Timeline for Proving, using Dedekind cuts, that $C(0)$ is an additive identity for addition on cuts.
Current License: CC BY-SA 4.0
19 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| May 10 at 13:34 | vote | accept | Vince Vickler | ||
| May 10 at 13:34 | vote | accept | Vince Vickler | ||
| May 10 at 13:34 | |||||
| May 6 at 22:05 | answer | added | Brian Moehring | timeline score: 1 | |
| May 6 at 21:39 | comment | added | Vince Vickler | @Arturo Magidin. Thanks for making all that clear. | |
| May 6 at 21:19 | answer | added | Villa | timeline score: 0 | |
| May 6 at 20:59 | comment | added | Arturo Magidin | @VinceVickler The cut $\{r\in\mathbb{Q}\mid r\lt 0\text{ or }(r\geq 0\text{ and }r^2\leq 2)\}$ is famously not of that form. | |
| May 6 at 20:57 | comment | added | Vince Vickler | @Arturo Magidin. Thanks for correcting my mistake as to the ( im)possibility of writing any cut as $C(r)$ for some rational $r$. I need to make a new post asking how a cut in $Q$ can fail to be of this form. | |
| May 6 at 20:53 | comment | added | Vince Vickler | @Brian Moering. Sincere thanks for having taken time to rewrite my post. | |
| May 6 at 20:49 | comment | added | Arturo Magidin | The first sentence is very weird. I suspect that the assertion is that if $r$ is a rational then the set $C(r)=\{x\in\mathbb{Q}\mid x\lt r\}$ is a Dedekind cut, but not that every Dedekind cut can be written that way. | |
| May 6 at 20:48 | comment | added | Arturo Magidin | (i) Do not use chatGPT for anything other than personal amusement when you are bored and want to waste time with a chatbot. (ii) That is a terrible use of Mathjax/Latex. | |
| May 6 at 20:46 | comment | added | Brian Moehring | I have re-written your post without using some much text in math mode. I have not fixed some of your errors. For instance, $$C(r) = \{x : x < r \text{ for some } r \in \mathbb{Q}$$ is not sensible for how $r$ is bound both to $C(r)$ and inside the set notation. I might guess you meant $$C \text{ can be written as } C(r) = \{x \in \mathbb{Q} : x < r\} \text{ for some } r \in \mathbb{Q}$$ but while this is notationally sensible, it's false in general. Please compare the text as it currently stands with your source material. | |
| May 6 at 20:41 | history | edited | Brian Moehring | CC BY-SA 4.0 |
Removed all that use of \text in math mode. Now it should be legible to more people.
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| May 6 at 20:30 | comment | added | Vince Vickler | I aimed at displaying a neat mathematical text with the natural language parts coded with the \text command. So I've provided ChatGPT with the above text with additional instructions , and I confess the result is not top level | |
| May 6 at 20:28 | history | edited | Vince Vickler | CC BY-SA 4.0 |
added 2 characters in body
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| May 6 at 20:25 | comment | added | Vince Vickler | My " prompt" : " * I now want to show that :* C inclued in C+C(0).// I first consider the case in which *C is a negative cut , meaning that * C subsetequal C(0)* Suppose that* x in C. Because multiplication by a positive real preserves order, x < r implies x/4 <r. Set c = x/4, we have : c<r* and c in C. *Also, set *c_0 =c = x/4. *This implies : c_0 in C* and consequently * c_0 in C(0). * It follows that * c+c_0= 2c = 2(x/4)=x * meaning that x can be written as the sum of two elements of ( respectively * C* and C(0 as desired.* | |
| May 6 at 20:21 | comment | added | Brian Moehring | Putting such large sections of text into math mode has made your post nearly illegible. Yours is not the first I've seen do this though, so I'm curious: did you actually put in the effort to manually type that out, or is it the output of some what-you-see-is-what-you-get program converting it to latex? | |
| May 6 at 20:15 | history | edited | Vince Vickler | CC BY-SA 4.0 |
added 211 characters in body
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| May 6 at 20:09 | history | edited | Vince Vickler | CC BY-SA 4.0 |
added 211 characters in body
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| May 6 at 20:02 | history | asked | Vince Vickler | CC BY-SA 4.0 |