay th q trong biu thc E bi biu thc q (tng ng vi q) ta c mt biu thc mi E' = q v p. Theo qui tc thay th 1 ta c:q v p q v p Qui tc 2 Gi s biu thc logic E l mt hng ng. Nu ta thay th mt bin mnh p bi mt biu thc logic tu th ta s c mt biu thc logic mi E' cng l mt hng ng. 11 V d: Ta c biu thc E(p,q) = (p q) ( p v q) l mt hng ng. Thay th bin q trong biu thc E bi biu thc q . r ta c biu thc logic mi: E'(p,q,r) = (p (q . r)) ( p v (q . r)) Theo qui tc thay th 2 ta c biu thc E'(p,q,r) cng l mt hng ng. 1.3.3. V d p dng V d 1: Chng minh rng (p q) ( q p). Chng minh : (p q) p v q (lut ko theo) q v p (lut giao hon) q v p (lut ph nh) q p (lut ko theo) V d 2: Chng minh rng biu thc ((p q) . p) q l mt hng ng. Chng minh. ((p q) . p) q ((p q) . p) v q (lut ko theo) ( (p q) v p) v q (lut De Morgan) (p q) v ( p v q) (lut kt hp) (p q) v (p q) (lut ko theo) 1 (lut v phn t b) Vy biu thc ((p q) . p) q l hng ng. V d 3: Chng minh rng biu thc p . q p l mt hng ng. Chng minh. 12 p . q p ( p . q) v p (lut ko theo) ( p v q) v p (lut De Morgan) ( q v p) v p (lut giao hon) q v ( p v p) (lut kt hp) q v 1 (lut v phn t b) 1 (lut n gin) Vy mnh p . q p l hng ng. V d 4: Chng minh rng biu thc p p v q l mt mnh hng ng. Chng minh. p p v q p v (pv q) (lut ko theo) ( p v p) v q (lut kt hp) 1 v q (lut v phn t b) 1 (lut n gin) Vy mnh p p v q l hng ng. Nhn xt: Cc v d trn cho ta thy mt quan h khc gia cc mnh phc hp hay cc mnh : quan h "suy ra". Khi mnh p q l hng ng, ta ni rng p suy ra q (v mt logic).Chngtasdngkhiuchquanh"suyra".Quanhsuyranyctnh truyn (hay bc cu), nhng khng c tnh cht i xng. 1.4. Cc dng chun tc Dngchuntc(chnhtc)ca1biuthclbiudinbiuthcvdngngin, ch bao gm cc php ton ph nh, hi tuyn ca cc mnh . 1.4.1. Chun tc tuyn Gi s p1, p2, , pn l cc bin mnh . Mt biu thc logic F theo cc bin mnh p1, p2, , pn c gi l mt biu thc hi c bn (hi s cp) nu n c dng sau: F = q1 . q2 . . qn vi qj = pj hoc qj = pj (j = 1, , n). 13 V d: Biu thc x . y . z l mt biu thc hi c bn theo 3 bin mnh x, y, z. BiuthclogicE(p1,p2,,pn)theoccbinmnhp1,p2,,pncnilcdng chnh tc tuyn khi E c dng:E = E1 v E2 v v Em Trong mi biu thc con Ei u c dng chnh tc tuyn theo cc bin p1, p2, , pn. V d: Cc biu thc sau y c dng chnh tc tuyn: E(x,y,z) = (x. y . z) v ( x . y . z) v (x . y . z) F(p1,p2,p3,p4) = (p1 . p2 . p3 . p4) v (p1 . p2 . p3. p4) nh l :MibiuthclogicE(p1,p2,,pn)ucthvitdidngchnhtctuyn(chuntc tuyn) duy nht, khng k s sai khc v th t trc sau ca cc biu thc hi c bn trong php tuyn). Ni mt cch khc, ta c duy nht mt tp hp cc biu thc hi c bn { E1, E2, , Em} sao cho biu thc E(p1, p2, , pn) tng ng logic vi biu thc E1 v E2 v v Em. 1.4.2. Chun tc hi Gi s p1, p2, , pn l cc bin mnh . Mt biu thc logic F theo cc bin mnh p1, p2, , pn c gi l mt biu thc tuyn c bn (tuyn s cp)nu n c dng sau: F = q1 v q2 v v qn vi qj = pj hoc qj = pj (j = 1, , n). V d: Biu thc x v y v z l mt biu thc tuyn c bn theo 3 bin mnh x, y, z. BiuthclogicE(p1,p2,,pn)theoccbinmnhp1,p2,,pncnilcdng chnh tc hi khi E c dng: E = E1 . E2 . . Em Trong mi biu thc con Ei u c dng chnh tc tuyn theo cc bin p1, p2, , pn. V d: Cc biu thc sau y c dng chun tc hi (chnh tc hi): E(x,y,z) = (x v y v z) . ( x v y v z) . (x v y v z) F(p1,p2,p3,p4) = (p1 v p2 v p3 v p4) . (p1 v p2 v p3 v p4) nh l : 14 Mi biu thc logic E(p1, p2, , pn) u c th vit di dng chnh tc hi duy nht, khng kssaikhcvthttrcsaucaccbiuthctuyncbntrongphphi).Nimt cch khc, ta c duy nht mt tp hp cc biu thc tuyn c bn{ E1, E2, , Em} sao cho biu thc E(p1, p2, , pn) tng ng logic vi biu thcE1 . E2 . . Em. 1.5. Quy tc suy din 1.5.1. i cng v quy tc suy din Mththngtonhcbaogmcctin,ccnhngha,vnhngkhinimkhng cnhngha.Cctincginhlng.Ccnhnghacsdngxy dng hay a ra nhng khi nim mi trn c s nhng khi nim c. Mt s thut ng, khinimskhngcnhngharrngnhngcngmnhnghabicctin. Trong mt h ton hc chng ta c th suy ra c cc nh l. Mt nh l l mt khng nh c chng minh l ng. Mt s loi nh l c xem l cc b , cc h qu. Mt lp lun (hay l lun) ch ra c tnh ng n ca mnh pht biu trong nh l cgilchngminh.Logiclmtcngcchovicphntchccchngminh.Trong phn ny chng ta s cp n vic xy dng mt chng minh ton hc. thc hin c mt lp lun hay mt chng minh chng ta cn hiu cc k thut v cc cng c c s dng xydngmtchngminh.Thngthngmtchngminhsbaogmnhiubcsuy lunmmibctain(haysuyra)mtskhngnhmitnhngkhngnh bit.V d v mt bc suy din: 1/NumtdanhschLlkhcrngthtacthlyraphntutrongdanhsch.V danh schL l rng nn theo s khng nh trn ta khng th lyra phn t u trong danh sch. 2/ Nu mt danh sch L l khc rng th ta c th ly ra phn t u trong danh sch.V ta khng th ly ra phn t u trong danh sch L nn danh sch L l danh sch rng. Trong2suydinvdtrnthsuydin2/lmtsuylunng,nhngsuydin1/l khng ng. Vy lm th no bit c mt suy din l ng hay sai ? Mt bc suy lun nhthphidatrnmtquitcsuydinhplnoncxemlmtsuylun ng. Cc qui tc suy din l c s tay bit c mt lp lun hay mt chng minh l ng hay sai. Trong cc mc tip theo chng ta s xem xt chi tit hn v cc qui tc suy din v giithiumtsquitcsuydncbnthngcdngtrongvicsuylunvchng minh. nh ngha qui tc suy din Tuy c nhiu k thut, nhiu phng php chng minh khc nhau, nhng trong chng minh trong ton hc ta thng thy nhng l lun dn xut c dng: 15 Nu p1 v p2 v . . . v pnth q. Dng l lun ny c xem l hp l (c chp nhn l ng) khi ta c biu thc(p1 . p2 . . . . . pn) ql hng ng. Ta gi dng l lun trn l mt lut suy din v ngi ta cng thng vit lut suy din trn theo cc cch sau y : Cch 1: Biu thc hng ng (p1 . p2 . . . . . pn) q 1 Cch 2: Dng suy din (p1 . p2 . . . . . pn) q Cch 3: M hnh suy din p1 . . . . pn ________ q Cc biu thc logic p1, p2, . . ., pn trong lut suydin trn c gi l gi thit (hay tin ), v biu thc q c gi l kt lun. y chng ta cng cn lu rng l lun trn ng khng c ngha l tacq ng v cng khng khng nhrng p1, p2, . . ., pn u ng.L lun ch mun khng nh rng nu nh ta c p1, p2, . . ., pn l ng th ta s c q cng phi ng. V d : Gi s p v q l cc bin logic. Xc nh xem m hnh sau y c phi l mt lut suy din hay khng? p q p --------- q Gii: Lp bng chn tr ta c: pqp q(p q). p((p q) . p) q 00101 01111 10001 16 11111 Bngchntrchothybiuthc((pq) .p)qlhngng.Do,mhnhsuylun trn ng l mt lut suy din. Tht ra, ta ch cn nhn vo cc ct chn tr ca p, q, v p q trong bng chn tr l ta c th kt lun c ri, v t bng chn tr trn ta thy rng nu cc gi thit p q v p ng (c gi tr bng 1) th kt lun q cng ng.Ta c th khng nh c m hnh suy lun trn l mt lut suy din m khng cn lp bng chn tr. Gi s p q v p ng. Khi q phi ng, bi v nu ngc li (q sai) th p cng phi sai (s mu thun vi gi thit). 1.5.2. Kim tra mt quy tc suy din kim tra mt suy lun c th l ng hay khng, tc l c "hp logic" hay khng, ta c th cn c vo cc qui tc suy din (lut suy din). Php suy lun c th c th c xem nh s suy din trn cc mnh phc hp. Cc mnh s cp c th (m chn tr c th ng hoc sai) trong php suy lun s c tru tng ha (thay th) bi cc bin logic. Nh th php suy lun c tru tng ha thnh mt qui tc suy din trn cc biu thc logic m ta c th kim tra xem qui tc suy din l ng hay khng. y chnh l bin php ta bit c mt suy lun c th l ng hay sai. V d 1: Xt s suy lun sau y: Nu mt danh sch L l khc rng th ta c th ly ra phn t u trong danh sch.V ta khng th ly ra phn t u trong danh sch L nn danh sch L l danh sch rng. Trong php suy lun, ta c cc mnh s cp "danh sch L l khc rng", "ta c th ly raphntu(tdanhschL)".Thaythccmnhscpnybiccbinlogicp,q tng ng th php suy lun c th trn s c tru tng ha thnh mt suy din trn cc biu thc logic nh sau: p q q p MhnhsuydinnychnhlquitcsuydinModusTollens,cbitlng. Vy php suy lun trn l suy lun ng. V d 2: Xt xem suy lun sau y c ng hay khng? Nul s hu t th phng trnh m2 = 2n2 c nghim nguyn dng m, n. Nu phng trnh m2 = 2n2 c nghim nguyn dng m v n th ta c mu thun. Vyl s v t. Tru tng ha cc mnh s cp "l s hu t", " phng trnh m2 = 2n2 c nghim nguyn dng m, n " thnh cc bin logic p, q tng ng th php suy lun trn c dng m hnh suy din 17 p q q 0 ----------- p Kim tra m hnh suy din ny ta s thy l ng. Nh th php suy lun trn l ng. 1.5.3. Cc quy tc suy din c bn Trong mc ny chng ta nu ln mt s qui tc suy din (ng) thng c s dng m ta c th kim tra chng bng cc phng php c trnh by trong mc trc. Qui tc Modus Ponens (p q) . p q hoc l vit di dng m hnh suy din p q p q Qui tc Modus Tollens (p q) . q p hoc l vit di dng m hnh suy din p q q p Tam on lun (p q) . (q r) . (p r) hoc l vit di dng m hnh suy din p q q r ------- p r Qui tc chng minh bng phn chng p q (p q) 0 Qui tc ny cho php ta chng minh (p q) 0 thay cho p q. Ni cch khc, nu ta thm gi thit ph vo tin p m chng minh c c s mu thun th ta c th kt lun q t tin p. Qui tc chng minh theo trng hp 18 (p1 q) . (p2 q) . . . . . (pn q) (p1 v p2 v . . . v pn) q hoc l vit di dng m hnh suy din p1 q p2 q . . . pn q -------------------- (p1 v p2 v . . . v pn) q 1.5.4. Cc v d p dng Di y ta trnh by chng minh ca mt s mnh m khng nu ln mt cch chi tit v cc qui tc suy din c p dng. Ngi c c th tm thy cc qui tc suy din c s dng trong chng minh mt cch d dng. Mnh 1: Vi mi s nguyn n, n3 + 2n chia ht cho 3. Suynghutinltathyrngkhngthtmthymtthas3trongbiuthcn3+2n. Nhng khi phn tch ra tha s th n3 + 2n = n(n2 + 2). Pht biu "n3 + 2n chia ht cho 3" s ngnunlbisca3.Cncctrnghpkhcthsao?.Tathphngphpphn chng. Chng minh:Ta c n3 + 2n = n(n2 + 2), v s t nhin n cmt trong 3 dng ng vi 3 trng hp di y: Trng hp 1: n = 3k, vi k l mt s nguyn. n3 + 2n = 3k(9k2 + 2) chia ht cho 3. Trng hp 