I'm not even sure this question is meaningful because any irrational number can be approximated by a rational as close as you want. But given that there are infinitely many irrationals between any two rationals, surely some must be closer to a rational number than others?

Is there a metric by which irrationals can be ranked, or can we compare two irrationals, based on how far they are from a rational?

I'm having trouble wrapping my head around this, can anyone assist?

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    $\begingroup$ If you take the time to adequately define "close" the answer should be immediately apparent. You say "there are infinitely many irrationals between any two rationals" but seem to ignore that this goes both ways... between any two irrationals there are also infinitely many rationals between them too. The keyword here is "dense." $\endgroup$
    – JMoravitz
    Commented Dec 30, 2020 at 22:31
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    $\begingroup$ Yes, indeed! You can google "Liouville numbers", for example. There is a whole body of ideas and work coming out of this. If no one else writes a reasonable answer, I'll maybe do so later. $\endgroup$ Commented Dec 30, 2020 at 22:31
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    $\begingroup$ As others have noted, then the answer is no. But there is one amusing concept that has been used here before. It has been commented before that the golden ratio $\phi$ is the "most irrational number" because it has the slowest-converging continued fraction of any irrational number. I am not aware of a rigorous concept of distance that uses this concept (the rate of convergence of the continued fraction), but one might exist. Of course, this is not what anyone typically means when they use the word "close." $\endgroup$
    – sasquires
    Commented Dec 30, 2020 at 22:34
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    $\begingroup$ You might look up irrationality measure. $\endgroup$ Commented Dec 30, 2020 at 22:39
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    $\begingroup$ well @Tumbleweed53, the answer is certainly "no" for some (perhaps most) reasonable definitions of "close", though it can be made into "yes" by taking a different definition of "close", as sasquires does. the question needs to be made precise before it can be answered :) . it is the case that, for any irrational number $\alpha$ and any $\epsilon>0$, there is a rational number $\beta_\epsilon$ such that $|\alpha-\beta_\epsilon|<\epsilon$, so in this sense every irrational number is arbitrarily close to a rational one, and in particular no irrational is "closer" to a rational than any other $\endgroup$ Commented Dec 30, 2020 at 22:47

4 Answers 4


This notion can be formalized in terms of distance between sets. In a metric space $(M,d)$ with $X,Y \subseteq M$, we can formalize the Hausdorff distance in particular by

$$d_{\mathrm H}(X,Y) := \max\left\{\,\sup_{x \in X} \left( \inf_{y \in Y} d(x,y)\right), \sup_{y \in Y} \left( \inf_{x \in X} d(x,y)\right) \right\}$$

In particular, if $\overline X = \overline Y$, then $d_{\rm H} = 0.$ This is the case for $\Bbb Q$ and $\Bbb R \setminus \Bbb Q$, which have closure $\Bbb R$, and so the distance between the rationals and irrationals is zero. In particular what this, more intuitively, claims is that for any irrational $x$, you can pick a rational $y$ that is within any distance you want from $x$.

For instance, take $y_n$ to be $x$, up to $n$ places after the decimal, i.e.

$$y_n := \frac{ \lfloor 10^n x \rfloor}{10^n}$$

Then $y_n \in \Bbb Q$ $\forall n \in \Bbb N$ while $x \in \Bbb R \setminus \Bbb Q$, but $y_n \to x$ as $n \to \infty$. Thus, the sequence $\{y_n\}_{n \in \Bbb N}$ gets "arbitrarily close" to $x$ as $n$ grows, at least within the Euclidean distance.

