Timeline for Slicing up $\mathbb{N}\setminus\{1\}$
Current License: CC BY-SA 4.0
8 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| May 12, 2020 at 6:20 | vote | accept | Dominic van der Zypen | ||
| May 12, 2020 at 1:21 | answer | added | Jan Kyncl | timeline score: 5 | |
| May 11, 2020 at 19:42 | comment | added | Gerhard Paseman | Yes, I misread, and so seriously edited my comment above. It may be possible to give a slicker proof than what you have posted as an answer, but I'm not seeing it. Gerhard "More Prime Powers To You" Paseman, 2020.05.11. | |
| May 11, 2020 at 19:34 | comment | added | LSpice | @GerhardPaseman, if each member of $B$ contains only one element, then, for each $B$, $\lvert B \cap p\mathbb N\rvert$ will equal $0$ for most primes $p$. On edit: it seems that your revised construction is just what I described, isn't it? | |
| May 11, 2020 at 19:33 | comment | added | Gerhard Paseman | Oops, I missed the every prime condition earlier. Start assembling sets of coprime integers. If you can't add a number to one of the existing sets, start a new set. In particular, each set will have at most one power of a given prime. Gerhard "Thinks Jumping Primes Are Cooler" Paseman, 2020.05.11. | |
| May 11, 2020 at 19:32 | answer | added | LSpice | timeline score: 4 | |
| May 11, 2020 at 19:06 | history | edited | Dominic van der Zypen | CC BY-SA 4.0 |
Title was misleading
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| May 11, 2020 at 18:59 | history | asked | Dominic van der Zypen | CC BY-SA 4.0 |