Find all bijections $ \mathbb{E}^1 \rightarrow \mathbb{E}^1 $ that preserve euclidian metric My question is: how do we find all bijections $\mathbb{E}^1\to\mathbb{E}^1$ that preserve the Euclidean metric?
If we have a metric-preserving bijective mapping $ f: \mathbb{E}^1 \rightarrow \mathbb{E}^1 $ then $ \forall x,y \in \mathbb{E}^1 \ |x-y| = |f(x) - f(y)| \Rightarrow f'(x)=\lim_{\delta \rightarrow 0}{\frac{f(x+\delta)-f(x)}{\delta}} = \pm 1 \ \forall x \in \mathbb{E}^1 $.  
It means that $f(x)$ is a shift by a constant (possibly with a reflection): $ f(x) = \pm x + c, \ c \in \mathbb{E}^1 $, because $ ( f(x) - (\pm x+a) )'=0 , \ a \in \mathbb{E}^1$.  
Is my solution correct? And could you propose a more geometric solution?
 A: More generally, any isometry of ${\mathbb R}^n$ to itself is an affine transformation.  
Suppose $f$ is such an isometry.  For any $x$ and $y$ in ${\mathbb R}^n$, and $0 < t < 1$, $t x + (1-t) y$ is the only point $p$ with $d(p,x) = (1-t) d(x,y)$ and $d(p,y) = t d(x,y)$, so we must have $f(t x + (1-t) y) = t f(x) + (1-t) f(y)$.  Extend that to all real $t$: e.g. if $t > 1$ and  $z = t x + (1-t) y$, then $x = (1/t) z + (1 - 1/t) y$.  Taking $y = 0$, we have $f(tx) = t f(x) + (1-t) f(0)$.  If $g(x) = f(x) - f(0)$, then $g(t x) = t g(x)$ and $g(a x + b y) = (g(2ax) + g(2by))/2 = a g(x) + b g(y)$, i.e. $g$ is linear and $f$ is affine. 
A: The proposed solution works, or can be made to work.  There is a bit of a problem in that the reasoning is not explained fully.  For example, the derivative is used in the argument. But it is conceivable that $f'$ does not exist, at least for some $x$.  
However, to me the main issue is that there are too many symbols, and too little geometry.  We are dealing with a very concrete problem, and a more concrete solution, if achievable, is better.  
So let us think about this mapping $f$.  Suppose that $f$ takes $0$ to $a$.   Let $g(x)=f(x)-a$.  Then $g(0)=0$, and $g$ is distance-preserving.   
We will show that $g(x)=x$ or $g(x)=-x$, from which it will follow that $f(x)=x+a$ or $f(x)=-x+a$.
Look at $g(1)$.  Because $g$ is distance-preserving, we have $g(1)=1$ or $g(1)=-1$.  We deal first with the case $g(1)=1$.
Case $g(1)=1$: Suppose that $g(1)=1$.  Let $x$ be any number other than $0$.  We show that $g(x)=x$.  This is clear, there is only one point at distance $|x|$ from $0$ and simultaneously at distance $|x-1|$ from $1$, and this point is $x$. For a "formal" verification, let's show that $-x$ doesn't work.  How can we have $|(-x)-1|=|x-1|$? We need either $-x-1=x-1$, which forces $x=0$, or $x+1=x-1$, which is impossible.
Case $g(1)=-1$: Let $h(x)=-g(x)$.  Then $h(1)=1$.  Since $h$ is distance-preserving, we have $h(x)=x$ for all $x$, and hence $g(x)=-x$.  
A: Another answer, expanding on my comment above:
In a normed space where the norm is generated by an inner-product, the isometries are precisely the maps that preserve the inner-product. In the reals, with $\Vert a-b \Vert = |a-b|$; this is the norm generated by the inner-product $\langle a,b \rangle :=ab$. If we are given an inner-product, we can find the norm generated by that inner-product, but we can also go in the opposite direction; given a norm (and knowing it is generated by an inner-product), we can find the inner-product that gives rise to the norm: 
$\Vert a-b \Vert := (a-b)^{1/2}$, means that $\Vert \cdot \Vert$ is generated by the inner product: $\langle a,b\rangle := a.b$
 (standard multiplication).
We then want to find all maps $f:\mathbb{R} \to \mathbb{R}$ that preserve $a.b$, i.e., we want to find all functions f with $a.b=f(a)f(b)$. This is not too hard : using $a=b=1$, we find $1= f(a)^2$, so that $f(a)= \pm 1$. Similarly, we have $f(0)=0$. Once we know $f(1)$, we are done; $a=a.1=
f(a).f(1)$, so $f(a)=\pm a$.
So we want to find $f$ with $(a-b)^{1/2} =(f(a)-f(b))^{1/2}$  Since $\mathbb {R}$ is a Hilbert space,
 its norm is generated by an inner-product, which we see is standard multiplication; this means that the maps that preserve distance are precisely those that preserve multiplication, which are the maps $f(x)=x$ and $f(x)=-x$.  
