P-adic Norms are Not Equivalent

Theorem
Let $p_1$ and $p_2$ be prime numbers such that $p_1 \neq p_2$.

Let $\norm {\,\cdot\,}_{p_1}$ and $\norm {\,\cdot\,}_{p_2}$ be the $p$-adic norms on the rationals $\Q$.

Then $\norm {\,\cdot\,}_{p_1}$ and $\norm {\,\cdot\,}_{p_2}$ are not equivalent norms.

That is, the topology induced by $\norm {\,\cdot\,}_{p_1}$ does not equal the topology induced by $\norm {\,\cdot\,}_{p_2}$.

Proof
Consider $p_1/p_2 \in \Q$.

With $\norm {\,\cdot\,}_{p_1}$:

On the other hand, with $\norm {\,\cdot\,}_{p_2}$:

By open unit ball equivalence, $\norm {\,\cdot\,}_{p_1}$ and $\norm {\,\cdot\,}_{p_2}$ are not equivalent norms.

By Equivalence of Definitions of Equivalent Division Ring Norms and topological equivalence then the topology induced by $\norm {\,\cdot\,}_{p_1}$ does not equal the topology induced by $\norm {\,\cdot\,}_{p_2}$.