# Chebyshev Distance on Real Number Plane is not Rotation Invariant

## Theorem

Let $r_\alpha: \R^2 \to \R^2$ denote the rotation of the Euclidean plane about the origin through an angle of $\alpha$.

Let $d_\infty$ denote the Chebyshev distance on $\R^2$.

Then it is not necessarily the case that:

$\forall x, y \in \R^2: \map {d_\infty} {\map {r_\alpha} x, \map {r_\alpha} y} = \map {d_\infty} {x, y}$

## Proof

Let $x = \tuple {0, 0}$ and $y = \tuple {1, 1}$ be arbitrary points in $\R^2$.

Then:

 $\ds \map {d_\infty} {x, y}$ $=$ $\ds \map {d_\infty} {\tuple {0, 0}, \tuple {1, 1} }$ Definition of $x$ and $y$ $\ds$ $=$ $\ds \max \set {\size {0 - 1}, \size {0 - 1} }$ Definition of Chebyshev Distance on Real Number Plane $\ds$ $=$ $\ds 1$

Now let $\alpha = \dfrac \pi 4 = 45 \degrees$.

 $\ds \map {d_\infty} {\map {r_\alpha} x, \map {r_\alpha} y}$ $=$ $\ds \map {d_\infty} {\tuple {0, 0}, \tuple {0, \sqrt 2} }$ Definition of Plane Rotation $\ds$ $=$ $\ds \max \set {\size {0 - 0}, \size {0 - \sqrt 2} }$ Definition of Chebyshev Distance on Real Number Plane $\ds$ $=$ $\ds \sqrt 2$ simplification $\ds$ $\ne$ $\ds 1$

$\blacksquare$