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  "body": "\n## Definition\n\nLet $p$ be a [[prime number|01826KMG2H]].\nFor a nonzero rational number $x$, let $v_p(x)$ be its [[p-adic valuation|03HDF50VN4]], the integer $k$ with $x = p^k a / b$ for integers $a, b$ not divisible by $p$.\nThe *$p$-adic absolute value* is the map $\\abs{\\cdot}_p \\colon \\Q \\to \\R$ given by\n$$\\abs{x}_p = p^{-v_p(x)} \\text{ for } x \\neq 0, \\qquad \\abs{0}_p = 0.$$\n\n## Remarks\n\nIt is an [[absolute value|0FFXKTKKQC]] on $\\Q$ in the sense of fields:\n\n1. $\\abs{x}_p = 0$ exactly when $x = 0$;\n2. $\\abs{xy}_p = \\abs{x}_p \\abs{y}_p$;\n3. $\\abs{x + y}_p \\le \\max(\\abs{x}_p, \\abs{y}_p)$, which is stronger than the triangle inequality.\n\nExample: $\\abs{12}_2 = 2^{-2} = 1/4$ and $\\abs{3/8}_2 = 2^{3} = 8$.\nNon-example: the map $x \\mapsto p^{v_p(x)}$, without the minus sign, is not an absolute value.\nIndeed, it gives $1$ to both $p^2 - 1$ and $1$, but $p^2$ to their sum $p^2$, so the triangle inequality fails.\nProduct formula: for every nonzero rational $x$, $\\abs{x} \\prod_p \\abs{x}_p = 1$, where $\\abs{x}$ is the usual absolute value.\n",
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