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suivant

7Q
Q' “ ' “
∗+ 2 ∗+ 2A = x∈Q ,x <2 B = x∈Q ,x >2 α = sup AQ
Q α∈Q
2
β =
α
2
x∈B ∈A
x
β B
β B Q
2α 62 †>0

2α =2 26∈Q
2β >2
α+β
γ =
2
2γ =2
2γ <2 γ6α
2γ >2
8
(A ,A ,A ,A ,A ) O0 1 2 3 4 −−→−→ →− →−(O, u, v) u = OA0
C ω ,ω ,ω ,ω ,ω A ,A ,A ,A ,A0 1 2 3 4 0 1 2 3 4
i A
1
A
2
A
O 1
A3
A
4
2007

6

2πi 5ω =e1
kω =ω k∈{0,1,2,3,4}k 1
2 3 41+ω +ω +ω +ω =01 1 1 1
2π 4π
1+2cos +2cos =0.
5 5
2 2πcos(2x) = 2cos x− 1 cos( )5
24z +2z−1=0

2π −1+ 5cos( )=5 4
B −1
2|ω +1|1
2πBA =2cos( )2 5
i 1I C I J
2 2
C [BI]
BI
BJ
BJ =BA2
5
a b c m
8
x − y + 2z = a<
(S) mx + (1−m)y + 2(m−1)z = b
: 2x + my − (3m+1)z = c
m=−1 (S) c=3a+b
m=−1 c=3a+b (S)
m = −1 (S)
m
0 1
1 −1 2
@ Am 1−m 2m−2
2 m −3m−1
2007

6

2 4 22x∈B x >2⇒ <1⇒ <2⇒ ∈A
2 2x x x
2 2 2
x∈B ∈A⇒ 6α⇒x> =β β
x x α
2 2
†>0 b=β+† b>β⇒ < =α α
b β
2 2
A x ∈ A < x < α b > > β
b x
2
∈B β B
x
∀† > 0,∃x∈ A,α−† < x6 α
2 2∀†>0,(α−†) <x <2
2†→0 α 62
√ √
2α =2⇒α= 2 α∈Q 26∈Q

2 2β >2 β >2 26∈Q
α+β 2α,β∈Q γ = ∈Q γ =2
2
α+β2γ < 2 γ ∈ A γ 6 α⇒ 6 α⇒ β 6 α
2
α<2 β >2
2γ > 2 γ
A Q
2πA A O1 0 5
i2π i2π
5 5ω =e z7→e z1
kω =ω k∈J0..4K Ak 01
51−ω2 3 4 1 51+ω +ω +ω +ω = =0 ω =11 1 1 1 11−ω1
2π 4π 6π 8π1+cos +cos +cos +cos =05 5 5 5
6π 2π 8π 4π 2π 4πcos =cos cos =cos 1+2cos +2cos =05 5 5 5 5 5
4π 2π2cos = 2cos − 1
5 5
2π 2π 2π 2π2 21+2cos +2(2cos −1)=0 4cos +2cos −1=0
5 5 5 5
2 2Δ=2 +4×4×1=20=2 ×5
√ √ √
−2−2 5 −1− 5 −1+ 5
z = = z = .1 28 4 4

−1+ 52πcos =5 4
4π 2π 2π 2π 2π 2π 2π 2π 2π2 i i i −i i i −i i −i5 5 5 5 5 5 5 5 5|ω +1|=|e +1|=|e (e +e )|=|e |×|e +e |=|e +e |=1 √
2π −1+ 5
|2cos |=
5 2
2007

2 2πBA =|ω −(−1)|=[ω +1|=2cos2 2 1 5 √
2 2OIB BI = OB +IB =r √q 5 511+ = =4 4 2
√ √
5 1 −1+ 5
BJ =BI−IJ = − =
2 2 2 √
−1+ 5
BJ =BA =2 2
O
(OA) (OB) I
[OA] I [IO] [BI]
J BJ B
A A [A A ]2 3 2 3
n
jk 2n=2 ×F ×F ×···×F F =2 +10 1 k j
F = 3 F = 5 F = 17 F = 257 F = 652370 1 2 3 4
m=−1
8
x − y + 2z = a<
−x + 2y − 4z = b L ←L +L2 2 1: 2x − y + 2z = c L ←L −2L3 3 1
8
x − y + 2z = a<
y − 2z = a+b
: y − 2z = −2a+c
L ←L −L 0=3a+b−c3 3 2
c=3a+b
‰ ‰
x − y + 2z = a x = 2a+b

y − 2z = a+b y = a+b+2z
z
A=(2a+b,a+b,0)
(0,2,1)
8
< x − y + 2z = a
mx + (1−m)y + 2(m−1)z = b L ←L −mL2 2 1:
2x + my − (3m+1)z = c L ←L −2L3 3 1
2007

6

8
x − y + 2z = a L ←L +L< 1 1 2
⇔ y − 2z = −ma+b
: (m+2)y − (3m+5)z = −2a+c L ←L −(m+2)L3 3 2
8
x = (1−m)a+b<
⇔ y − 2z = −ma+b
: 2−(1+m)z = (m +2m−2)a−(m+2)b+c
m=−1 1+m
8
x = (1−m)a+b>>>> 2< −3m −5m+4 3m+5 −2
y = a+ b+ c
m+1 m+1 m+1>>>>> 2> −m −2m+2 m+2 1: z = a+ b− c
m+1 m+1 m+1
0 1
1−m 1 0
B C
B C2B C−3m −5m+4 3m+5 −2B C.B Cm+1 m+1 m+1B C
B C
@ A2−m −2m+2 m+2 −1
m+1 m+1 m+1
2007