Show every dilation is a non-constant linear function.

Question

A dilation of reals is a function $f:\Re \mapsto \Re$ such that for some constant $c\neq0$ one has $|f(x)-f(y)|=c\ast|x-y|$ for all $x,y\in\Re$.

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

Answer 1

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 $x,y\in R$.

$Definition:$ A linear function is one where changing the input by any quantity changes the output by that quantity multiplied by a fixed number. Also, if a function is non-constant then, it has an end.

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

Using the definition of a linear function as described above, we can make a formula as shown: $f(x)=c\cdot x$

So if we have two functions that are non-constant [$f(x)$ and $f(y)$], $f(x)=c\cdot x$ 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. This shows us that every non constant linear function is a dilation.

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

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: $$f(x,y) = |f(x)-f(y)|=|c\cdot x| - |c\cdot y|$$

which is a non-constant linear function because it is the subtraction of two other non-constant linear functions. Which has shown us that every dilation is a non-constant linear function.