Timeline for Addition and multiplication are commutative but exponentiation and tetration are not. Do we know why?
Current License: CC BY-SA 4.0
15 events
| when toggle format | what | by | license | comment | |
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| Feb 15 at 8:09 | comment | added | Watson | This was asked on MSE in 2011 ... : math.stackexchange.com/questions/35598 | |
| Feb 14, 2023 at 2:41 | comment | added | Marco Ripà | Just found this topic. Although I think that the provided answers are satisfatory enough, I like to share here a quite surprising new result that has stolen years of life to be proven: tetration (hyper-$4$) over the positive integers is not commutative (indeed), but it is the only hyper-operator which is characterized by a constant congruence speed for any nontrivial base (i.e., in radix-$10$ the congruence speed is constant for every base that is not a multiple of $10$): arxiv.org/abs/2208.02622 and arxiv.org/abs/2210.07956 IMHO, this unique property is really fascinating! | |
| Feb 8, 2023 at 17:06 | history | edited | YCor |
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| Feb 8, 2023 at 0:03 | comment | added | Zach Hunter | perhaps it is more strange that multiplication is commutative... I remember some MO post saying this is very surprising when you look at it as stated in Peano arithmetic. | |
| Feb 7, 2023 at 23:27 | answer | added | Vincent | timeline score: 7 | |
| Feb 6, 2023 at 3:01 | comment | added | Taladris | @mz71. Nice remark. It is not surprising though, when one remark that $a\ast b = a^{ln(b)}$ is the operation induced on $(0,\infty)$ by $(\mathbb R,\times)$ and $f(x)=\exp(x)$, that is $a\ast b=f(f^{-1}(a)\times f^{-1}(b))$ for every $a,b>0$. | |
| Feb 6, 2023 at 1:46 | comment | added | Anixx | @mz71 $a^{\ln b}=e^{\ln a \ln b}$, not $e^{\ln(a+b)}$. | |
| Feb 6, 2023 at 0:39 | comment | added | Sidharth Ghoshal | That logarithmic tower gives you an infinite family in both directions… $\ln(e^x + e^y)$ is the commutative associative operation which addition distributes over | |
| Feb 5, 2023 at 23:46 | comment | added | mz71 | Not related to the problem at hand, but one curious fact that I saw on Wikipedia and I can vividly remember is that the list of commutative operations can be extended by taking $a^{\ln b}$ in place of exponentiation. As $a^{\ln b} = e^{\ln a+b} = b^{\ln a}$ this operation is commutative. | |
| Feb 5, 2023 at 23:32 | comment | added | Dave L Renfro | Possibly of interest is the discussion on p. 114 of An extended arithmetic of ordinal numbers by Doner/Tarski (1969). | |
| Feb 5, 2023 at 20:51 | history | became hot network question | |||
| Feb 5, 2023 at 16:37 | vote | accept | Rorsa | ||
| Feb 5, 2023 at 14:57 | history | edited | gmvh | CC BY-SA 4.0 |
Added missing right parenthesis
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| Feb 5, 2023 at 13:55 | answer | added | Gro-Tsen | timeline score: 34 | |
| Feb 5, 2023 at 12:50 | history | asked | Rorsa | CC BY-SA 4.0 |