# Euler's constant: irrationality and proof theory

Let $\gamma$ represent Euler's constant. Is there a real number $x$ such that there is a proof within Zermelo-Fraenkel set theory (ZF) that $x$ is irrational and there is also a proof within ZF that $\gamma + x$ is irrational?

• Assuming ZFC proves $\gamma$ is irrational, take $x$ to be $\gamma$. – Carl Mummert Nov 29 '15 at 12:57
• Really, asking about proofs in ZF is not likely to be interesting. There are many open questions about whether particular numbers are irrational - e.g., it is open AFIAK whether $\exp(\exp(\exp(10)))$ is irrational. But, when we can prove that a number is irrational, the proof will almost certainly work in ZF, just like almost every other proof in mathematics. So I think that asking about ZF is likely to be a red herring for this kind of question, unless there is some very specific reason to worry about whether the ordinary proof can be formalized in ZF. – Carl Mummert Nov 29 '15 at 13:08
• @ Carl Mummert I believe now that the appeal to ZF in the OP was due to a tentative to formalize the requirement (that I'm not able to formalize either): there is $x$ such that there is a proof that $x$ and $x+\gamma$ are both irrational, in contrast with there is a proof that there is $x$ such that $x$ and $x+\gamma$ are both irrational, which allow trivial answers (either $x=\sqrt2$ or $x=2\sqrt2$ have the above property). – Pietro Majer Jan 4 '16 at 23:56

## 3 Answers

Yes, let $x$ be Chaitin's constant. Then both $x$ and $\gamma + x$ are uncomputable, therefore irrational.

• I was thinking of this example, but I didn't post this, because I actually wasn't sure if $\gamma$ is uncomputable. For what I know, $\gamma$ might have some incredibly close dyadic fraction approximations, and they might be so close that to determine whether part of expansion is $01111...$ or $10000...$ might require uncomputable degree of accuracy. Although by no means I expect this to be the case, I don't see how such possibility could be excluded without some bounds on the approximation of $\gamma$ by rationals, and afaik none are known. – Wojowu Nov 29 '15 at 18:29
• @Wojowu: $\gamma$ is in fact computable (and this is what we need here); this means that a Turing machine can compute rational approximations to $\gamma$ with controlled errors, which is obviously the case here. – Christian Remling Nov 29 '15 at 19:34
• @ChristianRemling What do you mean by "controlled errors"? – Wojowu Nov 29 '15 at 19:35
• @Wojowu: $|a_n-\gamma|<2^{-n}$ – Christian Remling Nov 29 '15 at 19:37
• – Christian Remling Nov 29 '15 at 19:41

Yes, but this is nothing to do with $\gamma$. Let $a$ be any real number. Then there is $x$ so that $x$ and $a+x$ are both irrational. Proof (within ZF): the set of $x$ such that $x$ is rational is countable, the set of $x$ such that $a+x$ is rational is also countable. But $\mathbb R$ is uncountable.

• But this $x$ found by cardinality argument is not specified, what is probably required by OP. – Fedor Petrov Nov 29 '15 at 13:24
• The usual diagonal argument is constructive. (e.g. use the number whose $n$-th digit is 7 if the $n$-th rational's $n$-th digit is 3, and 3 otherwise.) – Noam D. Elkies Nov 29 '15 at 15:32
• @Noam D. Elkies: it think here the problem is that $x$ is not constructive, not the diagonal argument. For instance, it's also true that there is $x\in \{ \sqrt2,\ 2\sqrt2\}$ such that $x$ and $a+x$ are both irrational. – Pietro Majer Nov 30 '15 at 8:42
• (Was it clear what I wrote? For any number $a$ there is $x$ so that $x$ and $a+x$ are both irrational, and we can choose $x$ to be either $\sqrt{2}$ or $2\sqrt{2}$. Indeed $a+\sqrt{2}$ and $a+2\sqrt{2}$ can't be both rational, while $x$ is in any case irrational). – Pietro Majer Jan 4 '16 at 23:40

I got the feeling that maybe what the OP wanted was: Provide an explicit constructive example of an irrational $x$ such that $x+\gamma$ is irrational. In any case, under this respect the question seems interesting as well, so I would like to consider it. While the existence is quite clear (e.g. there exists $x\in\{\sqrt{2},2\sqrt{2} \}$ with this property, and also $x\in\{\sqrt{2},\gamma \}$), giving an explicit example is a bit less obvious.

As suggested in the various answers, we may consider the problem more in general than for the case where $\gamma$ is the Euler-Mascheroni constant. I'd like to formalize the question in the following Problem:

Split a given computable real number $\gamma$ into a sum of two irrational computable real numbers.

There is a cute construction that uses a "balanced factorial representation" of real numbers. Recall that any real number $\gamma$ can be written in the form $$\gamma=\sum_{k\ge 1}\frac{\gamma_k}{k!}$$ with coefficients $\gamma_k\in\mathbb{Z}$ verifying $$|\gamma_k |\le k-1 \ ,$$ for any $k\ge 2$. A number of this form is rational if and only if the sequence $\gamma_k$ is either eventually equal to $0$, or eventually equal to $k-1$, or eventually equal to $-(k-1)$. Moreover, $\gamma$ is a computable number if and only if the coefficients $\gamma_k$ are given by some computable function. (Note: as remarked in comments, this is generally not true for base systems with nonnegative coefficients: the binary expansion of a computable number may not be computable, which reflects the fact that one may not be able to decide whether $\gamma$ is larger or smaller than a given rational approximation of its, no matter how good).

Given the two facts above, the problem is then translated into: split computably the sequence $\gamma_k$ as a sum $\gamma_k=\alpha_k+\beta_k$, in such a way that $$0<|\alpha_k|<k-1$$ and $$0<|\beta_k|<k-1$$
for infinitely many $k$. Again, changing sign coefficients make everything quite easy to deal with: we can e.g. eventually take both $|\alpha_k|$ and $|\beta_k|$ in the interval $[k/4,3k/4]$.