Does the model $\mathbb{R}^2$ (The usual Cartesian plane.) with the distance defined as $d(P,Q) = | x_1-x_2| + |y_1-y_2|$ satisfy the ruler postulate?

The ruler postulate is defined as the following

For any line $l$ and any two distinct points $O$ and $P$ on $l$, there exists a bijection $c : l \rightarrow \mathbb{R}$ such that the following holds:

  1. $c(O)=0$ and $c(P) \geq 0$.
  2. $d(A,B)=|c(A)-c(B)|$, for all points $A$ and $B$ on $l$

Now the distance is defined in this space

  1. Symmetry: If $P(x_1,y_1)$ and $Q(x_2,y_2)$, then $$d(Q,P)=|x_2-x_1|+|y_2-y_1|=|x_1-x_2|+|y_1-y_2|=d(P,Q)$$

    1. Also $d(P,Q)$ is clearly nonnegative. If $d(P,Q)=|x_1-x_2|+|y_1-y_2|=0$, then $|x_1-x_2|=|y_1-y_2|=0$. This implies that $x_1=x_2$ and $y_1=y_2$, that is $P=Q$.
    2. Triangular inequality: If $P(x_1,y_1)$, $Q(x_2,y_2)$ and $R(x_3,y_3)$, then $$ d(P,R)=|x_1-x_3|+|y_1-y_3| \leqslant |x_1-x_2|+|x_2-x_3|+|y_1-y_2|+|y_2-y_3| = |x_1-x_2|+|y_1-y_2|+|x_2-x_3|+|y_2-y_3| = d(P,Q)+d(Q,R) $$

We can conclude that $d$ is indeed a distance.

How can I show that the distance exists therefore the ruler postulate will hold? It seems that since a distance exists there exists a way to measure it.

Edit: Would this mean a line in the usual sense of the solutions $(x,y)$ to some equation $ax+by = d$, where $(a,b) \not = (0,0)$ ?

  • $\begingroup$ What is a line in this geometry? $\endgroup$ Commented Feb 1, 2017 at 20:44
  • $\begingroup$ Wait, is this the hyperbolic plane? $\endgroup$ Commented Feb 1, 2017 at 21:23
  • $\begingroup$ This is very confusing. It clearly cannot work for circles since they have finite diameter in the defined distance. I think this quesiton needs work. $\endgroup$ Commented Feb 1, 2017 at 21:38
  • $\begingroup$ I'm sorry the line in this geometry is the stand lines and points. I apologize would that still mean it can't work? $\endgroup$ Commented Feb 2, 2017 at 0:15
  • $\begingroup$ I'm not very familiar with the notion of the ruler postulate, but it seems to me that this just means the distance function restricted to each (Euclidean) line being additive. And then clearly your metric satisfies the ruler postulate. $\endgroup$
    – cjackal
    Commented Feb 2, 2017 at 0:23

1 Answer 1


By subtracting the coordinates of $O$ from all points concerned, we may assume $O=(0,0).$ Then let $P=(a,b).$ Any point on line $l$ is now of the form $Q=(ka,kb)$ for some unique $k \in \mathbb{R}.$

Define $c(Q)=k(|a|+|b|).$ This maps line $l$ onto the reals. For point $P$ we have as required that $c(P)>0.$ Also if $A=(ka,kb),\ B=(k'a,k'b)$ we have $c(A)=k(|a|+|b|)$ and $c(B)=k'(|a|+|b|).$ If we compute the taxicab distance (your distance $d$) then we have $$d(A,B)=|k-k'|(|a|+|b|) = \\ |k(|a|+|b|)-k'(|a|+|b|) = |c(A)-c(B)|.$$


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