# Cardinality of transcendental numbers

#### topsquark

Math Team
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

#### Petek

Hint: The set of all real numbers is the union of the set of real algebraic numbers and the set of real transcendental numbers.

1 person

#### v8archie

Math Team
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.

2 people

#### topsquark

Math Team
Clever! Okay, thanks for sharing.

-Dan

#### al-mahed

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.

#### al-mahed

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
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".

1 person

#### Maschke

I have a diagram here
What are the "EL" numbers in that pic? I wasn't able to Google them.

#### Country Boy

Math Team
According to Closed-form expression - Wikipedia, the free encyclopedia
A narrower definition proposed in (Chow 1999, pp. 441â€“442), denoted E, and referred to as EL numbers, is the smallest subfield of C closed under exponentiation and logarithmâ€”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"

1 person

#### al-mahed

According to Closed-form expression - Wikipedia, the free encyclopedia
A narrower definition proposed in (Chow 1999, pp. 441â€“442), denoted E, and referred to as EL numbers, is the smallest subfield of C closed under exponentiation and logarithmâ€”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.