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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?

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    @gary: a$n$other elegant proof. Than$k$s.2011-06-24

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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 $f(x)=-x$.

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    Thanks for your suggestions; I will try to implement them. @Theo: I thought (maybe wrongly), that if the inner-product is preserved, then the same will be the case for the norm generated by the inner-product. I will double-check this argument, though.2011-06-26
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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.

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    To complement Robert's nice answer, I think it is worth pointing out that if $n\ge 2$ and $f:{\mathbb R}^n\to{\mathbb R}^n$ preserves distance 1, then it is an isometry. The earliest reference for this I know of is in the "Problem Book" of the *Students' Mathematical Society of the Jagiellonian University*, where it is credited to Sławomir Kołodziej.2011-06-23
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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 $g(x)=-x$.

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    thank you very $m$uch. Your solution is pure geo$m$etric and enlightening for me.2011-06-23