Are some irrational numbers closer to rational numbers than others?

Question

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?

Answer 1

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.

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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*}

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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.)

Answer 2

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 $\mathbb{R}$.

Answer 3

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.

Answer 4

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.