Timeline for Is this extension of the projectively extended real line, consistent?
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| Dec 12, 2022 at 20:42 | history | edited | Zuhair Al-Johar | CC BY-SA 4.0 |
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| Dec 1, 2022 at 11:54 | vote | accept | Zuhair Al-Johar | ||
| Nov 23, 2022 at 20:13 | history | undeleted | Zuhair Al-Johar | ||
| Nov 23, 2022 at 20:08 | history | deleted | Zuhair Al-Johar | via Vote | |
| Nov 23, 2022 at 18:55 | history | edited | Zuhair Al-Johar | CC BY-SA 4.0 |
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| Nov 23, 2022 at 18:49 | history | undeleted | Zuhair Al-Johar | ||
| Nov 23, 2022 at 17:59 | history | deleted | Zuhair Al-Johar | via Vote | |
| Nov 23, 2022 at 14:01 | answer | added | Zuhair Al-Johar | timeline score: 1 | |
| Nov 20, 2022 at 20:43 | comment | added | Zuhair Al-Johar | @MonroeEskew, but anyway, the idea of the projective extension is that $1/0$ represents a single number, and so it is not multi-valued, and so it cannot equal $0$, it is the point on the other side, so it must be distinct from zero, even at initial intuitive level. | |
| Nov 20, 2022 at 18:19 | comment | added | Zuhair Al-Johar | @MonroeEskew, be careful, what you wrote is not an arrow expression, what you've written (in your earlier comment) is $r/0 = 0$ this is an equation, this is straighforwardly inconsistent with the projectively extended real line, since $\infty +1 = \infty$ and $0 +1 \neq 0$. If we write $r/0 \to 0$, then this might work given that $r/0$ is multivalued. That is another story. You don't want to give up on associativity of mixed addition subtraction expressions, if that is an axiom, then of course my method here fails. | |
| Nov 20, 2022 at 17:47 | comment | added | Monroe Eskew | @ZuhairAl-Johar If we are dealing with multi-valued relations, it's not inconsistent to stipulate $\infty \to 0$. I'm not sure what the goals are. The idea of the projective real line is that going to infinity in either direction wraps around, correct? Maybe you could give some geometric motivation to these multi-valued relations you want to introduce? | |
| Nov 20, 2022 at 14:51 | comment | added | Zuhair Al-Johar | @MonroeEskew, yes, true, these equalities work here as well, the word defined in the conditions written correspond to mono-valuation here. This fails once this condition is not met. | |
| Nov 20, 2022 at 11:55 | comment | added | Monroe Eskew | According to the Wikipedia article you linked, the structure $\hat{\mathbb R}$ is supposed to satisfy associativity of addition. | |
| Nov 20, 2022 at 10:46 | comment | added | Zuhair Al-Johar | @MonroeEskew, in particular, this won't suit the purpose of this posting, namely to extend the projectively extended real line, here this leads to $\infty = 0$ which is clearly inconsistent. | |
| Nov 18, 2022 at 16:08 | comment | added | Zuhair Al-Johar | @MonroeEskew, well I need division to retain some of meanings of the usual division, we don't need the system to have very remote features, if $r/0=0$ then division is not the converse relation of multiplication, and so in this sense it seems distant from the usual intuition associated with division, which is indeed, as you said, unsatisfactory. | |
| Nov 18, 2022 at 14:17 | comment | added | Monroe Eskew | If all you want is some formal system in which an operation is defined, then you can just extend the usual operations on the reals by defining $r/0 = 0$ for all $r$. But this seems unsatisfying, right? | |
| Nov 18, 2022 at 13:59 | comment | added | Zuhair Al-Johar | @MonroeEskew, at $\infty$ and regarding opposing operators, sadly yes. I'm thinking of the simplest system that extends the reals and in which division by zero is definable | |
| Nov 18, 2022 at 8:37 | comment | added | Monroe Eskew | @ZuhairAl-Johar You really want to give up associativity? I thought this was trying to make sense of naive analysis arguments. What is the motivation? | |
| Nov 18, 2022 at 7:11 | comment | added | Zuhair Al-Johar | @MonroeEskew the expressions $(\infty + \infty) - \infty$ , $\infty + (\infty - \infty)$ do not necessary yield the same values. The first is multivalued (arrows every element of $\hat{\mathbb R}$), but the second is mono-valued. So they are not the same expressions. | |
| Nov 18, 2022 at 7:03 | comment | added | Monroe Eskew | @ZuhairAl-Johar I meant $(\infty +\infty) -\infty \to \infty -\infty$. So one expression “arrows” every element. | |
| Nov 18, 2022 at 6:56 | comment | added | Zuhair Al-Johar | @MonroeEskew, the step $\infty \to \infty - \infty$ is not well formed here, possibly you mean $\infty \pm r \to \infty - \infty$, but this is not among the rules here, and I don't think it's derivable here. So, as Noah mentioned this step breaks down. | |
| Nov 18, 2022 at 6:29 | comment | added | Zuhair Al-Johar | @NoahSchweber, the point that made me ask about consistency is actually $\infty + \infty = \infty$ and $\infty - \infty \to r$, I need to visualize this in some fragment of ZFC, I should confess that I can find some fragment with two distinct signed infinities with different consequences than those here, i.e. we extend the affinely extended real line, and as I said the results are different but I think I can find a model for it, but this projective one I need to figure it out. The consistency question is about the details of interpretation of these operations in ZFC. | |
| Nov 17, 2022 at 20:46 | comment | added | Zuhair Al-Johar | @NoahSchweber, I should say Thanks for directing me to Matt Baker's blog! That is at the heart of what I'm thinking of! | |
| Nov 17, 2022 at 20:29 | comment | added | Noah Schweber | @MonroeEskew Since operations are multivalued, rrows aren't reversible here. So indeed we have $\infty+\infty-\infty\rightarrow\infty$ and $\infty+\infty-\infty\rightarrow r$, but that doesn't mean that $\infty\rightarrow r$. (Precisely: the $\infty\rightarrow\infty-\infty$ step in your comment breaks down.) | |
| Nov 17, 2022 at 20:23 | comment | added | Monroe Eskew | So how do you make sense of $\infty + \infty - \infty \to \infty + r \to \infty \to \infty - \infty \to r$? | |
| Nov 17, 2022 at 20:17 | comment | added | Monroe Eskew | @NoahSchweber I guess you're right. My thought was more like, why worry about making sense of these things when we have non-archimedean fields to look at? | |
| Nov 17, 2022 at 19:36 | comment | added | Noah Schweber | The consistency question doesn't really make any sense - obviously since it has a model (that's how you're describing it!), all the rules of your system are consistent. You may be interested in the notion of hyperstructures and in particular hyperfields - see e.g. this blog post by Matt Baker. | |
| Nov 17, 2022 at 19:34 | comment | added | Noah Schweber | @MonroeEskew This has nothing to do with nonstandard analysis as far as I can tell. | |
| Nov 17, 2022 at 19:29 | comment | added | Monroe Eskew | Google "nonstandard analysis." | |
| Nov 17, 2022 at 18:58 | history | edited | Zuhair Al-Johar | CC BY-SA 4.0 |
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| Nov 17, 2022 at 17:53 | history | asked | Zuhair Al-Johar | CC BY-SA 4.0 |