# Distance on Real Numbers is Metric

## Theorem

Let $x, y \in \R$ be real numbers.

Let $\map d {x, y}$ be the distance between $x$ and $y$:

$\map d {x, y} = \size {x - y}$

Then $\map d {x, y}$ is a metric on $\R$.

Thus it follows that $\tuple {\R, d}$ is a metric space.

## Proof

We check the metric space axioms in turn.

### Axiom $(\text M 1)$

The statement of this axiom is:

$(\text M 1): \forall x \in X: \size {x - x} = 0$

This follows from the definition of absolute value.

$\Box$

### Axiom $(\text M 2)$

The statement of this axiom is:

$(\text M 2): \forall x, y, z \in X: \size {x - y} + \size {y - z} \ge \size {x - z}$

We have:

$\paren {x - y} + \paren {y - z} = \paren {x - z}$

The result follows from the Triangle Inequality for Real Numbers.

$\Box$

### Axiom $(\text M 3)$

The statement of this axiom is:

$(\text M 3): \forall x, y \in X: \size {x - y} = \size {y - x}$

As $x - y = -\paren {y - x}$, it follows from the definition of absolute value that $\size {x - y} = \size {y - x}$.

$\Box$

### Axiom $(\text M 4)$

The statement of this axiom is:

$(\text M 4): \forall x, y \in X: x \ne y \implies \size {x - y} > 0$

This follows from the definition of absolute value.

$\Box$

Having verified all the axioms, we conclude $d$ is a metric.

$\blacksquare$