Timeline for Dedekind-Peano axioms, but numbers have at most one successor
Current License: CC BY-SA 4.0
16 events
| when toggle format | what | by | license | comment | |
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| Jan 22, 2023 at 23:54 | comment | added | LSpice | @abo, exactly as you point out, there cannot be a statement that is true in all initial segments that implies infinitude of primes. My condition, for example, like the Bertrand's-postulate analogue proposed by the asker, is true in any initial segment of the natural numbers, and implies infinitude of the primes when the successor (and hence the product) operation is total. As to my being more knowledgeable—if at all, then not in this domain! | |
| S Jan 22, 2023 at 13:13 | vote | accept | James Propp | ||
| Jan 22, 2023 at 13:11 | comment | added | abo | I'm not making any claim of what someone knows or doesn't know, just a plea to be careful in a successorless world. For instance, LSpice (who is much more knowledgeable than me) doesn't seem to state a good infinitude condition, because the product of S = {top-1,top} is not defined and does not have a successor, but S is clearly finite. Also, Kevin's statement of Bertrand's Postulate is not the best, because whether 2k has a successor is neither here nor there, and the better condition is, "If 2k exists..." | |
| Jan 22, 2023 at 12:43 | comment | added | James Propp | Kevin gets where I’m coming from. One such P would be Bertrand’s Postulate (“If 2k has a successor, then there’s a prime between k and 2k”). It’s true in all the finite models as well as the infinite one, and if one singles out the infinite model by adding the assumption that every number has a successor, then one can easily deduce that the set of primes is infinite. So in that sense Bertrand’s Postulate is exactly the sort of thing I meant by a “size-neutral analogue” of the infinitude of the primes. (Is my stance incoherent in some way that I’m not noticing?) | |
| Jan 22, 2023 at 11:38 | answer | added | Ali Enayat | timeline score: 7 | |
| Jan 22, 2023 at 9:56 | comment | added | Kevin | @abo: I think OP is aware of that, and is asking for a lemma that is provably true of {0, 1, 2, 3, 4, 5}, but which could also usefully contribute to a proof of the infinitude of the primes in standard arithmetic. LSpice's comment is probably what OP wants, I think? | |
| S Jan 21, 2023 at 13:19 | vote | accept | James Propp | ||
| S Jan 22, 2023 at 13:13 | |||||
| Jan 21, 2023 at 11:52 | comment | added | abo | In a natural interpretation of "infinite," it is not possible to show there are an infinite number of primes in such a system, because it's not even true. For instance, if {0,1,2,3,4,5} are (all of) the natural numbers, then there are 3 prime numbers, and so the number of primes is a finite number (3). | |
| S Jan 21, 2023 at 10:52 | history | suggested | abo |
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| Jan 21, 2023 at 7:21 | review | Suggested edits | |||
| S Jan 21, 2023 at 10:52 | |||||
| Jan 21, 2023 at 2:32 | history | became hot network question | |||
| Jan 20, 2023 at 23:00 | answer | added | abo | timeline score: 10 | |
| Jan 20, 2023 at 22:37 | vote | accept | James Propp | ||
| S Jan 21, 2023 at 13:19 | |||||
| Jan 20, 2023 at 19:55 | answer | added | Joel David Hamkins | timeline score: 14 | |
| Jan 20, 2023 at 18:35 | comment | added | LSpice | Although multiplication isn't total, the notion of factorisation makes sense, right?, since one has a lower bound on size as witnessed by the number one is attempting to factor—and so primality should still make sense. One could encode an easy proof of infinitude of primes in the statement: if $S$ is the set of primes, then either the product of $S$ is not defined, or the product of $S$ does not have a successor. | |
| Jan 20, 2023 at 18:26 | history | asked | James Propp | CC BY-SA 4.0 |