Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Let $\nu \left( x \right) $ be the least number of steps that is required to construct a constructible length $x$, using compass and ruler in the well known fashion. Now, define the distance $d\left(x,y\right)$, of two constructible numbers $x$ and $y$ as the $\nu \left(|x -y|\right)$. Is this a metric for the space of all constructible numbers?

share|improve this question
I don't know. What do you think? –  Graphth Sep 11 '12 at 20:23
I think it is not. The third axiom for metric space should fail. –  Hooman Sep 11 '12 at 20:29
Well, it seems to me that $\nu(1)=0$, in which case the answer is no. –  Harald Hanche-Olsen Sep 11 '12 at 20:29
Actually, it depends on something subtle about knowledge. Years ago, i put in the final touch on proving Bolyai's construction in the hyperbolic plane of curvature $-1,$ that there are a countably infinite set of squares and circles, both constructible, with equal area, and showed there are no others. But I also showed there there are no one-way constructions there, it is not possible in $\mathbb H^2$ to square the circle or to circle the square. So there is something very tricky about knowledge that cannot be avoided. –  Will Jagy Sep 11 '12 at 20:59
Closer to what you are thinking about, T.-Y. Lam has pointed out that there is no canonical form in the constructible numbers. Given two expressions involving rational numbers, field operations, and square roots of positive elements, we can (I think) eventually decide whether they are the same number, I guess by showing they satisfy the same one-variable polynomial. –  Will Jagy Sep 11 '12 at 21:14

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.