Proof
(related to Proposition: The distance of real numbers makes real numbers a metric space.)
We have to prove that the absolute value of the difference \(xy\) is a metric function on the field of real numbers \((\mathbb R, + ,\cdot)\), i.e. to show that it fulfills in \(\mathbb R\) the general properties of any metric.
 Following the existence of negative numbers and their uniqueness it follows that \(xy=0\) if and only if \(x=y\), so the first property of a metric is fulfilled.
 To show the symmetry \(xy=yx\) as the second property of a metric, we assume without loss of generality that \(x\ge y\). Then we have by definition \(xy = xy\), which is equal \(y(x)=1(yx)\). Using the properties 1 and 2 of the absolute value, we get the desired result by \(1(yx)=1yx=1yx=yx\).
 In order to prove the triangle inequality: \(xz\le xy+yz\) for all \(x,y,z\in \mathbb R\), we observe that \((xy)\le xy\) and \((yz)\le yz\). Applying the rule 4 of calculation with inequalities we get
\[xz=xy+yz\le xy+yz \]
Because also \((xy)\le xy\) and \((yz)\le yz\), it follows with the same argument that
\[(xz)=(xy)+((yz))\le xy+yz. \]
By definition of absolute value, both results give us
\[xz\le xy+yz. \]∎
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References
Bibliography
 Forster Otto: "Analysis 1, Differential und Integralrechnung einer VerĂ¤nderlichen", Vieweg Studium, 1983