Oops, Square roots!

It was pointed out yesterday that I forgot to prove, or even mention the fact that \mathbb K is closed under taking square roots. That is, if k\in\mathbb K, then so is \sqrt k. So here goes.

Suppose we can construct k. Build a line segment AB of length k. Construct an additional segment of length 1 after it, so we have a segment AC of length k+1. Now draw a circle with this as the diameter. You should have something like this:

Great! Now we can drop a perpendicular at B, and see where it intersects the circle. Call this point D. Remember from my first post on constructibility that we can construct perpendiculars through a point. We now have:

Next we draw in lines AD and CD, and notice that the triangle \Delta ADC is a right triangle.

We can then call on some altitude on hypotenuse theorems. Namely, \frac{AB}{BD}=\frac{BD}{BC}, so BD=\sqrt{AB\cdot BC} But in our diagram, AB=k, and BC=1, so BD=\sqrt k, as desired.

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2 Responses to Oops, Square roots!

  1. Lauren says:

    In the last few lines of text, I think you mean BD = \sqrt{AB \cdot BC} (instead of BC = \sqrt{AB \cdot BC}) and BD = \sqrt{k} (instead of BC = \sqrt{k}, since BC=1).

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