It's known that imaginary numbers can be expressed as vectors such as $\alpha + \beta i$. My question is: Can you represent a computable real number with a similar algebraic definition?

For example does there exist some constants such that:

$$e =\alpha + \beta \sqrt{2}$$

Where $\alpha \in\Bbb{Z}$ and $\beta \in\Bbb{Q}$ and $e$ is Euler's number.

Computable reals are things that can be approached arbitrarily closely by an infinite sequence of rational numbers so is it possible to express any computable real in this way?

  • $\begingroup$ Presumably you want to fix the irrational number say $a$, and then write $c=n(c)+q(c) a$ where $n(c) \in \mathbb{Z}$ and $q(c) \in \mathbb{Q}$. But this is not possible, because there are computable irrational numbers within $1$ of each other which are not rational multiples of one another. Probably the easiest way to show this is to pick $c$ transcendental (if $a$ is algebraic) or algebraic (if $c$ is transcendental). $\endgroup$ – Ian Aug 30 '18 at 13:34
  • $\begingroup$ If you mean Euler's constant : This number is transcendental, so cannot be represented as desired. In the case of the Euler-Mascheroni-constant (not likely the intent of the question) : it is open whether it is even irrational. $\endgroup$ – Peter Aug 30 '18 at 13:34
  • 1
    $\begingroup$ The title and the text statements are different. In the title question you can use the integer "0" and the rational as "1". And your definition of computable seems fishy because every real can be approximated be an infinite sequence of rationals. $\endgroup$ – gammatester Aug 30 '18 at 13:39
  • $\begingroup$ @gammatester: Not all such infinite sequences of rationals are computable, though. $\endgroup$ – Henning Makholm Aug 30 '18 at 13:47
  • $\begingroup$ I read on wikipedia that e, euler's constant, was computable which is why I chose it. I know that the square root of 2 is computable by definition. $\endgroup$ – jake mckenzie Aug 30 '18 at 13:51

You can't have a single computable number that you can use to decompose all computable reals in that way.

You can't even have any finite number of "basis elements", even if you allow arbitrary rational functions of the basis elements rather than just the integer/rational combinations you describe.

This is because there exists a sequence of infinitely many algebraically independent computable numbers -- such as $$e^e, e^{e^2}, e^{e^3}, e^{e^4}, \ldots$$ thanks to the Lindemann-Weierstrass theorem. So the field of computable reals has transcendence degree $\aleph_0$ over $\mathbb Q$.

  • $\begingroup$ Does this extend to all transcendental numbers or are their some transcendentals for which there is some partial notion of decomposability, even if it is limited? $\endgroup$ – jake mckenzie Sep 1 '18 at 2:10

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.