Timeline for Particular problem solved by solving a more general problem.
Current License: CC BY-SA 2.5
8 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Apr 13, 2010 at 21:57 | history | edited | Cam McLeman | CC BY-SA 2.5 |
deleted 6 characters in body
|
| Apr 13, 2010 at 21:57 | comment | added | Cam McLeman | Ah, good point. Interesting. | |
| Apr 13, 2010 at 21:04 | comment | added | Qiaochu Yuan | You can delete the word "known": the two are equivalent, since if there's at least one prime in every arithmetic progression then given an arithmetic progression a mod n, there's a prime congruent to a+n mod n^2, a prime congruent to a+n^2 mod n^3, etc. | |
| Apr 13, 2010 at 17:01 | comment | added | villemoes | A good example of such a puzzle is this: Let x = sqrt(2) and y = 2+sqrt(2). Let X be the set { floor(nx) | n a positive integer }, and define Y similarly. Prove that X and Y are disjoint, and that their union is the set of positive integers. The not-so-obvious key is that x and y are (positive) irrational numbers satisfying 1/x + 1/y = 1; and in fact the statement holds (and is easier to prove) for all such pairs. | |
| Apr 13, 2010 at 16:39 | history | edited | Cam McLeman | CC BY-SA 2.5 |
deleted 94 characters in body
|
| Apr 13, 2010 at 16:38 | history | made wiki | Post Made Community Wiki by Ben Webster♦ | ||
| Apr 13, 2010 at 16:09 | comment | added | Sunni | When we come across a (rearch) problem, sometimes we may not see its full picture. And in each step for a good theorem, we may not need their full generality, but a special case is enough. | |
| Apr 13, 2010 at 15:49 | history | answered | Cam McLeman | CC BY-SA 2.5 |