Proof about isometries Suppose $f\colon\mathbb R\to\mathbb R$ is an isometry of the reals.  Prove $f$ is a non-trivial translation iff $f$ has no fixed points.
Assumption: $f$ is a non-trivial translation (trivial translation: $f$ s.t. $f(x)=x$ for all $x \in\mathbb  R$).
Conclusion: $f$ has no fixed points (definition of fixed point: $f(x)=x$ for all $x\in \mathbb R$).
I'm not sure how to use this information to get the proof started.

$f$ is an isometry so for all $x,y\in\Bbb R$ we have
$$|f(x)-f(y)|=|x-y|$$
so for $y=0$ we have
$$|f(x)-f(0)|=|x|$$
hence
$$f(x)=\pm x+a,\quad\forall x\in\Bbb R$$
and since $f$ hasn't a fixed point then the only possibility is
$$f(x)=x+a,\quad a\ne0$$
