Timeline for Applications of the Chinese remainder theorem
Current License: CC BY-SA 2.5
8 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jun 25, 2013 at 3:02 | review | Late answers | |||
| Jun 26, 2013 at 0:06 | |||||
| Apr 7, 2011 at 0:35 | comment | added | Burton Newman | That's a good point! | |
| Apr 4, 2011 at 22:57 | comment | added | darij grinberg | What we DO use here is the existence of a primitive $a_n$-th root of unity. This is somewhat tricky if we wish to avoid analysis, but many books do it right. | |
| Apr 4, 2011 at 22:56 | comment | added | darij grinberg | ... one doesn't. | |
| Apr 4, 2011 at 22:56 | comment | added | darij grinberg | While I absolutely don't want to ruin Doron's wonderful polemic, let me add that the "complex-analytic" proof he quotes is completely algebraic. Just look at it from the correct angle: We take the equation (MNDR) and rewrite it as $\frac{1}{1-z}-\sum_{i=1}^{n-1} \frac{z^{b_i}}{1-z^{a_i}} = \frac{z^{b_n}}{1-z^{a_n}}$. Bringing the left hand side to a common denominator, this denominator is going to be a polynomial which is nonzero on any primitive $a_n$-th root of unity, so if we multiply the equation with the $a_n$-th cyclotomic polynomial, the left hand side becomes $0$. But the right ... | |
| Apr 4, 2011 at 18:55 | history | undeleted | Burton Newman | ||
| Apr 4, 2011 at 18:53 | history | deleted | Burton Newman | ||
| Apr 4, 2011 at 18:52 | history | answered | Burton Newman | CC BY-SA 2.5 |