Les nombres réels

f:R→R
∀(x, y)∈R2,f(x+y) = f(x) + f(y);
∀(x, y)∈R2,f(xy) = f(x)f(y);
∃x∈R,f(x)6= 0.
f(0) f(1) f(−1)
f(x)x∈Zx∈Q
∀x≥0, f(x)≥0f
f= IdR
n∈N∗x1, . . . , xn∈R
n
X
k=1
xk=
n
X
k=1
x2
k=n
k∈ {1, . . . , n}xk= 1

f(0) = f(0 + 0) = f(0) + f(0) f(0) = 0
∀x∈R, f(x) = f(1.x) = f(1)f(x)
f f(1) = 1
f(1) + f(−1) = f(0) = 0 f(−1) = −1
n∈Nf(n) = n
f(−n) = f((−1) ×n) = f(−1) ×f(n) = −f(n) = −n
∀x∈Z, f(x) = x
x∈Qx=p
qp∈Z, q ∈N∗
f(x) = f(p×1
q) = f(p)×f(1 q)
f(p) = p
1 = f(1) = f(q×1
q) = f(q)×f(1 q) = q×f(1 q)
f(1
q) = 1 qf(x) = x
∀x≥0, f(x) = f(√x√x) = f(√x)2≥0
x, y ∈Rx≤y
f(y) = f(x+y−x) = f(x) + f(y−x)≥f(x)
f
x∈Rn∈N
b(nx)c
n≤x < b(nx)c+ 1 n
f
f(b(nx)c
n)≤f(x)< f(b(nx)c+ 1 n)
b(nx)c
n≤f(x)<b(nx)c+ 1 n
n→+∞x≤f(x)≤x f(x) = x
f= IdR
n
X
k=1
(xk−1)2=
n
X
k=1
x2
k−2
n
X
k=1
xk+
n
X
k=1
1=0
∀1≤k≤n, xk= 1