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Let $x$ be an irrational number, and $\beta$ strictly larger than its irrationality index, which means that for some $C>0$, for all $n\in \mathbb{Z}^*$, $$d(nx,\mathbb{Z})>C n^{-\beta}.$$ It is known that for a.e. irrational number $x$, the irrationality index is $1$. It is even known that some numbers satisf the above for $\beta=1$ (for instance, $x=\sqrt{2}$).

By arguments from measure theory, I have been able to prove that if $a_n$ satisfies $$\sum_{n=1}^{\infty}na_n<\infty,$$ almost every couple $(x,y)$ of $\mathbb{R}^2$ satisfies for some $C>0$ $$d(nx+my,\mathbb{Z})>Ca_{|n|+|m|},n,m\in \mathbb{Z}.$$ Ideally, I would like, as for a single number $x$, find irrational numbers $(x,y)$ such that this holds in the limiting case $a_n=n^{-2}$.

Has anyone an idea? Or has anyone a useful reference for such things?

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    $\begingroup$ I suspect the answer is no. Almost all real numbers $x$ have irrationality index $2$, and it seems very likely that one can a sum of two such numbers which is a Liouville number, for example. $\endgroup$
    – Greg Martin
    Commented Nov 12, 2019 at 21:20
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    $\begingroup$ E.g., if $\alpha$ is Liouville, then chances are that $\alpha+\sqrt2$ and $\alpha-\sqrt2$ are both index $2$ while their sum is $2\alpha$. $\endgroup$
    – Gerry Myerson
    Commented Nov 12, 2019 at 22:41
  • $\begingroup$ Ok thanks. So it is likely that even if I carefully choose $x$ and $y$, there will be no better bound than $0$? $\endgroup$
    – kaleidoscop
    Commented Nov 13, 2019 at 5:39
  • $\begingroup$ I completely modified the question as I have made some progress on my side. $\endgroup$
    – kaleidoscop
    Commented Nov 27, 2019 at 19:28
  • $\begingroup$ We must assume $(m,n) \neq (0,0)$, else $d(nx+my, {\bf Z}) = 0$ . . . $\endgroup$
    – Noam D. Elkies
    Commented Nov 28, 2019 at 17:44

1 Answer 1

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Yes, such $(x,y)$ exist; for example, $x = \root 3 \of 2$ and $y = x^2$. For $l,m,n \in \bf Z$, define $$ N(l,m,n) := l^3 + 2m^3 + 4n^3 - 6lmn \in {\bf Z}; $$ this is the algebraic norm $$ (l + mx + nx^2) (l + m\rho x + n \bar\rho x^2) (l + m\bar\rho x + n \rho x^2) $$ of $l + mx + nx^2$, where $\rho$ is the cube root of unity $e^{2\pi i/3} = (-1 + \sqrt{-3})/2$. But if $(m,n) \neq (0,0)$ then $l + mx + nx^2 \neq 0$, so $\left|N(l,m,n)\right| \geq 1$ and $\left|l + mx + nx^2\right| \gg (|l|+|m|+|n|)^{-2}$. Taking for $l$ the integer nearest to $-(mx+nx^2)$ we deduce that $d(mx+nx^2,{\bf Z}) \gg (|m|+|n|)^{-2}$, as claimed.

The same argument (which generalizes the familiar one for $\sqrt 2$) shows that in general if $x$ is an algebraic number of degree $D$ then $$ d\Bigl(\sum_{j=1}^{D-1} n_j x^j, {\bf Z}\Bigr) \gg \left( \sum_{j=1}^{D-1} \left| n_j \right| \right)^{1-D} $$ for $n_1,\ldots,n_{D-1} \in \bf Z$ not all zero. This is best possible up to the value of the implicit constant, because Dirichlet's celebrated "pigeonhole" argument shows that conversely for any $N$ one can find integers $n_1,\ldots,n_{D-1} \in [-N,N]$, not all zero, such that $d(\sum_{j=1}^{D-1} n_j x^j, {\bf Z}) \ll N^{1-D}$.

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  • $\begingroup$ Sorry I did not provide a reference; none of this is new, but it's easier to recall the proofs than to find them in the literature. $\endgroup$
    – Noam D. Elkies
    Commented Nov 28, 2019 at 17:46
  • $\begingroup$ I see, pretty neat, thanks! "Not new" but for me it is dropped from the sky... Do you know f by any chance such arguments extend even further to infinite sums? i.e. I'm looking for $x_j$ such that for integers $n_j$, the distance from $\sum_j x_j n_j$ to $\mathbb{Z}$ is bounded from below by the inverse value of some norm of $\|(n_j)\|$ on the space of integer sequences... $\endgroup$
    – kaleidoscop
    Commented Nov 29, 2019 at 21:04
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    $\begingroup$ Seems the best $C$ is $1/(6x) = 0.13228342\ldots$. For large $k$, if $l+mx+nx^2 = (x-1)^k$ then its product with $(|m|+|n|)^2$ depends on the phase of $(e^{2\pi i/3}x - 1)^k$, which can be made arbitrarily close to the optimum. As with $\sqrt2$ for $d(nx,{\bf Z})$, the choice $x = \root3\of 2$ here is the most familiar but not quite the best; taking for $x$ the real root of $x^3-x+1$, or the root $2\cos(2\pi/7)$ of $x^3+x^2-2x-1$, should yield somewhat larger $C$. To be sure the exact value of $C$ is partly an artifact of the choice of quadratic factor: [cont'd] $\endgroup$
    – Noam D. Elkies
    Commented Dec 9, 2019 at 21:34
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    $\begingroup$ changing $(|m|+|n|)^2$ to say $m^2+n^2$ or $\max(|m|,|n|)^2$ would change both the liminf and (probably) the sequence of algebraic units that attain it. But the liminf would remain finite and positive. $\endgroup$
    – Noam D. Elkies
    Commented Dec 9, 2019 at 21:36
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    $\begingroup$ Very cool. Thank you. For $x = \root 3 \of 2$ I could only get down to .132283780 before running out of patience. Two more digits and an ISC lookup and I may have got there. $\endgroup$
    – O. S. Dawg
    Commented Dec 10, 2019 at 5:11

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