(Trch bi ging: Cc phng php v k thut chng minh, chng trnh Gp g
Ton hc t chc ti HQG Tp HCM t ngy 25-31/1/2010)
Chng minh phn chng c th ni l mt trong nhng v kh quan trng ca ton
hc. N cho php chng ta chng minh s c th v khng c th ca mt tnh cht no
, n cho php chng ta bin thun thnh o, bin o thnh thun, n cho php
chng ta l lun trn nhng i tng m khng r l c tn ti hay khng. V d kinh
in nht v php chng minh phn chng thuc v Euclid vi php chng minh
nh l.Tn ti v s s nguyn t.
y, Euclid gi s ngc li rng tn ti hu hn s nguyn tp1,p2,...,pn. ng
xt tchN=p1p2...pn+1. N phi c t nht 1 c s nguyn t p. Khi ,
dop1,p2,...,pnl tt c cc s nguyn t nn tn ti i sao chop=pi. Nhng khi
p | 1, mu thun.
Bi tp
1. Chng minh rng tn ti v s s nguyn t dng 4k+3.2. Chng minh rng
tn ti v s s nguyn t dng 4k+1.
Mt chng minh ni ting khc bng phng php phn chng chnh l chng minh
ca Euler cho nh l nh Fermat vi trng hp n = 4.
nh l.Phng trnhx4+y4=z4(1) khng c nghim nguyn dng.
ng gi s rng phng trnh (1) c nghim nguyn dng. Khi , theo nguyn l
cc hn, tn ti nghim(x0,y0,z0)vix0+y0+z0nh nht. Sau , bng cch s dng
cu trc nghim ca phng trnh Pythagorex2+y2=z2, ng i n s tn ti ca mt
nghim(x1,y1,z1)cx1+y1+z1 B, ta c th chng minh ^B -> ^A . V mt bn
cht th hai php suy din ny c v ging nhau, nhng trong thc t th li kh
khc nhau. Ta th xem xt 1 vi v d.
V d 1.Chng minh rng hm sf(x)= l mt n nh t R vo R.
V d 2.Chng minh rng nu (p-1)! + 1 l s nguyn t th p l s nguyn
t.
Trong v d 1, r rng vic chng minhx1x2suy raf(x1)f(x2)kh khn hn
vic chng minhf(x1)=f(x2)suy rax1=x2, d rng v mt logic, hai iu ny l
tng ng.
Trong v d 2, gn nh khng c cch no khc ngoi cch chng minh nu p l
hp s, p = r.s th (p-1)! + 1 khng chia ht cho p.
Bi tp.
5. Cho hm s f: R -> R tho mn cc iu kin sau1) f n iu ;2)
f(x+y) = f(x) + f(y) vi mi x, y thuc R.Chng minh rng tn ti s thc a
sao cho f(x) = ax vi mi x thuc R.6. Cho a, b, c l cc s thc khng m
tho mn iu kina2+b2+c2+abc=4. Chng minh rnga+b+c3.
Trong vic chng minh mt s tnh cht bng phng php phn chng, ta c th
c thm mt s thng tin b sung quan trng nu s dng phn v d nh nht. tng l
chng minh mt tnh cht A cho mt cu hnh P, ta xt mt c trng f(P) ca P l
mt hm c gi tr nguyn dng. By gi gi s tn ti mt cu hnh P khng c tnh
cht A, khi s tn ti mt cu hnh P0 khng c tnh cht A vi f(P0) nh nht.
Ta s tm cch suy ra iu mu thun. Lc ny, ngoi vic chng ta c cu hnh P0
khng c tnh cht A, ta cn c mi cu hnh P vi f(P) < f(P0) u c tnh
cht A.
V d 3.Cho ng gic li ABCDE trn mt phng to c to cc nh u nguyn.a)
Chng minh rng tn ti t nht 1 im nm trong hoc nm trn cnh ca ng gic
(khc vi A, B, C, D, E) c to nguyn.b) Chng minh rng tn ti t nht 1 im
nm trong ng gic c to nguyn.c) Cc ng cho ca ng gic li ct nhau to ra
mt ng gic li nhA1B1C1D1E1bn trong. Chng minh rng tn ti t nht 1 im
nm trong hoc trn bin ng gic liA1B1C1D1E1.
