Corrigé TD Algèbre 1 : Logique, Ensembles, Applications

◦
∃x∈R:∀y∈R, x +y > 0
∀x∈R,∃y∈R:x+y > 0
∀x∈R,∀y∈R:x+y > 0
∃x∈R:∀y∈R, y2> x
∀x∈R,∃y∈Rx+y≤0x∈R
y≤ −x x +y≤0
x∈Ry > −x x +y > 0
∃x∈R,∀y∈R:x+y≤0
∃x∈R,∃y∈R:x+y≤0
x=−1∀y∈Ry2>−1
∀x∈R:∃y∈R, y2≤x
∀n∈N,∃m∈N:m > n
∃m∈N:∀n∈N, m > n
∀x∈Q∗
+,∃y∈Q: 0 < y < x
∀n∈N, n > 3⇒n > 6
∀n∈N,∃p∈N:n= 2p+ 1
∀x∈R, x < 2⇒x2<4
∀m, n ∈N,2|mn, 8|m2−n2
n m =n+ 1 ∃n∈N,∀m∈N:m≤n
∀m∈N:∃n∈N, m ≤n
y=1
2x∃x∈Q∗
+,∀y∈Q:y≤0x≤y
∃x∈Q∗
+:∀y∈Q, y > 0⇒x≤y
4>3 4 ≤6∃n∈N, n > 3n≤6
4∃n∈N,∀p∈N:n6= 2p+ 1
x=−3∃x∈R, x < 2x2≥4
2-mn 2-m2-n m = 2p+ 1 n= 2q+ 1
m2−n2= (2p+ 1)2−(2q+ 1)2= 4p2+ 4p−4q2−4p= 4(p(p+ 1) −q(q+ 1)) 2 |p(p+ 1)
2|q(q+ 1) 8 |m2−n2
P, Q, R
P⇒(Q⇒R)≡(P Q)⇒R
P Q P ⇒R Q ⇒R R
P Q (P⇒
Q) (( P)⇒Q)
P⇒(Q⇒R)≡P(Q⇒R)≡P(Q R)≡(P Q)R)
P Q ≡P Q
P⇒(Q⇒R)≡(P Q)R≡(P Q)⇒R
S T S S ⇒T T

IP P ⇒R R
IP P Q Q Q ⇒R R
S≡(P⇒Q) (P⇒Q)
IP P P ⇒Q S
IP P ⇒Q S
(P⇒Q) (P⇒Q)
I(P⇒Q) (P⇒Q)≡P Q P Q ≡(P P ) (Q Q) (P P )
(P⇒Q) (P⇒Q)
q∈Qx∈R\Qx+q∈R\Q
x+q∈Qx= (x+q)−q∈Q
q∈Q∗x∈R\Qx.q ∈R\Q
x.q ∈Qx=x.q
q∈Q
x∈R+\Qz=√x z ∈Qz2=x∈Q
a, n ana
ln(3)
ln(2) /∈Q
a ana0= 1
anan+1 =an.a
ln(3)
ln(2) =p q∈Qp, q ∈Zq6= 0 ln(3) ln(2) >0p, q > 0
pln(2) = qln(3) epln(2) =eqln(3) 2p= 3q2p3q
3
n≥10 2n≥n3n≥10 2 ≥(1 + 1/n)3
Sn=Pn
k=1 1
k(k+1) Sn=n n+1
un=1+3+5+... + (2n−1) un=n2n∈N∗
x0= 0 xn+1 =3
√xn+ 6 xn≤2,∀n∈N
n≥10 (1 + 1/n)3= 1 + 3/n + 3/n2+ 1/n3≤(1 + 3/10 + 3/100 + 1/1000) = 1000+300+30+1
1000 =
1331
1000 ≤2
∀n≥10 2n≥n3
n= 10 210 = 1024 ≥103= 1000 n= 10
n≥10 n2n+1 = 2n.2≥n3(1 + 1/n)3= (n+ 1)3
n= 1 Sn=1
2n= 1
n≥1Sn=n
n+1 Sn+1 =Pn+1
k=1 1
k(k+1) =Sn+1 (n+1)(n+2) Sn+1 =
n
n+1 +1 (n+1)(n+2) =n(n+2)+1 (n+1)(n+2) =n+1 n+2 n+ 1
un=Pn
k=1(2k−1) n= 1 u1= 1 = 12n= 1
n≥1n un+1 =un+ 2(n+ 1) −1 = un+ 2n+ 1 =
n2+ 2n+ 1 = (n+ 1)2
n= 0 x0= 0 ≤2n∈Nxn≤2xn+1 =3
√xn+ 6 ≤3
√2 + 6 = 3
√8 = 2
(un)n∈N
u0= 1 ∀n∈N, un+1 =un
1 + un

