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 \(|x-y|\) 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 \(|x-y|=0\) if and only if \(x=y\), so the first property of a metric is fulfilled.
- To show the symmetry \(|x-y|=|y-x|\) as the second property of a metric, we assume without loss of generality that \(x\ge y\). Then we have by definition \(|x-y| = x-y\), which is equal \(-y-(-x)=-1(y-x)\). Using the properties 1 and 2 of the absolute value, we get the desired result by \(-1(y-x)=|-1||y-x|=|1||y-x|=|y-x|\).
- In order to prove the triangle inequality: \(|x-z|\le |x-y|+|y-z|\) for all \(x,y,z\in \mathbb R\), we observe that \((x-y)\le |x-y|\) and \((y-z)\le |y-z|\). Applying the rule 4 of calculation with inequalities we get
\[x-z=x-y+y-z\le |x-y|+|y-z| \]
Because also \(-(x-y)\le |x-y|\) and \(-(y-z)\le |y-z|\), it follows with the same argument that
\[-(x-z)=-(x-y)+(-(y-z))\le |x-y|+|y-z|. \]
By definition of absolute value, both results give us
\[|x-z|\le |x-y|+|y-z|. \]∎
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References
Bibliography
- Forster Otto: "Analysis 1, Differential- und Integralrechnung einer Veränderlichen", Vieweg Studium, 1983
