Ostrowski's Theorem/Non-Archimedean Norm/Lemma 2.1

Theorem
Let $\norm {\, \cdot \,}$ be a non-trivial non-Archimedean norm on the rational numbers $\Q$.

Then:
 * $\exists n \in \N: 0 < \norm n < 1$.

Proof
Because $\norm {\, \cdot \,}$ is non-trivial:
 * $\exists \dfrac a b \in \Q : 0 < \norm {\dfrac a b} \mbox { and } \norm {\dfrac a b} \ne 1$

By Norm of Inverse then:
 * $\norm {\dfrac a b} > 1 \implies \norm {\dfrac b a} < 1$

Hence either $\norm {\dfrac a b} < 1$ or $\norm {\dfrac b a} < 1$.

assume $\norm {\dfrac a b} < 1$.

By Norm of Quotient then:
 * $\dfrac {\norm a} {\norm b} < 1$

Hence:
 * ${\norm a} < \norm b$

Let $n = \size a$ and $m = \size b$ where $\size {\,\cdot\,}$ is the absolute value on $\Q$.

Then $n, m \in \N$

By Norm of Negative then:
 * $\norm n = \norm a$
 * $\norm m = \norm b$

Hence:
 * $\norm n < \norm m$

Since $\norm {\, \cdot \,}$ is non-Archimedean then:
 * $\norm m \le 1$

Hence:
 * $\norm n < \norm m \le 1$