Timeline for Arriving at the same result with the opposite hypotheses
Current License: CC BY-SA 4.0
34 events
| when toggle format | what | by | license | comment | |
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| Oct 27, 2019 at 13:48 | comment | added | YCor | Just in case this elementary example fits the question: Proposition: Every group $G$ with at least 3 elements (and possibly infinite) has a nontrivial automorphism. Proof (a) $G$ is abelian with $G^2\neq 1$: take $x\mapsto x^{-1}$ (b) $G$ is 2-elementary abelian, thus a vector space of dimension $\ge 2$ over $Z/2$: use a nontrivial permutation of a basis to induce a nontrivial automorphism (c) $G$ is not abelian: use a nontrivial inner automorphism. Here both the statement "is abelian" and its negation are useful (to ensure that inversion is homomorphism, vs produce inner automorphism) | |
| Jun 6, 2019 at 16:55 | comment | added | Timothy Chow | @ManfredWeis : If RH had been proved decidable then it would have been reduced to a finite computation and we would all have heard about it. See mathoverflow.net/q/62144/3106 and mathoverflow.net/q/6250/3106 for some further discussion. | |
| Jun 6, 2019 at 7:56 | comment | added | Dan Romik | @ManfredWeis isn’t this essentially the case for the continuum hypothesis example in D.S.Lipham’s answer? I don’t see why a statement being undecidable should make the argument less valid. Also note that if RH is undecidable then it is also true (which means it will never be proved to be undecidable. It may be proved some day to be decidable but that hasn’t happened yet). | |
| Jun 6, 2019 at 5:30 | review | Close votes | |||
| Jun 7, 2019 at 12:43 | |||||
| Jun 6, 2019 at 4:58 | comment | added | Manfred Weis | Sorry for the naive question, but has the RH been proved decidable? According to Gödel there is more than just true or false. And I wonder what would happen if a proof were based on the assumed truth value of an actually undecidable conjecture. | |
| Jun 5, 2019 at 17:28 | answer | added | Daron | timeline score: 28 | |
| Jun 5, 2019 at 15:50 | comment | added | Michael | @NateEldredge, got it, thanks. | |
| Jun 5, 2019 at 1:29 | comment | added | user400188 | It seems that me that the italic text in this question just implies that the theory they refer to is independent of generalized RH. | |
| Jun 4, 2019 at 18:19 | comment | added | Nate Eldredge | @Michael: I think the difference is that, in your example, we know that the hypothesis $p = 4k+1$ is true for some $p$ and false for others, and so we really need both cases to prove the theorem for all $p$. In OP's example, the hypothesis is either universally true or universally false (there's no free variable like $p$), and hence one of the cases is completely unnecessary to treat, but we don't know which one. | |
| Jun 4, 2019 at 16:06 | comment | added | Michael | How is this situation essentially different from proving a result by considering special cases? Suppose you want to prove a result (say, quadratic reciprocity), for an arbitrary odd $p$, and you consider separately the case $p=4k+1$ and $p=4k+3$, the latter being the negation of former in the context of the problem. Both the assertion $p=4k+1$ and its negation $p=4k+3$ lead to the same conclusion, and both are essential. | |
| Jun 4, 2019 at 12:27 | vote | accept | CommunityBot | ||
| Jun 4, 2019 at 12:25 | comment | added | user141414 | @NateEldredge indeed, the answer appears to be relevant. The question is not, in my opinion. I would upvote the answer if it is reposted here. | |
| Jun 4, 2019 at 4:09 | history | edited | Martin Sleziak | CC BY-SA 4.0 |
since the question refers to Wikipedia article on RH, it seem reasonable to include also the link
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| Jun 4, 2019 at 4:04 | answer | added | D.S. Lipham | timeline score: 66 | |
| Jun 4, 2019 at 4:02 | comment | added | Nate Eldredge | @SamHopkins: In particular, your own answer to that question is a simple example of what OP wants, I think. | |
| Jun 3, 2019 at 22:37 | comment | added | Christian Remling | @Cutthewood: Yes, I know, I didn't mean to object to what you wrote, I just wanted to point out that the quote doesn't really let the reader share in the amazement. | |
| Jun 3, 2019 at 22:31 | comment | added | Pietro Majer | @Cutthewood My excuses, of course I like these arguments too, and I didn't mean to diminish their content. Just saying the form " A implies X and not A implies X as well" is not so strange | |
| Jun 3, 2019 at 22:25 | comment | added | user141414 | @PietroMajer I actually quite like such arguments, maybe I should try to join your field if such arguments are common. Anyway, all objections should be sent to Ireland & Rosen, not to me (I did not add any exclamation marks). | |
| Jun 3, 2019 at 22:24 | comment | added | Pietro Majer | Is this so amazing to deserve an exclamation mark? It is quite common to prove a result X by cases, and the fact that one case may be empty is not so relevant in order to prove X, after all ---(Sorry I didn't note that this comment had already been made) | |
| Jun 3, 2019 at 22:18 | comment | added | user141414 | @ChristianRemling I should note that the "truly amazing" part is due to the book of Ireland & Rosen, not due to me. Possibly you are better equipped than them to judge what is or is not truly amazing, I do not know. | |
| Jun 3, 2019 at 21:51 | comment | added | Christian Remling | One would have to see the actual proof I guess, but in this abstract description, nothing seems even remotely "truly amazing" to me. Rather, it sounds like doing an argument by distinguishing various (in this case, two) cases. | |
| Jun 3, 2019 at 21:39 | history | became hot network question | |||
| Jun 3, 2019 at 19:54 | answer | added | Izaak Meckler | timeline score: 34 | |
| Jun 3, 2019 at 17:43 | comment | added | user141414 | @ToddTrimble does this comment contribute anything to the discussion? | |
| Jun 3, 2019 at 17:37 | comment | added | Todd Trimble | meta.mathoverflow.net/questions/4200/flood-of-similar-new-users | |
| Jun 3, 2019 at 15:50 | answer | added | user36212 | timeline score: 30 | |
| Jun 3, 2019 at 15:38 | comment | added | user141414 | @DorianoBrogioli well, it does have something to do in the sense that the easiest (or the only known?) proof uses "decomposition along RH". If you have many interesting examples, post many interesting answers! Though I should add that there indeed was an implicit point that the proof using "decomposition along RH" should be somehow simpler than other proofs, or be the only known proof (which in my understanding is what is happening in the example mentioned in Ireland-Rosen). | |
| Jun 3, 2019 at 15:36 | comment | added | Doriano Brogioli | Maybe I misunderstood something. For what I understand, there are many theorems that can be proven both in case RH is true and false. For example, we can prove that the series of $1/n^2$ converges. Any theorem not related to RH can be proven. I would say that any theorem which is a consequence of RH and its negation simply does not have to do with RH, although, maybe, we are not able to see it as unrelated. | |
| Jun 3, 2019 at 15:24 | answer | added | Joel David Hamkins | timeline score: 13 | |
| Jun 3, 2019 at 14:11 | history | made wiki | Post Made Community Wiki by Todd Trimble | ||
| Jun 3, 2019 at 13:35 | review | First posts | |||
| Jun 3, 2019 at 13:40 | |||||
| Jun 3, 2019 at 13:34 | comment | added | user141414 | @SamHopkins I personally think that this question is not related to the one you have linked to and that this question is not nearly vague enough to be closed as "unclear". Opinions differ I guess. | |
| Jun 3, 2019 at 13:32 | comment | added | Sam Hopkins | Related (and closed) question: mathoverflow.net/questions/312439/… | |
| Jun 3, 2019 at 13:30 | history | asked | user141414 | CC BY-SA 4.0 |