# Taxicab Metric on Real Vector Space is Metric

## Theorem

The taxicab metric on the real vector space $\R^n$ is a metric.

## Proof 1

This is an instance of the taxicab metric on the cartesian product of $A_{1'}, A_{2'}, \ldots, A_{n'}$.

This is proved in Taxicab Metric is Metric.

$\blacksquare$

## Proof 2

The taxicab metric on $\R^n$ is:

$\displaystyle d_1 \left({x, y}\right) = \sum_{i \mathop = 1}^n \left\vert{x_i - y_i}\right\vert$

for $x = \left({x_1, x_2, \ldots, x_n}\right), y = \left({y_1, y_2, \ldots, y_n}\right) \in \R^n$.

### Proof of $M1$

 $\displaystyle d_1 \left({x, x}\right)$ $=$ $\displaystyle \sum_{i \mathop = 1}^n \left\vert{x_i - x_i}\right\vert$ Definition of $d_1$ $\displaystyle$ $=$ $\displaystyle \sum_{i \mathop = 1}^n 0$ $\displaystyle$ $=$ $\displaystyle 0$

So axiom $M1$ holds for $d_1$.

$\Box$

### Proof of $M2$

 $\displaystyle d_1 \left({x, y}\right) + d_1 \left({y, z}\right)$ $=$ $\displaystyle \sum_{i \mathop = 1}^n \left\vert{x_i - y_i}\right\vert + \sum_{i \mathop = 1}^n \left\vert{y_i - z_i}\right\vert$ Definition of $d_1$ $\displaystyle$ $=$ $\displaystyle \sum_{i \mathop = 1}^n \left({\left\vert{x_i - y_i}\right\vert + \left\vert{y_i - z_i}\right\vert}\right)$ $\displaystyle$ $\ge$ $\displaystyle \sum_{i \mathop = 1}^n \left\vert{x_i - z_i}\right\vert$ Triangle Inequality for Real Numbers $\displaystyle$ $=$ $\displaystyle d_1 \left({x, z}\right)$ Definition of $d_1$

So axiom $M2$ holds for $d_1$.

$\Box$

### Proof of $M3$

 $\displaystyle d_1 \left({x, y}\right)$ $=$ $\displaystyle \sum_{i \mathop = 1}^n \left\vert{x_i - y_i}\right\vert$ Definition of $d_1$ $\displaystyle$ $=$ $\displaystyle \sum_{i \mathop = 1}^n \left\vert{y_i - x_i}\right\vert$ $\displaystyle$ $=$ $\displaystyle d \left({y, x}\right)$ Definition of $d_1$

So axiom $M3$ holds for $d_1$.

$\Box$

### Proof of $M4$

 $\displaystyle x$ $\ne$ $\displaystyle y$ $\displaystyle \implies \ \$ $\, \displaystyle \exists i \in \left\{ {1, 2, \ldots, n}\right\}: \,$ $\displaystyle x_i$ $\ne$ $\displaystyle y_i$ $\displaystyle \implies \ \$ $\displaystyle \left\vert{x_i - y_i}\right\vert$ $>$ $\displaystyle 0$ Definition of Absolute Value $\displaystyle \implies \ \$ $\displaystyle \sum_{i \mathop = 1}^n \left\vert{x_i - y_i}\right\vert$ $>$ $\displaystyle 0$ $\displaystyle \implies \ \$ $\displaystyle d_1 \left({x, y}\right)$ $>$ $\displaystyle 0$ Definition of $d_1$

So axiom $M4$ holds for $d_1$.

$\blacksquare$