The field of constructible lengths, part 3

One of the rules for \mathbb K is that it is a field and it contains all of its positive square roots (though allowing complex numbers wouldn’t hurt, we won’t do so). We are going to show now that this is all it has.

If we have a quadratic equation ax^2+bx+c=0 which has a real solution, by the quadratic formula, the solution is

x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}.

So what? What’s so special about that? Well, notice that if ab, and c are in our field \mathbb K, then so is x, since \mathbb K is closed under square roots, as we showed last week.

This is an important fact, because, this is exactly how we construct \mathbb K. Each time we mark an intersection point, it is either the intersection of two lines, a line and a circle, or two circles. For each one, we can set up a linear, or quadratic equation.

This is a big step to answering our questions. For example, given a cube, can we build a cube with double the volume? Well, if we could, it would mean that \sqrt[3]{2} was constructible. This doesn’t look like it can be achieved via square roots to me! The answer is “no,” but we need to prove that \sqrt[3]2 can’t be written with only rationals and square roots. How will we do this?

We’ll look at the smallest field containing \mathbb Q and \sqrt[3]2. In a nice way, we can view it as a vector space over \mathbb Q, so we can see what its dimension is. It will turn out to have dimension 3, and that will turn out to be bad/impossible to construct.

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5 Responses to The field of constructible lengths, part 3

  1. Louis says:

    Andy! You never showed it was closed under sqrt! (Not that it’s hard, but…)

  2. Pingback: Oops, Square roots! « Andy Soffer

  3. Pingback: More awesomeness of the tower rule « Andy Soffer

  4. Pingback: The intermediate field lattice « Andy Soffer

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