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