Timeline for An easy-to-state elusive combinatorial problem (revisited)
Current License: CC BY-SA 3.0
5 events
| when toggle format | what | by | license | comment | |
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| Apr 13, 2017 at 12:57 | history | edited | CommunityBot |
replaced http://mathoverflow.net/ with https://mathoverflow.net/
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| Jun 30, 2013 at 3:50 | comment | added | The Masked Avenger | Yes, but if the first 2 don't work for x and the last 2 don't work for y, then 3 is not big enough. | |
| Jun 30, 2013 at 2:59 | comment | added | Maaz-ul-Haq | If $ry$ increases by $3(y/z)\ge 3$ wouldn't that imply an increment in $rx$ by $3(x/y)\ge 3$ as well? | |
| Jun 30, 2013 at 2:51 | comment | added | The Masked Avenger | I'm not convinced. For the values allowed for r, less than half might work for x, and a separate less than half might work for y. You need to argue in the main case that that does not happen (there is at least one r that works for both x and y), and I do not see that argument yet. | |
| Jun 30, 2013 at 2:30 | history | asked | Maaz-ul-Haq | CC BY-SA 3.0 |