P-adic Norm and Absolute Value are Not Equivalent/Proof 2

Theorem

Let $\norm {\,\cdot\,}_p$ be the $p$-adic norm on the rationals $\Q$ for some prime number $p$.

Let $\size {\,\cdot\,}$ be the absolute value on the rationals $\Q$.

Then $\norm {\,\cdot\,}_p$ and $\size {\,\cdot\,}$ are not equivalent norms.

That is, the topology induced by $\norm {\,\cdot\,}_p$ does not equal the topology induced by $\size {\,\cdot\,}$.

Proof

It is noted that:

$\sup \set {\size n: n \in \Z} = +\infty$

By a corollary of Characterisation of Non-Archimedean Division Ring Norms then $\size {\,\cdot\,}$ is Archimedean.

From P-adic Norm on Rational Numbers is Non-Archimedean Norm then $\norm {\,\cdot\,}_p$ is non-Archimedean.

By Equivalent Norms are both Non-Archimedean or both Archimedean, $\norm {\,\cdot\,}_p$ and $\size {\,\cdot\,}$ are not equivalent norms.

$\blacksquare$

Sources

• 1997: Fernando Q. Gouvea: p-adic Numbers: An Introduction: $\S 3.1.2$ Absolute Values on $\Q$, Problem 68