The Set of integer multiples of two irrational numbers is dense in reals?

Question

Let $\alpha , \beta $ be the linearly independent irrational numbers over $\mathbb Q$ with $\alpha > \beta > 0 $ , and $\mathrm A=\{n\alpha-m\beta \mid n,m \text{ are nonnegative integers} \}$

How to prove that $\mathrm A$ is dense in $\mathbb R$ ? Is it true?

This has been asked sooo often. For example: https://math.stackexchange.com/questions/136665/for-every-irrational-alpha-the-set-ab-alpha-a-b-in-mathbbz-is-den — amsmath, Oct 12 '17 at 20:17
@amsmath The only obstacle I see is that the OP wants $m,n\in\Bbb N$, not $\in\Bbb Z$. — Hagen von Eitzen, Oct 12 '17 at 20:19

score -1 · Answer 1 · edited Oct 12 '17 at 20:24

-1

Note that

$A$ is closed under addition
$\inf A=-\infty$
$\sup \{\,x\in A\mid x>0\,\}=0$.

The first is trivial, the second follows from $\beta>0$, and only the third involves the $\Bbb Q$-linear independence of $\alpha,\beta$.

edited Oct 12 '17 at 20:24

José Carlos Santos

427,504

answered Oct 12 '17 at 20:22

Hagen von Eitzen

374,180

The third does not make sense – amsmath Oct 12 '17 at 20:25
The third is wrong.. – merow Oct 12 '17 at 20:38
Do you mean $x < 0$? – amsmath Oct 12 '17 at 20:43
I'm not sure why you rejected my edit to the third bullet point: it makes no sense as written. – Steve D Oct 13 '17 at 04:24

The Set of integer multiples of two irrational numbers is dense in reals?

1 Answers1