Timeline for Is there an effective way to generalize this approach of affinely extending the number line?
Current License: CC BY-SA 4.0
12 events
| when toggle format | what | by | license | comment | |
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| Nov 27, 2022 at 6:12 | vote | accept | Zuhair Al-Johar | ||
| Nov 27, 2022 at 0:40 | comment | added | Zuhair Al-Johar | if we want to adopt that approach on the projectively extended reals, would it be the same but we change the definition of $\beta_i$ to $\beta_i:=f(|\alpha_i^1|,...,|\alpha_i^n|)$, where $||$ stands for the absolute value function, and of course extend $\mathbb R$ by just one point $\infty$? | |
| Nov 26, 2022 at 8:46 | comment | added | Zuhair Al-Johar | Just want to check really, is it the case that $(-1)^\infty \leadsto x \iff x \in \hat{\mathbb R}$? | |
| Nov 26, 2022 at 7:03 | comment | added | Zuhair Al-Johar | Nice! So, there is a negative real that has an even real root (since $\infty$ is divisible by $2$) that is a real. (real in the extended sense). | |
| Nov 26, 2022 at 6:34 | comment | added | Noah Schweber | @ZuhairAl-Johar (I presume you mean "$(-2)^\infty$") The latter, since both $-\infty$ and $\infty$ can be "achieved" by appropriate sequences. | |
| Nov 26, 2022 at 6:26 | comment | added | Zuhair Al-Johar | hmmm... Ok what is $-2^\infty$ should it be $-2^\infty = \infty$ (i.e. $\infty$ seen as an even number), or $-2^\infty \leadsto x \iff x \in \{-\infty, \infty\}$ ($\infty$ seen behaving as even and odd)? | |
| Nov 26, 2022 at 6:05 | comment | added | Noah Schweber | @ZuhairAl-Johar Those examples are easy to calculate: $1^\infty\leadsto x$ iff $x$ is nonnegative, and $0^{-\infty}\leadsto x$ iff $x\in\{-\infty, +\infty\}$. I don't think higher operators/complex numbers/etc. make things significantly harder; you just need to think about what the possible limiting behaviors are, and these are pretty tame. | |
| Nov 26, 2022 at 5:54 | comment | added | Zuhair Al-Johar | The only questions I'd think would be interesting are those sprinning from this method itself, a simple example: what are $1^\infty, 0^{-\infty}$?, it appears that applying $\infty$ to operator identities would cause difficulties and those do not appear to be trivial, and as we use higher operators those become more and more difficult. If we extend the complex plane, things become even murkier. Anyhow I don't have a question in mind outside this method, that serve as an application. | |
| Nov 25, 2022 at 21:58 | comment | added | Noah Schweber | @ZuhairAl-Johar Do you have any concrete test questions you think might be interesting? That's generally a good way to motivate an idea as worth pursuing. | |
| Nov 25, 2022 at 21:57 | history | edited | Noah Schweber | CC BY-SA 4.0 |
added 204 characters in body
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| Nov 25, 2022 at 8:40 | comment | added | Zuhair Al-Johar | I think determing the set of values an expression can take may be interesting, I noticed that this is more complex than with the projective extension of the reals, and also builds up in complexity with higher and higher operators. | |
| Nov 25, 2022 at 5:45 | history | answered | Noah Schweber | CC BY-SA 4.0 |