# Why is $\varphi$ called "the most irrational number"?

I have heard $\varphi$ called the most irrational number. Numbers are either irrational or not though, one cannot be more "irrational" in the sense of a number that can not be represented as a ratio of integers. What is meant by most irrational? Define what we mean by saying one number is more irrational than another, and then prove that there is no $x$ such that $x$ is more irrational than $\varphi$.

Note: I have heard about defining irrationality by how well the number can be approximated by rational ones, but that would need to formalized.

How well can a number $$\alpha$$ be approximated by rationals? Trivially, we can find infinitely many $$\frac pq$$ with $$|\alpha -\frac pq|<\frac 1q$$, so something better is needed to talk about a good approximation. For example, if $$d>1$$, $$c>0$$ and there are infinitely many $$\frac pq$$ with $$|\alpha-\frac pq|<\frac c{q^d}$$, then we can say that $$\alpha$$ can be approximated better than another number if it allows a higher $$d$$ than that other number. Or for equal values of $$d$$, if it allows a smaller $$c$$.
Intriguingly, numbers that can be approximated exceptionally well by rationals are transcendental (and at the other end of the spectrum, rationals can be approximated exceptionally poorly - if one ignores the exact approximation by the number itself). On the other hand, for every irrational $$\alpha$$, there exists $$c>0$$ so that for infinitely many rationals $$\frac pq$$ we have $$|\alpha-\frac pq|<\frac c{q^2}$$. The infimum of allowed $$c$$ may differ among irrationals and it turns out that it depends on the continued fraction expansion of $$\alpha$$. Especially, terms $$\ge 2$$ in the continued fraction correspond to better approximations than those for terms $$=1$$. Therefore, any number with infinitely many terms $$\ge 2$$ allows a smaller $$c$$ than a number with only finitely many terms $$\ge2$$ in the continued fraction. But if all but finitely many of the terms are $$1$$, then $$\alpha$$ is simply a rational transform of $$\phi$$, i.e. $$\alpha=a+b\phi$$ with $$a\in\mathbb Q, b\in\mathbb Q^\times$$.
Simple version, the convergent just before a large "digit" is a very good approximation, the relevant error being less than $$\frac{1}{q_n q_{n+1}}$$ where the $q$'s are the denominators. So, with a big digit, $q_n$ is of modest size but $q_{n+1}$ is quite large, so the error with denominator $q_n$ is small compared with $1/q_n^2.$