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I've been reading about Hilbert's Axioms of geometry, and I have two models on $\mathbb{R}^2$ which satisfy the usual axioms of incidence, betweenness, and congruence.

The first model has a distance function given by $$ d(A,B)=|a_1-b_1|+|a_2-b_2| $$ where $A=(a_1,a_2)$, $B=(b_1,b_2)$, and two segments $AB\cong CD$ if $d(A,B)=d(C,D)$.

The second model has a distance function given by $$ d(A,B)=\sup\{|a_1-b_1|,|a_2-b_2|\}, $$ and again $AB\cong CD$ if $d(A,B)=d(C,D)$.

The author claims that these two models are isomorphic in that there is a bijection between the sets of points in $\mathbb{R}^2$ that preserves lines, betweenness and congruence. I've been toying with it for a while, but it seems to me that the notions of congruence in particular are too different.

For example, in the first model, if $A=(0,0)$ and $B=(2,1)$, and $C=(2,0)$ and $D=(5/2,5/2)$, then $AB\cong CD$ since both segments have length $3$. To preserve congruence, I suppose two segments are congruent if the maximum of the displacement along the $x$-axis or $y$-axis by traveling along one segment is equal to the maximum of the displacement along either axis when traveling along the other segment. In this sense, how could you shape the segment $CD$ so that the maximum of the displacements along either axis is the same as the maximum of the displacement for whatever the image of $AB$ is?

In the first model, $AB$ is congruent to any segment which is the hypotenuse of triangle with legs that sum to $3$, and this sum could split into any two terms. I just don't see a function that could handle all these possibilities at once. Does anyone know of a particular isomorphism between these two? Thanks.

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In the second model, the locus of points at a fixed distance $d$ from a point $A$ is an axis-aligned square of side length $2d$ centered on $A$. In the first model, the locus is a square of side length $d\sqrt{2}$ and rotated $45^{\circ}$ from the axes, again centered on $A$. There is a combined rotation and dilation that maps the first model to the second.

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Thanks mjqxxxx. So a possible transformation from the first model to the second would be something like this:$$ \sqrt{2}\begin{bmatrix} \cos(\pi/4) & -\sin(\pi/4)\\ \sin(\pi/4) & \cos(\pi/4)\end{bmatrix}=\begin{bmatrix} 1 & -1\\ 1 & 1\end{bmatrix}$$, by applying it to any point $[x,y]$. –  yunone Feb 16 '11 at 8:41

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