unn
u0= 1 u1=1
2u2=1 3u3=1 4u4=1 5u5=1 6
un=1
n+1
n= 0 u0= 1
n∈Nn un+1 =un
1+un=
1
n+1
1+ 1
n+1
=1
n+2
n+ 1
∀n∈N, un=1
n+1
n
X
k=1
(−1)kk
un=Pn
k=1(−1)kk
n= 2m
un=u2m=
m
X
k=1
(−1)2k2k+
m
X
k=1
(−1)2k−1(2k−1) =
m
X
k=1
2k−
m
X
k=1
(2k−1) = m=n
2
n
un=un−1+ (−1)nn=un−1−n=n−1
2−n=−n+ 1 2
A B C E
A=B⇔A∪B=A∩B
A∆B=A∩B⇔A=B=∅
A∆B=A∆C⇔B=C
⇒A=B A ∪B=A∪A=A A ∩B=A∩A=A
⇐A∪B=A∩B A =B
A⊂A∪B=A∩B A ⊂A∩B A ∩B⊂B A ⊂B
B⊂A∪B=A∩B B ⊂A∩B A ∩B⊂A B ⊂A
⇒A∆B=A∩B(A∪B)\(A∩B)=(A∪B)∩(A∩B) = A∩B
A∩B=∅A∪B=∅A=B=∅
⇐⇒A∆B=A∆C B =C A∆A=∅
A∆(A∆B) = A∆(A∆C) (A∆A)∆B= (A∆A)∆C B =C
⇐
f:E→F A B E C D F
f(A∪B) = f(A)∪f(B)f(A∩B)⊂f(A)∩f(B)
−1
f(C∪D) = −1 f(C)∪−1 f(D)−1 f(C∩D) = −1 f(C)∩−1 f(D)
f⇔ ∀A, B ⊂E, f(A∩B) = f(A)∩f(B)
IA⊂A∪B f(A)⊂f(A∪B)f(B)⊂f(A∪B)f(A)∪f(B)⊂f(A∪B)
y∈f(A∪B)x∈A∪B y =f(x)
x∈A y ∈f(A)⊂f(A∪B)
x∈B y ∈f(B)⊂f(A∪B)
y∈f(A∪B)f(A∪B)⊂f(A)∪f(B)
IA∩B⊂A f(A∩B)⊂f(A)f(A∩B)⊂f(B)f(A∩B)⊂f(A)∩f(B)

