Scaled Euclidean Metric is Metric

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Theorem

Let $\R_{>0}$ be the set of (strictly) positive integers.

Let $\delta: \R_{>0} \times \R_{>0} \to \R$ be the metric on $\R_{>0}$ defined as:

$\forall x, y \in \R_{>0}: \map \delta {x, y} = \dfrac {\size {x - y} } {x y}$


Then $\delta$ is a metric.


Proof

Proof of Metric Space Axiom $(\text M 1)$

\(\ds \map \delta {x, x}\) \(=\) \(\ds \dfrac {\size {x - x} } {x^2}\) Definition of $\delta$
\(\ds \) \(=\) \(\ds 0\) as $\size {x - x} = 0$

So Metric Space Axiom $(\text M 1)$ holds for $\delta$.

$\Box$


Proof of Metric Space Axiom $(\text M 2)$: Triangle Inequality

\(\ds \map \delta {x, y} + \map \delta {y, z}\) \(=\) \(\ds \frac {\size {x - y} } {x y} + \dfrac {\size {y - z} } {y z}\) Definition of $\delta$
\(\ds \) \(=\) \(\ds \frac {z \size {x - y} + x \size {y - z} } {x y z}\) Sum of Quotients of Real Numbers
\(\ds \) \(=\) \(\ds \frac {\size {x z - y z} + \size {x y - x z} } {x y z}\) Valid, as $x, z > 0$
\(\ds \) \(\ge\) \(\ds \frac {\size {x z - y z + x y - x z} } {x y z}\) Triangle Inequality
\(\ds \) \(=\) \(\ds \frac {\size {x y - y z} } {x y z}\) simplifying
\(\ds \) \(=\) \(\ds \frac {\size {x - z} } {x z}\) simplifying further
\(\ds \) \(=\) \(\ds \map \delta {x, z}\) Definition of $\delta$

So Metric Space Axiom $(\text M 2)$: Triangle Inequality holds for $\delta$.

$\Box$


Proof of Metric Space Axiom $(\text M 3)$

\(\ds \map \delta {x, y}\) \(=\) \(\ds \frac {\size {x - y} } {x y}\) Definition of $\delta$
\(\ds \) \(=\) \(\ds \frac {\size {y - x} } {y x}\) Definition of Absolute Value and Real Multiplication is Commutative
\(\ds \) \(=\) \(\ds \map \delta {y, x}\) Definition of $\delta$

So Metric Space Axiom $(\text M 3)$ holds for $\delta$.

$\Box$


Proof of Metric Space Axiom $(\text M 4)$

\(\ds x\) \(\ne\) \(\ds y\)
\(\ds \leadsto \ \ \) \(\ds \size {x - y}\) \(>\) \(\ds 0\) Definition of Absolute Value
\(\ds \leadsto \ \ \) \(\ds \frac {\size {x - y} } {x y}\) \(>\) \(\ds 0\) as $x y > 0$
\(\ds \leadsto \ \ \) \(\ds \map \delta {x, y}\) \(>\) \(\ds 0\) Definition of $\delta$

So Metric Space Axiom $(\text M 4)$ holds for $\delta$.

$\blacksquare$