Timeline for Are the rationals on a unit circle dense?
Current License: CC BY-SA 3.0
11 events
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| Jan 15, 2017 at 22:31 | comment | added | Jeppe Stig Nielsen | @DerekElkins And there is a nice illustration in the subsection you link; and we can think of rational points on a straight line, if that is easier to "see" than a bunch of rational slopes. | |
| Jan 15, 2017 at 4:07 | comment | added | Derek Elkins left SE | Just for reference this is essentially an inverse stereographic projection. | |
| S Jan 14, 2017 at 21:35 | history | suggested | ruakh | CC BY-SA 3.0 |
enlarged the parentheses in a formula to match size of their contents
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| Jan 14, 2017 at 21:35 | review | Suggested edits | |||
| S Jan 14, 2017 at 21:35 | |||||
| Jan 14, 2017 at 21:13 | vote | accept | CommunityBot | moved from User.Id=4414 by developer User.Id=951463 | |
| Jan 14, 2017 at 21:13 | comment | added | user4414 | My bad, was short cutting rational arithmetic. gr8! | |
| Jan 14, 2017 at 21:12 | comment | added | Noah Schweber | I think I did it right. The slope is ${y_2-y_1\over x_2-x_1}$. Let $A=y_2-y_1, B=x_2-x_1$. $y_2={-2m\over m^2+1}$ and $y_1=0$, so $A={-2m\over m^2+1}$. $x_2={m^2-1\over m^2+1}$ and $x_1=1$, so $B={m^2-1\over m^2+1}-1={m^2-1\over m^2+1}-{m^2+1\over m^2+1}={-2\over m^2+1}$. Cancelling denominators, we get ${A\over B}={-2m\over -2}=m$. | |
| Jan 14, 2017 at 21:05 | comment | added | Noah Schweber | @j4nbur53 Fixed a (godawful) typo. But yes - $(m^2-1)^2=m^4-2m^2+1$, and $(-2m)^2=4m^2$. So adding the squares of the numerators we get $(m^4+1)^2$. But that's exactly the square of the denominator. | |
| Jan 14, 2017 at 21:04 | history | edited | Noah Schweber | CC BY-SA 3.0 |
deleted 2 characters in body
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| Jan 14, 2017 at 21:03 | comment | added | user4414 | Can you show that P_m is on the unit circle? | |
| Jan 14, 2017 at 20:51 | history | answered | Noah Schweber | CC BY-SA 3.0 |