2: n = 3k+1, vi k l mt s nguyn. n3 + 2n = (3k+1)((3k+1)2 + 2) = (3k+1)(9k2 +6k+3) = (3k+1)3(3k2 +2k+1) chia ht cho 3. Trng hp 3: n = 3k+2, vi k l mt s nguyn. n3 + 2n = (3k+2)((3k+2)2 + 2) = (3k+1)(9k2 +12k+6) = (3k+1)3(3k2 +4k+2) chia ht cho 3. Trong mi trng hp (c th c) ta u c n3 + 2n u chia ht cho 3.19 Vy ta kt lun n3 + 2n chia ht cho 3 i vi mi s nguyn n. Nhnxt: Chngminh trncthctrnhbyngngnhnbngcchsdngphp ng d modulo 3. Mnh 2:Nu n2 l mt s chn th n cng l mt s chn. Suy ngh: Gi s n2 = 2k (l s chn). Ta thy kh suy ra n l s chn. Nu bit thng tin g v n th suy ra iu g v n2 th d hn. Ta th phng php phn chng. Chng minh: Ta hy chng minh mnh "Nu n l th n2 l". Cho n l mt s l, ta c n = 2k+1 (k l mt s nguyn). Do n2 = (2k+1)2 = 4k2 + 4k + 1 l mt s l.Mnh trong cp nhy kp l ng nn mnh phn o ca n cng ng. Vy, nu n2 l mt s chn th n cng l mt s chn. Mnh 3:Nu p > 3 v p nguyn t th p2 -1 chia ht cho 3. Chng minh:Ta c (p-1), p, (p+1) l 3 s nguyn lin tip. Trong 3 s nguyn ny c mt s chia ht cho 3. Nhng s khng phi l p v p l s nguyn t ln hn 3. Do (p-1) chia ht cho 3 hoc (p+1) chia ht cho 3. Suy ra (p-1)(p+1) chia ht cho 3, tc l p2 -1 chia ht cho 3. Mnh 4:S lng cc s nguyn t l v hn. Chng minh: Gi s pht biu trong mnh l sai. Tc l ch c mt s hu hn, k, s nguyn t (dng). K hiu k s nguyn t l p1, p2, . . ., pk, y k l s nguyn dng.t n = p1p2 . . . pk + 1. S n ln hn tt c k s nguyn t nn n khng nguyn t.Do , t nh l c bn ca s hc , n phi c mt c s nguyn t p.p phi l mt trong k s nguyn t. Do p , ( p1p2 . . . pk). Suy ra p , (n - p1p2 . . . pk), hay p , 1.Nh th, ta c p l mt s nguyn t v p , 1. iu ny l khng th, hay ni cch khc, ta c mt mu thun. Vy, S lng cc s nguyn t l v hn. 20 1.6. V t, lng t 1.6.1. nh ngha v t v v d nh ngha:Mt v t l mt pht biu p(x, y, ) ph thuc theo cc bin x, y, ly gi tr trn cc min xcnh A, B, no . Khi thay th cc bin trong v t bi cc gi tr c th a, b, thuc cc min xc nh th ta c mt mnh p(a, b, ) c chn tr ng hoc sai. Gi B l tp hp gm c hai gi tr : Sai (k hiu bi 0), v ng (k hiu bi 1). Mt v t p(x, y, )V d1: P(n) "n l mt s nguyn t" l mt v t trn tp hp cc s t nhin (hoc trn tp hp cc s nguyn). Ta c th thy rng: P(1) = 0, tc l P(1) "1 l mt s nguyn t" l mt mnh sai. P(2) = 1, tc l P(2) "2 l mt s nguyn t" l mt mnh ng. P(12) = 0, tc l P(12) "12 l mt s nguyn t" l mt mnh sai. P(17) = 1, tc l P(17) "17 l mt s nguyn t" l mt mnh ng. V t "n l mt s nguyn t" c th c xem l mt nh x i t tp hp cc s t nhin N vo tp hp Boole B: P : N B V d2: p(m,n) "m l mt c s ca n", vi m v n l cc bin s t nhin, cho ta mt v t theo 2 bin m v n thuc tp hp cc s t nhin. Ta c:p(2,4) = 1, p(3,4) = 0. 1.6.2. Cc php ton trn v t Cho p(x, y, ) l mt v t theo cc bin x, y, . Ph nh ca p, k hiu l p, l mt v t m khi thay cc bin x, y, bi cc phn t c th a, b, tng ng th ta c mnh (p(a, b, )). Ni mt cch khc, v t p c nh ngha bi: ( p) (x, y, ) = (p(x, y, )) Cho p(x, y, ) v q(x, y, ) l cc v t theo cc bin x, y, . Php hi ca p v q, k hiu l p q, l mt v t m khi thay cc bin x, y, bi cc phn t c th a, b, tng ng 21 th ta c mnh p(a, b, ) q(a, b, ). Ni mt cch khc, v t p ?q c nh ngha bi:(p . q) (x, y, ) = p (x, y, ) . q (x, y, ) Mt cch tng t,ccphp ton tuyn, ko theo v tng ng ca 2v t p v q c th c nh ngha nh sau: (p v q) (x, y, )= p (x, y, )v q (x, y, ) (p q) (x, y, ) = p (x, y, ) q (x, y, ) (p q) (x, y, ) = p (x, y, ) q (x, y, ) 1.6.3. Lng t v mnh c lng t Ngoi vic thay th gi tr c th cho cc bin trong v t c mt mnh ta cn c mt cch quan trng khc chuyn t v t sang mnh . l cch s dng cc lng t "vi mi" v "tn ti" (hay "c t nht mt"). Lng t c s dng ni ln rng v t ng i vimigitrthucminxcnhhaychngvimtphnccgitrthucminxc nh. Cho P(n) l mt v t theo bin s t nhin n. Pht biu "vi mi n e N, P(n)" hay mt cch vn tt (hiu ngm min xc nh) l "vi mi n, P(n)" c ngha l P c gi tr ng trn ton b min xc nh. Ni cch khc, P l nh x hng c gi tr l 1. Ta s dng k hiu " " thay th cho lng t "vi mi".Pht biu "C (t nht) mt n e N, P(n)" hay mt cch vn tt (hiu ngm min xc nh) l "C (t nht) mt n , P(n)" c ngha l P c gi tr ng i vi mt hay mt s gi tr no thuc min xc nh. Ni cch khc, P khng phi l mt nh x hng 0. Ta sdng k hiu "- " thay th cho lng t "c t nht mt". Lng t ny cn c c mt cch khc l "tn ti". Trong trng hp tng qut, gi s P(x) l mt v t theo bin x (bin x ly gi tr thuc mt min xc nh bit no v min xc nh ny c th c hiu ngm, khng cn ghi r ra). Cc cch vit sau y: x : P(x) (1) - x : P(x)(2) ln lt c dng din t cho cc pht biu sau y: "Vi mi x (thuc min xc nh) ta c P(x) l ng" "C t nht m x (thuc min xc nh) sao cho P(x) l ng" 22 Cc pht biu (1) v (2) c chn tr hon ton xc nh. Ni cch khc chng l nhng mnh .Chntrcaccmnhnycxcnhmtcchtnhintheongnghathng thng ca cc lng t. Mnh (1) l ng khi v ch khi ng vi mi gi tr t y x thuc min xc nh ta u c mnh P(x) c chn tr ng. Mnh (2) l ng khi v ch khi c mt gi tr x no thuc min xc nh m ng vi gi tr x ta c P(x) c chn tr ng. Ghi ch:Pht biu " x : P(x)" v pht biu "- x : P(x)" khng phi l v t theo bin x na m l cc mnh c chn tr xc nh l ng hoc sai. Trong cc pht biu trn bin x c lng t ha v chn tr ca pht biu khng ph thuc theo bin x na. Ta cng ni rng bin x b buc bi lng t. i vi mt v t theo nhiu bin th ta c th lng t ha mt s bin no trong v t cmtvtmitheoccbincnli.Chnghn,nup(x,y,)lmtvttheocc bin x, y, th ta c biu thc q(y, ) x : p(x, y, ) s l mt v t theo cc bin y, . Nu tt c cc bin ca v t u c lng t ha th ta s c mt mnh .Chng hn, nu p(x, y) lmt v t theo 2 bin x, y th biu thc x, - y : p(x, y) s l mt mnh , tc l c chn tr xc nh v khng ph thuc vo cc bin x, y na. Trongnhiuphtbiungitacdngcmt"tntiduynht",khiubi$!,nhl mt s lng t ha c bit. V d 1: 1. Cho v t P(n) "n l mt s nguyn t". Mnh "Vi mi s t nhin n ta c n l nguyn t" c th c vit nh sau: n e N : P(n) v mnh ny c chn tr l 0 (sai). 2. Mnh "Vi mi s nguyn n ta c 2n-1 l mt s l" c th c vit di dng k hiu nh sau: n e Z : 2n-1 l 23 v mnh ny c chn tr l 1 (ng). 3. Mnh "Ta c x2 > 0, vi mi s thc x khc 0" c th c vit l x e R - { 0} : x2 > 0 v mnh ny c chn tr l 1 (ng). V d 2: Chng minh rng : Nu n l mt s chn th n2 l s chn. Mnh cn chng minh (l ng) c vit di dng n e Z : n chn n2 chn. T ta c th trnh by chng minh nh sau: Cho n l mt s nguyn t y . Ta c: n chn n = 2m, vi m l mt s nguyn no n2 = 4m2 n2 chn. Vy pht biu trn l ng. 1.6.4. Quy tc ph nh mnh c lng t Da vo cch xc nh chn tr ca cc mnh c lng t theo ng ngha t nhin ca cc pht biu, ta c cc qui tc ph nh mnh c lng t sau y: ( x : P(x)) - x : P(x)(1) (- x : P(x)) x : P(x)(2) V d 1: Tm ph nh ca mnh "tn ti mt s thc x sao cho x2 < 0". t P(x) "x2 < 0". Mnh cho c vit di dng k hiu nh sau:- x : P(x) p dng lut ph nh mnh c lng t, ta c mnh ph nh cn tm c dng : x : P(x).Vy mnh ph nh l: "Vi mi s thc x, x2 > 0". Ghi ch : T cc qui tc trn ta c th ni chung v qui tc ph nh mnh c lng t nh sau: Nutrongmtmnhclngttathaythlngtbilngt-,lngt-bi lng t , v biu thc v t c thay th bi ph nh ca n th ta s c mnh ph nh ca mnh c lng t ban u. Qui tc ny cng p dng c cho cc mnh vi nhiu lng t. V d 2:Cho p(x, y, z) l mt v t ph thuc vo bin b ba (x,y,z)e AxBxC. Min xc nh l tch -Cat ca 3 tp hp A, B, C. Trong trng hp ny ta ni v t p l mt v t theo 3 bin x, y, z. Min xc nh tng ng ca 3 bin ny l A, B, C. Hy tm ph nh ca mnh sau: 24 x e A, - y e B, - z e C : p(x,y,z) Theo qui tc chung ta c : ( x e A, - y e B, - z e C : p(x,y,z)) - x e A, y e B, z e C : p(x,y,z) Tht ra nu thc hin tng bc theo cc qui tc (1) v (2) ta cng t c mnh ph nh nh trn: ( x e A, - y e B, - z e C : p(x,y,z)) - x e A, (- y e B, - z e C : p(x,y,z)) - x e A, y e B, (- z e C : p(x,y,z)) - x e A, y e B, ze C : p(x,y,z) V d 3:Vi mt hm s f xc nh mt ln cn ca im a e R (a l mt s thc), ta c nh ngha s lin tc ca f ti a nh sau : f lin tc ti a nu v ch nu cho mt s dng c t y , ta c mt s dng o sao cho | x-a | < o | f(x) - f(a) | < c . Nh vy f lin tc ti a khi v ch khi mnh sau y ng: "cho s dng c t y , ta c mt s dng o sao cho vi mi x ta c| x-a | < o | f(x) - f(a) | < c ". Hy tm ph nh ca mnh trn. Mnh trn c vit l : c > 0, - o > 0 : ( x : | x-a | < o | f(x) - f(a) | < c ) Theo qui tc ph nh mnh c lng t, ph ca mnh trn l: - c > 0, o > 0 : ( x : | x-a | < o | f(x) - f(a) | < c ) - c > 0, o > 0 : (- x : (| x-a | < o | f(x) - f(a) | < c )) - c > 0, o > 0 : (- x : | x-a | < o . ( | f(x) - f(a) | < c )) - c > 0, o > 0 : (- x : | x-a | < o . | f(x) - f(a) | > c ) - c > 0, o > 0, - x : | x-a | < o . | f(x) - f(a) | > c Nh vy ta c th pht biu mnh ph nh nh sau:: "Tn ti mt s dng c sao cho ng vi s dng o t y c mt s thc x tho iu kin| x-a | < o v | f(x) - f(a) | > c ". Nh vy ta c th pht biu mnh ph nh nh sau: "Tn ti mt s dng c sao cho ng vi mi s dng o t y ta c mt s thc x tha iu kin | x-a | < o v | f(x) - f(a) |> c ". 