Steven Dutch, Natural and Applied Sciences, Universityof Wisconsin - Green Bay
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I will respond to questions and comments as time permits, but if you want to take issuewith any position ea
pressed here, you first have to answer this question:
What evidence would it take to prove your beliefs wrong?
I simply will not reply to challenges that do not address this question. Refutabilityis one of the classic determinants of whether a theory can be called scientific. Moreover,I have found it to be a great general-purpose cut-through-the-crap question to determinewhether somebody is interested in serious intellectual inquiry or just playing mind games.Note, by the way, that I am assuming the burden of proof here - all youhave to do is commit to a criterion for testing.It's easy to criticize science for being "closed-minded". Are you open-mindedenough to consider whether your ideas might be wrong?
The starting point for finding rational divisions of 360 degrees is the relationship
eia = cos(a) + i sin(a), where i = √(-1).
This leads to the well known construction for representing complex numbers where the real component is on the horizontal axis and the i ("imaginary") component is on the vertical axis.
If a = π, then we have eiπ = cos(π) + i sin(π) = -1, one of the most remarkable formulas in mathematics. Obviously
e2iπ = 1
Thus, constructing a polygon of n sides amounts to solving the equation an = 1, since we have
e2iπ/n = cos(2iπ/n) + i sin(2iπ/n)
To construct n-sided polygons, we need only consider prime values of n. Obviously even values of n merely requiring bisecting angles.
a3 -1 = 0 factors into
(a -1)(a2 + a + 1) = 0
Solving a2 + a + 1 = 0 we get a = [-1 (√(1-4)]/2 = -1/2 [√(-3)]/2 = -1/2 i[√(3)]/2
From this we can figure out (if we didn't already know it) that cos 120 = -1/2 and sin 120 = [√(3)]/2
However, it's instructive to factor a2 + a + 1:
a2 + a + 1 = [a + 1/2 + (√-3)/2][a + 1/2 - (√-3)/2]
The two solutions are obviously a =-1/2 i[√(3)]/2
a5 -1 = 0 factors into
(a -1)(a4 + a3 + a2 + a + 1) = 0
Solving a4 + a3 + a2 + a + 1 = 0
We first have to factor (a4 + a3 + a2 + a + 1)
There is a general method for solving quartic equations, but we're looking for a method that might lend itself more to extension. Since we solved n=3 by solving a quadratic equation, maybe we can find some quadratic equation within a quadratic equation.
[(a2 + ab + c)2 + d(a2 + ab + c) + e] = 1 or
(a2 + ab + c)2 + d(a2 + ab + c) + e - 1 = 0 We solve the quadratic in d and e to get
(a2 + ab + c) = [-d +/- √(d2 - 4e -4)]/2 or
a2 + ab + c + d +/- √(d2 - 4e -4)]/2, which we can then solve
a = -b +/- √(b2 - 4e -4)]/2
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Created 21 January, 2003, Last Update 24 May, 2020
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