How to prove that $$g:\Bbb R^3 \to \Bbb R^3 \in G = \{g \, | \, \text{ for each } g \text{ exists } n\in\mathbb{Z} : r(g(x),g(y)=2^n r(x,y) \}$$ for each $x,y\in \Bbb R^3, r$ is an euclidean distance) is a bijection? Have no idea.
I have the task to prove that $G$ is a group or isn't. Here is the task : $r$ is a euclidean metric of space $L = \Bbb R^3$. Does $G$ - multiplicity of transformations of $L$ (for each $g\in G$ exist $n\in\mathbb Z : r(g(x),g(y))=2^n r(x,y)$ for each $x,y\in L$) form a group?
My teacher said that first of all it's necessary to prove if it is a bijection or not, and if it is closed or not.
Let $g \in G$. First of all, $g$ is injective, since if $g(x) = g(y)$, then $$r(x,y) = 2^{-n}r(g(x),g(y)) = 0,$$ so $x = y$.
For surjectivity, define $f : \mathbb{R}^3 \to \mathbb{R}^3$ by $$f(x) = \frac{g(x)-g(0)}{2^n}.$$ To show that $g$ is surjective, it suffices to show that $f$ is. Now $f$ is constructed to satisfy $f(0) = 0$, and by homogeneity and translation invariance, $$\lVert f(x) -f(y) \rVert = r(f(x),f(y)) = 2^{-n}r(g(x),g(y)) = r(x,y) = \lVert x -y\rVert,$$ and so the claim follows by this answer to a similar question.