I'm having understanding the maths part of a certain chemistry concept. There's a formula that states:
$$\frac{A}{B}\sim v^{-3}.$$
From this I can say that as v gets bigger, the right side gets smaller, and because the two sides are proportional, the left side gets smaller as well. This happens because B gets bigger and A gets smaller.
Is all this correct?
On
The statement $$\frac{A}{B}\propto v^{-3}$$ means that for some non-identically-zero function $k$ of variables that don't have another dependence on $A,B$ or $v,$ $$\frac AB=\frac k{v^3}.$$
as $v$ gets bigger.... $B$ gets bigger and $A$ gets smaller.
Not necessarily. Suppose that $k=1$ and that $(A,B,v)$ changes from $(1,1,1)$ to $\left(\frac1{16},\frac12,2\right);$ as $v$ gets bigger, both $A$ and $B$ get smaller.
as $v$ gets bigger, because the two sides are proportional, $\dfrac AB$ gets smaller.
Not necessarily. Suppose that $k=-1$ and that $(A,B,v)$ changes from $(1,8,-2)$ to $(1,1,-1);$ as $v$ gets bigger, $\dfrac AB$ gets bigger.
The first concept is correct: as $v$ increases, the RHS decreases and thus the LHS must decrease as well.
However, how that decrease happens is not quite what you describe. Either $A$ must decrease or $B$ must increase. Either of those is sufficient.