Is $\{(m+ka,n+kb) : m,n,k\in\mathbb Z\}$ dense in $\mathbb R^2$ if $a,b\notin\mathbb Q$?

Question

Can anyone prove here that for $a,b\in\mathbb R\setminus\mathbb Q$ we have that the set $$ \{(m+ka,n+kb) : m,n,k\in\mathbb Z\} $$ is dense in $\mathbb R^2$? I know that the projections onto the coordinate axes both are dense in $\mathbb R$, but I cannot prove the density of the set in two dimensions.

EDIT: It was shown below that the set is not dense in the following cases: (1) $a+b\in\mathbb Q$ and (2) $\exists t\in\mathbb R : t(a,b)\in\mathbb Z^2$. The question is now: When is the set dense in $\mathbb R^2$?

When you project, you lose a lot of information. Try the simple(r) case of finding a sequence converging to $(0, 1/2)$. — user357980, Jun 17 '17 at 19:48
It is special in the way that $0$ is an integer and hence more comfortable. Basically the same form of this is $(0, p)$, where $p$ is any number. It is also sufficient.
Also, Tim has a good point. — user357980, Jun 17 '17 at 20:03

MaudPieTheRocktorate · Accepted Answer · 2017-06-17T20:07:51.840

2

Let $x\in\mathbb{R}\setminus\mathbb{Q}, a=1-x, b=x$, then the set is not dense.

In general, the set is not dense when $a+b\in \mathbb{Q}$.

When $a=b$, the set is also not dense. To see that, start with the square grid $\mathbb{Z}\times\mathbb{Z}$, and then to get $\{(m+ka,n+kb) : m,n,k\in\mathbb Z\}$, you take the grid and add copies of it shifted by the vector $(a,b)$. When $a=b$ the copies move along the diagonal, and so all the points are on the diagonal lines, so it's not dense.

edited Jun 17 '17 at 20:07

answered Jun 17 '17 at 19:58

MaudPieTheRocktorate

3,796
16
34

Can you show that it is not dense in the case $a = b$? I do not see it... – Friedrich Philipp Jun 17 '17 at 20:05
Thanks. You are right. So it depends on the angle of the line which is generated by $(a,b)$. I guess the set is dense if and only if this angle is an irrational multiple of $\pi$. – Friedrich Philipp Jun 17 '17 at 20:11
Sorry, I was wrong. I say that the set is not dense if and only if for some $t\in\mathbb R$ we have that $t(a,b)\in\mathbb Z^2$. – Friedrich Philipp Jun 17 '17 at 20:15
@FriedrichPhilipp That won't do. Let $a=\sqrt{2}/2, b=1-\sqrt{2}/2$, then there's no such $t\in\mathbb{R}$. – MaudPieTheRocktorate Jun 17 '17 at 20:29

Is $\{(m+ka,n+kb) : m,n,k\in\mathbb Z\}$ dense in $\mathbb R^2$ if $a,b\notin\mathbb Q$?

1 Answers1

Linked