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

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


Without loss of generality 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:

$\norm n < \norm m \le 1$

$\blacksquare$


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