We have been learning about isometries and how reflections, translations etc. and how they can affect a function. I was wondering if someone could help me with this proof by using the definition of isometries which I am struggling to understand.

If $f: \mathbb R\to \mathbb R$ is an isometry of the reals. How do I show that $f$ is a reflection in a point if and only if $f$ has a unique fixed point.

I must use definitions of isometries only, would that mean using other translations, reflections, and reflections? or how would I go about this.

I have been working through problems that practiced writing translations and reflections. For example, we have $g(x)=x+c$, $f(x)=2a-x$, and $h(x)=2b-x$, where $a$ and $b$ are the stationary points of $f$ and $h$ respectively, and $c$ is the amount we are translating by and the composition is $$ h(g(f(x))) = h(g(2a-x)) = h(2a-x+c) = 2b-(2a-x+c) = 2(b-a+\frac{1}{2}c)-x $$ so this is a reflection in the point $b-a+\frac{1}{2}c$.

But I am not quite sure this is the same work in this case.

  • $\begingroup$ Downvoted because no work shown $\endgroup$ – The_Sympathizer Jul 10 '15 at 2:55
  • $\begingroup$ You should show your work -- even if you could only get as far as the first couple of steps. What have you tried? $\endgroup$ – The_Sympathizer Jul 10 '15 at 2:57
  • $\begingroup$ Probably the reason for the downvotes and the vote to close the question is that this is phrased in language appropriate for assiging homework. That can make people suspect that you're copying a question without even understanding it. They should explain here in a comment what should be done, but usually they don't. $\endgroup$ – Michael Hardy Jul 10 '15 at 2:58
  • $\begingroup$ how do I delete this question? $\endgroup$ – user253595 Jul 10 '15 at 13:49
  • 2
    $\begingroup$ user253595: I think that the definition of isometry (that you are supposed to use) is the following. $f$ is an isometry, iff $|f(x)-f(y)|=|x-y|$ for all numbers $x,y$. If you have proven helpful results about isometries in class, then it is probably ok to use those results as well. $\endgroup$ – Jyrki Lahtonen Jul 10 '15 at 15:04

When tackling an "if and only if" proof, it will often seem that one direction is easier than the other.

Here we are asked to assume that $f(x)$ is an isometry, then show it is a reflection if and only if it has exactly one fixed point.

Which is the easier direction? It seems to me that proving that reflection of $\mathbb {R} $ in point $c $ has exactly one fixed point is very easy.

So let's focus on the more difficult implication, that if $f(x)$ is an isometry with one and only one fixed point, then it must be a reflection.

From already having the easy half done, we should expect that when $c $ is the unique fixed point, then the map $f:\mathbb {R}\to \mathbb {R} $ will turn out to be reflection in $c $. How can we prove this.

It helps to write down what we want to show, and sometimes it is just about clear after we do. Here the idea that $f $ is the reflection of real numbers in $c $ is expressed for all $x \in \mathbb {R} $:

$$ f(c+x) = c-x $$

Take a moment to think through why this identity expresses the notion of $f$ being a reflection in $c $.

Now try to argue, using the facts that $f $ is an isometry and that $c $ is the only fixed point, that this identity holds true for all real numbers $x $.

  • $\begingroup$ would i use the same process to show that any isometry of the reals is either a translation or a reflection in a point? Would i have to write a different form of f(x)? $\endgroup$ – user253595 Jul 10 '15 at 18:41
  • $\begingroup$ You could build on this approach. Consider an isometry of the real numbers that does not have exactly one fixed point. Either there will be more than one or less than one fixed point. Handling these cases as translations will involve some new ideas. $\endgroup$ – hardmath Jul 10 '15 at 20:07
  • $\begingroup$ Certainly a translation has a different form of $f(x)$, namely $f(x) = x + c$. After all the reflections and translations are distinct kinds of isometries. $\endgroup$ – hardmath Jul 10 '15 at 22:07

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