Which touches on an important point -- the question is somewhat ill-defined, in the sense of "how do we say two points or two sets are close? how do we quantify closeness?" Absent of any other definition, we can see that the rationals and irrationals are a distance zero from each other, but only under a particular definition. The Hausdorff distance is a metric motivated by the simpler distance $d(x,Y)$ between a point and a set, and a distance $d(X,Y)$ between sets, defined by

\begin{align*} d(x,Y)&:=\inf \{ d(x,y) \mid y \in Y \}\\ d(X,Y)&:=\sup \{ d(x,Y) \mid x \in X \} \end{align*}

Another interesting take, as given in the comments by Robert Israel, is the irrationality measure. If $x$ has irrationality measure $\mu$, then $\mu$ is the smallest number where $\exists Q \in \Bbb Z^+$ such that, $\forall \varepsilon > 0$ and $\forall q \ge Q$,

$$\left| x - \frac p q \right| > \frac{1}{q^{\mu + \varepsilon}}$$

which helps to encode how poorly one can approximate $x$ by rational numbers $p/q$, in the sense that higher $\mu$ implies it is easier to approximate. (Specifically, approximate $x$ by rational numbers which are not $x$ itself, since we may define $\mu$ for all $x \in \Bbb R$.)

  • $\pi$ is well-known for having somewhat decent approximations through its continued fraction. It's $\mu$ is known to be between $2$ and about $7.10320$.
  • Liouville numbers have $\mu = \infty$, implying they're well-approximated. (You can also see a section on the irrationality measure on Wikipedia via that link.) Pretty obvious considering how many zeroes they have.
  • $\varphi := (1+\sqrt 5)/2$ was claimed as "the most irrational number" in a video by Mathologer, because its continued fraction expansion converges slower than any other one to its corresponding constant -- which is part of why $\mu = 2$ for it, as it is for all other algebraic numbers. (Note that some transcendental numbers, like $e$, also have $\mu = 2$.) So I suppose one could say $\varphi$ is "tied" for the "most irrational number."

(Which I find quite interesting -- the definition of $\mu$ and the values it takes on $\Bbb Q$, the algebraic numbers, and the transcendental numbers imply that it's easier to approximate most transcendental numbers than it is algebraic numbers, or to approximate irrational numbers than it is to approximate rational numbers by other rational numbers.)

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    $\begingroup$ Indeed, a number is often proved transcendental by showing it's very closely approximable. By the way, the numbers tied with $\varphi$ are those of the form $\frac{a+b\varphi}{c+d\varphi}$ with integers $a,\,b,\,c,\, d$ where $ad\ne bc$. $\endgroup$
    – J.G.
    Commented Dec 30, 2020 at 23:26
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    $\begingroup$ Compare to the wiki page linked by @J.G. your claims about the irrationality measure are somewhat off. First the exact measure of $\pi$ is not known, it is only known to be between 2 and 7.1. Second I would add that almost all (ie density 1) numbers are Liouville numbers. Third, $e$ is not an algebraic number. Algebraic numbers are zeros of polynomicals with integer coefficients. This includes everything written in fractions with square roots and more stuff from roots of higher degree polynomials. $\endgroup$
    – quarague
    Commented Dec 31, 2020 at 8:35
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    $\begingroup$ @quarague I read "as it is for all other algebraic numbers among others like $e$" as implying some transcendental numbers, $e$ included, also fit the point being made. $\endgroup$
    – J.G.
    Commented Dec 31, 2020 at 8:52
  • $\begingroup$ Yeah, J.G. understood what I meant in that respect, though I misread the table as given approximate $\mu$ rather than upper bounds, so thanks for pointing that out. $\endgroup$ Commented Dec 31, 2020 at 17:35
  • $\begingroup$ Also isn't the Lebesgue measure of the set of Liouville numbers $0$, not $1$, as noted here on Wikipedia? $\endgroup$ Commented Dec 31, 2020 at 17:41

No, not with the usual metric. As the question stands, the answer is surely, “it depends on the metric”.

However, I’ll leave the rest of my answer up-it should be clear that I am working in the usual metric.

Between any two real numbers there are infinitely many rational and irrational numbers.

I should probably elaborate. To me, your question is basically, “which of the following is closer to a rational number: $\sqrt{2}, \pi, e, e^2$ and so on?”

The answer is that they’re all “infinitesimally close” to a rational number, although I’d rather not use this dodgy language.

Let $x$ be any of the irrational numbers above - or any other irrational number - and let $\varepsilon > 0$, no matter how small.

