Norms Equivalent to Absolute Value on Rational Numbers/Sufficient Condition

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 $\Q$.

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

Proof
Suppose $\alpha \le 1$.

It is shown that $\norm {\,\cdot\,}$ satisfies the norm axioms $(\text N 1)$-$(\text N 3)$.

Let $x \in \Q$.

Let $x, y \in \Q$.

Then:

Let $x, y \in \Q$.

, let $\norm y < \norm x$.

If $\norm x = 0$ then $\norm y = 0$.

By above:
 * $x = y = 0$

Hence:

If $\norm x > 0$ then:
 * $\norm x > 0 \leadstoandfrom \size x^\alpha > 0 \leadstoandfrom \size x > 0$

Hence: