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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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    The step leading to $f'(x)=1$ for every $x$ is not justified (and the result is wrong).2011-06-23
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    @Sergey: Many problems can be solved by reflecting.2011-06-23
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    @Didier: i have corrected it to +/-. Is it ok now?2011-06-23
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    @user6312: which reflecting? Geometrical one? I have updated my solution, please take a look.2011-06-23
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    @Sergey: Both. I could not resist the pun.2011-06-23
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    Now, how do you prove that $f'(x)$ exists?2011-06-23
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    $f'(x) = \lim_{\delta \rightarrow 0}{\frac{f(x+\delta)-f(x)}{(x + \delta) - x}} = \pm 1 $ because $ |(x+\delta)-x| = |f(x+\delta) - f(x)| $. Am I right ?2011-06-23
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    Even if it's much more than you're asking here, it is true that a *surjective* isometry $f: E \to F$ between normed vector spaces is affine. This is called the [Mazur-Ulam theorem](http://www.helsinki.fi/~jvaisala/mazurulam.pdf).2011-06-23
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    This is not at all enlightening, but it agrees with the other answers, which is a good sign: O(1,R)={1,-1}, where O(1,R) is the group of 1x1 orthogonal matrices.2011-06-23
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    Or, this other one: isometries in spaces whose norms are generated by inner-products (Hilbert Spaces) are maps that preserve inner-product. In $R^1$, the inner-product is standard multiplication. So we want to find all f with: a.b=f(a).f(b). So we start with a=b=1, so 1=$f(1)^2$, so f(1)=+/-1. It follows easily then that f(n)=n.f(1), so that f(n)=+/-n.2011-06-23
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    @Theo Buehler: thank you for your reply, i didn't know about Mazur-Ulam theorem before.2011-06-24
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    @gary: another elegant proof. Thanks.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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    thank you for your reply. Hint: your text will become more readable if you will use `$`-sings with EACH formula and use \rightarrow for '-->' and \cdot for '.', for example use `$` f: \mathbb{R} \rightarrow \mathbb{R} `$` for $ f: \mathbb{R} \rightarrow \mathbb{R} $2011-06-24
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    as a rule: enclose formulas inside TeX entirely, that is: write `$a^2 = b^2 = c^2$` to get $a^2 = b^2 = c^2$ *not* `$a^2$ = $b^2$ = $c^2$`.2011-06-24
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    @Theo: Thanks. It would be nice to get input on the reply, whether good or bad, so I can have some insights into it.2011-06-24
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    It's nice and clear, I think. However, you don't quite answer the question, as you're only considering functions that satisfy $\langle f(a), f(b) \rangle = ab$. What you really should do is considering functions that satisfy $\|a - b\| = \|f(a) - f(b)\|$.2011-06-24
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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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    @Sergey: I did the problem in gruesome detail, but the geometric idea is simple. Once you know where $0$ and $1$ go, everything is determined. The "tricks" $g(x)$, $h(x)$ are actually very geometric.2011-06-23
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    thank you very much. Your solution is pure geometric and enlightening for me.2011-06-23