Then $\exists \ q(\varepsilon) \in \mathbb{Q}$ such that $q \in (x-\varepsilon, x+ \varepsilon).$ In fact, there exist infinitely many such rational numbers in this interval.

Furthermore, let $y$ be any real number, rational or irrational.

Then $\exists \ p(\varepsilon) \in \mathbb{Q}$ such that $p \in (y-\varepsilon, y+ \varepsilon)$ (in fact, there are infinitely many rational numbers in this interval) and also $\exists \ j \in \mathbb{R} - \mathbb{Q}$ such that $j \in (y-\varepsilon, y+ \varepsilon)$ (in fact, there are infinitely many irrational numbers are in this interval).

Furthermore, this closeness relationship between $\mathbb{Q}$ and $\mathbb{R} - \mathbb{Q}$ can also be summarised as follows:

$ \mathbb{Q}$ is dense in $\mathbb{R}$, $\mathbb{R} - \mathbb{Q}$ is dense in $\mathbb{R}$, but $\mathbb{Q}$ and $\mathbb{R} - \mathbb{Q}$ are disjoint subsets of $\mathbb{R}$.

  • $\begingroup$ Maybe I misunderstand the question. Or I should elaborate. $\endgroup$ Commented Dec 30, 2020 at 23:23
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    $\begingroup$ True, the literal question "fails", but it is possible to reinterpret it in an interesting and useful way, as in the comments, and the answer by @EeveeTrainer, for example. $\endgroup$ Commented Dec 30, 2020 at 23:24
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    $\begingroup$ Oh yeah forgot about that lol. Thanks for pointing that out- I edited my answer. $\endgroup$ Commented Dec 30, 2020 at 23:51
  • $\begingroup$ I didn't change the question, I just worded it badly (very informally). Thanks! $\endgroup$ Commented Dec 31, 2020 at 0:38

I think you are confusing "a" rational number meaning a specific rational number with "a" rational number meaning any rational number.

Given the rational numb $3$ and the irrational numbers $e$ and $\pi$. $\pi$ is clearly closer to $3$ than to $e$ is. And there are irrational numbers closer. $\pi =0.14 = 3.0015926535897932384626433832795....$ is even closer to $3$.

And we can make a norm where metrics are ranked on how far they are from $3$. Via $|x|_n = |x-3|$.

That's easy but fairly pointless.

But what it sounds like what asking whether there are some irrational numbers that are closer to their closest rationals than other irrational to their closest rationals.

The answer to that is no. For any irrational we can find rationals arbitrarily close to it so there is not closest rational and there is no measurable distance between an irrational number and all the potential (infinitely many of them) rationals that can be arbitrarily close to it.

For any $\epsilon > 0$ there will always be (infinitely many) rationals within $\epsilon$ of the irrational.


Aside from a set of measure zero there is no much difference.
On the one hand, classical result due to Dirichlet asserts that for any irrational x (say x between 0 and 1), there are infinite many integers p,q such that $${|x - p/q|} < 1/q^{2}$$ On the other hand, for fixed $\epsilon$ > 0 the set of those numbers which can be approximated by order 2 + $\epsilon$ (denoted below by A) has measure zero. To see this define $$A_{k,N}^{\epsilon} = \{x; x\in [0,1], |x-k/N| < 1/N^{2 + \epsilon}, 1 \leqq k\leqq N\}$$ One has $$A = \bigcap_{n=1}^{^\infty}\bigcup_{k,N \geq n}^{\infty}A_{k,N}^{\epsilon}$$ Since $$|\bigcup_{k,N \geq n}^{\infty}A_{k,N}^{\epsilon}| \leqq \sum_{k,N\geq n}^{}|A_{k,N}^{\epsilon}|$$ and $$|A_{k,N}^{\epsilon}|\leqq 1/N^{1+\epsilon}$$ we get $$|\bigcup_{k,N \geq n}^{\infty}A_{k,N}^{\epsilon}| \leqq \sum_{k,N\geq n}^{}|A_{k,N}^{\epsilon}| \leqq \sum_{N=n}^{\infty}1/N^{1+\epsilon}$$ In words, A has really measure zero as advertised.


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