Multiplicative infinitesimals in q-analogs? Risking to be downvoted, here is a very lightweight question.
In various fields - say, algebraic geometry, nonstandard analysis, synthetic differential geometry - infinitely small quantities, i. e. those very-very close to zero, are represented by nilpotents, sometimes just square zero elements suffice to do quite a big portion of analysis. To give just one example, a generic $\varepsilon$ with $\varepsilon^2=0$ is used to represent things like tangent vectors, etc.
Now thinking about $q$-analogs, like $q$-derivative and similar gadgets, I am wondering what is the multiplicative analog of the above? That is, is there a way to capture algebraically quantities very-very close to 1? Or concisely -

"$\varepsilon$ very-very close to zero" is to "$\varepsilon^2=0$"
as
"$q$ very-very close to 1" is to what?

 A: There is more than one question that is being asked here so I will leave aside the one about $q$-analogues for the simple reason that one can take any question, say $X$, in mathematics, and ask for its $q$-analog, $X_q$, so things can get pretty monotonous.
As far as the multiplicative version of being "very small" is concerned, this can be expressed in Smooth Infinitesimal Analysis and algebraic geometry in terms of nilsquare infinitesimals, but it also has a straightforward meaning in the hyperreals as "being smaller than any positive real", so that one defines the additively invariant relation $x\approx y$ by requiring $x-y$ to be infinitesimal. Meanwhile, Fermat and Leibniz had a (multiplicatively invariant) relation that had more of a multiplicative character, which we will denote $x\;{}_{\ulcorner\!\urcorner}\; y$, which holds if and only if $\frac{x}{y}\approx 1$. The reason this is closer to Fermat's ideas is because Fermat would divide his adequalities at will by his $E$ which is of course impossible for the additively invariant relation.