1.6.5. Mt s quy tc dng trong suy lun Thay i th t lng t ha ca 2 bin 25 Cho mt v t p(x, y) theo 2 bin x, y. Nu lng t ha c 2 bin x, y trong ta lng t ha bin y trc v lng t ha bin x sau th s c 4 mnh sau y: x, y : p(x,y) - x, y : p(x,y) x, y : p(x,y) x, y : p(x,y) Tng t ta cng c 4 mnh lng t ha t v t p(x, y) trong ta lng t ha bin x trc v lng t ha bin y sau: y, x : p(x,y) - y, x : p(x,y) y, x : p(x,y) y, x : p(x,y) nhldiychotamtstnhchtlinquannthtcaviclngthacc bin trong cc mnh c lng t. nh l: Gi s p(x, y) l mt v t theo 2 bin x, y th cc mnh sau l ng ( x, y : p(x,y) ) ( y, x : p(x,y) ) (- x, - y : p(x,y) ) (- y, - x : p(x,y) ) (- x, y : p(x,y) ) ( y, - x : p(x,y) ) Qui tc c bit ha ph dng Qui tc:Gi s mt mnh c lng t trong bin x vi min xc nh l A, c lng t ha v b buc bi lng t ph dng , v mnh l ng. Khi nu thay th x bi a e A th ta s c mt mnh ng. V d1: Bit rng pht biu "mi s nguyn t ln hn 2 u l s l" l mt mnh ng. Cho a l mt s nguyn t ln hn 2 (c nh nhng t y ). Hy chng minh rng a l mt s l. Gii:t p(n) "n l s nguyn t ln hn 2", v q(n) "n l s l". Ta c p(n) v q(n) l cc v t theo bin s t nhin n, v ta c mnh ng sau y: n : p(n) q(n) Theo qui tc trn ta suy ra p(a) q(a) = 1. Theo gi thit ta cng c 26 p(a) = 1. Suy ra q(a) = 1 (qui tc suy din Modus Ponens). Vy ta c mnh "a l mt s l" l ng. V d 2: Trong cc nh l Ton hc ta thng thy cc khng nh l cc mnh lng t ha ph dng. V d nh cc trng hp bng nhau ca 2 tam gic bt k. Khi p dng ta s c bit ha cho 2 tam gic c th. Qui tc tng qut ha ph dng Qui tc: Nu ta thay th bin x trong v t P(x) bi mt phn t a c nh nhng t y thc min xc nh ca bin x m mnh nhn c c chn tr l ng, tc l P(a) = 1, th mnh lng t ha x : P(x) l mt mnh ng. Nhn xt: Nu xem v t P(x) nh l mt hm ly gi tr Bool trn min xc nh A ca bin x th ta c mnh lng t ha x : P(x) l mt mnh ng khi v ch khi P l hm hng 1. T cc qui tc trn ta c th chng minh c mt s tnh cht suydin c pht biu trong cc mnh sau y: Mnh 1: Cho p(x) v q(x) l cc v t theo bin x ly gi tr trong tp hp A (min xc nh ca bin x l tp hp A), v a l mt phn t c nh t y thuc A. Khi y ta c qui tc suy din sau y: x : p(x) q(x) p(a) q(a) Mnh 2: Cho p(x), q(x) v r(x) l cc v t theo bin x ly gi tr trong tp hp A (min xc nh ca bin x l tp hp A). Ta c qui tc suy din sau y: x : p(x) q(x) x : q(x) r(x) ------------------------- x : p(x) r(x) BI TP Cu1:Quitcsuydinlg?Chovdvmtquitcsuydinvpdngquitcsuy din trong mt suy lun c th. 27 Cu 2: Lit k nhng qui tc suy din thng dng v cho bit tn gi (nu c) ca mi qui tc suy din. Cu3:Nulnccphngphpkimtramtquitcsuydin.ngvimiphng php hy cho mt v d minh ha. Cu 4: V t l g? Cho v d. Cu 5: Pht biu qui tc ph nh mnh c lng t (hay mnh lng t ha) v cho v d c th. Cu 6: P(n) l v t "nu 4 , n th 2 , n". Cho bit chn tr ca cc mnh sau: a/ P(12)b/ P(10)c/ - n : P(n)d/ n : P(n) Cu7:Hychobitchntrcamimnhdiyvvitmnhphnhca mnh . a/ - x : x+3 = 5 b/ x : x+3 = 5 c/ - x, - y : x+y = 3 d/ - x, y : x+y = 3 e/ x, - y : x+y = 3 f/ x, y : x+y = 3 Trong cc mnh trn cc bin x v y l cc bin thc. Cu8:Hychobitchntrcamimnhdiyvvitmnhphnhca mnh . a/ - x, - y : (x2 = y2) (x = y) d/ - x, y : (x2 = y2) (x = y) e/ x, - y : (x2 = y2) (x = y) f/ x, y : (x2 = y2) (x = y) Cu 9: Hy s dng cc k hiu ton hc v logic vit li mnh sau y: 28 Vi mi s thc dng x, c mt s t nhin n sao cho x bng 2n hoc x nm gia 2n v 2n+1. Cho bit mnh ny ng hay sai, v vit ra mnh ph nh ca n. Cu 10: Trong bi tp ny k hiu n ch mt bin nguyn. Cho cc v t : P(n) "0 < n2 s 4" R(n) "0 < n3 s 8" S(n) "0 < n s 2" a/ngvimivttrnhychobittphpccgitrnlmchovtcchntrng (=1).b/ Trong cc v t trn, nhng v t no tng ng vi nhau. c/ Mnh " n : R(n) P(n)" l ng hay sai? 29 CHNG 2 CC PHNG PHP CHNG MINH 2.1. Cc phng php chng minh c bn 2.1.1. Khi nim v chng minh Mththngtonhcbaogmcctin,ccnhngha,vnhngkhinim khng c nh ngha. Cc tin c gi nhl ng. Cc nh ngha c s dng xydnghayaranhngkhinimmitrncsnhngkhinimc.Mtsthut ng, khi nim s khng c nh ngha r rng nhng c ngm nh ngha bi cc tin . Trong mt h ton hc chng ta c th suy ra c cc nh l. Mt nh l l mt mnh c chng minh l ng. Mt s loi nh l c xem l cc b , cc h qu.Mtlplun(hayllun)chractnhngncamnhphtbiu trongnhlcgilchngminh.Logiclcsthchinvicchngminh,c bitlcclutlogicvcclutsuydin.Trongphnnychngtascpnvicxy dngmtchngminhtonhc.Cccutrcchngminhthngcsdngl:chng minh trc tip, chng minh bng cch phn chia trng hp (phn chng), phn chng, phn v d, v chng minh qui np.2.1.2. Chng minh trc tip Chng minh trc tip lphng phpchng minh suydin trc tip dnt gi thit n kt lun thng qua vic p dng cc lut suy din (hay qui tc suy din), cc nh l, cc nguyn l v cc kt qu bit. y l mt kiu t duy gii bi ton rt t nhin v ngi ta thng xuyn s dng. Trong khi suy ngh tm ra cch chng minh theo phng php ny ngi ta thng phi t tr li cc cu hi sau y: Ta s dng lut suy din no? Cc nh l no, cc kt qua no c th s dngc ta suyra cmt iu g t nhng s kin, nhng yu t hin ang c? Vic p dng nh l c kh nng s dn n kt lun hay kt qu mong mun hay khng? Trong trng hp mt bc suy din no c nhiu nh l hay nhiu lut no c th pdngcvcngckkhnngsdnnktlunhayktqumongmunthtas chn ci no? n mtgiai on no , khi gp phi s b tc th ta s phi t hirng phi chng bi ton khng c li gii, hay v kin thc ca ta cha , hay ta phi s dng mt phng php chng minh no khc? Qu tht l khng th tr li c cc cu hi mt cch y v chnh xc. N ph thuc ch yu vo kin thc, kinh nghim ca ngi gii bi ton v c s nhy bn, tnh nng ng 30 sng to ca h. Tuy nhin Nhng cu hi trn cho ta mt s nh hng chung ca qu trnh suy ngh. Ngoi ra, cng cn ni thm rng chng l c s cho vic pht trin cc h chng trnh tr gip gii ton mt cch "thng minh" trn my tnh c thit k theo phng php chng minh ny.Di y, chng ta s xem xt mt 2 v d v phng php chng minh trc tip. V d 1: Gi s p, r, s, t, u l cc mnh sau cho ta c cc mnh sau y l ng: (1) p r (2) r s (3) t v s (4) t v u (5) u. Hy chng minh mnh p l sai, tc l chng minh mnh p l ng. Chng minh: p dng lut suy din tam on lun, t (1) v (2) ta suy ra: (6) p s p dng lut logic v php ton ko theo ta c th vit li (3) di dng: (7) s t p dng lut suy din tam on lun, t (6) v (7) ta suy ra: (8) p t p dng lut logic v php ton ko theo ta c th vit li (4) di dng: (9) t u p dng lut suy din tam on lun, t (8) v (9) ta suy ra: (10) p u p dng lut suy din Modus Tollens, t (10) v (5) ta suy ra: (11) p Vy mnh p l ng. 31 V d 2: Cho p(x), q(x) v r(x) l cc v t theo bin x (x e A), v a l mt phn t c nh nhng t y ca tp hp A. Gi s ta c cc mnh sau y l ng: (1) x e A : p(x) q(x) (2) x e A : q(x) r(x) (3) p(a) Chng minh rng mnh r(a) l ng. Chng minh: p dng kt qu trong mnh 2, Bi 2, mc 2.5, t (1) v (2) ta suy ra: (4) x e A : p(x) r(x) p dng kt qu trong mnh 1, Bi 2, mc 2.5, t (3) v (4) ta suy ra: (5) r(a) Vy mnh r(a) l ng. 2.1.3. Chng minh phn chng Phng php chng minh trc tip khng phi bao gi cng s dng c trong vic chng minh ngay c i vi nhng bi ton kh n gin nh bi ton sau y: Bi ton: Chng minh rng khng c hai s nguyn dng m v n sao cho Bng cch suy ngh tm mt cch chng minh trc tip ta s gp phi b tc: Vi q = m/n l mt s hu t cho trc (m v n l cc s nguyn dng) ta khng bit lm th no suy ra mt cch trc tip rng q2 = 2. bng phn chng mt khng nh hay mt mnh no , ta tm cch rt ra t mnh mtiurrnglvlhaymtsmuthun.Vmtkthuttathnggisrng mnh cn chng minh l sai ri t suy ra mt iu mu thun vi gi thit hay cc tin ca bi ton, t i n kt lun rng mnh l ng. Ngoi ra php chng minh phn chng cn c th c thc hin nh sau: ta gi s mnh cn chng minh l sai, kt hp vi gi thit cho suy ra c mt iu mu thun no ri t kt lun rng mnh l ng. C s cho phng php chng minh ny l qui tc chng minh phn chng: p q (p . q) 0 32 Tr li bi ton trn nh mt v d, chng ta c th thc hin vic chngminh mt cch d dng nh phng php chng minh phn chng. V d: Chng minh rng khng c hai s nguyn dng m v n sao cho Chng minh phn chng: Gi s ta c mnh ngc li ca iu cn phi chng minh, tc l gi s rng: C hai s nguyn dng m v n sao cho. V mt phn s c th vit di dng ti gin, nn ta c th gi thit thm rng cc s dng mvntrongmnh(1)nguyntc ngnhau,tcl:mvnkhngccschungln hn 1. Do m2 = 2n2, t (1) ta suy ra: C hai s nguyn dng m v n nguyn t c ng nhau sao cho m2 = 2n2. Vi m v n l hai s nguyn dng tha mn iu kin trong mnh (2) trn th ta d dng ln lt suy ra c cc khng nh sau y: m v n nguyn t c ng nhau. m2 = 2 n2. m l s chn. m l s chn, tc l m = 2k vi k l mt s nguyn dng. n2 = 2k2. n2 l s chn. n l s chn. 2 l mt c s chung ca m v n, v 2 > 1. S mu thun do (3) v (10). T lp lun trn ta i n kt lun: khng c hai s nguyn dng m v n sao cho. 33 Ghi ch: Ngoi cch chng minh phn chng ta cn c th thc hin php chng minh gin tip m v thc cht phng php ny l c ng loi vi phng php chng minh phn chng. Trong cch chng minh gin tip ngi ta thit lp s ng n ca mt mnh bng cch chng minh rng mnh ngc li (tc l mnh ph nh ca mnh ) l sai. 2.1.4. Chng minh bng cch phn chia trng hp Trong phng php chng minh bng cch phn chia trng hp, chng minh mt s khng nh no ta xem xt tt c cc trng hp cth xy ra i vi cc s kin hay cc yu t lin quan trong gi thit v chng minh rng trong mi trng hp ta ta u c mnh cn chng minh l ng, v t i n kt lun rng t gi thit ta c mnh cn chngminhlng.Cschophngphpchngminhnylquitcchngminhtheo trng hp: (p1 q) . (p2 q) . . . . . (pn q) (p1 v p2 v . . . v pn) q chng minh khng nh q (l ng), chng ta phn tch gi thit c c mt s khng nh ng di dng: p1 v p2 v . . . v pn ri tm cch chng minh rng t mnh pk suy ra c mnh q, ng vi mi i t 1 ti n. Chng ta hy tr li mt bi ton v s nguyn c trnh by trong phn trc v cc v d p dng ca logic trong vic lp lun v chng minh. V d: Cho mt s nguyn n, hy chng minh rng n3 + 2n chia ht cho 3. Tm tt chng minh: Vi mt s nguyn n, gi r l d s trong php chia n cho 3, ta c 3 trng hp: Trng hp 1: r = 0 Trng hp 1: r = 1 Trng hp 1: r = 2 Ni cch khc ta c: (r = 0) v (r = 1) v (r = 2). Trong mi trng hp ta u suy ra c n3 + 2n chia ht cho 3. y khng trnh by li chi tit vic tnh ton suy din trong mi trng hp. T i n kt lun: n3 + 2n chia ht cho 3. 