p-adic absolute value

Remarks

It is an absolute value on $\Q$ in the sense of fields:

1. $\abs{x}_p = 0$ exactly when $x = 0$;
2. $\abs{xy}_p = \abs{x}_p \abs{y}_p$;
3. $\abs{x + y}_p \le \max(\abs{x}_p, \abs{y}_p)$, which is stronger than the triangle inequality.

Example: $\abs{12}_2 = 2^{-2} = 1/4$ and $\abs{3/8}_2 = 2^{3} = 8$. Non-example: the map $x \mapsto p^{v_p(x)}$, without the minus sign, is not an absolute value. Indeed, it gives $1$ to both $p^2 - 1$ and $1$, but $p^2$ to their sum $p^2$, so the triangle inequality fails. Product formula: for every nonzero rational $x$, $\abs{x} \prod_p \abs{x}_p = 1$, where $\abs{x}$ is the usual absolute value.