Rational Numbers form Metric Space

Theorem

Let $\Q$ be the set of all rational numbers.

Let $d: \Q \times \Q \to \Q$ 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 $\Q$ and so $\left({\Q, d}\right)$ 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 a Metric to be a metric.

As the rational numbers form a vector space, it follows that the set of all rational numbers is a 1-dimensional Euclidean space.

$\blacksquare$