Timeline for Even XOR Odd Infinities?
Current License: CC BY-SA 4.0
15 events
when toggle format | what | by | license | comment | |
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S Oct 11 at 6:46 | history | edited | Daniele Tampieri | CC BY-SA 4.0 |
Hyperlinked relevant comments + removed unnecessary \text command
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S Oct 11 at 6:46 | history | suggested | Mohsen Shahriari | CC BY-SA 4.0 |
improved formatting
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Oct 11 at 5:15 | review | Suggested edits | |||
S Oct 11 at 6:46 | |||||
Jan 21, 2013 at 3:09 | history | edited | abo | CC BY-SA 3.0 |
added 134 characters in body
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Jan 21, 2013 at 1:02 | comment | added | Russell Easterly | I ran into this problem, too. Most of my attempts start by assuming -1 is not 0 (I exclude the trivial ring). | |
Jan 20, 2013 at 23:14 | comment | added | Eric Wofsey | The argument still doesn't work because $y=p$ is always a solution to $S(y+y)=p$. | |
Jan 20, 2013 at 22:59 | history | edited | abo | CC BY-SA 3.0 |
added 17 characters in body
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Jan 20, 2013 at 22:58 | comment | added | abo | Yes sorry. I'll edit... | |
Jan 20, 2013 at 22:57 | comment | added | François G. Dorais | By PA you mean MA? | |
Jan 20, 2013 at 22:54 | comment | added | abo | OK, I fixed the proof so it works in first-order PA. | |
Jan 20, 2013 at 22:49 | history | edited | abo | CC BY-SA 3.0 |
added 488 characters in body
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Jan 20, 2013 at 22:12 | comment | added | abo | In second-order PA, you can consider the ancestral of T, which gives you the normal ordering. The question, however, is confined to first-order PA, so this won't work. | |
Jan 20, 2013 at 22:00 | comment | added | François G. Dorais | "Then $T$ induces the normal ordering $\lt$ on $A$" - what does that mean? | |
Jan 20, 2013 at 21:58 | comment | added | abo | Sorry, I was working in 2nd-order PA. | |
Jan 20, 2013 at 21:37 | history | answered | abo | CC BY-SA 3.0 |