Cardinality of transcendental numbers

topsquark

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May 2013
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I was reading another thread and I came up with a question. I have no idea how to approach this.

A transcendental number is defined to be non-algebraic, that is, it cannot be formed as a root of a polynomial with rational coefficients. This covers complex numbers but let's just deal with the real transcendentals.

Do the real algebraic numbers have the same cardinality as the real transcendentals? How would we go about finding out?

-Dan
 
Nov 2010
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Hint: The set of all real numbers is the union of the set of real algebraic numbers and the set of real transcendental numbers.
 
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v8archie

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Dec 2013
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The algebraic numbers are countable. You can create an index $i$ for the roots of a polynomial $$p(x) = \sum_{k=0}^n a_kx^k \qquad (a_k \in \mathrm Z)$$ by $$i = n + \sum_{k=0}^n |a_n|$$
There are then countably many indexes (no more than the natural numbers) and finitely many roots having each index. Since and countable collection of countable sets is itself countable, the job is done.
 
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topsquark

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Clever! Okay, thanks for sharing.

-Dan
 
Dec 2007
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Interesting!

Alternatively, we can also transform all the equations themselves in natural numbers, since each equations is of the form

$ \pm a_nx^n\pm ...\pm a_2x^2 \pm a_1x\pm a_0=0 $

representing a set of constructible numbers, aka algebraic numbers, first lay down the exponents like this

$ \pm a_nxn\pm ... \pm a_2x2\pm a_1x\pm a_0=0 $

second, let $ x, +, -, = $ be new symbols representing each a numeral, them each equation turns into a numeral in the $ base_{14} $, so that

$ 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, x, +, -, = $

are the numerals in $ base_{14} $, now trivially any numerical base is denumerable with $ base_{10} $.

Example:

$ 3x^4-7x^2+1=0 $ is $3x4-7x2+1=0$ in $ base_{14} $ and so 1081600743426 in $ base_{10} $.

The algebraic numbers are countable. You can create an index $i$ for the roots of a polynomial $$p(x) = \sum_{k=0}^n a_kx^k \qquad (a_k \in \mathrm Z)$$ by $$i = n + \sum_{k=0}^n |a_n|$$
There are then countably many indexes (no more than the natural numbers) and finitely many roots having each index. Since and countable collection of countable sets is itself countable, the job is done.
 
Dec 2007
687
47
Just to correct my previous post, not all algebraic numbers are constructible by unruled straightedge and compass, e.g. \(\displaystyle \sqrt[3]{2}\) is a root of \(\displaystyle x^3-2=0\), is a counterexample, although all constructible numbers are algebraic.

Constructible Number -- from Wolfram MathWorld
 

CRGreathouse

Forum Staff
Nov 2006
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Just to correct my previous post, not all algebraic numbers are constructible by unruled straightedge and compass, e.g. \(\displaystyle \sqrt[3]{2}\) is a root of \(\displaystyle x^3-2=0\), is a counterexample, although all constructible numbers are algebraic.

Constructible Number -- from Wolfram MathWorld
I have a diagram here:
https://oeis.org/wiki/File:Number_classes.png
which is relevant. In particular you can follow the implication arrows from "constructible" to "algebraic".
 
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Country Boy

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According to Closed-form expression - Wikipedia, the free encyclopedia
A narrower definition proposed in (Chow 1999, pp. 441–442), denoted E, and referred to as EL numbers, is the smallest subfield of C closed under exponentiation and logarithm—this need not be algebraically closed, and correspond to explicit algebraic, exponential, and logarithmic operations. "EL" stands both for "Exponential-Logarithmic" and as an abbreviation for "elementary"
 
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Dec 2007
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According to Closed-form expression - Wikipedia, the free encyclopedia
A narrower definition proposed in (Chow 1999, pp. 441–442), denoted E, and referred to as EL numbers, is the smallest subfield of C closed under exponentiation and logarithm—this need not be algebraically closed, and correspond to explicit algebraic, exponential, and logarithmic operations. "EL" stands both for "Exponential-Logarithmic" and as an abbreviation for "elementary"
Isn't it one form of the definition of algebraically closed fields that they do not admit non trivial field extensions? If I'm correct, then E must be non alg. closed. In particular, no subfield of \(\displaystyle \mathbb{C}\) can be alg. closed.