About the Countability of the Algebraic Numbers
(Polynomials with integer coefficients, Prime Numbers, Rational Numbers and Transcendent Numbers)
JeanFrançois COLONNA
jeanfrancois.colonna@polytechnique.edu
CMAP (Centre de Mathématiques APpliquées) UMR CNRS 7641, Ecole Polytechnique, CNRS, 91128 Palaiseau Cedex, France
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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 nthdegree with integer coefficients:
P(X) = A_{n}*X^{n} + A_{n1}*X^{n1} + A_{n2}*X^{n2} + (...) + A_{2}*X^{2} + A_{1}*X^{1} + A_{0}*X^{0}
A_{i} E Z \/ i
A_{n} # 0
Then, using the n+1 polynomial coefficients, let's define the following unique rational number:
R = 2^{A0} * 3^{A1} * 5^{A2} * 7^{A3} (...)
where the numbers
{F_{1}=2,F_{2}=3,F_{3}=5,F_{4}=7,...,F_{n+1}}
are the n+1 first prime numbers F_{i}.
Then obviously:
R = a/b with a E N^{*},b E N^{*},HCF(a,b)=1
For example:

 
  
   
    
     
P(X) = X^{2}  X  1 = + 1*X^{2}  1*X^{1}  1*X^{0} ==> 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/ba E N^{*},b E N^{*},HCF(a,b)=1}
Conversely, any number R in Q' defines an unique polynomial with integer coefficients.
For example:

 
  
   
    
     
      
       
        
         
R = 22/7 = (2*11)/7 = 2^{+1} * 3^{0} * 5^{0} * 7^{1} * 11^{+1} ==> P(X) = + 1*X^{4}  1*X^{3} + 0*X^{2} + 0*X^{1} + 1*X^{0} = X^{4}  X^{3} + 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 socalled algebraic numbers.
Let's recall that a polynomial of the nthdegree 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 the roots (real or complex) of the polynom
P(X): {1,2,...,n}.
Then, M being the number of a certain root, one can add to the definition of R a new factor equal to F_{n+2}
to the power M.
This creates a bijection between the rational numbers and the set of the root "identity" (ie. their number)
of the polynomials with integer coefficients.
It is noteworthy to recall that the use of the root values (instead of their "identities") would imply the renouncement of
the bijection since any algebraic number can be obtained using an infinite number of ways;
for example, all the equations
K.X^{2}K.XK=0 with K#0
define the same two rational numbers
(1+sqrt(5))/2 (the golden ratio) and (1sqrt(5))/2.
Let's recall a consequence of this result: transcendent numbers are non countable and do exist. As a matter of fact:
RealNumbers = AlgebraicNumbers U 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) = X^{2} 
1/6: P(X) = X1 
1/7: P(X) = X^{3} 
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) = X^{2}+1 
2/6=1/3 
2/7: P(X) = X^{3}+1 

3/1: P(X) = +X 
3/2: P(X) = +X1 
3/3=1/1 
3/4: P(X) = +X2 
3/5: P(X) = X^{2}+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) = X^{2}+2 



5/1: P(X) = +X^{2} 
5/2: P(X) = +X^{2}1 
5/3: P(X) = +X^{2}X 
5/4: P(X) = +X^{2}2 




6/1: P(X) = +X+1 
6/2=3/1 
6/3=2/1 





7/1: P(X) = +X^{3} 
7/2: P(X) = +X^{3}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) = +X1 
2/3: P(X) = X+1 
1/4: P(X) = 2 




5/1: P(X) = +X^{2} 
4/2=2/1 
3/3=1/1 
2/4=1/2 
1/5: P(X) = X^{2} 



6/1: P(X) = +X+1 
5/2: P(X) = +X^{2}1 
4/3: P(X) = X+2 
3/4: P(X) = +X2 
2/5: P(X) = X^{2}+1 
1/6: P(X) = X1 


7/1: P(X) = +X^{3} 
6/2=3/1 
5/3: P(X) = +X^{2}X 
4/4=1/1 
3/5: P(X) = X^{2}+X 
2/6=1/3 
1/7: P(X) = X^{3} 

8/1: P(X) = +3 
7/2: P(X) = +X^{3}1 
6/3=2/1 
5/4: P(X) = +X^{2}2 
4/5: P(X) = X^{2}+2 
3/6=1/2 
2/7: P(X) = X^{3}+1 
1/8: P(X) = 3 
[*]: by convention
Copyright © JeanFrançois Colonna, 20202022.
Copyright © CMAP (Centre de Mathématiques APpliquées) UMR CNRS 7641 / Ecole Polytechnique, 20202022.