|Date||Tue, 30 Nov 2000 11:58:44 -0800 (PST)|
Since I like your site so much, I am inspired to offer some
The concept of "trisectable" is not like "constructible" in the sense that you suggest. Indeed, if two angles are constructible, then their sum or difference is also, SINCE YOU HAVE BOTH ANGLES IN HAND TO WORK WITH. However, if two angles are trisectable, then I see no reason why their sum or difference need be, since, given such a sum (say), you cannot assume that you have the original two angles IN HAND (since they are not necessarily constructible, even from this sum). Hence, you cannot use the two original angles and, by trisecting them, arrive at a trisection of the sum. Thus, I find the following Description misleading:
"The argument is completely general. Let A be a property of angles, such as being constructible or being trisectable. Then, schematically,
A - A = A
which simply says that the difference of two angles with property A also possesses that property. (Which is of course true for A being
either constructible or trisectable.)"
In the specific example, π/3 is actually constructible (from nothing), so you cannot construct from it π/21 or 2π/7 in order to do a trisection for π/3 from the ones you (hypothetically) have for them. In other words, I do not see that this argument can be made to work in the example either.
On the other hand, is u a primitive 7th root of unity, then u3 is also
a primitive 7th root of unity (since (u3)5 = (u14)×u = u. Thus, u3
is not constructible, while the field it generates over the constructibles contains its cube root. This was the first part of your piece. [In fact, you can see that u itself has a relatively constructible cube root, viz., u5, since (u3)5 = (u14)×u = u. In your terms, 10π/7 (= 5×2π/7) provides a relative trisection of 2π/7. One needn't worry that this angle seems large; you can get the smaller cube root by multiplying by an appropriate cube root of unity (2π/3 or -2π/3).] Now to find a non-constructible root of unity whose cube root is not relatively constructible, we need only note that the field of relatively constructible numbers will lie in a field which has degree a power of two over the field which contains the given initial root of unity (since one is taking iterated square roots by intersecting lines and circles). Taking v to be a primitive 42nd root of unity (angle π/21), the field generated by it has degree 12 over the rationals, while its cube root (being a primitive 126th root of unity) has degree 36 over the rationals. [The theorem is that any primitive nth root of unity has degree φ(n) over the rationals, where the function φ(n) is Euler's φ function.] But 36 is not 12 times a power of two, so your example works. But the reasoning (at least mine) is a bit more subtle than that given. This is nearly a smallest example, since you need at least 7 to prevent constructibility and a
factor of 3 to have the cube root pick up a factor of three in its degree over the intermediate field. It would seem that the lowest degree solution is given by a 21st root of unity (2π/21), but this does seem much more succinct when expressed as an angle (vs. π/21).
I hope I have been clear on this matter. Sorry it rambles on so much. Feel free to ask about anything that was not clear.
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