DS Groupe, Anneau, Arithmetique

2014/2015
f f(x) = √1+4x2
1 + x+√1 + x2
f f
f f
f+∞
+∞
f−∞
−∞
f α α = 0.14
10−2f(α) = 0.48 10−2
f f
n∈Nfn[0,+∞[f(x) = x5+n x −1
fnxnxn
n∈Nfn+1 (xn) (xn)n∈N
(xn)n∈N`
xn
1
n(n xn)n∈N
xn(n xn)n∈N
xn−1
n

2014/2015
(G, )H K (G, )
H K =z∈G
∃(x, y)∈H×K;z=x yK H =z∈G ∃(x, y)∈K×H;z=x y
z∈H K z−1K H
H K (G, )H K K H
H K (G, )H K K H
(Z,+)
(Z,+)
n∈ZnZ={x∈Z|∃k∈Z;x=nk}
n∈ZnZ(Z,+)
(n, p)∈Z2nZ=pZ⇐⇒ n=±p
(Z,+) nZ
H(Z,+) A=H∩N∗
H{0Z}
H{0Z}
A
A b
bZ⊂H
a∈H r a b r ∈H r = 0
H=bZ
(a, b)∈Z2
∃(d, p)∈N2|aZTbZ=pZaZ+bZ=dZ
d=a∧b p =a∨b

2014/2015
f f(x) = √1+4x2
1 + x+√1 + x2
∀x∈R,1 + x2>0,1+4x2>0−x6|x|=√x2<√1 + x21 + x+√1 + x2>0
fR
f√.
f√.
fR
1 + x+√1 + x2= 2 + x+1
2x2−1 8x4+◦(x4) = 2 ×1 + 1 2x+1 4x2−1 16 x4+◦(x4)
1
1 + x+√1 + x2=1
2×1−1
2x+1 8x3+◦(x4)
f(x) = 1
2×1−1
2x+1 8x3+◦(x4)×(1 + 2x2−2x4+◦(x4))
f(x) = 1
2−1 4x+x2−7 16x3−x4+◦(x4)
f Y =1
2−1 4X
t=1
xx+∞t
f(x) = f1
t=√4 + t2
1 + t+√1 + t2√4 + t2= 2 + 1
4t2−1 64t4+◦(t4)
f1
t=1 2×1−1 2t+◦(t2)×2 + 1 4t2+◦(t2)= 1 −1
2t+1
8t2+◦(t2)
f(x) = 1 −1
2x+1
8x2+◦1 x2
f Y = 1 +∞
t=1
xx−∞ t
f(x) = f1
t=−√4 + t2
1 + t−√1 + t2√4 + t2= 2 + 1
4t2+◦(t3) 1 + t−√1 + t2=t×
1−1
2t+1 8t3+◦(t3)
f1
t=−1 t×2 + 1 4t2+◦(t2)×1 + 1 2t+1 4t2+◦(t2)=−2 t−1−3 4t+◦(t)
f(x) = −2x−1−3
4x+◦1 x
f Y =−2X−1−∞
f α α = 0.14
10−2f(α) = 0.48 10−2
f f

2014/2015
x
f0(x)
f(x)
−∞ α+∞
−0+
+∞+∞
f(α)f(α)
11
1
2
n∈Nfn[0,+∞[f(x) = x5+n x −1
fnR+
R+fn(0) = −1
lim
x→+∞fn(x) = +∞
fnR+[−1,+∞[∃!xn∈R+|fn(xn) = 0
fn(0) = −1< fn(xn)<1
n5=fn1 nfn0< xn<1 n
xn
n∈Nfn+1 (xn) = xn>0fn+1 xn+1 < xn
(xn)n∈N
(xn)n∈N
(xn)n∈N`= 0
0< xn<1
n(n xn)n∈N
yn=n×xn1−yn=yn
n5(yn)n∈N1−yn
(n xn)n∈N
zn=yn−1 = nxn−11
n5(1 + zn)5+zn= 0 zn
zn∼−1
n5xn=1 n−1 n5+◦1 n5

2014/2015
H K (G, )H K =z∈G
∃(x, y)∈H×K;z=x y
z∈H K z =x y (x, y)∈H×K
z z−1=y−1x−1z−1∈K H
H K (G, )
z∈H K H K z−1H K
z−1K H z ∈K H
z∈K H z =x y x K y H
z z−1=y−1x−1∈H K H K
y−1x−1H K z ∈H K
H K =K H
H K K H
+H K G eG
+z∈H K K H H K H K
+z z0H K G z =x y z0=x0y0x
x0H y y0K z z0=x(y x0)y0y x0∈K H =H K
x00 ∈H y00 ∈K y x0=x00 y00 z z0= (x x00) (y00 y0)∈
H K H K
H K G
H K (G, )H K K H
(Z,+)
(Z,+)
n∈ZnZ={x∈Z|∃k∈Z;x=nk}
n∈ZnZnZ
nZ(Z,+)
(n, p)∈Z2nZ=pZn p
n=±p
n=±p n p n Z=pZ
nZ=pZ⇐⇒ n=±pH(Z,+) A=H∩N∗
H{0Z}H= 0Z
H{0Z}
H
AA
AN∗b
b∈H H
∀k∈N, k b ∈H−k b ∈HbZ⊂H
a∈H r q a b r =a−bq
a H b q b ZH H r∈H
r < b b A r A =H∩N∗r
H r /∈N∗r60r r >0
r= 0