34 2.1.5. Phn v d Ni mt cch tng qut, phn v d l vic ch ra mt tnh hung hay trng hp sai ca mt khng nh ph qut chng t rng khng nh ph qut l sai. Chng hn nh chng minh mnh x e A : P(x) l sai ta ch cn a ra mt phn t a c th thuc tp hp A m P(a) l sai. Tht ra lm nh vy tc l ta chng minh mnh - x e A : P(x) (c c ng chn tr vi mnh [ x e A : P(x)] ) l ng. Chng ta c th nu ln mt bi ton khc m i vi n ta phidng phn v d. l bi ton chng minh mt php suy din t p1, p2, , pn suy ra q l sai. chng minh php suy din l sai ta phi chng minh rng p1 . p2 . . . . . pn q khng phi l hng ng. lm iu ny chng ta ch cn tm v ch ra mt trng hp c th ca cc bin mnh m ng vi chng ta c cc tin p1, p2, , pn u ng nhng kt lun q l sai. V d: Hy kim tra suy lun sau y p r p r q ----------- q Sdngccphngphpkimtramtphpsuyluntacththycsuylun trn l khng ng. tm mt phn v d ta ch cnch ra mt trng hp v chn trca cc bin mnh sao cho cc tin trong php suy lun l ng cn kt lun l sai. V mt k thut ta s tm p, q, v r tha mn cc ng thc sau y: p r = 1 p = 1 r q = 1 35 q = 0 D dng tm thy mt trng hp phn v d l: p = 1, q = 0, r = 1. Vy suy lun cho l khng ng. 2.2. Nguyn l quy np 2.2.1. i cng v quy np 2.2.1.1. Tp hp s t nhin Tp hp gm tt c cc s t nhin (hay cc s m) c k hiu l N: N = { 0, 1, 2, . . . } Cc php ton: Trn tp hp s t nhin N ta c php ton cng (k hiu l +), v php ton nhn (k hiu l .).Phptoncng2stnhinavbchotatngscnglmtstnhincvitl a+b; cn php ton nhn 2 s t nhin a v b cho ta tch s cng l mt s t nhin c vit l a.b. Php ton cng v nhn cc s t nhin c cc tnh cht sau y: Php cng (+) v php nhn (.) c tnh giao hon v kt hp, ngha l vi mi s t nhin a, b, c ta c: a + b = b + a (a + b) + c = a + (b + c) a.b = b.a (a . b) . c = a . (b . c) Php nhn (.) phn phi i vi php cng (+), ngha l vi mi s t nhin a, b, c ta c: a . (b + c) = a . b + a . c y, php ton nhn c u tin cao hn php ton cng. Php cng v php nhn ln lt nhn 0 v 1 l phn t trung ha, ngha l ta c: a e N : a + 0 = 0 + a = a a e N : a . 1 = 1 . a = a Th t trn N: 36 Ngoi cc php ton, trn N cn c mt quan h th t ? c nh ngha nh sau: a s b - c e N : a + c = b y l mt quan h th t ton phn trn N, v theo quan h th t ny ta c: 0 < 1 < 2 < . . . < n < n+1 < . . . 0 l phn t nh nht ca tp N, nhng tp N khng c phn t ln nht. ThtstrntphpNcmttnhchtrtquantrngcphtbiutrongmnhsau y: Mnh:ChoAlmttphpccstnhinkhcrng.KhiAcphntnh nht, ngha l tn ti a e A sao cho: x e A, a s x Ghich:Phntatrongmnhtrnlduynhtvckhiulmin(A).Tnhcht c pht biu trong mnh l c s cho tnh ng n ca cc qui tc chng minh qui np s c trnh by trong mc sau. 2.2.2. Cc nguyn l quy np thng dng Cho n0 l mt s t nhin, v P(n) l mt v t theo bin t nhin n ? n0. Vn c t ra l chng minh tnh ng ng ca mnh sau y: n > n0 : P(n) Phng php chng minh qui np thng c s dng chng minh khng nh trn. Php chng minh qui np c thc hin da vo cc nguyn l qui np. Chng ta s nu ln hai dngnguynlquinpthngcsdng.Trongccpdngnguynlquinp,n0 thng l 0 hoc 1. Hai dng nguyn l qui np, c gi l dng qui np yu v dng qui np mnh s c vit di dng m hnh ca cc lut suy din. Nguyn l qui np dng yu: (c s)P(n0) (qui np) k > n0 : P(k) > P(k+1) ----------------------------------- n >n0 : P(n) Chng minh tnh ng n ca nguyn l qui np trn: 37 t A l tp hp cc s t nhin n > n0 m P(n) sai. Ta ch cn chng minh rng A = C (tp hprng)vigithitrngtachaikhngnhtrongphncsvphnquinptrong nguyn l trn. Ta s chng minh iu ny bng phng php phn chng. Gi s A = C . Theo tnh cht ca th t trn tp s t nhin N (xem mnh mc trn), A c phn t nh nht. Gi a l phn t nh nht ca tp hp A. V P(n0) ng nn a > n0+1, hay a-1>n0.Doa=min(A),nna-1eAvdoP(a-1)ng.VP(a-1)ngnntacngc P(a) ng theo khng nh phn qui np, ngha l ta cng c a e A. iu ny cho ta mt s mu thun (v a = min(A)). Vy A = C . Ta c iu cn chng minh. Nguyn l qui np dng mnh: (c s)P(n0) (qui np) k > n0 : P(n0) . P(n0+1) .. . . . P(k) P(k+1) ------------------------------------------------------------ n > n0 : P(n) Theo cc nguyn l trn, chng minh qui np bao gm 2 bc : bc c s v bc qui np. bc c s, ta phi kim chng khng nh P(n0) l ng. bc qui np, ng vi mt s t nhin k t y , ta phi chng minh mt mnh ko theo. Gi thit trong mnh ko theo bc 2 c gi l gi thit qui np. Gi thit qui np dng qui np yu l P(k), v dng mnh l P(n0) . P(n0+1) .. . . . P(k). Nguyn l qui np c rt nhiu bin th trong vic vn dng. Chng hn, t hai nguyn l trn ta c th rt ra mt nguyn l qui np c dng sau y: (c s)P(0) . P(1) (qui np) k > 1 : P(k-1) .P(k) . P(k+1) ------------------------------------------------------------ n > 0 : P(n) Trong chng minh mnh sau y, ta s dng dng qui np bin th ny. Mnh : Cho dy sx0, x1, . . ., xn, . . . c nh ngha bi : 38 x0 = 0; x1 = 1; v xn = 3xn-1 - 2xn-2 vi mi n > 2. Khi ta c: xn = 2n - 1, vi mi n > 0. Chng minh: t P(n) "xn = 2n - 1". D thy rng P(0) v P(1) l ng. By gi, ta ch cn thc hin bc qui np hon thnh php chng minh qui np. Gi s P(k-1) v P(k) ng vi mt s t nhin (t y ) k > 1. Th th xk-1 = 2k-1 - 1 v xk = 2k - 1. Do xk+1 = 3xk - 2xk-1
= 3(2k - 1) - 2(2k-1 - 1) = 3*2k - 3 - 2k - 2 = 2*2k - 1 = 2k+1 - 1 Suy ra P(k+1) ng. Vytheonguynlquinp(dngbinthcphbiu trn)taktlun:P(n)ngvi mi n > 0.2.2.3. Cc v d V d 1: Chng minh rng vi mi n > 1 (n e N) ta c: 12 + 22 + . . . + n2 = n(n+1)(2n+1) / 6 Chng minh: Vi mi s t nhin n > 1 (n e N), t P(n) "12 + 22 + . . . + n2 = n(n+1)(2n+1) / 6" Ta s chng minh mnh n > 1 : P(n) bngphngphpquinp(dngyu),nghalthchin2bcchngminhtrongphp chng minh qui np. 39 Bc c s: Khi n = 1 th P(1) l mnh "1 = 1.(1+1).(2.1+1) /6". V phi ca ng thc trong mnh tnh ra bng 1, nn ta c P(1) ng. Bc qui np: Cho n l mt s t nhin ty ln hn 0, ngha l n > 1, v gi s rng P(n) ng, tc l ta c: 12 + 22 + . . . + n2 = n(n+1)(2n+1) / 6 (GTQN) By gi ta s chng minh P(n+1) cng ng, tc l chng minh rng 12 + 22 + . . . + n2 + (n+1)2 = (n+1)(n+2)(2n+3) / 6 Tht vy, t (GTQN) ta suy ra 12 + 22 + . . . + n2 + (n+1)2 = n(n+1)(2n+1) / 6 + (n+1)2 = [(n+1) / 6] . [n(2n+1) + 6(n+1)] = [(n+1) / 6] . [2n2 + 7n + 6] = [(n+1) / 6] . (n+2) (2n+3) = n(n+1)(2n+1) / 6 Tc l ta suy ra c P(n+1) cng ng. Hai phn (phn c s v phn qui np) trong php chng minh qui np c chng minh. Vy theo nguyn l qui np ta kt lun rng vi mi n > 1 (n e N) ta c: 12 + 22 + . . . + n2 = n(n+1)(2n+1) / 6 V d 2: Chng minh nh l sau y: Cho a v b l 2 s nguyn t nhin, vi b > 0. Khi , c duy nht 2 s t nhin q v r tha 2 iu kin sau y: + a = q.b + r + 0 s r < b Chng minh: Tnhduynhtcaqvrtrongnhltrncthkimchngddng.ytaschng minh s tn ti ca q v r. Cho b l mt s t nhin khc 0 t y nhng c dnh. t P(a) "Tn ti cc s t nhin q v r tha mn 2 iu kin (1) v (2)" Ta s chng minh mnh sau y l ng: a e N : P(a) bng cch s dng nguyn l qui np dng mnh. 40 Bccs:Xttrnghpa=0,tathyviq=0vr=0thcciukin(1)v(2) c tha mn. Vy ta c P(0) ng. Bc qui np: Cho a l mt s t nhin t y v gi s rng cc mnh P(0), P(1), . . . , P(a) u ng (GTQN). Ta s chng minh P(a+1) cng ng bng cch xt 2 trng hp. + Trng hp 1: a+1 < b. Ta chn q = 0 v r = a+1 th ta c q v r tha cc iu kin a+1 = q.b + r 0 s r < b Vy trong trng hp 1 ny th P(a+1) ng. + Trng hp 2: a+1 > b. t a' = a+1-b, ta c 0 s a' s a. T (GTQN) ta suy ra P(a') ng, tc l c cc s tnhin q' v r' sao cho a' = q'.b + r', v0 s r' < b Suy ra cc s t nhin q = q'+1 v r = r' s tha mn 2 iu kin (1) a+1 = q.b + r (2) 0 s r < b Vy ta cng c P(a+1) ng. Tm li, ta chng minh phn c s v phn qui np trong nguyn l qui np. T c th kt lun rng: a e N : P(a) Ghich:nhltrnlcschovicnhnghaphptondiv(chialythngs)v php ton mod (chia ly d s). nh l ny cn c gi l thut chia Euclide. Xem xt vic chng minh ca nh l trn theo phng php qui np, chng ta c th rt ra mt thut ton (khi nim thut ton s c trnh by trong chng 3). Dng tng qut ca nh l, cho cc s nguyn, c pht biu trong bi tp 6. 41 CHNG 3 PHNG PHP M 3.1. Tp hp 3.1.1. Khi nim tp hp Khi nim tp hp c dng ch mt su tp hay mt nhm cc i tng no m ta ang quan tm xem xt, v su tp ny phi c xc nh tt. Cc i tng trong su tp hay trong nhm ny s c gi l cc phn t hay cc thnh vin ca tp hp. Tnh xc nhtt(haynivnttltnhxcnh)catphpchiutheonghalvimti tng no m ta ang quan tm th ta c th xc nh c ch xc rng trng hp no l ng trong hai trng hp sau y: Trng hp 1: i tng l mt phn t ca tp hp. Trong trng hp ny ta ni i tng thuc v tp hp. Trng hp 2: i tng khng phi l mt phn t ca tp hp. Trong trng hp ny ta ni i tng khng thuc v tp hp. thun tin cho vic cp n tp hp v sau, mi tp hp thng c t cho mt tn, chng hn nh A, B, C . Ta cng dng k hiu e din t quan h "thuc v" ca mt phn t i vi mt tp hp. Khi x l mt phn t thuc v tp hp A, th ta vit x e A vcl"xthucA",haycl"Achaphntx".Ngcli,nuxkhngphilmt phn t ca tp hp A th ta vit x e A v c l "x khng thuc A", hay c l "A khng cha phn t x". Tp hp bng nhau: Hai tp hp A v B s c xem la bng nhau khi chng c c ng cc phn t, tc l mi phn t thuc A u l phn t thuc B v ngc li. Khi y, ta vit l A = B. Tp hp rng: Tp hp khng c phn t no c gi l tp hp rng, v c k hiu l C . Cch xc nh mt tp hp: xc nh mt tp hp ta c th dng cc cch sau y: Cch lit k: Ta lit k tt c cc phn t ca tp hp gia 2 k hiu ngoc { v } . V d:A = { a, b, c } . B = { 0, 1 } . 