Timeline for Exactly two of the four roots must be greater than N/2
Current License: CC BY-SA 3.0
9 events
| when toggle format | what | by | license | comment | |
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| Jul 11, 2014 at 8:00 | comment | added | habillqabill | ok done. tq guys. | |
| Jul 11, 2014 at 7:59 | vote | accept | habillqabill | ||
| Jul 3, 2014 at 15:12 | comment | added | Thomas | @habillqabill If this answers your question, would you consider accepting it by clicking on the check mark on the left? | |
| Jul 1, 2014 at 14:50 | history | edited | Ilmari Karonen | CC BY-SA 3.0 |
add the symmetry argument to the answer; feel free to improve
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| Jul 1, 2014 at 10:30 | comment | added | habillqabill | oh yeah, that correct. n should be an odd number. By the way, i'll try to turn your argument into a mathematical statement. I see it all over the places, particularly when involving Rabin primitive, without any explanation. | |
| Jul 1, 2014 at 10:25 | comment | added | Thomas | @habillqabill Suppose $n$ is odd. Let $r$ be any integer between $1$ and $n$. Now suppose $r$ is less than $n/2$, then we are done. If $r$ is in fact greater than $n/2$, then $n - r < n/2$ and we are done. Basically, no matter $r$, either $r$ or $n - r$ will be less than $n/2$ (since both will fall on opposite sides of $n/2$). It's a symmetry argument. | |
| Jul 1, 2014 at 10:24 | history | edited | Thomas | CC BY-SA 3.0 |
deleted 3 characters in body
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| Jul 1, 2014 at 10:24 | comment | added | habillqabill | But, how we know for sure in each pair exactly one root will be greater than n/2. I;m sorry since i couldnt understand your point. For me, 'In each pair exactly one root will be greater than n/2' is still a plain statement. Dont u think so? | |
| Jul 1, 2014 at 8:17 | history | answered | Thomas | CC BY-SA 3.0 |