# Oops, Square roots!

October 11, 2011 2 Comments

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

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

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

Next we draw in lines and , and notice that the triangle is a right triangle.

We can then call on some altitude on hypotenuse theorems. Namely, , so But in our diagram, , and , so , as desired.

In the last few lines of text, I think you mean (instead of ) and (instead of , since ).

Yes, I did. And now I fixed it. Thanks!