Timeline for In which fixed-point free representations is the sum of every 3 elements invertible?
Current License: CC BY-SA 3.0
6 events
| when toggle format | what | by | license | comment | |
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| Mar 31, 2014 at 11:06 | vote | accept | A.B. | ||
| Mar 30, 2014 at 20:10 | answer | added | Geoff Robinson | timeline score: 9 | |
| Mar 30, 2014 at 19:30 | history | edited | A.B. | CC BY-SA 3.0 |
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| Mar 30, 2014 at 19:18 | comment | added | A.B. | I think it is equivalent because one can always divide by one of the elements since they are invertible. The product is invertible/0 iff it was already invertible/0, and the representation property gives this new sum. But sum of every 2 elements and the identity is actually what I'm interested in. | |
| Mar 30, 2014 at 19:13 | comment | added | Dima Pasechnik | you wrote "sum of every 3 elements", meaning "sum of every 2 elements and the identity"... | |
| Mar 30, 2014 at 18:55 | history | asked | A.B. | CC BY-SA 3.0 |