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I would like to know is there any notion of density over the rationals with which we could determine are rationals everywhere equally dense on the real line, because, for example, I am not sure would all the rationals in some $\delta>0$ neghborhood of, for example $\frac {1}{2}$, would be equally dense as in some $\delta>0$ neghborhood of zero, that is, would that definition of density give same results for rationals in $[\frac {1}{2}- \delta, \frac {1}{2}+\delta]$ as for those in $[0-\delta,0+\delta]$.

Because, although rationals are everywhere dense in the sense that between any two of them there is an infinite number of them it is probably true that irrationals are not self-similar in their positioning over the intervals on the real line, so if irrationals do not have that kind of self-similarity neither do rationals have, right?

So it would maybe be enough to define that notion of density over the irrationals and then study irrationals to know how exactly in the sense of that density the rationals are placed.

There probably have been ideas like this one before so someone can tell something about this task, I believe that this can be put on a firm ground.

What is known about this?

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    $\begingroup$ If $r$ is a rational and $k$ an integer, then $r+k$ is a rational. So the density is the same in one place as it is in any other. $\endgroup$
    – amrsa
    Commented Mar 31, 2018 at 15:32
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    $\begingroup$ What amrsa said, except pick k to be a rational. There is a lot of self symmetry in the order structure of the rationals. Gerhard "Wherever You Add, You're There" Paseman, 2018.03.31. $\endgroup$
    – Gerhard Paseman
    Commented Mar 31, 2018 at 15:50
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    $\begingroup$ It doesn’t make sense to ask how dense a countable set is: that only makes sense if you have a sequence (and you can ask how many of the first $n$ terms are within $r$ of $x$). $\endgroup$
    – Anthony Quas
    Commented Mar 31, 2018 at 15:58
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    $\begingroup$ In spite of the comments, I think this is an interesting question. For example, continued fractions provide a measure of “how far away” a given real number is from the “closest” rational number. It is an interesting and meaningful fact that the golden ratio is the “furthest away” from all rational numbers that it is possible to be. Even deeper is that if an irrational number is “too close” to the nearest rational number, then it must be transcendental. So it is as if the rationals are not homogeneous among the reals: they are “less dense” near algebraic numbers than in other regions. $\endgroup$
    – Theo Johnson-Freyd
    Commented Mar 31, 2018 at 19:22
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    $\begingroup$ @TheoJohnson-Freyd Indeed, one can, probably, make some non-trivial sense out of it. The problem is that (it seems like) Shalom is in a habit of throwing random puzzles at the MO public without thinking of them himself at all. This usually ends up in everybody getting irritated and voting to close without even reading the question to the end, and, alas, I can understand that perfectly well. Downvoting but not voting to close yet. $\endgroup$
    – fedja
    Commented Mar 31, 2018 at 19:37

2 Answers 2

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Rationals are equidistributed in the sense that If you take any "nice" function, then if you approximate the integral of $f$ over (say) $[0, 1]$ by $\frac{1}{N_B} \sum f(r),$ where $N_B$ is the number of rationals with denominator bounded by $B,$ and the sum is over all such rationals, then this converges to the actual integral.

This equidistribution is nontrivial, and is equivalent to the Prime Number Theorem. The optimal speed of the convergence is equivalent to the Riemann Hypothesis.

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  • $\begingroup$ What does nice mean in this context? $\endgroup$
    – Geoffrey Irving
    Commented Mar 31, 2018 at 22:46
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    $\begingroup$ @GeoffreyIrving Continuous will certainly be nice enough. $\endgroup$
    – Igor Rivin
    Commented Mar 31, 2018 at 23:53
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Consider four reals $x=\frac{a}{b},y=\frac{c}{d},z,w$ with $z,w$ irrational and $b<d$ (with the fractions in lowest terms). There is some sense in which the rationals are less dense at $x$ than at $y$, less dense at $y$ than at $z$ and (I think) about equally dense at $z$ and $w.$

Here is a picture of Ford Circles. A circle of radius $\frac1{q^2}$ tangent to the $x$ axis at $(\frac{p}q,0)$

enter image description here

It appears that the smaller $q$ is, the more other fractions are pushed away. It is not hard to see why that would be. For $x=\frac12$ and $q \gt 2$ even, the closest approximation is not in lowest terms. The next two closest are at distance $\frac1q$ (of course) SO they don't get counted until that is less than $\varepsilon.$ However, once that does start happening, both are in lowest terms. That perhaps explains the catching up. The case of odd $q$ is similar, to an extent.

Let $f(x,\varepsilon,N)$ be the number of rationals $\frac{p}{q}$ in lowest terms so that $q \leq N$ and $|x-\frac{p}{q}| \lt \varepsilon.$

Then one might want to say that the rationals are denser near $s$ than near $r$ if $F(r,\varepsilon,N) \lt f(s,\varepsilon,N)$ for all $\varepsilon,N.$ That won't quite work. If we fix $N$ and let $\varepsilon$ decrease to $0$ then eventually $f(x,\varepsilon,N)=0.$ If we fix $\varepsilon \gt 0$ and let $N$ grow then the counts start to differ by very little $f(r,\varepsilon,N)\sim f(s,\varepsilon,N).$ So the "right" definition should have $N$ grow as $\varepsilon$ decreases. Maybe something like $$\lim \frac{f(s,\frac1{N^2},N)}{f(r,\frac1{N^2},N)} \gt 1$$ would be a good model for "denser at $s$ than at $r$."

