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?
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4$\begingroup$ Assuming ZFC proves $\gamma$ is irrational, take $x$ to be $\gamma$. $\endgroup$– Carl MummertCommented Nov 29, 2015 at 12:57
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11$\begingroup$ 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. $\endgroup$– Carl MummertCommented Nov 29, 2015 at 13:08
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1$\begingroup$ @ 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). $\endgroup$– Pietro MajerCommented Jan 4, 2016 at 23:56
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3 Answers
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Yes, let $x$ be Chaitin's constant. Then both $x$ and $\gamma + x$ are uncomputable, therefore irrational.
answered Nov 29, 2015 at 15:19
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1$\begingroup$ 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. $\endgroup$– WojowuCommented Nov 29, 2015 at 18:29
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5$\begingroup$ @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. $\endgroup$ Commented Nov 29, 2015 at 19:34
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$\begingroup$ @ChristianRemling What do you mean by "controlled errors"? $\endgroup$– WojowuCommented Nov 29, 2015 at 19:35
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$\begingroup$ @Wojowu: $|a_n-\gamma|<2^{-n}$ $\endgroup$ Commented Nov 29, 2015 at 19:37
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3$\begingroup$ @Wojowu: en.wikipedia.org/wiki/Computable_number#Formal_definition $\endgroup$ Commented Nov 29, 2015 at 19:41
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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.
answered Nov 29, 2015 at 13:19
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2$\begingroup$ But this $x$ found by cardinality argument is not specified, what is probably required by OP. $\endgroup$ Commented Nov 29, 2015 at 13:24
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13$\begingroup$ 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.) $\endgroup$ Commented Nov 29, 2015 at 15:32
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2$\begingroup$ @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. $\endgroup$ Commented Nov 30, 2015 at 8:42
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2$\begingroup$ (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). $\endgroup$ Commented Jan 4, 2016 at 23:40
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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 $[k/4,3k/4]$.
answered Nov 30, 2015 at 7:16