# Algebraic properties of fundamental physical constants

A few weeks ago I started wondering if there's any reason for a given fundamental physical constant, like Boltzmann's one, to enjoy some algebraic properties, for example being rational/irrational or algebraic/transcendental, in a given system of units, of course. Note that, once every fundamental constant is a real number, every fundamental constant is algebraic over $\mathbb{R}[x]$: you can take the polynomial $x-a\in\mathbb{R}[x]$, where $a$ is the given fundamental constant. So the algebraic/transcendental mentioned before is over $\mathbb{Q}[x]$.

I know that, given a fundamental constant $a$ you can consider a system of units in which $a=1$. But the question is, given a system of units, MKS for instance, what can we say about the fundamental constants at that system of units: are they rational numbers? If not, are they algebraic numbers (over $\mathbb{Q}[x]$)?

I know that maybe this question is of no importance to physics (nature doesn't care if Planck's constant is rational or not), but I was just curious. If you guys don't think so, feel free to explain why.

• Physics does not make a difference between algebraic and transcendental numbers because any two such numbers which differ only by an amount that is smaller than the measurement errors of the best experiments determining them are in the equivalence class of all numbers that satisfy the hypothesis that they parametrize. $1$ and $1 + 10^{-20}*\pi$ are identical for physicists in all scenarios I am aware of. – CuriousOne Jun 12 '15 at 6:00

• it would be great if the fine structure constant turns out to be $1/137$ for some strange theoretical reason. (it's an adimensional constant) – Héctor Jun 12 '15 at 13:27