Proof of The Absolute Value Metric on the Real Numbers

By The Distance on the Real Numbers §distance, $d_{\mathbb{R}}$ is a map $\mathbb{R}\times\mathbb{R}\to\mathbb{R}$ with $d_{\mathbb{R}}(s,t)=|s-t|$. Since $\mathbb{R}$ is an ordered field by The Real Numbers, with the Natural Numbers, Integers and Rationals Identified with Subsets of the Reals, and Completeness §reals, Rules of Arithmetic and Order in an Ordered Field applies to it. Let $s,t,r\in\mathbb{R}$; we verify the conditions of Metric Space §metric.

Nonnegative. $0\le|s-t|=d_{\mathbb{R}}(s,t)$ by Rules of Arithmetic and Order in an Ordered Field §absolute-value, which is Metric Space §nonnegative.

Zero. By Rules of Arithmetic and Order in an Ordered Field §absolute-value, $|s-t|=0$ if and only if $s-t=0$. If $s=t$, then $s-t=t+(-t)=0$ by Negatives, Differences, Reciprocals and Quotients §negative. Conversely, if $s-t=0$, then by associativity of $+$ and $(-t)+t=0$ (Negatives, Differences, Reciprocals and Quotients §negative, with commutativity)

$$s=s+\bigl((-t)+t\bigr)=\bigl(s+(-t)\bigr)+t=0+t=t .$$

Hence $d_{\mathbb{R}}(s,t)=0$ if and only if $s=t$, which is Metric Space §zero.

Symmetric. $t-s=-(s-t)$ by Rules of Arithmetic in a Commutative Ring: Zero, Signs and Squares, and No Zero Divisors in a Field §signs, and $|-(s-t)|=|s-t|$ by Rules of Arithmetic and Order in an Ordered Field §absolute-value; so $d_{\mathbb{R}}(t,s)=d_{\mathbb{R}}(s,t)$, which is Metric Space §symmetric.

Triangle. By associativity and commutativity of $+$ and $(-t)+t=0$ (Negatives, Differences, Reciprocals and Quotients §negative), $(s-t)+(t-r)=s+\bigl((-t)+t\bigr)+(-r)=s-r$. Hence Rules of Arithmetic and Order in an Ordered Field §triangle gives

$$d_{\mathbb{R}}(s,r)=|(s-t)+(t-r)|\le|s-t|+|t-r|=d_{\mathbb{R}}(s,t)+d_{\mathbb{R}}(t,r),$$

which is Metric Space §triangle.

As $s,t,r$ were arbitrary, $d_{\mathbb{R}}$ is a metric on $\mathbb{R}$ by Metric Space §metric.