Timeline for Nonstandard proofs of the fundamental theorem of arithmetic
Current License: CC BY-SA 4.0
13 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jul 29, 2021 at 16:47 | vote | accept | James Propp | ||
| Jul 29, 2021 at 15:34 | answer | added | Steven Gubkin | timeline score: 22 | |
| Jul 29, 2021 at 7:32 | history | edited | Martin Sleziak | CC BY-SA 4.0 |
MathJax: \mid for divisibility
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| Jul 29, 2021 at 7:04 | comment | added | Laurent Moret-Bailly | The claim that a proof of Theorem A "does not use" Theorem B does not mean much if (as is the case here) A and B are easily deduced from each other. In fact, the proof under consideration here is similar to the "direct proof" of Euclid's lemma given in en.wikipedia.org/wiki/Euclid%27s_lemma. | |
| Jul 28, 2021 at 23:01 | review | Close votes | |||
| Jul 30, 2021 at 18:27 | |||||
| Jul 28, 2021 at 19:03 | comment | added | James Propp | Ah, Steven Gubkin has it right! It's the same as Zermelo's proof (thanks Ira), which Pete Clark calls attributes to Lindemann as well. Sorry to have wasted people's time on something that was on Wikipedia; I stopped reading too soon. (Steve, if you want the MathOverflow points, just repost your comment as an answer and I'll upvote and approve it.) | |
| Jul 28, 2021 at 16:22 | comment | added | Ira Gessel | Perhaps it's Zermelo's proof? planetmath.org/inductionproofoffundamentaltheoremofarithmetic. See also mathoverflow.net/questions/339853/…. | |
| Jul 28, 2021 at 16:12 | comment | added | Steven Gubkin | @MatthewvanEerde This is not obtained by Euclid's lemma, but rather the minimality of s. | |
| Jul 28, 2021 at 16:10 | comment | added | Steven Gubkin | You are looking at the wrong proof. Just below that one is another, called "Uniqueness without Euclid's Lemma" | |
| Jul 28, 2021 at 16:09 | comment | added | Matthew van Eerde | There are two proofs on that page; the second one does not use Euclid's lemma, or at least it claims not to. But it seems to me that it does, under the covers, by assuming that $p_1 | (q_1 - p_1) Q$ implies $p_1 | (q_1 - p_1)$ or $p_1 | Q$ | |
| Jul 28, 2021 at 16:06 | comment | added | James Propp | No; that proof says "We see $p_1$ divides $q_1 q_2 \cdots q_k$, so $p_1$ divides some $q_i$ by Euclid's lemma." | |
| Jul 28, 2021 at 15:29 | comment | added | Steven Gubkin | Perhaps it was the proof on Wikipedia? en.wikipedia.org/wiki/… | |
| Jul 28, 2021 at 15:18 | history | asked | James Propp | CC BY-SA 4.0 |