# Do these irrationals exist?

An irrational $$a$$ verifies : $$\{a\times n+k;(n,k)\in\mathbb Z^2 \}$$ is dense in $$\mathbb R$$.

If you take $$a$$ universe then : $$\forall b\in \mathbb N^*, \{a\times n^{b}+k;(n,k)\in\mathbb Z^2\}=A(a,b)$$ is dense in $$\mathbb R$$.

1. Does exists $$a$$ irrational with : $$\{a \times n^2+k ; (n,k) \in \mathbb Z^2 \}=A(a,2)$$ isn't dense in $$\mathbb R$$ ?
1. Let $$a \in \mathbb R -\mathbb Q$$. Is it easy to find $$C(a)=\{b\in\mathbb N^*,A(a,b) \text{ isn't dense in }\mathbb R$$} ?
• What does "if you take $a$ universe" mean? Sep 4, 2022 at 17:07
• @MattF. I think your latter question might be an issue on your end: I do see a definition of $C(a)$ instead of a blank space. Sep 4, 2022 at 17:07
• @Dattier Again, what does the term "universe" mean? (I think there might be a translation issue here.) Sep 4, 2022 at 17:18
• fr.wikipedia.org/wiki/Nombre_univers Sep 4, 2022 at 17:18
• I can't read French, but from google translate: "A universe number is a real number in the decimals of which one can find any succession of digits of finite length, for a given base." So universe-ness is a very weak form of normality (so weak, for instance, that comeager-many numbers have the universe property in contrast to the meagerness of the normal numbers). Sep 4, 2022 at 19:34

If I understand, your Question 1 asks "Does there exist an irrational number $$\alpha$$ such that $$\{n^2\alpha +k :n, k \in \mathbb Z\}$$ is not dense in $$\mathbb R$$?" The answer is no. This fact is usually deduced from a stronger condition on the sequence $$(n^2\alpha)_{n\in \mathbb N}$$, called uniform distribution mod $$1$$, cf. Wikipedia. When $$(x_n)_{n\in \mathbb N}$$ is uniformly distributed mod 1, the numbers $$\{x_n+k: n\in \mathbb N, k\in \mathbb Z\}$$ are dense in $$\mathbb R$$ (but the reverse implication can fail). The classic text of Kuipers and Niederreiter is the standard introduction - see Theorem 3.2 in Chapter 1. This may also answer your Question 2.

Kuipers, L.; Niederreiter, Harald, Uniform distribution of sequences, Pure and Applied Mathematics. New York etc.: John Wiley & Sons, a Wiley-Interscience Publication. xiv, 390 p. (1974). ZBL0281.10001.

I'm not sure I'm understanding your question (as for others, I'm confused about 'universe'), but for every irrational $$a$$ and positive integer $$b$$, your set $$A(a,b)$$ is dense. So $$C(a)$$ is always the empty set.

This seems to already be stated in your post, but maybe it's related to not knowing what 'universe' means.

EDIT: adding a few more details at request of the OP. The easiest self-contained proof uses the so-called van der Corput lemma, which states that if $$(x_n)$$ is a sequence in $$[0,1)$$ and for all $$h \in \mathbb{N}$$, the sequence $$y^{(h)}_n = (x_{n+h} - x_n) \pmod 1$$ is equidistributed, then $$(x_n)$$ itself is equidistributed.

Now you can easily prove the following by induction on the degree: for any non-trivial polynomial $$p(n)$$ with integer coefficients and any irrational $$a$$, the sequence $$p(n) a \pmod 1$$ is equidistributed. This obviously implies that $$\{p(n) a + k \ : \ k,n \in \mathbb{Z}\}$$ is dense in $$\mathbb{R}$$.

I'm sure there are many resources, but here is a nice expository paper on van der Corput and more: https://arxiv.org/pdf/1510.07332.pdf.

• Can you explain why ? Sep 4, 2022 at 17:17
• John Griesmer's answer below has some good sources. Two proof techniques are to prove directly from Weyl's equidistribution criterion or use dynamical systems called skew products. Sep 4, 2022 at 17:24