Timeline for Is $x + y \ne y+nx$ for $x \ne 0$ and $n \ge 2$ (in an ordered group)?
Current License: CC BY-SA 3.0
10 events
| when toggle format | what | by | license | comment | |
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| Feb 13, 2020 at 19:03 | history | edited | YCor |
edited tags
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| Oct 30, 2013 at 21:31 | vote | accept | Salvo Tringali | ||
| Oct 28, 2013 at 14:59 | answer | added | YCor | timeline score: 4 | |
| Oct 28, 2013 at 13:12 | comment | added | Salvo Tringali | Right, I was too hasty (and optimistic). But then, what if $z = 2x$? Actually, this is the case I am interested in (I edited the OP according to your comment). | |
| Oct 28, 2013 at 12:44 | history | edited | Salvo Tringali | CC BY-SA 3.0 |
Updated after a comment killing Q1
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| Oct 28, 2013 at 11:58 | comment | added | YCor | This is true only if $A$ is abelian. Otherwise, there exist two distinct conjugate elements $x,z$; since your order is total, we can suppose $x\preceq z$. So there exists $y$ such that $xy=yz$ (I avoid your additive notation which is confusing to me). | |
| Oct 28, 2013 at 11:09 | history | edited | Salvo Tringali | CC BY-SA 3.0 |
Removed an apparently useless remark
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| Oct 28, 2013 at 10:52 | comment | added | Salvo Tringali | Yes, I am. This is common in additive theory (e.g., see Ruzsa's survey Sumsets and structure). | |
| Oct 28, 2013 at 10:45 | comment | added | Tobias Kildetoft | Are you writing the group additively, even though it is not abelian? | |
| Oct 28, 2013 at 10:37 | history | asked | Salvo Tringali | CC BY-SA 3.0 |