For example, using Chaitin's constant as a base. I assume this wouldn't be very useful for representing any form of computable number, as any computable number would most likely become uncomputable and completely random. But can you use an uncomputable but definable base to actually do anything interesting? For example, does this mean that computability is relative to something and not absolute? Because Chaitin's constant would be trivially computable in a base Chaitin's constant number system, whilst 5 would be uncomputable I think, I cannot prove that however.

Edit:Chaitin's constant+1, Base n numbers where n is between -1 and 1 are even weirder and integers seem to become undefinable. And also to be specific(the "constant" isn't just 1 number) Lets just say, for example, in this case, The probability that a completely random,finite, Turing machine will halt on a completely random, finite input, but you could use any definable, uncomputable number for this.

  • $\begingroup$ Note that Chaitin's constant is only defined if we fix the computer we use. Additionally, this constant is usually between $0$ and $1$. How would you even define a base system with such a base ? Note that in this case , $b^n$ DECREASES with $n$ , how would we represent , say , $5$ in a $0.1$-base system ? $\endgroup$
    – Peter
    Jan 29, 2022 at 8:31
  • $\begingroup$ If you think about it definability is relative. You cannot define 1/2 with only addition and subtraction over natural numbers, for example. And you cannot define $\pi$ With only algebraic numbers in a neat formula. This means that 5 would be like $\aleph0$ in this number system, completely inaccessible and not definable from anything that comes before it. I could be completely wrong here however. Lets just say we use Chaitin's constant + 1 as a base, to sidestep that. $\endgroup$ Jan 29, 2022 at 20:56
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    $\begingroup$ Actually, thinking about it more, a base less then 1 makes total sense. Its just that the most significant figures would be the the right of the decimal point. A b^n where n is negative and b is less then 1 grows. $\endgroup$ Feb 1, 2022 at 23:24

1 Answer 1


Chaitin's number is uncomputable, therefore transcendental. From the viewpoint of number systems, all transcendental numbers are pretty much the same.

Assuming $c$ is the base, you're looking at the numbers of the form

$$a_n c^n + \dots + a_1 c + a_0 + a_{-1} c^{-1} + \dots + a_{-k} c^{-k}$$

where all digits $a_i$ are rationals (you could limit that further to integers, or just integers in a fixed range).

Because $c$ is transcendental, two numbers of this form are equal if and only if all of their coefficients are equal. For example, the only way to write $5$ is $a_0 = 5$ and all other $a_i$ being 0. If you limit $a_i$ to a range, say $\{0,1,2,3\}$, then the set will not be closed under addition $2 + 3$.

Assuming you allow any rational number (or any integer number) you can add and multiply numbers of this form, pretty much the same way you operate on polynomials. They are a ring isomorphic to Laurent polynomials.

This set is not closed under division. To be able to divide, you can consider expressions of the form $f(c)/g(c)$ where $f,g$ are polynomials with rational coefficients. The arithmetic on those expressions will follow the same laws as arithmetic on rational functions; the field $\mathbb Q(c)$ is isomorphic to the set of rational functions $\mathbb Q(x)$.

This is the reason why we don't use transcendental numbers as a base - it's really studying polynomials and rational functions.

All of the above applies to any transcendental number, e.g. you could use $c=\pi$ instead of $c=\text{Chaitin's constant}$. Unlike $\pi$, Chaitin's number is noncomputable. It is transcendental over computable numbers, and not just rational numbers. Therefore, for Chaitin's constant you could make $a_i$ computable, and not just rational, and the same reasoning would apply.


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