Are there different types of transcendental numbers?

233 Views Asked by At

For example, algebraic integers are a special case of an algebraic number. Does something similar happens with transcendental numbers? Is possible to a trans. num. be different the other transcendental?

3

There are 3 best solutions below

0
On

If $x\in\mathbb{R}$, we define its irrationality measure : $$ \mu(x):=\sup\left\{ \mu\in\mathbb{R},\text{ there exists an infinite number of } (p,q)\in\mathbb{Z}\times\mathbb{N}^*,0<\left|x-\frac{p}{q}\right|<\frac{1}{q^{\mu}}\right\} $$ First notice that $\mu(x)\geqslant 1$ for all $x\in\mathbb{R}$ because the above set contains $]-\infty,1[$. For instance, $\mu(x)=1$ if $x\in\mathbb{Q}$ and $\mu(x)\geqslant 2$ if $x\notin\mathbb{Q}$. Roth theorem states that $\mu(x)=2$ for all algebraic $x\in\mathbb{R}\setminus\mathbb{Q}$. Among the transcendental numbers, there are the Liouville numbers, they are the one such that their irrationnality measure is $+\infty$.

https://en.wikipedia.org/wiki/Liouville_number

0
On

In the following sense, all transcendental numbers "look the same" over $\Bbb{A}$, the field of algebraic numbers: for any two transcendental numbers $\alpha, \beta \in \Bbb{C}$, the field extensions $\Bbb{A}(\alpha)$ and $\Bbb{A}(\beta)$ are both isomorphic to $\Bbb{A}(x)$, the field of rational functions with algebraic coefficients.

The same holds true over the rationals: for any two transcendental numbers $\alpha, \beta \in \Bbb{C}$, the field extensions $\Bbb{Q}(\alpha)$ and $\Bbb{Q}(\beta)$ are both isomorphic to $\Bbb{Q}(x)$, the field of rational functions with rational coefficients.

So we can't "tell them apart" from the perspective of $\Bbb{Q}$ or its algebraic closure $\overline{\Bbb{Q}} = \Bbb{A}$. Any two transcendental numbers are "equally transcendental", that is, give rise to field extensions of $\Bbb{A}$ or $\Bbb{Q}$ with the same structure.

0
On

I think the most natural classification is:

  • Computable transcendental numbers (e.g. $\pi$)
  • Definable, but non-computable, real numbers (e.g. a Chaitin's constant)
  • Non-definable real numbers (no example can be given, because to give a number as an example, I would have to define it)

Notably, there are only countably many definable numbers. So the vast, vast majority of real numbers are non-definable.

So to round up, you can think of the following hierarchy of real numbers, each set including the last but also additional, more exotic and difficult numbers:

  • Natural numbers
  • Integers
  • Rational numbers
  • Constructible numbers
  • Algebraic numbers
  • Computable numbers
  • Definable numbers
  • Real numbers

Starting with the constructible numbers, you can think in terms of complex numbers instead of real numbers if you prefer.

Also, there are other interesting sets of numbers, and the hierarchy is not completely linear. For example, the algebraic integers you mentioned, would be on a side branch between integers and algebraic numbers, bypassing the sets of rational and constructible numbers.