P-adic Norm and Absolute Value are Not Equivalent/Proof 1

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

By definition of the $p$-adic norm:

$\norm p_p = \frac 1 p < 1$

By definition of the absolute value:

$\size p = p > 1$

By definition of open unit ball equivalence, $\norm {\,\cdot\,}_p$ and $\size {\,\cdot\,}$ are not equivalent norms.

By Equivalence of Definitions of Equivalent Division Ring Norms and the definition of topologically equivalent norms then the topology induced by $\norm {\,\cdot\,}_p$ does not equal the topology induced by $\size{\,\cdot\,}$.

$\blacksquare$

Sources

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