42 Cc nu c trng ca phn t:Theocchny,xcnhmttphpAtasnuln"tnhcht"dngxcnhxem phn t trong mt khng gian U c thuc v tp hp A hay khng: phn t x ca U s thuc Akhixtha"tnhcht",vxkhngthucAkhixkhngtha"tnhcht".T"tnhcht" thng c th hin di dng mt v t p(x) theo bin xe U. Khi y, tp hp A s c vit nh sau: A = { x e U : p(x) }hay vn tt (hiu ngm tp U) l A = { x : p(x) }V d: A = { n e N : n l s nguyn t } (khi nim s nguyn t c nh ngha trong mc III) B = { n e N : c mt s t nhin m sao cho n = m2 }Cch xc nh tp hp di dng nh ca mt tp hp khc A' qua mt php tng ng f m ng vi mi x e A' ta c mt phn t tng ng f(x) duy nht trong U. Khi y ta vit A = { f(x) : x e A' }Ghi ch: Php tng ng f c ni trn y chnh l mt nh x. Khi nim nh x s c nh ngha trong mc II. V d: B = { n2 : n e N }C = { (2n+1)2 : n e N }3.1.2. Quan h bao hm trong v tp hp con nh ngha:Cho A v B l hai tp hp m cc phn t ca chng u thuc mt tp hp ln U (hay cn gi l tp v tr). Ta ni tp A bao hm trong (hay cha trong) tp B nu mi phn t ca tp hp A u thuc tp hp B. Ta cng ni rng B bao hm A (hay B cha A), v vit l: A c B (hay BA). Khi A c B ta ni A l mt tp hp con ca tp hp B. 43 V d: { 0, 1, 2} c { n e N : n < 10}N c Z c Q c R c C, trong N #9; l tp hp cc s t nhin, Z l tp hp cc s nguyn, Q #9; l tp hp cc s hu t, R #9; l tp hp cc s thc, C #9; l tp hp cc s phc. Cho X l mt tp hp. Su tp tt c cc tp hp con ca X c k hiu lP(X). Ni mt cch khc, P(X) l mt tp hp m mi phn t ca n l mt tp hp con ca X. Tnh cht: C c A v A c A, vi mi tp hp A. (A c B) . (B c A) (A = B) (A c B) . (B c C) (A c C) X c Y P(X) c P(Y) Nu tp hp X c n phn t (n e N) th tp hp P(X) c 2n phn t. 3.1.3. Cc php ton trn tp hp Trong mc ny chng ta s nu ln nh ngha cc php ton tp hp trn cc tp hp con ca mt tp hp v tr U cho trc, v pht biu mt s tnh cht lin quan n cc php ton. nh ngha cc php ton: Giaoca2tphpAvB,khiubiAB,ltphpgmttccc phn t ca U m va thuc tp A va thuc tp B. A B = { x : (x e A) . (x e B) }Hi ca 2 tp hp A v B, k hiu bi AB, l tp hp gm tt c cc phn t ca U sao cho n thuc tp A hay thuc tp B. AB = { x : (x e A) v (x e B) }Hiu ca 2 tp hp A v B, k hiu bi A \ B (hay A - B), l tp hp gm tt c cc phn t ca U sao cho n thuc tp A v khng thuc tp B. 44 A - B = { x : (x e A) . (x e B) }PhnbcatpA(trongU),khiubiAc,ltphpttcccphnt ca U m khng thuc A. Ni cch khc, Ac = U - A Cc tnh cht ca cc php ton: Tnh giao hon: A B = B A AB = BA Tnh kt hp: A (B C) = (A B) C A(BC) = (AB)C Tnh phn b: A (BC) = (A B)(A C) A(B C) = (AB) (AC) Lut De Morgan: (A B)c = AcBc (AB)c = Ac Bc Phn t trung ha: AC = A A U = A Phn b: AAc = U A Ac = CTnh thng tr: AU = U A C = C45 3.1.4. Tch Decartes ca cc tp hp Tch Descartes ca 2 tp hp: Cho 2 tp hp A v B. Tch Descartes ca tp hp A v tp hp B, c k hiu bi AxB, l tp hp gm tt c cc cp (a,b) sao cho a e A v b e B. AxB = { (a,b) : a e A . b e B }Trong trng hp B = A, ta k hiu AxB l A2. Tch Descartes ca nhiu tp hp: ChontphpA1,A2,,An(n>1).TchDescartescantphpA1,A2,,An, c k hiu bi A1xA2x xAn, l tp hp gm tt c cc b n phn t (a1, a2, , an) vi ai e Ai vi mi i = 1, , n. Trong trng hp A1 = A2 = . . . = An = A th tp hp tch A1xA2x xAn s c vit l An. 3.2. Cc nguyn l m T lu, ngi ta nghin cu vic lit k, m cc phn t hay cc i tng c nhng tnh cht no gii quyt mt s vn cn thit c t ra. Chng hn, php m c s dng trong vic phn tch v nh gi phc tp ca thut ton. K thut m cn c s dng trong vic tnh ton xc sut ca cc s kin. Trong mc ny chng ta s trnh by cc qui tc c bn ca php m. Chng s gip ch rt nhiu cho vic gii nhiu vn lin quan n vic lit k, sp xp v m. 3.2.1. Php m Cho A l mt tp hp khc rng, xc nh s phn t ca tp hp A ta thng thc hin vic m bng cch ln lt gn cho cc phn t ca A cc s t nhin k tip nhau, v s t nhin u tin (c dng gn cho phn t u tin c xem xt) l 1. Nu qu trnh ny kt thc vi s t nhin n (c gn cho phn t cui c ng) th ta ni A l mt tp hp hu hn v c n phn t. Tht ra khi thc hin vic m nh th chnh l thit lp mt song nh t A vo tp hp { 1, 2, . . ., n} . T ta c th nh ngha php m nh sau: nh ngha:Cho A l mt tp hp khc rng. Nu tn ti mt s nguyn dng n v mt song nh f t A vo { 1, 2, . . ., n} th ta ni A l mt tp hp hu hn v A c n phn t. Khi song nhf : A { 1, 2, . . ., n}l s c xem l mt php m tp hp A. Tphprngcsphntl0,vcngcxemltphuhn.Sphntca mt tp hp hu hn A c k hiu l | A |. 46 Nu tp hp A khng hu hn, ta ni A l mt tp v hn v vit | A | = Ghich:khiquthakhinimsphntivicctphpt yvso snh lc lng ca cc tp hp ngi ta a ra nh ngha v quan h ng lc lng, v cc quan h so snh lc lng khc da vo khi nim nh x. Chng hn, hai tp hp A v B c ni l ng lc lng khi tn ti mt song nh f t A vo B. Tnh cht:Cho A v B l cc tp hp hu hn. Gi s tn ti n nh t A vo B. Khi y ta c: | A | s | B |. 3.2.2. Nguyn l cng C s ca nguyn l cng l mi lin h gia s phn t ca mt tp hp vi s phn t ca cc tp hp con to thnh phn hoch ca tp hp cho, c pht biu trong mnh sau y: Mnh : Cho A v B l 2 tp hp hu hn ri nhau, ngha l A B = C . Khi y ta c: | AB | = | A | + | B | Mt cch tng qut: Nu A1, A2, ..., An l cc tp hp hu hn ri nhau, ngha l phn giao ca hai tp hp bt k trong n tp hp l rng, th s phn t ca phn hi ca cc tp hp trn bng tng ca cc s lng phn t trong mi tp hp: | A1A2. . .An | = | A1 | + | A2 | + . . . + | An | chng minh mnh trn cho trng hp 2 tp hp A v B ta c th gi m l s phn t ca tp hp A v n l s phn t ca tp hp B. Sau , t vic gi s c cc song nh f : A { 1, 2, . . ., m} v g : B { 1, 2, . . ., n} , ta c th lp d dng mt song nh h : AB { 1, 2, . . ., m+n} i n kt lun | AB | = m+n. Trong trng hp tng qut ta c th s dng nguyn l qui np, vi bc c s l vic chng minh cho trng hp 2 tp hp va trnh by trn. Ghi ch: Trong trng hp i vi hai tp hp hu hn A v B t y th ta c: | AB | = | A | + | B | - | A B | 47 Tnh cht ny c th m rng cho trng hp i vi n tp hp t y A1, A2, ..., An nh sau: | A1A2. . .An | = E 1s rs n | Ar | - E 1s r< ss n | Ar As |+ E 1s r< s< t s n | Ar As At | - . . . + (-1)n | A1 A2 . . . An | Nguyn l cng : Gi s ta phi thc hin cng vic v thc hin cng vic ny tacthchnmttronghaibinphpkhcnhautheonghalcchthchinbin php th nht lun lun khc cch thc hin bin php th hai. Bin php th nht c n cch thc hin, v i vi bin php th hai ta c m cch thc hin. Vy ta c n+m cch thc hin cng vic. V d 1: Chng ta cn chn mt sinh vin ton nm th 3 hay nm th 4 i dmthingh.Hicbaonhiucchchnlamtsinhvinnhthbit rng c 100 sinh vin ton hc nm th 3 v 85 sinh vin ton hc nm th t ? Ligii:Tacththchinmttrong2vicchnlakhcnhau: chnmtsinhvintonnm3,hocchnmtsinhvintonnm4. thchincngvicthnhttac100cch,vthchincng vic th 2 ta c 85 cch. Vy chn mt sinh vin ton theo yu cu ta c 100+85 = 185 cch. Chng ta c th m rng nguyn l cng cho trng hp nhiu s chn la hn nh sau: Gi staphithchinmtcngvicbngcchchnmttrongmschnlaccbinphp khc nhau T1, T2, ..., Tm. thc hin Ti, 1 s i s m, ta c ni cch. Vy ta s cch thc hin cngvictrnln1+n2+...+nm.Nguynlcngdngtngqutnycthcchng minh bng qui np.V d 2: Mt sinh vin c th chn mt ti t mt trong 3 danh sch cc ti. S ti trong cc danh sch ti ln lt l 23, 15, 19. Hi sinh vin c bao nhiu cch chn mt ti. Li gii: Sinh vin c th chn mt ti trong danh sch th th nht theo 23 cch, trong danh sch th hai theo 15 cch, v trong danh sch th ba theo 19 cch. Do s cch chn ti l 23+15+19 = 57. V d 3: Xc nh gi tr ca k sau khi on chng trnh sau y c thc hin xong k := 0 for i1 := 1 to n1 do k := k + 1; for i2 := 1 to n2 do k := k + 1; . 48 . . for im := 1 to nm do k := k + 1; Ligii.Gitrcakbanul0.Saulmvnglpkhcnhau. Mi thao tc lp trong mt vng lp l cng thm 1 vo k. Vng lp th icnithaotc,vttcmvnglpkhngththchin2vnglp nomtcchngthi.Dosthaotcthchinxongon chngtrnhtrnln1+n2+...+nm.ycngchnhlgitrcui c ng ca k. 3.2.3. Nguyn l nhn Cscanguynlnhnlmilinhgiasphntcamttphptch Descartes vi s phn t ca cc tp hp thnh phn to nn tp hp tch, c pht biu trong mnh sau y: Mnh : Cho A v B l 2 tp hp hu hn ri nhau. Khi y ta c: | A x B | = | A | . | B | Mt cch tng qut: Nu A1, A2, ..., An l cc tp hp hu hn th s phn t ca tch Descartescacctphptrnbngtchcaccslngphntcacctphp trn: | A1 x A2 x . . . x An | = | A1 | . | A2 | . . . . . | An | chng minh mnh trn cho trng hp 2 tp hp A v B ta c th gi m l s phn t ca tp hp A v n l s phn t ca tp hp B. Sau , t vic gi s c cc song nh f : A { 1, 2, . . ., m} v g : B { 1, 2, . . ., n} , ta c th lp d dng mt song nh h : A x B { 1, 2, . . ., mn} i n kt lun | A x B | = mn. Trong trng hp tng qut ta c th s dng nguyn l qui np, vi bc c s l vic chng minh cho trng hp 2 tp hp va trnh by trn. Nguynlnhn:Gistaphithchinmtthtcbaogmhaicngvick tipnhau.thchincngvicthnhttacn1cch,vngvimicchchn thchincngvicthnhttacn2cchthchincngvicthhai.Vytacs cch thc hin th tc l n1 x n2. 49 Nguyn l nhn trn c th c m rng v c dng tng qut nh sau: Gi s mt th tc bao gm m cng vic k tip nhau T1, T2, . . ., Tm. Nu cng vic T1 c th c thc hin theo n1 cch, v sau khi chn cch thc hin cho T1 ta c n2 cch thc hin T2, v.v cho n cuic ng,saukhichncchthchincccngvicT1,T2,...,Tm-1 tacnmcchthc hin Tm. Vy ta c n1.n2. ... .nm cch thc hin th tc. Nguyn l nhn dng tng qut ny c th c chng minh bng qui np t qui tc nhn cho trng hp th tc gm 2 cng vic. V d 1: Cc gh ngi trong mt hi trng s c ghi nhn gm mt mu t v mt s nguyn dng khng ln hn 100. Hi s gh ti a c th c ghi nhn khc nhau l bao nhiu? Li gii. Th tc ghi nhn cho mt gh gm 2 vic : ghi mt trong 26 mu t v k tip l ghi mt trong 100 s nguyn dng. Qui tc nhn cho thy c 26 x 100 = 2600 cch khc nhau ghi nhn cho mt gh ngi. Do s gh ln nht c th c ghi nhn khc nhau l 2600. V d 2: Gi s ta phi i t mt a im A n mt a im C, ngang qua mt a im B. i t A n B ta c 8 cch i khc nhau, v c 6 cch i t B n C. Hi c bao nhiu cch i t A n C ? Li gii. Mt cch i t A n C gm 2 vic: i t A n B, ri i t B n C. Vic th nht (i t A n B) c 8 cch thc hin, vic th hai c 6 cch thc hin. vy, theo nguyn l nhn, s cch i t A n C l 8 x 6 = 48. V d 3: Hi c bao nhiu chui bit khc nhau c di 8 (tc l gm 8 bits) ? Li gii. Mi bit c th c chn theo 2 cch, v mi bit l 0 hoc 1. Do , qui tc nhn cho php ta kt lun rng c 28 = 256 chui bit c di 8. Vd4:Mtmbaogm6kt,tronggm3mutrin3ks thp phn. Hi c bao nhiu m khc nhau?Li gii. C 26 cch chn cho mi mu t v c 10 cch chn cho mi k s thp phn. Do , theo qui tc nhn, c tt c 26.26.26.10.10.10 = 17 576 000 m khc nhau. V d 5: C bao nhiu nh x i t mt tp hp gm m phn t vo mt tp hp gm n phn t ? Li gii. Mt nh x i t tp A gm m phn t vo mt tp hp B gm n phn t tng ng vi vic chn la mt trong n phn t ca B cho mi phn t ca A. Do , theo qui tc nhn, c n.n. ... .n = nm nh x t A vo B. V d 6: C bao nhiu n nh i t mt tp hp gm m phn t vo mt tp hp gm n phn t ? Ligii.Trchttanhnxtrngkhim>nthkhngcmtn nh no i t mt tp hp gm m phn t vo mt tp hp gm n phn 50 t. Vy, cho m s n. Gi s cc phn t trong min xc nh ca nh x l a1, a2, . . ., am. C n cch chn nh qua nh x cho phn t a1. V nh x l n nh nn i vi phn t a2 ta ch c n-1 cch chn nh tng ng (do gi tr nh c chn cho a1 khng th c chn li cho a2). Tng qut, gi tr nh ca phn t ak ch c th c chn theo n-k+1 cch. Theo qui tc nhn, c n.