How do you prove that a bijection ##f:\mathbb R^n\to\mathbb R^n## such that ##d(f(x),f(y))=d(x,y)## for all ##x,y\in\mathbb R^n## is a composition of a rotation and a translation? (d=Euclidean metric).(adsbygoogle = window.adsbygoogle || []).push({});

My first thought is to define ##g=f-f(0)##, prove that g is a bijection that preserves distances too, and then try to prove that g is a rotation. I recently learned how to prove that maps that take straight lines to straight lines are linear, so I'm guessing that the strategy involves proving that g takes straight lines to straight lines.

Maybe we can use that g preserves circles around 0 in the sense that if C is such a circle and ##x\in C##, then ##g(x)\in C##. I imagine that we would take an arbitrary line L, and then look at points where it intersects some circle around 0, but I don't see how to turn this idea into a proof. I'm not sure if I should try to prove that g takes straight lines to straight lines first, or if there's a more direct approach.

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# Distance-preserving maps are compositions of rotations and translations

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