Irrational Numbers form Metric Space

Theorem
Let $\mathbb I = \R \setminus \Q$ be the set of all irrational numbers.

Let $d: \mathbb I \times \mathbb I \to \R$ be defined as:


 * $d \left({x_1, x_2}\right) = \left|{x_1 - x_2}\right|$

where $\left|{x}\right|$ is the absolute value of $x$.

Then $d$ is a metric on $\mathbb I$ and so $\left({\mathbb I, d}\right)$ is a metric space.

Proof
From the definition of absolute value:


 * $\left|{x_1 - x_2}\right| = \sqrt {\left({x_1 - x_2}\right)^2}$

It is clear that this is the same as the euclidean metric, which is shown in Euclidean Metric is Metric to be a metric.