Timeline for Sets whose elements are mutually "weakly" coprime?
Current License: CC BY-SA 3.0
9 events
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| Dec 25, 2013 at 19:52 | comment | added | Seva | @Dustin G. Mixon: this seems to be easy to prove. Suppose that no $s\in S$ divides the product $P(s):=\prod_{t\in S\setminus\{s\}} t$. Then for any $s\in S$ there is a prime $p\in[1,n]$ such that, notation as in Greg Martin's comment, we have $\nu_p(s)>\nu_p(P(s))$. Now, if we had $|S|>\pi(n)$, then there would be two elements of $S$ associated with the same prime, which is a clear nonsense: if, say, $\nu_p(s_1)\le\nu_p(s_2)$, then $\nu_p(s_1)\le\nu_p(P(s_1))$ since $s_2\mid P(s_1)$. | |
| Dec 25, 2013 at 19:51 | history | edited | Konstantinos Gaitanas | CC BY-SA 3.0 |
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| Dec 25, 2013 at 19:45 | comment | added | Konstantinos Gaitanas | @DustinG.Mixon i made an edit ,here is the proof. | |
| Dec 25, 2013 at 19:43 | history | edited | Konstantinos Gaitanas | CC BY-SA 3.0 |
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| Dec 25, 2013 at 18:32 | comment | added | Dustin G. Mixon | @panoramix - Yes, I'd like to see the proof. :) | |
| Dec 25, 2013 at 15:54 | comment | added | Greg Martin | This doesn't really answer the posted question - I suspect a random set must be much smaller to have this property - but it is an interesting fact nonetheless. | |
| Dec 25, 2013 at 13:08 | comment | added | Konstantinos Gaitanas | At the beggining i thought that Erdos's theorem was included in "Topics in the theory of numbers"(Springer) but i could not find it right now.If you wish i could give the proof of the above statement. | |
| Dec 25, 2013 at 12:48 | comment | added | Dustin G. Mixon | Thanks - Can you include a reference? | |
| Dec 25, 2013 at 10:50 | history | answered | Konstantinos Gaitanas | CC BY-SA 3.0 |