Timeline for How many rearrangements must fail to alter the value of a sum before you conclude that none do?
Current License: CC BY-SA 3.0
7 events
| when toggle format | what | by | license | comment | |
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| S Jun 13, 2017 at 18:27 | history | suggested | CommunityBot | CC BY-SA 3.0 |
\operatorname{non}(M)
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| Jun 13, 2017 at 18:10 | review | Suggested edits | |||
| S Jun 13, 2017 at 18:27 | |||||
| Aug 20, 2015 at 18:47 | comment | added | Will Brian | @AndreasBlass: Thanks! Even in your jet-lagged state, it seems you're right. I've gone ahead and fixed the definition. | |
| Aug 20, 2015 at 18:45 | history | edited | Will Brian | CC BY-SA 3.0 |
Fixed a faulty definition
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| Aug 20, 2015 at 18:33 | comment | added | Andreas Blass | Maybe I'm just being dense (I'm still a little jet-lagged) but at the end of Part I of the proof, I don't see why $a_{n+1}>f_b(a_n)$ implies $b(a_{n+1})>b(a_n)$. I see that it implies that $b$ can't map $a_n$ up to $a_{n+1}$ and can't map $a_{n+1}$ down to $a_n$, but why can't it map $a_n$ up and $a_{n+1}$ down, to some in-between location where they're out of order? (If I'm not just being dense and this is really a problem, it can clearly be solved by modifying the definition of $f_b$, so the theorem survives.) | |
| Aug 20, 2015 at 17:16 | history | edited | Will Brian | CC BY-SA 3.0 |
corrected a typo
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| Aug 20, 2015 at 17:11 | history | answered | Will Brian | CC BY-SA 3.0 |