Norms Equivalent to Absolute Value on Rational Numbers/Necessary Condition

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Theorem

Let $\alpha \in \R_{> 0}$.

Let $\norm {\,\cdot\,}:\Q \to \R$ be the mapping defined by:

$\forall x \in \Q: \norm x = \size x^\alpha$

where $\size x$ is the absolute value of $x$ in $\Q$.

Let $\norm {\,\cdot\,}$ be a norm on $\Q$.


Then:

$\alpha \le 1$


Proof

The contrapositive is proved.

Let $\alpha > 1$.

The Norm Axiom $\text N 3$: Triangle Inequality is not satisfied:

\(\ds \norm {1 + 1}\) \(=\) \(\ds \size {1 + 1}^\alpha\)
\(\ds \) \(=\) \(\ds 2^\alpha\)
\(\ds \) \(>\) \(\ds 2\) Power Function on Base Greater than One is Strictly Increasing
\(\ds \) \(=\) \(\ds \size 1^\alpha + \size 1^\alpha\)
\(\ds \) \(=\) \(\ds \norm 1 + \norm 1\)

By Rule of Transposition the result follows.

$\blacksquare$


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