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, $\phi$?)