I've been thinking about numbers which have not yet been proven nor disproven to be transcendental, such as $e + \pi,\, \pi - e,\, \frac{\pi}{e},\, \gamma,\,\zeta(3),$ etc. Some of these numbers haven't even been proven to be irrational, so it naturally led to me questioning whether these numbers' transcendence could perhaps be independent of $\sf ZF$ or $\sf ZFC$. Do (or can) there exist numbers such that their transcendence or irrationality is independent of $\sf ZF$ or $\sf ZFC$?

I'm aware that this question may be unsolved, so relevent references would be appreciated as well!

Note that by number I mean one which is not defined conditionally using some other independent statement such as the continuum hypothesis or the axiom of choice. (Side question: what happens if we change this definition to demanding the number can be computed to arbitrary precision?)

  • 3
    $\begingroup$ Nice question. To me, it seems feasible that such numbers may exists. However, at least for your examples, I don't know of any technique that may be used to proof such a claim, since all of these questions are correctly answered within $L_{\omega+\omega}$. $\endgroup$ – Stefan Mesken May 11 '16 at 10:10
  • 3
    $\begingroup$ The number $V$ (de Vries' constant) is defined by $$V = \begin{cases} 1 &\text{if CH,}\\ \sqrt{2}&\text{else.}\end{cases}$$ Its irrationality is independent of ZFC. :-) $\endgroup$ – Mees de Vries May 11 '16 at 11:29
  • $\begingroup$ @MeesdeVries haha, while very true I wouldn't consider that to be an actual number (otherwise there exist infinitely many examples) $\endgroup$ – KoA May 11 '16 at 11:35
  • $\begingroup$ @KoA, of course my example is in jest, but with a serious point: you might want to specify which numbers you mean if you ask whether such numbers "can exist". Do you mean, for instance, a number for which each decimal place is explicitly specified by the definition under ZFC? $\endgroup$ – Mees de Vries May 11 '16 at 11:44
  • $\begingroup$ @MeesdeVries I've updated the question to perhaps be more specific. $\endgroup$ – KoA May 11 '16 at 11:52

You can code $\Pi^0_1$ statements to define a real as follows: Suppose $R(n)$ is a recursive predicate. Define $x_R = \sum \{2^{-n!} : (\forall m < n)R(m)\}$. Then it is not hard to check that $x_R$ is transcendental iff $(\forall n)R(n)$. Notice that using a computer program for $R(n)$, you can estimate $x_R$ within arbitrary precision. Since there are recursive predicates $R(n)$ (e.g., "$n$ does not code a proof of $0=1$ in ZFC") for which $(\forall n)R(n)$ is undecidable in ZFC, you have the sort of examples you are looking for.


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.