Dilation proof verification I am wanting to verify the proof below; can someone please tell me if they agree with the way I have argued this or if I have made any incorrect assumptions (and where they are).  Thanks!! 
A dilation of reals is a function $f:R\to R$ such that for some constant $c\ne 0$ one has
$$|f(x)-f(y)|=c\cdot |x-y|$$
for all $x,y\in R$.


*

*Show that every non-constant linear function is a dilation.

*Show that every dilation is a non-constant linear function.



$1.$ So if we have two functions that are non-constant $f(x)$ and $f(y)$, then $f(x)=c\cdot x$ and $f(y)=c\cdot y$.
If we subtract the two functions it gives us a new non-constant linear function:
$$f(x,y) = |f(x)-f(y)|=|c\cdot x| - |c\cdot y|$$
This can be reduced to the formula: 
$$f(x,y) = |f(x)-f(y)|=c\cdot |x-y|$$
which is the formula of a dilation of reals. Thus we have shown a non-constant linear function is a dilation.
$2.$ Now this can be reversed to show that every dilation is a non-constant linear function.
The formula of a dilation is shown as:
$$f(x,y) = |f(x)-f(y)|=c\cdot |x-y|$$
If you distribute the c it gives us the formula: 
$$f(x,y) = |f(x)-f(y)|=|c\cdot x - c\cdot y|$$
Then split the problem into $f(x)$ and $f(y)$.
Then we get: $f(x)=c\cdot x$ and $f(y)=c\cdot y$, which are both non-constant functions, based on the definition of a constant function.
Thus we have shown that every dilation is a non-constant linear function.
 A: For the first problem, it is not in general true that $|a-b|=|a|-|b|$. Watch out for that. Suppose $f(x)=ax+b$ for some $a\neq 0$ and $b$. Then we will have that $$|f(x)-f(y)|=|ax+b-(ay+b)|=|(ax-ay)+(b-b)|=|a(x-y)|=|a|\cdot|x-y|$$
Thus $f$ is a dilation with dilation constant $|a|$.
Your work on the 2nd problem needs work too. You cannot jump from $|f(x)-f(y)|=|cx-cy|$ to conclude that $f(x)=cx$ and $f(y)=cy$. For example, if $f(x)=cx+b$ and $f(y)=cy+b$ with $b\neq 0$.
Let $f(x)$ be a dilation of $\Bbb R$. It is clear its dilation constant $c$ is positive, for $|f(x)-f(y)|\geq 0$. Consider the function $g(x)=\frac{1}{c}\left(f(x)-f(0)\right)$. Then we have that $g(0)=0$ and for all $x$ we have that $$|g(x)|=|\frac{1}{c}\left(f(x)-f(0)\right)|=\frac{1}{c}|f(x)-f(0)|=\frac{1}{c}\cdot c|x-0|=|x|$$
From this, we can show that $g(x)=x$ or $g(x)=-x$. To go further however, we need to observe that $f$ is continuous, and thus so is $g$. The restriction that $|g(x)|=|x|$ for all $x$ implies that the only way $g$ could not be identically $x$ or $-x$ is if there at least two distinct points (both nonzero) with $g(x)=x$ and $g(y)=-y$. Can you see why this cause a problem with continuity?
Once we know that $g(x)=x$ or $g(x)=-x$, we will then know that $f(x)=\pm cx+f(0)$. And thus $f$ is affine (or linear).
