# Independent Transcendental Numbers

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?)

• 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 $L_{\omega+\omega}$. May 11 '16 at 10:10
• 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. :-) May 11 '16 at 11:29
• @MeesdeVries haha, while very true I wouldn't consider that to be an actual number (otherwise there exist infinitely many examples)
– KoA
May 11 '16 at 11:35
• @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? May 11 '16 at 11:44
• @MeesdeVries I've updated the question to perhaps be more specific.
– 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.