# Proof

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.

1. 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.
2. 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 $$-1(y-x)=|-1||y-x|=|1||y-x|=|y-x|$$.
3. 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 $|x-z|\le |x-y|+|y-z|.$

Github: ### References

#### Bibliography

1. Forster Otto: "Analysis 1, Differential- und Integralrechnung einer Veränderlichen", Vieweg Studium, 1983