(n-1). ... .(n-m+1) n nh i t mt tp hp gm m phn t vo mt tp hp gm n phn t. V d 7: Phng n nh s in thoi. Gi s mt s in thoi gm 10 k s c chia thnh 3 nhm: 2 nhm gm 3 ksvmtnhm4ks.Domtsldono,cmtshnchtrn cc k s ca s in thoi. xc nh dng hp l ca mt s in thoi. ta dng k hiu X ch mt k s c th ly gi tr t 0 n 9, N ch mt k st2n9,vYchmtksl0hoc1.Chngtac2phngn nh s in thoi : mt phng n c v mt phng n mi. Theo phng n c,sinthoicdngNYXNNXXXXX;vtheophngnmiths in thoi c dng NXX NXX XXXX. Hi s lng s in thoi khc nhau ca mi phng n l bao nhiu? Li gii. Do qui tc nhn, i vi phng n nh s in thoi c, s trnghpkhcnhaucaminhmkstrong3nhmlnltl : 8.2.10 = 160 (ng vi dng NYX), 8.8.10 = 640 (ng vi dng NNX), v 10.10.10.10 = 10000 (ng vi dng XXXX). Vy, trong phng n nh s in thoi c, s lng s in thoi l 160. 640.10000 = 1 024 000 000. Tng t S lng s in thoi trong phng n nh s mi l : (8.10.10).(8.10.10).(10.10.10.10) = 800.800.10000 = 6 400 000 000. V d 8: Cng theo qui tc nhn ta thy rng sau khi thc hin on chng trnh di y th gi tr ca bin k s l n1.n2. ... .nm. k := 0 for i1 = 1 to n1 do for i1 = 1 to n2 do . . . for i1 = 1 to nm do k := k + 1 V d 9: S dng qui tc nhn chng minh rng mt tp hp S hu hn c tt c 2|S| tp hp con khc nhau. Li gii. Cho S l mt tp hp hu hn, |S| = n. Lit k cc phn t ca S theo mt th t bt k. Ta c th thy rng c s tng ng mt-mt trn(songnh)giacctphpconcaSvtphpccchuibit gm n bits. Mt tp con ca S c cho tng ng vi mt chui bits c bit th i l 1 nu phn t th i trong danh sch lit k thuc tp hp 51 con, v bit th i l 0 trong trng hp ngc li. Bi qui tc nhn, c 2n chui bit gm n bits. Do , S c 2n tp hp con. Di y chng ta s xem xt mt s bi ton v php m phc tp hn. N i hi chng ta phi s dng c nguyn l cng ln nguyn l nhn. Vd10:TrongmtversioncangnngBASICtncamtbinl mt chuigm1hoc2kt,miktlmuthocksthplcphnv khngphnbitgiachinhoavchthng.Hnna,mttnbinphi bt u bi mt mu t v tn bin phi khc vi 5 chui gm 2 k t c dnhringchongnng.Hicbaonhiutnbinkhcnhautrongversion ny ca BASIC. Li gii. t V l s tn bin khc nhau trong version ny ca BASIC, V1 l s bin gm mt k t, v V2 l s bin gm hai k t. Theo qui tc cng ta cV = V1 + V2. V bin gm mt k t phi l mt mu t nn V1 = 26. Ngoi ra, theo qui tc nhn tac 26.36 chuic di 2 vi k t i u l mu t v k t k l mu t hoc k s thp phn. Tuy nhin, c 5 chui b loi ra nnV2 = 26.36 - 5 = 931. Vy c V = V1 + V2 = 26 + 931 = 957 tn khc nhau cho cc bin ca version ny ca BASIC. V d 11: Mi ngi s dng trn mt h thng my tnh c mt "password" di t 6 n 8 k t, trong mi k t l mt ch in hoa hoc l mt k s thpphn.Mi"password"phictnhtmtks.Hicbaonhiu password khc nhau ? Li gii. t P l s lng tt c cc "password", v P6, P7, P8 ln lt l s cc "password" c di 6, 7, 8. Do qui tc cng ta c P = P6 + P7 + P8. Chng ta s tnh P6, P7, v P8. Tnh trc tip P6 tng i kh. tnh P6 cho d, ta tnh s chui c di 6 gm cc ch in hoa hay k s thp phn, k c cc chui khng c k s thp phn, v tr cho s chui(vidi6)khngcksthpphn.Theoquitcnhn,s chui gm 6 k t l 366 v s chui khng c k s l 266. Suy ra P6 = 366 - 266 = 2 176 782 336 - 308 915 776= 1 867 866 560. Tng t, ta c th tnh ra c : P7 = 367 - 267 = 78 364 164 096 - 8 031 810 176 = 70 332 353 920. v P8 = 368 - 268 = 2 821 109 907 456 - 208 827 064 57652 = 2 612 282 842 880. T ta tnh c : P = P6 + P7 + P8 = 2 684 483 063 360. 3.2.4. Nguyn l b tr Khi hai cng vic c th c lm ng thi, ta khng th dng quy tc cng tnh s cch thc hin nhim v gm c hai vic. tnh ng s cch thc hin nhim v ny ta cng s cch lm mi mt trong hai vic ri tr i s cch lm ng thi c hai vic. Ta c th pht biu nguyn l m ny bng ngn ng tp hp. Cho A1, A2 l hai tp hu hn, khi |A1 A2| = |A1| + |A2| |A1 A2|. T vi ba tp hp hu hn A1, A2, A3, ta c: |A1 A2A3| = |A1| + |A2| + |A3| |A1 A2| |A2 A3| |A3 A1| + |A1 A2 A3|, v bng quy np, vi k tp hu hn A1, A2, ..., Ak ta c: | A1 A2...Ak| = N1 N2 + N3 ... + (1)k-1Nk, trong Nm (1 s m s k) l tng phn t ca tt c cc giao m tp ly t k tp cho, ngha l Nm =| ... |... 12 12 1 mmik i i ii iA A A s < < < s By gi ta ng nht tp Am (1 s m s k) vi tnh cht Am cho trn tp v tr hu hn U no v m xem c bao nhiu phn t ca U sao cho khng tha mn bt k mt tnh cht Am no. GiNl s cn m, N l s phn t ca U. Ta c: N= N | A1 A2...Ak| = N N1 + N2 ... + (1)kNk, trong Nm l tng cc phn t ca U tha mn m tnh cht ly t k tnh cht cho. Cng thcnycgilnguynlbtr.NchophptnhN quaccNmtrongtrnghp cc s ny d tnh ton hn. Th d 3: C n l th v n phong b ghi sn a ch. B ngu nhin cc l th vo cc phong b. Hi xc sut xy ra khng mt l th no ng a ch. Mi phong b c n cch b th vo, nn c tt c n! cch b th. Vn cn li l m s cch b th sao cho khng l th no ng a ch. Gi U l tp hp cc cch b th v Am l tnh cht l th th m b ng a ch. Khi theo cng thc v nguyn l b tr ta c: N= n! N1 + N2 ... + (1)nNn, trong Nm (1 s m s n) l s tt c cc cch b th sao cho c m l th ng a ch. Nhn xt rng, Nm l tng theo mi cch ly m l th t n l, vi mi cch ly m l th, c (n-m)! cch b m l th ny ng a ch, ta nhn c: Nm = mnC (n - m)! = nk!!vN= n!(1 11! + 12! ... + (1)n 1n!), 53 trong mnC= )! ( !!m n mn l t hp chp m ca tp n phn t (s cchchn mi tng trong n i tng ccho). T xc sut cn tm l:1 11! + 12! ... + (1)n 1n!. Mt iu l th l xc sut ny dn n e-1 (ngha l cn > 13) khi n kh ln. S Ntrong bi ton ny c gi l s mt th t v c k hiu l Dn. Di y l mt vi gi tr ca Dn, cho ta thy Dn tng nhanh nh th no so vi n: n234567891011 Dn 12944265185414833133496133496114684570 3.2.5. Nguyn l Dirichlet 3.2.5.1. M u Giscmtnchimbcubayvochung.Nuschimnhiuhnsngn chung th t nht trong mt ngn c nhiu hn mt con chim. Nguyn l ny d nhin l c th p dng cho cc i tng khng phi l chim b cu v chung chim. Mnh (Nguyn l): Nu c k+1 (hoc nhiu hn) vt c t vo trong k hp th tn ti mt hp c t nht hai vt. Chng minh: Gi s khng c hp no trong k hp cha nhiu hn mt vt. Khi tng s vt ccha trong cchp nhiu nht l bng k. iu nytrigi thit l c t nht k + 1 vt. Nguyn l ny thng c gi l nguyn l Dirichlet, mang tn nh ton hc ngi c th k 19. ng thng xuyn s dng nguyn l ny trong cng vic ca mnh. Th d 4: 1) Trong bt k mt nhm 367 ngi th no cng c t nht hai ngi c ngy sinh nht ging nhau bi v ch c tt c 366 ngy sinh nht khc nhau. 2) Trong k thi hc sinh gii, im bi thi c nh gi bi mt s nguyn trong khong t 0 n 100. Hi rng t nht c bao nhiu hc sinh d thi cho chc chn tm c hai hc sinh c kt qu thi nh nhau?TheonguynlDirichlet,shcsinhcntml102,vtac101ktquimthi khc nhau. 3)Trongsnhngngicmttrntrit,phitmchaingichmrngging nhau. Nu xem mi hm rng gm 32 ci nh l mt xu nh phn c chiu di 32, trong rng cn ng vi bit 1 v rng mt ng vi bit 0, th c tt c 232 = 4.294.967.296 hm rng khcnhau.Trongkhisngitrnhnhtinhnylvtqu5t,nntheonguynl Dirichlet ta c iu cn tm. 54 3.2.5.2. Nguyn l Dirichlet tng qut Mnh : Nu c N vt c t vo trongk hp th s tn ti mt hp cha t nht ]N/k[ vt. ( y, ]x[ l gi tr ca hm trn ti s thc x, l s nguyn nh nht c gi tr ln hn hoc bng x. Khi nim ny i ngu vi [x] gi tr ca hm sn hay hm phn nguyn ti x l s nguyn ln nht c gi tr nh hn hoc bng x.) Chng minh: Gi s mi hp u cha t hn ]N/k[ vt. Khi tng s vt l s k (] Nk[ 1) < k Nk = N. iu ny mu thun vi gi thit l c N vt cn xp. Th d 5: 1) Trong 100 ngi, c t nht 9 ngi sinh c ng mt thng. Xpnhngngisinhc ngthngvomtnhm.C12thngttc.Vytheo nguyn l Dirichlet, tn ti mt nhmc t nht ]100/12[= 9 ngi. 2) C nm loi hc bng khc nhau. Hi rng phi c t nht bao nhiu sinh vin chc chn rng c t ra l 6 ngi c ng nhn hc bng nh nhau. Gi N l s sinh vin, khi ]N/5[ = 6 khi v ch khi 5 < N/5 s 6 hay 25 < N s 30. Vy s N cn tm l 26. 