Corrigé du contrôle 1 - Jean

N
∀a∈N,∀b∈N∗,∃n∈N, bn > a.
P1n∈N3|n20|n
k∈Nn= 3k m ∈Nn= 20m3k= 20m3
20m3 20 m
m l ∈Nm= 3l
n= 20m= 20 ×3l= 60l60 n
3|n20|n60 n
P2n= 30 10|n6|n60
n P2
P1∀n∈N,60|n=⇒(3|n20|n).
P1∀n∈N, n n 20
n
P1∃n∈N,3|n, 20|n n
P1P1
P1
n= 10 m= 7 n2−m2= 100−49 = 51
n m n2−m2= 51
n= 26 m= 25 n2−m2= (n−m)(n+m) = (26 −25)(26 + 25) = 51
n m
(n, m)
n m n2−m2= 51 (n−m)(n+m) = 51 = 3 ×17
n m n+m= 17 n−m= 3
n= 10 m= 7
(n−m)(n+m) = 1 ×51 n m
n+m= 51 n−m= 1 n= 26 m= 25

N
(un)n∈Nn m n < m un≤um
(un)n∈NNA={un|n∈N}
NANu0∈A
A n ∈N
unA m ∈Nm > n um∈A un
A un≤um(un)n∈N
(un)n∈NN
∀n∈N, un≤u0−n.
n= 0 u0≤u0−0
n un≤u0−n
n+ 1
un+1 < un
un+1 ≤un−1un≤u0−n un+1 ≤u0−n−1 =
u0−(n+ 1) n+ 1
n∈Nn > u0u0−n
un≤u0−n unun∈N
N