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If $a$ is any real irrational, then the set of numbers of the form $ax+y$ with $x$ and $y$ co-prime integers is dense in $\mathbb{R}$. I managed to prove this, in what I suspect is an overly complicated way, in response to this question.

I think there must be a more perspicuous proof of this fundamental fact. Furthermore, I expect that the following more general assertion is true:

If $f\in \mathbb{R}[x]$ is any polynomial at least one of whose coefficients other than the constant term is irrational, then the numbers of the form $f(x)+y$ with $x$ and $y$ co-prime integers are dense in $\mathbb{R}$.

Now it seems to me that this ought to be a very well known fact or open problem (or maybe it is false!) but I can't find a single reference to it. Can anyone provide a proof or a reference?

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    $\begingroup$ I don't see any particular reason for this to be a very well known fact. The coprime condition doesn't seem very natural to me. $\endgroup$ Nov 25, 2013 at 18:10
  • $\begingroup$ Isn't this a consequence of Kronecker's (or Weyl or someone's) theorem on the equidistribution of fractional parts of integral multiples of an irrational number? $\endgroup$ Nov 25, 2013 at 18:25
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    $\begingroup$ @Avenger: The statement about polynomials (without the coprimality requirement) was first proved by Hardy and Littlewood. Shortly thereafter, Weyl's criterion on uniform distribution appeared, from which one can deduce a stronger result (uniform distribution mod 1, not merely density). But I don't see anything in either of these approaches that I can use to address the problem at hand. $\endgroup$ Nov 25, 2013 at 18:35
  • $\begingroup$ Does "one of whose coefficients" mean "at least one" or "exactly one"? $\endgroup$ Nov 25, 2013 at 22:08
  • $\begingroup$ @Gerry: I mean "at least one". $\endgroup$ Nov 26, 2013 at 1:14

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The continued fraction for $a$ gives rational approximations $$\frac{p_i}{q_i} \approx a$$ with $|q_ia-p_i| \lt \frac1{q_i}$. Furthermore, successive approximants satisfy $p_{i+1}q_q-p_iq_{i+1}=\pm1.$

Fix an index $i$ and consider the sequence of expressions $$r_k=x_ka+y_k=(q_{i-1}+kq_{i})a-(p_{i-1}+kp_{i})$$ where $k \in \mathbb{Z}$. The values $r_k$ form a two way infinite arithmetic progression with common difference $q_ia-p_i \lt \frac{1}{q_{i}}$ and thus allow us to approximate any desired real to an accuracy of better than $\frac{1}{2q_i}$. It remains only to note that

  • The denominators $q_i$ increase (exponentially) and thus allow us to get any accuracy we desire.
  • $x_{k+1}y_k-x_{k}y_{k+1}=p_{i+1}q_i-p_iq_{i+1}=\pm1$ so the pairs $(x_k,y_k)$ are relatively prime.
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  • $\begingroup$ Why are $a$ and $b$ coprime? And why is $ax+b$ close to $a'x+b'$? I like your idea of perturbing an approximation to get co-primality, but a bit of detail would be very helpful here. $\endgroup$ Nov 27, 2013 at 2:04
  • $\begingroup$ I changed the notation to match that of the question (though it makes your comment harder to follow) and simplified/corrected the proof while adding a few more details. $\endgroup$ Nov 29, 2013 at 9:45
  • $\begingroup$ Thank you. This looks like "the" way to do it. In terms of Farey fractions your idea is to trap $\alpha$ between succesive fractions $a/b$ and $c/d$ with large denominator. Coprimality is preserved by taking mediants $(a+kc)/(b+kd)$ but the values of $\alpha (b+kd)−(a+kc)$ will form an arithmetic progression with small common difference. Pity that this doesn't generalize (as far as I can see) to polynomials of degree greater than 1. $\endgroup$ Nov 30, 2013 at 3:59

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