Here is a table that uses $\varepsilon=\frac1{2000}.$

The first column is $N.$ The second is $f(x,\frac1{2000},N)$ for $[1,\frac12,\frac13,\frac15]$ and the fourth is $f(x,\frac1{2000},N)$ for $[e,\gamma,\pi,\tau]$ where $\gamma=0.5772\dots$ is Euler's gamma and $\tau=\frac{1+\sqrt{5}}2$ is the golden ratio.

In columns three and five are the quadruple to the left scaled by the first entry.

$ \begin {array}{ccccc} & [1,\frac12,\frac13,\frac15]&&[e,\gamma,\pi,\tau] \\500&[1,1,1,41]&[ 1.0, 1.0, 1.0, 41.0]&[ 76,78,75,76]&[ 1.0, 1.0263158, 0.98684211, 1.0]\\ 1000&[1,1,224,281]&[ 1.0, 1.0, 224.0, 281.0]&[303,306,294,305]&[ 1.0, 1.0099010, 0.97029703, 1.0066007]\\ 1500&[1,501,612 ,661]&[ 1.0, 501.0, 612.0, 661.0]&[683,686,679,684]&[ 1.0, 1.0043924, 0.99414348, 1.0014641]\\ 2000&[3,1003,1113,1176]&[ 1.0, 334.33333, 371.0, 392.0]&[1220,1220,1209,1215]&[ 1.0, 1.0, 0.99098361, 0.99590164]\\ 2500&[1003,1753,1834,1883 ]&[ 1.0, 1.7477567, 1.8285145, 1.8773679]&[1898,1900,1900,1900]&[ 1.0, 1.0010537, 1.0010537, 1.0010537]\\ 3000&[2003,2503, 2668,2712]&[ 1.0, 1.2496256, 1.3320020, 1.3539690]&[2738,2738,2732, 2732]&[ 1.0, 1.0, 0.99780862, 0.99780862]\\ 3500&[ 3003,3587,3646,3703]&[ 1.0, 1.1944722, 1.2141192, 1.2331002]&[3731, 3724,3715,3722]&[ 1.0, 0.99812383, 0.99571161, 0.99758778] \\ 4000&[4003,4669,4800,4840]&[ 1.0, 1.1663752, 1.1991007, 1.2090932]&[4864,4861,4864,4865]&[ 1.0, 0.99938322, 1.0, 1.0002056]\\ 4500&[5503,6003,6068,6122]&[ 1.0, 1.0908595, 1.1026713, 1.1124841]&[6157,6159,6150,6157]&[ 1.0, 1.0003248, 0.99886308, 1.0]\\ 5000&[7003,7337,7526, 7576]&[ 1.0, 1.0476938, 1.0746823, 1.0818221]&[7605,7601,7593,7600]&[ 1.0, 0.99947403, 0.99842209, 0.99934254]\\ 5500&[ 8503,9069,9132,9179]&[ 1.0, 1.0665647, 1.0739739, 1.0795014]&[9199, 9198,9188,9189]&[ 1.0, 0.99989129, 0.99880422, 0.99891293] \\ 6000&[10003,10803,10852,10925]&[ 1.0, 1.0799760, 1.0848745, 1.0921723]&[10941,10943,10941,10942]&[ 1.0, 1.0001828, 1.0 , 1.0000914]\\ 6500&[12171,12705,12794,12818]&[ 1.0, 1.0438748, 1.0511872, 1.0531591]&[12843,12845,12841,12847]&[ 1.0, 1.0001557, 0.99984427, 1.0003115]\\ 7000&[14337, 14603,14823,14862]&[ 1.0, 1.0185534, 1.0338983, 1.0366185]&[14894, 14902,14890,14895]&[ 1.0, 1.0005371, 0.99973144, 1.0000671] \\ 7500&[16503,16931,17001,17059]&[ 1.0, 1.0259347, 1.0301763, 1.0336908]&[17104,17107,17089,17102]&[ 1.0, 1.0001754, 0.99912301, 0.99988307]\\ 8000&[18671,19263,19379, 19439]&[ 1.0, 1.0317069, 1.0379198, 1.0411333]&[19459,19463,19448, 19449]&[ 1.0, 1.0002056, 0.99943471, 0.99948610]\\ 8500&[21337,21839,21865,21954]&[ 1.0, 1.0235272, 1.0247457, 1.0289169] &[21965,21962,21956,21962]&[ 1.0, 0.99986342, 0.99959026, 0.99986342] \\ 9000&[24003,24417,24559,24600]&[ 1.0, 1.0172478, 1.0231638, 1.0248719]&[24627,24617,24613,24625]&[ 1.0, 0.99959394, 0.99943152, 0.99991879]\\ 9500&[26671,27331,27407, 27417]&[ 1.0, 1.0247460, 1.0275955, 1.0279705]&[27439,27435,27431, 27436]&[ 1.0, 0.99985422, 0.99970844, 0.99989067]\\ 10000&[29337,30241,30344,30357]&[ 1.0, 1.0308143, 1.0343253, 1.0347684 ]&[30395,30398,30390,30398]&[ 1.0, 1.0000987, 0.99983550, 1.0000987] \end {array} $

Looking first at the rationals we see that by the time we get up to $N=10000$ the relative sizes differ by only about $3\%.$ The counts for $\frac12,\frac13,\frac15$ get an increasingly fast head start on those for $1.$ They stay ahead as far as shown but the lead drops both absolutely and relatively. I imagine one could explain why sometimes the count for $1$ jumps quite a bit more than that for $\frac12$ and other times it is the other way around. The particular $N$ values of $500t$ are rather arbitrary and perhaps not optimal.

The order chosen for the four irrational numbers is ad hoc. The differences in counts do not seem significant. They are all about the same and also always a bit bigger than the corresponding counts for rationals.

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