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.
Best Answer
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 $a\in\mathbb Q, b\in\mathbb Q^\times$.