It is often said that the golden mean $\phi$ is the "most irrational" number. However what I've never seen is an actual measure of irrationality which would make such a statement meaningful in a quantitative way.
Now I thought about what such a measure would look like. Obviously it would need to have the following properties:
* The irrationality of rational numbers is always $0$.
* The irrationality of irrational numbers is always strictly positive.
* The irrationality of $\phi$ is the maximal attainable value.
Clearly if the third point doesn't hold, it could still be a proper irrationality measure, it would just measure another aspect if irrationality that the one maximised by $\phi$.
Now I thought about it and came up with the following candidate:
$$i(x) = \inf_{m,n\in\mathbb Z, n\ne 0} \left|nx - m\right|$$
Obviously this fulfils the first objective: If $x=p/q$, then for $m=p, n=q$ the expression $nx-m$ gets $0$, and thus $i(x)=0$.
However I have no idea how to check the second objective, let alone the third. Nor have I any idea how one might actually calculate $i(x)$.
Now obviously each single absolute value would be larger than $0$, but then, an infimum of positive values may still be $0$, e.g. $\inf_{n\in\mathbb Z_{>0}} 1/n = 0$.
Indeed, I can't even tell if $i(x)$ will be nonzero for *any* irrational number.
Therefore my questions:
1. Is $i(x)$ a valid irrationality measure (i.e. does it give a strictly positive result for all irrational numbers)?
2. Does it give $\phi$ as maximally irrational, i.e. is $i(x)\le i(\phi)$ for all $x\in\mathbb R$?
3. Is there a way to actually calculate the value for at least some irrational numbers (in particular, $\phi$?)
celtschk
·
2025-10-19T12:26:59Z (11 months ago)