Cu a) c th gii quyt d dng nh nguyn l Dirichlet: V c 5 im nn tn
ti t nht 2 im X, Y m cp to (x, y) ca chng c cng tnh chn l (ta ch c
4 trng hp (chn, chn), (chn, l), (l, chn) v (l, l)). Trung im Z ca
XY chnh l im cn tm.
Sang cu b) l lun trn y cha , v nu XY khng phi l ng cho m l cnh
th Z c th s nm trn bin. Ta x l tnh hung ny nh sau. rng nu XY l mt
cnh, chng hn l cnh AB th ZBCDE cng l mt ng gic li c cc nh c to u
nguyn v ta c th lp li l lun nu trn i vi ng gic ZBCDE, Ta c th dng n
bin chng minh qu trnh ny khng th ko di mi, v n mt lc no s c 1 ng
gic c im nguyn nm trong.
Tuy nhin, ta c th trnh by li l lun ny mt cch gn gng nh sau: Gi s
tn ti mt ng gic nguyn m bn trong khng cha mt im nguyn no (phn v d).
Trong tt c cc ng gic nh vy, chn ng gic ABCDE c din tch nh nht (phn
v d nh nht). Nu c nhiu ng gic nh vy th ta chn mt trong s chng. Theo
l lun trnh by cu a), tn ti hai nh X, Y c cp to cng tnh chn l. Trung
im Z ca XY s c to nguyn. V bn trong ng gic ABCDE khng c im nguyn no
nn XY phi l mt cnh no . Khng mt tnh tng qut, gi s l AB. Khi ng gic
ZBCDE c to cc nh u nguyn v c din tch nh hn din tch ng gic ABCDE. Do
tnh nh nht ca ABCDE (phn v d nh nht pht huy tc dng!) nn bn trong ng
gic ZBCDE c 1 im nguyn T. iu ny mu thun v T cng nm trong ng gic
ABCDE.
Bi tp
7. Gii phn c) ca v d 3.8. (nh l Bezout) Chng minh rng nu (a, b)
= 1 th tn ti u, v sao cho au + bv = 1.
Phng php phn chng thng hay c s dng trong cc bi ton bt bin hoc bi
ton ph hnh chng minh s khng thc hin c. Sau y chng ta xem xt 2 v d
nh vy.
V d 4.Xt hnh vung 7 x 7 . Chng minh rng ta c th xo i mt phn cn
li khng th ph kn bng 15 qun trimino kch thc 1 x 3 v 1 qun trimino
hnh ch L.
V d 5.Cho trc cc hm sf1(x)=x2+2x,f2(x)=x+1/x,f3(x)=x22x. Cho php
thc hin cc php ton cng hai hm s, nhn hai hm s, nhn mt hm s vi mt
hng s tu . Cc php ton ny c th tip tc c thc hin nhiu ln trnfiv trn
cc kt qu thu c. Chng minh rng c th thu c hm s 1/x t cc hm
sf1,f2,f3bng cc s dng cc php ton trn nhng iu ny khng th thc hin c
nu thiu mt trong 3 hmf1,f2,f3.
Bi tp
9. Hnh vung 5 x 5 b i trung tm. Chng minh rng c th ph phn cn li
bng 8 qun trimino 1 x 3 nhng khng th ph c bng 8 qun trimino hnh ch
L. Tm tt c cc gi tr k sao cho c th ph phn cn li bng k qun trimino 1
x 3 v 8-k trimino hnh ch L.
10. Trn vng trn ban u theo mt th t tu c 4 s 1 v 5 s 0. khong gia
hai ch s ging nhau ta vit s 1 v khong gia hai ch s khc nhau ta vit
s 0. Cc s ban u b xo i. Hi sau mt s ln thc hin nh vy ta c th thu c
mt b gm 9 s 0?
