Can $|m\alpha+n\beta|$ be made arbitrarily small? 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!
 A: [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 .
A: 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.
