Logique et raisonnements - Page Personnelle de Jérôme Von Buhren

∃x∈R,(x+ 1 = 0 x+ 2 = 0)
(∃x∈R, x + 1 = 0) (∃x∈R, x + 2 = 0)
∀x∈R,(x+ 1 6= 0 x+ 2 6= 0)
∃x∈R∗,∀y∈R∗,∀z∈R∗, z −xy = 0
∀y∈R∗,∃x∈R∗,∀z∈R∗, z −xy = 0
∀y∈R∗,∀z∈R∗,∃x∈R∗, z −xy = 0
∃a∈R,∀ε > 0,|a|< ε
∀ε > 0,∃a∈R,|a|< ε
∀x∈R,∀y∈R, x =y.
∃x∈R,∃y∈R, x =y.
∃x∈R,∀y∈R, x =y.
∀x∈R,∃y∈R, x =y.
f:R→R
∀x∈R,∃y∈R, y =f(x)
∃y∈R,∀x∈R, y =f(x)
f:R→R
∀x∈R, f(x)6= 0.
∀M > 0,∃A > 0,∀x>A, f(x)> M.
∀x∈R, f(x)>0⇒x60.
f:R→R
f
f
fR
∀x∈R,∃y∈R, x =y.
n∈Nn2n
∀n∈N,n(n2+ 1)
2∈N.

x∈R
∀ε > 0,|x|< ε⇒x= 0.
√2
ln(2)/ln(3)
∀(p, q)∈Z2,√36=p+q√2.
n∈N∗06x06. . . 6
xn61 0 6i < j 6n xj−xi61/n
n∈N∗2n−16n!6nn
n∈N
x>−1 (1 + x)n>1 + nx
∀n∈N\ {0,1},1 + 1
22+··· +1 n2>3n 2n+ 1.
u0=u1= 1 (un)
∀n∈N, un+2 =un+1 +2
n+ 2un.
n∈N∗16un6n2
u0= 3, u1= 4 (un)
∀n∈N, un+2 =un+1 + 6un.
n∈Nun= 2 ×3n+ (−2)n
x∈R∗x+x−1∈N
∀n∈N, xn+1
xn∈N.
x∈R\Zx+x−1∈N
u0= 1 (un)
∀n∈N, un+1 =u0+u1+··· +un.
n∈N∗un= 2n−1
n∈N∗
n= 2p(2q+ 1) (p, q)∈N2
n∈N∗
2