Ix∈f−1(C∪D)⇔f(x)∈C∪D⇔f(x)∈C f(x)∈D⇔x∈f−1(C)x∈f−1(D)⇔x∈
f−1(C)∪f−1(D)f−1(C∪D) = f−1(C∪D)
Ix∈f−1(C∩D)⇔f(x)∈C∩D⇔f(x)∈C f(x)∈D⇔x∈f−1(C)x∈f−1(D)⇔x∈
f−1(C)∩f−1(D)f−1(C∩D) = f−1(C∩D)
⇒f x 6=y∈E f(x) = f(y) = z
A={x}B={y}A∩B=∅f(A)∩f(B) = {z}A, B ⊂E
f(A∩B)6=f(A)∩f(B)
⇐f f(A∩B)⊂f(A)∩f(B)
∀A, B ⊂E f(A)∩f(B)⊂f(A∩B)y∈f(A)∩f(B)y∈f(A)y∈f(B)
x∈A:y=f(x)x0∈B:y=f(x0)f x =x0∈A∩B
y=f(x)∈f(A∩B)
E, F f :E−→ F
A⊂E A ⊂−1 f(f(A)) B⊂F f(−1 f(B)) ⊂B
(i)f A ⊂E, −1 f(f(A)) ⊂A
(ii)f B ⊂F, B ⊂f(−1 f(B))
x∈A f(x)∈f(A)x∈−1 f(f(A)) A⊂−1 f(f(A))
y∈f(−1 f(B)) y∈B x ∈−1 f(B)y=f(x)x∈−1 f(B)
f(x)∈B y ∈B
f−1 f(f(A)) ⊂A x ∈−1 f(f(A)) f(x)∈f(A)
z∈A f(x) = f(z)f x =z x ∈A
A⊂E−1 f(f(A)) ⊂A f
x, z ∈E f(x) = f(z){x} ⊂ −1 f(f({x})) = −1 f(f({z})) = {z}x=z
f y ∈B x ∈E y =f(x)∈B x ∈−1 f(B)
y=f(x)∈f(−1 f(B))
B⊂F B ⊂f(−1 f(B)) y∈F B ={y}
B⊂f(−1 f(B)) −1 f(B)6=∅y
f:R→Rf(x) =|x|
f([1,2]), f(] −3,2]),−1 f(]0,1]),−1 f([−1,1]) −1 f([−2,0[)
g:R→Rg(x)x n ∈Z−1 g({n})
f([1,2]) = [1,2]
f(] −3,2]) = f(] −3,0] ∪[0,2]) = f(] −3,0]) ∪f([0,2]) = [0,3] ∪[0,2] = [0,3]
−1
f(]0,1]) = {x∈R: 0 <|x| ≤ 1}= [−1,0[∪]0,1] = [−1,1] \ {0}
−1
f([−1,1]) = [−1,1]
−1
f([−2,0[) = ∅
−1
g({n}) = {x∈R:g(x) = n}={x∈R:n≤x < n + 1}= [n, n + 1[
∀x∈R,x
1+|x|∈]−1,1[

f:R−→]−1,1[ f(x) = x
1+|x|f
f−1
|x|<|x|+ 1 0 ≤|x|
|x|+1 <1−1<x |x|+1 <1
f x1, x2∈Rf(x1) = f(x2)
x1
|x1|+1 =x2 |x2|+1
|x1|
|x1|+1 =|x2| |x2|+1 |x1|(|x2|+ 1) = |x2|(|x1|+ 1) |x1|=|x2|
x1x2x1=x2f
f y ∈]−1,1[ y=f(x) = x
|x|+1 y
x
y≥0y=x
x+1 yx +y=x x =y 1−y
y≤0y=x
−x+1 −yx +y=x x =y 1+y
x=y
1−|y|∈R∀y∈]−1,1[ f
f f−1(y) = y
1−|y|
R
∀x, y ∈R, xRy⇔cos2(x) + sin2(y) = 1
cl(0) R
xRy⇔cos2(x) + sin2(y) = 1 ⇔cos2(x) = 1 −sin2(y) = cos2(y)
f(x) = cos2(x)xRy⇔f(x) = f(y)R
cl(0) = {x∈R: cos2(x) = cos2(0) = 1}={x∈R: cos(x) = 1 cos(x) = −1}=πZ
R
∀x, y ∈R, xSy⇔x2−y2=x−y
x∈Rx
xSy⇔x2−y2=x−y⇔x2−x=y2−y
f(x) = x2−x xSy⇔f(x) = f(y)S
cl(x) = {y∈R:x2−y2=x−y}
x2−y2=x−y⇔x=y x =−y cl(x) = {x, −x}x6= 0 cl(0) = {0}
N
∀x, y ∈N, xTy⇔ ∃k∈N∗:y=xk
{2,3}
∀x∈Nx=x1xTxT
x, y ∈NxTy yTx k, m ∈N∗y=xkx=ymxkm =x
km = 1 k=m= 1 x=yT
x, y, z ∈NxTy yTz k, m ∈N∗y=xkz=ymz=xkm
xTzT2 3 2T3 3T2
M{2,3}k, m ∈N∗M= 2k= 3m
E, F f :E−→ F
A⊂E f(−1 f(f(A))) = f(A)
B⊂F−1 f(f(f
−1
(B))) = f
−1
(B)