My title - Departamento de Matemática da Universidade do Minho

4
NÚMEROS E DINÂMICA 30
Algebraic and transcendental numbers. Algebraic numbers are the roots of polynomials with
integer coefficients. The degree of the algebraic number x is the minimal degree of a polynomial
p ∈ Z[x] such that p(x) = 0. So, algebraic numbers of degree 1 are the rational numbers, algebraic
numbers of degree 2 are the quadratic irrationals, and so on. Numbers which are not algebraic are
called transcendental, and the natural problems are: do they exist could we write some of them
do we recognize them
Let us say that a number x is approximable to order n if there exists a constant λ(x) (which
may depend on x) such that the inequality
∣ x − p q ∣ < λ(x)
q n .
admits infinite rational solutions p/q.
Liouville theorem. A real algebraic number of degree n is not approximable to order > n.
Proof. Let x be an algebraic number of degree n, and f(x) = ax n + bx n−1 + · · · ∈ Z[x] be a
polynomial of degree n such that f(x) = 0. Let p/q be a rational approximation to x. We may
assume that p/q belongs to an interval I = (x − ε, x + ε) around x so small that the polynomial
f(x) has no other root than x, and has bounded derivative, say |f ′ (y)| < L if y ∈ I. Then
|f(p/q)| = |apn + bp n−1 q + . . . |
q n ≥ 1
q n
since the numerator is a positive integer. By the mean value theorem
|f(p/q)| = |f(x) − f(p/q)| = |f ′ (y)| · |x − p/q| ≤ L · |x − p/q|
for some y ∈ I. Putting together the two inequalities we get
|x − p/q| ≥ L−1
q n
Existence and construction of transcendental numbers. Liouville theorem shows that
transcendental numbers do exist! Moreover, it allows us to give some examples: it is sufficient to
produce numbers which are approximable to any order. The example in [HW59] is
0.1100010000000000000000010 · · · = 1
10 1! + 1
10 2! + 1
10 3! + 1
10 4! + . . .