Timeline for Does commutativity imply Associativity?
Current License: CC BY-SA 3.0
30 events
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| Apr 21, 2022 at 23:03 | comment | added | Atom | Good old mod arithmetic: Take $n\ge 2$. Then defining $x\ast y := (x\operatorname{mod} n^2)(y\operatorname{mod} n^2)$, we have that $0 = (n\ast n)\ast 1\ne n\ast(n\ast 1) = n^2$. | |
| Sep 18, 2019 at 14:38 | answer | added | CopyPasteIt | timeline score: 1 | |
| S Mar 4, 2018 at 18:50 | history | post merged (destination) | |||
| Nov 17, 2017 at 1:05 | answer | added | ಠ_ಠ | timeline score: 2 | |
| Sep 26, 2017 at 20:20 | vote | accept | SuperMathBrothers | ||
| S Mar 4, 2018 at 18:50 | |||||
| Sep 26, 2017 at 20:20 | vote | accept | SuperMathBrothers | ||
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| Sep 26, 2017 at 20:20 | vote | accept | SuperMathBrothers | ||
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| Sep 26, 2017 at 20:14 | answer | added | Jean Marie | timeline score: 8 | |
| Sep 26, 2017 at 20:13 | vote | accept | SuperMathBrothers | ||
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| Sep 26, 2017 at 20:13 | vote | accept | SuperMathBrothers | ||
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| Sep 26, 2017 at 20:13 | vote | accept | SuperMathBrothers | ||
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| Sep 26, 2017 at 20:12 | answer | added | quasi | timeline score: 5 | |
| Sep 26, 2017 at 20:08 | answer | added | Bram28 | timeline score: 9 | |
| Sep 26, 2017 at 20:00 | answer | added | Kenny Lau | timeline score: 3 | |
| Oct 17, 2016 at 20:05 | answer | added | Paul Orland | timeline score: 64 | |
| Jan 22, 2016 at 15:48 | history | edited | Martin Sleziak |
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| Nov 30, 2015 at 17:27 | answer | added | bof | timeline score: 16 | |
| Nov 30, 2015 at 16:57 | answer | added | John Coleman | timeline score: 67 | |
| Sep 14, 2012 at 15:54 | history | edited | Bill Dubuque | CC BY-SA 3.0 |
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| Jun 22, 2012 at 21:31 | comment | added | Dave L. Renfro | See my 3 February 2009 sci.math post A natural example of a commutative, non-associative operator (see Google archive version or Math Forum archive version) for some examples and references. | |
| Jun 20, 2012 at 22:14 | comment | added | Zhen Lin | The interchange law $(x * y) \cdot (z * w) = (x \cdot z) * (y \cdot w)$, in the presence of a two-sided common unit element, implies commutativity and associativity of $*$ and $\cdot$. (In fact, they have to be the same operation!) | |
| Jun 20, 2012 at 22:10 | vote | accept | Eugene | ||
| Jun 20, 2012 at 21:58 | comment | added | Bruno Stonek | Not even in the presence of an identity element and an opposite, see this. In fact, William's answer is already in that post ;) | |
| Jun 20, 2012 at 20:41 | comment | added | André Nicolas | Commutative operations that are associative are the exception. But an important exception! Let $x\ast y=|x-y|$. | |
| Jun 20, 2012 at 20:36 | answer | added | William | timeline score: 17 | |
| Jun 20, 2012 at 20:35 | answer | added | Thomas Andrews | timeline score: 70 | |
| Jun 20, 2012 at 20:33 | history | edited | Eugene |
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| Jun 20, 2012 at 20:32 | answer | added | marlu | timeline score: 131 | |
| Jun 20, 2012 at 20:30 | answer | added | Will Jagy | timeline score: 28 | |
| Jun 20, 2012 at 20:27 | history | asked | Eugene | CC BY-SA 3.0 |