Timeline for An integrality question about expressing an integer as a product of numbers below $n$
Current License: CC BY-SA 3.0
18 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Apr 13, 2017 at 12:58 | history | edited | CommunityBot |
replaced http://mathoverflow.net/ with https://mathoverflow.net/
|
|
| Dec 15, 2013 at 11:28 | vote | accept | Lucia | ||
| Dec 15, 2013 at 10:19 | answer | added | Ilya Bogdanov | timeline score: 18 | |
| Dec 14, 2013 at 22:37 | history | edited | Lucia | CC BY-SA 3.0 |
added 398 characters in body
|
| Dec 14, 2013 at 21:19 | history | edited | Lucia | CC BY-SA 3.0 |
added 368 characters in body
|
| Dec 14, 2013 at 20:44 | history | edited | Lucia | CC BY-SA 3.0 |
added 141 characters in body
|
| Dec 14, 2013 at 20:20 | answer | added | Greg Martin | timeline score: 8 | |
| Dec 14, 2013 at 20:20 | history | edited | Lucia | CC BY-SA 3.0 |
added 228 characters in body
|
| Dec 14, 2013 at 20:15 | comment | added | Lucia | @TheMaskedAvenger -- I don't want it to be that subtle! Assume that the $x_j$'s are all rational and add up to an integer. Does it then follow that there is an integral representation? Or simply assume that $N$ is composed only of primes up to $n$. | |
| Dec 14, 2013 at 20:14 | comment | added | The Masked Avenger | From Gelfond Schneider, I expect not, but I am out of (err) my field here. | |
| Dec 14, 2013 at 20:10 | comment | added | The Masked Avenger | Some of the subtleties are striking me now. Here is an important case to consider. Let N be so represented, except that the (set of) primes with nonzero exponent are disjoint from the (set of) prime factors of N. Is it even possible that the sum of the exponents is rational, let alone equal to the sum from a standard factorization? | |
| Dec 14, 2013 at 20:03 | history | edited | Lucia | CC BY-SA 3.0 |
added 38 characters in body
|
| Dec 14, 2013 at 20:01 | comment | added | Lucia | If you use $1^{1/4}$ then $2$ can indeed just be written as $2^1$ (all other exponents are zero) and the exponents do add up. | |
| Dec 14, 2013 at 19:59 | comment | added | The Masked Avenger | Uh, what prevents $1^{1/4}$ to appear in Christian's example? Can't you tweak the exponent of 1 to meet your conditions? | |
| Dec 14, 2013 at 19:30 | history | edited | Lucia | CC BY-SA 3.0 |
added 213 characters in body
|
| Dec 14, 2013 at 19:25 | history | undeleted | Lucia | ||
| Dec 14, 2013 at 18:59 | history | deleted | Lucia | via Vote | |
| Dec 14, 2013 at 17:52 | history | asked | Lucia | CC BY-SA 3.0 |