Norms Equivalent to Absolute Value on Rational Numbers

Theorem
Let $\alpha \in \R_{\gt 0}$.

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 $\Q$.

Then:
 * $\norm{\,\cdot\,}$ is a norm on $\Q$ $\,\,\alpha \le 1$

Necessary Condition
Suppose $\alpha \le 1$.

It is shown that $\norm{\,\cdot\,}$ satisfies the norm axioms (N1)-(N3).

(N1) Positive Definiteness
Let $x \in \Q$.

(N2) Multiplicativity
Let $x, y \in \Q$.

If $x = 0$ then $xy = 0$ and:

Similarly if $y = 0$.

If $x \neq 0$ and $y \neq 0$ then:

(N3) Triangle Inequality
Let $x, y \in \Q$.

let $\norm y \lt \norm x$.

Case 1
Let $\norm x = 0$.

Then $\norm y = 0$ and by (N1) above:
 * $x = y = 0$.

Hence:

Case 2
Let $\norm x \gt 0$.

Then:

Sufficient Condition
The contrapositive is proved.

Let $\alpha \gt 1$.

The norm axiom (N3) (Triangle Inequality) is not satisfied:

By Rule of Transposition the result follows.