Timeline for Are there any interesting or lesser known proofs related to Bertrand's Postulate
Current License: CC BY-SA 3.0
7 events
| when toggle format | what | by | license | comment | |
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| Jul 5, 2018 at 17:25 | comment | added | Benjamin Dickman | @ĐàoThanhOai At least one of the numbers in that list is divisible by a prime $\geq k+1$; but, every number in that list is $\leq 2k$. So the number divisible by a new prime is itself prime: if it had additional prime factors, it would exceed $2k$, which it does not. | |
| Jul 5, 2018 at 16:53 | comment | added | Đào Thanh Oai | Can You help me why? "The theorem implies immediately that for any positive integer $k$, one of $k+1, k+2, \ldots, 2k$ is a prime (since one of these integers must be divisible by a prime $\geq k+1).$" | |
| Aug 14, 2013 at 20:41 | history | edited | Benjamin Dickman | CC BY-SA 3.0 |
added link to new AMM article on Ramanujan's proof of Bertrand's Postulate
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| Sep 6, 2012 at 1:17 | comment | added | Larry Freeman | Here's an online link to the Sylvester paper: gdz.sub.uni-goettingen.de/dms/load/img/… | |
| Aug 7, 2012 at 18:02 | comment | added | Charles | J. J. Sylvester. On arithmetical series. Messenger Math. 21 (1892), pp. 1-19, 87-120, 192. (See p. 4) | |
| Aug 2, 2012 at 7:04 | vote | accept | Larry Freeman | ||
| Aug 2, 2012 at 0:35 | history | answered | Benjamin Dickman | CC BY-SA 3.0 |