srn347 comments on Second-Order Logic: The Controversy - LessWrong

24 Post author: Eliezer_Yudkowsky 04 January 2013 07:51PM

You are viewing a comment permalink. View the original post to see all comments and the full post content.

Comments (188)

You are viewing a single comment's thread. Show more comments above.

Comment author: [deleted] 04 January 2013 11:07:14PM 0 points [-]

Alternatively, one could start by asking whether 2^50 is a real number or not, if yes go up to 2^75, if no go to 2^25, and in up to 7 steps find a real number that, when doubled, ceases to be a real number. There may be impractical or even noncomputable numbers, but continuity holds that doubling a real number always yields a real number.

Comment author: Qiaochu_Yuan 04 January 2013 11:09:38PM 5 points [-]

I think the point of the fable is that Yesenin-Volpin was counting to each number in his head before declaring whether it was 'real' or not, so if you asked him whether 2^50 was 'real' he'd just be quiet for a really really long time.

Comment author: [deleted] 04 January 2013 11:13:30PM *  3 points [-]

But wouldn't that disprove ultrafinitism? All finite numbers, even 3^^^3, can be counted to (in the absence of any time limit, such as a lifespan), there's just no human who really wants to.

Comment author: Qiaochu_Yuan 04 January 2013 11:18:51PM 11 points [-]

All finite numbers, even 3^^^3, can be counted to

As I understand it, this is precisely the kind of statement that ultrafinitists do not believe.

Comment author: MrMind 08 January 2013 01:45:30PM 0 points [-]

If that's true, then Yesenin-Volpin was just playin the role of a Turing machine trying to determine if a certain program halts (the program that counts from 0 to the input). If 3^^^3 (say) for a finitist doesn't exists, then s/he really has a different concept of number than you have (for example, they are not axiomatized by the Peano Arithmetic). It's fun to observe that ultrafinitism is axiomatic: if it's a coherent point of view, it cannot prove that a certain number doesn't exists, only assume it. I also suspect (but finite model theory is not my field at all) that they have an 'inner' model that mimics standard natural numbers...

Comment author: Larks 04 January 2013 11:19:59PM 5 points [-]

Well, that's what the anti-ultrafinitists say. It is precisely the contention of the ultrafinitists that you couldn't "count to 3^^^3", whatever that might mean.

Comment author: SecondWind 05 January 2013 02:55:47AM 0 points [-]

Hmm.

So, it's not sufficient to define a set of steps that determine a number... it must be possible to execute them? That's a rather pragmatic approach. Albeit it one you'd have to keep updating if our power to compute and comprehend lengthier series of steps grows faster than you predict.

Comment author: Larks 05 January 2013 02:01:54PM 0 points [-]

No, ultrafinitism is not a doctrine about our practical counting capacities. Ultrafinitism holds that you may not have actually denoted a number by '3^^^3', because there is no such number.

Comment author: Peterdjones 05 January 2013 02:13:34PM 3 points [-]

Utlrafrinitists tend no to specfify the highest number, to prevent people adding one to it.

Comment author: Larks 05 January 2013 09:31:31PM 2 points [-]

Hence "may not"

Comment author: wuncidunci 05 January 2013 02:57:53AM *  0 points [-]

I would have done the following if I had been asked that: calculate which numbers I would have time to count up to before I was thrown out/got bored/died/earth ended/universe ran out of negentropy. I would probably have to answer I don't know, or I think X is a number for some of them, but it's still an answer, and until recently people could not say wether "the smallest n>2 such that there are integers a,b,c satisfying a^n + b^n = c^n" was a number or not.

I'm not advocating any kind of finitism, but I agree that the position should be taken seriously.