Suppose that the real valued function $g$ is defined by $g(x) = f(kx)$ with number $k\gt 0$.
As it is well known, in that case, the graph of $g$ can be obtained by streching or shrinking the graph of $f$ horizontally - depending on whether $k\gt 1$ ( in which case the horizontal transformation results in a stretch) or $k\in (0,1)$ ( in which case the horizontal transformation results in a shrink).
If function $f$ and $g$ are considered as sets of ordered pairs, one can express the same fact by saying that :
$$g = \{ (\frac xk , y) | (x,y) \in f\}$$.
( which expresses the fact that the transformation leaves the $y$-value unchanged : " Horizontal stretches and shrinks, respectively, horizontally pull the base graph, or push it together, while leaving the y-intercept unchanged to anchor the graph" - Biomath, Biology project, Uni. Arizona http://www.biology.arizona.edu/biomath/tutorials/Transformations/HorizontalStretchesShrinks.html)
- My question is : how to prove this in a purely algebraic way?
In other words , how to go algebraically from
$$g = \{ (x , y) | y= f(kx)\} , k\gt 0$$ .
or, equivalently , from :
$$g(x)= f(kx), k\gt 0$$.
to
$$g = \{ (\frac xk , y) | (x,y) \in f\}$$.
Note : there are good intuitive explanations of this fact on MSE, but what I am looking for is an algebraic one.
Note : I would be specially interested in an answer using the set builder notation, that is , starting from :
$$g = \{ (x , y) | y= f(kx)\} , k\gt 0$$ .
By definition of $g$, for all $x$ you have $g(\tfrac xk)=f(x)$ and hence $$g(\tfrac xk)=y\qquad\Leftrightarrow\qquad f(x)=y,$$ and therefore $$\{(x,y):\ y=g(x)\}=\{(\tfrac xk,y):\ y=g(\tfrac xk)\}=\{(\tfrac xk,y):\ y=f(x)\}.$$