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I wondered is that always true that if $\alpha$ and $\beta$ are non-zero real numbers, then can we make $|m\alpha+n\beta|$ arbitrarily close to zero, for some non-zero integers $m$ and $n$. My guess would be "no", but I couldn't come up with a simple counter example. Sorry in advance if there is a very simple one. (That's easily seen to be true when $\alpha$ and $\beta$ are rational, since it can be made equal to zero! But, I somehow feel like it shouldn't be true for all reals. I first tried to prove it's true considering density of rationals, but I think I failed, that's why I believe it's not true.) Thanks a lot!

vgmath
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  • I'd focus on the density. A real number is arbitrarilly close to any rational number you want, so you can get your two numbers really close to rationals that you can then work to minimize – Alan Jul 08 '15 at 10:16
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    You get zero non-trivially if the ratio $\alpha/\beta$ is rational. When the ratio is irrational you can try and adapt [Mariano's answer to an earlier question](http://math.stackexchange.com/a/73272/11619). – Jyrki Lahtonen Jul 08 '15 at 10:29

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[Standard Argument]: Suppose, to the contrary, that you can not get near zero (non-trivially). We want to prove that the ratio $\frac{\alpha}{\beta}$ is rational. Look at the set $$S = \{m\alpha + n\beta, m,n \in \mathbb Z\}$$ Clearly that set is closed under addition and closed under multiplication by integers. We note that if there are any real accumulation points for S then 0 is an accumulation point. (Pf: differences between elements of S are also elements of S and if some sequence drawn from S converged to L, say, the elements in that sequence would draw arbitrarily close to each other.). Thus if S does not have 0 as an accumulation point there is a least positive element of S, call it A. But then all elements of S are multiples of A (else the division algorithm would hand us a smaller positive element of S). in particular $\alpha$ = MA and $\beta$ = NA for integers M and N. But then $\frac{\alpha}{\beta}$ = M/N .

lulu
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  • I think you should put the modulus of $m\alpha+ n\beta$ in $S$. That is, $S = \{|m\alpha + n\beta|, m,n \in \mathbb Z\}$. – Rafael Deiga Dec 19 '17 at 12:57
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    @RafaelDeiga Why? Your set is just the non-negative elements of mine. Moreover, $s\in S \iff -s \in S$. If, in my set, I have a set of non-zero elements which approach $0$ in a limit then I know I have a set of positive elements which approach $0$. – lulu Dec 19 '17 at 13:20
  • Because I think that to guarantee the existence of A, we need to use that there is an infimum for all nonempty set which is bounded below. However, this is kind of implicit in your answer. – Rafael Deiga Dec 19 '17 at 22:28
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This is not true in general.

Take $\alpha=\beta=1$ for a counterexample. You can't find a nonzero sequence of elements of the set converging to zero.

However it is true if you take for example $\alpha=1$ and $\beta=\sqrt{2}$.

$$G=\{m \alpha + n \beta | (n,m) \in \mathbb Z^2\}$$ is a subgroup of the reals. It is a classical result that the subgroups of the real are either discrete or dense.

mathcounterexamples.net
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    For $\alpha=\beta=1$, take $m=1, n=-1$, this makes the expression zero? – vgmath Jul 08 '15 at 10:21
  • I think "arbitrarily close to zero" certainly includes "equal to zero". Just recall the definition of the limit. – Vim Jul 08 '15 at 10:36
  • Formally, I agree with your comments. However, I think that the interesting topic is: can a sequence of non zero elements of the set converge to zero? – mathcounterexamples.net Jul 08 '15 at 10:45
  • If I understood your question, take, for example, the sequence $\frac{1}{n}$, it's never equal to zero, but converges to zero. – vgmath Jul 08 '15 at 10:52