Show every dilation is a non-constant linear function. 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 $x,y\in\Re$.


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

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

 A: 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.


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*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.


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*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.
