Timeline for A prime number determined by its congruence relation?
Current License: CC BY-SA 3.0
10 events
| when toggle format | what | by | license | comment | |
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| Jul 13, 2017 at 17:13 | history | edited | Gerhard Paseman | CC BY-SA 3.0 |
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| Jul 12, 2017 at 23:46 | history | edited | Gerhard Paseman | CC BY-SA 3.0 |
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| Jul 12, 2017 at 23:40 | review | Low quality posts | |||
| Jul 12, 2017 at 23:47 | |||||
| Jul 12, 2017 at 23:39 | comment | added | user56983 | Where does $i$-th prime number not mean the sequence of primes? | |
| Jul 12, 2017 at 23:38 | comment | added | Gerhard Paseman | Are you given n? Are you given k(n)? Can you infer the information I mention? If you are the original poster, you can say. Otherwise the best you can do is interpret. Gerhard "Has Given His Own Interpretation" Paseman, 2017.07.12. | |
| Jul 12, 2017 at 23:32 | comment | added | user56983 | How is it not? That's what it usually means. | |
| Jul 12, 2017 at 23:30 | history | edited | Gerhard Paseman | CC BY-SA 3.0 |
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| Jul 12, 2017 at 23:30 | comment | added | mesel | By saying that $p_i$ is $i$'th prime, is it not clear that $p_1=2,p_2=3,p_3=5 ...$. | |
| Jul 12, 2017 at 23:29 | comment | added | user56983 | There is no $m$. $k(n)$ was the upper limit for $i$ in the residue classes for which $\{x_i \pmod{p_i}\}$ was known if $p_i$ was the $i$-th prime number. See my counter example. | |
| Jul 12, 2017 at 23:21 | history | answered | Gerhard Paseman | CC BY-SA 3.0 |