How to prove that every rigid motion $F:\,\mathbb{R}\to \mathbb{R}$ is 1-to-1?

Question

A function $F:\,\mathbb{R}\to \mathbb{R}$ is a rigid motion if for all $x,y\in\mathbb{R}$ with $x\neq y$, $\vert x-y\vert = \vert F(x)-F(y)\vert$.

Using this definition of rigid motion, prove that every rigid motion is 1-to-1.

Any hint is highly appreciated.

What is there to prove? If $x\neq y$, then $|x-y|\neq0$ and therefore $\bigl|F(x)-F(y)\bigr|\neq0$. — José Carlos Santos, Jun 06 '18 at 14:21
No, that's not the definition of injectivity. What you wrote is what being a function means. — José Carlos Santos, Jun 06 '18 at 14:30
Ok, I got what you mean. I've updated the question. Hope it's clearer now. — utobi, Jun 06 '18 at 14:32

score 1 · Accepted Answer · answered Jun 06 '18 at 14:28

1

Hint:

To show that $F$ is 1-to-1, you need to show that if $F(x)=F(y)$ then $x=y$.

answered Jun 06 '18 at 14:28

Thanks, I know the definition. But I don't know how to apply it in this case. – utobi Jun 06 '18 at 14:30
1

@utobi Ok, so first you have to assume that $F(x)=F(y)$, what can you say about $x,y$ then? Look at the definition of your function. – cansomeonehelpmeout Jun 06 '18 at 14:31

1 Answers1