# Ruler-and-Compass metric

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?

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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

Some thoughts:

I started thinking of this as a path metric (in which context Will Jagy's comment is important). Then I realised that's the wrong idea with the metric as defined. It's more of a norm metric.

Check the conditions of a metric:

$d(x,y)=\nu(|x-y|)\geq 0$ because it is given by counting.

If $x=y$ then $|x-y|=0$, which is defined to be constructed in $0$ steps, so $d(x,x)=0$. Conversely, if $d(x,y)=0$ then length $|x-y|$ can be constructed in $0$ steps, and definition says the only such length is $0$. Hence $x=y$.

$d(x,y)=\nu(|x-y|)=\nu(|y-x|)=d(y,x)$.

(Note that so far we haven't used much in the way of specific details.)

Finally, consider $x$, $y$, $z$, such that length $|x-y|$ can be constructed in $n$ steps and length $|y-z|$ can be constructed in $m$ steps. Is it true that length $|x-z|$ can be constructed in at most $m+n$ steps? Note that if one of the two lengths is $0$ then this is immediate. So what's the answer if neither is $0$?

To get this, I'd like to say that I can make a length $|x-y|$ along the positive $X$-axis of $\mathbb{R}^2$, and $|y-z|$ along the positive or negative $X$-axis as required to give the two points as $|x-z|$ apart. I'd say would be reasonable to say we can do all constructions steps in mirror image if we wish, so I can do positive/negative as I want. The problem is the idea of getting them actually on the axis. I don't have any reason to think that the quickest way to construct a point will leave it on the $X$-axis.

One idea would be to buy a spare move by saying I'd only need to construct the point $1$ once, and then I can use it for both constructions. But that would not fit with the 'do the mirror image' idea.

Another (obvious) idea is that I don't really care if it's the $X$-axis, so long as I can get them on the same line. So I can assume, without loss of generality, that $|x-y|$ is constructed (in $\mathbb{R}^2$). How quickly can I construct $|y-z|$ on the axis?

First assume the length $|y-z|$ has one endpoint at $0$. Then I can draw a circle with centre at $0$ and radius $|y-z|$. Then I can draw in the $X$--axis (if I haven't already done so) to find the intersection point in the right place. So I've constructed the right length, but I've used a couple too many steps, (how many too many depending on exactly how we count, but it's fixed and finite).

Instead then, what if neither end is at $0$? What exactly are the rules of the game? Can I set my compasses using that length and draw a circle with that radius around $0$? Or do I need to try and move the actual line? If the former, I'm in the same situation as before. In the latter case, how would I move the line? Hmm... Can I find the fourth corner of a parallelogram when I'm given three corners? Hmm... Can I draw a line parallel to one I'm given? Google says 'yes': http://www.mathopenref.com/constparallel.html

In conclusion: I can't see an obvious reason why the triangle inequality should hold. However, it seems reasonable that tweaking the metric (eg. adding $13$ to all non-zero distances) will give something that works.

(For preservation purposes, here is Will Jagy's comment:)

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.

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

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