3) S m v ng cn thit nh nht phi l bao nhiu m bo 25 triu my in thoi trong nccsinthoikhcnhau,misc9chs(gissinthoicdng0XX- 8XXXXX vi X nhn cc gi tr t 0 n 9). C 107 = 10.000.000 s in thoi khc nhau cdng 0XX- 8XXXXX.V vy theo nguynlDirichlettngqut,trongs25triumyinthoitnhtc ]25.000.000/10.000.000[ = 3 c c ng mt s. m bo mi my c mt s cn c t nht 3 m v ng. 3.2.5.3. Mt s ng dng ca nguyn l Dirichlet. TrongnhiungdngthvcanguynlDirichlet,khinimvtvhpcn phi c la chn mt cch khn kho. Trong phn nay c vi th d nh vy. Thd6:1)Trongmtphnghpcnngi,baogicngtmc2ngicsngi quen trong s nhng ngi d hp l nh nhau. S ngi quen ca mi ngi trong phng hp nhn cc gi tr t 0 n n 1. R rng trong phng khng th ng thi c ngi c s ngi quen l 0 (tc l khng quen ai) v c ngi c s ngi quen l n 1 (tc l quen tt c). V vy theo s lng ngi quen, ta ch c th phn n ngi ra thnh n 1 nhm. Vy theo nguyn l Dirichlet tn tai mt nhm c t nht 2 ngi, tc l lun tm c t nht 2 ngi c s ngi quen l nh nhau. 55 2) Trong mt thng gm 30 ngy, mt i bng chuyn thi u mi ngy t nht 1 trn nhng chi khng qu 45 trn. Chng minh rng tm c mt giai on gm mt s ngy lin tc no trong thng sao cho trong giai on i chi ng 14 trn. Gi aj l s trn m i chi t ngy u thng n ht ngy j. Khi 1 s a1 < a2 < ... < a30 < 45 15 s a1+14 < a2+14 < ... < a30+14 < 59. Su mi s nguyn a1, a2, ..., a30, a1+ 14, a2 + 14, ..., a30+14 nm gia 1 v 59. Do theo nguyn l Dirichlet c t nht 2 trong 60 s ny bng nhau. V vy tn ti i v j sao cho ai = aj + 14 (j < i). iu ny c ngha l t ngy j + 1 n ht ngy i i chi ng 14 trn. 3) Chng t rng trong n + 1 s nguyn dng khng vt qu 2n, tn ti t nht mt s chia ht cho s khc. Tavitmisnguyna1,a2,...,an+1didngaj= jk2 qjtrongkjlsnguyn khng m cn qj l s dng l nh hn 2n. V ch c n s nguyn dng l nh hn 2n nn theo nguyn l Dirichlet tn ti i v j sao cho qi = qj = q. Khi ai= ik2 q v aj = jk2 q. V vy, nu ki s kj th aj chia ht cho ai cn trong trng hp ngc li ta c ai chia ht cho aj. Thdcuic ngtrnhbycchpdngnguynlDirichletvolthuytthpmvn quen gi l l thuyt Ramsey, tn ca nh ton hc ngi Anh. Ni chung, l thuyt Ramsey gii quyt nhng bi ton phn chia cc tp con ca mt tp cc phn t.Th d 7. Gi s trong mt nhm 6 ngi mi cp hai hoc l bn hoc l th . Chng t rng trong nhm c ba ngi l bn ln nhau hoc c ba ngi l k thln nhau. Gi A l mt trong 6 ngi. Trong s 5 ngi ca nhm hoc l c t nht ba ngi l bncaAhocctnhtbangilkth caA,iunysuyratnguynlDirichlet tngqut,v]5/2[=3.TrongtrnghputagiB,C,DlbncaA.nutrongba nginychaingilbnthhc ngviAlpthnhmtbbangibnlnnhau, ngc li, tc l nu trong ba ngi B, C, D khng c ai l bn ai c th chng t h l b ba ngi thln nhau. Tng t c th chng minh trong trng hp c t nht ba ngi l k thca A. 56 CHNG 4 QUAN H 4.1. Quan h hai ngi 4.1.1. nh ngha quan h v v d Giaccphnttrongmttphpnomchngtaangquantmthngc nhng mi lin h hay nhng quan h. V d: quan h ln hn gia cc s thc, quan h "anh em"giangivingi,quanhngdnggiacctamgic,v.v....Miquanhtrong mt tp hp c c trng bng mt hay mt s tiu chun no thhin ng nghaca quan h. y chng ta ch cp n nhng quan h, c gi l nhng quan h 2 ngi, ni ln s lin h gia mi phn t vi cc phn t khc trong tp hp. Khi ta ang xem xt mt quan h nh th, th vi hai phn t x, y t y trong tp hp chng s c : hoc l x c quan h vi y, hoc l x khng c quan h vi y. Ni nh vy cng c ngha l tp hp cc cp (x, y) gm 2 phn t c quan h c th xc nh c quan h ang xt trn tp hp. V mt ton hc, mt quan h 2 ngi c nh ngha nh sau: nh ngha :Cho mt tp hp X khc rng. Mt quan h 2 ngi trn X l mt tp hp con R ca X2. Cho 2 phn t x v y ca X, ta ni x c quan h R vi y khi v ch khi (x,y)e R, v vit l x R y. Nh vy: x R y (x,y) e R Khi x khng c quan h R vi y, ta vit: xy. V d: 1. Trn tp hp X = { 1,2,3,4} , xt quan h 2 ngi R c nh ngha bi: R = { (1,1), (1,3), (2,2), (2,4), (3,1), (3,3), (4,2), (4,4)}Vi quan h ny ta c: 2 R 4, nhng 23. 2. Trn tp hp cc s nguyn Z ta nh ngha mt quan h 2 ngi R nh sau: x R y nu v ch nu x-y l s chn. hay ni cch khc: R = { (x,y) e Z2 | x-y = 2k vi k e Z }Quan h R ny chnh l quan h ng d modulo 2. 57 3. Cho n l mt s nguyn dng. Nhc li rng quan h ng d modulo n trn tp hp cc s nguyn Z, k hiu bi (mod n), c nh ngha nh sau: a b (mod n) - k e Z : (a - b) = k.n Quan h ny l mt quan h 2 ngi trn Z. 4. Quan h s trn tp hp cc s thc R cng l mt quan h 2 ngi. 5. Cho E l mt tp hp, t X = P(E). Mi phn t thuc X l mt tp hp con ca E. Trn E c cc quan h quen thuc sau y: - quan h bao hm, k hiu bi c- quan h cha, k hiu bi - quan h bng nhau, k hiu bi =Ghi ch :Ngi ta cn nh ngha mt quan h (2 ngi) gia mt tp hp A v mt tp hp B l mt tp hp con ca AxB. V d: A = { 1, 2, 3, 4, 5} , B = { 0, 1} . Ta c R = { (1,1), (2,0), (3,1), (4,0), (5,0)} l mt quan h gia A v B. Tng qut hn, ta c th nh ngha mt quan h gia cc tp hp A1, A2, . . ., An l mt tp hp con ca A1 x A2 x . . . x An (tch Descartes ca cc tp hp A1, A2, . . ., An). Nh vy, khi R l mt quan h gia cc tp A1, A2, . . ., An th mi phn t ca R l t b n (a1, a2, . . ., an) vi ai e Ai (i=1, , n). Cch xc nh mt quan h: Da vo cc phng php xc nh mt tp hp, ta c th xc nh mt quan h bng cc phng php sau y: Lit k: lit k tt c cc cp hay b phn t c quan h R (tc l thuc R). Trong v d 1 trn, quan h R c cho theo cch lit k. Nu tnh cht c trng cho quan h R, tc l tnh cht hay tiu chun xc nh cc phn t thuc R hay khng. Trong cc v d 2 v 3 trn, quan h R c cho bng cch nu ln tnh cht xc nh quan h. 4.1.2. Cc tnh cht ca quan h Mtquanh2ngitrnmttphpcthcmtstnhchtnolmchotphpc mt cu trc nht nh. Di y l nh ngha mt s tnh cht thng c xt i vi mt quan h 2 ngi. nh ngha : Gi s R l mt quan h 2 ngi trn mt tp hp X. Ta ni quan h R c tnh phn x (reflexive) nu v ch nu58 x R x vi mi x e X. Ta ni quan h R c tnh i xng (symmetric) nu v ch nux R y y R x vi mi x,y e X. Ta ni quan h R c tnh phn xng (antisymmetric) nu v ch nu(x R y v y R x) x = y vi mi x,y e X. Ta ni quan h R c tnh truyn hay bc cu (transitive) nu v ch nu(x R y v y R z) x R z vi mi x,y,z e X. V d: Trong v d ny chng ta cp n mt s quan h c nu ln trong cc v d ca mc 1.1 trn, v pht biu cc tnh chtca chng. Vic kim chng cc tnh cht ny kh d dng. 1. Quan h ng d modulo n trn Z c 3 tnh cht: phn x, i xng, truyn. 2. Quan h s trn tp hp cc s thc c 3 tnh cht: phn x, phn xng, truyn. 3. Cho E l mt tp hp. Quan h _ trn P(E) c 3 tnh cht: phn x, phn xng, truyn. 4.1.3. Biu din quan h Ngoi phng php biu din mt quan h 2 ngi di dng tp hp cc cp phn t ngi ta cn c th s dng ma trn biu din cho quan h trong trng hp cc tp hp l hu hn. Khi nim ma trn s c nh ngha v kho st chi tit hn trong phn "i s Tuyn tnh". y chng ta ch cn hiu ma trn mt cch n gin l mt bng lit k cc phn t thnh cc dng v cc ct. V d, bng lit k 6 s nguyn thnh 2 dng v 3 ct sau y l mt ma trn: Mt ma trn M gm m dng, n ct s c gi l mt ma trn c cp mxn. Nu m = n th ta ni M l mt ma trn vung cp n. Gi s R l mt quan h 2 ngi gia mt tp hp hu hn A = { a1, a2, ... , am } v mt tp hu hn B = {b1, b2, ... , bm} . Quan h R c th c biu din bi ma trn MR = [mij] gm m dng v n ct (tc l ma trn cp mxn), trong mij = 1 nu (ai , bj) e R mij = 0 nu (ai , bj) e R 59 Ta gi ma trn MR l ma trn biu din ca quan h R. V d: Vi A = { 1,2,3} v B = { a, b, c} , th cc quan h sau y: R = { (1,a), (1,b), (1,c)}S = { (1,a), (1,b), (1,c), (2,b), (2,c), (3,c)}c cc ma trn biu din l MR = MS = Trong trng hp R l mt quan h 2 ngi trn mt tp X hu hn v c n phn t th ma trn biu din ca R l mt ma trn c n dng v n ct (tc l ma trn vung cp n). Ghi ch: Ngoi cch biu din quan h di dng ma trn ta cn biu (dng th) biudinquanh.Cchbiudinnyscxtntrongphnsau,khinivbiu Hasse ca mt cu trc th t. 4.2. Quan h tng ng 4.2.1. Khi nim quan h tng ng nh ngha:Mt quan h 2 ngi R trn mt tp hp X c gi l mtquan h tng ng nu v ch nu n tha 3 tnh cht: phn x, i xng, truyn.V d:Mtvdquantrngvquanhtngnglquanhngdmodulon trn Z. Ta bit quan h ny c 3 tnh cht phn x, i xng, truyn. Quan h s trn Z khng phi l mt quan h tng ng v n khng c tnh cht i xng. 4.2.2. Lp tng ng v tp hp tng ng nh ngha:Vi mi phn t xe X, ta nh ngha lp tng ng cha x, k hiu, l tp hp tt c nhng phn t (thuc x) c quan h R vi x: 60 = { y e X : y R x }NhvymilptngnglmttphpconcaX.Ngitachngminhrngtp hp cc lp tng ng ca quan h tng ng R trn X to thnh mt "phn hoch" ca tp hp X, tc l su tp cc lp tng ng khc nhau cho ta mt h cc tp con ca X ri nhau i mt v c phn hi bng X. TphpcclptngngcaquanhtngngRtrnXny(lmttpconca P(X))cgiltphpthng(caquanhtngngRtrnX).Quanhngd modulo n trn Z c tp hp thng tng ng, c k hiu l Zn, gm n phn t : Zn = {}trong (ke Z) l tp hp tt c nhng s nguyn ng d vi k modulo n. Ngi ta cn chng minh c rng vic xc nh mt quan h tng ng trn mt tp hp X tng ng vi vic xc nh mt phn hoch ca tp hp X, tc l c mt song nh gia tp hp tt c cc quan h tng ng trn X v tp hp tt c cc phn hoch ca tp X. 4.3. Quan h th t 4.3.1. Cc nh ngha inh ngha 1:Mt quan h 2 ngi R trn mt tp hp X (khc rng) c gi l mt quan h th t (hayvntt,lmttht)nuvchnunc3tnhcht:phnx,phnxng, truyn. Khi ta cng ni tp hp X l mt tp c th t. Nu c thm tnh cht: vi mi x, y e X ta c xRy hay yRx th ta ni R l mt quan h th t ton phn trn X. Ghi ch :Trong trng hp trn X c nhiu quan h th t th khi xt n th t trn X ta phi nirthtno,vtathngvittphpXcthtdidngmtcp(X,R); trong R l quan h th t ang xt trn X. Vi 2 tp hp c th t X v Y ta c th nh ra mt th t trn tch Descartes XxY da vo cc th t trn X v trn Y. T ta XxY tr thnh mt tp hp th t (xem phn bi tp). V d: 1.QuanhstrntphpccsthcRlmtquanhthtton phn. 2.ChoElmttphp.QuanhctrnP(E)lmtquanhtht. Nu E c nhiu hn 2 phn t th th t ny khng phi l th t ton phn. Vic kim chng iu ny c dnh cho ngi c. 3.TrntphpccsnguynZ,xtqnah"chiaht"hay"cs ca", k hiu l | , c nh ngha nh sau: 61 a| b - ke Z : a = k.b Ddngkimchngrngquanh|c3tnhcht:phnx,phn xng, truyn. T ta c (Z,| ) l mt tp hp c th t. Ta c 2 s nguyn 2 v 3, khng c quan h vi nhau theo quan h | . Do | khng phi l th t ton phn trn Z. Nhn xt: Nu (X,R) l mt tp hp c th t v A c X th quan h th R thu hp trn tp A,cngckhiulR(nukhnggyranhmln),lmtquanhthttrnA.Ni mt cch khc, ta c: (X,R) th t v A c X (A,R) th t ivimttphpcthtthviccpncckhinimnh"phntnhnht", "phn t ln nht", ... l iu rt t nhin. Di y, chng ta s gii thiu mt s khi nim quan trng khi xt mt tp hp c th t. nh ngha 2: Cho (X, s ) l mt tp hp c th t, v A c X. Tagi mt phn t aeA l mt phn t nh nht ca tp hp A nu v ch nu vi mi x e A ta c : a s x. TagimtphntaeAlmtphntlnnhtcatphpAnuvch nu vi mi x e A ta c : x s a. Ta gi mt phn t a e A l mt phn t ti tiu ca tp hp A nu v ch nu khng tn ti x e A sao cho x = a v x s a. Ta gi mt phn t a e A l mt phn t ti i ca tp hp A nu v ch nu khng tn ti x e A sao cho x = a v a s x. 9;Nhn xt:(1) Phn t nh nht (ln nht) ca mt tp hp, nu c, l duy nht.Ta k hiu phn t nh nht ca mt tp hp A l min A hay min (A), v k hiu phn t ln nht ca A l max A hay max (A). (2) Phn t ti tiu (ti i) ca mt tp hp c th t khng nht thit l duy nht. V d: xt tp hp X = { 1,2,3} vi quan h 2 ngi c chobi={(1,1),(2,2),(3,3),(1,2),(3,2)}.Chngtacthkim chng rng (X, ) l mt tp hp c th t. Vi th t ny, X c 2 phn t ti tiu l 1 v 3. (3) Phn t ln nht (nh nht) ca mt tp hp, nu c, l phn t ti i (ti tiu) duy nht ca tp hp . V d:62 1. Trong tp hp c th t (Z, s ), tp hp A = { me Z| m2 < 100} c phn t nh nht l -9, v phn t ln nht l 9. Ta c th vit: min(A) = -9; max(A) = 9. 