About the Countability of the Algebraic Numbers
(Polynomials with integer coefficients, Prime Numbers, Rational Numbers and Transcendent Numbers)
CMAP (Centre de Mathématiques APpliquées) UMR CNRS 7641, École polytechnique, Institut Polytechnique de Paris, CNRS, France
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[en français/in french]
Abstract: How to "count" the polynomials with integer coefficients?
A relationship between prime numbers and transcendent numbers.
Keywords: Polynomials, Polynômes, Prime Numbers, Nombres Premiers, Rational Numbers, Nombres Rationnels, Algebraic Numbers, Nombres Algébriques, Transcendent Numbers, Nombres Transcendants.
Let P(X) be a polynomial of the nth-degree with integer coefficients:
P(X) = A0*X0 + A1*X1 + A2*X2 + (...) + An-2*Xn-2 + An-1*Xn-1 + An*Xn
Ai ∈ Z ∀ i
An # 0
Then, using the n+1 polynomial coefficients, let's define the following unique rational number:
R = 2A0 * 3A1 * 5A2 * 7A3 * (...) * Fn+1An
where the numbers
{F1=2,F2=3,F3=5,F4=7,...,Fn+1}
are the n+1 first prime numbers Fi.
Then obviously:
R = a/b with a ∈ N*,b ∈ N*,HCF(a,b)=1
For example:
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P(X) = X2 - X - 1 = + 1*X2 - 1*X1 - 1*X0 ==> R = 2-1 * 3-1 * 5+1 = 5/(2*3) = 5/6
(by the way, the positive root of the equation P(X)=0 defines the golden ratio).
The number R belongs therefore to the following set:
Q' = {a/b|a ∈ N*,b ∈ N*,HCF(a,b)=1}
Conversely, any number R in Q' defines an unique polynomial with integer coefficients.
For example:
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R = 22/7 = (2*11)/7 = 2+1 * 30 * 50 * 7-1 * 11+1 ==> P(X) = + 1*X4 - 1*X3 + 0*X2 + 0*X1 + 1*X0 = X4 - X3 + 1
This process defines a bijection between Q' and the set of the polynomials with integer coefficients.
Q' being a subset of Q (the rational numbers) and
Q being countable,
then the set of the polynomials with integer coefficients is countable.
At last, the real roots of the polynomials with integer coefficients are defining the so-called algebraic numbers.
Let's recall that a polynomial of the nth-degree has n complex roots and then at the most n real roots.
Then the algebraic numbers are countable (a well known result, obtained here by means of the prime numbers and of the rational numbers).
Nota: One can go a step further numbering
{1,2,...,n}
the roots (real or complex) of the polynom P(X).
Then, M being the number of a certain root, one can add to the definition of R a new factor equal to Fn+2
to the power M:
R = 2A0 * 3A1 * 5A2 * 7A3 * (...) * Fn+1An * Fn+2M
Then, in order to be able to define a bijection, the following condition should be added:
HCF(A0,A1,A2,(...),An)=1
As a matter of fact, K being an arbitrary non nul integer number, the polynoms P(X) and K.P(X), have the same roots.
For example, all the equations
K.X2-K.X-K=0 with K#0
define the same two rational numbers
(1+sqrt(5))/2 (the golden ratio) and (1-sqrt(5))/2.
This creates a bijection between the rational numbers and the set of the root "identity" (ie. their number)
of the polynomials with integer coefficients.
Let's recall a consequence of this result: transcendent numbers are non countable and do exist. As a matter of fact:
RealNumbers = AlgebraicNumbers ∪ TranscendentNumbers
AlgebraicNumbers ∩ TranscendentNumbers = 0
Real Numbers are not countable and Algebraic Numbers are countable
then:
Transcendent Numbers are not countable and do exist
Annexe:
Here are the first positive Rational Numbers using the same order than the one used for
the demonstration of their countability:
| 1/1: P(X) = 0 [*] | 1/2: P(X) = -1 | 1/3: P(X) = -X | 1/4: P(X) = -2 | 1/5: P(X) = -X2 | 1/6: P(X) = -X-1 | 1/7: P(X) = -X3 | 1/8: P(X) = -3 |
| 2/1: P(X) = +1 | 2/2=1/1 | 2/3: P(X) = -X+1 | 2/4=1/2 | 2/5: P(X) = -X2+1 | 2/6=1/3 | 2/7: P(X) = -X3+1 | |
| 3/1: P(X) = +X | 3/2: P(X) = +X-1 | 3/3=1/1 | 3/4: P(X) = +X-2 | 3/5: P(X) = -X2+X | 3/6=1/2 | | |
| 4/1: P(X) = +2 | 4/2=2/1 | 4/3: P(X) = -X+2 | 4/4=1/1 | 4/5: P(X) = -X2+2 | | | |
| 5/1: P(X) = +X2 | 5/2: P(X) = +X2-1 | 5/3: P(X) = +X2-X | 5/4: P(X) = +X2-2 | | | | |
| 6/1: P(X) = +X+1 | 6/2=3/1 | 6/3=2/1 | | | | | |
| 7/1: P(X) = +X3 | 7/2: P(X) = +X3-1 | | | | | | |
| 8/1: P(X) = +3 | | | | | | | |
or again:
| 1/1: P(X) = 0 [*] | | | | | | | |
| 2/1: P(X) = +1 | 1/2: P(X) = -1 | | | | | | |
| 3/1: P(X) = +X | 2/2=1/1 | 1/3: P(X) = -X | | | | | |
| 4/1: P(X) = +2 | 3/2: P(X) = +X-1 | 2/3: P(X) = -X+1 | 1/4: P(X) = -2 | | | | |
| 5/1: P(X) = +X2 | 4/2=2/1 | 3/3=1/1 | 2/4=1/2 | 1/5: P(X) = -X2 | | | |
| 6/1: P(X) = +X+1 | 5/2: P(X) = +X2-1 | 4/3: P(X) = -X+2 | 3/4: P(X) = +X-2 | 2/5: P(X) = -X2+1 | 1/6: P(X) = -X-1 | | |
| 7/1: P(X) = +X3 | 6/2=3/1 | 5/3: P(X) = +X2-X | 4/4=1/1 | 3/5: P(X) = -X2+X | 2/6=1/3 | 1/7: P(X) = -X3 | |
| 8/1: P(X) = +3 | 7/2: P(X) = +X3-1 | 6/3=2/1 | 5/4: P(X) = +X2-2 | 4/5: P(X) = -X2+2 | 3/6=1/2 | 2/7: P(X) = -X3+1 | 1/8: P(X) = -3 |
[*]: by convention
Copyright © Jean-François COLONNA, 2020-2025.
Copyright © CMAP (Centre de Mathématiques APpliquées) UMR CNRS 7641 / École polytechnique, Institut Polytechnique de Paris, 2020-2025.
