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

Question

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 $$d(Q,P)=|x_2-x_1|+|y_2-y_1|=|x_1-x_2|+|y_1-y_2|=d(P,Q)$$

   2. 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 $P=Q$.
   3. 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 $(a,b) \not = (0,0)$ ?

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 $$d(A,B)=|k-k'|(|a|+|b|) = \\ |k(|a|+|b|)-k'(|a|+|b|) = |c(A)-c(B)|.$$