Rational Numbers form Metric Space

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Theorem

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

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


Proof

From the definition of absolute value:

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

This is the same as the Euclidean metric.

This is shown in Euclidean Metric on Real Vector Space is Metric to be a metric.

From Rational Numbers form Vector Space, it follows that the set of all rational numbers is a 1-dimensional Euclidean space.

$\blacksquare$


Sources