2.Trongtphpctht(R,s),tphpA={xeR|x2 (xA y) V (xA z) x V (y A z) s (xVy) A (xVz) 2. (x s z) (xV(y A z) s (xVy) A z) 71 CHNG 5 I S BOOL 5.1. Cc php ton 5.1.1. Cc nh ngha Rtnhiutphpquenthucucnhngphptontrn.Trntphpccs nguyn c php ton cng, php ton nhn. Trn tp hp cc s thc khc khng c php ton nhnvcnccphplynghcho.Miphptontrnmttphpchotamtquitc nhm to ra mt phn t t mt hay nhiu phn t no . Php cng thc hin trn 2 s 5 v 7 cho ra s 12; php ly nghch o ca s 2 cho kt qu l s 0.5. Khi php ton thc hin trn 2 phn t ta ni l php ton 2 ngi, php ton thc hin trn mt phn t th c gi l php ton mt ngi. Di y ta s nu ln nh ngha ton hc ca cc php ton. nh ngha 1. (php ton 2 ngi) Cho X l mt tp hp khc rng. Mt php ton hai ngi trn tp hp X l mt nh x T i t XxX vo X. K hiu ca nh x c gi l k hiu ca php ton hay l mt ton t. nh ca cp(a,b)quanhxTcgilktquthchinphptonTtrn2phntavb,v thng c vit l a T b. Nh vy, nu T l mt php ton 2 ngi trn X th ta c nh x: T : XxX X (a,b)T(a,b) = a T b nh ngha 2. (php ton 1 ngi) Cho X l mt tp hp khc rng. Mt php ton 1 ngi trn tp hp X l mt nh x T i t X vo X. K hiu ca nh x c gi l k hiu ca php ton hay l mt ton t. nh ca a qua nh x T c gi l kt qu thc hin php ton T trn phn t a. V d:1.TrncctphpsN,Z,Q,R,Ccccphpton+(cng)v* (nhn). 2.Trntphpccmatrnsthcvungcpncccphpton:+ (cng matrn) v * (nhn ma trn). 3. Cho E l mt tp hp. t X = P(E). Trn X c cc php ton tp hp thng thng : 72 php giao hai tp hp, c k hiu l . php hi hai tp hp, c k hiu l U . phplybcamttphp,ckhiul c.Theokhiu ny, phn b ca tp A X l Ac. Php ton v U l cc php ton 2 ngi, php ton c l php ton 1 ngi. 4. Cho E l mt tp hp khc rng. t X l tp hp cc nh x i t E vo E. Trn X c mt php ton nh x thng thng; l php hp ninhx.Phptonnyckhiulo.ylmtphpton2 ngi trn X. 5. t X l tp hp tt c cc chui k t. Php ni 2 chui p t l mt php ton 2 ngi trn X.6.TphpB={0,1}gm2phntidincho2chntrsaiv "ng.TabittrntphpBcccphptonlogic:v(hay),. (v), (ph nh), (ko theo). Trong cc php ton trn ch c php ton(phnh)lphpton1ngi,cnccphptonkhcul cc php ton 2 ngi. 7. Cho (L, ) l mt dn. Khi vi mi cp (a,b) gm cc phn t ca L ta c hai phn t tng ng l inf(a,b) v sup(a,b), c k hiu ln lt l a b v ab. Nh vy trn mi dn ta c hai php ton 2 ngi l v v . . Php ton v thc hin trn cc phn t a, b s cho kt qu l chn trn nh nht ca a v b. Php ton . thc hin trn cc phn t a, b s cho kt qu l chn di ln nht ca a v b. Ghi ch : 1. Mt cch tng qut, ta c th nh ngha php ton n-ngi trn mt tp hp X l mt nh x i t Xn vo X.ng vi mi b n phn t (a1, ... , an) php ton s cho ta mt phn t kt qu thuc X. 2. Trong trng hp tp hp X l hu hn th ngi ta cth nh ngha hay xcnhphptonbngcchlitkktquthchinphptonchomi trng hp c th c. V d X = { a1, ... , an } gm n phn t. Gi s * l mt phpton2ngitrnX.Khi,phpton*cthcxcnhbibng sau y: *a1 aj an a1: :: ai ai* aj : an73 Bng trn c gi l bng Cayley ca php ton 2 ngi. Nh vy ng vi mi php ton 2 ngi trn X ta c mt ma trn c cp n vi phn t dng i ct j bng ai* aj. V sau, nhiu tnh cht ca php ton s c xem xt thng qua ma trn ny. Chng ta tng thy nhng php ton 2 ngi c nh ngha bng bng nh th; l cc php ton logic v (hay), . (v), (ko theo). V d: Cho n l mt s nguyn ln hn 1. Trn tp hp Zn = { } , tp hp thng theo quan h ng d modulo n trn tp hp cc s nguyn Z, ta nh ngha 2 php ton + v * nh sau: , " a,b Z , " a,b Z nh ngha trn da vo phn t i din ca lp tng ng. Tuy nhin ta c th kim chng d dng rng nh ngha trn l hp l. Trng hp n = 3, php + trn Z3 c bng Cayley nh sau : +012 0012 1120 2201 trong ta vit a thay chovi mi a = 0, 1, 2. 5.1.2. Cc tnh cht ca php ton hai ngi Trongmcnychngtanulnmtstnhchtisthngcxemxtivicc php ton 2 ngi. nh ngha 3. 1. Ta ni mt php ton 2 ngi T trn mt tp hp X c tnh giao hon nu" x,y e X : x T y = y T x 2. Ta ni mt php ton 2 ngi T trn mt tp hp X c tnh kt hp nu" x,y,z e X : x T (y T z) = (x T y) T z V d: 1. Cc php ton + v * trn cc tp hp s u c tnh cht kt hp v giao hon. 2. Php ton + (cng) trn tp hp cc ma trn thc vung cp n, k hiu Mn(R), c tnh cht giao hon v kt hp. Nhng php ton * (nhn ma trn) c tnh kt hp nhng khng c tnh 74 giao hon. Php nhn trn Mn(R) ch c tnh giao hon khi n = 1. (nh ngha ca cc php ton ma trn c th c tham kho phn i s tuyn tnh) 3. Php chia trn tp hp cc s hu t khc 0 khng c tnh kt hp vcng khng c tnh giao hon. Bi v 1 / 2 2 / 1, v (8/4) / 2 = 1 4 = 8/(4/2).4. t X l tp hp tt c cc nh x i t mt tt hp E vo chnh n. Php ton o (hp ni nh x) trn X c tnh kt hp. Bi v vi 3 nh x f, g, h t E vo E, theo tnh cht v nh x hp, ta c fo(goh) = (fog)oh. Nichungphptonokhngctnhgiaohon;phptonchctnhgiaohontrong trng hp E c mt phn t. Tht vy, nu E c nhiu hn 1 phn t th ta c th chn ra 2 phn t khc nhau a, b trong E. Xt 2 nh x f v g nh sau: f(x) = a vi mi x e E, v g(x) = b vi mi x e E. Ta c fog(x) = a vi mi xe E, v gof(x) = b vi mi xe E; nn fog gof. 5. Trn mt dn L, cc php ton 2 ngi v c tnh cht giao hon v kt hpCho * l mt php ton 2 ngi trn mt tp hp X = { a1, ... , an} . By gi chng ta s xem xt cc tnh cht giao hon v kt hp ca php ton * c lin quan nh th no n cc tnh cht ca ma trn M = (mij) e Mn(X) lin kt vi php ton. Nhc li l mij = ai* aj.Phpton*giaohonkhivchkhiai*aj=aj*aivimi i,j.iunytngngvi iu kin : mij = mj i vi mi i, j. Vy php ton giao hon khi v ch khi ma trn lin kt vi php ton l i xng. i vi tnh cht kt hp chng ta c th gi s rng X = { 1, 2, ... , n} ,v ta s xy dng mt hm ly gi tr Bool ASSOC vi i l mt ma trn M c cp n thuc Mn(X) sao cho ASSOC(M) = 1 (True) php ton 2 ngi c linkt vi ma trn M c tnh kt hp NhnxtrngviMeMn(X),M[i,j] X={1,2,...,n},hmASSOCcthcvit di dng m gi nh sau: Function ASSOC (M) : Boole begin T 1 for i=1 to n do for j=1 to n do for k=1 to n do if M[M[i,k],j] M[i, M[k,j]] then begin 75 T 0 goto OUT end OUT: return T end nh ngha 4. Cho X l mt tp hp khc rng, * l mt php ton 2 ngi trn X. 1. php ton * c gi l ly ng khi v ch khix * x = x , vi mi xe X 2. Mt phn t ee X c gi l phn t trung ha ca php ton * trn X khi v ch khi x * e = e * x = x , vi mi xe X 3. Gi s php ton * c phn t trung ha l e. Ta ni mt phn t xe X l kh nghch (hay c nghch o) khi v ch khi $ xe X : x * x = e = x * x Nhn xt :Nu php ton c tnh kt hp th phn t trung ha (nu c) l duynht, v trong trng hp tng qut ngi ta cn gi l "phn t n v". Khi php ton c tnh kt hp v c phn t trung ha, vi mi phn t x kh nghch,phntxtrongnhnghatrnlduynht.Tagixlphnt nghch o ca x, v k hiu l x-1. Ghi ch :1. Trong trng hp php ton c k hiu l . (du nhn) th phn t trung ha ca php ton thng c k hiu 1, v c gi l "n v".2. Khi php ton c k hiu l + (du cng) th phn t trung ha ca php ton thng c k hiu 0, v c gi l "zero". Trong trng hp ny, mi phntxthaiukinkhnghchscgil"ci,vphntx trong nh ngha c gi l phn t i ca x. Ta k hiu phn t i ca x l -x. V d: 1. Trn tp B = { 0,1} (gm 2 gi tr Boole), c cc php ton v . Php ton V c cc tnh cht : giao hon, kt hp, c trung ha l 0, ly ng. Phn t 1 khng kh nghch i vi php v 1 x = 1 0 vi mi xe B. Php ton c cc tnh cht : giao hon, kt hp, c trung 76 ha l 1, ly ng. Phn t 0 khng kh nghch i vi php v 0 x = 0 1 vi mi xe B. 2. Php ton + (cng) trn cc tp hp s N, Z, Q, R, C c cc tnh cht : giao hon, kt hp, c trung ha (hay phn t zero) l 0. Trong cc tp hp Z, Q, R, C mi s u c i. Php ton * (nhn) trn cc tp hp sN,Z,Q,R,Cccctnhcht:giaohon,kthp,ctrungha (hay phn t n v) l 1. Trong cc tp hp Q, R, C mi s khc 0 u kh nghch. 3. Cho (L, ) l mt dn, chng ta bit rng trn L c hai php ton 2 ngi c k hiu l v . Cc php ton ny ngoi tnh cht giao hon v kt hp cn c tnh cht ly ng na.4.ChoEl mttphp.CcphptonUvtrnP(E)uctnh cht ly ng. Php ton U c phn t trung ha l . Php ton c phn t trung ha l E.5. t X = Mn(R), tp hp cc ma trn thc vung cp n. K hiu php nhn ma trn l *, ta c php ton * c tnh kt hp v c phn t trung ha l ma trn n v 9; 9; 9; Php ton + trn X c tnh giao hon, kt hp, c trung hal ma trn 0, mi ma trn u c ma trn i tng ng. Chng ta bit rng trn mt cu trc dn ta c mt cu trc i s vi 2 php ton v ; ccphptonnyccctnhcht:kthp,giaohon,lyng.Ngcli,trongmts trng hp, trn cc cu trc i s ta c th xy mt cu trc th t nh trong nh l di y. nh l 1.Cho * l mt php ton 2 ngi giao hon, kt hp v ly ng trn mt tp hp X. Khi , nu R l quan h 2 ngi trn X c nh ngha bi (a R b) (a * b = b) th ta c : 1. R l mt quan h th t. 2. " a,b e X : sup(a,b) = a * b. Chng minh: 77 1. Quan h R phn x do tnh ly ng ca php ton *. Nu a R b v b R a th ta c : b = a * b = b * a = a
