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.

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2 Answers2


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

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Hagen von Eitzen
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The measure is indeed what you think. The simple continued fraction has "digits" which are all 1's. See, for example, Khinchin's little book on continued fractions.

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 $1/q_n^2.$

Will Jagy
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