Timeline for For consecutive primes $a\lt b\lt c$, prove that $a+b\ge c$.
Current License: CC BY-SA 3.0
9 events
| when toggle format | what | by | license | comment | |
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| Feb 10, 2013 at 4:13 | comment | added | Charles | (Of course 2.001 can be replaced by 2 unconditionally.) | |
| Feb 10, 2013 at 0:54 | history | edited | user9072 |
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| Dec 10, 2012 at 23:42 | vote | accept | Bavid | ||
| Nov 20, 2012 at 0:40 | answer | added | José Hdz. Stgo. | timeline score: 13 | |
| Nov 20, 2012 at 0:10 | comment | added | GH from MO | Just to complement the responses below: The prime number theorem says that the $n$-th prime is asymptotically $n\log n$, whence your sum $a+b$ is asymptotically $2c$. So your inequality holds for large $c$ without any calculation, in fact $2.001 c>a+b>1.999 c$ for large $c$. | |
| Nov 19, 2012 at 23:52 | answer | added | user9072 | timeline score: 17 | |
| Nov 19, 2012 at 22:15 | history | edited | Gerry Myerson | CC BY-SA 3.0 |
formatting, incorporated question into body
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| Nov 19, 2012 at 16:27 | answer | added | Tony Huynh | timeline score: 44 | |
| Nov 19, 2012 at 16:12 | history | asked | Bavid | CC BY-SA 3.0 |