$\def\prop#1#2#3{#1:#2:#3}$Let's say I have a proportion, $\prop 135$
And then a set of other ones, $$ \begin{array}{l} \prop 235\\ \prop 145 \\ \prop 136 \end{array} $$
Which one is closer to $\prop 135$?
How to define it closer with mathematics?
(If you think it needs editing, edit away or comment)
This depends on which metric you define. Possible metrics would, for example, result in converting the $a:b:c$ to a vector $$\left( \begin{matrix}a\\b\\c\end{matrix} \right)$$ And using some standard $\mathbb R^3$ metrics, like $$d(x,y) := \Vert x-y \Vert_p$$ For $1\leq p\leq \infty$. The results vary with the choice of $p$. For $p=\infty$, for example we have with enumeration $y_1, y_2, y_3$ of your alternatives and $x$ the first: $$\begin{align*} d(x,y_1) & = 1 \\ d(x,y_2) & = 1 \\ d(x,y_3) & = 1 \end{align*}$$ so they are all "equally close".
However, any of these $p$ will yield a distance of $1$, since the changes are always to add one to one "component".
Another choice for converting $a:b:c$ to a vector containing "proportions", would be $$\vec{x} = \left(\begin{matrix}a/b\\b/c\\c/a\end{matrix}\right)$$
With the second option and $p=2$ we get, according to MATLAB with the code
x = [1 3 5];Y = [x;x;x] + eye(3);
xx = x./x([2 3 1]);
YY = Y./Y(:,[2 3 1]);
Z = [norm(xx - YY(1,:)); norm(xx - YY(2,:)); norm(xx - YY(3,:))]
we have $$Z \approx \left(\begin{matrix}2.5221\\0.2167\\1.0050\end{matrix}\right)$$ So $y_2$ is "closest" in this interpretation.