Guess, you can't have that. Points in a line sengment are not countable - so every point can't be rational. Correct?

Also, what do you mean by tangent of the angle?

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- Dec 2nd 2009, 09:24 AMaman_cc
- Dec 2nd 2009, 09:35 AMUnbeatable0
Sure you're right, I meant the vertices are in http://www.mathhelpforum.com/math-he...fe4dca53-1.gif, my mistake for no paying enough attention to wording.

If a is an interior angle of the triangle, tan(a) is what I meant by the tangent of the angle, and the proposition claims that it is rational.

In the original case of equilateral triangle, it turns out no equilateral triangle

exist with vertices in http://www.mathhelpforum.com/math-he...fe4dca53-1.gif, as tan(60 deg) = sqrt(3) - Dec 2nd 2009, 02:01 PMDrexel28
That is precisely why! When speaking of triangles, we must first talk about the edges of the triangle. These edges are connected and may be interpreted as being homeomorphic to some interval in and since any interval in is uncountable we can't speak of lines in a countable spac such as

P.S. I didn't check that my language above was precise, but you get the idea. - Dec 3rd 2009, 08:05 AMUnbeatable0
I almost forgot about the first problem you stated.

The solution:

__Spoiler__: - Dec 3rd 2009, 12:57 PMDrexel28
- Dec 4th 2009, 02:32 AMUnbeatable0
Actually, from the proposition I posted before, it follows that for any regular n-sided polygon, if is irrational, then it can't have three vertices in http://www.mathhelpforum.com/math-he...fe4dca53-1.gif, two of which are adjacent.

Nobody is up to prove the claim in post #14? - Dec 4th 2009, 07:53 AMDrexel28
- Dec 6th 2009, 05:41 AMUnbeatable0
- Dec 7th 2009, 10:02 PMDrexel28