Ging vin: ThS. Trn Quang Khi

TON RI RC

Chng 3:Suy lun Chng minh

Ton ri rc: 2011-2012

Ni dung

1. Gii thiu.

2. Cc quy tc suy lun

3. Phng php chng minh.

Quy np ton hc.

4. Pht biu quy.

5. Bi tp Hi p.

Chng 3: Suy lun - Chng minh 2

Ton ri rc: 2011-2012

Gii thiu

Chng 3: Suy lun - Chng minh 3

Hai vn trong ton hc:

1. Khi no mt suy lun ton hc l NG?

2. PHNG PHP no xy dng cc suylun ton hc?

Ton ri rc: 2011-2012

Gii thiu Trong ton hc

Chng 3: Suy lun - Chng minh 4

Ton ri rc: 2011-2012

OK

Gii thiu - Trong tin hc

Chng 3: Suy lun - Chng minh 5

ProgramD liu 1 Kt qu 1

ProgramD liu 2 Kt qu 2

ProgramD liu n Kt qu n

Hmmm!Tm by!

OK

Ton ri rc: 2011-2012

Cc khi nim

nh l: theorem = a TRUE statement

mt pht biu hoc cng thc c suy lun ra t cc tin da vo cc quy tc suy lun s chng minh.

Tin (Axiom cn gi l nh )

mt mnh khng ph thuc vo s chng minh.

gi thit c s ca cc cu trc ton hc.

Gi thit (Hypothesis)

Nhng mnh /pht biu ng c s dng tranh lun hoc nghin cu.

Chng 3: Suy lun - Chng minh 6

Ton ri rc: 2011-2012

Chng minh l g?

Chng 3: Suy lun - Chng minh 7

Quy tc suy lun nh l

nh l c CM

Tin Gi thit

ca nh l

Quy tc suy lun = c ch rt ra kt lun t nhng iu c khng nh khc.

S chng minh c th thc hin bng vic kt hp ccbc chng minh.

Ton ri rc: 2011-2012

Cc quy tc suy lun (1)

Chng 3: Suy lun - Chng minh 8

Simplification(Lut rt gn)

Addition(Lut cng)

Modus ponens(Lut tch ri)

Ton ri rc: 2011-2012

Cc quy tc suy lun (2)

Chng 3: Suy lun - Chng minh 9

Hypothetical syslogism (Tam on

lun gi nh)

Disjunctive syslogism(Tam on lun

tuyn)

Modus tollens

Ton ri rc: 2011-2012

V d

1. Kaka tng ot qu bng vng Th Gii. Do Kaka tng ot qu bng vng Th Gii hoc gii hc sinh gii ton ri rc cp phng.

2. Tri th nng nc v bn ang qung bom. Do bn ang qung bom.

3. Nu bn chm gi th bn ca bn cm lnh. Nu bn ca bn cm lnh th bn y ht x. Vy nu bn chm gi th bn ca bn ht x.

4. Nu ln bit lp trnh th g bit chi Game. G khng bit chi game. Vy ln bit lp trnh.

Chng 3: Suy lun - Chng minh 10

Ton ri rc: 2011-2012

Quy tc suy lun vi lng t

Chng 3: Suy lun - Chng minh 11

Universal instantiation(S c th ha )

Universal generalization(S tng qut ha )

Existential instantiation(S c th ha )

Existantial generalization(S tng qut ha )

vi bt k

vi mt s

vi mt s

Ton ri rc: 2011-2012

Phng php chng minh

1. Chng minh trc tip (direct).

2. Chng minh gin tip (indirect).

3. Chng minh bng phn chng (contradiction).

4. Chng minh quy np (inductive).

Chng 3: Suy lun - Chng minh 12

Ton ri rc: 2011-2012

1. Chng minh trc tip

Chng minh p q bng cch ch ra:

Nu p l ng th q phi ng.

V d: Nu n l s l th n2 cng l s l

CM: gi s n l th n = 2k + 1

n2 = (2k + 1)2

= 4k2 + 4k + 1

= 2(k2+2k) + 1 (l s l)

Chng 3: Suy lun - Chng minh 13

Ton ri rc: 2011-2012

2. Chng minh gin tip

Chng minh p q bng cch:

thc hin chng minh trc tip q p.

s dng (p q) (q p).

