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  "body": "\n## Definition\n\nLet $X$ be a set and let $\\langle \\cdot, \\cdot, \\cdot \\rangle \\colon X \\times X \\times X \\to X$ be a ternary operation.\nThe pair $(X, \\langle \\cdot, \\cdot, \\cdot \\rangle)$ is a *median algebra* if for all $x, y, z, w \\in X$:\n\n1. $\\langle x, y, y \\rangle = y$;\n2. $\\langle x, y, z \\rangle = \\langle z, x, y \\rangle$;\n3. $\\langle x, y, z \\rangle = \\langle x, z, y \\rangle$;\n4. $\\langle \\langle x, w, y \\rangle, w, z \\rangle = \\langle x, w, \\langle y, w, z \\rangle \\rangle$.\n\n## Remarks\n\nClauses 2 and 3 together say that $\\langle x, y, z \\rangle$ does not change under any permutation of $x, y, z$.\nClause 4 says that for fixed $w$, the binary operation $(x, y) \\mapsto \\langle x, w, y \\rangle$ is associative.\n\nExample: $X = \\R$ with $\\langle x, y, z \\rangle$ the median of $x, y, z$, which equals $\\max(\\min(x, y), \\min(y, z), \\min(z, x))$.\nMore generally, every [[distributive lattice|0HM0HQCK6N]] is a median algebra with $\\langle x, y, z \\rangle = (x \\wedge y) \\vee (y \\wedge z) \\vee (z \\wedge x)$.\nFor clause 1, $(x \\wedge y) \\vee y \\vee (y \\wedge x) = y$.\n\nNon-example: a set $X$ with at least two elements and $\\langle x, y, z \\rangle = x$.\nClause 1 fails: for $x \\ne y$, $\\langle x, y, y \\rangle = x \\ne y$.\n",
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