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In a paper by Scott Aaronson, I read that Busy Beaver of 7918 can't be computed from ZFC. But $BB(7918)$ is a specific finite natural number, call it $k$. So, using $k$ applications of the successor function starting from $0$, we can define it. Granted, there is not enough space in the observable universe nor enough time until the heat death of the universe to define it, but it doesn't matter, it can be defined in principle. So, am I misunderstanding something? Please correct my misunderstanding if I am wrong.

user107952
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  • Who is Scott Aaraonson (is the spelling correct?)? What is the paper? – markvs Oct 07 '20 at 01:58
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    @JCAA https://www.scottaaronson.com/blog/?p=4916 the article appears to be https://www.scottaaronson.com/papers/bb.pdf The first name in the Acknowledgements is Bill Gasarch, I was an undergraduate with a person of that name, i think it is the same person – Will Jagy Oct 07 '20 at 02:05
  • So the spelling is wrong as I suspected. – markvs Oct 07 '20 at 02:05
  • @QiaochuYuan that should be an answer. – Noah Schweber Oct 07 '20 at 02:09
  • @WillJagy: The last name should be Aharonson. But certainly not Aaraonson as in OP. There are streets in Israel named after people with this last name. The blog post contains references to Covid-19 and Trump among other things. The first person who came up with the idea about unreachable numbers was A. Zinoviev, a Russian philosopher, logician and political activist. The Soviet government put him in a psychiatric prison but not for his logic. – markvs Oct 07 '20 at 02:33
  • @JCAA perhaps it is https://en.wikipedia.org/wiki/Alexander_Zinoviev – Will Jagy Oct 07 '20 at 02:41
  • Yes. that is him. – markvs Oct 07 '20 at 02:42
  • @JCAA I can't find a reference for Zinoviev discussing unreachable numbers; do you have one? (Maybe it's Esenin-Volpin instead?) – Noah Schweber Oct 07 '20 at 03:58
  • Could be Esenin-Volpin (I often mix the two). – markvs Oct 07 '20 at 03:59
  • @NoahSchweber: You are right: https://en.wikipedia.org/wiki/Alexander_Esenin-Volpin#Mathematical_work – markvs Oct 07 '20 at 04:00

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Converting my comment into an answer and elaborating: "computed" is sort of a misleading way to say it. The theorem is that ZFC doesn't decide the value of this busy beaver number; that is, there is no $n$ such that ZFC proves $BB(7918) = n$.

The problem is not even that $n$ is very big, just that it encodes a question ZFC can't answer. Here is a simpler example: consider the number which is either equal to $0$ if ZFC is inconsistent or $1$ if ZFC is consistent. Then the incompleteness theorem says exactly that ZFC doesn't decide the value of this number, despite the fact that by definition it is either $0$ or $1$!

"Number" is also sort of a misleading way to say it; you'll get very confused thinking about this if you don't distinguish carefully between a number and a description of a number. What I gave above is a description of a number, and it's a description that evaluates to a different number in different models of ZFC depending on whether they do or don't believe that ZFC is consistent.

$BB(7918)$, similarly, is a description of a number, and to say that ZFC doesn't decide its value is exactly to say that it will evaluate to different numbers in different models of ZFC (by the completeness theorem).

(I guess there might be an additional subtlety here about what it even means to compare the values of different numbers in different models of ZFC. But I think in this case we're okay.)

Qiaochu Yuan
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    "it will evaluate to different numbers in different models of ZFC" Not sure this is the right way to put it. I don't think it would evaluate to any number under any model of ZFC. It's more that for TMs of 7918 states, ZFC will most likely be able to prove some fraction of them halt, it will be able to prove some number of them will never halt, but then there will be some set of those machines for which ZFC cannot produce a proof either that they halt or run forever. It evaluating to different numbers would imply there is a proof in ZFC of those different numbers which would be a contradiction. – Shufflepants Dec 03 '21 at 15:27
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    @Shufflepants No, the answer is correct. Each model $\mathcal{M}$ of $\mathsf{ZFC}$ has what it thinks is the Busy Beaver function, call it "$BB^\mathcal{M}$," and can make sense of $BB^\mathcal{M}(x)$ for any natural number $x$. The fact that $\mathsf{ZFC}$ doesn't decide $BB(7918)$ amounts (via Godel's completeness theorem, and that's not a typo) exactly to the fact that different models of $\mathsf{ZFC}$ disagree about $BB(7918)$. – Noah Schweber Dec 31 '21 at 22:19
  • Just out of curiosity: is it that every uncomputable function has some input such that the value of which is independent of $\mathsf{ZFC}$? – Jianing Song Nov 11 '23 at 10:38
  • @Jianing: yes. It's easier to think about the contrapositive: if ZFC (or any other computably axiomatizable theory) decides every value of a function $f$ (from $\mathbb{N}$ to $\mathbb{N}$, to be concrete), then it's computable. This is because you can compute $f(n)$ by searching through proofs in ZFC that $f(n) = m$, and by hypothesis this search always terminates. – Qiaochu Yuan Nov 12 '23 at 20:30
  • @QiaochuYuan Thanks! – Jianing Song Nov 15 '23 at 18:13