V d: Nu 3n+2 l s l th n l s l

CM: Gi s n chn (kt lun trn l FALSE): n = 2k

3n + 2 = 6k + 2 = 2(3k + 1) (chn)

Vy gi thit l FALSE.

nh l c chng minh.

Chng 3: Suy lun - Chng minh 14

Ton ri rc: 2011-2012

3. Chng minh bng phn chng

M t:

Cn chng minh pht biu p l T.

Gi s tm c mu thun q sao cho p q l T.

Tc (p F) l T. Khi p phi l F th p l T.

c s dng khi c th tm c mu thun dng r r, tc mnh p (r r) l T.

Chng 3: Suy lun - Chng minh 15

Ton ri rc: 2011-2012

3. Chng minh bng phn chng

V d: Chng minh l s v t

Gi s l s hu t, tc trong av b khng c c chung (phn s ti gin)

Khi hay .

Suy ra a2 l s chn hay a cng l s chn.

Ta t vy suy ra b l s chn.

Vy phn s a/b l khng ti gin Mu thun

Chng 3: Suy lun - Chng minh 16

2

2b

a2

2

2

2b

a 222 ab

ca 2 22 42 cb

)( rrp

Ton ri rc: 2011-2012

4. Chng minh bng quy np

Tnh c sp tt: mt tin c bn trn tp cc s nguyn

Chng 3: Suy lun - Chng minh 17

Mi tp hp khng rng cc s nguyn khng mlun lun c phn t nh nht.

}3,9,15,2,4,1{

}9,7,5,3,1{

2

1

S

S

Ton ri rc: 2011-2012

4. Chng minh bng quy np

Chng 3: Suy lun - Chng minh 18

Hai bc chng minh:

1. Bc c bn: Chng minh l TRUE.

2. Bc quy np: CM l TRUE

)1(P

)1()( nPnPn

Php chng minh quy np thng dng chngminh mnh dng

S dng tnh c sp tt ca tp hp.)(nPn

Ton ri rc: 2011-2012

4. Chng minh bng quy np

V d: Tng ca n s nguyn l khng m u tin l n2.

CM:

1. Bc c bn: vi n = 1 ta thy P(1) l TRUE.

2. Bc quy np: gi s ta c gi thit P(n) l TRUE

khi

Tc l P(n+1) l TRUE nu P(n) l TRUE.

Chng 3: Suy lun - Chng minh 19

2)12(...531 nn

12)12()12(...531 2 nnnn2)1( n

Ton ri rc: 2011-2012

quy (Recursion)

Recursive definition (nh ngha quy):

i khi kh nh ngha mt i tng mt cch tng minh.

nh ngha i tng bng chnh n.

V d:

Bn tng qu sinh nht cho bn mnh:

Qu tng l ci hp qu ng ci hp qu.

Chng 3: Suy lun - Chng minh 20

Ton ri rc: 2011-2012

quy (Recursion)

Chng 3: Suy lun - Chng minh 21

Ton ri rc: 2011-2012

quy (Recursion)

Chng 3: Suy lun - Chng minh 22

Ton ri rc: 2011-2012

nh ngha quy

Chng 3: Suy lun - Chng minh 23

Hai bc:

1. Cho gi tr ca hm ti 0.

2. Cng thc tnh gi tr hm ti s nguyn nt cc gi tr hm ti cc s nh hn.

Cn gi l nh ngha quy np.

Ton ri rc: 2011-2012

nh ngha quy

V d:

1. Hm giai tha

D thy

V

Nn

2. Dy Fibonacci:

Chng 3: Suy lun - Chng minh 24

!)( nnF

1)0( F

)1(!)1(...3.2.1)!1( nnnnn

)1).(()!1()1( nnFnnF

21

1

0

1

0

nnn fff

f

f

Ton ri rc: 2011-2012

Thut ton quy

Chng 3: Suy lun - Chng minh 25

Gii bi ton ban u bng cch rt gn n thnh bi ton ging nh vy nhng c d liu u vo nh hn.

V d: thut ton quy tm UCLN(a,b)

int UCLN(int a, int b){

if(a == 0) return b;

else return UCLN(b mod a, a);

}

Ton ri rc: 2011-2012

Bi tp Hi p

1. Chng minh nu a2 l s chn th a cng l s chn.

2. Vit hm quy (ngn ng C) tnh s Fibonacci th n.

Chng 3: Suy lun - Chng minh 26