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  "body": "\n## Statement\n\nLet $n \\ge 2$ and let $f \\colon \\R^n \\to \\R^n$ be any map.\nAssume that for all $x, y \\in \\R^n$ with $\\norm{x - y} = 1$ one has $\\norm{f(x) - f(y)} = 1$.\nThen $f$ is an isometry: $\\norm{f(x) - f(y)} = \\norm{x - y}$ for all $x, y \\in \\R^n$.\n\n## Remarks\n\nNo continuity, injectivity or surjectivity of $f$ is assumed; preserving the single distance $1$ is enough.\nThe theorem was published by F. S. Beckman and D. A. Quarles in 1953.\nThe proof uses rigid configurations made of unit segments, such as unit equilateral triangles and simplices.\nThey show that $f$ also preserves other distances, and then that it preserves all distances.\nSince an isometry of [[Euclidean space|0XPR1E85W0]] is affine, $f$ is then of the form $x \\mapsto Ax + b$ with $A$ orthogonal.\n\nThe hypothesis $n \\ge 2$ is necessary.\nOn $\\R$, let $h \\colon [0, 1) \\to [0, 1)$ be any bijection, for instance $h(t) = t^2$, and set $f(x) = \\floor{x} + h(x - \\floor{x})$.\nIf $\\abs{x - y} = 1$, then $x$ and $y$ have the same fractional part and their integer parts differ by $1$, so $\\abs{f(x) - f(y)} = 1$.\nBut $f(0) = 0$ and $f(1/2) = 1/4$, so $f$ is not an [[isometry|02HPP5VCZH]].\n\n## Explanations\n\n### L1 human\n\nSuppose a map of the plane keeps every pair of points at distance $1$ at distance $1$.\nNothing is said about other distances, and the map need not even be continuous.\nThe theorem says that the map must then keep all distances, so it is just a rigid motion.\nIn the plane, unit triangles and other rigid unit frameworks lock everything in place.\n\n### L2 human\n\nThe theorem is a rigidity result in [[Euclidean distance|0NPPGHQNXQ]] geometry: one preserved distance forces the whole metric to be preserved.\nIt contrasts with the classical characterization of isometries, which assumes that all distances are preserved.\nIt is connected with unit distance graphs: a unit-distance-preserving map is a graph homomorphism of the unit distance graph of $\\R^n$ to itself.\nThe one-dimensional counterexample shows that the plane's richer supply of unit triangles is what makes the result work.\n",
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